\section{Basic Definitions}

Let $A$ and $B$ be two subsets of a ring $(\, R, \, + , \, \cdot\,)$. We shall use $A+B$ and $AB$ to denote the following subsets of $R$ 
$$A+B:=\{\, a+b \, | \, a\in A, \, b\in B \, \}, \quad 
AB:=\{\, ab \, |\, a\in A, \, b\in B \,\}.$$

\medskip
We now introduce a class of noncommutative rings in the following 

\medskip
\begin{definition}\label{def2.1} A ring $R$ with a multiplication $\cdot$ is called a {\bf Hu-Liu triring} if the following three properties hold.
\begin{description}
\item[(i)] There exist a commutative subring $R_0$ of the ring $(R,\, +,\, \cdot)$ and a subgroups $R_1$ of the additive group $(R,\, +)$, called the {\bf even part} and {\bf odd part} of $R$ respectively, such that 
$R=R_0\oplus R_1$ (as Abelian groups) and 
\begin{equation}\label{eq1.1}
R_0R_0\subseteq R_0,\quad R_0R_1+R_1R_0\subseteq R_1, \quad R_1R_1=0;
\end{equation}
\item[(ii)] There exists a  binary operation $\sharp$ on the odd part $R_1$ such that 
$(\,R_1 , \, + , \, \,\sharp\, \,)$ is a commutative ring and the two associative products $\cdot$ and $\,\sharp\,$ satisfy the {\bf triassociative law}:
\begin{eqnarray}
\label{eq2} x (\alpha \,\sharp\, \beta) &=&(x \alpha) \,\sharp\, \beta, \\
\label{eq3} (\alpha \,\sharp\, \beta ) x &=&\alpha \,\sharp\, (\beta x), 
\end{eqnarray}
where $x\in R$ and $\alpha$, $\beta\in R_1$;
\item[(iii)] For each $x_0\in R_0$, we have
\begin{equation}\label{eq6}
R_1x_0=x_0R_1.
\end{equation}
\end{description}
\end{definition}

\medskip
A Hu-Liu triring $R=R_0\oplus R_1$, which clearly has the partially commutative property, is sometimes denoted by 
$(\,R=R_0\oplus R_1 , \, + , \, \cdot ,\,  \,\sharp\, \,)$,
where the associative product $\sharp$ on the odd part $R_1$ is called the {\bf local product}, and the identity $1^\sharp$ of the ring $(\, R_1 , \, + , \, \,\sharp\,\,)$ is called the {\bf local identity} of the triring $R$. If 
$(\,R=R_0\oplus R_1 , \, + , \, \cdot ,\,  \,\sharp\, \,)$ is a Hu-Liu triring, then $R$ is 
the trivial extension of a ring $R_0$ by a $R_0$-bimodule $R_1$, and the triassociative law interweaves the noncommutative ring structure on the  trivial extension with the commutative ring 
structure on the odd part.

\medskip
Since a commutative ring is a Hu-Liu triring with zero odd part, the concept of Hu-Liu trirings naturally generalizes the concept of commutative rings. The first and the most important example of Hu-Liu trirings which is not a commutative ring is triquaternions whose definition is given in the following example.

\medskip
\noindent
{\bf Example} Let 
${\bf Q}=\mathcal{R}\,1\oplus \mathcal{R}\,i\oplus\mathcal{R}\,j\oplus\mathcal{R}\,k$ be a 
$4$-dimensional real vector space, where $\mathcal{R}$ is the field of real numbers. Then 
$(\,{\bf Q}={\bf Q}_0\oplus {\bf Q}_1, \, +, \,\cdot, \, \,\sharp\, \,)$ is a Hu-liu triring, where ${\bf Q}_0=\mathcal{R}\,1\oplus \mathcal{R}\,i$ is the even part, 
${\bf Q}_1=\mathcal{R}\,j\oplus\mathcal{R}\,k$ is the odd part, the ring multiplication $\cdot$ and the local product $\sharp$ are defined by the following multiplication tables:
\begin{center}
{\huge\begin{tabular}{|c||c|c|c|c|}
\hline
$\cdot$&$1$&$i$&$j$&$k$\\
\hline\hline
$1$&$1$&$i$&$j$&$k$\\
\hline
$i$&$i$&$-1$&$k$&$-j$\\
\hline
$j$&$j$&$-k$&$0$&$0$\\
\hline
$k$&$k$&$j$&$0$&$0$\\
\hline
\end{tabular}}
\qquad\qquad
{\huge\begin{tabular}{|c||c|c|}
\hline
$\sharp$&$j$&$k$\\
\hline\hline
$j$&$j$&$k$\\
\hline
$k$&$k$&$-j$\\
\hline
\end{tabular}}
\end{center}
The Hu-Liu triring ${\bf Q}$ is called the {\bf triquaternions}.

\hfill\raisebox{1mm}{\framebox[2mm]{}}

\medskip
\begin{definition}\label{def2.2} Let $(\,R=R_0\oplus R_1, \, +, \,\cdot, \, \,\sharp\, \,)$ be a Hu-Liu triring with the identity $1$, and let $I$ be a subgroup of the additive group of $R$.
\begin{description}
\item[(i)] $I$ is called a {\bf triideal} of $R$ if $IR+RI\subseteq I$, 
$I=(R_0\cap I)\oplus (I\cap R_1)$, and $I\cap R_1$ is an ideal of the ring 
$(\,R_1, \, +, \, \,\sharp\, \,)$.
\item[(ii)] $I$ is called a {\bf subtriring} of $R$ if $1\in I$, $II\subseteq I$,  
$I=(R_0\cap I)\oplus (I\cap R_1)$ and $I\cap R_1$ is a subring of the ring 
$(\,R_1, \, +, \, \,\sharp\, \,)$.
\end{description}
\end{definition}

\medskip
Let $I$ be a triideal of a Hu-Liu triring $(\,R=R_0\oplus R_1, \, +, \,\cdot, \, \,\sharp\, \,)$. It is clear that $\displaystyle\frac{R}{I}=\left(\displaystyle\frac{R}{I}\right)_0\oplus 
\left(\displaystyle\frac{R}{I}\right)_1$  with
$\left(\displaystyle\frac{R}{I}\right)_i:=\displaystyle\frac{R_i+I}{I}$ for $i=0$ and $1$. We now define a local product on $\left(\displaystyle\frac{R}{I}\right)_1$ by
\begin{equation}\label{eq5}
(\alpha +I)\,\sharp\, (\beta +I):=\alpha\,\sharp\,\beta +I 
\quad\mbox{for $\alpha$, $\beta\in R_1$.}
\end{equation}
Then the local product defined by (\ref{eq5}) is well-defined, and the triassociative law holds. 
Therefore, $\displaystyle\frac{R}{I}$ becomes a Hu-Liu triring, which is called the {\bf quotient Hu-Liu triring} of $R$ with respect to the triideal $I$.

\medskip
\begin{definition}\label{def1.4} Let $R=R_0\oplus R_1$ and 
$\overline{R}=\overline{R}_0\oplus \overline{R}_1$ be Hu-Liu trirings. A map 
$\phi: R\to \overline{R}$ is called a {\bf triring homomorphism} if 
$$
\phi (x+y)=\phi (x)+\phi (y), \quad \phi (xy)=\phi (x)\phi (y), \quad
\phi (1_R)=1_{\overline{R}},
$$$$
\phi (R_0)\subseteq \overline{R}_0, \quad \phi (R_1)\subseteq \overline{R}_1,
$$$$
\phi (\alpha \,\sharp\, \beta )=\phi (\alpha) \,\sharp\, \phi (\beta), \quad
\phi(1^{\sharp})=\bar{1}^{\sharp}
$$
where $x$, $y\in R$, $\alpha$, $\beta\in R_1$, $1_R$ and $1_{\overline{R}}$ are the identities of $R$ and $\overline{R}$ respectively, and $1^{\sharp}$ and $\bar{1}^{\sharp}$ are the local identities of $R$ and $\overline{R}$ respectively. A bijective triring homomorphism is called a {\bf triring isomorphism}. We shall use $R\simeq \overline{R}$ to indicate that there exists a triring isomorphism from $R$ to $\overline{R}$.
\end{definition}

\medskip
Let  $\phi$ be a  triring homomorphism from a Hu-Liu triring $R$ to a Hu-Liu triring $\overline{R}$. the {\bf kernel} $Ker\phi$ and the {\bf image} $Im\phi$ of $\phi$ are defined by
$$ Ker\phi :=\{\, x \,|\, \mbox{$a\in R$ and $\phi (x)=0$} \, \}$$ 
and
$$ Im\phi :=\{\, \phi (x) \,|\, \mbox{$x\in R$} \, \}.$$ 
Clearly, $Ker\phi$ is a triideal of $R$, $Im\phi$ is a subtriring of $\overline{R}$ and 
$$\overline{\phi}: x+Ker\phi \to \phi (x) \quad\mbox{for $x\in R$}$$
is a triring isomorphism from the quotient Hu-Liu triring $\displaystyle\frac{R}{Ker\phi}$ to the subtriring $Im\phi$ of $\overline{R}$. If $I$ is triideal of $R$, then 
$$ \nu : x\mapsto x+I \quad\mbox{for $x\in R$}$$
is a surjective triring homomorphism from $R$ the quotient tiring $\displaystyle\frac{R}{I}$ with kernel $I$. The map $\nu$ is called the {\bf natural triring homomorphism}.

\bigskip
\begin{proposition}\label{pr1.1}  Let $\phi$ be a surjective homomorphism from a Hu-Liu triring $R=R_0\oplus R_1$ to a Hu-Liu triring $\overline{R}=\overline{R}_0\oplus \overline{R}_1$.
\begin{description}
\item[(i)] $\phi (R_i)=\overline{R}_i$ for $i=0$ and $1$.
\item[(ii)] Let
$$
\mathcal{S}: =\{\, I \,|\, \mbox{$I$ is a triideal of $R$ and $I \supseteq Ker \phi $} \, \}
$$
and
$$
\overline{\mathcal{S}}: =\{\, \overline{I} \,|\, \mbox{$\overline{I}$ is a triideal of $\overline{R}$} \, \}.
$$
The map 
$$
\Psi : I \mapsto \phi (I): =\{\, \phi (x) \,|\, x\in I \, \}
$$
is a bijection from $\mathcal{S}$ to $\overline{\mathcal{S}}$, and the inverse map $\Psi ^{-1} : \overline{\mathcal{S}} \to \mathcal{S}$ is given by
$$
\Psi ^{-1} : \overline{I} \mapsto \phi ^{-1} (\overline{I}) : = \{\, x \,|\, \mbox{$x\in R$ and $\phi (x)\in \overline{I}$ \, \}.}
$$
\item[(iii)] If $I$ is a triideal of $R$ containing $Ker \phi$, then the map
$$
x+I \mapsto \phi (x) + \phi (I) \quad\mbox{for $x\in R$}
$$
is a triring isomorphism from the quotient triring $\displaystyle\frac{R}{I}$ onto the quotient triring $\displaystyle\frac{\overline{R}}{\phi (I)}$.
\end{description}
\end{proposition}

\medskip
\noindent
{\bf Proof} A routine check.

\hfill\raisebox{1mm}{\framebox[2mm]{}}

\medskip
We now prove a basic property for Hu-Liu trirings.

\begin{proposition}\label{pr2.1}  Let $(\,R=R_0\oplus R_1, \, +, \,\cdot, \, \,\sharp\, \,)$ be a Hu-Liu triring. If $x_i\in R_i$  for $i\in\{\, 0, \, 1\,\}$, then both $Rx_0=R_0x_0\oplus R_1x_0$ and $R_1\,\sharp\, x_1$ are triideals of $R$.
\end{proposition}

\medskip
\noindent
{\bf Proof} Since $(R_1, \, +, \, \sharp)$ is a commutative ring, $R_1\,\sharp\, x_1$ is clearly a triideal of $R$.

\medskip
Using the triassociative law and (\ref{eq6}), we have
\begin{equation}\label{eq7}
\Big(R_i(R_0x_0+R_1x_0)\Big)\bigcup
\Big(R_0x_0+R_1x_0)R_i\Big)\subseteq R_0x_0+R_1x_0
\end{equation}
for $i=0$, $1$. By (\ref{eq7}), $R_0x_0\oplus R_1x_0$ is an ideal of the ring $(\,R, \, +, \, \cdot\,)$. Also, $R_1x_0=R_1\,\sharp\, (1^{\sharp}x_0)$ is obviously an ideal of the commutative ring $(\,R_1, \, +, \, \sharp\,)$. Thus, $R_0x_0\oplus R_1x_0$ is a triideal of $R$.

\hfill\raisebox{1mm}{\framebox[2mm]{}}

\medskip
The next position gives some operations about triideals in a Hu-Liu triring.

\medskip
\begin{proposition}\label{pr2.2}  Let $I$, $J$, $I_{\lambda}$ with $\lambda\in \Lambda$ be triideals of a Hu-Liu triring $R$.
\begin{description}
\item[(i)] The intersection $I\cap J$ and the sum 
$\displaystyle\sum_{\lambda\in \Lambda}I_{\lambda}$ are triideals of $R$.Moreover, we have
$(I\cap J)_i=I_i\cap J_i$ and 
$\left(\displaystyle\sum_{\lambda\in \Lambda}I_{\lambda}\right)_i=
\displaystyle\sum_{\lambda\in \Lambda}(I_{\lambda})_i$ for $i=0$, $1$.
\item[(ii)] The {\bf mixed product} 
$I\,\stackrel{\cdot}{\sharp}\,J:=(I\,\stackrel{\cdot}{\sharp}\,J)_0\oplus 
(I\,\stackrel{\cdot}{\sharp}\,J)_1$ of $I$ and $J$ is a triideal, where 
$(I\,\stackrel{\cdot}{\sharp}\,J)_0=I_0J_0$ and $(I\,\stackrel{\cdot}{\sharp}\,J)_1=I_1\,\sharp\,J_1$.
\end{description}
\end{proposition}

\medskip
\noindent
{\bf Proof} (i) It is clear.

\medskip
(ii) By triassociative law, we have
$$
R_0(I\,\stackrel{\cdot}{\sharp}\,J)\subseteq 
R_0I_0J_0+ R_0(I_1\,\sharp\,J_1)\subseteq I_0J_0+ (R_0I_1)\,\sharp\,J_1\subseteq I\,\stackrel{\cdot}{\sharp}\,J,
$$
$$
(I\,\stackrel{\cdot}{\sharp}\,J)R_0\subseteq 
I_0J_0R_0+(I_1\,\sharp\,J_1)R_0\subseteq I_0J_0+ I_1\,\sharp\,(J_1R_0)\subseteq I\,\stackrel{\cdot}{\sharp}\,J,
$$
$$
R_1(I\,\stackrel{\cdot}{\sharp}\,J)+(I\,\stackrel{\cdot}{\sharp}\,J)R_1\subseteq
R_1I_0J_0+I_0J_0R_1\subseteq I_1J_0+I_0J_1\subseteq I\,\stackrel{\cdot}{\sharp}\,J
$$
and $R_1\,\sharp\,(I_1\,\sharp\,J_1)\subseteq
R_1\,\sharp\,I_1\,\sharp\,J_1\subseteq I\,\stackrel{\cdot}{\sharp}\,J$. This proves that (ii) holds.

