Pierce stalks in preprimal varieties

D. Vaggione and W. J. Zuluaga Botero

Abstract

An algebra $\mathbf{P}$ is called preprimal if $\mathbf{P}$ is finite and $\mathop{\mathrm{C}lo}(\mathbf{P})$ is a maximal clone. A preprimal variety is a variety generated by a preprimal algebra. After Rosenberg’s classification of maximal clones [13]; we have that a finite algebra is preprimal if and only if its term operations are exactly the functions preserving a relation of one of the following seven types:

1.

Permutations with cycles all the same prime length,
2.

Proper subsets,
3.

Prime-affine relations,
4.

Bounded partial orders,
5.

$h$ -adic relations,
6.

Central relations $h\geq 2$ ,
7.

Proper, non-trivial equivalence relations.

In [10] Knoebel studies the Pierce sheaf of the different preprimal varieties and he asks for a description of the Pierce stalks. He solves this problem for the cases 1.,2. and 3. and left open the remaining cases. In this paper, using central element theory we succeeded in describing the Pierce stalks of the cases 6. and 7..

1 Introduction

An algebra $\mathbf{P}$ is called preprimal if $\mathbf{P}$ is finite and $\mathop{\mathrm{C}lo}(\mathbf{P})$ is a maximal clone. A preprimal variety is a variety generated by a preprimal algebra.

After Rosenberg’s classification of maximal clones [13]; we have that a finite algebra is preprimal if and only if its term operations are exactly the functions preserving a relation of one of the following seven types:

1.

Permutations with cycles all the same prime length,
2.

Proper subsets,
3.

Prime-affine relations,
4.

Bounded partial orders,
5.

$h$ -adic relations,
6.

Central relations $h\geq 2$ ,
7.

Proper, non-trivial equivalence relations.

In [10] Knoebel studies the Pierce sheaf ([12], [6]) of the different preprimal varieties and he asks for a description of the Pierce stalks. He solves this problem for the cases 1.,2. and 3. and left open the remaining cases.

In this paper, using central element theory we succeeded in describing the Pierce stalks of the cases 6. and 7..

2 Notation and some basic results

Let $\{\mathbf{A}_{i}\}_{i\in I}$ be a family of algebras of the same type. If there is no place to confusion, we write $\prod\mathbf{A}_{i}$ in place of $\prod_{i\in I}\mathbf{A}_{i}$ . For $x,y\in\prod A_{i}$ , let $E(x,y)=\left\{i\in I:x(i)=y(i)\right\}$ . A subdirect product $\mathbf{A}\leq\prod\mathbf{A}_{i}$ is global [11] if there is a topology $\tau$ on $I$ such that $E(x,y)\in\tau$ , for every $x,y\in A$ , and

-

(Patchwork Property) For every $\{U_{r}:r\in R\}\subseteq\tau$ and $\{x_{r}:r\in R\}\subseteq A$ , if $\bigcup\{U_{r}:r\in R\}=I$ and $U_{r}\cap U_{s}\subseteq E(x_{r},x_{s})$ , for every $r,s\in R$ , then there exists $x\in A$ such that $U_{r}\subseteq E(x,x_{r})$ , for every $r\in R$ .

Global subdirect products were introduced in [11] as a universal algebra counterpart to sheaves, where it is proved that a global subdirect product and an algebra of global sections of a sheaf are the same thing.

For an algebra $\mathbf{A}$ we use $\nabla^{\mathbf{A}}$ to denote the universal congruence on $\mathbf{A}$ and $\Delta^{\mathbf{A}}$ to denote the trivial congruence on $\mathbf{A}$ . We use $\mathrm{Con}(\mathbf{A})$ to denote the congruence lattice of $\mathbf{A}$ . We use $\theta^{\mathbf{A}}(a,b)$ to denote the principal congruence of $\mathbf{A}$ generated by $(a,b)$ . If $\vec{a},\vec{b}\in A^{n}$ , we use $\theta^{\mathbf{A}}(\vec{a},\vec{b})$ to denote the congruence $\bigvee_{k=1}^{n}\theta^{\mathbf{A}}(a_{k},b_{k})$ .

Lemma 1

If $\sigma:\mathbf{A}\rightarrow\mathbf{B}$ is a homomorphism and $(x,y)\in\theta^{\mathbf{A}}(\vec{a},\vec{b})$ , then $(\sigma(x),\sigma(y))\in\theta^{\mathbf{B}}(\sigma(\vec{a}),\sigma(\vec{b}))$ .

Proof. Folklore.

If $\mathbf{A}\leq\prod\mathbf{A}_{i}$ , for each $i\in I$ , let $\pi_{i}^{\mathbf{A}}:\mathbf{A}\rightarrow\mathbf{A}_{i}$ be the canonical projection. We use $\theta_{i}^{\mathbf{A}}$ to denote the congruence $\ker\pi_{i}^{\mathbf{A}}$ .

