\section{Introduction}

The $F$-pure threshold was introduced by Musta\c t\u a, Takagi, and Watanabe~\cite{TW, MTW}; it is a positive characteristic invariant, analogous to log canonical thresholds in characteristic zero. We calculate the possible values of the $F$-pure threshold of the affine cone over a Calabi-Yau hypersurface, and relate the threshold to the order of vanishing of the Hasse invariant, and to a numerical invariant introduced by van~der~Geer and Katsura in~\cite{GK}.

\begin{theorem}
\label{thm:intro}
Suppose $R=K[x_0,\dots,x_n]$ is a polynomial ring over a field $K$ of characteristic $p>n+1$, and $f$ is a homogeneous polynomial in $R$ defining a smooth Calabi-Yau hypersurface $X=\operatorname{Proj} R/fR$. Then the $F$-pure threshold of $f$ has the form
\[
\mathrm{fpt}(f)\ =\ 1-\frac{h}{p}\,,
\]
where $h$ is an integer with $0\leqslant h\leqslant\dim X$. If $p\geqslant n^2-n-1$, then the integer $h$ equals the order of vanishing of the Hasse invariant on the versal deformation space of $X \subset \mathbf{P}^n$.
\end{theorem}

Hern\'andez has computed $F$-pure thresholds for binomial hypersurfaces~\cite{Hernandez1} and for diagonal hypersurfaces~\cite{Hernandez2}. The $F$-pure threshold is computed for a number of examples in~\cite[Section~4]{MTW}. Example~4.6 of that paper computes the $F$-pure threshold in the case of an ordinary elliptic curve, and raises the question for supersingular elliptic curves; this is answered by the above theorem.

The theory of $F$-pure thresholds is motivated by connections to log canonical thresholds; for simplicity, let $f$ be a homogeneous polynomial with rational coefficients. Using~$f_p$ for the corresponding prime characteristic model, one has
\[
\mathrm{fpt}(f_p)\leqslant\mathrm{lct}(f)\qquad\text{ for all }p\gg0\,,
\]
where $\mathrm{lct}(f)$ denotes the log canonical threshold of $f$, and
\[
\lim_{p\longrightarrow\infty}\mathrm{fpt}(f_p)=\mathrm{lct}(f)\,,
\]
see~\cite[Theorems~3.3,~3.4]{MTW}; this builds on the work of a number of authors, primarily Hara and Yoshida~\cite{Hara-Yoshida}. It is conjectured that $\mathrm{fpt}(f_p)$ and $\mathrm{lct}(f)$ are equal for infinitely many primes; see~\cite{MS} for more in this direction.

The $F$-pure threshold is known to be rational in a number of cases, including for principal ideals in an excellent regular local ring of prime characteristic~\cite{KLZ:JALG}. Other results on rationality include~\cite{BMS:MMJ, BMS:TAMS, BSTZ, Hara, ST}. For more on $F$-pure thresholds, we mention~\cite{BHMM, HMTW, MSV, Mustata, MY}.
 
\section{The $F$-pure threshold}

In~\cite{TW} the $F$-pure threshold is defined for a pair $(R,\mathfrak{a})$, where $\mathfrak{a}$ is an ideal in an $F$-pure ring of prime characteristic. The following special case is adequate for this paper:

\begin{definition}
Let $(R,\mathfrak{m})$ be a regular local ring of characteristic $p>0$. For an element $f$ in $\mathfrak{m}$, and integer $q=p^e$, we define
\[
\mu_f(q)\ :=\ \min\big\{k\in\mathbf{N}\mid f^k\in\mathfrak{m}^{[q]}\big\}\,,
\]
where $\mathfrak{m}^{[q]}$ denotes the ideal generated by the elements $r^q$ for $r\in\mathfrak{m}$. Note that $\mu_f(1)=1$, and that $1\leqslant\mu_f(q)\leqslant q$. Moreover, $f^{\mu_f(q)}\in\mathfrak{m}^{[q]}$ implies that $f^{p\mu_f(q)}\in\mathfrak{m}^{[pq]}$, and it follows that $\mu_f(pq)\leqslant p\mu_f(q)$. Thus,
\[
\Big\{\frac{\mu_f(p^e)}{p^e}\Big\}_{e\ge0}
\]
is a non-increasing sequence of positive rational numbers; its limit is the \emph{$F$-pure threshold} of $f$, denoted $\mathrm{fpt}(f)$.
\end{definition}

By definition, $f^{\mu_f(q)-1}\notin\mathfrak{m}^{[q]}$. Taking $p$-th powers, and using that $R$ is $F$-pure, we get
\[
f^{p\mu_f(q)-p}\ \notin\ \mathfrak{m}^{[pq]}\,.
\]
Combining with the observation above, one has
\begin{equation}
\label{eqn:mu}
p\mu_f(q)-p+1\ \leqslant\ \mu_f(pq)\ \leqslant\ p\mu_f(q)\,.
\end{equation}
Note that this implies
\[
\mu_f(q)\ =\ \left\lceil{\frac{\mu_f(pq)}{p}}\right\rceil\qquad\text{ for each }q=p^e\,.
\]

The definition is readily adapted to the graded case where $R$ is a polynomial ring with homogeneous maximal ideal $\mathfrak{m}$, and $f$ is a homogeneous polynomial.

\begin{remark}
\label{rmk:fptthickenings}
The numbers $\mu_f(p^e)$ may be interpreted in terms of thickenings of the hypersurface $f$ as follows. Let $K$ be a field of characteristic $p>0$, and $f$ a homogeneous polynomial of degree $d$ in $R=K[x_0,\dots,x_n]$. Fix integers $q=p^e$ and $t\leqslant q$. The Frobenius iterate $F^e\colon R/fR\longrightarrow R/fR$ lifts to a map $R/fR\longrightarrow R/f^qR$; composing this with the canonical surjection $R/f^qR\longrightarrow R/f^tR$, we obtain a map
\[
\widetilde{F^e_t}\colon R/fR\longrightarrow R/f^tR\,.
\]