\hfill\raisebox{1mm}{\framebox[2mm]{}}


\bigskip
\section{Prime Triideals}

\medskip
We begin this section by introducing the notion of prime triideals.

\begin{definition}\label{def5.1} Let $(\, R=R_0\oplus R_1, \, + , \, \cdot\, , \,\sharp\, \,)$ be a Hu-Liu triring. An triideal $P=P_0\oplus P_1$ of $R$ is called a {\bf prime triideal} if $P\ne R$ and 
\begin{eqnarray}
\label{eq15}x_0y_0\in P_0  &\Rightarrow&  \mbox{$x_0\in P_0$ or $y_0\in P_0$},\\
\label{eq16}x_0y_1\in P_1  &\Rightarrow&  \mbox{$x_0\in P_0$ or $y_1\in P_1$},\\
\label{eq17}x_1y_0\in P_1  &\Rightarrow&  \mbox{$x_1\in P_1$ or $y_0\in P_0$},\\
\label{eq18}x_1\,\sharp\, y_1\in P_1 &\Rightarrow&  \mbox{$x_1\in P_1$ or $y_1\in P_1$},
\end{eqnarray}
where $x_i$, $y_i\in R_i$ for $i=0$ and $1$.
\end{definition}

\medskip
Let $(\, R=R_0\oplus R_1, \, + , \, \cdot\, \,\sharp\, \,)$ be a Hu-Liu triring. The set of all prime triideals of $R$ is called the {\bf trispectrum} of $R$ and denoted by $Spec^{\sharp}R$. It is clear that
$$Spec^{\sharp}R=Spec^{\sharp}_0R\cup Spec^{\sharp}_1R\quad\mbox{and}\quad 
Spec^{\sharp}_0R\cap Spec^{\sharp}_1R=\emptyset,$$
where
$$Spec^{\sharp}_0R:=\{\, P \, |\, \mbox{$P\in Spec^{\sharp}R$ and $P\supseteq R_1$}\,\}$$
is called the {\bf even trispectrum} of $R$ and 
$$Spec^{\sharp}_1R:=\{\, P \, |\, \mbox{$P\in Spec^{\sharp}R$ and $P\not\supseteq R_1$}\,\}$$
is called the {\bf odd trispectrum} of $R$.  

\medskip
Let $P=P_0\oplus P_1$ be a triideal of a Hu-Liu triring $R$. Then 
$P\in Spec^{\sharp}_0R$ if and only if $P_1=R_1$ and $P_0$ is a prime ideal of the commutative ring 
$(\, R_0, \, + , \, \cdot\, \,)$. It is also obvious that $P\in Spec^{\sharp}_1R$ if and only if
$P_0$ is a prime ideal of the commutative ring $(\, R_0, \, + , \, \cdot\, \,)$, $P_1$ is a prime ideal of the commutative ring $(\, R_1 , \, + , \, \,\sharp\, \, )$, and the $\left(\displaystyle\frac{R_0}{P_0}, \displaystyle\frac{R_0}{P_0}\right)$-bimodule $\displaystyle\frac{R}{P}$ is faithful as both left module and right module, where the left $\displaystyle\frac{R_0}{P_0}$-module action on $\displaystyle\frac{R}{P}$ is defined by 
$$
(x_0+P_0)(y+P):= x_0y+P \quad\mbox{for $x_0\in P_0$ and $y\in R$}
$$
and the right $\displaystyle\frac{R_0}{P_0}$-module action  on $\displaystyle\frac{R}{P}$ is defined by 
$$
(y+P)(x_0+P_0):= yx_0+P \quad\mbox{for $x_0\in P_0$ and $y\in R$}.
$$

\medskip
Clearly, the even trispectrum $Spec^{\sharp}_0R$ of a Hu-Liu triring $R$ is not empty. A basic property of Hu-Liu trirings is that the odd trispectrum $Spec^{\sharp}_1R$ is always not empty provided $R_1\ne 0$. This basic fact is a corollary of the following

\begin{proposition}\label{pr5.1} Let $(\, R=R_0\oplus R_1, \, + , \, \cdot\, \,\sharp\, \,)$ be a Hu-Liu triring with $R_1\ne 0$. If $P_1$ is a prime ideal of the commutative ring 
$(\, R_1 , \, + , \, \,\sharp\, \, )$, then there exists prime ideal $P_0$ of the commutative ring $(\, R_0 , \, + , \, \,\cdot\, \, )$ such that $P:=P_0\oplus P_1$ is a prime triideal of $R$, and $P_0$ contains every ideal $I_0$ of the ring  $(\, R_0 , \, + , \, \,\cdot\, \, )$ which has the property: $R_1I_0\subseteq P_1$.
\end{proposition}

\medskip
\noindent
{\bf Proof} Consider the set $\Omega$ defined by
$$
\Omega :=\{\,I_0\,|\, \mbox{$I_0$ is an ideal of $(\, R_0 , \, + , \, \,\cdot\, \, )$ and $R_1I_0\subseteq P_1$}\,\}.
$$

Clearly, $1\not\in I_0$ if $I_0\in \Omega$. Since $0\in \Omega$, $\Omega$ is not empty. The relation of inclusion, $\subseteq$, is a partial order on $\Omega$. Let $\Delta$ be a non-empty totally ordered subset of $\Omega$. Let $J_0:=\displaystyle\bigcup _{I_0\in\Delta} I_0$. Then $J_0\in\Omega$. Thus $J_0$ is an upper bound for $\Delta$ in $\Omega$. By Zorn's Lemma, the partial order set $(\Omega,\,\subseteq)$ has a maximal element $P_0$. We are going to prove that $P:=P_0\oplus P_1$ is a prime triideal of $R$. 
Clearly, $P=P_0\oplus P_1$ is a triideal satisfying (\ref{eq18}). Let $x_i$, $y_i\in R_i$ for $i=0$, $1$. 

\medskip
If $x_1y_0\in P_1$ and $x_1\not\in P_1$, then 
$x_1\,\sharp\,(1^{\sharp}y_0)=x_1y_0\in P_1$, which implies that $1^{\sharp}y_0\in P_1$. Hence, we have
\begin{eqnarray}\label{eq19}
R_1(P_0+R_0y_0)&\subseteq& R_1P_0+R_1R_0y_0\subseteq P_1+R_1y_0\nonumber\\
&=&P_1+R_1\,\sharp\,(1^{\sharp}y_0)\subseteq P_1+R_1\,\sharp\,P_1\subseteq P_1.
\end{eqnarray}
Using (\ref{eq19}) and the fact that $P_0+R_0y_0$ is an ideal of $R_0$, we get
$P_0+R_0y_0\in \Omega$. Since $P_0+R_0y_0\supseteq P_0$, we have to have 
$P_0+R_0y_0= P_0$, which implies that $y_0\in P_0$. This proves that (\ref{eq17}) holds.

\medskip
If $x_0y_1\in P_1$ and $y_1\not\in P_1$, then 
$(x_01^{\sharp})\,\sharp\,y_1=x_0y_1\in P_1$, which implies that $x_01^{\sharp}\in P_1$. It follows from this fact and (\ref{eq6}) that
\begin{eqnarray}\label{eq20}
R_1(P_0+R_0x_0)&\subseteq& R_1P_0+R_1R_0x_0\subseteq P_1+R_1x_0= P_1+x_0R_1\nonumber\\
&=&P_1+(x_01^{\sharp})\,\sharp\,R_1\subseteq P_1+P_1\,\sharp\,R_1\subseteq P_1.
\end{eqnarray}
Using (\ref{eq20}) and the fact that $P_0+R_0x_0$ is an ideal of $R_0$, we get
$P_0+R_0x_0\in \Omega$. Since $P_0+R_0x_0\supseteq P_0$, we have to have 
$P_0+R_0x_0= P_0$, which implies that $x_0\in P_0$. This proves that (\ref{eq16}) holds.

\medskip
Finally, if $x_0y_0\in P_0$, then 
$(1^{\sharp}x_0)\,\sharp\,(1^{\sharp}y_0)=1^{\sharp}x_0y_0\in P_1$. Thus, 
$(1^{\sharp}x_0)\in P_1$ or $(1^{\sharp}y_0)\in P_1$, which implies that either $x_0\in P_0$ or 
$y_0\in P_0$ by the argument above. This proves that (\ref{eq15}) holds.

\medskip
Summarizing what we have proved, $P=P_0\oplus P_1$ is a prime triideal of $R$. 

\hfill\raisebox{1mm}{\framebox[2mm]{}}

\medskip
The following proposition gives another characterization of prime triideals.

\begin{proposition}\label{pr5.2} Let $(\, R=R_0\oplus R_1, \, + , \, \cdot , \, \,\sharp\, \,)$ be a Hu-Liu triring. The following are equivalent.
\begin{description}
\item[(i)] $P$ is a prime triideal.
\item[(ii)] For two triideals $I$, $J$ of $R$, $I\,\stackrel{\cdot}{\sharp}\,J\subseteq P$ implies that $I\subseteq P$ or $J\subseteq P$.
\end{description}
\end{proposition}


\medskip
\noindent
{\bf Proof} (i) $\Rightarrow$ (ii): Assume that $I\,\stackrel{\cdot}{\sharp}\,J\subseteq P$ and $I\not\subseteq P$. Then either $I_0\not\subseteq P_0$ or $I_1\not\subseteq P_1$. Let 
$y_i\in J_i$ with $i=0$, $1$.

\medskip
If $I_0\not\subseteq P_0$, then there exists $x_0\in P_0\setminus I_0$. Clearly, 
$x_0y_i\in I\,\stackrel{\cdot}{\sharp}\,J\subseteq P$ for $i=0$, $1$. Since $P$ is a prime triideal, we get $y_i\in P$ for $i=0$, $1$ by (\ref{eq15}) and (\ref{eq16}). Thus $J\subseteq P$ in this case. 

\medskip
If $I_1\not\subseteq P_1$, then there exists $x_1\in P_1\setminus I_1$. Since 
$x_1y_0\in I\,\stackrel{\cdot}{\sharp}\,J\subseteq P$ and 
$x_1\,\sharp\, y_1\in \,\stackrel{\cdot}{\sharp}\,J\subseteq P$, we get $y_i\in P$ for $i=0$, $1$ by (\ref{eq17}) and (\ref{eq18}). Thus we also get $J\subseteq P$ in this case. 

\bigskip
(ii) $\Rightarrow$ (i): Let $x_i$, $y_i\in P_i$ with for $i=0$, $1$. By Proposition \ref{pr2.1}, both $R_0x_0\oplus R_1x_0$ and $R_0y_0\oplus R_1y_0$ are triideals. By the definition of the mixed product of two triideals in Proposition \ref{pr2.2}, we have
$$(R_0x_0\oplus R_1x_0)\,\stackrel{\cdot}{\sharp}\,(R_0y_0\oplus R_1y_0)
=R_0x_0R_0y_0+(R_1x_0)\,\sharp\, (R_1y_0).$$ 
Since $(R_1x_0)\,\sharp\, (R_1y_0)=R_1x_0y_0$, we get
\begin{equation}\label{eq21}
(R_0x_0\oplus R_1x_0)\,\stackrel{\cdot}{\sharp}\,(R_0y_0\oplus R_1y_0)
\subseteq R_0x_0y_0+R_1x_0y_0.
\end{equation}

\medskip
If $x_0y_0\in P_0$ and $x_0\not\in P_0$, then 
\begin{equation}\label{eq22}
R_0x_0\oplus R_1x_0\not\subseteq P.
\end{equation}
It follows from  (\ref{eq21}) and (\ref{eq22}) that $R_0y_0\oplus R_1y_0\subseteq P$, which implies that $y_0\in P_0$. Thus, (\ref{eq15}) holds.

\medskip
If $x_0y_1\in P_1$ and $x_0\not\in P_0$, then (\ref{eq22}) holds.
By (\ref{eq6}), $1^{\sharp}x_0=x_0z_1$ for some $z_1\in R_1$. Hence, we have
\begin{eqnarray}\label{eq23}
&&(R_0x_0\oplus R_1x_0)\,\stackrel{\cdot}{\sharp}\,(R_1\,\sharp\, y_1)= (R_1x_0)\,\sharp\,(R_1\,\sharp\, y_1)\subseteq 
R_1\,\sharp\, (1^{\sharp}x_0)\,\sharp\, y_1\nonumber\\
&\subseteq &R_1\,\sharp\, (x_0z_1)\,\sharp\, y_1
\subseteq R_1\,\sharp\, (x_01^{\sharp})\,\sharp\,z_1\,\sharp\, y_1
\subseteq R_1\,\sharp\, z_1\,\sharp\,(x_01^{\sharp})\,\sharp\, y_1\nonumber\\
&\subseteq& R_1\,\sharp\, \big(x_0(1^{\sharp}\,\sharp\,y_1)\big)
\subseteq R_1\,\sharp\, (x_0y_1).
\end{eqnarray}
It follows from  (\ref{eq21}) and (\ref{eq23}) that $R_1\,\sharp\, y_1\subseteq P$, which implies that $y_1\in P$. Thus, (\ref{eq16}) holds.

\medskip
If $x_1y_0\in P_1$ and $x_1\not\in P_1$, then
\begin{equation}\label{eq24} 
R_1\,\sharp\, x_1\not\subseteq P.
\end{equation}

It follows from $x_1y_0\in P_1$ that
\begin{eqnarray}\label{eq25}
&&(R_1\,\sharp\, x_1)\,\stackrel{\cdot}{\sharp}\,(R_0y_0+R_1y_0)
=(R_1\,\sharp\, x_1)\,\sharp\,(R_1y_0)=R_1\,\sharp\, x_1\,\sharp\,R_1\,\sharp\,(1^{\sharp}y_0)
\nonumber\\
&=&R_1\,\sharp\, x_1\,\sharp\,(1^{\sharp}y_0)=R_1\,\sharp\, (x_1\,\sharp\,1^{\sharp})y_0
=R_1\,\sharp\, (x_1y_0)\subseteq P.
\end{eqnarray}
By (\ref{eq24}) and (\ref{eq25}), we get $R_0y_0+R_1y_0\subseteq P$, which implies that 
$y_0\in P$. Thus, (\ref{eq17}) holds.

\medskip
If $x_1\,\sharp\,y_1\in P_1$ and $x_1\not\in P_1$, then (\ref{eq24}) holds and
\begin{equation}\label{eq26} 
(R_1\,\sharp\, x_1)\,\stackrel{\cdot}{\sharp}\,(R_1\,\sharp\, y_1)
=R_1\,\sharp\, x_1\,\sharp\,y_1\subseteq P.
\end{equation}
By (\ref{eq24}) and (\ref{eq26}), $R_1\,\sharp\, y_1\subseteq P$, which implies that 
$y_1\in P$. Thus, (\ref{eq18}) holds.

