Pierce sheaves of pseudo EMV-algebras

Abstract

We study a Pierce sheaf representation of pseudo EMV-algebras which are a non-commutative generalization of MV-algebras, pseudo MV-algebras and of generalized Boolean algebras, so that the top element is not assumed a priori. We present one sheaf using a Boolean type of representation and the main results are concerning the Hausdorff sheaf representation of representable pseudo EMV-algebras. For this aim, we study also the space of maximal ideals not necessarily normal and the space of minimal prime ideals to give conditions when these spaces are compact/locally compact in the hull-kernel topology and when they imply existence of a top element.

Appendix

In this section, we present a representable pseudo EMV-algebra without top element having countably many maximal ideals and each maximal ideal contains exactly one minimal prime ideal.

(1) Let $M_1,\ldots ,M_n$ be linearly ordered pseudo MV-algebras. Set $M=M_1\times \cdots \times M_n$ and look at M as at a pseudo EMV-algebra. Let $I_i$ be a maximal ideal of $M_i$. Then $\{0_i\}$ is a unique minimal prime ideal of $M_i$. We assert that M has exactly n maximal ideals ${{\hat{I}}}_i=M^i_1\times \cdots \times M^i_{n}$, where $M^i_k=M_k$ if $i\ne k$ otherwise $M^i_i=I_i$, where $i=1,\ldots ,n$. Every ${{\hat{I}}}_i$ is normal.

Using the criterion (Dvurečenskij and Zahiri 2019d, Prop 5.13(iv)), we show that M has exactly n minimal prime ideals $P_i=A^i_1\times \cdots \times A^i_{n}$, where $A^i_k=M_k$ if $i\ne k$ and $A^i_i=\{0_i\}$ for $i=1,\ldots ,n$: Clearly every $P_i$ is a prime ideal, and using the criterion, we see that every $P_i$ is a minimal prime ideal.

Let P be a minimal prime ideal of M. There is a unique maximal ideal ${{\hat{I}}}_i$, say ${{\hat{I}}}_1$, such that $P\subseteq {{\hat{I}}}_1$. We show that $P=P_1$. Suppose the converse. Then there is $y=(y_1,\ldots ,y_n)\in P_1\setminus P$ with $y_1=0$. We claim that if $(x_1,\ldots ,x_n)\in P$ and $x_1>0$, then there are $x_k>0$ for all $k=2,\ldots ,n$. If $x_k=0$ for all $k=2,\ldots ,n$, then for $x=(1_1,0_2,\ldots ,0_n)$, we have $x, y\notin P$ but $x\wedge y=0$, contradicting that P is prime, so that at least one $x_k>0$. Assume that for some $k=2,\ldots ,n$, we have $x_k=0$. Then either, for each $x_1>0$, $x_k=0$, or there is $x_k>0$. In the first case, we have $P\subseteq {{\hat{I}}}_k$, which is impossible because, ${{\hat{I}}}_1$, is a unique maximal ideal of M containing P. Therefore, there exists $x_k\in M_k$ such that $x_k>0$. Whence, for each $k=2,\ldots ,n$, there exist $x^k_1>0$ in $M_1$ and $x^k_k>0$ in $M_k$ such that ${{\hat{x}}}_k=(x^k_1,\ldots ,x^k_k,\ldots , x^k_n)\in P$. Putting $x={{\hat{x}}}_1\vee \cdots \vee {{\hat{x}}}_k$, we have $x\in P$ and all its coordinates are strictly positive, as claimed.

If $z=(z_1,\ldots ,z_n)\in M$ is such that $z\in x^\bot$, we have $z=(0_1,\ldots ,0_n)\in P$ contradicting by Dvurečenskij and Zahiri (2019d, Prop 5.13(iv)) that P is minimal prime. Whence, $P=P_1$, and $P_1,\ldots ,P_n$ are only minimal prime ideals of M.

Let Q be a prime ideal of M, then there is a unique $i=1,\ldots ,n$ such that $Q\subseteq {{\hat{I}}}_i$, consequently $P_i\subseteq Q\subseteq {{\hat{I}}}_i$, so that $Q\in \{P_i,{{\hat{I}}}_i\}$ and $Spec(M)= MaxI(M)\cup MinP(M)$, where MinP(M) is the set of minimal prime ideals of M.