Consider the commutative diagram with exact rows
\[
\CD
0@>>>R(-d)@>f>>R@>>>R/fR@>>>0\phantom{\,,}\\
@. @VVf^{q-t}F^eV @VVF^eV @VV\widetilde{F^e_t}V \\
0@>>>R(-dt)@>f^t>>R@>>>R/f^tR@>>>0\,,
\endCD
\]
and the induced diagram of local cohomology modules
\begin{equation}
\label{equation:cd}
\CD
0@>>>H^n_\mathfrak{m}(R/fR)@>>>H^{n+1}_\mathfrak{m}(R)(-d)@>f>>H^{n+1}_\mathfrak{m}(R)@>>>0\phantom{\,.}\\
@. @VV\widetilde{F^e_t}V @VVf^{q-t}F^eV @VVF^eV \\
0@>>>H^n_\mathfrak{m}(R/f^tR)@>>>H^{n+1}_\mathfrak{m}(R)(-dt)@>f^t>>H^{n+1}_\mathfrak{m}(R)@>>>0\,.
\endCD
\end{equation}
Since the vertical map on the right is injective, it follows that $\widetilde{F^e_t}$ is injective if and only if the middle map is injective, i.e., if and only if the element
\[
f^{q-t}F^e\left(\left[\frac{1}{x_0\cdots x_n}\right]\right)\ =\ \left[\frac{f^{q-t}}{x_0^q\cdots x_n^q}\right]
\]
is nonzero, equivalently, $f^{q-t}\notin\mathfrak{m}^{[q]}$. Hence $\widetilde{F^e_t}\colon H^n_\mathfrak{m}(R/fR)\longrightarrow H^n_\mathfrak{m}(R/f^tR)$ is injective if and only if $\mu_f(q)>q-t$.
\end{remark}

We note that the generating function of the sequence $\{\mu_f(p^e)\}_{e\ge1}$ is a rational function:

\begin{theorem}
Let $(R,\mathfrak{m})$ be a regular local ring of characteristic $p>0$, and let $f$ be an element of $\mathfrak{m}$. Then the generating function
\[
G_f(z)\ :=\ \sum_{e\geqslant 0}\mu_f(p^e)z^e
\]
is a rational function of $z$ with a simple pole at $z=1/p$; the $F$-pure threshold of $f$ is
\[
\mathrm{fpt}(f)\ =\ \lim_{z\longrightarrow 1/p}(1-pz)\,G_f(z)\,.
\]
\end{theorem}

\begin{proof}
Since the numbers $\mu_f(p^e)$ are unchanged when $R$ is replaced by its $\mathfrak{m}$-adic completion, there is no loss of generality in assuming that $R$ is a complete regular local ring; the rationality of $\mathrm{fpt}(f)$ now follows from~\cite[Theorems~3.1,~4.1]{KLZ:JALG}. Let $\mathrm{fpt}(f)=a/b$ for integers $a$ and $b$. By~\cite[Proposition~1.9]{MTW}, one has
\[
\mu_f(p^e)\ =\ \lceil p^e\,\mathrm{fpt}(f)\rceil\ =\ \left\lceil{\frac{ap^e}{b}}\right\rceil
\qquad\text{for each }q=p^e\,.
\]
Suppose $ap^{e_0}\equiv ap^{e_0+e_1}\mod b$ for integers $e_0$ and $e_1$. Then $ap^{e_0}\equiv ap^{e_0+ke_1}\mod b$ for each integer~$k\ge0$. Hence there exists an integer $c$ such that
\begin{align*}
H(z)\ := \ \sum_{k\ge0}\mu_f(p^{e_0+ke_1})z^{e_0+ke_1}\ &=\ \sum_{k\ge0}\left\lceil{\frac{ap^{e_0+ke_1}}{b}}\right\rceil z^{e_0+ke_1}\\
&=\ \sum_{k\ge0}\frac{ap^{e_0+ke_1}+c}{b}\ z^{e_0+ke_1}\,,
\end{align*}
is a rational function of $z$ with a simple pole at $z=1/p$. Moreover,
\[
\lim_{z\longrightarrow 1/p}(1-pz)\,H(z)\ =\ \frac{a}{be_1}\,.
\]
Partitioning the integers $e\geqslant e_0$ into the congruence classes module $e_1$, it follows that $G_f(z)$ is the sum of a polynomial in $z$ and $e_1$ rational functions of the form
\[
\sum_{k\geqslant 0}\mu_f(p^{\ell+ke})z^{\ell+ke}\,.
\]
The assertions regarding the pole and the limit now follow.
\end{proof}

The theorem holds as well in the graded setting.

\section{Preliminary results}

We record some elementary calculations that will be used later. Here, and in the following sections, $R$ will denote a polynomial ring $K[x_0,\dots,x_n]$ over a field $K$ of characteristic~$p>0$, and $\mathfrak{m}$ will denote its homogeneous maximal ideal. By the \emph{Jacobian ideal} of a polynomial~$f$, we mean the ideal generated by the partial derivatives
\[
f_{x_i}\ :=\ \partial\!f/\partial x_i\quad\text{ for }\ 0\leqslant i\leqslant n\,.
\]
If $f$ is homogeneous of degree coprime to $p$, then the Euler identity ensures that $f$ is an element of the Jacobian ideal; this is then the defining ideal of the singular locus of $R/fR$.

\begin{lemma}
\label{lemma:jacobian}
Let $f$ be a homogeneous polynomial of degree $d$ in $K[x_0,\dots,x_n]$ such that the Jacobian ideal $J$ of $f$ is $\mathfrak{m}$-primary. Then
\[
\mathfrak{m}^{(n+1)(d-2)+1}\ \subseteq\ J\,.
\]
\end{lemma}

\begin{proof}
Since $J$ is $\mathfrak{m}$-primary, it is a complete intersection ideal. As it is generated by forms of degree $d-1$, the Hilbert-Poincar\'e series of $R/J$ is
\[
P(R/J,t)\ =\ \frac{(1-t^{d-1})^{n+1}}{(1-t)^{n+1}}\ =\ (1+t+t^2+\dots+t^{d-2})^{n+1}\,.
\]
It follows that $R/J$ has no nonzero elements of degree greater than $(n+1)(d-2)$.
\end{proof}

\begin{lemma}
\label{lemma:colon}
Let $R=K[x_0,\dots,x_n]$ and $\mathfrak{m}^{[q]}=(x_0^q,\dots,x_n^q)$. Then
\[
\mathfrak{m}^{[q]}:_R\mathfrak{m}^{(n+1)(d-2)+1}\ \subseteq\ \mathfrak{m}^{[q]}+\mathfrak{m}^{(n+1)(q-d+1)}\,,
\]
where $\mathfrak{m}^i=R$ for $i\le0$.
\end{lemma}

\begin{proof}
We prove, more generally, that
\[
\mathfrak{m}^{[q]}:_R\mathfrak{m}^k\ =\
\begin{cases}
\mathfrak{m}^{[q]}+\mathfrak{m}^{nq+q-n-k}&\text{ if }\ 0\leqslant k\leqslant nq+q-n\,,\\
R&\text{ if }\ k\geqslant nq+q-n\,.
\end{cases}
\]
Suppose $r$ is a homogeneous element of $\mathfrak{m}^{[q]}:_R\mathfrak{m}^k$. Computing the local cohomology module~$H^{n+1}_\mathfrak{m}(R)$ via a \v Cech complex on $x_0,\dots,x_n$, the element
\[
\left[\frac{r}{x_0^q\cdots x_n^q}\right]\ \in\ H^{n+1}_\mathfrak{m}(R)
\]
is annihilated by $\mathfrak{m}^k$, and hence lies in ${\left[H^{n+1}_\mathfrak{m}(R)\right]}_{\geqslant -n-k}$. If $r\notin\mathfrak{m}^{[q]}$, then
\[
\deg r-(n+1)q\ \geqslant\ -n-k\,,
\]
i.e., $r\in\mathfrak{m}^{nq+q-n-k}$. The pigeonhole principle implies that $\mathfrak{m}^{nq+q-n}$ is contained in $\mathfrak{m}^{[q]}$, which gives the rest.
\end{proof}