\medskip
This proves that $P$ is a prime triideal.

\hfill\raisebox{1mm}{\framebox[2mm]{}}

\bigskip
\section{Trinilradicals}

Let $R=R_0\oplus R_1$ be a Hu-Liu triring with the local identity $1^\sharp$. For 
$\alpha \in R_1$, the {\bf local $n$th power} $\alpha ^{\sharp n}$ is defined by:
$$
\alpha^{\sharp n} :=\left\{ \begin{array}{ll}
1^\sharp, & \qquad\mbox{if $n=0$;}\\
\underbrace{\alpha \,\sharp\, \alpha  \,\sharp\, \cdots  \,\sharp\, \alpha}_n , & 
\qquad\mbox{if $n$ is a positive integer.} \end{array}\right.
$$
The products $(x^m)(\alpha^{\sharp n})$ and $(\alpha^{\sharp n})(x^m)$ will be denoted by $x^m\alpha^{\sharp n}$ and $\alpha^{\sharp n}x^m$ respectively, where $x\in R$ and $\alpha \in R_1$.

\begin{proposition}\label{pr6.1} Let $(\, R=R_0\oplus R_1, \, + , \, \cdot , \, \,\sharp\, \,)$ be a Hu-Liu triring with the local identity $1^\sharp$. 
If $x$, $y\in R$, $\alpha$, $\beta\in R_1$ and $m\in\mathcal{Z}_{>0}$, then
\begin{equation}\label{eq27}
(x\alpha)\,\sharp\, (y\beta)=(xy)(\alpha\,\sharp\, \beta), \qquad
(\alpha x)\,\sharp\, (\beta y)=(\alpha\,\sharp\, \beta)xy
\end{equation}
and
\begin{equation}\label{eq28}
(x\alpha)^{\sharp m}=x^m \alpha ^{\sharp m}, \qquad 
(\alpha x)^{\sharp m}=\alpha ^{\sharp m}x^m.
\end{equation}
\end{proposition}

\medskip
\noindent
{\bf Proof} By the triassociative law, we have
$$
(x\alpha )\,\sharp\, (y\beta)=x\left(\alpha\,\sharp\, (y\beta)\right)=
x\left((y\beta)\,\sharp\, \alpha\right)=(xy)(\beta\,\sharp\, \alpha)=(xy)(\alpha\,\sharp\, \beta)
$$
and
$$
(\alpha x)\,\sharp\, (\beta y)=\left((\alpha x)\,\sharp\, \beta)\right)y=
\left((\beta)\,\sharp\, (\alpha x)\right)y=(\beta\,\sharp\, \alpha)xy=
(\alpha\,\sharp\, \beta)xy.
$$
Hence, (\ref{eq27}) holds. Clearly, (\ref{eq28}) follows from (\ref{eq27}).

\hfill\raisebox{1mm}{\framebox[2mm]{}}

\medskip
\begin{definition}\label{def6.1} Let $(\, R=R_0\oplus R_1, \, + , \, \cdot , \, \,\sharp\, \,)$ be a Hu-Liu triring. 
\begin{description}
\item[(i)] An element $x$ of $R$ is said to be {\bf  trinilpotent} if 
\begin{equation}\label{eq29}
\mbox{$x_0^m=0$ and $x_1^{\,\sharp\, n}=0$ for some $m$, $n\in\mathcal{Z}_{>0}$,}
\end{equation}
where $x_0$ and $x_1$ are the even component and the old component of $x$ respectively.
\item[(ii)] The set of all graded nilpotent elements of $R$ is called the {\bf trinilradical} of $R$ and denoted by $nilrad ^{\sharp}(R)$ or $\sqrt[\sharp]{0}$.
\end{description}
\end{definition}

\medskip
The ordinary nilradical of a ring $(\, A, \, + , \, \cdot  \,)$ is denoted by $nilrad(A)$ or
$nilrad(A, + , \cdot)$; that is,
$$
nilrad(A):=\{\, x \,|\, \mbox{$x^m=0$ for some $m\in\mathcal{Z}_{>0}$}\,\}.
$$

If $(\, R=R_0\oplus R_1, \, + , \, \cdot , \, \,\sharp\, \,)$ is a Hu-Liu triring, then
$$
nilrad(R, +, \cdot)=nilrad(R_0, + , \cdot)\oplus R_1
$$
and
\begin{equation}\label{eq30}
nilrad^{\sharp}R=nilrad(R_0, + , \cdot)\oplus nilrad(R_1, +, \,\sharp\,).
\end{equation}
Hence, $nilrad^{\sharp}R\subseteq nilrad(R, +, \cdot)$, i.e., the trinilradical of a Hu-Liu triring $R$ is smaller than the ordinary nilradical of the ring $(R,\,+, \,\cdot)$.

\begin{proposition}\label{pr6.2} Let $(\, R=R_0\oplus R_1, \, + , \, \cdot , \, \,\sharp\, \,)$ be a Hu-Liu triring.
\begin{description}
\item[(i)] The trinilradical $nilrad ^{\sharp}(R)$ is a triideal of $R$.
\item[(ii)] $nilrad ^{\sharp}\left( \displaystyle\frac{R}{nilrad ^{\sharp}(R)}\right)=0.$
\end{description}
\end{proposition}

\medskip
\noindent
{\bf Proof} (i) By the definition of trinilradicals, we have 
$$
(nilrad^{\sharp}R)\cap R_0=nilrad(R_0, + , \cdot)\quad\mbox{and}\quad
(nilrad^{\sharp}R)\cap R_1=nilrad(R_1, +, \,\sharp\,).
$$
Using the fact above and (\ref{eq30}), we need only to prove
\begin{equation}\label{eq31}
\{xa,\, ax\}\subseteq nilrad ^{\sharp}R\quad\mbox{for $x\in R$ and $a\in nilrad ^{\sharp}R$}.
\end{equation}

Let $x=x_0+x_1\in R$ and $a=a_0+a_1\in nilrad ^{\sharp}R$, where $x_i$, $a_i\in R_i$ for $i=0$, $1$. Then we have $a_0^m=a_1^{\,\sharp\, m}=0$ for some $m\in\mathcal{Z}_{>0}$. Since
\begin{eqnarray}
\label{eq32} xa&=&(x_0+x_1)(a_0+a_1)=x_0a_0+(x_0a_1+x_1a_0),\\
(x_0a_0)^m&=&x_0^ma_0^m=x_0^m0=0,\nonumber\\
(x_0a_1)^{\sharp m}&=&x_0^ma_1^{\sharp m}=x_0^m0=0,\nonumber\\
(x_1a_0)^{\sharp m}&=&(x_1a_0)^{\sharp m}=x_1^{\sharp m}a_0^m=x_1^{\sharp m}0=0,\nonumber
\end{eqnarray}
we get
\begin{equation}\label{eq33}
x_0a_0\in nilrad(R_0, \,+, \,\cdot) \quad\mbox{and}\quad 
x_0a_1+x_1a_0\in nilrad ^{\sharp}(R_1, \,+,\, \sharp).
\end{equation}

It follows from (\ref{eq32}) and (\ref{eq33}) that $xa\in nilrad ^{\sharp}R$. Similarly, we have $ax\in nilrad ^{\sharp}R$. This proves (i).

\medskip
(ii) If $x+nilrad ^{\sharp}R\in \displaystyle\frac{R}{nilrad ^{\sharp}R}$, then there exist positive integers $m$ and $n$ such that
$$x_0^m+nilrad ^{\sharp}R=\left(x_0+nilrad ^{\sharp}R\right)^m=nilrad ^{\sharp}R$$
and
$$x_1^{\,\sharp\, n}+nilrad ^{\sharp}R=\left((x_1+nilrad ^{\sharp}R)_1\right)^{\,\sharp\, n}=nilrad ^{\sharp}R,$$
where $x=x_0+x_1$ and $x_i\in R_i$ for $i=0$, $1$. Hence, we get
$$ x_0^m\in nilrad ^{\sharp}R \quad\mbox{and}\quad x_1^{\,\sharp\, n}\in nilrad ^{\sharp}R,$$
which imply that
$$x_0^{mu}=(x_0^m)^u=0  \quad\mbox{and}\quad 
x_1^{\,\sharp\, (nv)}=(x_1^{\,\sharp\, n})^{\,\sharp\, v}=0$$
for some $u$, $v\in \mathcal{Z}_{>0}$. Thus, $x=x_0+x_1\in nilrad ^{\sharp}R$ or 
$x+nilrad ^{\sharp}R$ is the zero element of $\displaystyle\frac{R}{nilrad ^{\sharp}R}$. This proves (ii).

\hfill\raisebox{1mm}{\framebox[2mm]{}}

\medskip
If $I$ is a triideal of a Hu-Liu triring $R=R_0\oplus R_1$, then the {\bf trinilradical} $\sqrt[\sharp]{I}$ of $I$ is defined by
$$
\sqrt[\sharp]{I}:=\{\, x\in R \, | \, \mbox{$x_0^m\in I_0$ and $x_1^{\,\sharp\, n}\in I_1$ for some $m$, $n\in\mathcal{Z}_{>0}$} \,\},
$$
where $x=x_0+x_1$, $x_i\in R_i$ and $I_i=I\cap R_i$ for $i=0$, $1$. Since
$$
nilrad ^{\sharp}\left(\displaystyle\frac{R}{I}\right)=\displaystyle\frac{\sqrt[\sharp]{I}}{I},
$$
$\sqrt[\sharp]{I}$ is a triideal of $R$. A triideal $I$ of a Hu-Liu triring $R$ is called a {\bf radical triideal} if $\sqrt[\sharp]{I}=I$.

\medskip
We now characterize the trinilradical of a Hu-Liu triring by using prime triideals. 

\begin{proposition}\label{pr6.3} Let $(\, R=R_0\oplus R_1, \, + , \, \cdot\, , \,\sharp\, \,)$ be a Hu-Liu triring. The trinilradical of $R$ is the intersection of the prime triideals of $R$.
\end{proposition}

\medskip
\noindent
{\bf Proof}  Let $x=x_0+x_1$ be any element of $nilrad ^{\sharp}(R)$, where $x_0\in R_0$ and $x_1\in R_1$. Then $x_0^m=0$ and $x_1^{\,\sharp\, n}=0$ for some $m$, $n\in\mathcal{Z}_{>0}$. Let $P=P_0\oplus P_1$ be any prime triideal of $R$. Since $x_0^m=0\in P_0$, we have 
\begin{equation}\label{eq34} 
x_0\in P_0
\end{equation}
by (\ref{eq15}).

If $P\not\supseteq R_1$, then $P_1$ is a prime ideal of the commutative ring 
$(\, R_1, \, + , \, \,\sharp\, \,)$. Using this fact and $x_1^{\,\sharp\, n}=0$, we get
\begin{equation}\label{eq35} 
x_1\in P_1.
\end{equation}
If $P\supseteq R_1$, then (\ref{eq35}) is obviously true. It follows from (\ref{eq34}) and (\ref{eq35}) that $x=x_0+x_1\in P$. This proves that
\begin{equation}\label{eq36} 
nilrad ^{\sharp}(R)\subseteq \bigcap_{P\in Spec^{\sharp}R} P.
\end{equation}

\medskip
Conversely, we prove that
\begin{equation}\label{eq37} 
z\not\in nilrad ^{\sharp}(R) \,\Rightarrow\, z\not\in \bigcap_{P\in Spec^{\sharp}R} P.
\end{equation}

\medskip
\underline{\it Case 1:} $z^m\ne 0$ for all $m\in \mathcal{Z}_{>0}$, in which case, 
$z^m\not\in R_1$ for all $m\in \mathcal{Z}_{>0}$. Hence, $z+ R_1$ is not a nilpotent element of the commutative ring $\displaystyle\frac{R}{R_1}$; that is,
$$
z+ R_1\ne nilrad\left(\displaystyle\frac{R}{R_1} \right)
=\bigcap _{\frac{I}{R_1}\in Spec \left( \frac{R}{R_1}\right)} 
\left(\displaystyle\frac{I}{R_1}\right),
$$
where $Spec \left( \displaystyle\frac{R}{R_1}\right)$ is the ordinary spectrum of the commutative ring
$\left(\displaystyle\frac{R}{R_1},\, +,\, \cdot\right)$.
Hence, there exists a prime ideal $\displaystyle\frac{I}{R_1}$ of the commutative ring $\displaystyle\frac{R}{R_1}$ such that $z\not\in I$. Since $I$ is a prime triideal of $R$, (\ref{eq37}) holds in this case.

\medskip
\underline{\it Case 2:} $z^m= 0$ for some $m\in \mathcal{Z}_{>0}$. Let $z=z_0+z_1$ with 
$z_0\in R_0$ and $z_1\in R_1$. Then $0=z^m=z_0^m+z_0^{m-1}r_1$ for some $r_1\in R_1$ by (\ref{eq6}). Thus, $z_0^m=0$, which implies that  $z_1^{\,\sharp\, n}\ne 0$ for all 
$n\in \mathcal{Z}_{>0}$ in this case.
We  now consider the following set
$$
T:=\left\{\, J \, \left| \, \mbox{$J$ is a triideal of $R$ and $z_1^{\,\sharp\, n}\not\in J$ for all $n\in \mathcal{Z}_{>0}$}\right.\right\}.
$$
Since $\{0\}\in T$, $T$ is nonempty. Clearly, $(\,T, \, \subseteq \,)$ is a partially order set, where $\subseteq $ is the relation of set inclusion. If $\{\, J_{\lambda} \, |\, \lambda\in \Lambda \,\}$ is a nonempty totally ordered subset of $T$, then 
$\bigcup_{\lambda\in \Lambda} J_{\lambda}$ 
is an upper bound of 
$\{\, J_{\lambda} \, |\, \lambda\in \Lambda \,\}$ in $T$. By Zorn's Lemma, the partially ordered set $(\,T, \, \subseteq \,)$ has a maximal element $P$. We are going to prove that $P$ is a prime triideal of $R$.