By Proposition 6.3, every mapping $\phi _x:S(x)\rightarrow \Gamma (x)$ is continuous.

(2) Let $(M_n)_n$ be a sequence of linearly ordered pseudo MV-algebras. Define a sum $M=\sum _n M_n$, which consists of all sequences $(x_n)_n\in \prod _n M_n$ which have all $x_n$’s but finitely many nonzero elements. Then M is a pseudo EMV-algebra, see Dvurečenskij and Zahiri (2019e), which is representable. Every $M_n$ has a unique maximal ideal $J_n$ which is also normal. Given $n\ge 1$, let $s_n$ be a unique state-morphism on $M_n$ whose kernel is $J_n$. Let $\pi _i:M \rightarrow M_i$ be the ith projection. Then $\mu _i((x_n)_n)=s_i(\pi _i((x_n)_n))$ is a state-morphism on M and its kernel $I_i=\{x=(x_n)_n\in M:s_i(\pi _i(x))=0\}$ is a maximal and normal ideal of M. It is possible to show that $I_i=\sum _n M^i_n$, where $M^i_n = M_n$ if $i\ne n$, otherwise, $M^i_n=J_i$.

We assert that $MaxI(M)=\{I_i :i \in {\mathbb {N}}\}$. Let $\mu$ be a state-morphism on M. There is an element $x_0=(x_n)_n$ such that $\mu (x)=1$. Since there is $n_0$ such that $x_n=0$ for each $n>n_0$, we can assume that $x_i=1_i$ for all $i=1,\ldots ,n_0$, where $1_i$ is a top element in $M_i$. Define the direct product $M_0=\prod _{n=1}^{n_0}M_n$. By (1), it has exactly $n_0$ maximal ideals ${{\hat{I}}}_i=M^i_1\times \cdots \times M^i_{n_0}$, where $M^i_n=M_n$ if $n\ne i$ otherwise $M^i_n=I_n$ for $n=1,\ldots ,n_0$, and $M_0$ has exactly $n_0$ state-morphisms $t_i$, and every state on $M_0$ is a convex combination of $t_1,\ldots ,t_{n_0}$.

Since $\mu$ is a state-morphism, we have $\mu (x\wedge x_0)=\mu (x)$ for each $x\in M$, i.e., we can assume that $x=(x_n)_n$ has all $x_n=0$ for each $n\ge n_0$. Whence $\mu$ induces a state-morphism ${\hat{\mu }}$ on $M_0$ such that ${\hat{\mu }}(y_1,\ldots ,y_{n_0})=\mu (y_1,\ldots ,y_{n_0},0,\ldots ,0, \ldots )$. Then ${\hat{\mu }}$ has to coincide with a unique $t_{i_0}$ for some $i_0=1,\ldots ,n_0$. Let $x=(x_1,\ldots ,x_{n_0}, 0,\ldots ,0, \ldots )$ and define ${{\hat{x}}}_i$ as a sequence where ith coordinate is $x_i$ and all other coordinates are zeros, $i=1,\ldots ,n_0$. Then $x={{\hat{x}}}_1\vee \cdots \vee {{\hat{x}}}_{n_0}$, so that $\mu (x)=\bigvee _{i=1}^{n_0} \mu ({{\hat{x}}}_i)$. But $\mu ({{\hat{x}}}_i)=0$ for $i\ne i_0$ and $\mu ({{\hat{x}}}_{i_0})= t_{i_0}({{\hat{x}}}_{i_0})=\mu _{i_0}({{\hat{x}}}_{i_0})= \mu _{i_0}(x)$ and $\mu = \mu _{i_0}$.

We note that $M_0$ has also $n_0$ minimal prime ideals $P^{n_0}_i=A^i_1\times \cdots \times A^i_{n_0}$, where $A^i_n=M_n$ if $n\ne i$ and $A^i_i=\{0_i\}$ for $i=1,\ldots ,n_0$, and, they are only minimal prime ideals of $M_0$, see part (1).