\begin{lemma}
\label{lemma:bound}
Let $f$ be a homogeneous polynomial of degree $d$ in $K[x_0,\dots,x_n]$, such that the Jacobian ideal of $f$ is $\mathfrak{m}$-primary. If $\mu_f(q)$ is not a multiple of $p$, then 
\[
\mu_f(q)\ \geqslant\ \frac{(n+1)(q+1)-nd}{d}\,.
\]
\end{lemma}

\begin{proof}
Set $k:=\mu_f(q)$, i.e., $k$ is the least integer such that
\[
f^k\ \in\ \mathfrak{m}^{[q]}\,.
\]
Applying the differential operators $\partial/\partial x_i$ to the above, we see that
\[
kf^{k-1}f_{x_i}\ \in\ \mathfrak{m}^{[q]}\qquad\text{ for each }i\,,
\]
since $\partial/\partial x_i$ maps elements of $\mathfrak{m}^{[q]}$ to elements of $\mathfrak{m}^{[q]}$. As $k$ is nonzero in $K$, one has
\[
f^{k-1}J\ \subseteq\ \mathfrak{m}^{[q]}\,,
\]
where $J$ is the Jacobian ideal of $f$. Lemma~\ref{lemma:jacobian} now implies that
\[
f^{k-1}\mathfrak{m}^{(n+1)(d-2)+1}\ \subseteq\ \mathfrak{m}^{[q]}\,.
\]
By Lemma~\ref{lemma:colon}, we then have
\[
f^{k-1}\ \in\ \mathfrak{m}^{[q]}+\mathfrak{m}^{(n+1)(q-d+1)}\,.
\]
But $f^{k-1}\notin\mathfrak{m}^{[q]}$ by the minimality of $k$, so $f^{k-1}$ has degree at least $(n+1)(q-d+1)$, i.e.,
\[
d(k-1)\ \geqslant\ (n+1)(q-d+1)\,;
\]
rearranging the terms, one obtains the desired inequality
\[
k\ \geqslant\ \frac{(n+1)(q+1)-nd}{d}\,.
\qedhere
\]
\end{proof}

\begin{lemma}
\label{lemma:patterns}
Let $f$ be a homogeneous polynomial of degree $d$ in $K[x_0,\dots,x_n]$, such that the Jacobian ideal of $f$ is $\mathfrak{m}$-primary. 
\begin{enumerate}[\quad\rm(1)]
\item If \ $\displaystyle{\frac{\mu_f(q)-1}{q-1}=\frac{n+1}{d}}$ \ for some $q=p^e$, then \ $\displaystyle{\frac{\mu_f(pq)-1}{pq-1}=\frac{n+1}{d}}$.

\item Suppose $p\geqslant nd-d-n$. If \ $\displaystyle{\frac{\mu_f(q)}{q}<\frac{n+1}{d}}$ \ for some $q=p^e$, then $\mu_f(pq)=p\mu_f(q)$.
\end{enumerate}
\end{lemma}

\begin{proof}
(1) Since $f^{\mu_f(q)-1}$ has degree $(q-1)(n+1)$ and is not an element of $\mathfrak{m}^{[q]}$, it must generate the socle in $R/\mathfrak{m}^{[q]}$. But then
\[
\left(f^{\mu_f(q)-1}\right)^{\frac{pq-1}{q-1}}
\]
generates the socle in $R/\mathfrak{m}^{[pq]}$, so
\[
\mu_f(pq)-1\ =\ \big(\mu_f(q)-1\big)\left(\frac{pq-1}{q-1}\right)\,.
\]

For (2), suppose that $\mu_f(pq)<p\mu_f(q)$. Then $\mu_f(pq)$ is not a multiple of $p$ by~\eqref{eqn:mu}. Lemma~\ref{lemma:bound} thus implies that
\[
(n+1)(pq+1)-nd\ \leqslant\ d\mu_f(pq)\,.
\]
Combining with $\mu_f(pq)\leqslant p\mu_f(q)-1$ and $d\mu_f(q)\leqslant q(n+1)-1$, we obtain
\[
p\ \leqslant\ nd-d-n-1\,,
\]
which contradicts the assumption on $p$.
\end{proof}

We next prove a result on the injectivity of the Frobenius action on negatively graded components of local cohomology modules:

\begin{theorem}
\label{thm:highdeg}
Let $K$ be a field of characteristic $p>0$. Let $f$ be a homogeneous polynomial of degree $d$ in $R=K[x_0,\dots,x_n]$, such that the Jacobian ideal of $f$ is primary to the homogeneous maximal ideal $\mathfrak{m}$ of $R$. 

If $p\geqslant nd-d-n$, then the Frobenius action below is injective:
\[
F\colon{\left[H^n_\mathfrak{m}(R/fR)\right]}_{<0}\longrightarrow{\left[H^n_\mathfrak{m}(R/fR)\right]}_{<0}\,.
\]
\end{theorem}

