\section{Introduction} 
\label{intro}

Although there are no elliptic curves $E/\mathbb{Q}$ with everywhere good reduction, there are curves such that $[E(\mathbb{Q}_p) : E_0(\mathbb{Q}_p)] = 1$ for all primes $p$, where $E_0(\mathbb{Q}_p)$ is the subgroup consisting of the nonsingular points of $E(\mathbb{Q}_p)$ after reduction modulo $p$. For example, the curve
$$ E/\mathbb{Q} : y^2 = x^3 + 3x + 1,$$
which has discriminant $-2^4 \cdot 3^3 \cdot 5$, has $[E(\mathbb{Q}_2) : E_0(\mathbb{Q}_2)] = [E(\mathbb{Q}_3) : E_0(\mathbb{Q}_3)] = [E(\mathbb{Q}_5) : E_0(\mathbb{Q}_5)] = 1$. We refer to such curves as \textit{Tamagawa trivial curves.} Recent work by Griffin, Ono, and Tsai \cite[Corollary 1.2]{og} establishes that over half of the elliptic curves over $\mathbb{Q}$ are Tamagawa trivial. More precisely, the proportion of elliptic curves in short Weierstrass form that are Tamagawa trivial is $0.5054\ldots$ when the curves are ordered by height. 

For every elliptic curve over $\mathbb{Q}$, we associate the {\it Tamagawa product}
\[
\mathrm{Tam}(\mathbb{Q}; E) := \prod_{p \text{ prime}} c_p,
\]
where $c_p := [E(\mathbb{Q}_p) : E_0(\mathbb{Q}_p)]$ is the {\it Tamagawa number} at $p$. Obviously, $E/\mathbb{Q}$ is Tamagawa trivial if and only if $\mathrm{Tam}(\mathbb{Q};E) = 1$. It is known that $\mathrm{Tam}(\mathbb{Q}; E)$ can be arbitrarily large (see, for instance,~\cite[Table C.15.1]{Silverman}), and so it is natural to ask whether there is an average Tamagawa product for $E/\mathbb{Q}$. The numerics by Balakrishnan et al. ~\cite[Figure A.14]{Balakrishnan4} suggest that the average Tamagawa product over $\mathbb{Q}$ exists and is in the neighborhood of $1.82$. This speculation was confirmed by Griffin et al.~\cite[Theorem 1.3]{og}, who constructed a new $L$-function and computed the exact average to be $L_\mathrm{Tam}(-1) = 1.8183\dots$.

It is natural to ask about the values of such arithmetic statistics in an arbitrary number field $K$. To this end, we define the Tamagawa product $\mathrm{Tam}(K;E)$ for an elliptic curve $E/K$. We let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_K$, the ring of integers of $K$, that lies above a rational prime $p$. Recall that there is a unique extension $v := v_{\mathfrak{p}}$ to $K$ corresponding to $\mathfrak{p}$. We let $K_v$ be the completion of $K$ with respect to $v$. Select a uniformizer $\pi$ and denote the corresponding valuation on $K_v$ by $v_\pi$. The Tamagawa product for elliptic curves $E/K$ is
\begin{equation}
\mathrm{Tam}(K;E) := \prod_{p \text{ prime}} \ \prod_{\mathfrak{p} \mid (p)} c_\mathfrak{p},
\end{equation}
where $c_\mathfrak{p} := [E(K_v) : E_0(K_v)]$ is the Tamagawa number at $\mathfrak{p}$. 

Generalizing the work of Griffin et al.~\cite{og}, we compute the arithmetic statistics of Tamagawa products over arbitrary number fields $K$. Specifically, we compute the proportion of curves with fixed $\mathrm{Tam}(K;E)$ over short Weierstrass curves
\begin{equation}
    E = E(a_4, a_6) : y^2 = x^3 + a_4 x + a_6,
\end{equation}
where $a_4, a_6 \in \mathcal{O}_{K_v}.$ To compute the proportion of curves with fixed $\mathrm{Tam}(K;E)$, we require a consistent way to count sets of elliptic curves. To do so, we order $E/K$ by their height. Recall that the height of $E/K$ is
\[ \mathrm{ht}(K; E) := \prod_{\mathfrak{p} \in M_K}\max\left\{4|a_4|_\mathfrak{p}^3, 27|a_6|^2_\mathfrak{p}, 1 \right\},\] 
where $M_K$ contains all Archimedean and non-Archimedean places on $K.$ To count the number of $E/K$ with height $\leq X$, we introduce: 
\begin{equation}
    \mathcal{N}(K; X) := \#\{E := E(a_4, a_6) : \mathrm{ht}(K; E) \le X \}.
\end{equation} Similarly, to count the number of $E/K$ with Tamagawa product $m$ and height $\leq X$, we define 
\begin{equation}
    \mathcal{N}_{m}(K; X) := \#\{E := E(a_4, a_6) : \mathrm{ht}(K; E) \le X \textrm{ with } \mathrm{Tam}(K; E) = m\}.
\end{equation}
We now formally define the proportion of elliptic curves $E(a_4, a_6)$ with Tamagawa number $m$ to be 
\begin{equation}
\label{eq:count}
P_{\mathrm{Tam}}(K; m):=\lim_{X\rightarrow +\infty}\frac{\mathcal{N}_{m}(K; X)}{\mathcal{N}(K; X)}.
\end{equation}

We compute the global statistic $P_{\mathrm{Tam}}(K;m)$ by computing the local statistics of Tamagawa numbers at each $\mathfrak{p}.$ Namely, we let $\delta_{K,\mathfrak{p}}(c)$ be the local proportion of elliptic curves with Tamagawa number $c$ at $\mathfrak{p}$ when the elliptic curves are ordered by height. Using $\delta_{K,\mathfrak{p}}(c)$, we define an analogue of the $L-$function as presented in \cite{og}.




\begin{theorem} 
\label{l-series}
If $K$ is a number field, then $P_\mathrm{Tam}(K;m)$ are the Dirichlet coefficients of
 \[L_{\mathrm{Tam}}(K; s):=\sum_{m=1}^{\infty}\frac{P_{\mathrm{Tam}}(K; m)}{m^s}=\prod_{p \ \text{prime}} \ \prod_{\mathfrak{p} \mid 
 (p)} \left(\frac{\delta_{K,\mathfrak{p}}(1)}{1^s}+\frac{\delta_{K,\mathfrak{p}}(2)}{2^s}+\frac{\delta_{K,\mathfrak{p}}(3)}{3^s}+\dots\right).\]
 The exact values of $\delta_{K,\mathfrak{p}}(c)$ are given in \Cref{p=5_total}, \Cref{p=3_total}, and \Cref{p=2_total}.
\end{theorem}
\begin{remark}
\Cref{l-series} gives $P_\mathrm{Tam}(K; m)$ for all number fields $K$ and every positive integer $m$. In particular, the theorem makes no assumption on the class number $h_K$, the structure of the units in $\mathcal{O}_K^\times$, as well as the possible splitting types of primes in $K$. 
\end{remark}
\begin{corollary}
If $K$ is a number field, then the following are true.
\begin{enumerate}
    \item We have \[P_\mathrm{Tam}(K;1) = \prod_{p \ \text{prime}} \ \prod_{\mathfrak{p} \mid (p)} \delta_{K,\mathfrak{p}}(1) \in (0,1).\]
 
    \item The average Tamagawa product $L_\mathrm{Tam}(K;-1)$ is well-defined by absolute convergence.
\end{enumerate}
\end{corollary}

In the following example, we illustrate the results of \Cref{l-series} by computing $P_\mathrm{Tam}(K;1)$ and $L_\mathrm{Tam}(K;-1)$ for all imaginary quadratic fields $\mathbb{Q}(\sqrt{-D})$ with class number 1. For the values of $P_{\mathrm{Tam}}(\mathbb{Q}(\sqrt{-D});m)$ with $m \geq 2$, refer to \Cref{examples}. For further examples, also refer to \Cref{examples}, where we compute $P_\mathrm{Tam}(K;m)$ and $L_\mathrm{Tam}(K;-1)$ for real quadratic fields $\mathbb{Q}(\sqrt{D})$ with squarefree $D < 10^4$ and a number field with Galois group $S_4$.

\begin{example}
\label{imag_fields}
Tables \ref{trivtable} and \ref{avgtable} illustrate the convergence to $P_\mathrm{Tam}(\mathbb{Q}(\sqrt{-D}); 1)$ and $L_{\mathrm{Tam}}(\mathbb{Q}(\sqrt{-D}); -1)$ for class number 1 quadratic number fields.
\end{example}

{\footnotesize
\begin{center}
\begin{table}[H]
\begin{tabular}{| c | c | c | c | c | c | c | c | c | c |}
\hline
\multicolumn{10}{|c|}{$\mathcal{N}_{1}(\mathbb{Q}(\sqrt{-D}); X)/\mathcal{N}(\mathbb{Q}(\sqrt{-D}); X)$} \\
\hline
 $X$ & $\sqrt{-1}$ & $\sqrt{-2}$ & $\sqrt{-3}$ & $\sqrt{-7}$ & $\sqrt{-11}$ & $\sqrt{-19}$ & $\sqrt{-43}$ & $\sqrt{-67}$ & $\sqrt{-163}$   \\ \hline
 
 
 
 $10^4$ & $0.542$ & $0.488$ & $0.663$ & $0.357$ & $0.609$ & $0.620$ & $0.657$ & $0.560$  & $0.450$ \\
 
 $10^5$ & $0.539$ & $0.460$ & $0.665$ & $0.359$ & $0.599$ & $0.678$ & $0.716$ & $0.711$ & $0.636$ \\ 
 
 $10^6$ & $0.528$ & $0.468$ & $0.660$ & $0.343$ & $0.586$ & $0.667$ & $0.726$ & $0.744$ & $0.728$ \\ 
 
 $\vdots$ & $\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ \\ 
 
 $\infty$ & $0.529$ & $0.468$ & $0.661$ & $0.349$ & $0.581$ & $0.665$ & $0.733$ & $0.750$ & $0.763$ \\
 \hline
\end{tabular}
\caption{Convergence to $P_\mathrm{Tam}(\mathbb{Q}(\sqrt{-D}); 1)$.}
\label{trivtable}
\end{table}
\end{center}
}

{\footnotesize
\begin{center}
\begin{table}[H]
    \centering
    \begin{tabular}{| c | c | c | c | c | c | c | c | c | c |}
    \hline
\multicolumn{10}{|c|}{$\sum_{\mathrm{ht}(\mathbb{Q}(\sqrt{-D}); E) \le X}\mathrm{Tam}(\mathbb{Q}(\sqrt{-D})); E)/N(\mathbb{Q}(\sqrt{-D}); X)$} \\
\hline
        $X$ & $\sqrt{-1}$ & $\sqrt{-2}$ & $\sqrt{-3}$ & $\sqrt{-7}$ & $\sqrt{-11}$ & $\sqrt{-19}$ & $\sqrt{-43}$ & $\sqrt{-67}$ & $\sqrt{-163}$   \\ \hline
  $10^4$ & $1.751$ & $2.054$ & $1.589$ & $2.570$ & $1.850$ & $1.718$ & $1.535$ & $1.698$ & $2.017$ \\
  $10^5$ & $1.720$ & $1.979$ & $1.538$ & $2.417$ & $1.763$ & $1.537$ & $1.393$ & $1.403$ & $1.612$ \\
   $10^6$ & $1.708$ & $1.946$ & $1.508$ & $2.418$ & $1.723$ & $1.519$ & $1.361$ & $1.333$ & $1.372$ \\
 
 $\vdots$ & $\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ \\ 
 
 $\infty$ & $1.678$ & $1.904$ & $1.487$ & $2.376$ & $1.708$ & $1.480$ & $1.331$ & $1.300$ & $1.277$ \\
 \hline
    \end{tabular}
    \caption{Convergence to $L_\mathrm{Tam}(\mathbb{Q}(\sqrt{-D}); -1)$.}
    \label{avgtable}
\end{table}
\end{center}
}

In \Cref{trivtable}, $P_{\mathrm{Tam}}(\mathbb{Q}(\sqrt{-D}); 1)$ is noticeably smaller when $D=7$ and noticeably larger when $D=163$. On the other hand, in \Cref{avgtable}, $L_{\mathrm{Tam}}(\mathbb{Q}(\sqrt{-D}); -1)$ is noticeably larger when $D=7$ and noticeably smaller when $D=163$. The variance in $P_\mathrm{Tam}(\mathbb{Q}(\sqrt{-D});1)$ and $L_{\mathrm{Tam}}(\mathbb{Q}(\sqrt{-D}); -1)$ is due to the splitting type of small primes, since $\delta_{K,\mathfrak{p}}(1)$ is smaller and $\sum_{m=1}^\infty \delta_{K,\mathfrak{p}}(m)m$ is larger when $\mathfrak{p}$ has small norm (see \Cref{p=5_total}, \Cref{p=3_total}, and \Cref{p=2_total}). Indeed, $2$ splits only in $\mathbb{Q}(\sqrt{-7})$, while $2,3,\dots,37$ are all inert in $\mathbb{Q}(\sqrt{-163})$. For general number fields $K$, the possible splitting types of primes are determined by $d := \deg K$. It is then natural to ask whether $P_{\mathrm{Tam}}(K; 1)$ and $L_{\mathrm{Tam}}(\mathbb{Q}(\sqrt{-D}); -1)$ can be uniformly bounded as a function of $d$. We answer this question in the following corollary, which makes use of the Riemann zeta-function $\zeta(s)$ and Bernoulli numbers.


\begin{corollary}
\label{d_range}
If $K$ has degree $d$, then we have the following uniform bounds on $P_\mathrm{Tam}(K;1)$:
\[(0.5054)^d < P_{\mathrm{Tam}}(\mathbb{Q}; 1)^d \leq P_{\mathrm{Tam}}(K;1) <  (-1)^{d+1}\frac{2(2d)!}{B_{2d}(2\pi)^{2d}} = \frac{1}{\zeta(2d)}. \]
Moreover, we have the following uniform bounds on the average Tamagawa number $L_\mathrm{Tam}(K;-1)$:
\[ \frac{\zeta(2d)}{\zeta(4d)} = (-1)^d\frac{B_{2d}(4d)!}{B_{4d}(2d)!(2\pi)^{2d}} < L_\mathrm{Tam}(K; -1) \leq L_\mathrm{Tam}(\mathbb{Q};-1)^d < (1.8184)^d.\]
\end{corollary}

As $d\to \infty$ in \Cref{d_range}, the given lower and upper bounds for $P_{\mathrm{Tam}}(K;1)$ tend to 0 and 1, respectively. 
We can then ask whether $P_{\mathrm{Tam}}(K;1)$ can be arbitrarily close to $0$ or arbitrarily close to $1$ as $d \to \infty$. More formally, we define 
\begin{equation}
    t^{-}(d) := \inf_{\deg K = d} \left\{ P_{\mathrm{Tam}}(K;1) \right\} \quad \textrm{ and } \quad t^{+}(d) := \sup_{\deg K = d}\left\{ P_{\mathrm{Tam}}(K;1) \right\}
\end{equation}
to be the infimum and supremum of the Tamagawa trivial proportion over number fields $K$ with degree $d$. In a similar vein, one can see in \Cref{d_range} that as $d\to \infty$, the given lower and upper bounds for $L_\mathrm{Tam}(K;-1)$ tend to 1 and $\infty$, respectively. To this end, we similarly define the quantities
\begin{equation}
    \mu^-(d) := \inf_{\deg K = d} \{L_\mathrm{Tam}(K;-1)\} \quad \text{and} \quad \mu^+(d) := \sup_{\deg K=d} \{L_\mathrm{Tam}(K;-1)\}
\end{equation}
to be the infimum and supremum of the average Tamagawa product over number fields $K$ with degree $d$.

It is a conjecture of Ono \cite{onoconj} that as $d \to \infty,$ the limit infimum of $t^{-}(d)$ is 0, the limit supremum of $t^{+}(d)$ is 1, the limit infimum of $\mu^{-}(d)$ is 1, and the limit supremum of $\mu^{+}(d)$ is $\infty$.
With the following theorem, we confirm the conjecture.


\begin{theorem}
\label{d1d2}
 We have
\[
\liminf_{d\rightarrow +\infty} t^-(d) = 0 \quad \textrm{ and } \quad \limsup_{d\rightarrow +\infty} t^+(d) = 1.
\]
Moreover, we have
\[\liminf_{d\to+\infty} \mu^-(d) = 1 \quad \text{and} \quad \limsup_{d\to+\infty} \mu^+(d) = \infty.\]
\end{theorem}

\begin{remark}
The proof of \Cref{d1d2} is constructive. Namely, we provide a family of multiquadratic fields $K$ for which $P_{\mathrm{Tam}}(K;1) \to 0$ and $L_{\mathrm{Tam}}(K;-1) \to \infty$, and a family of cyclotomic fields $K$ for which $P_{\mathrm{Tam}}(K;1) \to 1$ and $L_{\mathrm{Tam}}(K;-1) \to 1$. In \Cref{examples}, we compute $P_{\mathrm{Tam}}(K; 1)$ and $L_{\mathrm{Tam}}(K; -1)$ for example fields within these families.
\end{remark}

The remainder of the paper is structured as follows: In \Cref{prelim}, we introduce Tate's algorithm, a recursive procedure that computes the local invariants for elliptic curves, including the Tamagawa number of an elliptic curve $E/K$ at $\mathfrak{p}$. Running Tate's algorithm at $\mathfrak{p} \nmid (6)$ is relatively straightforward, but additional challenges arise at $\mathfrak{p} \mid (3)$ and $\mathfrak{p} \mid (2)$ since $E/K$ is cubic in $x$ and quadratic in $y$. Therefore, we begin by running Tate's algorithm at $\mathfrak{p} \nmid (6)$ in \Cref{TateP5}, and then run Tate's algorithm for $\mathfrak{p} \mid (3)$ and $\mathfrak{p} \mid (2)$ in Sections \ref{TateP3} and \ref{TateP2}, respectively. In \Cref{proofs}, we prove our main theorems. In \Cref{examples}, we compute $P_{\mathrm{Tam}}(K; m)$ and $L_{\mathrm{Tam}}(K; -1)$ for example number fields to illustrate \Cref{l-series}. For a classification of non-minimal short Weierstrass models at primes above $2$ and $3$, refer to \Cref{appendix}. For the exact proportions of Tamagawa numbers for primes above $2$ and $3$ with large ramification indices, refer to \Cref{charts}. 

\section*{Acknowledgements}
We are grateful for the generous support of the National Science Foundation (Grants DMS 2002265 and DMS 205118), National Security Agency (Grant H98230-21-1-0059), the Thomas Jefferson Fund at the University of Virginia, and the Templeton World Charity Foundation. The authors would like to thank Ken Ono and Wei-Lun Tsai for many helpful suggestions and conversations. 

\section{Tate's Algorithm Over $\mathcal{O}_K$}
\label{prelim}

Tate's algorithm (see \cite[Section 4.9]{SilvermanAdvanced}) is an iterative process that returns the Kodaira type and the Tamagawa number of an elliptic curve $E/K$ at $\mathfrak{p}$. As such, Tate's algorithm allows us to compute $\delta_{K,\mathfrak{p}}(c).$ A single iteration of the algorithm consists of 11 steps. If the model for $E$ has minimal $v_\pi(\Delta)$ over the $\pi-$integral models for $E$, then the algorithm terminates during the first ten steps, which correspond to each of the ten Kodaira types. We refer to such elliptic curves as \emph{$\mathfrak{p}$-minimal models}. However, if $E$ is not $\mathfrak{p}-$minimal, then, at Step 11 of Tate's algorithm, we scale $(x , y) \mapsto (\pi^2 x, \pi^3 y).$ The substitutions from Step 1 through 10 of Tate's algorithm guarantee that $E$ is $\pi-$integral after Step 11. A non-minimal model $E$ then loops back into Step 1 of Tate's algorithm. Tate's algorithm eventually terminates as the scaling at Step 11 decreases $v_{\pi}(\Delta)$ by $12$. At the step in which the algorithm terminates, the Tamagawa number of $E/K$ at $\mathfrak{p}$ is determined. After determining the proportion of elliptic curves with a fixed Tamagawa number that terminate at each step, we sum these proportions over all steps to compute $\delta_{K,\mathfrak{p}}(c)$.

In \cite{og}, Griffin et al.~classify elliptic curves into cases depending on $v_{\pi}(a_4), v_{\pi}(a_6),$ and $v_{\pi}(\Delta).$ They then apply distinct linear shifts to the curves in each case and parametrize $a_4$ and $a_6$ in terms of these shifts prior to running Tate's algorithm. Finally, they run the algorithm on each case separately. In our paper, we do not classify elliptic curves into cases before running Tate's algorithm. Instead, for each $\mathfrak{p},$ we run Tate's algorithm altogether for $\mathfrak{p}-$minimal elliptic curves in the short Weierstrass form. These $\mathfrak{p}-$minimal elliptic curves are guaranteed to terminate during the first iteration of the algorithm. But when $E$ is non-minimal, $E$ passes through Step 11, then loops back into Step 1. For prime ideals $\mathfrak{p} \nmid (6),$ a non-minimal $E(a_4, a_6)$ is still in the short Weierstrass form after Step 11. But when $\mathfrak{p} \mid (3),$ the coefficient $a_2$ may be non-zero after Step 11, since $E$ is cubic in $x$. Therefore, when $\mathfrak{p} \mid (3)$, we must study Tate's algorithm over 
\begin{equation}
\label{E246}
    E(a_2, a_4, a_6) : \ \ y^2 = x^3 + a_2x^2 + a_4x + a_6
\end{equation} to understand how non-minimal curves loop back into Tate's algorithm. When $\mathfrak{p} \mid (2)$, the coefficients $a_1$ and $a_3$ may be nonzero after Step 11, since $E$ is quadratic in $y$. Likewise, we must study Tate's algorithm over 
\begin{equation}
\label{E1346}
    E(a_1,a_3,a_4,a_6) :\ \ y^2 +a_1xy+a_3y = x^3 + a_4x + a_6. 
\end{equation} Therefore, to study how Tate's algorithm acts up on non-minimal $E(a_4, a_6)$, we should study how Tate's algorithm acts upon $E(a_1, a_2, a_3, a_4, a_6).$ 

We compute $\delta_{K,\mathfrak{p}}(c)$ by first studying the $\mathfrak{p}-$minimal curves and then the non-minimal curves. To study the $\mathfrak{p}-$minimal curves, we define $\delta'_{K,\mathfrak{p}}(T, c; \alpha_1, \alpha_2, \alpha_3)$ to be the proportion of $\mathfrak{p}-$minimal models $E(a_1,a_2,a_3,a_4,a_6)$ with $v_\pi(a_i) = \alpha_i$ for $i=1,2,3$, Kodaira type $T$, and Tamagawa number $c$. In \Cref{p=5_pipelines}, \Cref{p=3_pipelines}, and \Cref{p=2_pipelines}, we compute $\delta'_{K,\mathfrak{p}}(T, c;\infty,\infty,\infty)$ for $\mathfrak{p} \nmid (6)$, $\delta'_{K, \mathfrak{p}} (T, c; \infty, \alpha_2, \infty)$ for $\mathfrak{p} \mid (3)$, and $\delta'_{K, \mathfrak{p}}(T, c; \alpha_1,\infty,\alpha_3)$ for $\mathfrak{p} \mid (2)$, respectively. These values of $\delta'$ accounts for the potentially nonzero coefficients after Step 11.

Now, we study the form of non-minimal curves after Step 11. When $\mathfrak{p} \nmid (6),$ the elliptic curves that loop back are still in the short Weierstrass form, which we visualize in a simple Markov chain in \Cref{markovp5}. When $\mathfrak{p} \mid (3)$ (resp.~$\mathfrak{p} \mid (2)$) however, a non-minimal elliptic curve after Step 11 may not be short anymore, and so the underlying Markov chain structure is more complex as in \Cref{markovp3} (resp.~\Cref{markovp2e1}). Families, which are sets of elliptic curves that act as nodes in these Markov chain, are defined in \Cref{family3} and \Cref{family2}. The edges are determined for $\mathfrak{p} \mid(3)$ (resp. $\mathfrak{p} \mid (2)$) in \Cref{p=3_markov} and \Cref{p=3_corollary} (resp. Lemmas \ref{p=2_markov}, \ref{p=2_markov2}, \ref{p=2_markov3}, \ref{p=2_markov4}).

Our analysis of Tate's algorithm boils down to studying congruences in terms of the coefficients $a_1, a_2, a_3, a_4, a_6$ modulo bounded powers of $\pi$. We often note that, when certain quantities like $a_2$, $a_4$, or $a_6 \pmod{\pi}$ are fixed, there are a fixed number of choices for $a_4$ and $a_6$ modulo bounded powers of $\pi$. As such, an important quantity throughout our calculation is the ideal norm $q := N_{K/\mathbb{Q}}(\mathfrak{p})$, or the number of distinct residues in $\mathcal{O}_{K_v}$ modulo $\pi$. To further illustrate the connection between Tate's algorithm and these modular congruences, we generalize the results of Griffin et al.~\cite[Lemmas 2.2, 2.3]{og} by classifying and counting the non-minimal short Weierstrass models at primes $\mathfrak{p} \mid (6)$ in \Cref{appendix}.

\begin{comment}
We refer to the $\pi-$adic digits of various quantities of the elliptic curve. However, these are not the typical $p$-adic digits $\{0,1,\dots,p-1\}$. Instead, in the absence of a canonical way to represent the isomorphism $K_v/(\pi) \cong \mathbb{F}_{p^f}$, the $\pi-$adic digits are abstract representatives of cosets in $K_v/(\pi)$. 