\medskip
Let $x=x_0+x_1$ and $y=y_0+y_1$ be two elements of $R$, where $x_i$, $y_i\in R_i$ for $i=0$, $1$. First, if $x_0\not\in P_0$ and $y_0\not\in P_0$, then 
\begin{equation}\label{eq38}
P\subset P+Rx_0\quad\mbox{and}\quad P\subset P+Ry_0.
\end{equation}
Since both $P+Rx_0$ and $P+Ry_0$ are triideals of $R$ by Proposition~\ref{pr2.1}, (\ref{eq38}) implies that
$$
z_1^{\,\sharp\, u}\in P+Rx_0\quad\mbox{and}\quad z_1^{\,\sharp\, v}\in P+Ry_0
$$
or
$$
z_1^{\,\sharp\, u}\in P_1+R_1x_0\quad\mbox{and}\quad z_1^{\,\sharp\, v}\in P_1+R_1y_0\quad\mbox{for some $u$, $v\in \mathcal{Z}_{>0}$.}
$$
Thus, we have
\begin{eqnarray*}
&&z_1^{\,\sharp\, (u+v)}=z_1^{\,\sharp\, u}\,\sharp\, z_1^{\,\sharp\, v}\in (P_1+R_1x_0)\,\sharp\, (P_1+R_1y_0)\\&&\\
&\subseteq& 
\underbrace{P_1\,\sharp\, P_1+ P_1\,\sharp\, (R_1y_0)+
(R_1x_0)\,\sharp\, P_1}_{\mbox{This is a subset of $P$}}+ (R_1x_0)\,\sharp\, (R_1y_0)\\&&\\
&\subseteq& P+R_1x_0y_0,
\end{eqnarray*}
which implies that $x_0y_0\not\in P$. This proves that
\begin{equation}\label{eq39} 
\mbox{$x_0\not\in P_0$ and $y_0\not\in P_0$} \,\Rightarrow\, x_0y_0\not\in P_0.
\end{equation}

Next, if $x_0\not\in P_0$ and $y_1\not\in P_1$, then 
\begin{equation}\label{eq40}
P\subset P+Rx_0\quad\mbox{and}\quad P\subset P+R_1\,\sharp\,y_1.
\end{equation}
Since both $P+Rx_0$ and $P+R_1\,\sharp\,y_1$ are triideals of $R$ by Proposition~\ref{pr2.1}, (\ref{eq40}) implies that there exist some positive integers $s$ and $t$ such that
$$
z_1^{\,\sharp\, s}\in P+Rx_0\quad\mbox{and}\quad z_1^{\,\sharp\, t}\in P+ R_1 \,\sharp\, y_1
$$ 
or
\begin{equation}\label{eq41} 
z_1^{\,\sharp\, s}\in P_1+R_1x_0\quad\mbox{and}\quad z_1^{\,\sharp\, t}\in 
P_1+ R_1 \,\sharp\, y_1.
\end{equation}
It follows that
\begin{eqnarray*}
&&z_1^{\,\sharp\, (s+t)}=z_1^{\,\sharp\, s}\,\sharp\, z_1^{\,\sharp\, t}\in (P_1+R_1x_0)\,\sharp\, (P_1+ R_1 \,\sharp\, y_1)\nonumber\\&&\nonumber\\
&\subseteq& 
\underbrace{P_1\,\sharp\, P_1+ P_1\,\sharp\, R_1 \,\sharp\, y_1+
(R_1x_0)\,\sharp\, P_1}_{\mbox{This is a subset of $P_1$}}
+ (R_1x_0)\,\sharp\, R_1\,\sharp\, y_1\nonumber\\
&\subseteq& P_1+ R_1\,\sharp\,(x_0y_1),
\end{eqnarray*}
which implies that $x_0y_1\not\in P_1$. This proves that
\begin{equation}\label{eq44} 
\mbox{$x_0\not\in P_0$ and $y_1\not\in P_1$} \,\Rightarrow\, x_0y_1\not\in P_1.
\end{equation}

\medskip
Similarly, we have
\begin{equation}\label{eq47} 
\mbox{$y_1\not\in P_1$ and $x_0\not\in P_0$} \,\Rightarrow\, y_1x_0\not\in P_1
\end{equation}
and
\begin{equation}\label{eq48} 
\mbox{$x_1\not\in P_1$ and $y_1\not\in P_1$} \,\Rightarrow\, x_1y_1\not\in P_0.
\end{equation}

\medskip
By (\ref{eq39}), (\ref{eq44}), (\ref{eq47}) and (\ref{eq48}), $P$ is a prime triideal. Since $z_1\not\in P$, (\ref{eq37}) also holds in Case 2.

\medskip
It follows from (\ref{eq36}) and (\ref{eq37}) that Proposition~\ref{pr6.3} is true.

\hfill\raisebox{1mm}{\framebox[2mm]{}}

\bigskip
The next proposition is a corollary of Proposition~\ref{pr6.3}.

\begin{proposition}\label{pr6.4} If $I$ is a triideal of a Hu-Liu triring $R$ and $I\ne R$, then
$$\sqrt[\sharp]{I}=\bigcap _{\mbox{$P\in Spec^{\sharp}R$ and $P\supseteq I$}} P.$$
\end{proposition}

\medskip
\noindent
{\bf Proof} By Proposition~\ref{pr6.3}, we have
\begin{eqnarray*}
x\in \sqrt[\sharp]{I}
&\Leftrightarrow& x+I\in nilrad ^{\sharp}\left(\frac{R}{I}\right)
=\bigcap _{\mbox{$\frac{P}{I}\in Spec^{\sharp}\left(\frac{R}{I}\right)$}}\frac{P}{I}\\
&\Leftrightarrow& x\in \bigcap _{\mbox{$P\in Spec^{\sharp}R$ and $P\supseteq I$}} P.
\end{eqnarray*}

\hfill\raisebox{1mm}{\framebox[2mm]{}}

\bigskip
\section{Extended Zariski Topology}

Let $(\, R=R_0\oplus R_1, \, + , \, \cdot , \, \,\sharp\, \,)$ be a Hu-Liu triring. For a triideal $I$ of $R$, we define a subset $\mathcal{V}(I)$ of $Spec^{\sharp}R$ by
\begin{equation}\label{eq49} 
V^{\sharp}(I): =\{\, P \, |\, \mbox{$P\in Spec^{\sharp}R$ and $P\supseteq I$} \,\}.
\end{equation}

\begin{proposition}\label{pr7.1} Let $R$ be a Hu-Liu triring.
\begin{description}
\item[(i)] $V^{\sharp}(0)=spec^{\sharp}R$ and $V^{\sharp}(R)=\emptyset$.
\item[(ii)] $V^{\sharp}(I)\cup V^{\sharp}(J)
=V^{\sharp}(I\cap J)=V^{\sharp}(I\,\stackrel{\cdot}{\sharp}\,J)$, where $I$ and $J$ are two triideals of $R$.
\item[(iii)] $\displaystyle\bigcap _{\lambda \in \Lambda} 
V^{\sharp}\left(I_{(\lambda)}\right)=
V^{\sharp}\left(\displaystyle\sum _{\lambda \in \Lambda}I_{(\lambda)}\right)$, where 
$\left\{\left.\,I_{(\lambda)}\, \right| \, \lambda \in \Lambda\, \right\}$ is a set of triideals of $R$.
\end{description}
\end{proposition}

\medskip
\noindent
{\bf Proof} Since (i) and (iii) are clear, we need only to prove (ii).

\medskip
Since $I\,\stackrel{\cdot}{\sharp}\,J\subseteq I\cap J\subseteq I$, we get $V^{\sharp}(I\,\stackrel{\cdot}{\sharp}\,J)\supseteq V^{\sharp}(I\cap J)
\supseteq V^{\sharp}(I)$. Similarly, we have $V^{\sharp}(I\,\stackrel{\cdot}{\sharp}\,J)\supseteq V^{\sharp}(I\cap J)\supseteq V^{\sharp}(J)$. Thus, we get
\begin{equation}\label{eq50} 
V^{\sharp}(I)\cup V^{\sharp}(J)\subseteq V^{\sharp}(I\cap J)
\subseteq V^{\sharp}(I\,\stackrel{\cdot}{\sharp}\,J).
\end{equation}

\medskip
Conversely, if $P\in V^{\sharp}(I\,\stackrel{\cdot}{\sharp}\,J)$, then $I\,\stackrel{\cdot}{\sharp}\,J\subseteq P$. By Proposition~\ref{pr5.2}, we get $I\subseteq P$ or 
$J\subseteq P$. Hence, $P\in V^{\sharp}(I)\cup V^{\sharp}(J)$. This proves that
\begin{equation}\label{eq51} 
V^{\sharp}(I\,\stackrel{\cdot}{\sharp}\,J)\subseteq  V^{\sharp}(I)\cup V^{\sharp}(J).
\end{equation}

\medskip
It follows from (\ref{eq50}) and (\ref{eq51}) that (ii) is true.

\hfill\raisebox{1mm}{\framebox[2mm]{}}

\bigskip
Let $(\, R=R_0\oplus R_1, \, + , \, \cdot\, , \,\sharp\, \,)$ be a Hu-Liu triring. By Proposition~\ref{pr7.1}, the collection
$$
V^{\sharp}: =\{\, V^{\sharp}(I) \, |\, \mbox{$I$ is a triideal of $R$} \, \}
$$
of subsets of $Spec^{\sharp}R$ satisfies the axioms for closed sets in a topological space. The topology on $Spec^{\sharp}R$ having the elements of $V^{\sharp}$ as closed sets is called the {\bf extended Zariski topology}. The collection
$$
D^{\sharp}: =\{\, D^{\sharp}(I) \, |\, \mbox{$I$ is a triideal of $R$} \, \}
$$
consists of the open sets of the extended Zariski topology on $Spec^{\sharp}R$, where
$$
D^{\sharp}(I): =Spec^{\sharp}R\setminus V^{\sharp}(I)=
\{\, P \, |\, \mbox{$P\in Spec^{\sharp}R$ and $P\not\supseteq I$} \, \}.
$$

\medskip
For $x_i\in R_i$ with $i\in \{0,\, 1\}$, both $Rx_0$ and $R_1\,\sharp\,x_1$ are triideals of $R$ by Proposition~\ref{pr2.1}. Let
$$
D^{\sharp}(x_0): =D^{\sharp}(Rx_0), \qquad D^{\sharp}(x_1): =D^{\sharp}(R_1\,\sharp\,x_1).
$$
If $I_0$ and $I_1$ are the even part and odd part of an triideal $I$, then 
$$D^{\sharp}(I)=\bigcup _{x_i\in I_i,\, i=0,1}D^{\sharp}(x_i).$$
Thus, $\{\, D^{\sharp}(x_i)\,|\,\mbox{$x_i\in R_i$ with  $i=0,\, 1$}\,\}$ forms an open base for the extended Zariski topology on $Spec^{\sharp}R$. Each $D^{\sharp}(x_i)$ is called a {\bf basic open subset} of $Spec^{\sharp}R$. Clearly, $D^{\sharp}(0)=\emptyset$, $D^{\sharp}(1)=Spec^{\sharp}R$, $D^{\sharp}(1^{\sharp})=Spec^{\sharp}_1R$, and $D^{\sharp}(x_1)\subseteq  Spec^{\sharp}_1R$ for $x_1\in R_1$.

\begin{proposition}\label{pr7.2} Let $I$ and $J$ be triideals of a Hu-Liu triring. 
\begin{description}
\item[(i)] $V^{\sharp}(I)\subseteq V^{\sharp}(J)$ if and only if 
$\sqrt[\sharp]{J}\subseteq \sqrt[\sharp]{I}$.
\item[(ii)] $V^{\sharp}(I)=V^{\sharp}(\sqrt[\sharp]{I})$.
\end{description}
\end{proposition}

\medskip
\noindent
{\bf Proof} (i) If $V^{\sharp}(I)\subseteq V^{\sharp}(J)$, then $P\supseteq I$ implies that $P\supseteq J$ for $P\in Spec^{\sharp}R$. Thus, we have
\begin{eqnarray*}
&&\{\,P\,|\,\mbox{$P\in Spec^{\sharp}R$ and $P\supseteq J$}\,\}\\
&=&\{\,P\,|\,\mbox{$P\in Spec^{\sharp}R$ and $P\supseteq I$}\,\}\bigcup
\{\,P\,|\,\mbox{$P\in Spec^{\sharp}R$, $P\supseteq J$ and $P\not\supseteq I$}\,\}.
\end{eqnarray*}

By Proposition~\ref{pr6.4}, we get
\begin{eqnarray*}
\sqrt[\sharp]{J}&=&\bigcap _{\begin{array}{c}P\in spec^{\sharp}R\\P\supseteq J\end{array}} P=
\Bigg(\bigcap _{\begin{array}{c}P\in spec^{\sharp}R\\P\supseteq I\end{array}} P\Bigg)\bigcap
\Bigg(\bigcap _{\begin{array}{c}P\in spec^{\sharp}R\\P\supseteq J,\,P\not\supseteq I\end{array}} P\Bigg)\\
&\subseteq &\bigcap _{\begin{array}{c}P\in spec^{\sharp}R\\P\supseteq I\end{array}} P=
\sqrt[\sharp]{J}.
\end{eqnarray*}

\medskip
Conversely, if $\sqrt[\sharp]{J}\subseteq \sqrt[\sharp]{I}$, then for any 
$Q\in V^{\sharp}(I)$, we have $Q\supseteq I$ and
$$
Q\supseteq \bigcap _{\begin{array}{c}P\in Spec^{\sharp}R\\P\supseteq I\end{array}} P
=\sqrt[\sharp]{I}\supseteq \sqrt[\sharp]{J}=
\bigcap _{\begin{array}{c}P\in Spec^{\sharp}R\\P\supseteq J\end{array}} P\supseteq J,
$$
which proves that $Q\in V^{\sharp}(J)$. Thus, we get $V^{\sharp}(I)\subseteq V^{\sharp}(J)$.

\medskip
(ii) Since $\sqrt[\sharp]{\sqrt[\sharp]{I}}=\sqrt[\sharp]{I}$, (ii) follows from (i).