\begin{proof}
Using~\eqref{equation:cd} in the case $t=1$ and $e=1$, and restricting to the relevant graded components, we have the diagram with exact rows
\[
\CD
0@>>>{\left[H^n_\mathfrak{m}(R/fR)\right]}_{\leqslant-1}@>>>{\left[H^{n+1}_\mathfrak{m}(R)\right]}_{\leqslant-d-1}@>>>\cdots\phantom{\,.}\\
@. @VVFV @VVf^{p-1}FV\\
0@>>>{\left[H^n_\mathfrak{m}(R/fR)\right]}_{\leqslant-p}@>>>{\left[H^{n+1}_\mathfrak{m}(R)\right]}_{\leqslant-d-p}@>>>\cdots\,.
\endCD
\]
Thus, it suffices to prove the injectivity of the map
\[
f^{p-1}F\colon{\left[H^{n+1}_\mathfrak{m}(R)\right]}_{\leqslant-d-1}\longrightarrow{\left[H^{n+1}_\mathfrak{m}(R)\right]}_{\leqslant-d-p}\,.
\]
A homogeneous element of ${[H^{n+1}_\mathfrak{m}(R)]}_{\leqslant-d-1}$ may be written as
\[
\left[\frac{g}{(x_0\cdots x_n)^{q/p}}\right]
\]
for some $q$, where $g\in R$ is homogeneous of degree at most $(n+1)q/p-d-1$. Suppose
\[
f^{p-1}F\left( \left[\frac{g}{(x_0\cdots x_n)^{q/p}}\right] \right)\ =\ 0\,,
\]
then it follows that $f^{p-1}g^p\in\mathfrak{m}^{[q]}$. Let $k$ be the least integer with
\[
f^kg^p\ \in\ \mathfrak{m}^{[q]}\,,
\]
and note that $0\leqslant k\leqslant p-1$. If $k$ is nonzero, then applying $\partial/\partial x_i$ we see that
\[
f^{k-1}g^pJ\ \subseteq\ \mathfrak{m}^{[q]}\,.
\]
Lemma~\ref{lemma:jacobian} and Lemma~\ref{lemma:colon} show that
\[
f^{k-1}g^p\ \in\ \mathfrak{m}^{[q]}+\mathfrak{m}^{(n+1)(q-d+1)}\,.
\]
Since $f^{k-1}g^p\notin\mathfrak{m}^{[q]}$, we must have
\[
\deg f^{k-1}g^p\ \geqslant\ (n+1)(q-d+1)\,.
\]
Using $k\leqslant p-1$ and $\deg g^p\leqslant q(n+1)-pd-p$, this gives
\[
nd-d-n-1\ \geqslant\ p\,,
\]
contradicting the assumption on $p$. It follows that $k=0$, i.e., that $g^p\in\mathfrak{m}^{[q]}$. But then
\[
\left[\frac{g}{(x_0\cdots x_n)^{q/p}}\right]\ =\ 0
\]
in $H^{n+1}_\mathfrak{m}(R)$, which proves the desired injectivity.
\end{proof}

\begin{remark}
Theorem~\ref{thm:highdeg} is equivalent to the following geometric statement: if $X$ is a smooth hypersurface of degree $d$ in $\mathbf{P}^n$, then the map
\[
H^{n-1}(X,{\mathcal{O}}_X(j))\longrightarrow H^{n-1}(X,{\mathcal{O}}_X(jp))\,,
\]
induced by Frobenius map on $X$, is injective for $j<0$ and $p\geqslant nd-d-n$. This statement indeed admits a geometric proof based on the Deligne-Illusie method \cite{DI}. One views the de Rham complex $\Omega^*_{X/K}$ as an ${\mathcal{O}}_{X^{(1)}}$-complex, where $X^{(1)}$ is the Frobenius twist of $X$ over~$K$, and twists it over the latter with ${\mathcal{O}}_{X^{(1)}}(j)$. For $p>n-1$, the Deligne-Illusie decomposition $\Omega^*_{X/K} \simeq \oplus_i \Omega^i_{X^{(1)}/K}[-i]$, which is available as $X$ clearly lifts to $W_2(K)$, reduces the above injectivity statement to proving $H^{n-1-i}(X,\Omega^i_{X/K}(jp)) = 0$ for $i > 0$ and $j < 0$. If $p \geqslant nd-d-n$, this vanishing can be proven using standard sequences (details omitted).
\end{remark}

\section{Calabi-Yau hypersurfaces}

We get to the main theorem; see below for the definition of the Hasse invariant.

\begin{theorem}
\label{thm:CalabiYau}
Let $K$ be a field of characteristic $p>0$, and $n$ a positive integer. Let $f$ be a homogeneous polynomial of degree $n+1$ in $R=K[x_0,\dots,x_n]$, such that the Jacobian ideal of~$f$, i.e., the ideal $(f_{x_0},\dots,f_{x_n})$, is primary to the homogeneous maximal ideal of $R$. Then:
\begin{enumerate}[\quad\rm(1)]
\item $\mu_f(p)=p-h$, where $h$ is an integer with $0\leqslant h\leqslant n-1$,
\item $\mu_f(pq)=p\mu_f(q)$ for all $q=p^e$ with $q\geqslant n-1$.
\item If $p\geqslant n-1$, then $\displaystyle{G_f(z)=\frac{1-hz}{1-pz}}$ \ and \ $\displaystyle{\mathrm{fpt}(f)=1-\frac{h}{p}}$\,, where $0\leqslant h\leqslant n-1$.
\item Set $X=\operatorname{Proj} R/fR$. If $p\geqslant n^2-n-1$, then the integer $h$ in $(1)$ is the order of vanishing of the Hasse invariant on the versal deformation space of $X \subset \mathbf{P}^n$.
\end{enumerate}
\end{theorem}

The deformation space in (4) refers to embedded deformations of $X\subset\mathbf{P}^n$; if $n\geqslant 5$, this coincides with the versal deformation space of $X$ as an abstract variety (see Remark~\ref{rmk:VersalDefHyp}). The following example from~\cite{Hernandez2} shows that all possible values of $h$ from (1) above are indeed attained:

\begin{example}
\label{ex:CalabiYau}
Consider $f=x_0^{n+1}+\dots+x_{n+1}^{n+1}$ over a field of prime characteristic $p$ not dividing $n+1$. Let $h$ be an integer such that $p\equiv h+1\mod n+1$ and $0\leqslant h\leqslant n-1$. Then
\[
\mathrm{fpt}(f)\ =\ 1-h/p\,,
\]
for a proof, see~\cite[Theorem~3.1]{Hernandez2}.
\end{example}

\begin{proof}[Proof of Theorem~\ref{thm:CalabiYau}]
If $\mu_f(p)=p$, then Lemma~\ref{lemma:patterns}\,(1) shows that $\mu_f(q)=q$ for all $q$, and assertions (1--3) follow. Assume that $\mu_f(p)<p$. Lemma~\ref{lemma:bound} gives $\mu_f(p)\geqslant p-n+1$, which proves (1). As $\mu_f(p)\leqslant p-1$, it follows~that
\[
\mu_f(q)\ \leqslant\ q-q/p\qquad\text{ for each }q=p^e\,.
\]
If $\mu_f(pq)<p\mu_f(q)$, then $\mu_f(pq)$ is not a multiple of $p$ by~\eqref{eqn:mu}. Lemma~\ref{lemma:bound} now implies
\[
\mu_f(pq)\ \geqslant\ pq-n+1\,,
\]
and combining with $\mu_f(pq)\leqslant p\mu_f(q)-1\leqslant pq-q-1$, we see that $pq-n+1\leqslant pq-q-1$, i.e., that $q\leqslant n-2$. This completes the proof of (2), and then (3) follows immediately.