Section $2$ of \cite{og} outlines a method to calculate the proportions of elliptic curves in short Weierstrass form with a specific Tamagawa number at primes with residue field $\mathbb{F}_p$. Griffin et al. defines explicit models and invariants for each class of curves. We generalize their results to the residue field $K_v/(\pi)$ by instead investigating the restrictions imposed on the $\pi-$adic digits of $a_4$ and $a_6$ during each step of Tate's algorithm.

Before running Tate's algorithm, \cite{og} classifies the $(a_4, a_6)$ of non-minimal elliptic curves a priori. To generalize, in \Cref{p=3_lemma} (resp.~\Cref{p=2_lemma}), we completely characterize $(a_4, a_6)$ that admits a $\mathfrak{p}-$non-minimal $E(a_4, a_6)$ for $\mathfrak{p} \mid (3)$ (resp.~$(2)$). But unlike in \cite{og}, the non-minimal models above $\mathfrak{p} \mid (6)$ are not well-behaved. To bypass this hurdle, we classify elliptic curves under a new standard: the $\pi-$adic valuation of its coefficients. After running the algorithm once, we reduce and reclassify the non-minimal models. This structure naturally hints at a Markov chain, in which the nodes are the classes of elliptic curves depending on the valuation of its coefficients. The nodes are then enriched with directed edges. The weight associated to an edge is the proportion of the non-minimal elliptic curves from the starting node that, after an iteration of Tate's algorithm, belong to the ending node. Under this setup, we analyze how short Weierstrass models travel through each node before the algorithm terminates. Formally, we want to calculate $\delta_{K,\mathfrak{p}}(c_{\mathfrak{p}})$, the proportion of $E_{K,\mathfrak{p}}$ with Tamagawa number $c_{\mathfrak{p}}$. We similarly define $\delta_{K,\mathfrak{p}}(T ,c_{\mathfrak{p}})$ to be the proportion of elliptic curves with Kodaira type $T$ and Tamagawa number $c_{\mathfrak{p}}.$ Along the way, to track the proportion of Tamagawa numbers and Kodaira types in a given node per iteration of Tate's algorithm, we define $\delta'_{K,\mathfrak{p}}(T, c_{\mathfrak{p}}, \alpha_1, \alpha_2, \alpha_3)$ to be the proportion of $\mathfrak{p}-$minimal elliptic curves in $E$ with $v_\pi(a_i) = \alpha_i$ for $i=1,2,3$, Kodaira type $T$, and Tamagawa number $c_{\mathfrak{p}}.$
\end{comment}

\section{Classification for $\mathfrak{p} \nmid (6)$}\label{TateP5}

For reasons as described above, the required analysis of Tate's algorithm is significantly more involved for $\mathfrak{p} \mid (6)$. Therefore, we first calculate $\delta_{K,\mathfrak{p}}(c)$ for $\mathfrak{p}$ above $p \geq 5.$ We realize upon running Tate's algorithm for non-minimal short Weierstrass models (see \Cref{p=5_pipelines}) that the coefficients $a_1,$ $a_2,$ and $a_3$ remain invariant at $0$. (This confirms the converse of non-minimality for $\operatorname{char} k \neq 2,3$ in~\cite{Silverman}.) Hence, we define $\varphi_{K,\mathfrak{p}}(T, c_{\mathfrak{p}}) := \delta'_{K,\mathfrak{p}}(T, c_{\mathfrak{p}}, \infty, \infty, \infty)$. The rest of this section is as follows: \Cref{p=5_pipelines} unveils $\varphi_{K,\mathfrak{p}}(T, c_\mathfrak{p})$ and the structure of the associated Markov chain. Then, in \Cref{p=5_total}, we calculate $\delta_{K,\mathfrak{p}}(c_{\mathfrak{p}})$ in \Cref{p=5_total} using the Markov chain and $\varphi_{K,\mathfrak{p}}(T, c_\mathfrak{p})$ from \Cref{p=5_pipelines}. 



\begin{lemma}
\label{p=5_pipelines}
Suppose that $\mathfrak{p} \nmid (6)$ is a prime ideal in $K$. Consider the family of Weierstrass models $E(a_4,a_6) : y^2 = x^3 + a_4x + a_6$. Then the local densities $\varphi_{K,\mathfrak{p}}(T, c)$ are provided in \Cref{pgeq5_varphi}.
\end{lemma}

\setlength{\tabcolsep}{6pt} 
\renewcommand{\arraystretch}{2.}
\begin{center}
\begin{table}[h!]
\begin{tabular}{|c|c|c|||c|c|c|||c|c|c|}
 \hline
Type & $c_\mathfrak{p}$ & $\varphi_{K,\mathfrak{p}}(T,c_{\mathfrak{p}})$ & Type & $c_\mathfrak{p}$ & $\varphi_{K,\mathfrak{p}}(T,c_{\mathfrak{p}})$ &Type &$c_\mathfrak{p}$ &$\varphi_{K,\mathfrak{p}}(T,c_{\mathfrak{p}})$ \\ \hline \hline \hline
$I_0$ & $1$ & $\frac{q-1}{q}$ & $I_0^*$ & $1$ & $\frac{1}{3}\frac{(q^2-1)}{q^7}$ &$III$ & $2$ & $\frac{(q-1)}{q^4}$\\ \hline
$I_1$ & $1$ & $\frac{(q-1)^2}{q^3}$ & $I_0^*$ & $2$ & $\frac{1}{2}\frac{(q-1)}{q^6}$ &  $III^*$ & $2$ & $\frac{(q-1)}{q^9}$\\ \hline
$I_2$ & $2$ & $\frac{(q-1)^2}{q^4}$ & $I_0^*$ & $4$ & $\frac{1}{6}\frac{(q-1)(q-2)}{q^7}$  & $IV$ & $1$ & $\frac{1}{2}\frac{(q-1)}{q^5}$\\ \hline
$I_{n\geq3}$ & $\varepsilon(n)$ & $\frac{1}{2}\frac{(q-1)^2}{q^{n+2}}$ & $I^*_{n\geq 1}$ & $2$ & {\color{black}$\frac{1}{2}\frac{(q-1)^2}{q^{7+n}}$} & $IV$ & $3$ & $\frac{1}{2}\frac{(q-1)}{q^5}$\\ \hline
$I_{n\geq3}$ & $n$ & $\frac{1}{2}\frac{(q-1)^2}{q^{n+2}}$ & $I^*_{n\geq 1}$ & $4$ & {\color{black}$\frac{1}{2}\frac{(q-1)^2}{q^{7+n}}$} & $IV^*$ & $1$ & $\frac{1}{2}\frac{(q-1)}{q^8}$\\ \hline
$II$ & $1$ & $\frac{(q-1)}{q^3}$ & $II^*$ & $1$ & $\frac{(q-1)}{q^{10}}$   & $IV^*$ & $3$ & $\frac{1}{2}\frac{(q-1)}{q^8}$  \\ \hline
\end{tabular}
\smallskip
\caption{The $\varphi_{K,\mathfrak{p}}(T, c)$ for $\mathfrak{p} \nmid (6)$. (Note: $\varepsilon(n):=((-1)^n+3)/2.$)}
\label{pgeq5_varphi}
\end{table}
\end{center}

\begin{remark}
The above local proportions $\varphi_{K,\mathfrak{p}}(T, c)$ exactly match the local proportions $\delta'_p(T, c)$ for $p \geq 5$ in Griffin et al.~\cite[Table 5]{og} after replacing the rational prime $p$ with the ideal norm $q$.
\end{remark}

\begin{proof}
We run through Tate's algorithm over $E(a_4, a_6)$ to compute $\varphi_{K, \mathfrak{p}}(T, c)$. Recall that $q = N_{K/\mathbb{Q}}(\mathfrak{p})$ is the ideal norm of $\mathfrak{p}$.

\begin{comment}
\begin{table}
    \centering
\setlength{\tabcolsep}{6pt} 
\renewcommand{\arraystretch}{2.}
\begin{tabular}{|c|}\hline
\makecell{$a_4$ : \underline{Step 9} \ \underline{Step 6, 7} \ \underline{Step 4} \ \underline{Step 1, 2} \\
$a_6$ : \underline{Step 10} \ \underline{Step 8} \ \underline{Step 6, 7} \ \underline{Step 5} \ \underline{Step 3} \ \underline{Step 1, 2}}\\ \hline
\end{tabular}

\medskip

    \caption{The $\pi-$adic digits of $(a_4,a_6)$ that determines the termination at each step of Tate's algorithm for $\mathfrak{p} \nmid (6)$.}
    \label{fig:digit_fix_p5}
\end{table}
\end{comment}

\noindent \textbf{Case 1.} $E$ terminates if $\pi \nmid \Delta =  - 16 (4 a_4^3+27 a_6^2)$ or when $(a_4, a_6) \not\equiv (-3w^2, 2w^3) \pmod{\pi}.$ Therefore, we have $q^2-q$ choices of $(a_4, a_6)$ modulo $\pi.$ Resultantly, $\varphi_{K,\mathfrak{p}}(I_{0}, 1) = \frac{q-1}{q}.$   

\noindent\textbf{Case 2.} Suppose that $E$ is singular at $(u,0)$ after reduction. We shift $x \mapsto x + u$; the new model is
\begin{equation}
    y^2 = (x+u)^3 + a_4(x+u) + a_6 \label{shift1p5}.
\end{equation}
We stop if $u \not\equiv 0 \pmod{\pi}$. We check that exactly half of the choices of $u$ result in $T^2 - 3u$ splitting. By Hensel's lemma, these curves have a $\frac{(q-1)^2}{q^{n+2}}$ chance of satisfying $n = v_{\pi}(\Delta)$. Hence, we have $\delta_{K,\mathfrak{p}}(I_n, n) = \delta_{K,\mathfrak{p}}(I_n, \varepsilon(n)) = \frac{(q-1)^2}{2q^{n+2}}$. Henceforth, assume $u = 0$, which implies $\pi \mid a_4, a_6$.

\noindent\textbf{Case 3.} We stop if $\pi^2 \nmid a_6.$ There is one choice for $a_4 \pmod{\pi}$ and $q-1$ choices for $a_6 \pmod{\pi^2}$, whence $\delta_{K,\mathfrak{p}}(II, 1) = \frac{q-1}{q^3}.$ Henceforth, assume $\pi^2 \mid a_6$.

\noindent\textbf{Case 4.} We stop when $\pi^2 \nmid a_4$. Thus, we have $q-1$ choices for $a_4 \pmod{\pi^2}$ and one choice for $a_6 \pmod{\pi^2}$, so $\delta_{K,\mathfrak{p}}(III, 1) = \frac{q-1}{q^4}.$ Henceforth, assume $\pi^2 \mid a_4$.

\noindent\textbf{Case 5.} We stop at Step 5 if $\pi^3 \nmid a_6.$ Thus, we have one choice for $a_4 \pmod{\pi^2}$ and $q-1$ choices for $a_6 \pmod{\pi^3}$. The Tamagawa number is $3$ if $Y^2 - (\pi^{-2} a_6)^2$ splits modulo $\pi$, and $1$ otherwise. Hence, $\delta_{K,\mathfrak{p}}(IV, 1) = \delta_{K,\mathfrak{p}}(IV, 3)= \frac{q-1}{2q^5}.$ Henceforth, assume $\pi^3 \mid a_6$.

\noindent\textbf{Case 6.} In Step 6 to 8, we study the polynomial $P(T) = T^3 + \frac{a_4}{\pi^2} T + \frac{a_6}{\pi^3} =: T^3 + A_4 T + A_6$. Note that $A_4$ and $A_6$ are equally distributed among the residues in $k$. The cubic $P(T)$ has discriminant $4A_4^3 + 27A_6^2$. We stop if $P(T)$ has three distinct roots, i.e., if $4A_4^3 + 27A_6^2 \not\equiv 0 \pmod{\pi}$. This accounts for $q^2 - q$ residue pairs as $(A_4, A_6) \not\equiv (-3w^2, 2w^3)$ modulo $\pi$.

We now classify $P(T)$ based on the number of its roots in $k$. First, if $c_{\mathfrak{p}} = 1$, then $P(T)$ must be irreducible. The linear map $\operatorname{Tr} : \mathbb{F}_{q^3} \to \mathbb{F}_q$ is surjective, so there are $q^3 / q = q^2$ traceless elements of $\mathbb{F}_{q^3}$, one of which is in $\mathbb{F}_q$. Thus, there are $\frac{q^2-1}{3}$ such cubics. Next, if $c_{\mathfrak{p}} = 2$, then $P(T)$ factors into a linear term and an irreducible quadratic. There are $\frac{q^2-q}{2}$ irreducible quadratics and upon fixing this quadratic, the linear term is fixed as $P(T)$ is traceless. Finally, if $c_{\mathfrak{p}} = 4$, then $P(T)$ has three roots in $k$. There are $\binom{q}{3}$ ways to select three roots and $1/q$ of them have trace 0, so we have $\frac{(q-1)(q-2)}{6}$ such $P(T)$. In all, we have $\delta_{K,\mathfrak{p}}(I_0^*, 1) = \frac{q^2-1}{3q^7}$, $\delta_{K,\mathfrak{p}}(I_0^*, 2) = \frac{q-1}{2q^6}$, and $\delta_{K,\mathfrak{p}}(I_0^*, 4) = \frac{(q-1)(q-2)}{6q^7}$.

\noindent\textbf{Case 7.} We stop when $4A_4^3 + 27A_6^2 \equiv 0 \pmod{\pi}$ but $(A_4,A_6) \not\equiv (0,0) \pmod{\pi}$. Let $r := \sqrt{A_4/3}$ that serves as the double root of $P(T)$. We accordingly shift $x \mapsto x+\pi r$: 
\begin{equation}
    y^2 = (x+\pi r)^3 + a_4(x+\pi r) + a_6 = x^3 + a_2'x^2 + a_6'.\label{shift2p5}
\end{equation}
Ultimately, the Kodaira type depends on $n := v_\pi(a_6') - 3$, which occurs with proportion $\frac{(q-1)^2}{q^{n+7}}$. The Tamagawa number depends on whether $\frac{a_6'}{\pi^{v_\pi(a'_6)}}$ is a quadratic residue, which happens half of the time. Hence, $\delta_{K,\mathfrak{p}}(I_n^*, 2) = \delta_{K,\mathfrak{p}}(I_n^*, 4) = \frac{(q-1)}{2q^{n+7}}$.

\noindent\textbf{Case 8.} $P(T)$ is traceless. Therefore, its triple root must be $0$, so henceforth $\pi^3 \mid a_4$ and $\pi^4 \mid a_6$. We stop if $Y^2 - \pi^{-4} a_6$ has distinct roots, i.e., if $\pi^{-4} a_6 \not\equiv 0\pmod{\pi}$. We have one choice for $a_4 \pmod{\pi^3}$ and $q-1$ choices for $a_6 \pmod{\pi^5}$, half of which cause $Y^2 - \pi^{-4}a_6$ to split. Thus, $\varphi_{K,\mathfrak{p}}(IV^*, 1) = \varphi_{K,\mathfrak{p}}(IV^*, 3) = \frac{q-1}{2q^8}$.

\noindent\textbf{Case 9.} We terminate if $\pi^4 \nmid a_4$. There are $q-1$ choices for $a_4 \pmod{\pi^4}$ and one choice for $a_6 \pmod{\pi^5}$, whence $\varphi_{K,\mathfrak{p}}(III^*, 2) = \frac{q-1}{q^9}$.

\noindent\textbf{Case 10.} We stop if $\pi^6 \nmid a_6$. There is one choice for $a_4 \pmod{\pi^4}$ and $q-1$ choices for $a_6 \pmod{\pi^6}$, whence $\varphi_{K,\mathfrak{p}}(II^*, 1) = \frac{q-1}{q^{10}}$.

\noindent\textbf{Case 11.} For $E$ not to be $\mathfrak{p}-$minimal, it must be that $\pi^4 \mid a_4$ and $\pi^6 \mid a_6$, so the proportion of non-minimal $E$ is $\frac{1}{q^{10}}.$
\end{proof}

From \Cref{p=5_pipelines}, a non-minimal curve $E(a_4,a_6)$ transforms into $E'(a_4,a_6) := E(\pi^{-4}a_4, \pi^{-6}a_6)$. As we run through the non-minimal $E$, the family of $E'$ is equivalent to that of $E$. Thus, we may rerun \Cref{p=5_pipelines} on the new family $E'$. \Cref{markovp5} illustrates the resultant Markov chain structure.

\begin{figure}
\begin{center}
\begin{minipage}{.5\textwidth}
\begin{tikzpicture}[->, >=stealth', auto, semithick, node distance=3.5cm]
\node[inner sep=0pt] (tate) at (-4.75,-0.5)
    {\includegraphics[width=.20\textwidth]{diagram2.png}};
\tikzstyle{every state}=[draw, rectangle, minimum height = 1cm, minimum width = 1cm, fill=white,draw=black,thick,text=black,scale=1]

\node[state]    (A)                     {Terminate};
\node[state]    (D)[left of=A]   {};

\path
(D) edge                node{$1 - \frac{1}{q^{10}}$}           (A)

(D) edge[loop above]    node{$\frac{1}{q^{10}}$}     (D);

\end{tikzpicture}
\end{minipage}
\end{center}

\caption{The Markov chain structure for short Weierstrass curves when $\mathfrak{p} \nmid (6)$.}
\label{markovp5}
\end{figure}

\begin{proposition}
\label{p=5_total}
If $\mathfrak{p} \nmid (6)$ is a prime ideal in $K$ and $n \geq 1,$ then letting $q := N_{K/\mathbb{Q}}(\mathfrak{p})$ we have
$$
\delta_{K,\mathfrak{p}}(c)=
\begin{cases}
\displaystyle\ 1-\frac{q(6q^7+9q^6+9q^5+7q^4+8q^3+7q^2+9q+6)}{6(q+1)^2(q^8+q^6+q^4+q^2+1)} \ \ \ \ &{\text if }\ n=1,\\[14pt]
\displaystyle \ \ \ \frac{q(2q^7+2q^6+q^5+q^4+2q^3+q^2+2q+2)}{2(q+1)^2(q^8+q^6+q^4+q^2+1)} \ \ \ \  &{\text if}\  n=2,\\[14pt]
\displaystyle\ \ \ \frac{q^2(q^4+1)}{2(q+1)(q^8+q^6+q^4+q^2+1)} \ \ \ \ &{\text if}\ n=3,\\[14pt]
\displaystyle\ \ \ \frac{q^3(3q^2-{2q}-1)}{6(q+1)(q^8+q^6+q^4+q^2+1)} \ \ \ \ &{\text if}\  n=4,\\[14pt]
\displaystyle \ \ \ \frac{q^{10}-2q^9+q^8}{2q^{n}(q^{10}-1)} \ \ \ \ &{\text if }\ n\geq 5.\\
\end{cases}
$$
\end{proposition}

\begin{remark}
When $K = \mathbb{Q}$, then $\mathfrak{p} = (p)$ for rational primes $p$ and we recover the formulae over rationals found in~\cite{og}. Conversely, the expressions for the proportions over general $K$ match that of $\mathbb{Q}$, except the rational prime $p$ is replaced with the prime ideal norm $q$.
\end{remark}

\begin{proof}
\Cref{p=5_pipelines} reveals the formulae for $\varphi_{K,\mathfrak{p}}(T, c_\mathfrak{p})$. Following the Markov chain as illustrated in \Cref{markovp5}, we may explicitly compute
\begin{equation}
\delta_{K,\mathfrak{p}}(c) = \left(1 +\frac{1}{q^{10}}+ \frac{1}{q^{20}}+ \ldots\right) \sum_{c_\mathfrak{p} = n} \varphi_{K,\mathfrak{p}}(T, c_\mathfrak{p}). \qedhere
\end{equation}
\end{proof}



\section{Classification for $\mathfrak{p} \mid (3)$}
\label{TateP3}

In this section, we derive $\delta_{K,\mathfrak{p}}(c)$ for $\mathfrak{p} \mid (3).$ Unlike in \Cref{TateP5}, Tate's algorithm may introduce a non-zero $a_2-$coefficient on non-minimal $E(a_4, a_6)$ that loops back into the algorithm. Therefore, to determine $\delta_{K,\mathfrak{p}}(c),$ we study the action of Tate's algorithm on a larger class of elliptic curves---namely, on $E(a_2, a_4, a_6)$, defined in \Cref{E246}. Upon running Tate's algorithm, we find that the quantity $\alpha_2 := v_\pi(a_2)$ influences whether or not $E$ passes a given step. Moreover, the short Weierstrass elliptic curves are exactly the $E(a_2,a_4,a_6)$ with $\alpha_2 = \infty$. Naturally, we group elliptic curves into families depending on $\alpha_2$ as follows.


\begin{definition}\label{family3}
The \emph{3-family} $F(\alpha_2)$ refers to the set of models
\[F(\alpha_2) := \{E(a_2, a_4, a_6) : v_\pi(a_2) = \alpha_2; \ a_4, a_6 \ \text{integral}\}.\]
The 3-family $F(\geq \alpha_2)$ refers to the set $\bigsqcup_{\alpha \geq \alpha_2} F(\alpha)$.
\end{definition}


For brevity, we refer to 3-families as families for this section. The rest of the section is structured as follows. In \Cref{p=3_pipelines}, we run Tate's algorithm to calculate $\chi_{K,\mathfrak{p}}(T, c; \alpha_2) := \delta'_{K,\mathfrak{p}}(T, c;\infty, \alpha_2, \infty)$, which is the proportion of $\mathfrak{p}-$minimal models with Kodaira type $T$ and Tamagawa number $c$ for each family $F(\alpha_2)$. Then, we show in \Cref{p=3_markov} and \Cref{p=3_corollary} that the non-minimal models from a given family may themselves be viewed as a family. Finally, in \Cref{p=3_total}, we leverage these lemmas to form a Markov chain whose nodes are families and whose edges represent the reclassification of non-minimal models, which we use to compute the local proportion $\delta_{K,\mathfrak{p}}(c)$. 



\begin{lemma}
\label{p=3_pipelines}

Suppose that $\mathfrak{p} \subseteq K$ is above $3$ with ramification index $e$. Then for $F(\alpha_2)$, the local densities $\chi_{K,\mathfrak{p}}(T, c; \alpha_2)$ is as provided in Table~\ref{p3chi}.