For every $i\ge 1$, let $P_i=\sum _n Q^i_n$, where $Q^i_n=M_n$ if $n\ne i$ and $Q^i_i=\{0_i\}$. We assert that each $P_i$ is a minimal prime ideal. Let $x=(x_n)_n,y=(y_n)_n\in M$ and $x\wedge y \in P_i$. We can assume that for each $n>n_0$, $x_n=y_n=0$. Then $(x_1,\ldots ,x_{n_0})\wedge (y_1,\ldots ,y_{n_0})\in P^{n_0}_i$, so that one of them belongs to $P^{n_0}_i$, because it is a prime ideal of $M_0$. Consequently, x or y belongs to $P_i$. Now, we show that it is minimal prime. Using criterion (Dvurečenskij and Zahiri 2019d, Prop 5.13(iv)), let $x =(x_n)_n\in P_i$ and $y=(y_n)_n\in x^\bot$, and again, let $x_n=0=y_n$ for each $n\ge 0$. Since $P^{n_0}_i$ is minimal prime, we find $(y_1,\ldots ,y_{n_0})\in (x_1,\ldots ,x_{n_0})^\bot$ such that $(y_1,\ldots ,y_{n_0})\notin P^{n_0}_i$. Whence, there is $y\in x^\bot$ such that $y\notin P_i$.

Let P be an arbitrary minimal prime ideal of M. There is a maximal ideal $I_i$ of M, say $I_1$, containing P. We show that $P=P_1$. Let $n_0$ be any natural number and let $P_{n_0}$ be the projection of P onto $M_1\times \cdots \times M_{n_0}$. Then $P_{n_0}$ is a prime ideal of $M_0$ and it is embedded into the maximal ideal $I^{n_0}_1$, which yields $P^{n_0}_1 \subseteq P_{n_0}$. Let $P'$ be the set of all $x=(x_n)_n\in M$ whose projection is in $P^{n_0}_1$. It is an ideal of M. We show that it is prime. Indeed, let $x=(x_n)$ and $y=(y_n)_n$ be such that $x\wedge y = 0=(0_n)_n$. Then the projection of x and y is $(0_1,\ldots ,0_{n_0})$, so that either $(x_1,\ldots ,x_{n_0})$ or $(y_1,\ldots ,y_{n_0})$ lies in $P^{n_0}_1$. Consequently, x or y lies in $P'$ and $P'$ is a prime and contained in the minimal prime P, whence $P'=P$. This is true for each $n_0$, so that the projection of P onto each $M_1\times \cdots \times M_{n_0}$ is $P^{n_0}_1$ which means that $P=P_1$.

In particular, we have that each maximal ideal $I_i$ contains only one minimal prime ideal, $P_i$, for each $i\ge 1$.

Now, let Q be a prime ideal of M. There is a unique integer $i\ge 1$ such that $P_i\subseteq Q \subseteq I_i$. Then $Q\in \{P_i,I_i\}$ so that $Spec(M)=MaxI(M)\cup MinP(M)$.

Take $x=(x_n)_n \in M$. There is $n_0$ such that $x_n=0$ for each $n\ge n_0$. Take a maximal ideal $I_i\in MaxI(M)$. Then $x\notin I_i$ iff $x_i\in M_i\setminus J_i$ for some $i=1,\ldots ,n_0$. Whence, the set $\{I\in MaxI(M):x\notin I\}$ is finite. In addition a minimal prime ideal $P_i$ satisfies $x\notin P_i$ iff $x_i>0$ and $i=1,\ldots ,n_0$. Therefore, $S(x)=\{Q \in Spec(M):x\notin Q\}$ is finite.

Let $\phi :Spec(M)\rightarrow MaxI(M)$ be a mapping such that $\phi (Q)$ is a unique maximal ideal of M containing Q. Then $\phi (P_n)=I_n$ and $\phi (I_n)=I_n$ for each $n\ge 1$. Whence, if $c\in M$ and let $U(c)=\{I\in MaxI(M) :c\notin I\}$ be given; it is an open set in MaxI(M). Without loss of generality, we can assume that $c=(c_n)_n$ with $c_n=0$ for each $n\ge n_0$. Then $\phi ^{-1}(U(c))$ is a finite subset of Spec(M), so it is finite. By Lemma 4.1, S(x) is clopen and compact, so that $\phi ^{-1}(U(c))$ is open proving $\phi$ is continuous.