\hfill\raisebox{1mm}{\framebox[2mm]{}}

\bigskip
\begin{definition}\label{def7.1} Let $X$ be a topological space.
\begin{description}
\item[(i)] A closed subset $F$ of $X$ is {\bf reducible} if $F=F_{(1)}\cup F_{(2)}$ for proper closed subsets $F_{(1)}$, $F_{(2)}$ of $X$. We call a closed subset $F$ {\bf irreducible} if it is not reducible.
\item[(ii)] $X$ is {\bf quasicompact} if given an arbitrary open covering 
$\{\, U_{(i)}\,|\, i\in I\,\}$ of $X$, there exists a finite subcovering of $X$, i.e., there exist finitely many members $U_{(i_1)}$, $\dots$, $U_{(i_n)}$ of $\{\, U_{(i)}\,|\, i\in I\,\}$ such that $X=U_{(i_1)}\cup\cdots\cup U_{(i_n)}$.
\end{description}
\end{definition}

\medskip
Using the topological concepts above, we have the following

\medskip
\begin{proposition}\label{pr7.3}  Let $(\, R=R_0\oplus R_1, \, + , \, \cdot , \, \,\sharp\, \,)$ be a Hu-Liu triring.
\begin{description}
\item[(i)] The trispectrum $Spec^{\sharp}R$ is quasicompact.
\item[(ii)] Both $Spec^{\sharp}_0R$ and $Spec^{\sharp}_1R$ are quasicompact subsets of $Spec^{\sharp}R$.
\item[(iii)] If $I$ is a triideal of $R$, then the closed subset $V^{\sharp}(I)$ of $Spec^{\sharp}R$ is irreducible if and only if $\sqrt[\sharp]{I}$ is a prime triideal.
\end{description}
\end{proposition}

\medskip
\noindent
{\bf Proof} (i) Let $\{\, D^{\sharp}(I_{(i)})\,|\, i\in\Delta\,\}$ be an open covering of $Spec^{\sharp}R$, where $I_{(i)}=I_{(i)0}\oplus I_{(i)1}$ is a triideal with the even part $I_{(i)0}$ and the odd part $I_{(i)1}$ for each $i\in \Delta$. Thus, 
$Spec^{\sharp}R=\bigcup_{i\in\Delta}D^{\sharp}(I_{(i)})=
D^{\sharp}(\sum_{i\in\Delta}I_{(i)})$. Hence, $V^{\sharp}(\sum_{i\in\Delta}I_{(i)})=\emptyset$. If $1\not\in (\sum_{i\in\Delta}I_{(i)})_0=\sum_{i\in\Delta}I_{(i)0}$, then $(\sum_{i\in\Delta}I_{(i)})_0$ is a proper ideal of the commutative ring 
$( R_0,\, +,\, \cdot)$. Hence, there exists a maximal ideal $M_0$ of the ring 
$( R_0,\, +,\, \cdot)$ such that $(\sum_{i\in\Delta}I_{(i)})_0\subseteq M_0$. Since 
$M_0\oplus R_1$ is a prime triideal of $R$ and $\sum_{i\in\Delta}I_{(i)}\subseteq M_0\oplus R_1$, we get that $M_0\oplus R_1\in V^{\sharp}(\sum_{i\in\Delta}I_{(i)})=\emptyset$, which is impossible. Therefore, $1 \in (\sum_{i\in\Delta}I_{(i)})_0=\sum_{i\in\Delta}I_{(i)0}$, which implies that $x_{(i_1)0}+x_{(i_2)0}+\cdots +x_{(i_n)0}=1$ for some positive integer $n$ and 
$x_{(i_k)0}\in I_{(i_k)0}$ with $i_k\in\Delta$ and $n\ge k\ge 1$. It follows that
$D^{\sharp}(x_{(i_k)0})\subseteq D^{\sharp}(I_{(i_k)})$ and
\begin{eqnarray*}
&&Spec^{\sharp}R=\bigcup_{i\in\Delta}D^{\sharp}(I_{(i)})\supseteq \bigcup_{k=1}^nD^{\sharp}(I_{(i_k)})\supseteq \bigcup_{k=1}^nD^{\sharp}(x_{(i_k)0})\\
&=&\bigcup_{k=1}^nD^{\sharp}(Rx_{(i_k)0})
=D^{\sharp}\left(\sum_{k=1}^nRx_{(i_k)0}\right)=D^{\sharp}(R)=Spec^{\sharp}R,
\end{eqnarray*}
which implies that $Spec^{\sharp}R=\bigcup_{k=1}^nD^{\sharp}(I_{(i_k)})$.

\bigskip
(ii) Note that a closed subset of a quasicompact topological space is a quasicompact subset. Since  $Spec^{\sharp}_0R=V^{\sharp}(R_1)$ is a closed subset of the quasicompact topological space $Spec^{\sharp}R$, $Spec^{\sharp}_0R$ is a quasicompact subset.

\medskip
It is well-known that a subset $C$ of a topological space $X$ is a quasicompact subset of $X$ if and only if every covering of $C$ by open subsets of $X$ has a finite subcovering. Hence, in order to prove that $Spec^{\sharp}_1R$ is a quasicompact subsets of $Spec^{\sharp}R$, it suffices to prove that
if $Spec^{\sharp}_1R=\bigcup_{j\in\Gamma}D^{\sharp}(J_{(j)})$ for triideals $J_{(j)}$ of $R$, then there exists a positive integer $m$ such that $Spec^{\sharp}_1R=\bigcup_{k=1}^mD^{\sharp}(J_{(j_k)})$ for some $j_1$, $\cdots$, $j_k\in\Gamma$.

\medskip
Since $Spec^{\sharp}_1R=\bigcup_{j\in\Gamma}D^{\sharp}(J_{(j)})=D^{\sharp}(\sum_{j\in\Gamma}J_{(j)})$, we have $V^{\sharp}(\sum_{j\in\Gamma}J_{(j)})=Spec^{\sharp}R\setminus D^{\sharp}(\sum_{j\in\Gamma}J_{(j)})=Spec^{\sharp}_0R$. If $(\sum_{j\in\Gamma}J_{(j)})_1\ne R_1$, then there exists a maximal ideal $N_1$ of the commutative ring 
$( R_1,\, +,\, \sharp)$ such that $(\sum_{j\in\Gamma}J_{(j)})_1\subseteq N_1$. By Proposition~\ref{pr5.1}, there exists an ideal $N_0$ of the commutative ring 
$( R_0,\, +,\, \cdot)$ such that $N_0\supseteq (\sum_{j\in\Gamma}J_{(j)})_0$ and $N_0\oplus N_1$
is a prime triideal of $R$. Thus, 
$N_0\oplus N_1\in V^{\sharp}(\sum_{j\in\Gamma}J_{(j)})=Spec^{\sharp}_0R$, which is impossible because $N_1\ne R_1$. This proves that $(\sum_{j\in\Gamma}J_{(j)})_1= R_1$. Hence, we have
$y_{(j_1)1}+y_{(j_2)1}+\cdots +y_{(j_m)1}=1^{\sharp}$ for some positive integer $m$ and $y_{(j_k)1}\in J_{(j_k)1}$ with $j_k\in\Gamma$ and $m\ge k\ge 1$. It follows that 
$D^{\sharp}(y_{(j_k)1})\subseteq D^{\sharp}(J_{(j_k)})$ and
\begin{eqnarray*}
&&Spec^{\sharp}_1R=\bigcup_{j\in\Gamma}D^{\sharp}(J_{(j)})
\supseteq \bigcup_{k=1}^mD^{\sharp}(y_{(j_k)1})
=\bigcup_{k=1}^mD^{\sharp}(R_1\,\sharp\,y_{(j_k)1})\\
&=&D^{\sharp}\left(\sum_{k=1}^m R_1\,\sharp\,y_{(j_k)1}\right)
=D^{\sharp}(R_1)=Spec^{\sharp}_1R,
\end{eqnarray*}
which implies that $Spec^{\sharp}_1R=\bigcup_{k=1}^mD^{\sharp}(y_{(j_k)1})$.

\bigskip
(iii) By Proposition~\ref{pr7.2}, we may assume $I=\sqrt[\sharp]{I}$ in the following proof.
First, we prove that if $V^{\sharp}(I)$ is irreducible , then $I$ is a prime triideal. 

\medskip
Suppose that $a_0b_0\in I$ for some $a_0$, $b_0\in R_0$. Let 
$$J_{(1)}=I+Ra_0=(I_0+R_0a_0)\oplus (I_1+R_1a_0)$$
and
$$K_{(1)}=I+Rb_0=(I_0+R_0b_0)\oplus (I_1+R_1b_0).$$
Then both $J_{(1)}$ and $K_{(1)}$ are triideals of $R$ and
\begin{eqnarray*}
J_{(1)}\,\stackrel{\cdot}{\sharp}\, K_{(1)}&=&(I_0+R_0a_0)(I_0+R_0b_0)+
(I_1+R_1a_0)\,\sharp\,(I_1+R_1b_0)\\
&\subseteq& I+R_0a_0R_0b_0+(R_1a_0)\,\sharp\,(R_1b_0)\subseteq
I+R_0a_0b_0+R_1a_0b_0\subseteq I.
\end{eqnarray*}
Hence, we get $V^{\sharp}(J_{(1)})\cup V^{\sharp}(K_{(1)})=
V^{\sharp}(J_{(1)}\,\stackrel{\cdot}{\sharp}\, K_{(1)})\supseteq V^{\sharp}(I)$ by Proposition~\ref{pr7.1} (ii). It is clear that 
$V^{\sharp}(J_{(1)})\subseteq V^{\sharp}(I)$ and $V^{\sharp}(K_{(1)})\subseteq V^{\sharp}(I)$. Hence, we get that $V^{\sharp}(J_{(1)})\cup V^{\sharp}(K_{(1)})\subseteq  V^{\sharp}(I)$. Thus we have  $V^{\sharp}(J_{(1)})\cup V^{\sharp}(K_{(1)})= V^{\sharp}(I)$. Since $V^{\sharp}(I)$ is irreducible, $V^{\sharp}(J_{(1)})=V^{\sharp}(I)$ or $V^{\sharp}(K_{(1)})=V^{\sharp}(I)$, which imply that $I= \sqrt[\sharp]{J_{(1)}}\supseteq J_{(1)}\ni a_0$ or
$I= \sqrt[\sharp]{K_{(1)}}\supseteq K_{(1)}\ni b_0$. This proves that
\begin{equation}\label{eq52} 
a_0b_0\in I\Longrightarrow \mbox{$a_0\in I$ or $b_0\in I$ for $a_0$, $b_0\in R_0$}.
\end{equation}

\medskip
Suppose that $a_0b_1\in I$ for some $a_0\in R_0$ and $b_1\in R_1$. Using the triideals 
$$J_{(2)}=I+Ra_0=(I_0+R_0a_0)\oplus (I_1+R_1a_0)$$ and 
$$K_{(2)}=I+R_1\,\sharp\, b_1=I_0\oplus (I_1+R_1\,\sharp\, b_1),$$ 
we have
\begin{eqnarray*}
J_{(2)}\,\stackrel{\cdot}{\sharp}\, K_{(2)}&=&(I_0+R_0a_0)I_0+
(I_1+R_1a_0)\,\sharp\,(I_1+R_1\,\sharp\, b_1)\\
&\subseteq& I+(R_1a_0)\,\sharp\,R_1\,\sharp\, b_1\subseteq
I+(a_0R_1)\,\sharp\,b_1\,\sharp\, R_1\subseteq I+a_0(R_1\,\sharp\,b_1)\,\sharp\, R_1\\
&\subseteq& I+a_0(b_1\,\sharp\,R_1)\,\sharp\, R_1\subseteq I+(a_0b_1)\,\sharp\,R_1\,\sharp\, R_1
\subseteq I,
\end{eqnarray*}
which implies that  $I= \sqrt[\sharp]{J_{(2)}}\supseteq J_{(2)}\ni a_0$ or
$I= \sqrt[\sharp]{K_{(2)}}\supseteq K_{(2)}\ni b_1$. This proves that
\begin{equation}\label{eq53} 
a_0b_1\in I\Longrightarrow \mbox{$a_0\in I$ or $b_1\in I$ for $a_0\in R_0$ and $b_1\in R_1$}.
\end{equation}

\medskip
Similarly, we have
\begin{equation}\label{eq54} 
a_1b_0\in I\Longrightarrow \mbox{$a_1\in I$ or $b_0\in I$ for $a_1\in R_1$ and $b_0\in R_0$}
\end{equation}
and
\begin{equation}\label{eq55} 
a_1\,\sharp\,b_1\in I\Longrightarrow \mbox{$a_1\in I$ or $b_1\in I$ for $a_1$, $b_1\in R_1$}.
\end{equation}

\medskip
By (\ref{eq52}), (\ref{eq53}), (\ref{eq54}) and (\ref{eq55}), $I$ is a prime triideal. 

\bigskip
Next, we prove that if $I$ is a prime triideal, then $V^{\sharp}(I)$ is irreducible. Suppose that 
$V^{\sharp}(I)=V^{\sharp}(J)\cup V^{\sharp}(K)$, where $J$ and $K$ are triideals of $R$. Using Proposition~\ref{pr7.2} (ii), we can assume that $\sqrt[\sharp]{J}=J$ and $\sqrt[\sharp]{K}=K$. In this case, we have
$$
V^{\sharp}(J)\subseteq V^{\sharp}(I)\Longrightarrow I=
\sqrt[\sharp]{I}\subseteq \sqrt[\sharp]{J}=J
$$
and
$$
V^{\sharp}(K)\subseteq V^{\sharp}(I)\Longrightarrow I=
\sqrt[\sharp]{I}\subseteq \sqrt[\sharp]{K}=K.
$$
By Proposition~\ref{pr7.1} (ii), we have 
$V^{\sharp}(I)=V^{\sharp}(J)\cup V^{\sharp}(K)=V^{\sharp}(J\,\stackrel{\cdot}{\sharp}\,K)$. This fact and Proposition~\ref{pr7.2} (i) give $J\,\stackrel{\cdot}{\sharp}\,K\subseteq \sqrt[\sharp]{I}=I$. Since $I$ is a prime triideal, we get $J\subseteq I$ or $K\subseteq I$ by 
Proposition~\ref{pr5.2}. Hence, $I=J$ or $I=K$. Thus, $V^{\sharp}(I)=V^{\sharp}(J)$ or 
$V^{\sharp}(I)=V^{\sharp}(K)$. This proves that $V^{\sharp}(I)$ is irreducible.

\hfill\raisebox{1mm}{\framebox[2mm]{}}


\bigskip
\section{Localization of Hu-Liu Trirings}

Let $(\, R=R_0\oplus R_1, \, + , \, \cdot , \, \,\sharp\, \,)$ be a Hu-Liu triring with the identity $1$ and the local identity $1^{\sharp}$. A subset $S$ of $R$ is called a 
{\bf multiplicative subset} if 
$$
0\not\in S, \quad 1\in S_0:=S\cap R_0, \quad 1^{\sharp}\in S_1:=S\cap R_1, \quad 
S=S_0\cup S_1
$$ 
and
\begin{equation}\label{eq11}
s_0t_0\in S_0, \quad s_0s_1\in S_1, \quad s_1s_0\in S_1,\quad s_1 \,\sharp\, t_1\in S_1,
\quad s_0S_1=S_1s_0,
\end{equation}
where $s_i$, $t_i\in S_i$ and $i=0$, $1$.

\medskip
Given a  multiplicative subset $S$ of a Hu-Liu triring $R$, we define a relation in the Cartesian product $(R_0\times S_0)\times (R_1\times S_1)$ as follows:
\begin{eqnarray}
&&\left((a_0, s_0),\, (a_1, s_1) \right)\sim \left((b_0, t_0),\, (b_1, t_1)\right)\nonumber\\
\label{eq12}&\Leftrightarrow& u_0(a_0t_0-b_0s_0)=0 \quad\mbox{and}\quad u_1\,\sharp\,(a_1\,\sharp\,t_1\, -b_1\,\sharp\,s_1)=0
\end{eqnarray}
for some $u_i\in S_i$, where $a_i$, $b_i\in R_i$, $s_i$, $t_i\in S_i$ and $i=0$, $1$.