The proof of (4) and the surrounding material occupy the rest of this section.
\end{proof}

\subsection*{The Hasse invariant}

We briefly review the construction of the Hasse invariant for suitable families of varieties in characteristic $p$. Fix a proper flat morphism $\pi\colon{\mathcal{X}}\longrightarrow S$ of relative dimension $N$ between noetherian $\mathbf{F}_p$-schemes. Assume that the formation of~$\mathrm{R}^i\pi_*{\mathcal{O}}_{\mathcal{X}}$ is compatible with base change, and that $\omega := \omega_{{\mathcal{X}}/S} := \mathrm{R}^N\pi_*{\mathcal{O}}_{\mathcal{X}}$ is a line bundle; the key example is a family of degree $(n+1)$ hypersurfaces in $\mathbf{P}^n$. The standard diagram of Frobenius twists of $\pi$ takes the shape
\[
\xymatrix{
{\mathcal{X}} \ar@/_/[ddr]_{\pi} \ar@/^/[drr]^{\mathrm{Frob}_{\mathcal{X}}} \ar[dr]^-{\mathrm{Frob}_\pi} & & \\
& {\mathcal{X}}^{(1)} \ar[d]^-{\pi^{(1)}} \ar[r]^-{\mathrm{Frob}_S} & {\mathcal{X}} \ar[d]^-{\pi} \\
& S \ar[r]^-{\mathrm{Frob}_S} & S\,,}
\]
where the square is Cartesian. Our assumption on $\pi$ shows that
\[
\omega_{{\mathcal{X}}^{(1)}/S}\ :=\ \mathrm{R}^N \pi_*{\mathcal{O}}_{{\mathcal{X}}^{(1)}}\ \simeq\ \mathrm{Frob}_S^*\omega\ \simeq\ \omega^p\,.
\]
Using this isomorphism, we define:

\begin{definition}
The \emph{Hasse invariant} $H$ of the family $\pi$ is the element in 
\[
\mathrm{Hom}(\omega_{{\mathcal{X}}^{(1)}/S},\omega)\ \simeq\ \mathrm{Hom}(\mathrm{Frob}_S^*\omega,\omega)\ \simeq\ \mathrm{Hom}(\omega^p,\omega)\ \simeq\ H^0(S,\omega^{1-p})\,,
\]
defined by pullback along the relative Frobenius map $\mathrm{Frob}_\pi\colon{\mathcal{X}}\longrightarrow{\mathcal{X}}^{(1)}$.
\end{definition}

\begin{remark} 
\label{rmk:hasseflatbasechange}
The formation of the relative Frobenius map $\mathrm{Frob}_\pi\colon{\mathcal{X}}\longrightarrow{\mathcal{X}}^{(1)}$ is compatible with base change on $S$. It follows by our assumption on $\pi$ that the formation of $H$ is also compatible with base change. In particular, given a flat morphism $g\colon S' \longrightarrow S$ and a point $s'\in S'$, the order of vanishing of $H$ at $s'$ coincides with that at $g(s')$. Thus, in proving Theorem~\ref{thm:CalabiYau}\,(4), we may assume that $K$ is perfect.
\end{remark}

To analyze $H$, fix a point $s$ in $S$ and an integer $t\ge0$. Write $t[s]\subset S$ for the order $t$ neighbourhood of $s$, and let $t\!{\mathcal{X}}_s\subset{\mathcal{X}}$ and $t\!{\mathcal{X}}^{(1)}_s\subset{\mathcal{X}}^{(1)}$ be the corresponding neighbourhoods of the fibres of $\pi$ and $\pi^{(1)}$. The map $\mathrm{Frob}_\pi$ induces maps $t\!{\mathcal{X}}_s\longrightarrow t\!{\mathcal{X}}^{(1)}_s$, and hence maps
\[
\phi_t\colon H^N\big(t\!{\mathcal{X}}_{s}^{(1)},\ {\mathcal{O}}_{t\!{\mathcal{X}}_s^{(1)}}\big)\longrightarrow H^N\big(t\!{\mathcal{X}}_s,\ {\mathcal{O}}_{t\!{\mathcal{X}}_s}\big)\,.
\]
The order of vanishing of $H$ at $s$ is, by definition, the maximal $t$ such that this map is zero. In favourable situations, one can give a slightly better description of this integer:

\begin{lemma}
\label{lem:identifyhasse}
If the map
\[
\psi_t\colon H^N\big({\mathcal{X}}_s,\ {\mathcal{O}}_{X_s}\big) \longrightarrow H^N\big(t\!{\mathcal{X}}_s,\ {\mathcal{O}}_{t\!{\mathcal{X}}_s}\big)
\]
induced by $\mathrm{Frob}_{{\mathcal{X}}}$ is nonzero for some $t\leqslant p$, then the minimal such $t$ is $\mathrm{ord}_s H+1$.
\end{lemma}

\begin{proof}
For $t \leqslant p$, by the base change assumption on $\mathrm{R}^N\pi_*{\mathcal{O}}_{{\mathcal{X}}}$, one has 
\[
H^N\big({\mathcal{X}}_s,\ {\mathcal{O}}_{X_s}\big)\otimes_{\kappa(s)}{\mathcal{O}}_{t[s]}\ \simeq\ H^N\big(t\!{\mathcal{X}}^{(1)}_s,\ {\mathcal{O}}_{t\!{\mathcal{X}}^{(1)}_s}\big)\,,
\]
where ${\mathcal{O}}_{t[s]}$ is viewed as $\kappa(s)$-algebra via the composite
\[
\CD
\kappa(s)@>{\mathrm{Frob}_S}>>{\mathcal{O}}_{\mathrm{Frob}_S^{-1}[s]}@>{\mathrm{can}}>>{\mathcal{O}}_{t[s]}\,,
\endCD
\]
and the isomorphism is induced by the base change ${\mathcal{X}}^{(1)} \longrightarrow {\mathcal{X}}$ of $\mathrm{Frob}_S\colon S \longrightarrow S$. Hence, for such $t$, by adjunction, the map $\phi_t$ induced by $\mathrm{Frob}_\pi$ is nonzero if and only the map $\psi_t$ induced by $\mathrm{Frob}_{\mathcal{X}}$ is nonzero. But $\mathrm{ord}_s H$ is the maximal integer $t$ with $\phi_t = 0$.
\end{proof}

It is typically hard to calculate $H$, or even bound its order of vanishing. However, for families of Calabi-Yau hypersurfaces, we have the following remarkable theorem due to Deuring and Igusa; see~\cite{Igusa} for $n=2$, and Ogus~\cite[Corollary~3]{Ogus} in general:

\begin{theorem}
\label{thm:ogusigusadeuring}	
Let $\pi\colon{\mathcal{X}}\longrightarrow\mathrm{Hyp}_{n+1}$ be the universal family of Calabi-Yau hypersurfaces in $\mathbf{P}^n$. For any point $[Y]\in\mathrm{Hyp}_{n+1}(K)$ corresponding to a smooth hypersurface $Y\subset\mathbf{P}^n$, we have $\mathrm{ord}_{[Y]}(H)\leqslant n-1$ if $n\leqslant p$.
\end{theorem}