{\tiny 
\setlength{\tabcolsep}{6pt} 
\renewcommand{\arraystretch}{2.}
\begin{center}
\begin{table}[ht!]
\begin{tabular}{|c|c||c|c||c|c|c|}
 \hline
  & & \multicolumn{2}{c||}{$e=1$}  &  \multicolumn{3}{c|}{$e\geq 2$}   \\ \hline
 Type & $c_\mathfrak{p}$ & $\alpha_2 = 0$ & $\alpha_2 \geq 1$ & $\alpha_2 = 0$ & $\alpha_2 = 1$ & $\alpha_2 \geq 2$ \\ \hline \hline
 $I_0$ & $1$ & $(q-1)/q$ & $(q-1)/q$ &$(q-1)/q$  & $(q-1)/q$ & $(q-1)/q$ \\ \hline
 $I_1$ & $1$ & $(q-1)/q^2$ & $0$ & $(q-1)/q^2$ & $0$ & $0$ \\ \hline
 $I_2$ & $2$ & $(q-1)/q^3$ & $0$ & $(q-1)/q^3$ & $0$ & $0$ \\ \hline
 $I_{n\ge 3}$ & $n$ & $(q-1)/2q^{n+1}$ & $0$ & $(q-1)/2q^{n+1}$ & $0$ & $0$ \\ \hline
 $I_{n\ge 3}$ & $\varepsilon(n)$ & $(q-1)/2q^{n+1}$ & $0$ & $(q-1)/2q^{n+1}$ & $0$ & $0$ \\ \hline
 $II$ & $1$ & $0$ & $(q -1)/q^2$ & $0$ & $(q-1)/q^2$ & $(q-1)/q^2$ \\ \hline
 $III$ & $2$ & $0$ & $(q-1)/q^3$ & $0$ & $(q-1)/q^3$ & $(q-1)/q^3$ \\ \hline
 $IV$ & $1$ & $0$ & $(q-1)/2q^4$ & $0$ & $(q-1)/2q^4$ & $(q-1)/2q^4$ \\ \hline
 $IV$ & $3$ & $0$ & $(q-1)/2q^4$ & $0$ & $(q-1)/2q^4$ & $(q-1)/2q^4$ \\ \hline
 $I_0^*$ & $1$ & $0$ & $(q^2-1)/3q^6$ & $0$ & $1/3q^5$  & $(q-1)/3q^5$   \\ \hline
 $I_0^*$ & $2$ & $0$ & $(q-1)/2q^5$ & $0$ &  $(q-1)/2q^6$ & $(q-1)/2q^5$ \\ \hline
 $I_0^*$ & $4$ & $0$ & $(q-1)(q-2)/6q^6$ & $0$ & $(q-3)/6q^6$  & $(q-1)/6q^5$ \\ \hline
 $I_n^*$ & $2$ & $0$ & $(q-1)^2/2q^{6+n}$ & $0$ & $(q-1)/2q^{5+n}$ & $0$ \\ \hline
 $I_n^*$ & $4$ & $0$ & $(q-1)^2/2q^{6+n}$ & $0$ & $(q-1)/2q^{5+n}$ & $0$ \\ \hline
 $IV^*$ & $1$ & $0$ & $(q-1)/2q^7$ & $0$ & $0$ & $(q-1)/2q^6$ \\ \hline
 $IV^*$ & $3$ & $0$ & $(q-1)/2q^7$ & $0$ & $0$ & $(q-1)/2q^6$ \\ \hline
 $III^*$ & $2$ & $0$ & $(q-1)/q^8$ & $0$ & $0$ & $(q-1)/q^7$ \\ \hline
 $II^*$ & $1$ & $0$  & $(q-1)/q^9$ & $0$ & $0$ & $(q-1)/q^8$ \\ \hline
 
 
\end{tabular}
\smallskip
\caption{The $\chi_{K,\mathfrak{p}}(T, c; \alpha_2)$ for $\mathfrak{p} \mid (3).$}
\label{p3chi}
\end{table}
\end{center}
}
\end{lemma}


\begin{proof}

\begin{comment}
\begin{table}[ht!]
    \centering
    {\tiny
\setlength{\tabcolsep}{6pt} 
\renewcommand{\arraystretch}{2.}
\begin{tabular}{|c||c|}\hline
\makecell{$\alpha_2 = 0$} & \makecell{$e=1,$ $\alpha_2 \geq 1$} \\ \hline
\makecell{$a_4$ : \underline{ } \ \underline{ } \ \underline{ } \ \underline{ } \\
$a_6$ : \underline{ } \ \underline{ } \ \underline{ } \ \underline{ } \ \underline{ } \ \underline{Step 1}} & 
\makecell{$a_4$ : \underline{Step 9} \ \underline{$A_4$} \ \underline{Step 4} \ \underline{Step 1} \\
$a_6$ : \underline{Step 10} \ \underline{Step 8} \ \underline{$A_6$} \ \underline{v} \ \underline{Step 3} \ \underline{$A_2$}} \\ \hline\hline
\makecell{$e=2,$ $\alpha_2 = 1$} & \makecell{$e=2,$ $\alpha_2 \geq 2$} \\ \hline
\makecell{$a_4$ : \underline{ } \ \underline{$A_4$} \ \underline{Step 4} \ \underline{Step 1} \\
$a_6$ : \underline{ } \ \underline{ } \ \underline{ } \ \underline{$A_6$} \ \underline{Step 3} \ \underline{}} & 
\makecell{$a_4$ : \underline{Step 9} \ \underline{$A_4$} \ \underline{Step 4} \ \underline{Step 1} \\
$a_6$ : \underline{Step 10} \ \underline{Step 8} \ \underline{v} \ \underline{$A_6$} \ \underline{Step 3} \ \underline{$u$}} \\ \hline
\end{tabular}
}

\medskip

    \caption{Digits fixed at each step.}
    \label{fig:digit_fix_p3}
\end{table}
\end{comment} 

We run Tate's algorithm over $F(\alpha_2)$ to compute $\chi_{K,\mathfrak{p}}(T, c;\alpha_2).$

\noindent \textbf{Case 1.} Suppose that $E$ terminates at Step 1. If $E \in F(0)$, then $\Delta \equiv -a_2^3 a_6 + a_2^2 a_4^2 - a_4^3 \pmod{\pi}$. Thus, for $\pi \nmid \Delta,$ there are $q-1$ choices for $a_6$ modulo $\pi$ for fixed $a_2$ and $a_4$. If $E \in F(1)$ or $E \in F(\geq 2)$, then $\Delta \equiv -a_4^3 \pmod{\pi}.$ Therefore, there are $q-1$ choices for $a_4$ modulo $\pi$ for fixed $a_2$ and $a_6$. Thus, $\chi_{K,\mathfrak{p}}(I_{0}, 1;0) = \chi_{K,\mathfrak{p}}(I_{0}, 1; 1) = \chi_{K,\mathfrak{p}}(I_{0}, 1;\geq 2) = \frac{q-1}{q}.$ Moving forward, for curves that pass Step 1, in $F(0),$ the units digit of $a_6$ is fixed for fixed $a_4$ and $a_2$, and in $F(\geq 1)$ the units digit of $a_4$ is $0$.

\noindent
\textbf{Case 2.} Suppose that the singular point of $E$ after reduction by $\pi$ is at $(u,0)$. Then, the translation $x \mapsto x + u$ yields the new model
\begin{equation}
    y^2 = (x+u)^3 + a_2(x+u)^2 + a_4(x+u) + a_6.
\end{equation}
Curves in $F(0)$ always terminate as $\pi \nmid b_2 = a_2 + 3u$. By Hensel's lemma, exactly $\frac{q-1}{q^{n+1}}$ of curves within $F(0)$ satisfy $v_\pi(\Delta) = n$ as $a_2$ varies. We also check that half of these values make $a_2$ a quadratic residue modulo $\pi$. Thus, $\chi_{K,\mathfrak{p}}(I_n, n; 0) = \chi_{K,\mathfrak{p}}(I_n, \varepsilon(n); 0) = \frac{q-1}{2q^{n+1}}$. 

On the other hand, if $E \in F(\geq 1)$, we always pass this step, so $\chi_{K,\mathfrak{p}}(I_n, n; 0) = \chi_{K,\mathfrak{p}}(I_n, \varepsilon(n); 0) = 0$. Since $(u,0)$ lies on the curve after reduction and $v_\pi(a_4) \geq 1$ from Case $1$, note $u^3 + a_6 \equiv 0 \pmod{\pi}$.

\noindent \textbf{Case 3.} If we stop at Step 3, $\pi^2 \nmid a_6 + a_4 u + a_2 u^2 + u^3.$ From Case $2,$ we have that $\pi \mid a_6 + a_4u + a_2u^2 + u^3.$ After fixing $u,a_2,a_4$, we have $q-1$ choices for $a_6 \pmod{\pi^2}$. As a result, $\chi (II, 1; \geq 1) = \frac{q-1}{q^2}$.

{
\noindent
\textbf{Case 4.} To terminate at Step 4, we must have $\pi^3 \nmid 4(a_2+3u)(a_6+a_4u+a_2u^2+u^3) - (3u^2+3a_2u+a_4)^2$. From Steps 2 and 3 respectively, we have that $\pi \mid a_2+3u$ and $\pi^2 \mid a_6 + a_4u + a_2u + u^3$. As such, we stop if $\pi^2 \nmid 3u^2 + 3a_2u + a_4$. We find that for fixed $a_2$ and $u$, there are $q(q-1)$ choices for $a_4 \pmod{\pi^2}$ and one choice for $a_6 \pmod{\pi^2}$. Hence, $\chi_{K,\mathfrak{p}}(III, 2; \geq 1) = \frac{q-1}{q^3}$.

\noindent \textbf{Case 5.} To stop at Step $5,$ $\pi \nmid \pi^{-2}(a_6 + a_4 u + a_2 u^2 + u^3)$. For fixed $a_2$ and $u,$ we have $q$ choices for $a_4 \pmod{\pi^2}$ and $q-1$ choices for $a_6 \pmod{\pi^3}$, half of which make $\pi^{-2}(a_6 + a_4 u + a_2 u^2 + u^3)$ a quadratic residue. Hence, $\chi_{K,\mathfrak{p}}(IV, 1; \geq 1) = \chi_{K,\mathfrak{p}}(IV, 3; \geq 1) = \frac{q-1}{2q^4}.$ Moving forward, we have $q$ choices for $a_4 \pmod{\pi^2}$ and $a_6 \pmod{\pi^3}$ for each choice of $a_2$, $u$, and the $\pi^2-$digit of $a_4$.

\noindent \textbf{Case 6.} We begin by writing $P(T) = T^3 + \frac{3u + a_2}{\pi} T^2 + \frac{3u^2 + 2 a_2 u + a_4}{\pi^2} T + \frac{u^3 + a_2 u^2 + a_4 u + a_6}{\pi^3} =: T^3 + A_2 T^2 + A_4 T + A_6 \pmod{\pi}.$ Note that fixing $A_4, A_6 \pmod{\pi}$ and a choice of $a_2$, $u$, and the $\pi^3-$digit of $a_4$ uniquely determines $a_4 \pmod{\pi^3}$ and $a_6 \pmod{\pi^4}$.

Suppose that we fix $A_2$ and consider $P(T)$ across $(A_4, A_6)$ modulo $\pi.$ For $P(T)$ to have $3$ distinct roots, $P(T)$ and $P'(T) = 2 A_2 x + A_4 \pmod{\pi}$ must not have a shared root. In other words, $A_4^3 + 2 A_2^2 A_4^2 + A_6 A_2^3 \not\equiv 0 \pmod{\pi}.$ As such, for a fixed $A_2$ modulo $\pi,$ there are $q$ choices for $A_4$ and $q-1$ choices of $A_6$ modulo $\pi$ such that $P(T)$ has three distinct roots. Suppose further that all three of $P(T)$'s roots are in $\mathbb{F}_{q}.$ If $P(T)$ is traceless, i.e., $A_2 \equiv 0 \pmod{\pi},$ there are $q$ choices for the first root and $q-1$ choices for the second root. Then, the third root is guaranteed to be different from the first two. If $P(T)$ has a non-zero trace, however, then there are $q$ choices for the first root and $q-3$ choices for the second root to guarantee that the fixed third root is distinct from the first two. Now, suppose that $P(T)$ has exactly one root in $\mathbb{F}_{q}.$ Then, once we fix an irreducible quadratic, the linear term is fixed. Therefore, no matter the trace, the number of $P(T)$ with exactly one root in $\mathbb{F}_{q}$ across $(A_4, A_6)$ modulo $\pi$ is $\frac{q^2-q}{2}.$ The last case is when $P(T)$ is irreducible. In $\mathbb{F}_{q^3},$ there are $\frac{q^2}{3}$ elements of a given trace as the linear map $\operatorname{Tr} : \mathbb{F}_{q^3} \to \mathbb{F}_q$ is surjective. The elements of $\mathbb{F}_{q}$ all have trace 0. Therefore, the number of irreducible $P(T)$ with zero trace is $\frac{q^2-q}{3}$, and the number of that with a non-zero trace is $\frac{q^2-q}{3}$. 

We first discuss $e=1$. Here, a fixed $A_2$ modulo $\pi$ uniquely determines $u$---the units digit of $a_6.$ We thus conclude from the aforementioned analysis that for $e=1$, $\chi_{K,\mathfrak{p}}(I_0^{*}, 1; \geq 1) = \frac{q^2-1}{3 q^6},$ $\chi_{K,\mathfrak{p}}(I_0^{*}, 2; \geq 1) = \frac{q-1}{2 q^5},$ and $\chi_{K,\mathfrak{p}}(I_0^{*}, 4; \geq 1) = \frac{(q-1)(q-2)}{6 q^6}.$ We now repeat for $e \geq 2$ and $E \in F(1)$. Here, the trace is fixed and necessarily non-zero. As such, $\chi_{K,\mathfrak{p}}(I_0^{*}, 1; 1) = \frac{1}{3 q^5},$ $\chi_{K,\mathfrak{p}}(I_0^{*}, 2; 1) = \frac{q-1}{2 q^6},$ and $\chi_{K,\mathfrak{p}}(I_0^{*}, 4; 1) = \frac{q-3}{6 q^6}.$ Finally, we discuss $e\geq 2$ and $E \in F(\geq 2).$ Here, the trace is necessarily zero. Thus, $\chi_{K,\mathfrak{p}}(I_0^{*}, 1; \geq 2) = \frac{q-1}{3 q^5},$ $\chi_{K,\mathfrak{p}}(I_0^{*}, 2; \geq 2) = \frac{q-1}{2 q^5},$ and $\chi_{K,\mathfrak{p}}(I_0^{*}, 4; \geq 2) = \frac{q-1}{6 q^5}.$ 

\noindent \textbf{Case 7.} If $E$ stops at Step 7, $P(T)$ has a double root that is not a triple root. This implies that $A_2 \not\equiv 0 \pmod{\pi}$, but that $A_2^3 + 2A_2^2A_4^2 + A_6A_2^3 \equiv 0 \pmod{\pi}$. For $A_2 \not\equiv 0 \pmod{\pi},$ $\alpha_2 = 1$ and $e \geq 2$, or $\alpha_2 \geq 1$ and $e=1$. Hence, $\chi_{K,\mathfrak{p}}(I_n^*, 2; \geq 1) = \chi_{K,\mathfrak{p}}(I_n^*, 4; \geq 1) = \frac{(q-1)^2}{2q^{6+n}}$ for $e=1$. If $E \in F(1)$ and $e \geq 2$, $E$ terminates at Step 7. Now suppose that $E \in F(\geq 2)$ and $e=1$. Then, for $A_2^3 + 2A_2^2A_4^2 + A_6A_2^3 \equiv 0 \pmod{\pi},$ we have $q-1$ choices for $A_6$ for each $A_4$. We then shift the double root of $P(T)$ to $0.$ The constant term of the shifted model is surjective over $a_6$. Therefore, the proportion of $E$ with valuation $3+n$ is $\frac{q-1}{q^n}$, half of which has $n=2$ and half of which has $n = 4$. Hence, $\chi_{K,\mathfrak{p}}(I_n^*, 2; 1) = \chi_{K,\mathfrak{p}}(I_n^*, 4; 1) = \frac{q-1}{2q^{5+n}}$ and $\chi_{K,\mathfrak{p}}(I_n^*, 2; \geq 2) = \chi_{K,\mathfrak{p}}(I_n^*, 4; \geq 2) = 0$ for $e \geq 2$.

\noindent \textbf{Case 8.} If $E$ reaches Step 8, $P(T)$ has a triple root. If so, then $e\geq 1$ and $E \in F(1)$, or $e\geq 2$ and $E \in F(\geq 2).$ Let the triple root of $P(T)$ be $v$, so $v^3 \equiv A_6 \pmod{\pi}$. We shift the triple root of $P(T)$ to $0$ via $x \mapsto x+v\pi$. Letting $s := u + v\pi$, $E$ becomes
\begin{equation}
    y^2 = (x+s)^3 + a_2(x+s)^2 + a_4(x+s) + a_6.\label{shift2p3}
\end{equation}
We stop if $\pi \nmid \pi^{-4}(s^3 + a_2s^2 + a_4s + a_6)$. We count by fixing $a_2$ and $s \pmod{\pi^2}$, the latter of which we have $q$ choices for $e = 1$ and $q^2$ choices for $e \geq 2$. Resultantly, there is one choice for $a_4 \pmod{\pi^3}$ and $q-1$ choices for $a_6 \pmod{\pi^5}$, half of which make $\pi^{-4}(s^3 + a_2s^2 + a_4s + a_6)$ a quadratic residue. Hence, we have $\chi_{K,\mathfrak{p}}(IV^*, 1; \geq 1) = \chi_{K,\mathfrak{p}}(IV^*, 3; \geq 1) = \frac{q-1}{2q^7}$ for $e=1$, and $\chi_{K,\mathfrak{p}}(IV^*, 1; \geq 2) = \chi_{K,\mathfrak{p}}(IV^*, 3; \geq 2) = \frac{q-1}{2q^6}$ for $e \geq 2$.

\noindent \textbf{Case 9.} Suppose that $E$ reaches Step 9. We stop if $\pi^4 \nmid 3s^2 + 2a_2s + a_4$. As in Case 8, we first choose $a_2$ and $s \pmod{\pi^2}$, the latter of which we have $q$ choices for $e = 1$ and $q^2$ choices for $e \geq 2$. These choices allow for $q-1$ choices of $a_4 \pmod{\pi^4}$ and a unique choice of $a_6 \pmod{\pi^5}$ Hence, we have $\chi_{K,\mathfrak{p}}(III^*, 2; \geq 1) = \frac{q-1}{q^8}$ for $e=1$, and $\chi_{K,\mathfrak{p}}(III^*, 2; \geq 2) = \frac{q-1}{q^7}$ for $e \geq 2$.

\noindent \textbf{Case 10.} We stop at Step 10 if $\pi \nmid \pi^{-5}(s^3 + a_2s^2 + a_4s + a_6)$. Once again, we first choose $a_2$ and $s$, the latter of which we have $q$ choices for $e = 1$ and $q^2$ choices for $e \geq 2$. These choices give a unique choice of $a_4 \pmod{\pi^4}$ and $q-1$ choices of $a_6 \pmod{\pi^6}$. Hence, we have $\chi_{K,\mathfrak{p}}(II^*, 1; \geq 1) = \frac{q-1}{q^9}$ for $e=1$, and $\chi_{K,\mathfrak{p}}(II^*, 1; \geq 2) = \frac{q-1}{q^8}$ for $e \geq 2$.

\noindent \textbf{Case 11.} Suppose $E$ is non-minimal. If $s$ is fixed, then $(a_4, a_6)$ is fixed up to modulo $\pi^4$ and $\pi^6$ respectively. When $e= 1$ and $E \in F(\geq 1)$, there are $q$ choices of $s$ modulo $\pi^2$. Therefore, $\frac{1}{q^9}$ of the curves are non-minimal. When $e \geq 2$ and $E \in F(\geq 2)$, there are $q^2$ choices for $s.$ Hence, $\frac{1}{q^8}$ curves are non-minimal.}
\end{proof}


We now justify how we reclassify the non-minimal models of one family as another family of curves. We first show that, for $\alpha_2 < e$, the local densities at the non-minimal models of a family $F(\alpha_2)$ exactly match the local densities at the family $F(\alpha_2-2)$. To do this, we establish a map which sends a non-minimal model in $F(\alpha_2)$ to another isomorphic model in $F(\alpha_2-2)$, induced by the transformation at Step 11.

\begin{lemma} \label{p=3_markov}
Let $\alpha_2 < e$. There is a surjective, $q^2$-to-$1$ map between the set of non-minimal models in $F(\alpha_2)$ and the set $F(\alpha_2-2)$ that sends $E$ to its transformation $E'$ after passing Step 11.
\end{lemma}

\begin{proof}
Fix a set of $q^2$ representatives to be the residues modulo $\pi^2$. Recall that for non-minimal $E(a_2, a_4, a_6)$, Tate's algorithm produces a unique residue $s \pmod{\pi^2}$ for which
\[y^2 = (x+s)^3 + a_2(x+s)^2 + a_4(x+s) + a_6\]
has the coefficient of $x^i$ divisible by $\pi^i$ for $i=2,4,6$. Hence, each non-minimal model $E(a_2, a_4, a_6) \in F(\alpha_2)$ is sent to 
\[E\left(\frac{3s+a_2}{\pi^2}, \frac{3s^2+2a_2s+a_4}{\pi^4}, \frac{s^3+a_2s^2+a_4s+a_6}{\pi^6}\right) \in F(\alpha_2-2)\] and the map is well-defined.

Conversely, given a model $E'(a_2', a_4', a_6') \in F(\alpha_2-2)$, choosing $s$ uniquely determines $a_2, a_4, a_6$, and moreover $v_\pi(a_2) = \alpha_2$. Hence, $E'$ has exactly $q^2$ preimages in $F(\alpha_2)$, as we had sought.
\end{proof}


It remains to classify the non-minimal models in $F(\geq e)$, and in particular, the non-minimal short Weierstrass models in $F(\infty)$. In fact, models in $F(\geq e)$ and $F(\infty)$ have the same local properties in the following sense. Note that linear transformations do not change the local data of an elliptic curve. As such, instead of computing local densities on short Weierstrass models $F(\infty)$, we may compute the local densities on the set
\begin{equation}
\{\left(E : y^2 = (x+t)^3 + a_4(x+t) + a_6\right) : t, a_4, a_6 \ \text{integral} \}.
\label{eq:reshiftp3}
\end{equation}
By the surjectivity of $a_4$ and $a_6$, the set in \cref{eq:reshiftp3} is precisely $F(\geq e)$. Thus, we instead begin with the curves in $F(\geq e)$ moving forward.


Now, in the manner of \Cref{p=3_markov}, we demonstrate that the local densities at the non-minimal models of $F(\geq e)$ match the local densities at $F(\geq e-2)$. The proof of the following corollary is analogous to that of \Cref{p=3_markov}.

\begin{corollary} \label{p=3_corollary}
There is a surjective, $q^2$-to-$1$ map between the set of non-minimal models in $F(\geq e)$ and the set $F(\geq e-2)$ that sends $E$ to its transformation $E'$ after passing Step $11$.
\end{corollary}

Equipped with our lemmas, we complete our classification for $\mathfrak{p} \mid (3)$ in \Cref{p=3_total}. A key ingredient is the underlying Markov chain structure that helps us study how non-minimal curves loop back into Tate's algorithm. Note that Lemmas \ref{p=3_markov} and \ref{p=3_corollary} allows us to identify the non-minimal curves of $F(0), F(1), \dots, F(\geq e)$ to be identified with other families. In particular, we see from \Cref{p=3_markov} that the non-minimal curves in $F(\alpha)$ for $\alpha < e$ are always transformed into curves in $F(\alpha-2)$. We also see from \Cref{p=3_corollary} that the non-minimal curves in $F(\geq e)$ with proportion $\frac{1}{q^2}$ loop back to itself, with proportion $\frac{q-1}{q^2}$ loop to $F(e-1),$ and with proportion $\frac{q-1}{q}$ transform to $F(e-2).$ Finally, \Cref{p=3_pipelines} determines the local densities at each $F(\alpha_2)$. We collate all of this information in Figure~\ref{markovp3}.


\begin{figure}
\begin{minipage}{.3\textwidth}
\centering
\scalebox{0.9}{\begin{tikzpicture}[->, >=stealth', auto, semithick, node distance=3.4cm]
\node[inner sep=0pt] (tate) at (-3.7,2)
    {\includegraphics[width=.25\textwidth]{diagram2.png}};
\tikzstyle{every state}=[draw, rectangle, minimum height = 1cm, minimum width = 1cm, fill=white,draw=black,thick,text=black,scale=1]

\node[state]    (A)                     {Terminate};
\node[state]    (B)[below left of=A]   {$F(0)$};
\node[state]    (C)[above left of=A]   {$F(\ge 1)$};
\path
(B) edge                node{$1$}           (A)
(C) edge[loop above]    node{$\frac{1}{q^{10}}$}     (C)
    edge[bend right]    node{$\frac{q-1}{q^{10}}$}      (B)
    edge
    node{$1 - \frac{1}{q^{9}}$}         (A);
\end{tikzpicture}}
\end{minipage}
\quad \quad
\begin{minipage}{.3\textwidth}
\centering
\scalebox{0.77}{\begin{tikzpicture}[->, >=stealth', auto, semithick, node distance=1.8cm]
\node[inner sep=0pt] (tate) at (-4.55,7.3)
    {\includegraphics[width=.25\textwidth]{diagram2.png}};
\tikzstyle{every state}=[draw, rectangle, minimum height = 1cm, minimum width = 1cm, fill=white,draw=black,thick,text=black,scale=1]

\node[state]    (A)                     {Terminate};
\node[state]    (D)[above left of=A, xshift=-2cm, yshift=1cm]   {$F(e-3)$};
\node[state]    (B)[above of=D]   {$F(e-2)$};
\node[state]    (F)[above of=B]   {$F(e-1)$};
\node[state]    (G)[above of=F]   {$F(\ge e)$};
\node[state]    (C)[below left of=A, xshift=-2cm, yshift=-1cm]   {$F(2)$};
\node[state]    (I)[below of=C]   {$F(1)$};
\node[state]    (J)[below of=I]   {$F(0)$};
\path (D) -- node[auto=false]{\Huge\mydots} (C);

\path
(I) edge[bend right=45, swap]                node{$1$}           (A)
(J) edge[bend right=65, swap]               node{$1$}           (A)

(G) edge[loop above]    node{$\frac{1}{q^{10}}$}     (G)
    edge[bend left]    node{$\frac{q-1}{q^{10}}$}      (F)
    edge[bend left=80]     node{$\frac{q-1}{q^{9}}$}      (B)
    edge[bend left=63]
    node{$1 - \frac{1}{q^{8}}$}         (A)
(B) edge[bend left=15]
    node{$1 - \frac{1}{q^{8}}$}         (A)
(D) edge[bend right] 
    node{$1 - \frac{1}{q^{8}}$}         (A)
    
(F) edge[bend right=85]    node{$\frac{1}{q^{8}}$}      (D)
    edge[bend left=40]
    node{$1 - \frac{1}{q^{8}}$}         (A)
(C) edge[bend right=85]  node{$\frac{1}{q^{8}}$}      (J)
    edge[bend left, swap]
    node{$1 - \frac{1}{q^{8}}$}         (A);

\path (B)  edge[bend right=60] node{$\frac{1}{q^{8}}$} ([xshift=-12mm]$ (C) !.60! (D) $);

\path ([xshift=-12mm]$ (C) !.40! (D) $) edge[bend right=60] node{$\frac{1}{q^{8}}$} (I);