It is clear that $\sim$ is an equivalence relation. Let 
$\left(\displaystyle\frac{a_0}{s_0},  \,\displaystyle\frac{a_1}{s_1}\right)$ be the equivalence class containing $\left((a_0, s_0),\, (a_1, s_1) \right)$; that is,
$$
\left(\displaystyle\frac{a_0}{s_0},  \,\displaystyle\frac{a_1}{s_1}\right):=
\left\{\, \left((a_0', s_0'),\, (a_1', s_1') \right) \, \left| \, \begin{array}{c} 
\mbox{$(a_i', s_i')\in R_i\times S_i$,}\\
\mbox{$u_0(a_0s_0'-s_0a_0')=0$,}\\
\mbox{ $u_1\,\sharp\,(a_1\,\sharp\,s_1'\, -s_1\,\sharp\,a_1')=0$ }\\
\mbox{for some $u_i\in S_i$ with $i=0$, $1$.}
\end{array}\right.\right\}.
$$

\medskip
Let
$$
S^{\stackrel{\sharp}{-}1}R:=\left\{\, \left.\left(\displaystyle\frac{a_0}{s_0},  \,\displaystyle\frac{a_1}{s_1}\right) \, \right | \, 
\left((a_0, s_0),\, (a_1, s_1)\right)\in (R_0\times S_0)\times (R_1\times S_1)\right\}
$$
be the set of all equivalence classes. Then 
$S^{\stackrel{\sharp}{-}1}R=(S^{\stackrel{\sharp}{-}1}R)_0\oplus 
(S^{\stackrel{\sharp}{-}1}R)_1$ is a Hu-Liu triring, where the even part $\Big(S^{\stackrel{\sharp}{-}1}R\Big)_0$ and odd part 
$\Big(S^{\stackrel{\sharp}{-}1}R\Big)_1$ are given by
$$
\Big(S^{\stackrel{\sharp}{-}1}R\Big)_0:=\left\{\left.\left(\displaystyle\frac{a_0}{s_0}, \,\displaystyle\frac{0}{1^\sharp}\right) \, \right| \, (a_0, s_0)\in R_0\times S_0 \,\right\},
$$
$$
\Big(S^{\stackrel{\sharp}{-}1}R\Big)_1:=\left\{\left.\left(\displaystyle\frac{0}{1}, \,\displaystyle\frac{a_1}{s_1}\right) \, \right| \, (a_1, s_1)\in R_1\times S_1 \,\right\}
$$
and the addition $+$, the ring multiplication $\cdot$ and the local product $\,\sharp\,$ are given by
$$
\left(\displaystyle\frac{a_0}{s_0},  \,\displaystyle\frac{a_1}{s_1}\right)+
\left(\displaystyle\frac{b_0}{t_0},  \,\displaystyle\frac{b_1}{t_1}\right):=
\left(\displaystyle\frac{a_0t_0+s_0b_0}{s_0t_0},  
\,\displaystyle\frac{a_1\,\sharp\, t_1+s_1\,\sharp\, b_1}{s_1\,\sharp\, t_1}\right),
$$
$$
\left(\displaystyle\frac{a_0}{s_0},  \,\displaystyle\frac{a_1}{s_1}\right)\cdot
\left(\displaystyle\frac{b_0}{t_0},  \,\displaystyle\frac{b_1}{t_1}\right):=
\left(\displaystyle\frac{a_0b_0}{s_0t_0},  \,\displaystyle\frac{(a_0b_1)\,\sharp\,(s_1t_0)+(s_0t_1)\,\sharp\,(a_1b_0)}{(s_0t_1)\,\sharp\, (s_1t_0)}\right)
$$
and
$$
\left(\displaystyle\frac{0}{1}, \,\displaystyle\frac{a_1}{s_1}\right) \,\sharp\,
\left(\displaystyle\frac{0}{1}, \,\displaystyle\frac{b_1}{t_1}\right):=
\left(\displaystyle\frac{0}{1}, \,\displaystyle\frac{a_1\,\sharp\,b_1}{s_1\,\sharp\,t_1}\right).
$$
Clearly, $\left(\displaystyle\frac{1}{1}, \,\displaystyle\frac{0}{1^{\sharp}}\right)$ is the identity of 
the ring $S^{\stackrel{\sharp}{-}1}R$, and 
$\left(\displaystyle\frac{0}{1},  \,\displaystyle\frac{1^{\sharp}}{1^{\sharp}}\right)$ is the local identity.

\medskip
Let $i^{\sharp S}_R: R\to S^{\stackrel{\sharp}{-}1}R$ be the map defined by
$$
i^{\sharp S}_R: a=a_0+a_1\mapsto \left(\displaystyle\frac{a_0}{1}, \,\displaystyle\frac{a_1}{1^{\sharp}}\right),
$$
where $a_i\in R_i$ for $i=0$ and $1$. Then $i^{\sharp S}_R$ is a triring homomorphism from $R$ to 
$S^{\stackrel{\sharp}{-}1}R$.

\medskip
The pair $\Big(S^{\stackrel{\sharp}{-}1}R, \, i^{\sharp S}_R \Big)$ constructed above is called the {\bf localization} of $R$ with respect to $S$, and $i^{\sharp S}_R$ is called the 
{\bf canonical triring homomorphism} from $R$ to $S^{\stackrel{\sharp}{-}1}R$. The next proposition gives the universal mapping characterization of the localization 
$\Big(S^{\stackrel{\sharp}{-}1}R, \, i^{\sharp S}_R \Big)$.

\begin{proposition}\label{pr3.1} Let $R=R_0\oplus R_1$ be a Hu-Liu triring. If 
$S=S_0\cup S_1$ is a multiplicative subset of $R$, then the 
localization $\Big(S^{\stackrel{\sharp}{-}1}R, \, i^{\sharp S}_R \Big)$ of $R$ with respect to $S$ has the following two properties.
\begin{description}
\item[(i)] For any $s_i\in S_i$ with $i=0$ and $1$, $i^{\sharp S}_R(s_0)$ is invertible in the commutative ring $\left(\Big(S^{\stackrel{\sharp}{-}1}R\Big)_0, +, \cdot \right)$ and $i^{\sharp S}_R(s_1)$ is invertible in the commutative ring $\left(\Big(S^{\stackrel{\sharp}{-}1}R\Big)_1, +, \,\sharp\, \right)$.
\item[(ii)] If $\psi: R\to \overline{R}$ is a triring homomorphism from the Hu-Liu triring $R$ to a Hu-Liu triring $\overline{R}$ such that $\psi (s_0)$ is invertible in the commutative ring 
$(\overline{R}_0, +, \cdot )$ and $\psi (s_1)$ is invertible in the commutative ring 
$(\overline{R}_1, +, \,\sharp\, )$ for any $s_i\in S_i$ with $i=0$ and $1$, then there exists a unique triring homomorphism
$\overline{\psi}: S^{\stackrel{\sharp}{-}1}R\to \overline{R}$ such that $\psi=\overline{\psi}\,i^{\sharp S}_R$.
\end{description}
\end{proposition}

\medskip
\noindent
{\bf Proof} A direct computation.

\hfill\raisebox{1mm}{\framebox[2mm]{}}

\bigskip
Let  $R=R_0\oplus R_1$ be a Hu-Liu triring. The following three types of localizations of $R$ are very useful in the study of Hu-Liu trirings: 
\begin{description}
\item[Type 1.] If $P=P_0\oplus P_1$ is a triideal of $R$ with $P_1\ne R_1$, then $(R_0\setminus P_0)\bigcup (R_1\setminus P_1)$ is a multiplicative subset of $R$. The localization of $R$ with respect to $(R_0\setminus P_0)\cup (R_1\setminus P_1)$ is called the {\bf localization} of $R$ at $P$ and is denoted by $R^{\sharp}_P$. 
\item[Type 2.] If $f_0\in R_0$ and $D^{\sharp}(f_0)\bigcap spec^{\sharp}_1R\ne\emptyset$, then
$T(f_0)=T_0(f_0)\cup T_1(f_0)$ is a multiplicative subset of $R$, where 
$T_0(f_0):=\{\, f_0^n\,|\, n\in\mathcal{Z}_{\ge 0}\,\}$
and
$$T_1(f_0):=\{\, z_1f_0^n\,|\,\mbox{ $n\in\mathcal{Z}_{\ge 0}$ and $z_1\in R_1\setminus P$ for all $P\in D^{\sharp}(f_0)\bigcap spec^{\sharp}_1R$}\,\}.$$
The localization of $R$ with respect to $T(f_0)=T_0(f_0)\cup T_1(f_0)$ is called the {\bf localization of $R$ at $f_0$} and is denoted by $R_{f_0}^{\sharp}$. Clearly, we have
$$
R_{f_0}^{\sharp}=\left\{\,\left.\frac{a_0}{f_0^n}+\frac{a_1}{z_1f_0^m}\,\right|\, \mbox{$m$, $n\ge 0$, $a_i\in R_i$ for $i=0$, $1$ and $z_1f_0^m\in T_1(f_0)$}\,\right\}.
$$
\item[Type 3.] If $f_1\in R_1$ and $f_1$ is not trinilpotent, then 
$\{\, 1\}\bigcup \{\, f_1^{\sharp\,m}\,\}_{m\ge 0}$
is multiplication subset of $R$. The localization of $R$ with respect to 
$\{\, 1\}\bigcup \{\, f_1^{\sharp\,m}\,\}_{m\ge 0}$ is called the {\bf localization of $R$ at $f_1$} and is denoted by $R_{f_1}^{\sharp}$. Thus, we have

$$
R_{f_1}^{\sharp}=\left\{\,\left.a_0+\frac{a_1}{f_1^{\sharp\,m}}\,\right|\, \mbox{$m$, $n\ge 0$ and $a_i\in R_i$ for $i=0$, $1$}\,\right\}.
$$
\end{description}

\bigskip
\section{Sheaf Structures on Trispectra}

We begin this section with the following

\begin{definition}\label{def8.1} Let $X$ be a topological space. A {\bf presheaf} $\mathcal{P}$ of Hu-Liu trirings on $X$ assigns a Hu-Liu triring $\mathcal{P}(U)$ to each open set $U$ of $X$ and a triring homomorphism: $\rho _{_{UV}}: \mathcal{P}(U)\to \mathcal{P}(V)$, called the {\bf restriction map}, to each inclusion of open sets $V\subseteq U$, subject to the conditions:
\begin{description}
\item[(i)] $\mathcal{P}(0)=\emptyset$;
\item[(ii)] $\rho _{_{UU}}: \mathcal{P}(U)\to \mathcal{P}(U)$ is the identity map;
\item[(iii)] If $W\subseteq V\subseteq U$ are three open subsets, then 
$\rho _{_{UW}}=\rho _{_{VW}}\rho _{_{UV}}$.
\end{description}
\end{definition}

\medskip
If $\mathcal{P}$ is a presheaf of Hu-Liu trirings on a  topological space $X$, then an element $s\in \mathcal{P}(U)$ is called a {\bf section} of $\mathcal{P}$ over the open set $U$.

\medskip
\begin{definition}\label{def8.2} A presheaf $\mathcal{P}$ of Hu-Liu trirings on a  topological space $X$ is called a {\bf sheaf} of  Hu-Liu trirings if it satisfies the following two conditions for an arbitrary open subset $U$ of $X$ and an arbitrary open covering 
$\{\,U_{(i)}\,|\, i\in\Delta\,\}$ of $U$:
\begin{description}
\item[(i)] If $s\in \mathcal{P}(U)$ and $\rho _{_{UU_{(i)}}}(s)=0$ for all $i\in\Delta$, then $s=0$;
\item[(ii)] If we have elements $s_{(i)}\in \mathcal{P}(U_{(i)})$ for each $i\in\Delta$ having the property that 
$\rho _{_{U_{(i)}, U_{(i)}\cap U_{(j)}}}(s_{(i)})=
\rho _{_{U_{(j)}, U_{(i)}\cap U_{(j)}}}(s_{(j)})$ for $i$, $j\in \Delta$, then there is an element $s\in \mathcal{P}(U)$ such that $\rho _{_{UU_{(i)}}}(s)=s_{(i)}$ for each $i\in\Delta$.
\end{description}
\end{definition}

\medskip
Let $(\, R=R_0\oplus R_1, \, + , \, \cdot , \, \,\sharp\, \,)$ be a Hu-Liu triring with the identity $1$ and the local identity $1^{\sharp}$. If $f_0\in R_0$ and $f_1\in R_1$, then we define
$$
\mathcal{O}(D^{\sharp}(f_0)):=\left\{\begin{array}{ll}
(R_0)_{f_0}\oplus R_0\, &\,\mbox{if $f_0\not\in \sqrt[\sharp]{0}$ and $D^{\sharp}(f_0)\bigcap spec^{\sharp}_1R=\emptyset$},\\
R_{f_0}^{\sharp}\oplus R_{1^{\sharp}f_0}^{\sharp}\, &\,\mbox{if $f_0\not\in \sqrt[\sharp]{0}$ and $D^{\sharp}(f_0)\bigcap spec^{\sharp}_1R\ne\emptyset$},\\
0\, &\,\mbox{if $f_0\in \sqrt[\sharp]{0}$}\end{array}\right.
$$
and
$$
\mathcal{O}(D^{\sharp}(f_1)):=\left\{\begin{array}{ll}R_{f_1}^{\sharp}\, &\,\mbox{if $f_1\not\in \sqrt[\sharp]{0}$},\\
0\, &\,\mbox{if $f_1\in \sqrt[\sharp]{0}$}\end{array}\right. ,
$$
where $(R_0)_{f_0}$ is the ordinary localization of the commutative ring $(R_0, \, +,\, \cdot)$ with respect to the multiplication set $\{\,f^n_0\,|\, n\in\mathcal{Z}_{\ge 0}\,\}$, and 
$(R_0)_{f_0}\oplus R_0$ is a Hu-Liu triring with zero odd part. We then have the following result.

\medskip
\begin{proposition}\label{pr8.1} Let $(\, R=R_0\oplus R_1, \, + , \, \cdot , \, \,\sharp\, \,)$ be a Hu-Liu triring. If $f_i$, $g_i\in R_i$  for $i=0$, $1$ and 
$D^{\sharp}(g_i)\subseteq D^{\sharp}(f_j)$ for $i$, $j\in\{\, 0,\, 1\,\}$, then there exist a triring homomorphism $\rho _{_{f_j,g_i}}: \mathcal{O}(D^{\sharp}(f_j))\to \mathcal{O}(D^{\sharp}(g_i))$ such that
\begin{description}
\item[(i)] $\rho _{_{f_j, f_j}}=id_{_{D^{\sharp}(f_j)}}$,
\item[(ii)] $\rho _{_{f_j, h_k}}=\rho _{_{g_i, h_k}}\rho _{_{f_j, g_i}}$ provided 
$D^{\sharp}(h_k)\subseteq D^{\sharp}(g_i)\subseteq D^{\sharp}(f_j)$,
\end{description}
where $i$, $j$, $k\in\{\, 0, \, 1\,\}$.
\end{proposition}

\medskip
\noindent
{\bf Proof} If $g_i\in \sqrt[\sharp]{0}$, then we define $\rho _{_{f_j,g_i}}: \mathcal{O}(D^{\sharp}(f_j))\to \mathcal{O}(D^{\sharp}(g_i))$ to be the zero homomorphism, i.e.,
\begin{equation}\label{eq56} 
\rho _{_{f_j,g_i}}(x):=0\quad\mbox{for $x\in \mathcal{O}(D^{\sharp}(f_j))$ and 
$g_i\in \sqrt[\sharp]{0}$}.
\end{equation}

\medskip
We now define $\rho _{_{f_j,g_i}}: \mathcal{O}(D^{\sharp}(f_j))\to \mathcal{O}(D^{\sharp}(g_i))$ for $g_i\not\in \sqrt[\sharp]{0}$ and $f_j\not\in \sqrt[\sharp]{0}$ by cases.