Ogus's proof relies on crystalline techniques: he relates $\mathrm{ord}_{[Y]}(H)$ to the relative position of the conjugate and Hodge filtrations on a crystalline cohomology group of $Y$ (following an idea of Katz), and then exploits the natural relation between the Hodge filtration and deformation theory of $Y$. His result will not be used in proving Theorem~\ref{thm:CalabiYau}; in fact, our methods yield an alternative proof of Theorem~\ref{thm:ogusigusadeuring} avoiding crystalline methods under a mild additional constraint on the prime $p$, see Remark~\ref{remark:Ogus}.

\begin{remark}
\label{rmk:VersalDefHyp}
The universal family $\pi\colon{\mathcal{X}}\longrightarrow\mathrm{Hyp}_{n+1}$ is, in fact, versal at $[X]$ if $n \geqslant 5$ so $\mathrm{ord}_{[X]}(H)$, i.e., the order of vanishing of $H$ at $[X]\in\mathrm{Hyp}_{n+1}(K)$, is completely intrinsic to $X$. To see versality, it suffices to show that the map $\mathrm{Hom}(I_X/I_X^2, {\mathcal{O}}_X)\longrightarrow H^1(X,T_X)$ obtained from the adjunction sequence 
\[
\CD
0@>>>I_X/I_X^2@>>>\Omega^1_{\mathbf{P}^n}|_X@>>>\Omega^1_X@>>>0
\endCD
\]
is surjective, and that $H^2(X,T_X)=0$. By the long exact sequence, it suffices to show the vanishing of $H^1(X,T_{\mathbf{P}^n}|_X)$ and $H^2(X,T_X)$. The Euler sequence
\[ 
\CD
0@>>>{\mathcal{O}}_{\mathbf{P}^n}@>>>{\mathcal{O}}_{\mathbf{P}^n}(1)^{\oplus n+1}@>>>T_{\mathbf{P}^n}@>>>0
\endCD
\]
restricted to $X$ immediately shows that $H^i(X,T_{\mathbf{P}^n}|_X) = 0$ for $i = 1,2$ if $n \geqslant 5$; here we use that $H^i(X,{\mathcal{O}}_X(j)) = 0$ for $0 < i < n-1$ and all $j$. The cohomology sequence for the dual of the adjunction sequence then shows that $H^2(X,T_X)=0$.
\end{remark}

\subsection*{The universal family}

Fix notation as in Theorem~\ref{thm:CalabiYau}, with $K$ a perfect field. Let
\[
\mathrm{Hyp}_{n+1}\ := \ \mathbf{P}\big(H^0(\mathbf{P}^n,{\mathcal{O}}_{\mathbf{P}^n}(n+1))^\vee\big)
\]
be the space of hypersurfaces of degree $(n+1)$ in $\mathbf{P}^n$; we follow Grothendieck's conventions regarding projective bundles. Let $\pi\colon{\mathcal{X}}\longrightarrow\mathrm{Hyp}_{n+1}$ be the universal family, and let $\mathrm{ev}\colon{\mathcal{X}}\longrightarrow\mathbf{P}^n$ be the evaluation map. Informally, ${\mathcal{X}}$ parametrizes pairs $(x,Y)$ where $x\in\mathbf{P}^n$ and $Y \in \mathrm{Hyp}_{n+1}$ is a degree $(n+1)$ hypersurface containing $x$. This description shows that $\mathrm{ev}\colon{\mathcal{X}}\longrightarrow\mathbf{P}^n$ is a projective bundle, and we can formally write it as $\mathbf{P}({\mathcal{K}}^\vee)\longrightarrow\mathbf{P}^n$, where ${\mathcal{K}}\in\mathrm{Vect}(\mathbf{P}^n)$ is defined as 
\[
{\mathcal{K}}\ :=\ \ker\big(H^0(\mathbf{P}^n,{\mathcal{O}}_{\mathbf{P}^n}(n+1))\otimes{\mathcal{O}}_{\mathbf{P}^n}\ \longrightarrow\ {\mathcal{O}}_{\mathbf{P}^n}(n+1)\big)\,,
\]
and the map is the evident one. The resulting map
\[
{\mathcal{X}}\longrightarrow\mathbf{P}\big(H^0(\mathbf{P}^n,{\mathcal{O}}_{\mathbf{P}^n}(n+1))^\vee\big)
\]
is identified with $\pi$. Our chosen hypersurface $X \in \mathbf{P}^n$ gives a point $[X] \in \mathrm{Hyp}_{n+1}(K)$ with $X' := \pi^{-1}([X])$ mapping isomorphically to $X$ via $\mathrm{ev}$. For an integer $t \geqslant 1$, let $tX \subset\mathbf{P}^n$ and $tX' \subset {\mathcal{X}}$ be the order $t$ neighbourhoods of $X \subset\mathbf{P}^n$ and $X' \subset {\mathcal{X}}$ respectively. 

\begin{lemma}
\label{lem:pullbackevinjective}
The map $H^{n-1}(tX,{\mathcal{O}}_{tX}) \longrightarrow H^{n-1}(tX',{\mathcal{O}}_{tX'})$ is injective for all $t$.
\end{lemma}

\begin{proof}
Let $V = H^0(\mathbf{P}^n,{\mathcal{O}}_{\mathbf{P}^n}(1))$, so $\mathbf{P}^n = \mathbf{P}(V)$ and $\mathrm{Hyp}_{n+1} = \mathbf{P}(\mathrm{Sym}^{n+1}(V)^\vee)$. For each~$t$, the sheaf ${\mathcal{O}}_{tX}$ admits a filtration defined by powers of the ideal defining $X \subset tX$, and similarly for $tX'$. The map $tX' \longrightarrow tX$ is compatible with this filtration as it sends $X' \subset tX'$ to~$X \subset tX$. Hence, it suffices to check that the induced map
\[
\phi_j\colon H^{n-1}(X,I_X^j/I_X^{j+1})\longrightarrow H^{n-1}(X',I_{X'}^j/I_{X'}^{j+1})
\]
is injective for each $j \geqslant 0$. Fix an isomorphism $\det(V)\simeq K$ and $f \in R_{n+1}$ defining $X$. These choices determine isomorphisms $I_X \simeq {\mathcal{O}}_{\mathbf{P}^n}(-n-1) \simeq K_{\mathbf{P}^n}$, and hence an isomorphism
\[
H^{n-1}(X,O_X)\ \simeq\ H^n(\mathbf{P}^n,I_X)\ \simeq\ H^n(\mathbf{P}^n,K_{\mathbf{P}^n})\ \simeq\ K\,.
\]
Tensoring the exact sequence $0\longrightarrow I_X\longrightarrow{\mathcal{O}}_{\mathbf{P}^n}\longrightarrow{\mathcal{O}}_X\to0$ with $I_X^j$ and using Serre duality shows that
\begin{align*}
H^{n-1}\big(X,I_X^j/I_X^{j+1}\big)\ &=\ \ker\Big(H^n\big(\mathbf{P}^n,I_X^{j+1}\big)\longrightarrow H^n\big(\mathbf{P}^n,I_X^j\big)\Big)\\
&=\ \mathrm{coker}\Big(H^0\big(\mathbf{P}^n,{\mathcal{O}}_{\mathbf{P}^n}((j-1)(n+1))\big)\longrightarrow H^0\big(\mathbf{P}^n,{\mathcal{O}}_{\mathbf{P}^n}(j(n+1))\big)\Big)^\vee\\ 
&=\ \mathrm{coker}\Big(\mathrm{Sym}^{(j-1)(n+1)}(V) \stackrel{f}{\longrightarrow} \mathrm{Sym}^{j(n+1)}(V)\Big)^\vee. 
\end{align*}