Given a class $\mathcal{K}$ of algebras, we use $\mathbb{I}(\mathcal{K})$ , $\mathbb{H}(\mathcal{K})$ , $\mathbb{S}(\mathcal{K})$ , $\mathbb{P}_{u}(\mathcal{K})$ and $\mathbb{V}(\mathcal{K})$ to denote the classes of isomorphic images, homomorphic images, subalgebras, ultraproducts of elements of $\mathcal{K}$ and the variety generated by $\mathcal{K}$ . For a variety $\mathcal{V}$ , we write $\mathcal{V}_{SI}$ and $\mathcal{V}_{DI}$ to denote the classes of subdirectly irreducible and directly indecomposable members of $\mathcal{V}$ .

Given a variety $\mathcal{V}$ and a set $X$ of variables we use $\mathbf{F}_{\mathcal{V}}(X)$ for the free algebra of $\mathcal{V}$ freely generated by $X$ .

Lemma 2

Let $\mathcal{V}$ be a variety and let $p_{k}(\vec{z},\vec{w}),q_{k}(\vec{z},\vec{w})$ , $k=1,...,n$ , be terms in the language of $\mathcal{V}$ . Let $X=\{x,y,\vec{z},\vec{w}\}$ ,

\theta=\bigvee_{k=1}^{n}\theta^{\mathbf{F}_{\mathcal{V}}(X)}(p_{k}^{\mathbf{F}_{\mathcal{V}}(X)}(\vec{z},\vec{w}),q_{k}^{\mathbf{F}_{\mathcal{V}}(X)}(\vec{z},\vec{w}))

and $\mathbf{H}=\mathbf{F}_{\mathcal{V}}(X)/\theta$ . Let $\mathbf{A}\in\mathcal{V}$ and suppose that $p_{k}^{\mathbf{A}}(\vec{e},\vec{a})=q_{k}^{\mathbf{A}}(\vec{e},\vec{a})$ , for $k=1,...,n$ . Then, for every $a,b\in A$ , there exists a unique homomorphism $\Omega:\mathbf{H}\rightarrow\mathbf{A}$ , such that $\Omega(x/\theta)=a$ , $\Omega(y/\theta)=b$ , $\Omega(\vec{z}/\theta)=\vec{e}$ and $\Omega(\vec{w}/\theta)=\vec{a}$ .

Proof. The proof is straightforward. The details are left to the reader.

A variety $\mathcal{V}$ has the Fraser-Horn property (FHP, for short) [8] if for every $\mathbf{A}_{1},\mathbf{A}_{2}\in\mathcal{V}$ , it is the case that every congruence $\theta$ in $\mathbf{A}_{1}\times\mathbf{A}_{2}$ is the product congruence $\theta_{1}\times\theta_{2}$ for some congruences $\theta_{1}$ of $\mathbf{A}_{1}$ and $\theta_{2}$ of $\mathbf{A}_{2}$ .

Let $\mathcal{L}$ be a first order language. If a $\mathcal{L}$ -formula $\varphi(\vec{x})$ has the form

{\exists}{\vec{w}}\bigwedge_{j=1}^{n}p_{j}(\vec{x},\vec{w})=q_{j}(\vec{x},\vec{w}),

for some positive number $n$ and terms $p_{j}(\vec{x},\vec{w})$ and $q_{j}(\vec{x},\vec{w})$ in $\mathcal{L}$ , then we say that $\varphi(\vec{x})$ is a ( $\exists\bigwedge p=q$ )-formula. In a similar manner we define ( $\forall\bigwedge p=q$ ) and ( $\forall\exists\bigwedge p=q$ )-formulas.

Let $\mathcal{L}$ be a first order language and $\mathcal{K}$ be a class of $\mathcal{L}$ -structures. If $R\in\mathcal{L}$ is an $n$ -ary relation symbol, we say that a formula $\varphi(x_{1},...,x_{n})$ defines $R$ in $\mathcal{K}$ if

\mathcal{K}\vDash\varphi(\vec{x})\leftrightarrow R(\vec{x})\text{.}

If $\mathcal{L}\subseteq\mathcal{L}^{\prime}$ are first order languages, then for a $\mathcal{L}^{\prime}$ -structure $\mathbf{A}$ , we use $\mathbf{A}_{\mathcal{L}}$ to denote the reduct of $\mathbf{A}$ to the language $\mathcal{L}$ . If $\vec{a}_{i}=(a_{i1},...,a_{in})\in A_{i}^{n}$ with $1\leq i\leq m$ , then we write $[\vec{a}_{1},...,\vec{a}_{m}]$ to denote the $n$ -tuple $((a_{11},...,a_{m1}),...,(a_{1n},...,a_{mn}))\in(\prod A_{i})^{n}$ .

Lemma 3

Let $\mathcal{L}$ be a language of algebras, $R$ be a $n$ -ary relation symbol and consider $\mathcal{L}^{\prime}=\mathcal{L}\cup\{R\}$ . Let $\mathcal{K}$ be any class of $\mathcal{L}^{\prime}$ -structures. The following are equivalent:

1.