\end{tikzpicture}}
\end{minipage}
\caption{The Markov chain structure when $\mathfrak{p} \mid (3)$ for $e = 1$ (left) and $e \ge 2$ (right).}
\label{markovp3}
\end{figure}

\begin{proposition
\label{p=3_total}
If $\mathfrak{p} \mid (3)$ is a prime ideal in $K$ and $n \geq 1,$ then letting $q := N_{K/\mathbb{Q}}(\mathfrak{p})$ and $e = 1$ we have
\[
\delta_{K,\mathfrak{p}}(c)=
{\small
\begin{cases}
\displaystyle\ 1 - \frac{(q - 1)(6q^{10} + 9q^{9} + 7q^{8} + 8q^7 + 7q^6 + 9q^5 + 6q + 3)}{6q^2(q + 1)(q^{10} - 1)} \ \ \ \ &{\text if }\ n=1,\\[14pt]
\displaystyle \ \ \ \frac{(q - 1)(2q^{11} + 2q^{10} + q^9 + 2q^8 + q^7 + 2q^6 + 2q^5 + 2q^2 - 1)}{2q^3(q + 1)(q^{10}-1)}  &{\text if}\  n=2,\\[14pt]
\displaystyle\ \ \  \frac{(q - 1)(q^{10} + q^7 + q - 1)}{2q^4(q^{10}-1)}   \ \ \ \ &{\text if}\ n=3,\\[14pt]
\displaystyle\ \ \  \frac{(q - 1)(q^{10} + q^9 + 3q - 3)}{6q^5(q^{10}-1)}  \ \ \ \ &{\text if}\  n=4,\\[14pt]
\displaystyle \ \ \  \frac{(q - 1)^2}{2q^{n+1}(q^{10}-1)}   \ \ \ \ &{\text if }\ n\geq 5.\\
\end{cases}}
\]

If $e \ge 2$ is even, we have
\[
\delta_{K,\mathfrak{p}}(c)=
{\small
\begin{cases}
 \displaystyle\ 1 - (q - 1) \Bigg[\displaystyle\ \frac{\left(6q^{14} + 9q^{13} + 13q^{12} + 16q^{11} + 22(q^{10} + q^{9} + q^{8})\right)}{6(q + 1)(q^2 + 1)(q^4 + 1)(q^{10} - 1)} + O\left(\frac{1}{q^{10}}\right) \Bigg]&{\text if}\  n=1,\\\\[6pt]

\displaystyle \ \ \ \frac{(q - 1)(2q^{13} + 3q^{11} + 5q^{9} + 5q^{7} + 5q^{5} + 3q^{3} + 2q)}{2(q^2 + 1)(q^4 + 1)(q^{10} - 1)} + O\left( \frac{1}{q^{4e+3}}\right)\ \ \ \ &{\text if}\  n=2,\\\\[6pt]
\displaystyle\ \ \  \frac{(q - 1)(q^{4e+8} + q^{4e+6} + q^{4e+4} + q^{4e+2} + q^{4e} - q^6 + q^4 - q^2)}{2q^{4e - 2}(q^4 + 1)(q^{10}-1)}   \ \ \ \ &{\text if}\ n=3,\\\\[6pt]
\displaystyle\ \ \  \frac{(q-1)(q^{11} + q^{9} + q^{7} + q^{5} + q^{3})}{6(q^2 + 1)(q^4 + 1)(q^{10}-1)} + O\left(\frac{1}{q^{4e+4}} \right)  \ \ \ \ &{\text if}\  n=4,\\\\[6pt]
\displaystyle \ \ \  \frac{(q - 1)^2}{2q^{4e + n - 8}(q^{10}-1)}   \ \ \ \ &{\text if }\ n\geq 5. \\
\end{cases}}
\]

If $e > 2$ is odd, we have
$$
\delta_{K,\mathfrak{p}}(c)=
{\small
\begin{cases}
 \displaystyle\ 1 - (q - 1) \Bigg[\displaystyle\ \frac{\left(6q^{14} + 9q^{13} + 13q^{12} + 16q^{11} + 22(q^{10} + q^{9} + q^{8}) \right)}{6(q + 1)(q^2 + 1)(q^4 + 1)(q^{10} - 1)} + O\left(\frac{1}{q^{10}} \right) \Bigg]\ \ \ \ &{\text if}\  n=1, \\\\[6pt] 

\displaystyle \ \ \ \frac{(q - 1)(2q^{13} + 3q^{11} + 5q^{9} + 5q^{7} + 5q^{5} + 3q^{3})}{2(q^2 + 1)(q^4 + 1)(q^{10} - 1)} + O\left(\frac{1}{q^{4e-2}} \right) \ \ \ \ &{\text if}\  n=2, \\\\[6pt]
\displaystyle\ \ \  \frac{(q - 1)(q^{10} + q^{8} + q^{6} + q^{4} + q^{2})}{2(q^4 + 1)(q^{10}-1)} + O\left(\frac{1}{q^{4e-1}} \right) \ \ \ \ &{\text if}\  n=3, \\\\[6pt]
\displaystyle\ \ \  \frac{(q-1)(q^{11} + q^{9} + q^{7} + q^{5} + q^{3})}{6(q^2 + 1)(q^4 + 1)(q^{10}-1)} + O\left(\frac{1}{q^{4e+1}} \right) \ \ \ \ &{\text if}\  n=4, \\\\[6pt]
\displaystyle \ \ \  \frac{(q - 1)^2}{2q^{4e + n - 3}(q^{10}-1)}   \ \ \ \ &{\text if }\ n\geq 5.\\
\end{cases}
}
$$

The exact proportions are given in \Cref{charts}.
\end{proposition}



\begin{proof

Refer to Figure~\ref{markovp3}. Because we begin with a short Weierstrass form, we start at the node $F(\geq e)$. Fix a Kodaira type $T$ and Tamagawa number $c$ over which we compute the local density of curves with this data. We perform casework on the family we terminate in.

First, suppose that $e = 1$. The proportion of curves that terminate in $F(\geq 1)$ with Kodaira type $T$ and Tamagawa number $c$ is 
\[\left(1+ \frac{1}{q^{10}} + \frac{1}{q^{20}} + \ldots\right)\chi_{K,\mathfrak{p}}(T, c, \geq 1).\] 
On the other hand, the proportion of curves that terminate in $F(0)$ with our prescribed local data is 
\[\left(1+ \frac{1}{q^{10}} + \frac{1}{q^{20}} + \ldots\right)\frac{q-1}{q^{10}}\chi_{K,\mathfrak{p}}(T, c, 0).\]

Now, suppose that $e \geq 2$. The proportion of curves that terminate in $F(2), F(3), \dots, F(\geq e)$ with Kodaira type $T$ and Tamagawa number $c$ is
\[\left( 1 + \frac{1}{q^{10}} + \frac{1}{q^{20}} + \dots \right) \left(1 + \sum_{k=0} ^{\floor{(e-1)/2}} \frac{q-1}{q^{8k+10}} + \sum_{k=0} ^{\floor{(e-2)/2}} \frac{q-1}{q^{8k+9}} \right)\chi_{K,\mathfrak{p}}(T, c, \geq2).\]
Second, the proportion of curves that terminate in $F(1)$ with our prescribed local data is 
\[\begin{cases}\left(1+\frac{1}{q^{10}} + \frac{1}{q^{20}} \dots\right) \frac{q-1}{q^{4e+2}} \chi_{K,\mathfrak{p}}(T, c, 1) & \text{if }2 \mid e \\
\left(1+\frac{1}{q^{10}} + \frac{1}{q^{20}} \dots\right)\frac{q-1}{q^{4e-3}}\chi_{K,\mathfrak{p}}(T, c, 1) & \text{if }2 \nmid e\end{cases}.\] Finally, the proportion of curves that terminate in $F(0)$ with our prescribed local data is \[\begin{cases}\left(1+\frac{1}{q^{10}} + \frac{1}{q^{20}} \dots\right)\frac{q-1}{q^{4e+1}}\chi_{K,\mathfrak{p}}(T, c, 0) & \text{if }2 \mid e \\
\left(1+\frac{1}{q^{10}} + \frac{1}{q^{20}} \dots\right)\frac{q-1}{q^{4e+6}}\chi_{K,\mathfrak{p}}(T, c, 0) & \text{if }2 \nmid e
\end{cases}.\]

The values $\delta_{K,\mathfrak{p}}(T, c)$ are provided in \Cref{p=2_pipelines}. We sum these proportions over all Kodaira types $T$ with Tamagawa number $c$ to get the total density $\delta_{K,\mathfrak{p}}(c)$.
\end{proof}


\section{Classification for $\mathfrak{p} \mid (2)$}
\label{TateP2}

In this section, we calculate $\delta_{K,\mathfrak{p}}(c_{\mathfrak{p}})$ for $\mathfrak{p} \mid (2).$ Unlike in \Cref{TateP5} and in \Cref{TateP3}, Tate's algorithm may introduce a non-zero $a_1,$ $a_2,$ and $a_3$ coefficient on non-minimal $E(a_4, a_6)$ that loops back into the algorithm. However, even if $E$ is transformed after Step 11 into $E'(a_1', a_2', a_3', a_4', a_6')$, as in \Cref{E1346}, with $a_2' \neq 0$, the translation $x \mapsto x - a_2/3$ eliminates the $a_2$ coefficient of $E'$ without changing the local data of the curve. Therefore, to study how $\mathfrak{p}-$non-minimal elliptic curves loop back into Tate's algorithm, we study the action of Tate's algorithm on the larger class of elliptic curves $E = E(a_1, a_3, a_4, a_6)$, defined as in \Cref{E1346}. By convention, we re-eliminate the $a_2$ coefficient after passing Step 11 before re-running Tate's algorithm. 

Upon running Tate's algorithm, we find that the sets of elliptic curves $E(a_1, a_3, a_4, a_6)$ across $(a_4, a_6)$ with fixed $a_1$ and $a_3$ behave similarly if they have the same $\alpha_1 := v_\pi(a_1)$ and $\alpha_3 := v_\pi(a_3)$. Moreover, the short Weierstrass elliptic curve are exactly the $E(a_1,a_3,a_4,a_6)$ with $\alpha_1 = \alpha_3 = \infty$. We thus group elliptic curves entering Tate's algorithm into families depending on $\alpha_1$ and $\alpha_3$ as follows.

\begin{definition}\label{family2}
The \emph{2-family} $F(\alpha_1, \alpha_3)$ refers to the set of models
\[F(\alpha_1, \alpha_3) := \{E(a_1, a_3, a_4, a_6) : v_\pi(a_1) = \alpha_1, v_{\pi}(a_3) = \alpha_3; \ a_4, a_6 \ \text{integral}\}.\]
The 2-family $F(\geq \alpha_1, \geq \alpha_3)$ refers to the set $\bigsqcup_{\alpha \geq \alpha_1} \bigsqcup_{\beta \geq \alpha_3} F(\alpha, \beta)$.
\end{definition}


For brevity, we refer to 2-families as families for the rest of this section. The rest of the section is structured similarly to \Cref{TateP3}. In \Cref{p=2_pipelines}, we run Tate's algorithm to calculate $\psi_{K,\mathfrak{p}}(T, c;\alpha_1,\alpha_3) := \delta'_{K,\mathfrak{p}}(T, c;\alpha_1,\infty,\alpha_3)$, which is the proportion of $\mathfrak{p}-$minimal models with Kodaira type $T$ and Tamagawa number $c$ for each family $F(\alpha_1,\alpha_3)$. Then, we show in Lemmas \ref{p=2_markov}, \ref{p=2_markov2}, \ref{p=2_markov3}, and \ref{p=2_markov4} that the non-minimal models from certain families may themselves be viewed as a family. The analysis of non-minimal models in this section is more involved than the analysis in \Cref{TateP3} for two reasons: there are now two relevant valuations $\alpha_1$ and $\alpha_3$, and we have to incorporate the shift $x \mapsto x - a_2/3$ after Step 11. Finally, in \Cref{p=2_total}, we leverage these lemmas to form a Markov chain whose nodes are families and whose edges represent the reclassification of non-minimal models, which we use to compute the local proportion $\delta_{K,\mathfrak{p}}(c)$.

\begin{lemma}
\label{p=2_pipelines}

Suppose that $\mathfrak{p} \subseteq K$ is above $2$ with ramification index $e$. Then for $F(\alpha_1, \alpha_3)$, the local densities $\psi_{K,\mathfrak{p}}(T, c_{\mathfrak{p}}, \alpha_1, \alpha_3)$ is as provided in Table~\ref{p2psi1} for $e= 1, 2$ and Table~\ref{p2psi2} for $e \geq 3$. 

{\tiny 
\setlength{\tabcolsep}{6pt} 
\renewcommand{\arraystretch}{2.}
\begin{center}
\begin{table}[ht!]
\begin{tabular}{|c|c||c|c|c||c|c|c|c|c|}
 \hline
  & & \multicolumn{3}{c||}{$e=1$}  &  \multicolumn{5}{c|}{$e=2$} \\ \hline
 Type & $c_\mathfrak{p}$ & $\alpha_1 = 0$ & \makecell{$\alpha_1 \geq 1$\\$\alpha_3 = 0$} & \makecell{$\alpha_1 \geq 1$ \\ $\alpha_3 \geq 1$} & $\alpha_1=0$ & \makecell{$\alpha_1 \geq 1$\\$\alpha_3 = 0$} & \makecell{$\alpha_1 = 1$\\$\alpha_3 \geq 1$} & \makecell{$\alpha_1 \geq 2$ \\ $\alpha_3 = 1$} & \makecell{$\alpha_1 \geq 2$ \\ $\alpha_3 \geq 2$} \\ \hline \hline
 $I_0$ & 1 & $\frac{q-1}{q}$ & $1$ & $0$ & $\frac{q-1}{q}$ & $1$ & $0$ & $0$ & $0$ \\\hline
 $I_1$ & 1 & $\frac{q-1}{q^2}$ & 0 & 0 & $\frac{q-1}{q^2}$ & 0 & 0 & 0 & 0 \\\hline
 $I_2$ & 2 & $\frac{q-1}{q^3}$ & 0 & 0 & $\frac{q-1}{q^3}$ & 0 & 0 & 0 & 0 \\\hline
 $I_{n\ge3}$ & n & $\frac{q-1}{2q^{n+1}}$ & 0 & 0 & $\frac{q-1}{2q^{n+1}}$ & 0 & 0 & 0 & 0 \\\hline
 $I_{n\ge3}$ & $\varepsilon(n)$ & $\frac{q-1}{2q^{n+1}}$ & 0 & 0 & $\frac{q-1}{2q^{n+1}}$ & 0 & 0 & 0 & 0 \\\hline
 $II$ & 1 & 0 & 0 & $\frac{q-1}q$ & 0 & 0 & $\frac{q-1}q$ & $\frac{q-1}q$ & $\frac{q-1}q$ \\ \hline 
 $III$ & 2 & 0 & 0 & $\frac{q-1}{q^2}$ & 0 & 0 & $\frac{q-1}{q^2}$ & $\frac{q-1}{q^2}$ & $\frac{q-1}{q^2}$ \\ \hline
 $IV$ & 1 & 0 & 0 & $\frac{q-1}{2q^3}$ & 0 & 0 & $\frac{q-1}{2q^3}$ & $\frac{1}{2q^2}$ & 0 \\ \hline
 $IV$ & 3 & 0 & 0 & $\frac{q-1}{2q^3}$ & 0 & 0 & $\frac{q-1}{2q^3}$ & $\frac{1}{2q^2}$ & 0 \\ \hline
 $I_0^*$ & 1 & 0 & 0 & $\frac{q^2-1}{3q^5}$ & 0 & 0 & $\frac{q^2-1}{3q^5}$ & 0 & $\frac{q^2-1}{3q^4}$ \\ \hline
 $I_0^*$ & 2 & 0 & 0 & $\frac{q-1}{2q^4}$ & 0 & 0 & $\frac{q-1}{2q^4}$ & 0 & $\frac{q-1}{2q^3}$ \\ \hline
 $I_0^*$ & 4 & 0 & 0 & $\frac{(q-1)(q-2)}{6q^5}$ & 0 & 0 & $\frac{(q-1)(q-2)}{6q^5}$ & 0 & $\frac{(q-1)(q-2)}{6q^4}$ \\ \hline
 $I_n*$ & 2 & 0 & 0 & $\frac{(q-1)^2}{2q^{5+n}}$ & 0 & 0 & $\frac{(q-1)^2}{2q^{5+n}}$ & 0 & $\frac{(q-1)^2}{2q^{4+n}}$ \\ \hline
 $I_n*$ & 4 & 0 & 0 & $\frac{(q-1)^2}{2q^{5+n}}$ & 0 & 0 & $\frac{(q-1)^2}{2q^{5+n}}$ & 0 & $\frac{(q-1)^2}{2q^{4+n}}$ \\ \hline
 $IV*$ & 1 & 0 & 0 & $\frac{q-1}{2q^6}$ & 0 & 0 & $\frac{q-1}{2q^6}$ & 0 & $\frac{q-1}{2q^5}$ \\ \hline
 $IV*$ & 3 & 0 & 0 & $\frac{q-1}{2q^6}$ & 0 & 0 & $\frac{q-1}{2q^6}$ & 0 & $\frac{q-1}{2q^5}$ \\ \hline
 $III*$ & 2 & 0 & 0 & $\frac{q-1}{q^7}$ & 0 & 0 & $\frac{q-1}{q^7}$ & 0 & $\frac{q-1}{q^6}$ \\ \hline 
 $II*$ & 1 & 0 & 0 & $\frac{q-1}{q^8}$ & 0 & 0 & $\frac{q-1}{q^8}$ & 0 & $\frac{q-1}{q^7}$ \\ \hline 


\end{tabular}
\smallskip
\caption{The $\psi_{K,\mathfrak{p}}(T, c;\alpha_1, \infty, \alpha_3)$ for $\mathfrak{p} \mid (2).$}
\label{p2psi1}
\end{table}
\end{center}
}

{\tiny 
\setlength{\tabcolsep}{6pt} 
\renewcommand{\arraystretch}{2.}
\begin{center}
\begin{table}[ht!]
\begin{tabular}{|c|c||c|c|c|c|c|c|c|}
 \hline
  \multicolumn{9}{|c|}{$e\geq 3$} \\ \hline
 Type & $c_\mathfrak{p}$ & $\alpha_1=0$ & \makecell{$\alpha_1 \geq 1$\\$\alpha_3 = 0$} & \makecell{$\alpha_1 = 1$\\$\alpha_3 \geq 1$} & \makecell{$\alpha_1 \geq 2$ \\ $\alpha_3 = 1$} & \makecell{$\alpha_1 = 2$ \\ $\alpha_3 \geq 2$} & \makecell{$\alpha_1 \geq 3$ \\ $\alpha_3 = 2$} & \makecell{$\alpha_1 \geq 3$ \\ $\alpha_3 \geq 3$} \\ \hline \hline
 $I_0$ & 1 & $\frac{q-1}q$ & 1 & 0 & 0 & 0 & 0 & 0 \\\hline
 $I_1$ & 1 & $\frac{q-1}{q^2}$ & 0 & 0 & 0 & 0 & 0 & 0 \\\hline
 $I_2$ & 2 & $\frac{q-1}{q^3}$ & 0 & 0 & 0 & 0 & 0 & 0 \\\hline
 $I_{n\ge3}$ & n & $\frac{q-1}{2q^{n+1}}$ & 0 & 0 & 0 & 0 & 0 & 0 \\\hline
 $I_{n\ge3}$ & $\varepsilon(n)$ & $\frac{q-1}{2q^{n+1}}$ & 0 & 0 & 0 & 0 & 0 & 0 \\\hline
 $II$ & 1 & 0 & 0 & $\frac{q-1}q$ & $\frac{q-1}q$ & $\frac{q-1}q$ & $\frac{q-1}q$ & $\frac{q-1}q$\\\hline
 $III$ & 2 & 0 & 0 & $\frac{q-1}{q^2}$ & $\frac{q-1}{q^2}$ & $\frac{q-1}{q^2}$ & $\frac{q-1}{q^2}$ & $\frac{q-1}{q^2}$\\\hline
 $IV$ & 1 & 0 & 0 & $\frac{q-1}{2q^3}$ & $\frac{1}{2q^2}$ & 0 & 0 & 0\\\hline 
 $IV$ & 3 & 0 & 0 & $\frac{q-1}{2q^3}$ & $\frac{1}{2q^2}$ & 0 & 0 & 0\\\hline

 $I_0^*$ & 1 & 0 & 0 & $\frac{q^2-1}{3q^5}$ & 0 & $\frac{q^2-1}{3q^4}$ & $\frac{q^2-1}{3q^4}$ & $\frac{q^2-1}{3q^4}$\\\hline
 $I_0^*$ & 2 & 0 & 0 & $\frac{q-1}{2q^4}$ & 0 & $\frac{q-1}{2q^3}$ & $\frac{q-1}{2q^3}$ & $\frac{q-1}{2q^3}$\\\hline
 $I_0^*$ & 4 & 0 & 0 & $\frac{(q-1)(q-2)}{6q^5}$ & 0 & $\frac{(q-1)(q-2)}{6q^4}$ & $\frac{(q-1)(q-2)}{6q^4}$ & $\frac{(q-1)(q-2)}{6q^4}$\\\hline
 $I_n*$ & 2 & 0 & 0 & $\frac{(q-1)^2}{2q^{5+n}}$ & 0 & $\frac{(q-1)^2}{2q^{4+n}}$ & $\frac{(q-1)^2}{2q^{4+n}}$ & $\frac{(q-1)^2}{2q^{4+n}}$\\\hline
 $I_n*$ & 4 & 0 & 0 & $\frac{(q-1)^2}{2q^{5+n}}$ & 0 & $\frac{(q-1)^2}{2q^{4+n}}$ & $\frac{(q-1)^2}{2q^{4+n}}$ & $\frac{(q-1)^2}{2q^{4+n}}$\\\hline
 $IV*$ & 1 & 0 & 0 & $\frac{q-1}{2q^6}$ & 0 & $\frac{q-1}{2q^5}$ & $\frac{1}{2q^4}$ & 0\\\hline
 $IV*$ & 3 & 0 & 0 & $\frac{q-1}{2q^6}$ & 0 & $\frac{q-1}{2q^5}$ & $\frac{1}{2q^4}$ & 0\\\hline
 $III*$ & 2 & 0 & 0 & $\frac{q-1}{q^7}$ & 0 & $\frac{q-1}{q^6}$ & 0 & $\frac{q-1}{q^5}$ \\\hline
 $II*$ & 1 & 0 & 0 & $\frac{q-1}{q^8}$ & 0 & $\frac{q-1}{q^7}$ & 0 & $\frac{q-1}{q^6}$ \\\hline
 

\end{tabular}
\smallskip
\caption{The $\psi_{K,\mathfrak{p}}(T, c; \alpha_1, \infty, \alpha_3)$ for $\mathfrak{p} \mid (2).$}
\label{p2psi2}
\end{table}
\end{center}
}
\end{lemma}

{
\begin{proof}

\begin{comment}
\begin{table}[ht!]
    \centering
    {\tiny
\setlength{\tabcolsep}{6pt} 
\renewcommand{\arraystretch}{2.}
\begin{tabular}{|c||c||c|}\hline
\makecell{$e = 1$, $v_\pi(a_1) \geq 1$, $v_\pi(a_3) \geq 1$} & \makecell{$e=2$, $v_\pi(a_1) \geq 2$, $v_\pi(a_3) \geq 2$} & \makecell{$e\geq 3$, $v_\pi(a_1) \geq 3$, $v_\pi(a_3) \geq 3$} \\ \hline
\makecell{$a_4$ : \underline{Step 9} \ \underline{$\triangle$} \ \underline{Step 4} \ \underline{$s$} \\
$a_6$ : \underline{Step 10} \ \underline{$w$} \ \underline{$\square$} \ \underline{Step 8} \ \underline{Step 3} \ \underline{Step 5}} & 
\makecell{$a_4$ : \underline{Step 9} \ \underline{$\triangle$} \ \underline{Step 4} \ \underline{$s$} \\
$a_6$ : \underline{Step 10} \ \underline{$w$} \ \underline{$\square$} \ \underline{$v$} \ \underline{Step 3} \ \underline{Step 8}} &
\makecell{$a_4$ : \underline{Step 9} \ \underline{$\triangle$} \ \underline{Step 4} \ \underline{$s$} \\
$a_6$ : \underline{Step 10} \ \underline{$w$} \ \underline{$\square$} \ \underline{$v$} \ \underline{Step 3} \ \underline{$u$}}\\ \hline\hline
\makecell{$e = 1$, $v_\pi(a_1) \geq 1$, $v_\pi(a_3) \geq 1$} & \makecell{$e=2$, $v_\pi(a_1) \geq 2$, $v_\pi(a_3) \geq 2$} & \makecell{$e\geq 3$, $v_\pi(a_1) \geq 3$, $v_\pi(a_3) \geq 3$} \\ \hline
\makecell{$a_4$ : \underline{Step 9} \ \underline{$\triangle$} \ \underline{Step 4} \ \underline{$s$} \\
$a_6$ : \underline{Step 10} \ \underline{$w$} \ \underline{$\square$} \ \underline{Step 8} \ \underline{Step 3} \ \underline{Step 5}} & 
\makecell{$a_4$ : \underline{Step 9} \ \underline{$\triangle$} \ \underline{Step 4} \ \underline{$s$} \\
$a_6$ : \underline{Step 10} \ \underline{$w$} \ \underline{$\square$} \ \underline{$v$} \ \underline{Step 3} \ \underline{Step 8}} &
\makecell{$a_4$ : \underline{Step 9} \ \underline{$\triangle$} \ \underline{Step 4} \ \underline{$s$} \\
$a_6$ : \underline{Step 10} \ \underline{$w$} \ \underline{$\square$} \ \underline{$v$} \ \underline{Step 3} \ \underline{$u$}}\\ \hline
\end{tabular}
}

\medskip

    \caption{Digits fixed at each step.}
    \label{fig:digit_fix_p2}
\end{table}
\end{comment}

We run through Tate's algorithm to compute $\psi_{\mathfrak{p},K}(T, c_{\mathfrak{p}}, \alpha_1, \alpha_3).$ Recall that $q = N_{K/\mathbb{Q}}(\mathfrak{p})$ is the norm of $\mathfrak{p}$.