As $X' \subset {\mathcal{X}}$ is a fibre of $\pi$, one has $tX' = \pi^{-1}(t[X])$, where $t[X] \subset \mathrm{Hyp}_{n+1}$ is the order $t$ neighbourhood of $[X] \in \mathrm{Hyp}_{n+1}(K)$. Using flatness of $\pi$ and the aforementioned isomorphism $H^{n-1}(X,O_X) \simeq K$, we get
\[
H^{n-1}(X',\ I_{X'}^j/I_{X'}^{j+1})\ =\ \big(\mathrm{Sym}^j(\mathrm{Sym}^{n+1}(V)/(f))\big)^\vee.
\]
One can check that the pullback $\phi_j$ above is dual to the map induced by the composition map $\mathrm{Sym}^j(\mathrm{Sym}^{n+1}(V)) \longrightarrow \mathrm{Sym}^{j(n+1)}(V)$ by passage to the appropriate quotients. In particular, the dual map is surjective, so $\phi_j$ is injective.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:CalabiYau}\,(4)]
By Remark~\ref{rmk:hasseflatbasechange}, we may assume that the field $K$ is perfect. Fix some $1\leqslant t\leqslant p$. We get a commutative diagram
\[
\CD
X' @>i>> tX' @>j>> pX' @>a>> \mathrm{Frob}_{\mathcal{X}}^*X' @>c>> X'\\
@VVV @VVV @VVV @VVV @VVV\\
X @>k>> tX @>\ell>> pX @>b>> \mathrm{Frob}_{\mathbf{P}^n}^*X @>d>> X
\endCD
\]
Here all vertical maps are induced by $\mathrm{ev}\colon{\mathcal{X}}\longrightarrow\mathbf{P}^n$, the maps $c$ and $d$ are induced by the Frobenius maps on ${\mathcal{X}}$ and $\mathbf{P}^n$ respectively, and $i$, $j$, $k$, $\ell$, $a$ and $b$ are the evident closed immersions; the map $b$ is an isomorphism as $X \subset \mathbf{P}^n$ is a Cartier divisor. In particular, the composite map $X' \longrightarrow X'$ and $X \longrightarrow X$ obtained from each row are the Frobenius maps on~$X'$ and $X$ respectively. Passing to cohomology gives a commutative diagram
\[
\CD
H^{n-1}(X,{\mathcal{O}}_X) @= H^{n-1}(X',{\mathcal{O}}_{X'})\\
@VV{a_t}V @VV{b_t}V \\
H^{n-1}(tX,{\mathcal{O}}_{tX}) @>{c_t}>> H^{n-1}(tX',{\mathcal{O}}_{tX'})
\endCD
\]
where $a_t$ and $b_t$ are induced by the Frobenius maps on $\mathbf{P}^n$ and ${\mathcal{X}}$ respectively, while $c_t$ is injective by Lemma~\ref{lem:pullbackevinjective}. To finish the proof, observe that Lemma~\ref{lem:identifyhasse} shows that $h+1$ is the minimal value of $t$ for which $b_t$ is injective. Since $c_t$ is injective as well, this is also the minimal value of $t$ for which $a_t$ is injective. It follows by Remark~\ref{rmk:fptthickenings} and Theorem~\ref{thm:highdeg} that $\mu_f(p) = p - (h+1) + 1 = p-h$. 
\end{proof}

\begin{remark}
\label{remark:Ogus}
We recover Ogus's result, Theorem~\ref{thm:ogusigusadeuring}, for primes $p\geqslant n^2-n-1$; this is immediate from Theorem~\ref{thm:CalabiYau}\,(1) and (4).
\end{remark}

We conclude by giving a de~Rham interpretation for $\mathrm{fpt}(f)$; write $\mathrm{Fil}^\mathrm{conj}_\bullet$ and $\mathrm{Fil}^\bullet_H$ for the increasing conjugate and decreasing Hodge filtrations on $H^{n-1}_\mathrm{dR}(X)$ respectively. Then:

\begin{corollary}
One has $\mathrm{fpt}(f)=1-a/p$, where $a$ is the largest $i$ such that 
\[
\mathrm{Frob}_X^* H^{n-1}_\mathrm{dR}(X)\ \simeq\ \mathrm{Fil}^\mathrm{conj}_0(H^{n-1}_\mathrm{dR}(X))\ \subset\ \mathrm{Fil}^i_H (H^{n-1}_\mathrm{dR}(X))\,.
\]
\end{corollary}

\begin{proof}
This follows from Theorem~\ref{thm:CalabiYau} and Ogus's result~\cite[Theorem~1]{Ogus}.
\end{proof}

Note that the integer $a$ appearing above is the $a$ number defined by van~der~Geer and Katsura~\cite{GK} for the special case of a Calabi-Yau family.

\section{Quartic hypersurfaces in $\mathbf{P}^2$}

Our techniques also yield substantive information for hypersurfaces other than Calabi-Yau hypersurfaces; as an example, we include here the case of quartic hypersurfaces in~$\mathbf{P}^2$.

When $f$ defines a Calabi-Yau hypersurface $X$, it is readily seen that the Frobenius action on the vector space $H^{\dim X}(X,{\mathcal{O}}_X)$ is injective if and only if $\mathrm{fpt}(f)=1$, i.e., if and only
\[
\mathrm{fpt}(f)=\mathrm{lct}(f)\,.
\]
For hypersurfaces $X$ of general type, the injectivity of the Frobenius on $H^{\dim X}(X,{\mathcal{O}}_X)$, or even the ordinarity of $X$ in the sense of Bloch and Kato~\cite[Definition~7.2]{Bloch-Kato}---a stronger condition---does not imply the equality of the $F$-pure threshold and the log canonical threshold: for example, for each $f$ defining a quartic hypersurface in $\mathbf{P}^2$ over a field of characteristic $p\equiv 3\mod 4$, we shall see that $\mathrm{fpt}(f)<\mathrm{lct}(f)$; we emphasize that this includes the case of generic hypersurfaces, and that these are ordinary in the sense of Bloch and Kato by a result of Deligne; see~\cite{Illusie}. More generally:

\begin{lemma}
\label{lemma:pigeon}
Let $f$ be a homogeneous polynomial of degree $d$ in $K[x_0,\dots,x_n]$. Then:
\begin{enumerate}[\quad\rm(1)]
\item For each $q=p^e$, one has $\displaystyle{\mu_f(q)\ \leqslant\ \left\lceil\frac{nq+q-n}{d}\right\rceil}\,,$