\medskip
\underline{\it Case 1}: $i=0$, $j=0$, in which case, we have
\begin{eqnarray}\label{eq57}
&&D^{\sharp}(Rg_0)=D^{\sharp}(g_0)\subseteq D^{\sharp}(f_0)=D^{\sharp}(Rf_0)\nonumber\\
&\Longrightarrow& V^{\sharp}(Rg_0)\supseteq V^{\sharp}(Rf_0)
\Longrightarrow \sqrt[\sharp]{Rg_0}\subseteq \sqrt[\sharp]{Rf_0}
\quad\mbox{(by Proposition~\ref{pr7.2} (i))}\nonumber\\
&\Longrightarrow& g_0^u=r_0f_0\quad\mbox{for some $u\in\mathcal{Z}_{>0}$ and $r_0\in R_0$.}
\end{eqnarray}

Since $D^{\sharp}(g_0)\bigcap spec^{\sharp}_1R\subseteq 
D^{\sharp}(f_0)\bigcap spec^{\sharp}_1R$, we have three subcases. 

\medskip
\underline{\it Case 1(i)}: $D^{\sharp}(g_0)\bigcap spec^{\sharp}_1R=\emptyset$ and
$D^{\sharp}(f_0)\bigcap spec^{\sharp}_1R=\emptyset$. In this case,
using (\ref{eq57}) and universal property of the ordinary localization $(R_0)_{f_0}$, we  get a triring homomorphism 
$$\rho _{_{f_0,g_0}}: \mathcal{O}(D^{\sharp}(f_0))=(R_0)_{f_0}\oplus R_0\to (R_0)_{g_0}\oplus R_0=\mathcal{O}(D^{\sharp}(g_0))$$
such that
\begin{equation}\label{eq591} 
\rho _{_{f_0,g_0}}\left(\frac{a_0}{f_0^n},\, b_0\right)
=\left(\frac{a_0r_0^n}{g_0^{nu}},\, b_0\right)
\qquad\mbox{for $a_0,\, b_0\in R_0$}.
\end{equation}

\medskip
\underline{\it Case 1(ii)}: $D^{\sharp}(g_0)\bigcap spec^{\sharp}_1R=\emptyset$ and
$D^{\sharp}(f_0)\bigcap spec^{\sharp}_1R\ne\emptyset$. In this case, we have a triring homomorphism 
$$\rho _{_{f_0,g_0}}: \mathcal{O}(D^{\sharp}(f_0))=R_{f_0}^{\sharp}\oplus R_{1^{\sharp}f_0}^{\sharp}\to (R_0)_{g_0}\oplus R_0=\mathcal{O}(D^{\sharp}(g_0))$$
such that
\begin{equation}\label{eq592} 
\rho _{_{f_0,g_0}}\left(\left(\frac{a_0}{f_0^n}+\frac{a_1}{z_1f_0^m},\,\,
b_0+\frac{b_1}{1^{\sharp}f_0^k}\right)\right)
=\left(\frac{a_0r_0^n}{g_0^{nu}},\, b_0\right),
\end{equation}
where $z_1f_0^m\in T_1(f_0)$, $n$, $m$, $k\in \mathcal{Z}_{\ge 0}$, $a_i$, $b_i\in R_i$ and $i=0$, $1$.

\medskip
\underline{\it Case 1(iii)}: $D^{\sharp}(g_0)\bigcap spec^{\sharp}_1R\ne\emptyset$ and
$D^{\sharp}(f_0)\bigcap spec^{\sharp}_1R\ne\emptyset$.
Using (\ref{eq57}) and universal property of the localization $R_{f_0}^{\sharp}$ and $R_{1^{\sharp}f_0}^{\sharp}$, 
we get two triring homomorphism $\overline{\phi}: R_{f_0}^{\sharp}\to R_{g_0}^{\sharp}$ and
$\overline{\psi}: R_{1^{\sharp}f_0}^{\sharp}\to R_{1^{\sharp}g_0}^{\sharp}$
such that
\begin{equation}\label{eq58} 
\overline{\phi}\left(\frac{a_0}{f_0^n}+\frac{a_1}{z_1f_0^m}\right)=
\frac{a_0r_0^n}{g_0^{nu}}+\frac{a_1r_0^m}{z_1g_0^{mu}}
\end{equation}
and
\begin{equation}\label{eq59} 
\overline{\psi}\left(b_0+\frac{b_1}{1^{\sharp}f_0^k}\right)=
b_0+\frac{b_1r_0^k}{1^{\sharp}g_0^{ku}},
\end{equation}
where $z_1f_0^m\in T_1(f_0)$, $n$, $m$, $k\in \mathcal{Z}_{\ge 0}$, $a_i$, $b_i\in R_i$ and $i=0$, $1$.

By (\ref{eq58}) and (\ref{eq59}), we get a triring homomorphism 
$$\rho _{_{f_0,g_0}}: \mathcal{O}(D^{\sharp}(f_0))=R_{f_0}^{\sharp}\oplus R_{1^{\sharp}f_0}^{\sharp}\to 
R_{g_0}^{\sharp}\oplus R_{1^{\sharp}g_0}^{\sharp}=\mathcal{O}(D^{\sharp}(g_0))$$ 
such that
\begin{eqnarray}\label{eq60}
&&\rho _{_{f_0,g_0}}\left(\left(\frac{a_0}{f_0^n}+\frac{a_1}{z_1f_0^m},\,\,
b_0+\frac{b_1}{1^{\sharp}f_0^k}\right)\right)\nonumber\\
&=&\left(\overline{\phi}\left(\frac{a_0}{f_0^n}+\frac{a_1}{z_1f_0^m}\right),\,\,
\overline{\psi}\left(b_0+\frac{b_1}{1^{\sharp}f_0^k}\right)\right)\nonumber\\
&=&\left(\frac{a_0r_0^n}{g_0^{nu}}+\frac{a_1r_0^m}{z_1g_0^{mu}},\,\,
b_0+\frac{b_1r_0^k}{1^{\sharp}g_0^{ku}}\right),
\end{eqnarray}
where $z_1f_0^m\in T_1(f_0)$, $n$, $m$, $k\in \mathcal{Z}_{\ge 0}$, $a_i$, $b_i\in R_i$ and $i=0$, $1$.

\medskip
\underline{\it Case 2}: $i=0$ and $j=1$, in which case, we have 
\begin{equation}\label{eq61} 
D^{\sharp}(g_0)\subseteq D^{\sharp}(f_1)\Longrightarrow g_0\in \sqrt[\sharp]{0}. 
\end{equation}
Hence, $\rho _{_{f_1,g_0}}=0$ in this case according to the definition given at the beginning.

\medskip
\underline{\it Case 3}: $i=1$ and $j=0$, in which case, 
\begin{eqnarray}\label{eq62}
&&D^{\sharp}(g_1)\subseteq D^{\sharp}(f_0)\nonumber\\
&\Longrightarrow& g_1^{\sharp v}=r_1f_0
\quad\mbox{for some $v\in\mathcal{Z}_{>0}$ and $r_1\in R_1$}
\end{eqnarray}
and $D^{\sharp}(f_0)\bigcap spec^{\sharp}_1R\supseteq D^{\sharp}(g_1)\ne\emptyset$
It follows that we have a triring homomorphism 
$$\rho _{_{f_0,g_1}}: \mathcal{O}(D^{\sharp}(f_0))
=R_{f_0}^{\sharp}\oplus R_{1^{\sharp}f_0}^{\sharp}\to 
R_{g_1}^{\sharp}=\mathcal{O}(D^{\sharp}(g_1))$$ 
such that
\begin{equation}\label{eq63}
\rho _{_{f_0,g_0}}\left(\left(\frac{a_0}{f_0^n}+\frac{a_1}{z_1f_0^m},\,\,
b_0+\frac{b_1}{1^{\sharp}f_0^k}\right)\right)
=b_0+\frac{b_1\,\sharp\,r_1^{\sharp k}}{g_1^{\sharp kv}},
\end{equation}
where $z_1f_0^m\in T_1(f_0)$, $n$, $m$, $k\in \mathcal{Z}_{\ge 0}$, $a_i$, $b_i\in R_i$ and $i=0$, $1$.

\medskip
\underline{\it Case 4}: $i=1$ and $j=1$, in which case, 
\begin{eqnarray}\label{eq64}
&&D^{\sharp}(g_1)\subseteq D^{\sharp}(f_1)\nonumber\\
&\Longrightarrow& g_1^{\sharp \theta}=t_1\,\sharp\, f_1
\quad\mbox{for some $\theta\in\mathcal{Z}_{>0}$ and $t_1\in R_1$.}
\end{eqnarray}
It follows that we have a triring homomorphism 
$$\rho _{_{f_1,g_1}}: \mathcal{O}(D^{\sharp}(f_1))=R_{f_1}^{\sharp}
\to R_{g_1}^{\sharp}=\mathcal{O}(D^{\sharp}(g_1))$$ 
such that
\begin{equation}\label{eq65}
\rho _{_{f_1,g_1}}\left(b_0+\frac{b_1}{f_1^{\sharp n}}\right)
=b_0+\frac{b_1\,\sharp\,t_1^{\sharp n}}{g_1^{\sharp \theta n}},
\end{equation}
where $n\in \mathcal{Z}_{\ge 0}$, $b_i\in R_i$ and $i=0$, $1$.

\medskip
It is clear that the triring homomorphisms $\rho _{_{f_j,g_i}}$ defined by (\ref{eq56}), (\ref{eq60}), (\ref{eq63}) and (\ref{eq65}) has the property (i).

\medskip
Note that if $h_k\in \sqrt[\sharp]{0}$, then 
the property (ii) holds because $\rho _{_{f_j, h_k}}=\rho _{_{g_i, h_k}}=0$. 
If $h_k\not\in \sqrt[\sharp]{0}$, then we have $k\ge i\ge j$ by (\ref{eq61}). Therefore, there are only four cases:
$$
(k, i, j)=(0, 0, 0)\,\, \mbox{or}\,\, (1, 0, 0)\,\, \mbox{or}\,\, (1, 1, 0)\,\, 
\mbox{or}\, (1, 1, 1).
$$
One can check that  the property (ii) also holds  for each of the four cases above.

\hfill\raisebox{1mm}{\framebox[2mm]{}}

\bigskip
Let $U$ be a nonempty open subset of $Spec^{\sharp}R$, and let
$$S_U:=\{\, D^{\sharp}(f_{(\alpha)}) \,|\, \emptyset\ne D^{\sharp}(f_{(\alpha)})\subseteq U\quad\mbox{and}\quad \alpha \in \Lambda _U\,\}$$
be the set of the nonempty basic open subsets contained in $U$. The set $\Lambda _U$ is a partial order set with respect to the following partial order:
$$
\alpha \ge \beta\quad\mbox{if and only if}\quad D^{\sharp}(f_{(\alpha)})\supseteq D^{\sharp}(f_{(\beta)}).
$$
$\left(\mathcal{O}(D^{\sharp}(f_{(\alpha)})), \, \rho _{_{f_{(\alpha)}, f_{(\beta)}}}\right)$ is an 
{\bf inverse system} on the set $\Lambda _U$, where the triring homomorphism 
$$\rho _{_{f_{(\alpha)}, f_{(\beta)}}}: \mathcal{O}(D^{\sharp}(f_{(\alpha)}))\to \mathcal{O}(D^{\sharp}(f_{(\beta)}))\quad\mbox{for $\alpha \ge \beta$}
$$
is defined by Proposition~\ref{pr8.1}. We define
\begin{equation}\label{eq66}
\mathcal{O}(U):=\lim_{\longleftarrow _{\Lambda _U}} \mathcal{O}(D^{\sharp}(f_{(\alpha)})),
\end{equation}
where $\displaystyle\lim_{\longleftarrow _{\Lambda _U}} \mathcal{O}(D^{\sharp}(f_{(\alpha)}))$ is the {\bf inverse limit} of the inverse system \newline
$\left(\mathcal{O}(D^{\sharp}(f_{(\alpha)})), \, \rho _{_{f_{(\alpha)}, f_{(\beta)}}}\right)$.
The inverse limit is a subtriring of the direct product 
$\displaystyle\prod _{\alpha \in \Lambda _U}\mathcal{O}(D^{\sharp}(f_{(\alpha)}))$, which is given by
\begin{equation}\label{eq67}
\displaystyle\lim_{\longleftarrow _{\Lambda _U}} \mathcal{O}(D^{\sharp}(f_{(\alpha)}))=
\left\{\,(x_{(\alpha)})_{\alpha \in \Lambda _U}\,\left| \begin{array}{c}
x_{(\alpha)}\in\mathcal{O}(D^{\sharp}(f_{(\alpha)}))\\\mbox{and 
$x_{(\beta)}=\rho _{_{f_{(\alpha)}, f_{(\beta)}}}(x_{(\alpha)})$}\\ \mbox{whenever 
$\alpha \ge \beta$}\end{array}
\right.\right\}.
\end{equation}
The even part and odd part of 
$\displaystyle\lim_{\longleftarrow _{\Lambda _U}} \mathcal{O}(D^{\sharp}(f_{(\alpha)}))$ are given by
\begin{equation}\label{eq68}
\left(\displaystyle\lim_{\longleftarrow _{\Lambda _U}} \mathcal{O}(D^{\sharp}(f_{(\alpha)}))\right)_i=
\left\{\,(x_{(\alpha)i})_{\alpha \in \Lambda _U}\,\left| \begin{array}{c}
x_{(\alpha)i}\in\left(\mathcal{O}(D^{\sharp}(f_{(\alpha)}))\right)_i\\\mbox{and 
$x_{(\beta)i}=\rho _{_{f_{(\alpha)}, f_{(\beta)}}}(x_{(\alpha)i})$}\\ \mbox{whenever 
$\alpha \ge \beta$}\end{array}
\right.\right\},
\end{equation}
where $i=0$, $1$. For each $\alpha \in \Lambda _U$, let 
$p^U_{(\alpha)}: \displaystyle\lim_{\longleftarrow _{\Lambda _U}} \mathcal{O}(D^{\sharp}(f_{(\alpha)}))\to \mathcal{O}(D^{\sharp}(f_{(\alpha)}))$ be the triring homomorphism defined by
\begin{equation}\label{eq69}
p^U_{(\alpha)}\left((x_{(\gamma)})_{\gamma \in \Lambda _U}\right):=x_{(\alpha)}.
\end{equation}
Clearly, we have
\begin{equation}\label{eq70}
p^U_{(\beta)}=\rho _{_{f_{(\alpha)}, f_{(\beta)}}}p^U_{(\alpha)}\quad\mbox{if $\alpha \ge \beta$ and $\alpha$, $\beta\in \Lambda _U$ }.
\end{equation}
The pair $\left(\displaystyle\lim_{\longleftarrow _{\Lambda _U}} \mathcal{O}(D^{\sharp}(f_{(\alpha)})),\, \{\,p^U_{(\alpha)}\,|\, \alpha\in \Lambda _U\,\}\right)$
has the following universal property:
\begin{quote}
Let $(\,X,\, \{\,q_{(\alpha)}\,|\, \alpha\in \Lambda _U\,\})$ be a pair consisting of a triring $X$ and a family of triring homomorphism 
$$\{\,q_{(\alpha)}: X\to \mathcal{O}(D^{\sharp}(f_{(\alpha)}))\,|\,\alpha\in \Lambda _U \,\}.$$ If 
$q_{(\beta)}=\rho _{_{f_{(\alpha)}, f_{(\beta)}}}q_{(\alpha)}$ whenever $\alpha \ge \beta$, then there exists a unique triring homomorphism $\sigma : X\to  \displaystyle\lim_{\longleftarrow _{\Lambda _U}} \mathcal{O}(D^{\sharp}(f_{(\alpha)}))$ such that $q_{(\alpha)}=p^U_{(\alpha)}\sigma$ for each $ \alpha\in \Lambda _U$.
\end{quote}
If $V\subseteq U$, then $S_V\subseteq S_U$ and $\Lambda _V\subseteq \Lambda _U$. Consider the pair 
$$
\left(\displaystyle\lim_{\longleftarrow _{\Lambda _U}} \mathcal{O}(D^{\sharp}(f_{(\alpha)})),\, \{\,p^U_{(\alpha)}\,|\, 
\alpha\in \Lambda _V\,\}\right).
$$
Note that (\ref{eq70}) is clearly true for $\alpha$, 
$\beta\in \Lambda _V\subseteq \Lambda _U$ with $\alpha \ge \beta$. By the universal property of the following pair
$$
\left(\displaystyle\lim_{\longleftarrow _{\Lambda _V}} \mathcal{O}(D^{\sharp}(f_{(\alpha)})),\, \{\,p^U_{(\alpha)}\,|\, 
\alpha\in \Lambda _V\,\}\right),
$$
we have a unique triring homomorphism
\begin{equation}\label{eq71}
\rho _{_{U, V}}: \mathcal{O}(U)=\lim_{\longleftarrow _{\Lambda _U}} \mathcal{O}(D^{\sharp}(f_{(\alpha)}))\to
\lim_{\longleftarrow _{\Lambda _V}} \mathcal{O}(D^{\sharp}(f_{(\alpha)}))=\mathcal{O}(V)
\end{equation}
such that 
\begin{equation}\label{eq72}
p^U_{(\alpha)}=p^V_{(\alpha)}\rho _{_{U, V}}\qquad\mbox{for $\alpha\in \Lambda _V$.}
\end{equation}
It follows from (\ref{eq71}) and (\ref{eq72}) that
\begin{equation}\label{eq73}
\rho _{_{U, U}}=id_{\mathcal{O}(U)}\qquad\mbox{for each open subset $U$ of $Spec^{\sharp}_R$}
\end{equation}
and
\begin{equation}\label{eq74}
\rho _{_{U, W}}=\rho _{_{V, W}}\rho _{_{U, V}}\qquad\mbox{for three open subsets 
$W\subseteq V\subseteq U$.}
\end{equation}