\noindent\textbf{Case 1.} $E$ terminates at Step 1 if $\pi \nmid \Delta$. By definition, $\Delta \equiv b_2^2b_8 + b_6^2 + b_2b_4b_6 \pmod{\pi},$ where $b_2 \equiv a_1^2 \pmod{\pi}$, $b_4 \equiv a_1a_3 \pmod{\pi},$ $b_6 \equiv a_3^2 \pmod{\pi},$ and $b_8 \equiv a_1^2a_6 - a_1a_3a_4 - a_4^2 \pmod{\pi}.$ Therefore, $\Delta \equiv a_1^6a_6 - a_1^5a_3a_4 - a_1^4a_4^2 + a_3^4 + a_1^3a_3^3 \pmod{\pi}.$ If $\alpha_1 \geq 1,$ then $\Delta \equiv a_3^4 \pmod{\pi}.$ As such, $\psi_{K,\mathfrak{p}}(I_0, 1; \alpha_1, \alpha_3) = 1$ if $\alpha_3 = 0$ and $0$ if $\alpha_3 \geq 1$. Now, suppose that $\alpha_1 = 0,$ in which case $\Delta$ is then linear in terms of $a_6$. Therefore, for each $a_1$ and $a_3,$ there is one choice of $a_6$ modulo $\pi$ such that $E$ terminates at Step 1. We thus have $\psi_{K,\mathfrak{p}} (I_0, 1; 0 , \geq 0) = \frac{q-1}{q}.$ 

\noindent\textbf{Case 2.} Suppose that the singular point of $E$ is at $(s,u)$ after reduction by $\pi$; accordingly, we shift the singular point to $(0, 0)$ by $(x,y) \mapsto (x+s, y+u)$. Our model is now: 
\begin{equation}
\label{eq:p=2_step2}
    (y+u)^2 + a_1(x+s)(y+u) + a_3(y+u) = (x+s)^3 + a_4(x+s) + a_6.
\end{equation}
To stop at Step 2, we require that $\pi \nmid b_2 = a_1^2 + 12 s$. Therefore, if $\alpha_1 = 0$, then we always stop. By Hensel's lemma, exactly $\frac{q-1}{q^{n+1}}$ of curves have $v_{\pi}(\Delta) = n$. Also, for exactly half of $E,$ $T^2 +  a_1T - 3s$ splits in $k$. Hence, $\psi_{K,\mathfrak{p}}(I_n, n; 0, \geq 0) = \psi_{K,\mathfrak{p}}(I_n, \varepsilon(n); 0, \geq 0) =\frac{q-1}{2q^{n+1}}$. If $\alpha_1 \geq 1$, then $\alpha_3 \geq 1$ by Case 1. In this case, we always pass. Thus, $\psi_{K,\mathfrak{p}}(I_n, c_{\mathfrak{p}}; \alpha_1, \alpha_3) = 0$ for all $\alpha_1,\alpha_3\geq 1$. Henceforth, $\alpha_1, \alpha_3 \geq 1$. By taking partial derivatives of \Cref{eq:p=2_step2}, we find that $s^2 \equiv a_4 \pmod{\pi}$ and $u^2 \equiv a_6 \pmod{\pi}.$

\noindent\textbf{Case 3.} Suppose that $E$ reaches Step 3. We stop if $\pi \nmid 
\pi^{-1}(s^3 + a_4s + a_6 - u^2 - a_1su - a_3u)$. For fixed $a_1$, $a_3$, and $a_4,$ there are $q(q-1)$ choices of $a_6$ modulo $\pi^2$. Hence, $\psi_{K,\mathfrak{p}}(II, 1; \alpha_1, \alpha_3) = \frac{q-1}{q}$ for each $\alpha_1,\alpha_3 \geq 1.$ 

\noindent\textbf{Case 4.} $E$ terminates at this step if $\pi^3 \nmid (3s)(2u + a_1s+a_3)^2-(3s^2 + a_4 - a_1 u)^2.$ By Step 2, we know that $\pi^2 \mid (3s)(2u + a_1s+a_3)^2-(3s^2 + a_4 - a_1 u)^2.$ Thus, we want $\pi \nmid \pi^{-2}(3s (2u+a_1s+a_3)^2 + (3s^2 + a_4 - a_1u)^2)$. Therefore, for fixed $a_1$ and $a_3,$ there are $q(q-1)$ choices for $a_4$ modulo $\pi^2$ and $q$ choices for $a_6$ modulo $\pi^2.$ Thus, $\psi_{K,\mathfrak{p}}(III, 2; \alpha_1,\alpha_3) = \frac{q-1}{q^2}$ for $\alpha_1,\alpha_3 \geq 1.$ 

\noindent\textbf{Case 5.} For an elliptic curve to terminate at Step 5, it must be that $\pi \nmid \pi^{-1}(2u + a_3 + a_1s)$. If $e=1$, for fixed $a_1$ and $a_3,$ we have $q$ choices of $a_4$ modulo $\pi^2$ and $q-1$ choices of $a_6$ modulo $\pi^2$. Now, the Tamagawa number depends on whether the polynomial $Y^2 + \frac{2u+a_3+a_1}{\pi} Y -\frac{a_6 + a_4 s +s^3 - u^2 -a_1su-a_3u}{\pi^2}$ modulo $\pi$ factors over $\mathbb{F}_{q}.$ To count, we will fix $a_1,$ $a_3,$ and $a_4$ and count over the $q(q-1)$ possible $a_6$ modulo $\pi^3$. Suppose that the polynomial has a root in $\mathbb{F}_{q}$ and fix one of the roots. Then, since the trace is fixed, the other root is fixed. Because $\pi^{-1}(2u+a_3+a_1)$ is non-zero modulo $\pi,$ the two roots must be distinct. Therefore, there are $\frac{q(q-1)}{2}$ choices of $a_6$ modulo $\pi^3$ that each results in Tamagawa number 1 and 3. Thus, $\psi_{K,\mathfrak{p}}(IV, 1; \geq 1, \geq 1) = \psi_{K,\mathfrak{p}}(IV, 3; \geq 1, \geq 1) = \frac{q-1}{2q^3}$ when $e=1.$ Now, suppose that $e \geq 2.$ First, if $\alpha_1 =1,$ then we have $q-1$ choices of $a_4$ modulo $\pi^2$ and $q$ choices for $a_6$ modulo $\pi^2.$ With the same argument as above, we conclude that for $\frac{q^2}{2}$ choices of $a_6$ modulo $\pi^3,$ the Tamagawa number is 1, and for the same number of choices, the Tamagawa number is 3. Now, suppose that $\alpha_1 \geq 2.$ If $\alpha_3 = 1,$ then the elliptic curves always terminate at this step. Again, we have that for half of the choices of $a_6,$ the Tamagawa number is 1 and that for the other half, the Tamagawa number is 3. Therefore, we have that $\psi_{K,\mathfrak{p}}(IV, 1; \geq 2, 1) = \psi_{K,\mathfrak{p}}(IV, 3; \geq 2, 1) = \frac{1}{2q^2}.$  Conversely, if $\alpha_3 \geq 2,$ then no curves terminate at this step. Therefore, $\psi_{K,\mathfrak{p}}(IV, 1; \geq 2, \geq 2) = \psi_{K,\mathfrak{p}}(IV, 3; \geq 2, \geq 2) = 0.$ 

\noindent\textbf{Case 6.} Let $t^2 \equiv s \pmod{\pi}$ and $\beta^2 \equiv \pi^{-2}(s^3+a_4s+a_6-u^2-a_1su-a_3u) \pmod{\pi}$, and define $v := u + \beta \pi$. Following the shifts outlined in Step 6 of Tate's algorithm, we have \begin{equation}
    (y + tx + v)^2 + a_1(x+t^2)(y+tx+v) + a_3(y+tx+v) = (x+t^2)^3 + a_4(x+t^2) + a_6.\label{shift2p2}
\end{equation} 
We study $P(T) = T^3 + \frac{2t^2 - a_1t}{\pi}T^2 + \frac{3t^4 + a_4 - 2tv - a_1t^3 - a_1v - a_3t}{\pi^2}T + \frac{t^6 + a_4t^2 + a_6 - v^2 - a_1t^2v - a_3v}{\pi^3}.$ Define $A_2, A_4,$ and $A_6$ such that $P(Y) \equiv T^3 + A_2 T^2 + A_4 T + A_6 \pmod{\pi}$. Now, suppose that we fix $a_4$ modulo $\pi^2$ and $a_6$ modulo $\pi^3.$ There then exists a bijective map between the $\pi$ possible values of $a_4$ modulo $\pi^3$ and $A_4$ modulo $\pi$ and between the $\pi$ possible values of $a_6$ modulo $\pi^4$ and $A_6$ modulo $\pi.$ For $E$ to terminate at Step 6, $P(T)$ must have three distinct roots. If so, $P(T)$ and $P'(T) \equiv 3 T^2 + A_4 \equiv 0 \pmod{\pi}$ should not have shared roots. Therefore, for $P(T)$ to have three distinct roots, $A_2 A_4 \not\equiv A_6 \pmod{\pi}.$ Thus, for each $A_2$ modulo $\pi,$ there are $q(q-1)$ choices of $(A_4, A_6)$ modulo $\pi.$ We also note that when $P(T)$ has three distinct roots, none of the roots can be $A_2$ as if so, the remaining two roots must be the same---a contradiction to $P(T)$ having distinct roots. 

We now fix $A_2$ and count the number of $P(T)$ with three distinct roots that have three, one, and no roots in $\mathbb{F}_q$ over $(A_4, A_6)$ modulo $\pi.$ Because we fix $A_2$ modulo $\pi,$ the trace of $P(T)$ is fixed. We first count the number of $P(T)$ that have all three roots in $\mathbb{F}_q$ with fixed trace. We have $q-1$ choices for the first root, $q-2$ choices for the second root, and a fixed choice for the third root, because as long as none of the roots are congruent to $A_2$ modulo $\pi,$ the three roots are distinct. Therefore, for fixed $A_2,$ there are $\frac{(q-1)(q-2)}{6}$ choices of $(A_4, A_6)$ modulo $\pi$ that allows for $P(T)$ to have three distinct roots, all of which are in $\mathbb{F}_q.$ We now proceed to count the number of $P(T)$ with three distinct roots with exactly one root in $\mathbb{F}_q$ with fixed $a_2.$ We start by choosing one of $\frac{q^2-q}{2}$ irreducible quadratics. The root in $\mathbb{F}_q$ is then fixed as the trace of $P(T)$ is fixed. Therefore, there are a total of $\frac{q^2-q}{2}$ choices of $P(T)$ with three distinct roots, exactly one root of which is in $\mathbb{F}_q.$ Lastly, we count the number of irreducible cubics with three distinct roots. Out of the $q^3$ elements in $\mathbb{F}_{q^3},$ $q$ are in $\mathbb{F}_q.$ Because the traces are equally distributed, for a fixed trace, there are $\frac{q^3-q}{q} = q^2-1$ elements with that fixed trace. Since $P(T)$ is a cubic, there are $\frac{q^2-1}{3}$ irreducible cubics with trace $A_2$. 

Now, suppose that $e = 1.$ From Steps 1 and 2, we have that $\alpha_1, \alpha_3 \geq 1.$ Now, for each fixed $(a_4, \frac{a_1}{\pi})$ modulo $\pi,$ we have $\frac{(q-1)(q-2)}{6}$ $P(T)$ with three distinct roots all in $\mathbb{F}_q$, $\frac{q(q-1)}{2}$ $P(T)$ with exactly one of the three distinct roots in $\mathbb{F}_q,$ and $\frac{q^2-1}{2}$ $P(T)$ with three distinct roots, none of which are in $\mathbb{F}_q.$ Thus, we have that for $\psi_{K,\mathfrak{p}}(I_0^*, 4; \geq 1, \geq 1) = \frac{(q-1)(q-2)}{6q^5},$ $\psi_{K,\mathfrak{p}}(I_0^*, 2; \geq 1, \geq 1) = \frac{(q-1)}{2q^4},$ and $\psi_{K,\mathfrak{p}}(I_0^*, 1; \geq 1, \geq 1) = \frac{q^2-1}{3q^5}.$ Suppose that $e \geq 2.$ If $\alpha_1 = 1,$ then $\alpha_3 \geq 1$ from Step 1. Then, $A_2$ modulo $\pi$ forms a bijective map with $\frac{a_1}{\pi}$ modulo $\pi.$ Therefore, by our aforementioned counting of $P(T)$ with three distinct roots, a fixed number of which are in $\mathbb{F}_q,$ we have that for $\psi_{K,\mathfrak{p}}(I_0^*, 4; 1, \geq 1) = \frac{(q-1)(q-2)}{6q^5},$ $\psi_{K,\mathfrak{p}}(I_0^*, 2; 1, \geq 1) = \frac{(q-1)}{2q^4},$ and $\psi_{K,\mathfrak{p}}(I_0^*, 1; 1, \geq 1) = \frac{q^2-1}{3q^5}.$ If $\alpha_1 \geq 2,$ then $\alpha_3 \geq 2$ from Step 6. Then, $A_2 \equiv 0 \pmod{\pi}.$ Therefore, by our aforementioned counting of $P(T)$ with a fixed number of roots in $\mathbb{F}_q,$ we have that for $\psi_{K,\mathfrak{p}}(I_0^*, 4; \alpha_1,\alpha_3) = \frac{(q-1)(q-2)}{6q^4},$ $\psi_{K,\mathfrak{p}}(I_0^*, 2; \alpha_1,\alpha_3) = \frac{(q-1)}{2q^3},$ and $\psi_{K,\mathfrak{p}}(I_0^*, 1; \alpha_1, \alpha_3) = \frac{q^2-1}{3q^4}$ for $\alpha_1,\alpha_3 \geq 2.$

\noindent\textbf{Case 7.} $E$ terminates at Step 7 if $A_2A_4 \equiv A_6 \pmod{\pi}$ and $(A_4, A_6) \not\equiv (A_2^2, A_2^3) \pmod{\pi}$. We study the interaction between quadratics $R(Y) = Y^2 + a'_{3,*}Y - a'_{6,*}$ and $S(X) = a'_{2,*} X^2 + a'_{4,*} X + a'_{6,*}$, translating the curve as we move between them. Note that by varying $a_6$, the quantity $a'_{6,*}$ is surjective modulo $\pi$. As before, since $a_2, a_3, a_4, a_6$ are equidistributed, by Hensel's lemma, there are $\frac{q-1}{q^n}$ residues for which we have Kodaira type $I_n$, and moreover, half of these cause the quadratic in question to split. Hence, $ \psi_{K,\mathfrak{p}}(I_n^*, 1, \geq 1, \geq 1) =  \psi_{K,\mathfrak{p}}(I_n^*, 3, \geq 1, \geq 1) = \frac{q-1}{2q^{5+n}}$ for $e=1$ and $ \psi_{K,\mathfrak{p}}(I_n^*, 1, \geq 2, \geq 2) = \psi_{K,\mathfrak{p}}(I_n^*, 3, \geq 2, \geq 2) = \frac{q-1}{2q^{4+n}}$ for $e \geq 2$.

\noindent \textbf{Case 8.} Suppose that $E$ reaches Step 8. Then $(A_2, A_4, A_6) \equiv (A_2, A_2^2, A_2^3) \pmod{\pi}$. Perform $x\mapsto x + \pi A_2 = x + (2t^2 - a_1t)$ and let $v' := v + 2t^3 - a_1t^2$ to get the penultimate model
\begin{equation}
    (y+tx+v')^2 + a_1(x+3t^2-a_1t)(y+tx+v') + a_3(y+tx+v') = (x+3t^2-a_1t)^3 + a_4(x+3t^2-a_1t) + a_6.
\end{equation}
We stop if $\pi \nmid \pi^{-2}(2v'+3a_1t^2 - a_1^2 t + a_3)$. We notice that if we fix $a_1,$ $a_4,$ and $a_6,$ then $a_3$ modulo $\pi^2$ is fixed such that $2v'+3a_1t^2 - a_1^2 t + a_3$ is a multiple of $\pi^2.$ We then notice that $\pi^{-2}(2v'+3a_1t^2 - a_1^2 t + a_3)$ modulo $\pi$ forms a bijective map with the $q$ possible values of $a_3$ modulo $\pi^3.$  

First, suppose that $e= 1.$ From Steps 1 and 2, we have that $\alpha_1, \alpha_3 \geq 1.$ For each $a_3$ modulo $\pi^2,$ we see that for $\frac{q}{2}$ possible values of $a_3$ modulo $\pi^3,$ $E$ terminates with Tamagawa number $1$ and that for $\frac{q}{2}$ choices for $a_3$ moudlo $\pi^3,$ $E$ terminates with Tamagawa number 3. Therefore, when $e \geq 1,$ $\psi_{K,\mathfrak{p}}(IV^*, 1; \geq 1, \geq 1) = \psi_{K,\mathfrak{p}}(IV^*, 3; \geq 1, \geq 1) = \frac{q-1}{2q^6}.$ For the same reasons, we conclude that when $e=2$ and $\alpha_1 = 1$ and $\alpha_3 \geq 1,$  $\psi_{K,\mathfrak{p}}(IV^*, 1; 1, \geq 1) = \psi_{K,\mathfrak{p}}(IV^*, 3; 1, \geq 1) = \frac{q-1}{2q^6}.$ We also see in the same way that $\psi_{K,\mathfrak{p}}(IV^*, 1; \geq 2, \geq 2) = \psi_{K,\mathfrak{p}}(IV^*, 3; \geq 2, \geq 2) = \frac{q-1}{2q^5}.$

Now, suppose that $e \geq 3.$ When $\alpha_1=1$ and $\alpha_3 \geq 1,$ we conclude as we did in the previous paragraph that $\psi_{K,\mathfrak{p}}(IV^*, 1; 1, \geq 1) = \psi_{K,\mathfrak{p}}(IV^*, 3; 1, \geq 1) = \frac{q-1}{2q^6}.$ We also see in the same way that $\psi_{K,\mathfrak{p}}(IV^*, 1; \geq 2, \geq 2) = \psi_{K,\mathfrak{p}}(IV^*, 3; \geq 2, \geq 2) = \frac{q-1}{2q^6}.$ Similarly, when $\alpha_1=2$ and $\alpha_3 \geq 2,$ we have that $\psi_{K,\mathfrak{p}}(IV^*, 1; 2, \geq 2) = \psi_{K,\mathfrak{p}}(IV^*, 3; 2, \geq 2) = \frac{q-1}{2q^5}.$ But when $\alpha_1 \geq 3$ and $\alpha_3 = 2,$ then $E$ necessarily terminates at Case 8. Then depending on $\frac{a_3}{\pi^2},$ $E$ has Tamagawa number 1 and 3 with equal proportions. Therefore, we have that $\psi_{K,\mathfrak{p}}(IV^*, 1; \geq 3, 2) = \psi_{K,\mathfrak{p}}(IV^*, 3; \geq 3, 2) = \frac{1}{2q^4}.$ If $\alpha_1, \alpha_3 \geq 3,$ however, no $E$ terminates at this step. Therefore, $\psi_{K,\mathfrak{p}}(IV^*, 1; \geq 3, 2) = \psi_{K,\mathfrak{p}}(IV^*, 3; \geq 3, 2) = 0.$

\noindent \textbf{Case 9.} Let $w^2 \equiv \pi^{-4}((3t^2-a_1t)^3 + a_4(3t^2 - a_1t) + a_6 - v'^2 + a_1^2 t v' -a_3v') \pmod{\pi}$ and let $w  := \pi^2 w' + v'.$ Then, we have the final model 
\begin{equation}
\label{p=2_final_eq}
    (y+tx+w)^2 + a_1(x+3t^2-a_1t)(y+tx+w) + a_3(y+tx+w) = (x+3t^2-a_1t)^3 + a_4(x+3t^2-a_1t) + a_6.
\end{equation}
We terminate at this case if $\pi^4 \nmid -2tw -a_1w - a_1^2t^2 - a_3t + 3(3t^2-a_1t)^2 + a_4.$ From Step 8, we have that $\pi^3 \mid -2tw -a_1w - a_1^2t^2 - a_3t + 3(3t^2-a_1t)^2 + a_4.$ Therefore, we want $\pi \mid \pi^{-3}(-2tw -a_1w - a_1^2t^2 - a_3t + 3(3t^2-a_1t)^2 + a_4)$. For fixed $a_4$ modulo $\pi^3,$ $\pi^{-3}(-2tw -a_1w - a_1^2t^2 - a_3t + 3(3t^2-a_1t)^2 + a_4)$ forms a bijective map with the $q$ possible values of $a_4$ modulo $\pi^4.$ When $e=1,$ $\alpha_1, \alpha_3 \geq 1$ from Steps 1 and 2. Therefore, $\psi_{K,\mathfrak{p}}(III^*, 2; \geq 1, \geq 1) = \frac{q-1}{q^7}.$ When $e=2,$ we have from Steps 1, 2, and 5 that either $\alpha_1 = 1$ and $\alpha_3 \geq 1$ or $\alpha_1, \alpha_3 \geq 2.$ We thus conclude, $\psi_{K,\mathfrak{p}}(III^*, 2; 1, \geq 1) =\frac{q-1}{q^7}$ and $ \psi_{K,\mathfrak{p}}(III^*, 2; \geq 2, \geq 2) = \frac{q-1}{q^6}.$ Lastly, when $e \geq 3,$ we have that either $\alpha_1 = 1$ and $\alpha_3 \geq 1,$ $\alpha_1 =2$ and $\alpha_3 \geq 2,$ and $\alpha_1 \geq 3$ and $\alpha_3 \geq 3.$ We similarly conclude that $\psi_{K,\mathfrak{p}}(III^*, 2; 1, \geq 1) =\frac{q-1}{q^7},$ $\psi_{K,\mathfrak{p}}(III^*, 2; 2, \geq 2) =\frac{q-1}{q^6},$ and $ \psi_{K,\mathfrak{p}}(III^*, 2; \geq 3, \geq 3) = \frac{q-1}{q^5}.$ 

\noindent \textbf{Case 10.} $E$ terminates at Step 10 if $\pi^6 \nmid -w^2+a_1^2tw - a_3w + (3t^2-a_1t)^3 -a_1a_4t + a_6.$ From Step 9, we have that $\pi^5 \mid -w^2+a_1^2tw - a_3w + (3t^2-a_1t)^3 -a_1a_4t + a_6.$ Therefore, we want that $\pi \mid \pi^{-5}(-w^2+a_1^2tw - a_3w + (3t^2-a_1t)^3 -a_1a_4t + a_6)$. Fix $a_6$ modulo $\pi^5.$ Then, note that $\pi^{-5}(-w^2+a_1^2tw - a_3w + (3t^2-a_1t)^3 -a_1a_4t + a_6)$ modulo $\pi$ forms a bijective map with the $q$ possible values of $a_6$ modulo $\pi^6.$ For $q-1$ of the $q$ possible values of $a_6$ modulo $\pi^6,$ $E$ terminates at Case 10. When $e=1,$ $\alpha_1, \alpha_3 \geq 1$ from Steps 1 and 2. Therefore, $\psi_{K,\mathfrak{p}}(II^*, 1; \geq 1, \geq 1) = \frac{q-1}{q^8}.$ When $e=2,$ we have from Steps 1, 2, and 5 that either $\alpha_1 = 1$ and $\alpha_3 \geq 1$ or $\alpha_1, \alpha_3 \geq 2.$ We thus conclude, $\psi_{K,\mathfrak{p}}(II^*, 1; 1, \geq 1) =\frac{q-1}{q^8}$ and $ \psi_{K,\mathfrak{p}}(II^*, 2; \geq 2, \geq 2) = \frac{q-1}{q^7}.$ Lastly, when $e \geq 3,$ we have that either $\alpha_1 = 1$ and $\alpha_3 \geq 1,$ $\alpha_1 =2$ and $\alpha_3 \geq 2,$ and $\alpha_1 \geq 3$ and $\alpha_3 \geq 3.$ We similarly conclude that $\psi_{K,\mathfrak{p}}(II^*, 1; 1, \geq 1) =\frac{q-1}{q^8},$ $\psi_{K,\mathfrak{p}}(II^*, 1; 2, \geq 2) =\frac{q-1}{q^7},$ and $ \psi_{K,\mathfrak{p}}(II^*, 1; \geq 3, \geq 3) = \frac{q-1}{q^6}.$  