\item If $nq+q$ is congruent to any of $1,2,\dots,n\mod d$ \ for some $q$, then $\displaystyle{\mathrm{fpt}(f)<\frac{n+1}{d}}$.
\end{enumerate}
\end{lemma}

For quartics in $\mathbf{P}^2$, one has $n=2$ and $d=4$; thus, if $p\equiv 3\mod 4$, then $np+p\equiv1\mod d$, so the $F$-pure threshold is strictly smaller than the log canonical threshold by (2).

\begin{proof}
The pigeonhole principle implies that $f^k\in\mathfrak{m}^{[q]}$ whenever $dk\geqslant (n+1)(q-1)+1$, which proves (1). For (2), suppose $nq+q$ is congruent to any of $1,2,\dots,n\mod d$. Then
\[
\left\lceil\frac{nq+q-n}{d}\right\rceil\ \leqslant \ \frac{nq+q-n+(n-1)}{d}\ =\ \frac{nq+q-1}{d}\,,
\]
and it follows using (1) that $\mu_f(q)<(nq+q)/d$. Thus, 
\[
\mathrm{fpt}(f)\ \leqslant\ \mu_f(q)/q\ <\ (n+1)/d\,.\qedhere
\]
\end{proof}

\begin{theorem}
\label{theorem:quartic}
Let $K$ be a field of characteristic $p> 2$. Let $f$ be a homogeneous polynomial of degree $4$ in $K[x_0,x_1,x_2]$, such that the Jacobian ideal of $f$ is $\mathfrak{m}$-primary. Then the possible values for $\mu_f(q)$ and the $F$-pure threshold are:
\begin{align*}
p&\equiv 1\mod 4:\qquad \mu_f(q)=
\begin{cases}
\frac{q(3p-3)}{4p}&\text{ for all }q,\quad\mathrm{fpt}(f)=\frac{3p-3}{4p}\,,\\
\frac{3q+1}{4}&\text{ for all }q,\quad\mathrm{fpt}(f)=\frac{3}{4}\,,\\
\end{cases}\\
p&\equiv 3\mod 4:\qquad \mu_f(q)=
\begin{cases}
\frac{q(3p-5)}{4p}&\text{ for all }q,\quad\mathrm{fpt}(f)=\frac{3p-5}{4p}\,,\\
\frac{q(3p-1)}{4p}&\text{ for all }q,\quad\mathrm{fpt}(f)=\frac{3p-1}{4p}\,.\\
\end{cases}
\end{align*}
\end{theorem}

\begin{proof}
Lemma~\ref{lemma:bound} and Lemma~\ref{lemma:pigeon}\,(1) provide the respective inequalities
\[
\left\lceil\frac{3p-5}{4}\right\rceil\ \leqslant\ \mu_f(p)\ \leqslant\ \left\lceil\frac{3p-2}{4}\right\rceil\,.
\]

If $p\equiv 3\mod 4$, this reads
\[
\frac{3p-5}{4}\ \leqslant\ \mu_f(p)\ \leqslant\ \frac{3p-1}{4}\,,
\]
so there are two possible values for the integer $\mu_f(p)$. The sequence $\{\mu_f(q)/q\}_q$ is constant by Lemma~\ref{lemma:patterns}\,(2), which completes the proof in this case.

If $p\equiv 1\mod 4$, the inequalities read
\[
\frac{3p-3}{4}\ \leqslant\ \mu_f(p)\ \leqslant\ \frac{3p+1}{4}\,.
\]
Again, there are two choices for $\mu_f(p)$. If $\mu_f(p)=(3p-3)/4$, then $\{\mu_f(q)/q\}_q$ is a constant sequence by Lemma~\ref{lemma:patterns}\,(2), whereas if $\mu_f(p)=(3p+1)/4$, then Lemma~\ref{lemma:patterns}\,(1) implies that $\mu_f(q)=(3q+1)/4$.
\end{proof}

\begin{remark}
Similarly, for a quintic $f$ in $K[x_0,x_1,x_2]$ with an $\mathfrak{m}$-primary Jacobian ideal, the possibilities for $\mathrm{fpt}(f)$ are easily determined; we do not list the corresponding $\mu_f(q)$ as these are dictated by $\mathrm{fpt}(f)$. We assume below that $p>5$.
\begin{alignat*}7
p&\equiv1\mod 5:\quad && \mathrm{fpt}(f):\quad &&(3p-3)/5p\,,\ && \text{ or }\ &&4/5\,,\\
p&\equiv2\mod 5:\quad && \mathrm{fpt}(f):\quad &&(3p-6)/5p\,,\ && \text{ or }\ &&(3p-1)/5p\,,\\
p&\equiv3\mod 5:\quad && \mathrm{fpt}(f):\quad &&(3p-4)/5p\,,\ && \text{ or }\ &&(3p^2-7)/5p^2\,,\ && \text{ or }\ &&(3p^2-2)/5p^2\,,\\
p&\equiv4\mod 5:\quad && \mathrm{fpt}(f):\quad &&(3p-7)/5p\,,\ && \text{ or }\ &&(3p-2)/5p\,.
\end{alignat*}
\end{remark}

The log canonical threshold of a smooth quartic in $\mathbf{P}^2$ is $3/4$; except for the case where it equals $3/4$, the denominator of $\mathrm{fpt}(f)$ in Theorem~\ref{theorem:quartic} is $p$. For a quintic as above, if $\mathrm{fpt}(f)\neq\mathrm{lct}(f)$, then the denominator of $\mathrm{fpt}(f)$ is a power of $p$. More generally:

\begin{proposition}
Let $K$ be a field of characteristic $p>0$. Let $f$ be a homogeneous polynomial of degree $d$ in $K[x_0,\dots,x_n]$ with an $\mathfrak{m}$-primary Jacobian ideal. If $p\geqslant nd-d-n$, then either $\mathrm{fpt}(f)=(n+1)/d$, or else the denominator of $\mathrm{fpt}(f)$ is a power of $p$.
\end{proposition}

\begin{proof}
If $\mathrm{fpt}(f)<(n+1)/d$, then there exists an integer $q$ such that $\mu_f(q)/q<(n+1)/d$. But then $\mathrm{fpt}(f)=\mu_f(q)/q$ by Lemma~\ref{lemma:patterns}\,(2).
\end{proof}

\section*{Acknowledgments}

We thank Mircea Musta\c t\u a for raising the question for elliptic curves at the AMS-MRC workshop on Commutative Algebra, Snowbird, 2010, the workshop participants, and the American Mathematical Society. We also thank Johan de Jong for a useful conversation.