By (\ref{eq73}) and (\ref{eq74}), $\mathcal{O}$ is a presheaf with the restriction map 
$\rho _{_{U, V}}$ given by (\ref{eq71}). This  presheaf is called the {\bf structure presheaf}
on $Spec^{\sharp}R$.

\medskip
\begin{proposition}\label{pr8.2} If $(\, R=R_0\oplus R_1, \, + , \, \cdot , \, \,\sharp\, \,)$ is a Hu-Liu triring, then the structure presheaf on $Spec^{\sharp}R$ is a sheaf of Hu-Liu trirings.
\end{proposition}

\medskip
\noindent
{\bf Proof} According to the definition of the structure presheaf on $Spec^{\sharp}R$, it suffices to prove that the properties (i) and (ii) in Definition~\ref{def8.2} holds for an arbitrary basic open set and an arbitrary open covering which consists of basic open sets.
Let $D^{\sharp}(f_i)$ be an arbitrary basic open set of $Spec^{\sharp}R$. Consider the following 
arbitrary open covering  of $D^{\sharp}(f_i)$
\begin{equation}\label{eq75}
D^{\sharp}(f_i)=\bigg(\bigcup _{\alpha\in\Lambda _0}D^{\sharp}(g_{(\alpha)0})\bigg)\bigcup
\bigg(\bigcup _{\beta\in\Lambda _1}D^{\sharp}(g_{(\beta)1})\bigg),
\end{equation}
where $f_0$, $g_{(\alpha)0}\in R_0$ for $\alpha\in\Lambda _0$, 
$f_1$, $g_{(\beta)1}\in R_1$ for $\beta\in\Lambda _1$, $\Lambda _0$ and $\Lambda _1$ are two index sets.

\bigskip
To prove the properties (i) in Definition~\ref{def8.2}, we need to prove that if 
$s\in \mathcal{O}(D^{\sharp}(f_i))$ satisfies
\begin{equation}\label{eq76}
\rho _{_{f_{i}, g_{(\alpha)0}}}(s)=0\qquad\mbox{for $\alpha\in\Lambda _0$}
\end{equation}
and
\begin{equation}\label{eq77}
\rho _{_{f_{i}, g_{(\beta)1}}}(s)=0\qquad\mbox{for $\beta\in\Lambda _1$},
\end{equation}
then $s=0$.

\medskip
\underline{\it Case 1}: $i=0$, in which case, $\Lambda _0\ne \emptyset$. Since 
$s\in \mathcal{O}(D^{\sharp}(f_0))$, we have
\begin{equation}\label{eq78}
s=\left(\frac{a_0}{f_0^n}+\frac{a_1}{z_1f_0^m},\,\,
b_0+\frac{b_1}{1^{\sharp}f_0^k}\right)
\end{equation}
for some $n$, $m$, $k\in \mathcal{Z}_{\ge 0}$, $z_1f_0^m\in T_1(f_0)$, $a_i$, $b_i\in R_i$ and $i=0$, $1$. Since 
$D^{\sharp}(g_{(\alpha)0})\subseteq D^{\sharp}(f_0)$, there exist some $r_{(\alpha)0}\in R_0$ and $u_{(\alpha)}\in\mathcal{Z}_{>0}$ such that
\begin{equation}\label{eq79}
g_{(\alpha)0}^{u_{(\alpha)}}=r_{(\alpha)0}f_0\quad\mbox{for each $\alpha\in\Lambda _0$. }
\end{equation}

$\bullet$ If $D^{\sharp}(f_{0})\bigcap Spec^{\sharp}_1R\ne\emptyset$, then for each 
$\alpha\in\Lambda _0$, we have 
$$\mbox{either}\quad
D^{\sharp}(g_{(\alpha)0})\bigcap Spec^{\sharp}_1R\ne\emptyset\quad \mbox{or}\quad
D^{\sharp}(g_{(\alpha)0})\bigcap Spec^{\sharp}_1R=\emptyset.
$$
Using (\ref{eq60}), (\ref{eq592}), (\ref{eq76}) and (\ref{eq79}), we have
\begin{eqnarray}\label{eq80}
&&0=\rho _{_{f_{0}, g_{(\alpha)0}}}(s)=
\rho _{_{f_{0}, g_{(\alpha)0}}}\left(\left(\frac{a_0}{f_0^n}+\frac{a_1}{z_1f_0^m},\,\,
b_0+\frac{b_1}{1^{\sharp}f_0^k}\right)\right)\nonumber\\
&=& \left\{\begin{array}{ll}
\left(\frac{a_0r_{(\alpha)0}^n}{g_0^{nu_{(\alpha)}}}+
\frac{a_1r_{(\alpha)0}^m}{z_1g_0^{mu_{(\alpha)}}},\,\,
b_0+\frac{b_1r_{(\alpha)0}^k}{1^{\sharp}g_0^{ku_{(\alpha)}}}\right)&\mbox{if $D^{\sharp}(g_{(\alpha)0})\bigcap Spec^{\sharp}_1R\ne\emptyset$}\\&\\
\left(\frac{a_0r_{(\alpha)0}^n}{g_0^{nu_{(\alpha)}}},\,\,b_0\right)&
\mbox{if $D^{\sharp}(g_{(\alpha)0})\bigcap Spec^{\sharp}_1R=\emptyset$}
\end{array}\right.
\end{eqnarray}

$\bullet$ If $D^{\sharp}(f_{0})\bigcap Spec^{\sharp}_1R=\emptyset$, then we have 
$D^{\sharp}(g_{(\alpha)0})\bigcap Spec^{\sharp}_1R=\emptyset$ for each 
$\alpha\in\Lambda _0$. By (\ref{eq592}), (\ref{eq76}) and (\ref{eq79}), we have
\begin{eqnarray}\label{eq80a}
0&=&\rho _{_{f_{0}, g_{(\alpha)0}}}(s)=
\rho _{_{f_{0}, g_{(\alpha)0}}}\left(\left(\frac{a_0}{f_0^n}+\frac{a_1}{z_1f_0^m},\,\,
b_0+\frac{b_1}{1^{\sharp}f_0^k}\right)\right)\nonumber\\
&=& \left(\frac{a_0r_{(\alpha)0}^n}{g_0^{nu_{(\alpha)}}},\,\,b_0\right)\quad
\mbox{for each $\alpha\in\Lambda _0$}
\end{eqnarray}

Note  that
\begin{equation}\label{eq80b} 
D^{\sharp}(g_{(\alpha)0})\bigcap Spec^{\sharp}_1R=\emptyset\Longrightarrow
1^{\sharp}g_{(\alpha)0}^b=0\quad\mbox{for some $b\in \mathcal{Z}_{\ge 0}$}
\end{equation}


It follows from (\ref{eq80}), (\ref{eq80a}), (\ref{eq80b}) and Proposition~\ref{pr7.2} (i) that there exist 
$t$, $d$, $\delta _{\alpha _i}\in \mathcal{Z}_{>0}$,
$ \alpha _i\in \Lambda _0$ and $h_{(\alpha _i)0}\in R_0$  such that
\begin{equation}\label{eq81} 
f_0^t=\sum_{i=1}^d h_{(\alpha _i)0}\,\,g_{(\alpha _i)0}^{\delta _{\alpha _i}}
\end{equation}
and
\begin{equation}\label{eq82} 
f_0^ta_0=(1^{\sharp}f_0^t)\,\sharp\, a_1=b_0=(1^{\sharp}f_0^t)\,\sharp\, b_1=0.
\end{equation}

By (\ref{eq82}), $\displaystyle\frac{a_0}{f_0^n}+\displaystyle\frac{a_1}{1^{\sharp}f_0^m}=0$ and 
$b_0+\displaystyle\frac{b_1}{1^{\sharp}f_0^k}=0$. This proves $s=0$. 

\medskip
\underline{\it Case 2}: $i=1$, in which case, $s\in \mathcal{O}(D^{\sharp}(f_1))$. Thus 
$s=a_0+\displaystyle\frac{a_1}{f_1^{\sharp m}}$ for some $a_i\in R_i$ and 
$m\in\mathcal{Z}_{\ge 0}$. By $(\ref{eq61})$, $D^{\sharp}(g_{(\alpha)0)})=\emptyset $ for 
$\alpha\in\Lambda _0$ in this case. Thus, (\ref{eq75}) becomes
\begin{equation}\label{eq83} 
D^{\sharp}(f_1)=\bigcup _{\beta\in\Lambda _1}D^{\sharp}(g_{(\beta)1}).
\end{equation}
Since $D^{\sharp}(g_{(\beta)1})\subseteq D^{\sharp}(f_1)$, there exist some $r_{(\beta)1}\in R_1$ and $\theta _{(\beta)}\in\mathcal{Z}_{>0}$ such that
\begin{equation}\label{eq84}
g_{(\beta)1}^{\theta _{(\beta)}}=r_{(\beta)1}\,\sharp\,f_1\quad\mbox{for each 
$\beta\in\Lambda _1$. }
\end{equation}

Using (\ref{eq65}) and (\ref{eq84}), we have
$$
0=\rho _{_{f_1, g_{(\beta)1}}}(s)
=\rho _{_{f_1, g_{(\beta)1}}}\left(a_0+\displaystyle\frac{a_1}{f_1^{\sharp m}}\right)
=a_0+\displaystyle\frac{a_1\,\sharp\,r_{(\beta)1}^{\sharp m}}
{g_{(\beta)1}^{\sharp m\theta _{(\beta)}}},
$$
which implies that there exist 
$t'$, $d'$, $\delta _{\beta _i}'\in \mathcal{Z}_{>0}$,
$ \beta _i\in \Lambda _1$ and $h_{(\beta _i)0}\in R_0$  such that
\begin{equation}\label{eq85} 
f_1^{t'}=\sum_{i=1}^{d'} h_{(\beta _i)1}\,\,g_{(\beta _i)1}^{\delta _{\beta _i}'}
\end{equation}
and
\begin{equation}\label{eq86} 
a_0=f_1^{\sharp t'}\,\sharp\, a_1=0.
\end{equation}
Hence, $s=0$ by (\ref{eq86}).

\medskip
This proves that the properties (i) in Definition~\ref{def8.2} holds for the basic open set $D^{\sharp}(f_i)$ and the open covering (\ref{eq75}). 

\bigskip
In order to prove that the properties (ii) in Definition~\ref{def8.2} holds for the basic open set $D^{\sharp}(f_i)$ and its open covering
\begin{equation}\label{eq87}
\{\, D^{\sharp}(g_{(\alpha)0})\,|\, \alpha\in\Lambda _0\,\}\bigcup
\{\, D^{\sharp}(g_{(\beta)1})\,|\, \beta\in\Lambda _1\,\},
\end{equation}
it is sufficient to prove that the properties (ii) in Definition~\ref{def8.2} holds for the basic open set $D^{\sharp}(f_i)$ and a finite subcovering of the open covering given by (\ref{eq87}). A direct computation proves  that this fact is indeed true. 

\medskip
This completes the proof of  Proposition~\ref{pr8.2}.

\hfill\raisebox{1mm}{\framebox[2mm]{}}


\bigskip