\noindent \textbf{Case 11.} For $E$ to reach Step 11, it must not have terminated at a previous step. Therefore, we check that when $e=1,$ $\alpha_1, \alpha_3 \geq 1,$ the proportion of non-minimal curves is $\frac{1}{q^8},$ when $e=2,$ $\alpha_1 =1,$ and $\alpha_3 \geq 1,$ the proportion of non-minimal curves is $\frac{1}{q^8}$ as well, and that when $e=2$ and $\alpha_1,\alpha_3 \geq 2,$ the proportion of non-minimal curves is $\frac{1}{q^7}.$ When $e \geq 3,$ the proportion of non-minimal curves equal $\frac{1}{q^8}$ when $\alpha_1=1$ and $\alpha_3 \geq 1,$ $\frac{1}{q^7}$ when $\alpha_1=2$ and $\alpha_3 \geq 2,$ and $\frac{1}{q^6}$ when $\alpha_1, \alpha_3 \geq 3.$ \end{proof}}


We now show how we reclassify the non-minimal models of one family as another family of curves. We first show that, for $\alpha_3 < \alpha_1 < e$, the local densities at the non-minimal models of a family $F(\alpha_1,\alpha_3)$ exactly match the local densities at the family $F(\alpha_1-1,\alpha_3-3)$. To do this, we establish a map which sends a non-minimal model in $F(\alpha_1,\alpha_3)$ to another isomorphic model in $F(\alpha_1-1,\alpha_3-3)$, induced by the transformation at Step $11$ followed by the shift $x \mapsto x - a_2/3$.


\begin{lemma} \label{p=2_markov}
Let $\alpha_3 < \alpha_1 < e$. There is a surjective, $q^4$-to-$1$ map between the set of non-minimal models in $F(\alpha_1,\alpha_3)$ and the set $F(\alpha_1-1,\alpha_3-3)$ that sends $E$ to its transformation $E'$ after passing Step 11.
\end{lemma}

\begin{proof}
Recall from Step 11 of Tate's algorithm that for non-minimal $E(a_1, a_2, a_3, a_4, a_6)$, Tate's algorithm produces a unique residue $t \pmod{\pi}$ and $w \pmod{\pi^3}$ for which
\begin{align*}
    \widehat{E}(\widehat{a_1}, \widehat{a_2}, \widehat{a_3}, \widehat{a_4}, \widehat{a_6}) := \  & (y+tx+w)^2 + a_1(x+3t^2-a_1t)(y+tx+w) + a_3(y+tx+w) \\ & = (x+3t^2-a_1t)^3 + a_4(x+3t^2-a_1t) + a_6
\end{align*}
has the coefficient of $y$ and $xy$ divisible by $\pi$ and $\pi^3$, respectively, and the coefficient of $x^i$ divisible by $\pi^i$ for $i=2,4,6$. Hence, each non-minimal model $E(a_1, a_2, a_3, a_4, a_6) \in F(\alpha_1, \alpha_3)$ is sent to
\begin{align*}
    & \widehat{E} \left(\frac{a_1+2t}{\pi}, \frac{8t^2-4a_1t}{\pi^2}, \frac{2w+3a_1t^2-a_1^2t+a_3}{\pi^3},\right. \\ &\qquad\qquad\frac{-2tw-a_1w-3a_1t^3+a_1^2t^2-a_3t+3(3t^2-a_1t^2)+a_4}{\pi^4}, \\ &\qquad\qquad\left.\frac{-w^2-a_1w(3t^2-a_1t)-a_3w+(3t^2-a_1t)^3 + a_4(3t^2-a_1t)+a_6}{\pi^6}\right),
\end{align*}
with $v_{\pi}(2t + a_1) = v_{\pi}(a_1) = \alpha_1$ and $v_{\pi}(2w+3a_1t^2 - a_1^2t +a_3) = v_{\pi}(a_3) = \alpha_3$. Now, we perform $x \to x -\frac{\widehat{a_2}}{3}$ to $\widehat{E}$ and transform $\widehat{E}$ to $E':$
\begin{equation*}
    E'\left(a_1', a_3', a_4', a_6'\right) := y^2 + \widehat{a_1} \left(x-\frac{\widehat{a_2}}{3}\right) y + \widehat{a_3} y = \left(x-\frac{\widehat{a_2}}{3}\right)^3 + \widehat{a_2}\left(x-\frac{\widehat{a_2}}{3}\right)^2 + \widehat{a_4}\left(x-\frac{\widehat{a_2}}{3}\right) + \widehat{a_6}
\end{equation*}
We now have 
\begin{align*}
    E'\left(\widehat{a_1},-\frac{\widehat{a_1}\widehat{a_2}}{3}+\widehat{a_3}, \frac{-\widehat{a_2}^2}{3}+\widehat{a_4}, \frac{2\widehat{a_2}^3}{27}-\frac{\widehat{a_2}\widehat{a_4}}{3}+\widehat{a_6}\right) \in F(\alpha_1-1, \alpha_3-3).
\end{align*}
The two transformations are well-defined, so the map is also well-defined.

Conversely, given a model $E'(a_1', a_3', a_4', a_6') \in F(\alpha_1-1,\alpha_3-3)$, pick a pair of residues $t \pmod{\pi}$ and $w \pmod {\pi^3}$. Then, there is a unique choice of $\widehat{a_2}$ for which
\[\widehat{E}(\widehat{a_1}, \widehat{a_2}, \widehat{a_3}, \widehat{a_4}, \widehat{a_6}) : y^2 + a_1' \left(x+\frac{\widehat{a_2}}{3}\right) y + a_3' y = \left(x+\frac{\widehat{a_2}}{3}\right)^3 + a_4'\left(x-\frac{\widehat{a_2}}{3}\right) + a_6'\]
and $(\widehat{a_1}, \widehat{a_2}) = \left( \frac{a_1+2t}{\pi}, \frac{8t^2 - 4a_1t}{\pi^2} \right)$ for some $a_1$; this ensures that $\widehat{E}$ has some preimage $E \in F(\alpha_1,\alpha_3)$. In fact, there is a unique preimage $E$ for which $E \mapsto \widehat{E}$ after Step 11. Hence, after varying $t$ and $w$ across a set of $q$ and $q^3$ representatives, respectively, we have shown the aforementioned map is $q^4$-to-$1$, as we had sought.
\end{proof}



On the other hand, in the case of $\alpha_1 < e$ but $\alpha_3 \geq \alpha_1$, we may reclassify the non-minimal models of $F(\alpha, \geq \alpha)$ as exactly the models of $F(\alpha-1, \geq \alpha-3)$. Thus, the local densities of curves for each of the two sets are the same. The proof is analogous to that of \Cref{p=2_markov} and is thus omitted.

\begin{lemma} \label{p=2_markov2}
Let $\alpha < e$. There is a surjective, $q^4$-to-$1$ map between the set of non-minimal models in $F(\alpha,\geq \alpha)$ and the set $F(\alpha-1,\geq \alpha-3)$ that sends $E$ to its transformation $E'$ after passing Step 11.
\end{lemma}


Similarly, in the case of $\alpha_1 \geq e$ but $\alpha_3 < e$, we may reclassify the non-minimal models of $F(\geq e, \alpha_3)$ as exactly the models of $F(\geq e-1, \alpha_3-3)$. The proof is analogous to that of \Cref{p=2_markov} and is thus omitted.

\begin{lemma} \label{p=2_markov3}
Let $\alpha < e$. There is a surjective, $q^4$-to-$1$ map between the set of non-minimal models in $F(\alpha,\geq \alpha)$ and the set $F(\alpha-1,\geq \alpha-3)$, sending $E$ to its transformation $E'$ after passing Step 11.
\end{lemma}


Finally, it remains to classify the non-minimal models in the case of $\alpha_1, \alpha_3 \geq e$, and in particular, the case of short Weierstrass models $F(\infty, \infty)$. As in \Cref{TateP3}, it turns out the local data of the family $F(\infty, \infty)$ exactly matches that of $F(\geq e, \geq e)$ in the following sense. Note, once again, that linear transformations do not change the local data of an elliptic curve. Therefore, without loss of generality, instead of computing the local density on short Weierstrass forms $F(\infty, \infty)$, we can compute the local density at the set 
\begin{equation}\label{reshiftp2}
    \left\{ \left(E : \left(y +  tx + s\right)^2 =  \left(x+\frac{t^2}{3}\right)^3 + a_4\left(x+\frac{t^2}{3}\right) + a_6 \right) : s, t, a_4, a_6 \ \text{integral} \right\}.
\end{equation}
The set in (\ref{reshiftp2}) is exactly the set $F(\geq e, \geq e)$, up to equal multiplicity at each model, so we work with these models instead. We show that the non-minimal models of $F(\geq e, \geq e)$ are exactly the models of $F(\geq e-1, \geq e-3)$ by establishing the aforementioned map. The proof is analogous to that of \Cref{p=2_markov} and is thus omitted.


\begin{lemma} \label{p=2_markov4}
There is a surjective, $q^4$-to-$1$ map between the non-minimal models in $F(\geq e, \geq e)$ and the set $F(\geq e-1,\geq e-3)$, sending $E$ to its transformation $E'$ after passing Step 11.
\end{lemma}


We finish by using our lemmas to compute $\delta_{K,\mathfrak{p}}(c)$ by forming a Markov chain amongst families of curves. Lemmas~\ref{p=2_markov}, \ref{p=2_markov2}, \ref{p=2_markov3}, and \ref{p=2_markov4} establish the edges between these families, while Lemma~\ref{p=2_pipelines} establishes the local densities at each node. This information is collated in Figures~\ref{markovp2e1}, \ref{markovp2e2}, and \ref{markovp2e3}.

{
\begin{figure}[ht!]
\centering
\hspace*{1cm}
\begin{tikzpicture}[->, >=stealth', auto, semithick, node distance=3.5cm,  no arrow/.style={-,every loop/.append style={-}}]
\node[inner sep=0pt] (tate) at (-3.05,1.23)
    {\includegraphics[width=.05\textwidth]{diagram2flip.png}};
\tikzstyle{every state}=[draw, rectangle, minimum height = 1cm, minimum width = 1cm, fill=white,draw=black,thick,text=black,scale=1,align=center]

\node[state]    (A) {Terminate};
\node[state]    (D)[below left of=A, xshift=-4.4cm, yshift=1.25cm]   {$v_\pi(a_1) = 0$\\ $v_\pi(a_3) \ge 0$};
\node[state]    (B)[above left of=D]   {$v_\pi(a_1) \ge 1$\\ $v_\pi(a_3) = 0 $};
\node[state]    (C)[above right of=D]   {$v_\pi(a_1) \ge 1$\\ $v_\pi(a_3) \ge 1$};

\path
(C) edge[loop above]    
    node{$\frac{1}{q^{10}}$}     (C)
    edge[swap]
    node{$\frac{q-1}{q^{10}}$}      (B)
    edge
    node{$\frac{q-1}{q^{9}}$}      (D)
    edge[bend left=50, swap]
    node[xshift=0.35cm]{$1 - \frac{1}{q^{8}}$}         (A)
(B) edge[bend left=75]
    node{$1$}         (A)
(D) edge[bend right=10]
    node{$1$}         (A);
\draw[no arrow] (-12.8,0.73) -- (-2.7,0.73);
\draw[no arrow] (-12.8,-1.75) -- (-2.7,-1.75);
\node[text width=2cm] at (-11.7,-1.5) {$v_\pi(a_1) = 0$};
\node[text width=2cm] at (-11.7,0.98) {$v_\pi(a_1) \ge 1$};
\end{tikzpicture}
\caption{The Markov Chain structure when $\mathfrak{p} \mid (2)$ for $e=1.$}
\label{markovp2e1}
\end{figure}

\begin{figure}[ht!]
\centering
\begin{minipage}{.5\textwidth}
\hspace*{-3cm}
\begin{tikzpicture}[->, >=stealth', auto, semithick, node distance=3.5cm,  no arrow/.style={-,every loop/.append style={-}}]
\node[inner sep=0pt] (tate) at (-2.73,1.4)
    {\includegraphics[width=.09\textwidth]{diagram2flip.png}};
\tikzstyle{every state}=[draw, rectangle, minimum height = 1cm, minimum width = 1cm, fill=white,draw=black,thick,text=black,scale=1,align=center]

\node[auto=false]    (A) {};
\node[state]    (C)[left of=A, xshift=-3cm, yshift=-1cm]   {$v_\pi(a_1) = 1$\\ $v_\pi(a_3) \ge 1$};
\node[state]    (D)[below left of=C]   {$v_\pi(a_1) = 0$\\ $v_\pi(a_3) \ge 0$};
\node[state]    (B)[above left of=D]   {$v_\pi(a_1) = 1$\\ $v_\pi(a_3) = 0 $};
\node[state]    (E)[above left of=B]   {$v_\pi(a_1) \ge 2$\\ $v_\pi(a_3) = 0$};
\node[state]    (F)[above right of=B]   {$v_\pi(a_1) \ge 2$\\ $v_\pi(a_3) = 1$};
\node[state]    (G)[above right of=C]   {$v_\pi(a_1) \ge 2$\\ $v_\pi(a_3) \ge 2$};

\path
(G) edge[loop above]    
    node{$\frac{1}{q^{10}}$}     (G)
    edge[swap]
    node{$\frac{q-1}{q^{10}}$}      (F)
    edge[bend right=20, swap]
    node{$\frac{q-1}{q^{9}}$}      (E)
    edge
    node{$\frac{q-1}{q^{9}}$}      (C)
    edge[bend left=8, swap]
    node[xshift=0.25cm, yshift=-0.13cm]{$\frac{(q-1)^2}{q^{9}}$}      (B)
(C) edge node{$\frac{1}{q^8}$} (D);

\draw[no arrow] (-17.2,0.95) -- (-2.6,0.95);
\draw[no arrow] (-17.2,-1.5) -- (-2.6,-1.5);
\draw[no arrow] (-17.2,-4) -- (-2.6,-4);

\node[text width=2cm] at (-16,1.25) {$v_\pi(a_1) \ge 2$};
\node[text width=2cm] at (-16,-1.2) {$v_\pi(a_1) = 1$};
\node[text width=2cm] at (-16,-3.7) {$v_\pi(a_1) = 0$};

\end{tikzpicture}
\end{minipage}
\caption{The Markov Chain structure when $\mathfrak{p} \mid (2)$ for $e=2.$}
\label{markovp2e2}
\end{figure}}

\begin{comment}
\begin{center}
\hspace*{-2.5cm}
\scalebox{0.7}{\begin{tikzpicture}[->, >=stealth', auto, semithick, node distance=3.5cm,  no arrow/.style={-,every loop/.append style={-}}]
\tikzstyle{every state}=[draw, rectangle, minimum height = 1cm, minimum width = 1cm, fill=white,draw=black,thick,text=black,scale=1,align=center]

\node[state, fill=Goldenrod]    (I7) {};
\node[state, fill=Goldenrod]    (I8)[below of=I7] {};
\node[state, fill=Goldenrod]    (H7)[xshift=2.75cm, right of=I8] {};
\node[state, fill=Goldenrod]    (H6)[xshift=2.75cm, right of=I7] {};
\node[state, fill=ProcessBlue]    (B2)[xshift=0.75cm, right of=H7] {};
\node[state, fill=LimeGreen]    (C3)[xshift=0.75cm, right of=H6] {};
\node[state, fill=black]    (A)[yshift=-1cm, above of=I7] {};
\node[state, fill={rgb:red,1;green,2;blue,5}]    (B)[right of=A] {};
\node[state, fill=magenta]    (C)[right of=B] {};
\node[state, fill=Orchid]    (D)[right of=C] {};

\path
(B) edge[color=Gray]
    node{1} ([xshift=15mm, yshift=-8mm]B);
    

\path
(A) edge[color=Gray]
    node{1} ([xshift=15mm, yshift=-8mm]A);
    
\path
(C) edge[color=Gray]
    node{1} ([xshift=15mm, yshift=-8mm]C);
    
\path
(D) edge[color=Gray]
    node{1} ([xshift=15mm, yshift=-8mm]D);
    
\path
(B2)    edge[bend left=40, color=Sepia]
        node{\tiny$\frac{1}{q^8}$}  ([xshift=-18.25mm, yshift=-21.25mm]B2)
        edge[color=Gray]
        node{\tiny$1 - \frac{1}{q^8}$} ([xshift=15mm, yshift=-8mm]B2);

\path
(C3)    edge[bend left=40, color=MidnightBlue]
        node{\tiny$\frac{1}{q^8}$}  ([xshift=-18.25mm, yshift=-21.25mm]C3)
        edge[bend left=0, color=MidnightBlue]
        node[yshift=2mm,xshift=2mm]{\tiny$\frac{(q-1)}{q^8}$}  ([xshift=-36.5mm, yshift=-21.25mm]C3)
        edge[color=Gray]
        node{\tiny$1 - \frac{1}{q^7}$} ([xshift=15mm, yshift=-8mm]C3);
        
\path
(H7)    edge[bend left=40, color=RubineRed]
        node{\tiny$\frac{1}{q^8}$}  ([xshift=-18.25mm, yshift=-21.25mm]H7)
        edge[bend left=0, color=RubineRed]
        node[yshift=2mm,xshift=2mm]{\tiny$\frac{(q-1)}{q^8}$}  ([xshift=-36.5mm, yshift=-21.25mm]H7)
        edge[color=RubineRed, swap]
        node[yshift=-2mm]{\tiny$\frac{(q-1)}{q^7}$}  ([xshift=-54.75mm, yshift=-21.25mm]H7)
        edge[color=Gray]
        node{\tiny$1 - \frac{1}{q^6}$} ([xshift=15mm, yshift=-8mm]H7);
        
\path
(H6)    edge[bend left=0, color=RoyalPurple, swap]
        node[yshift=-1mm]{\tiny$\frac{1}{q^6}$}  ([xshift=-54.75mm, yshift=-21.25mm] H6)
        edge[color=Gray]
        node{\tiny$1 - \frac{1}{q^6}$} ([xshift=15mm, yshift=-8mm]H6);
        
\path
(I7)    edge[bend right, color=ForestGreen, swap]
        node{\tiny$\frac{1}{q^7}$}  ([xshift=-54.75mm, yshift=4.5mm] I7)
        edge[bend left=0, color=ForestGreen, swap]
        node[yshift=-2mm]{\tiny$\frac{(q-1)}{q^7}$}  ([xshift=-54.75mm, yshift=-21.25mm] I7)
        edge[color=Gray]
        node{\tiny$1 - \frac{1}{q^6}$} ([xshift=15mm, yshift=-8mm]I7);
        
\path
(I8)    edge[loop above, color=Bittersweet]
        node{\tiny$\frac{1}{q^{10}}$}     (I8)
        edge[bend left=40, color=Bittersweet]
        node{\tiny$\frac{(q-1)}{q^9}$}  ([xshift=-18.25mm, yshift=-21.25mm]I8)
        edge[bend left=0, color=Bittersweet]
        node[yshift=3mm,xshift=1mm]{\tiny$\frac{(q-1)^2}{q^9}$}  ([xshift=-36.5mm, yshift=-21.25mm]I8)
        edge[color=Bittersweet, swap]
        node[yshift=-2mm]{\tiny$\frac{(q-1)^2}{q^8}$}  ([xshift=-54.75mm, yshift=-21.25mm]I8)
        edge[bend right, color=Bittersweet]
        node{\tiny$\frac{1}{q^9}$}  ([xshift=-18.25mm, yshift=2.25mm] I8)
        edge[bend right, color=Bittersweet]
        node{\tiny$\frac{(q-1)}{q^9}$}  ([xshift=-36.5mm, yshift=4.5mm] I8)
        edge[bend right, color=Bittersweet, swap]
        node{\tiny$\frac{(q-1)}{q^8}$}  ([xshift=-54.75mm, yshift=4.5mm] I8)
        edge[color=Gray]
        node{\tiny$1 - \frac{1}{q^6}$} ([xshift=15mm, yshift=-8mm]I8);
\end{tikzpicture}}
\end{center}
\end{comment}

\begin{figure}[ht!]
{\tiny 
\setlength{\tabcolsep}{6pt} 
\renewcommand{\arraystretch}{2.}
\begin{center}
\begin{tabular}{|c||c|c|c|c|c|c|c|}
 \hline
 Case & $v(a_1)=0$ & \makecell{$v(a_1) \geq 1$\\$v(a_3) = 0$} & \makecell{$v(a_1) = 1$\\$v(a_3) \geq 1$} & \makecell{$v(a_1) \geq 2$ \\ $v(a_3) = 1$} & \makecell{$v(a_1) = 2$ \\ $v(a_3) \geq 2$} & \makecell{$v(a_1) \geq 3$ \\ $v(a_3) = 2$} & \makecell{$v(a_1) \geq 3$ \\ $v(a_3) \geq 3$} \\ \hline \hline
 Color & \cellcolor{black} & \cellcolor{rgb:red,1;green,2;blue,5} & \cellcolor{ProcessBlue} & \cellcolor{magenta} & \cellcolor{LimeGreen} & \cellcolor{Orchid} & \cellcolor{Goldenrod} \\\hline
\end{tabular}
\smallskip
\end{center}
}
\vspace*{1cm}
\begin{center}
\hspace*{-2.5cm}
\scalebox{0.7}{\begin{tikzpicture}[->, >=stealth', auto, semithick, node distance=3.5cm,  no arrow/.style={-,every loop/.append style={-}}]
\tikzstyle{every state}=[draw, rectangle, minimum height = 1cm, minimum width = 1cm, fill=white,draw=black,thick,text=black,scale=1,align=center]

\node[state, fill=Goldenrod]    (I7) {};
\node[state, fill=Goldenrod]    (I8)[below of=I7] {};
\node[state, fill=Goldenrod]    (H7)[xshift=2.75cm, right of=I8] {};
\node[state, fill=Goldenrod]    (H6)[xshift=2.75cm, right of=I7] {};
\node[state, fill=ProcessBlue]    (B2)[xshift=0.75cm, right of=H7] {};
\node[state, fill=LimeGreen]    (C3)[xshift=0.75cm, right of=H6] {};
\node[state, fill=black]    (A)[yshift=-1cm, above of=I7] {};
\node[state, fill={rgb:red,1;green,2;blue,5}]    (B)[right of=A] {};
\node[state, fill=magenta]    (C)[right of=B] {};
\node[state, fill=Orchid]    (D)[right of=C] {};

\path
(B) edge[color=Gray]
    node{1} ([xshift=15mm, yshift=-8mm]B);
    

\path
(A) edge[color=Gray]
    node{1} ([xshift=15mm, yshift=-8mm]A);
    
\path
(C) edge[color=Gray]
    node{1} ([xshift=15mm, yshift=-8mm]C);
    
\path
(D) edge[color=Gray]
    node{1} ([xshift=15mm, yshift=-8mm]D);
    
\path
(B2)    edge[bend left=40, color=Sepia]
        node{\tiny$\frac{1}{q^8}$}  ([xshift=-18.25mm, yshift=-21.25mm]B2)
        edge[color=Gray]
        node{\tiny$1 - \frac{1}{q^8}$} ([xshift=15mm, yshift=-8mm]B2);

\path
(C3)    edge[bend left=40, color=MidnightBlue]
        node{\tiny$\frac{1}{q^8}$}  ([xshift=-18.25mm, yshift=-21.25mm]C3)
        edge[bend left=0, color=MidnightBlue]
        node[yshift=2mm,xshift=2mm]{\tiny$\frac{(q-1)}{q^8}$}  ([xshift=-36.5mm, yshift=-21.25mm]C3)
        edge[color=Gray]
        node{\tiny$1 - \frac{1}{q^7}$} ([xshift=15mm, yshift=-8mm]C3);
        
\path
(H7)    edge[bend left=40, color=RubineRed]
        node{\tiny$\frac{1}{q^8}$}  ([xshift=-18.25mm, yshift=-21.25mm]H7)
        edge[bend left=0, color=RubineRed]
        node[yshift=2mm,xshift=2mm]{\tiny$\frac{(q-1)}{q^8}$}  ([xshift=-36.5mm, yshift=-21.25mm]H7)
        edge[color=RubineRed, swap]
        node[yshift=-2mm]{\tiny$\frac{(q-1)}{q^7}$}  ([xshift=-54.75mm, yshift=-21.25mm]H7)
        edge[color=Gray]
        node{\tiny$1 - \frac{1}{q^6}$} ([xshift=15mm, yshift=-8mm]H7);
        
\path
(H6)    edge[bend left=0, color=RoyalPurple, swap]
        node[yshift=-1mm]{\tiny$\frac{1}{q^6}$}  ([xshift=-54.75mm, yshift=-21.25mm] H6)
        edge[color=Gray]
        node{\tiny$1 - \frac{1}{q^6}$} ([xshift=15mm, yshift=-8mm]H6);
        
\path
(I7)    edge[bend right, color=ForestGreen, swap]
        node{\tiny$\frac{1}{q^7}$}  ([xshift=-54.75mm, yshift=4.5mm] I7)
        edge[bend left=0, color=ForestGreen, swap]
        node[yshift=-2mm]{\tiny$\frac{(q-1)}{q^7}$}  ([xshift=-54.75mm, yshift=-21.25mm] I7)
        edge[color=Gray]
        node{\tiny$1 - \frac{1}{q^6}$} ([xshift=15mm, yshift=-8mm]I7);
        
\path
(I8)    edge[loop above, color=Bittersweet]
        node{\tiny$\frac{1}{q^{10}}$}     (I8)
        edge[bend left=40, color=Bittersweet]
        node{\tiny$\frac{(q-1)}{q^9}$}  ([xshift=-18.25mm, yshift=-21.25mm]I8)
        edge[bend left=0, color=Bittersweet]
        node[yshift=3mm,xshift=1mm]{\tiny$\frac{(q-1)^2}{q^9}$}  ([xshift=-36.5mm, yshift=-21.25mm]I8)
        edge[color=Bittersweet, swap]
        node[yshift=-2mm]{\tiny$\frac{(q-1)^2}{q^8}$}  ([xshift=-54.75mm, yshift=-21.25mm]I8)
        edge[bend right, color=Bittersweet]
        node{\tiny$\frac{1}{q^9}$}  ([xshift=-18.25mm, yshift=2.25mm] I8)
        edge[bend right, color=Bittersweet]
        node{\tiny$\frac{(q-1)}{q^9}$}  ([xshift=-36.5mm, yshift=4.5mm] I8)
        edge[bend right, color=Bittersweet, swap]
        node{\tiny$\frac{(q-1)}{q^8}$}  ([xshift=-54.75mm, yshift=4.5mm] I8)
        edge[color=Gray]
        node{\tiny$1 - \frac{1}{q^6}$} ([xshift=15mm, yshift=-8mm]I8);
\end{tikzpicture}}
\end{center}
\begin{center}
\hspace*{-3.5cm}
\scalebox{0.55}{\begin{tikzpicture}[->, >=stealth', auto, semithick, node distance=2.25cm,  no arrow/.style={-,every loop/.append style={-}}]
\tikzstyle{every state}=[draw, rectangle, minimum height = 1cm, minimum width = 1cm, fill=white,draw=black,thick,text=black,scale=1,align=center]

\node[state,fill=black]    (A)   {};
\node[state,fill={rgb:red,1;green,2;blue,5}]    (B1)[above of=A]   {};
\node[state,fill=ProcessBlue]    (B2)[right of=B1]   {};
\node[state,fill={rgb:red,1;green,2;blue,5}]    (C1)[above of=B1]   {};
\node[state,fill=magenta]    (C2)[above of=B2]   {};
\node[state,fill=LimeGreen]    (C3)[right of=C2]   {};
\node[state,fill={rgb:red,1;green,2;blue,5}]    (D1)[above of=C1]   {};
\node[state,fill=magenta]    (D2)[above of=C2]   {};
\node[state,fill=Orchid]    (D3)[above of=C3]   {};
\node[state,fill=Goldenrod]    (D4)[right of=D3]   {};
\node[state,fill={rgb:red,1;green,2;blue,5}]    (E1)[above of=D1, yshift=1cm]   {};
\node[state,fill=magenta]    (E2)[above of=D2, yshift=1cm]   {};
\node[state,fill=Orchid]    (E3)[above of=D3, yshift=1cm]   {};
\node[state,fill=Goldenrod]    (E4)[right of=E3, xshift=2cm]   {};
\node[state,fill={rgb:red,1;green,2;blue,5}]    (F1)[above of=E1]   {};
\node[state,fill=magenta]    (F2)[above of=E2]   {};
\node[state,fill=Orchid]    (F3)[above of=E3]   {};
\node[state,fill=Goldenrod]    (F4)[above of=E4]   {};
\node[state,fill=Goldenrod]    (F5)[right of=F4]   {};
\node[state,fill={rgb:red,1;green,2;blue,5}]    (G1)[above of=F1]   {};
\node[state,fill=magenta]    (G2)[above of=F2]   {};
\node[state,fill=Orchid]    (G3)[above of=F3]   {};
\node[state,fill=Goldenrod]    (G4)[above of=F4]   {};
\node[state,fill=Goldenrod]    (G5)[above of=F5]   {};
\node[state,fill=Goldenrod]    (G6)[right of=G5]   {};
\node[state,fill={rgb:red,1;green,2;blue,5}]    (H1)[above of=G1]   {};
\node[state,fill=magenta]    (H2)[above of=G2]   {};
\node[state,fill=Orchid]    (H3)[above of=G3]   {};
\node[state,fill=Goldenrod]    (H4)[above of=G4]   {};
\node[state,fill=Goldenrod]    (H5)[above of=G5]   {};
\node[state,fill=Goldenrod]    (H6)[above of=G6]   {};
\node[state,fill=Goldenrod]    (H7)[right of=H6]   {};
\node[state,fill={rgb:red,1;green,2;blue,5}]    (I1)[above of=H1]   {};
\node[state,fill=magenta]    (I2)[above of=H2]   {};
\node[state,fill=Orchid]    (I3)[above of=H3]   {};
\node[state,fill=Goldenrod]    (I4)[above of=H4]   {};
\node[state,fill=Goldenrod]    (I5)[above of=H5]   {};
\node[state,fill=Goldenrod]    (I6)[above of=H6]   {};
\node[state,fill=Goldenrod]    (I7)[above of=H7]   {};
\node[state,fill=Goldenrod]    (I8)[right of=I7]   {};

\path (D1) -- node[auto=false]{\Huge\mydots} (E1);
\path (D2) -- node[auto=false]{\Huge\mydots} (E2);
\path (D3) -- node[auto=false]{\Huge\mydots} (E3);
\path (E3) -- node[auto=false]{\Huge\myhdots} (E4);
\path (F3) -- node[auto=false]{\Huge\myhdots} (F4);
\path (G3) -- node[auto=false]{\Huge\myhdots} (G4);
\path (H3) -- node[auto=false]{\Huge\myhdots} (H4);
\path (I3) -- node[auto=false]{\Huge\myhdots} (I4);
\path (D4) -- node[auto=false, rotate=-35]{\Huge\mydots} (E4);

\path
(I8)    edge[loop above, color=Bittersweet]
        node{}     (I8)
        edge[bend right, color=Bittersweet]
        node{}  (I7)
        edge[bend right, color=Bittersweet]
        node{}  (I6)
        edge[bend right, color=Bittersweet]
        node{}  (I5)
        edge[bend left, color=Bittersweet]
        node{}  (H7)
        edge[bend left=0, color=Bittersweet]
        node{}  (H6)
        edge[color=Bittersweet]
        node{}  (H5)
(I7)    edge[bend right, color=ForestGreen]
        node{}  (I4)
        edge[bend left=0, color=ForestGreen]
        node{}  (H4)
(I6)    edge[bend right=40, color=ForestGreen]
        node{}  ([yshift=4mm]$ (I4) !.20! (I3) $)
        edge[bend left=2, color=ForestGreen]
        node{}  ([yshift=5mm]$ (H4) !.20! (H3) $)
(I5)    edge[bend right=60, color=ForestGreen]
        node{}  ([yshift=4mm]$ (I4) !.30! (I3) $)
        edge[color=ForestGreen]
        node{}  ([yshift=5mm]$ (H4) !.30! (H3) $)
(I4)    edge[bend right=80, color=ForestGreen]
        node{}  ([yshift=4mm]$ (I4) !.40! (I3) $)
        edge[color=ForestGreen]
        node{}  ([yshift=5mm]$ (H4) !.40! (H3) $)

(H7)    edge[bend left, color=RubineRed]
        node{}  (G6)
        edge[bend left=0, color=RubineRed]
        node{}  (G5)
        edge[color=RubineRed]
        node{}  (G4)
(H6)    edge[bend left=2, color=RoyalPurple]
        node{}  ([yshift=5mm]$ (G4) !.20! (G3) $)
        
(H5)    edge[bend left=0, color=RoyalPurple]
        node{}  ([yshift=5mm]$ (G4) !.30! (G3) $)
(H4)    edge[bend left=0, color=RoyalPurple]
        node{}  ([yshift=5mm]$ (G4) !.40! (G3) $)        

(G6)    edge[bend left, color=RubineRed]
        node{}  (F5)
        edge[bend left=0, color=RubineRed]
        node{}  (F4)
        edge[bend left=2, color=RubineRed]
        node{}  ([yshift=5mm]$ (F4) !.20! (F3) $)
(G5)    edge [color=RoyalPurple]
        node{}  ([yshift=5mm]$ (F4) !.30! (F3) $)
(G4)    edge[bend left=0, color=RoyalPurple]
        node{}  ([yshift=5mm]$ (F4) !.40! (F3) $)

(F5)    edge[bend left, color=RubineRed]
        node{}  (E4)
        edge[color=RubineRed]
        node{}  ([yshift=4mm]$ (E4) !.20! (E3) $)
        edge[bend right=5, color=RubineRed]
        node{}  ([yshift=4mm]$ (E4) !.30! (E3) $)
(F4)    edge[bend left=0, color=RoyalPurple]
        node{}  ([yshift=4mm]$ (E4) !.40! (E3) $)
        
($ (E4) !.55! (D4) $)   edge[bend left, color=RubineRed] 
                        node{} (D4)
([xshift=-3mm]$ (E4) !.55! (D4) $)   edge[bend left=0, color=RubineRed] 
                        node{} (D3)
([xshift=-8mm]$ (E4) !.55! (D4) $)   edge[bend right=5, color=RubineRed]
                        node{} (D2)
([yshift=-18mm]$ (E4) !.8! (E3) $)   edge[bend right=4, color=RoyalPurple] 
                        node{} (D1)
                        
([yshift=5mm]$ (I4) !.80! (I3) $)   edge[bend right=30, color=ForestGreen] 
                        node{} (I1)
                        
([yshift=5mm]$ (I4) !.70! (I3) $)   edge[bend right=40, color=ForestGreen] 
                        node{} (I2)
([yshift=5mm]$ (I4) !.60! (I3) $)   edge[bend right=60, color=ForestGreen] 
                        node{} (I3)

([yshift=-5mm]$ (I4) !.80! (I3) $)   edge[bend right=4, color=ForestGreen] 
                        node{} (H1)
([yshift=-5mm]$ (I4) !.70! (I3) $)   edge[bend right=7, color=ForestGreen] 
                        node{} (H2)
([yshift=-5mm]$ (I4) !.60! (I3) $)   edge[bend right=7, color=ForestGreen] 
                        node{} (H3)
([yshift=-5mm]$ (H4) !.80! (H3) $)   edge[bend right=4, color=RoyalPurple] 
                        node{} (G1)
([yshift=-5mm]$ (H4) !.70! (H3) $)   edge[bend right=7, color=RoyalPurple] 
                        node{} (G2)
([yshift=-5mm]$ (H4) !.60! (H3) $)   edge[bend right=7, color=RoyalPurple] 
                        node{} (G3)
([yshift=-5mm]$ (G4) !.80! (G3) $)   edge[bend right=4, color=RoyalPurple] 
                        node{} (F1)
([yshift=-5mm]$ (G4) !.70! (G3) $)   edge[bend right=7, color=RoyalPurple] 
                        node{} (F2)
([yshift=-5mm]$ (G4) !.60! (G3) $)   edge[bend right=7, color=RoyalPurple] 
                        node{} (F3)
([yshift=-5mm]$ (F4) !.80! (F3) $)   edge[bend right=4, color=RoyalPurple] 
                        node{} (E1)
([yshift=-5mm]$ (F4) !.70! (F3) $)   edge[bend right=7, color=RoyalPurple] 
                        node{} (E2)
([yshift=-5mm]$ (F4) !.60! (F3) $)   edge[bend right=7, color=RoyalPurple] 
                        node{} (E3)
(E4)    edge[bend left, color=RubineRed]
        node{}  ($ (E4) !.42! (D4) $)
        edge[bend left=0, color=RubineRed]
        node{}  ([yshift=-14mm]$ (E4) !.30! (E3) $)
        edge [color=RubineRed]
        node{}  ([yshift=-14mm]$ (E4) !.40! (E3) $)

(D4)    edge[bend left, color=RubineRed]
        node{} (C3)
        edge[bend left=0, color=RubineRed]
        node{} (C2)
        edge [color=RubineRed]
        node{} (C1)
        
(C3)    edge[bend left, color=MidnightBlue]
        node{} (B2)
        edge [color=MidnightBlue]
        node{} (B1)

(B2)    edge[bend left, color=Sepia]
        node{}  (A);


\end{tikzpicture}}
\caption{The Markov Chain structure when $\mathfrak{p} \mid (2)$ for $e\ge 3.$}
\label{markovp2e3}
\end{center}
\end{figure}


\begin{proposition}
\label{p=2_total}
For $\mathfrak{p} \mid (2)$ is a prime ideal in $K$ and $n \geq 1,$ let $q := N_{K/\mathbb{Q}}(\mathfrak{p}).$ If $e = 1$, we have
$$
\delta_{K,\mathfrak{p}}(c)=
{\small
\begin{cases}
\displaystyle\ 1 - \frac{(q - 1)(6q^{10} + 9q^{9} + 7q^{8} + 8q^7 + 7q^6 + 9q^5 + 6q^4 + 6q + 3)}{6q(q + 1)(q^{10} - 1)} \ \ \ \ &{\text if }\ c=1,\\\\[6pt]
\ \ \ \displaystyle\ \frac{(q - 1)(2q^{11} + 2q^{10} + q^9 + 2q^8 + q^7 + 2q^6 + 2q^5 + 2q^2 - 1)}{2q^2(q + 1)(q^{10}-1)}  &{\text if}\  c=2,\\\\[6pt]
\displaystyle\ \ \  \frac{(q - 1)(q^{10} + q^7 + q - 1)}{2q^3(q^{10}-1)}   \ \ \ \ &{\text if}\ c=3,\\\\[6pt]
\displaystyle\ \ \  \frac{(q - 1)(q^{10} + q^9 + 3q - 3)}{6q^4(q^{10}-1)}  \ \ \ \ &{\text if}\  c=4,\\[14pt]
\displaystyle \ \ \  \frac{(q - 1)^2}{2q^{c}(q^{10}-1)}   \ \ \ \ &{\text if }\ c\geq 5.\\
\end{cases}
}
$$

If $e = 2,$ we have
$$
\delta_{K,\mathfrak{p}}(c)=
{\small \begin{cases}
\displaystyle\ 1 - \frac{(q - 1)(6q^{18} + 10q^{17} + 8q^{16} + 7q^{15} + 9q^{14} + 6q^{13} + 6q^{10} + 9q^9)}{6q^9(q + 1)(q^{10} - 1)} + O\left(\frac{1}{q^{11}}\right)
\ \ \ \ &{\text if }\ c=1,\\[14pt]
 \ \ \ \\\ \displaystyle \frac{(q - 1)(2q^{19} + 3q^{18} + 2q^{17} + q^{16} + 2q^{15} + 2q^{14} + 2q^{11} + 2q^{10}))}{2q^{10}(q + 1)(q^{10}-1)} + O\left(\frac{1}{q^{11}} \right) \ \ \ \  &{\text if}\  c=2, \\\\ 
\displaystyle\ \ \  \frac{(q - 1)(q^2 + 1)(q^4 - q^2 + 1)(q^{10} + q - 1)}{2q^{11}(q^{10}-1)}   \ \ \ \ &{\text if}\ c=3, \\\\
\displaystyle\ \ \  \frac{(q - 1)(q^{19} + q^{18} + q^{10} - q^8 + 3q - 3)}{6q^{12}(q^{10}-1)}  \ \ \ \ &{\text if}\  c=4, \\\\
\displaystyle \ \ \  \frac{(q - 1)^2}{2q^{8+c}(q^{10}-1)}   \ \ \ \ &{\text if }\ c\geq 5.\\
\end{cases}}
$$
The exact proportions are given in \Cref{charts}. For $e\ge 3,$ refer to \Cref{egeq3} and \Cref{uglyformulae}. These values are not written here due to their length and complexity. 

\end{proposition}

{
\begin{proof}
The proof is very similar to \Cref{p=3_total}: we compute the proportion of curves which reach each of the $(e+1)(e+2)/2$ non-terminal nodes, then sum and scale the proportions by $\frac{q^{10}}{q^{10}-1}$ to account for curves which initially loop back to $F(\geq e, \geq e)$. For sake of brevity, the exact computations and proportions are provided in \Cref{egeq3} and \Cref{uglyformulae}.
\end{proof}}



\section{Proofs of the Main Results}
\label{proofs}



In this section, we make use of the computed local densities to prove our main results. We prove \Cref{l-series} by analyzing the convergence of $P_\mathrm{Tam}(K; m),$ thus establishing the absolute convergence of $L_\mathrm{Tam}(K; -1).$ For \Cref{d_range}, we use the formulae from \Cref{p=5_total}, \Cref{p=3_total}, and \Cref{p=2_total} to uniformly bound $P_\mathrm{Tam}(K;1)$ and $L_\mathrm{Tam}(K;-1).$ Finally, we prove \Cref{d1d2} by constructing a family of multiquadratic fields and a family of cyclotomic fields. 




\begin{proof}[Proof of \Cref{l-series}]
All of the counts in this paper assume that the elliptic curves are ordered by height. But the congruence conditions are over bounded powers of $\pi$ which are pairwise relatively prime for different prime ideals, so we can compute the $L$-series with a simple multiplicativity argument. Namely, we have that
\[
L_{\mathrm{Tam}}(K; s):=\sum_{m=1}^{\infty}\frac{P_{\mathrm{Tam}}(K; m)}{m^s}=\prod_{p \ \text{prime}} \prod_{\mathfrak{p} \mid p} \left(\frac{\delta_{K,\mathfrak{p}}(1)}{1^s}+\frac{\delta_{K,\mathfrak{p}}(2)}{2^s}+\frac{\delta_{K,\mathfrak{p}}(3)}{3^s}+\dots\right).
\]
It is sufficient to verify the convergence of the Dirichlet coefficients $P_{\mathrm{Tam}}(K; m)$ to show that they are well-defined. Thus, we make use of Corollary~\ref{p=5_total} (3) for a prime ideal $\mathfrak{p}$ such that $\mathfrak{p} \nmid (6).$ In particular, we note that $1 - \frac{1}{q^2} < \delta_{K,\mathfrak{p}}(1) < 1,$ where $q := N_{K/\mathbb{Q}}(\mathfrak{p}).$ Therefore, we see that convergence follows with the comparison to the Dedekind zeta function associated with the number field $K$ at $s=2,$ making use of the well-known fact that $\zeta_K(s)$ is convergent for $\mathrm{Re}(s) > 1.$ 


To find the ``average value'' of $\mathrm{Tam}(K; E)$, given by
\begin{equation}
\label{avg}
    L_{\mathrm{Tam}}(K; -1):=\sum_{m=1}^{\infty}P_{\mathrm{Tam}}(K; m)m =\prod_{p \ \text{prime}} \prod_{\mathfrak{p} \mid p} \left(\delta_{K,\mathfrak{p}}(1)+2\delta_{K,\mathfrak{p}}(2)+3\delta_{K,\mathfrak{p}}(3)+\dots\right),
\end{equation}
we once again use Corollary~\ref{p=5_total}. 
For prime ideal $\mathfrak{p}$ with $\mathfrak{p} \nmid (6)$ and $q := \mathcal{N}_{K/\mathbb{Q}}(\mathfrak{p}),$ we have $\delta_{K,\mathfrak{p}}(1) = 1 - \frac{1}{q^2} + o(1/q^2)$ and $0 < n\delta_{K,\mathfrak{p}}(c) < \frac{n}{q^n}$ for $n \ge 2.$ Since $\sum_{n=1}^\infty \frac{n}{q^n} = \frac{q}{(q-1)^2}$, we obtain
\[
\sum_{n=1}^\infty n\delta_{K,\mathfrak{p}}(c) = 1 - \frac{1}{q^2} + o\left(\frac{1}{q^2} \right).
\]
For $\mathfrak{p} \mid (2)$ and $\mathfrak{p} \mid (3)$, letting $q := N_{K/\mathbb{Q}}(\mathfrak{p})$, we have 
\[\sum_{n\ge 1} n\delta_{K,\mathfrak{p}}(c) \le \delta_{K,\mathfrak{p}}(1) + \sum_{k = 1}^\infty (4+k)(1 - \delta_{K,\mathfrak{p}}(1))\frac{q-1}{q^{k}} = 5 - 4\delta_{K,\mathfrak{p}}(1) + \frac{(1 - \delta_{K,\mathfrak{p}}(1))}{q-1}.
\]
The explicit values of $\delta_{K,\mathfrak{p}}(1)$ can be found in \Cref{p=3_total} and \Cref{p=2_total}.
By multiplicativity, (\ref{avg}) must converge, and we are thus done.
\end{proof}

{
\begin{proof}[Proof of \Cref{d_range}]
We first note that for $\mathfrak{p} \mid (2)$ and $\mathfrak{p} \mid (3),$ the value $\delta_{K,\mathfrak{p}}(1)$ when $\mathfrak{p}$ ramifies is greater than the square of the value
when $\mathfrak{p}$ splits completely, as can be seen in \Cref{p=3_total} and \Cref{p=2_total}. For $\mathfrak{p} \nmid (6),$ $\delta_{K,\mathfrak{p}}(1)$ is similarly minimized when $\mathfrak{p}$ splits, as can be seen from the formulae in \Cref{p=5_total}. 

We now prove the uniform bounds of $P_\mathrm{Tam}(K;1).$ Recall 
\[ P_\mathrm{Tam}(K; 1) = \prod_{p \text{ prime}}\,\prod_{\mathfrak{p} \mid (p)} \delta_{K,\mathfrak{p}}(1).\]
To bound $P_\mathrm{Tam}(K; 1)$ from below, we consider the case where each $\mathfrak{p}$ splits completely; specifically, there are $d$ distinct prime ideals lying above it, where each prime ideal has ramification index and inertial degree 1. Then, $q = \mathcal{N}_{K/\mathbb{Q}}(\mathfrak{p}) = p$ and we have
\[
 \prod_{p \text{ prime}} \delta_{\mathbb{Q}, p}(1)^d = P_\mathrm{Tam}(\mathbb{Q}; 1) \le P_\mathrm{Tam}(K; 1).
\]
We make use of the computation $P_\mathrm{Tam}(\mathbb{Q}; 1) = 0.5054\ldots$ in~\cite{og} to obtain the bound
\[
(0.5054)^{d} < P_\mathrm{Tam}(\mathbb{Q}; 1)^{d} \le P_\mathrm{Tam}(K; 1)^{d},
\]
where equality only holds when $d = 1.$

We now bound $P_\mathrm{Tam}(K; 1)$ from above by supposing each $\mathfrak{p}$ is inert, meaning $\mathfrak{p}$ has ramification index 1 and inertial degree $d$. In this case, $q := N_{K/\mathbb{Q}}(\mathfrak{p}) = p^d.$ Using \Cref{pgeq5_varphi} and \Cref{p3chi}, we see
$\delta_{K, \mathfrak{p}}(1) \le 1 - \frac{1}{q^2} + \frac{1}{q^3} = 1 - \frac{1}{p^{2d}} + \frac{1}{p^{3d}}$ for every $\mathfrak{p} \mid (3)$ and $\mathfrak{p} \nmid (6).$ From \Cref{p2psi1} for $\mathfrak{p} \mid (2),$  we see $\delta_{K, \mathfrak{p}}(1) \le 1 - \frac{1}{q}+ \frac{1}{q^2} = 1 - \frac{1}{2^d}+ \frac{1}{2^{2d}}.$ Hence, we have
\[
P_\mathrm{Tam}(K; 1) \le \left(1 - \frac{1}{2^d} + \frac{1}{2^{2d}}\right)\prod_{p\ge 3 \text{ prime}}\left(1 - \frac{1}{p^{2d}} + \frac{1}{p^{3d}}\right).
\]
We wish to prove the following inequality:
\[
\left(1 - \frac{1}{2^d} + \frac{1}{2^{2d}}\right)\prod_{p\ge 3 \text{ prime}}\left(1 - \frac{1}{p^{2d}} + \frac{1}{p^{3d}}\right) \le \prod_{p \text{ prime}}\left(1 - \frac{1}{p^{2d}} \right) = \frac{1}{\zeta(2d)},
\]
which can be rewritten as
\[
\prod_{p\ge 3 \text{ prime}}\left(1 + \frac{1}{p^{3d}-p^d} \right) \le 1 + \frac{2^d - 2}{2^{2d}-2^d + 1}.
\]
We proceed as follows:
\[\prod_{p\ge 3 \text{ prime}}\left(1 + \frac{1}{p^{3d}-p^d} \right) \le  \prod_{p\ge 3 \text{ prime}}\left(1 + \frac{1}{p^{3d-1}} \right) \le \zeta(3d - 1).\]
We then have the bounds
\[
\zeta(3d - 1) \le 1 + \frac{1}{2^{3d-1}} + \int_2^\infty \frac{1}{x^{3d-1}}\,dx = 1 + \frac{1}{2^{3d-1}} + \frac{1}{3d \cdot 2^{3d-2}- 2^{3d-1}}
\le 1 + \frac{2^d - 2}{2^{2d}-2^d + 1}.
\]
We have therefore shown that
\[
P_\mathrm{Tam}(K; 1) \le \frac{1}{\zeta(2d)} = (-1)^{d+1}\frac{2(2d)!}{B_{2d}(2\pi)^{2d}}.
\]
The last equality is a well-known relationship due to Euler between the Riemann zeta function $\zeta(2n)$ and the Bernoulli numbers ${B_{2n}}$. Thus, we have bounded $P_\mathrm{Tam}(K; 1)$ depending on $d = [K : \mathbb{Q}],$ as desired.

In a similar fashion as above, we prove the uniform bounds for $L_\mathrm{Tam}(K;-1)$. Recall
\[
L_\mathrm{Tam}(K; -1) = \prod_{p \text{ prime}} \prod_{\substack{\mathfrak{p} \mid (p)}}  \sum_{m=1}^\infty \delta_{K,\mathfrak{p}}(m)m.
\]
In order to bound $L_\mathrm{Tam}(K;-1)$ from below, we must again consider the case where every $\mathfrak{p}$ is inert, meaning $q:=N_{K/\mathbb{Q}} = p^d.$ We first note 
$
\sum_{m=1}^\infty \delta_{K,\mathfrak{p}}(m)m \ge \delta_{K,\mathfrak{p}}(1) + 2(1 - \delta_{K,\mathfrak{p}}(1)) = 2 - \delta_{K,\mathfrak{p}}(1).
$
Recalling the previous bounds of $\delta_{K,\mathfrak{p}}(1)$ for $\mathfrak{p}$ inert,
we have
\[
\prod_{p \text{ prime}} \prod_{\substack{\mathfrak{p} \mid (p)}}(2 - \delta_{K,\mathfrak{p}}(1)) \ge \left(1 + \frac{1}{2^d} - \frac{1}{2^{2d}} \right)\prod_{p\ge3 \text{ prime}} \left(1 + \frac{1}{p^{2d}} - \frac{1}{p^{3d}} \right).
\]
We wish to prove that 
\[
\left(1 + \frac{1}{2^d} - \frac{1}{2^{2d}} \right)\prod_{p\ge3 \text{ prime}} \left(1 + \frac{1}{p^{2d}} - \frac{1}{p^{3d}} \right) \ge \prod_{p \text{ prime}} \left(1 + \frac{1}{p^{2d}} \right) = \frac{\zeta(2d)}{\zeta(4d)},
\]
which can be restated as
\[
\prod_{p\ge3 \text{ prime}} \left(1 - \frac{1}{p^{3d}+p^{d}}\right) \ge 1 - \frac{2^d - 2}{2^{2d}+2^d-1}.
\]
We proceed by bounding as follows:
\[
\prod_{p\ge3 \text{ prime}} \left(1 - \frac{1}{p^{3d}+p^{d}}\right) \ge \prod_{p\ge3 \text{ prime}}\left(1 - \frac{1}{p^{3d-1}}\right) \ge \frac{1}{\zeta(3d-1)}.
\]
We make use of our previous upper bound for $\zeta(3d-1)$ to see
\[
    \frac{1}{\zeta(3d-1)} \ge \frac{1}{1+\frac{1}{2^{3d-1}}+ \frac{1}{2^{3d-2}(3d-2)}}
    = 1 - \frac{3d}{2^{3d-1}(3d-2)+3d} \\
    \ge 1 - \frac{2^d - 2}{2^{2d}+2^d-1},
\]
as desired.
Thus, using the aforementioned relationship between the Riemann zeta function $\zeta(2n)$ and the Bernoulli numbers $B_{2n}$, we have the lower bound
\[
\frac{\zeta(2d)}{\zeta(4d)} = (-1)^d\frac{B_{2d}(4d)!}{B_{4d}(2d)!(2\pi)^{2d}} \le L_\mathrm{Tam}(K; -1).
\]
To now bound $L_\mathrm{Tam}(K;-1)$ from above, we once again consider the case where each $\mathfrak{p}$ splits completely. Thus, $q := N_{K/\mathbb{Q}}(\mathfrak{p}) = p$ and we have
\[
L_\mathrm{Tam}(K;-1) \le \prod_{p \text{ prime}} \sum_{m=1}^\infty \delta_{\mathbb{Q},p}(m)m = L_\mathrm{Tam}(\mathbb{Q}; -1).
\]
The value of $L_\mathrm{Tam}(\mathbb{Q}; -1)$ is computed in \cite{og} to be $1.8184\ldots$. Hence, we obtain the bound
\[
L_\mathrm{Tam}(K;-1) \le L_\mathrm{Tam}(\mathbb{Q}; -1) < (1.8184)^d,
\]
and we are therefore done. 
\end{proof}
}

{
\begin{proof}[Proof of \Cref{d1d2}] To prove this theorem, we explicitly construct a family of multiquadratic fields whose Tamagawa trivial proportions tend to $0$ and a family of cyclotomic fields whose proportions tend to $1$.

We begin by proving $\liminf_{d\to +\infty} t^{-}(d) = 0$. Let $p_1 < p_2 < p_3 < \dots$ be an infinite sequence of $1 \pmod{8}$ primes. Consider the sequence of multiquadratic field extensions
\begin{align*}
    K_1 &= \mathbb{Q}(\sqrt{p_1}), \\
    K_2 &= \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}), \\
    K_3 &= \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \sqrt{p_3}), \\
    &\vdots
\end{align*}
Fix a field $K_i$. Since $2$ is a quadratic residue modulo $p_i$ for $1 \leq j \leq i$, $(2)$ splits completely in the number fields $\mathbb{Q}(\sqrt{p_j})$. Hence the ideal $(2)$ also splits completely in the composite field $K_i$. In particular, there are $2^i$ distinct prime ideals above $(2)$, each with inertial degree and ramification index $1$. Recall from \Cref{p=2_total} that for norm 2 unramified prime ideals $\mathfrak{p}$ we have $\delta_{K,\mathfrak{p}}(1) = 241/396$. Hence, we have an upper bound
\[t^-(2^i) \leq P_\mathrm{Tam}(K_i;1) = \prod_{p \text{ prime}}\prod_{\mathfrak{p} \mid (p)} \delta_{K_i,\mathfrak{p}}(1) \leq \prod_{\mathfrak{p} \mid (2)} \delta_{K_i,\mathfrak{p}}(1) = (241/396)^{2^i}\]
which tends to $0$ as $i$ grows large. Thus, $\liminf_{d\to +\infty} t^-(d) = 0$.

In fact, this sequence of fields also show that $\limsup_{d\to+\infty}\mu^+(d) = \infty$. For each $\mathfrak{p}$ above 2, we may compute $\sum_{m=1}^\infty \delta_{K,\mathfrak{p}}(m)m > 1.49$, whence
\[\mu^+(2^i) \geq \prod_{p \text{ prime}}\prod_{\mathfrak{p} \mid (p)} \sum_{m=1}^\infty \delta_{K_i,\mathfrak{p}}(m)m \geq \prod_{\mathfrak{p} \mid (2)} \sum_{m=1}^\infty \delta_{K_i,\mathfrak{p}}(m)m > (1.49)^{2^i}.\]
The right-hand side tends to infinity as $i \to \infty$, from which the claim follows.

We now prove $\limsup_{d\rightarrow +\infty} t^+(d) = 1$. Let $a$ be an arbitrary prime and $K=\mathbb{Q}(\zeta_a)$ be the $a$-th cyclotomic field. For primes $p \neq a$, the ideal $(p)$ factors in $K$ as the product of $(a-1)/\operatorname{ord}_a(p)$ prime ideals of norm $p^{\operatorname{ord}_a(p)}$. On the other hand, $(a) = \mathfrak{a}^{a-1}$ where $\mathfrak{a} = (1+\zeta_a)$.

By Propositions~\ref{p=5_total}, \ref{p=3_total}, and \ref{p=2_total}, a prime ideal $\mathfrak{p}$ satisfies the bound $\delta_{K,\mathfrak{p}}(1) \geq 1 - (N_{K/\mathbb{Q}}(\mathfrak{p}))^{-1}$. Hence, we have the lower bound
\[t^+(a) \geq P_\mathrm{Tam}(K;1) = \prod_{p \text{ prime}} \prod_{\substack{\mathfrak{p} \mid (p)}} \delta_{K,\mathfrak{p}}(1) \geq \delta_{K,\mathfrak{a}}(1) \cdot \prod_{p \neq a \text{ prime}} \left(1 - p^{-\operatorname{ord}_a(p)}\right)^{\frac{a-1}{\operatorname{ord}_a(p)}}. \]
We claim that as $a$ grows large, the right-hand side converges to $1$. This is because we have $\delta_{\mathfrak{a},K}(1) \to 1$, and moreover by the naive bound $\operatorname{ord}_a(p) \geq \max\{2,\log_p(a)\}$ we have
\[1 \geq \lim_{a\to \infty} \prod_{p \text{ prime}} \left(1 - p^{-\operatorname{ord}_a(p)}\right)^{\frac{a-1}{\operatorname{ord}_a(p)}} \geq \lim_{a\to\infty} \prod_{p \ \text{prime}} (1-\max\{a,p^2\}^{-1})^{\frac{a-1}{\max\{2,\log_p(a)\}}} = 1.\]
Thus, $\limsup_{d\to +\infty} t^+(d) = 1$.

We finish by showing $\liminf_{d\to+\infty}\mu^{-}(d) = 1$. Borrow the fields $K = \mathbb{Q}(\zeta_a)$ from the above proof. Then by Propositions~\ref{p=5_total}, \ref{p=3_total}, and \ref{p=2_total}, we have the loose bound $\sum_{m=1}^\infty \delta_{K,\mathfrak{p}}(m)m \leq 1 + 100(N_{K/\mathbb{Q}}(\mathfrak{p}))^{-1}$. Thus,
\[
\mu^-(a) \leq \prod_{p \text{ prime}} \prod_{\substack{\mathfrak{p} \mid (p)}} \sum_{m=1}^\infty \delta_{K,\mathfrak{p}}(m)m
\leq \sum_{m=1}^\infty \delta_{K,\mathfrak{a}}(m)m \cdot \prod_{p \neq a \text{ prime}} \left(1 + 100 p^{-\operatorname{ord}_a(p)}\right)^{\frac{a-1}{\operatorname{ord}_a(p)}}.
\]
As $a \to \infty$, the right-hand side converges to $1$. This is because $\sum_{m=1}^\infty \delta_{K,\mathfrak{a}}(m)m \to 1$, and moreover by the naive bound $\operatorname{ord}_a(p) \geq \max\{\log_p(a),2\}$ we have
\[1 \leq \lim_{a\to \infty} \prod_{p \text{ prime}}\left(1 + 100 p^{-\operatorname{ord}_a(p)}\right)^{\frac{a-1}{\operatorname{ord}_a(p)}} \leq \lim_{a\to\infty} \prod_{p \text{ prime}} (1+100\max\{a,p^2\}^{-1})^{\frac{a-1}{\max\{\log_p(a),2\}}} = 1.\]
We have proved both the limit infimum and limit supremum bounds, and we are thus done.
\end{proof}

\section{Examples} \label{examples}

In this section, we offer numerical examples which illustrate the results in this paper.

\begin{example}
There are finitely many imaginary quadratic fields. As briefly discussed in the introduction, in \Cref{trivtable2} we calculated the following values for their proportion of Tamagawa trivial curves. Then we have similar Tables \ref{tamprop2} and \ref{tamprop3}, showing the convergence to $P_{Tam}(\mathbb{Q}(\sqrt{-D}), 2)$ and $P_{Tam}(\mathbb{Q}(\sqrt{-D}),3)$ respectively. Lastly, in \Cref{avgtable2} we have average Tamagawa product for each of the imaginary quadratic fields, as shown below. Note that in \Cref{trivtable2}, $P_{Tam}(\mathbb{Q}(\sqrt{-D}, 1)$ is noticeably smaller when $D = 7 $, and noticeably larger when $D = 163$. This is due to the fact that $2$ splits over $\mathbb{Q}(\sqrt{-7})$, and primes up to $37$ are inert in $\mathbb{Q}(\sqrt{-163})$. Similarly in \Cref{avgtable2}, notice that $\mathbb{Q}(\sqrt{-7})$ has a noticeably high average Tamagawa product, which is related to the noticeably low percentage of Tamagawa trivial curves. On the opposite end, we see that $\mathbb{Q}(\sqrt{-163})$ has a noticeably low average Tamagawa product, which corresponds to its high percentage of Tamagawa trivial curves. 

\renewcommand{\arraystretch}{1}
{\footnotesize
\begin{center}
\begin{table}[H]
\begin{tabular}{| c | c | c | c | c | c | c | c | c | c |}
\hline
\multicolumn{10}{|c|}{$\mathcal{N}_{1}(\mathbb{Q}(\sqrt{-D}); X)/\mathcal{N}(\mathbb{Q}(\sqrt{-D}); X)$} \\
\hline
 $X$ & $\sqrt{-1}$ & $\sqrt{-2}$ & $\sqrt{-3}$ & $\sqrt{-7}$ & $\sqrt{-11}$ & $\sqrt{-19}$ & $\sqrt{-43}$ & $\sqrt{-67}$ & $\sqrt{-163}$   \\ \hline
 
 
 
 $10^4$ & $0.542$ & $0.488$ & $0.663$ & $0.357$ & $0.609$ & $0.620$ & $0.657$ & $0.560$  & $0.450$ \\
 
 $10^5$ & $0.539$ & $0.460$ & $0.665$ & $0.359$ & $0.599$ & $0.678$ & $0.716$ & $0.711$ & $0.636$ \\ 
 
 $10^6$ & $0.528$ & $0.468$ & $0.660$ & $0.343$ & $0.586$ & $0.667$ & $0.726$ & $0.744$ & $0.728$ \\ 
 
 $\vdots$ & $\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ \\ 
 
 $\infty$ & $0.529$ & $0.468$ & $0.661$ & $0.349$ & $0.581$ & $0.665$ & $0.733$ & $0.750$ & $0.763$ \\
 \hline
\end{tabular}
\caption{Convergence to $P_\mathrm{Tam}(\mathbb{Q}(\sqrt{-D}); 1)$.}
\label{trivtable2}
\end{table}
\end{center}
}



{\footnotesize

\begin{center}

\begin{table}[H]
\begin{tabular}{| c | c | c | c | c | c | c | c | c | c |}
\hline
\multicolumn{10}{|c|}{$\mathcal{N}_2(X, K)/\mathcal{N}(X, K)$} \\
\hline
 $X$ & $\sqrt{-1}$ & $\sqrt{-2}$ & $\sqrt{-3}$ & $\sqrt{-7}$ & $\sqrt{-11}$ & $\sqrt{-19}$ & $\sqrt{-43}$ & $\sqrt{-67}$ & $\sqrt{-163}$   \\ \hline
 
 
 $10^4$ & $0.353$ & $0.351$ & $0.248$ & $0.354$ & $0.255$ & $0.278$ & $0.260$ & $0.284$  & $0.283$ \\
 
 $10^5$ & $0.360$ & $0.387$ & $0.253$ & $0.369$ & $0.277$ & $0.231$ & $0.220$ & $0.218$ & $0.207$ \\ 
 
 $10^6$ & $0.377$ & $0.382$ & $0.266$ & $0.382$ & $0.298$ & $0.256$ & $0.228$ & $0.211$ & $0.215$ \\ 
 
 $\vdots$ & $\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ \\
 
 $\infty$ & $0.378$ & $0.384$ & $0.264$ & $0.370$ & $0.299$ & $0.265$ & $0.226$ & $0.216$ & $0.206$ \\
 \hline
 
 
\end{tabular}

\caption{Convergence to $P_\mathrm{Tam}(\mathbb{Q}(\sqrt{-D}); 2)$.}
\label{tamprop2}
\end{table}
\end{center}
}

{\footnotesize
\begin{center}

\begin{table}[H]
\begin{tabular}{| c | c | c | c | c | c | c | c | c | c |}
\hline
\multicolumn{10}{|c|}{$\mathcal{N}_3(X, K)/\mathcal{N}(X, K)$} \\
\hline
 $X$ & $\sqrt{-1}$ & $\sqrt{-2}$ & $\sqrt{-3}$ & $\sqrt{-7}$ & $\sqrt{-11}$ & $\sqrt{-19}$ & $\sqrt{-43}$ & $\sqrt{-67}$ & $\sqrt{-163}$   \\ \hline
 

 $10^4$ & $0.008$ & $0.023$ & $0.015$ & $0.074$ & $0.018$ & $0.027$ & $0.019$ & $0.069$  & $0.100$ \\
 
 $10^5$ & $0.015$ & $0.026$ & $0.029$ & $0.078$ & $0.035$ & $0.029$ & $0.035$ & $0.040$ & $0.089$ \\ 
 
 $10^6$ & $0.017$ & $0.025$ & $0.027$ & $0.078$ & $0.035$ & $0.024$ & $0.021$ & $0.025$ & $0.031$ \\  
 
  
 $\vdots$ & $\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ \\
 
 $\infty$ & $0.018$ & $0.026$ & $0.032$ & $0.082$ & $0.038$ & $0.028$ & $0.024$ & $0.024$ & $0.024$ \\
 \hline
\end{tabular}

\caption{Convergence to $P_\mathrm{Tam}(\mathbb{Q}(\sqrt{-D}); 3)$.}
\label{tamprop3}
\end{table}
\end{center}
}

{\footnotesize
\begin{center}
\begin{table}[H]
    \centering
    \begin{tabular}{| c | c | c | c | c | c | c | c | c | c |}
    \hline
\multicolumn{10}{|c|}{$\sum_{\mathrm{ht}(\mathbb{Q}(\sqrt{-D}); E) \le X}\mathrm{Tam}(\mathbb{Q}(\sqrt{-D})); E)/N(\mathbb{Q}(\sqrt{-D}); X)$} \\
\hline
        $X$ & $\sqrt{-1}$ & $\sqrt{-2}$ & $\sqrt{-3}$ & $\sqrt{-7}$ & $\sqrt{-11}$ & $\sqrt{-19}$ & $\sqrt{-43}$ & $\sqrt{-67}$ & $\sqrt{-163}$   \\ \hline
  $10^4$ & $1.751$ & $2.054$ & $1.589$ & $2.570$ & $1.850$ & $1.718$ & $1.535$ & $1.698$ & $2.017$ \\
  $10^5$ & $1.720$ & $1.979$ & $1.538$ & $2.417$ & $1.763$ & $1.537$ & $1.393$ & $1.403$ & $1.612$ \\
   $10^6$ & $1.708$ & $1.946$ & $1.508$ & $2.418$ & $1.723$ & $1.519$ & $1.361$ & $1.333$ & $1.372$ \\
 
 $\vdots$ & $\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ &$\vdots$ \\ 
 
 $\infty$ & $1.678$ & $1.904$ & $1.487$ & $2.376$ & $1.708$ & $1.480$ & $1.331$ & $1.300$ & $1.277$ \\
 \hline
    \end{tabular}
    \caption{Convergence to $L_\mathrm{Tam}(\mathbb{Q}(\sqrt{-D}); -1)$.}
    \label{avgtable2}
\end{table}
\end{center}
}


\end{example}

\begin{example}
We computed the Tamagawa trivial proportions for some real quadratic fields. Below we have \Cref{figrealquads} and \Cref{figrealquadsavg} demonstrating the spread of $P_{\mathrm{Tam}}(\mathbb{Q}(\sqrt{D}),1)$ a, for square-free $2 \leq D < 10^4$. 

\begin{figure}[H]
    \centering
    \includegraphics[height=6cm]{realquads.png}
    \caption{Distribution of $P_\mathrm{Tam}(K;1)$ for real quadratic number fields.}
    \label{figrealquads}
\end{figure}

\begin{figure}[H]
    \centering
    \includegraphics[height=6cm]{realquadsavg.png}
    \caption{Distribution of $L_\mathrm{Tam}(K;-1)$ for real quadratic number fields.}
    \label{figrealquadsavg}
\end{figure}

In \Cref{figrealquads}, one can see that $P_{\mathrm{Tam}}(\mathbb{Q}(\sqrt{D}), 1)$ does not have a normal distribution, nor does it resemble a skewed normal distribution. Instead, there appear to be three distinct sections, from $\approx 0.26$ to $\approx 0.36$, from $\approx 0.41$ to $\approx 0.57$, and from $\approx 0.58$ to $\approx 0.75$. These three distinct regions correspond to how 2 behaves at any particular field. The left-most section corresponds to fields where 2 splits, the middle section corresponds to fields where 2 ramifies, and the right-most section corresponds to fields where 2 is inert. This leads us to many possible questions. Is the distribution uniform or random among each section? Are the proportions dense on any interval? How might these generalize over different degree fields? Are there always gaps between the sections? In \Cref{figrealquadsavg}, we see that $L_{\mathrm{Tam}}(Q(\sqrt{D}),-1)$ also does not have a normal distribution, but instead has two distinct sections. This leads to further questions, such as how do these sections relate to those for $P_{\mathrm{Tam}}(Q(\sqrt{D}),1)$? Though we do not answer these questions in our paper, they are a noteworthy topic for future study.
\end{example}

\begin{example}
From \Cref{d1d2} we have
\[\liminf_{d\rightarrow +\infty} t^-(d) = 0 \quad \text{and} \quad \limsup_{d\rightarrow +\infty} t^+(d) = 1.\]
Moreover, we have
\[\liminf_{d\to+\infty} \mu^-(d) = 1 \quad \text{and} \quad \limsup_{d\to+\infty} \mu^+(d) = \infty.\]

Below in \Cref{multiquad}, \Cref{zetas}, and \Cref{zetasavg} we have the trivial Tamagawa proportions and average Tamagawa products for sequences of fields that demonstrate these limits.


\renewcommand{\arraystretch}{1.3}
\begin{center}
\begin{table}[H]
\begin{tabular}{| c || c | c | c | c | }
\hline
$K$ & $\mathbb{Q}(\sqrt{17})$ & $\mathbb{Q}(\sqrt{17}, \sqrt{41})$ & $\mathbb{Q}(\sqrt{17}, \sqrt{41}, \sqrt{73})$ & $\mathbb{Q}(\sqrt{17}, \sqrt{41}, \sqrt{73}, \sqrt{89})$ \\ \hline
     $P_{\mathrm{Tam}}(K; 1)$ & $0.35585$  & $0.13273$ & $0.01778$ & $0.00031$ \\ \hline
     $L_{\mathrm{Tam}}(K;-1)$ & $2.32335$ & $5.14423$ & $26.22779$ & $686.87874$ \\ \hline
\end{tabular}
\smallskip
\caption{Tamagawa trivial proportions of multiquadratic fields.}
\label{multiquad}
\end{table}
\end{center}

\renewcommand{\arraystretch}{2.1}

In \Cref{multiquad} we have an example of a sequence of fields with Tamagawa trivial proportion decreasing to zero, and the corresponding average Tamagawa products which head off to infinity. The sequence is constructed by the methods described in the proof of \Cref{d1d2}.

\begin{figure}[H]
    \centering
    \includegraphics[height=6cm]{cyclotomic.png}
    \caption{$P_{\mathrm{Tam}}(\mathbb{Q}(\zeta_p),1)$, for cyclotomic fields with $p < 258$.}
    \label{zetas}
\end{figure}

\begin{figure}[H]
    \centering
    \includegraphics[height=6cm]{cyclotomicavg.png}
    \caption{$L_{\mathrm{Tam}}(\mathbb{Q}(\zeta_p),-1)$, for cyclotomic fields with $p < 258$.}
    \label{zetasavg}
\end{figure}


Similarly, in \Cref{zetas}, we have an example of a sequence of fields with Tamagawa trivial proportion tending to $1$. In \Cref{zetasavg} we have the corresponding average Tamagawa products similarly converging to $1$. The sequence is created by taking $\mathbb{Q}(\zeta_p)$ for primes $p$. Note that in both \Cref{zetas} and \Cref{zetasavg} the data points from $p = 5, 7, 31,$ and $127$ have been omitted due to being outliers.
\begin{table}[H]
    \centering
    \begin{tabular}{|c||c|c|c|c|}
    \hline
        $p$ & $5$ & $7$ & $31$ & $127$ \\ \hline
        $P_{\mathrm{Tam}}(\mathbb{Q}(\zeta_p),1)$ & $0.867\dots$ & $0.753\dots$ & $0.827\dots$ & $0.868\dots$ \\ \hline
         $L_{\mathrm{Tam}}(\mathbb{Q}(\zeta_p),-1)$ & $1.155\dots$ & $1.309\dots$ & $1.205\dots$ & $1.151\dots$ \\ \hline
    \end{tabular}
    \caption{Omitted values for $P_{\mathrm{Tam}}(\mathbb{Q}(\zeta_p),1)$ and $L_{\mathrm{Tam}}(\mathbb{Q}(\zeta_p),-1)$}
    \label{tab:my_label}
\end{table} 
\end{example}

\begin{example}
Many of the above examples are of simpler fields, however in our paper we are able to calculate trivial Tamagawa proportions and average Tamagawa products for all fields, regardless of class number or degree. Thus, even for a more complicated field such as $\mathbb{Q}(x^4+5x^2-6x+3)$ with Galois group $S_4$, we can even determine the trivial Tamagawa proportion. Note that $\mathbb{Q}(x^4+5x^2-6x+3)$ has $\Delta = 32880$, and thus the primes $2, 3, 5,$ and $137$ all ramify. More specifically, we have that $2 = (-8\alpha^3-\alpha^2-37\alpha+47)^2$, $3 = (-\alpha)^2(2\alpha^3+9\alpha-13)(\alpha-1)$, $5 = (-\alpha^3+\alpha-1)(-2\alpha^2+2\alpha-1)^2$, and $137 = (44\alpha^3+33\alpha^2 +235\alpha-100)(-\alpha^3-4\alpha^2-9\alpha-17)^2$. By knowing how these primes ramify, we can calculate that $P_{\mathrm{Tam}}(\mathbb{Q}(x^4+5x^2-6x+3), 1) = 0.526\dots$.
\end{example}