There is an existential positive $\mathcal{L}$ -formula which defines $R$ in $\mathcal{K}$ .
2.

For all $\mathbf{A},\mathbf{B}\in\mathbb{P}_{u}(\mathcal{K})$ and all homomorphisms $\sigma:\mathbf{A}_{\mathcal{L}}\rightarrow\mathbf{B}_{\mathcal{L}}$ , we have that $\sigma:\mathbf{A}\rightarrow\mathbf{B}$ is a homomorphism.

Moreover, if $\mathcal{K}$ is closed under the formation of finite direct products, then $1.$ and $2.$ are equivalent to

3.

$R$ is definable in $\mathcal{K}$ by a $(\exists\bigwedge p=q)$ -formula.

Proof. The proof of the equivalence of $1.$ and $2.$ can be obtained from $(3)$ of Theorem 6.2 in [5]. Of course $3.$ implies $1.$ . We will show that $1.$ implies $3.$ . Suppose

\phi(\vec{z})=\exists\vec{w}\bigvee_{j=1}^{m}\bigwedge_{k=1}^{n_{j}}p_{jk}(\vec{z},\vec{w})=q_{jk}(\vec{z},\vec{w})

defines $R$ in $\mathcal{K}$ . We will prove that for some $j\in\{1,...,m\}$ , the formula $\psi_{j}(\vec{z})=\exists\vec{w}\bigwedge_{k=1}^{n_{j}}p_{k}(\vec{z},\vec{w})=q_{k}(\vec{z},\vec{w})$ defines $R$ in $\mathcal{K}$ . We proceed by contradiction. So suppose that no formula $\psi_{j}(\vec{z})$ defines $R$ in $\mathcal{K}$ . Then for every $j$ there exists $\mathbf{A}_{j}\in\mathcal{K}$ and $\vec{e}_{j}\in R^{\mathbf{A}_{j}}$ , such that $\mathbf{A}_{j}\models\lnot\psi_{j}(\vec{e}_{j})$ . Let $\vec{e}=[\vec{e}_{1},...,\vec{e}_{m}]$ . Since $\vec{e}\in R^{\Pi\mathbf{A}_{j}}$ and $\prod\mathbf{A}_{j}\in\mathcal{K}$ , we have that $\prod\mathbf{A}_{j}\models\phi(\vec{e})$ . Naturally, this says that $\prod\mathbf{A}_{j}\models\psi_{j}(\vec{e})$ , for some $j$ and hence $\mathbf{A}_{j}\models\psi_{j}(\vec{e}_{j})$ , for some $j$ , which produces a contradiction.

3 Central elements

In a universal-algebraic setting, one key concept for the study of the Pierce sheaf is that of central element. This tool can be developed fruitfully in varieties with $\vec{0}$ and $\vec{1}$ , which we now define.

A variety with $\vec{0}$ and $\vec{1}$ is a variety $\mathcal{V}$ in which there are 0-ary terms $0_{1},\ldots,0_{N},$ $1_{1},\ldots,1_{N}$ such that $\mathcal{V}\vDash\vec{0}=\vec{1}\rightarrow x=y$ , where $\vec{0}=(0_{1},\ldots,0_{N})$ and $\vec{1}=(1_{1},\ldots,1_{N})$ . The terms $\vec{0}$ and $\vec{1}$ are analogue, in a rather general manner, to identity (top) and null (bottom) elements in rings (lattices), and its existence in a variety, when the language has at least a constant symbol, is equivalent to the fact that no non-trivial algebra in the variety has a trivial subalgebra (see Proposition 2.3 of [4]).

If $\mathbf{A}\in\mathcal{V}$ , then we say that $\vec{e}\in A^{N}$ is a central element of $\mathbf{A}$ if there exists an isomorphism $\mathbf{A}\rightarrow\mathbf{A}_{1}\times\mathbf{A}_{2}$ such that $\vec{e}\rightarrow[\vec{0},\vec{1}]$ . We use $Z(\mathbf{A})$ to denote the set of central elements of $\mathbf{A}$ .

Central elements are a generalization of both central idempotent elements in rings with identity and neutral complemented elements in bounded lattices. In these classical cases it is well known that the central elements concentrate the information concerning the direct product representations. This happens when $\mathcal{V}$ has the Fraser-Horn property [17]¹¹1In [14] it is solved the problem of characterizing those varieties with $\vec{0}$ and $\vec{1}$ in which central elements determine the direct product representations. These are the varieties with definable factor congruences or equivalently, the varieties with Boolean factor congruences. See [1] for a non constructive short proof of this fact.. It is well known that the set of factor congruences of an algebra $\mathbf{A}$ in a variety with the Fraser-Horn property forms a Boolean algebra $FC(\mathbf{A})$ which is a sublattice of $\mathrm{Con}(\mathbf{A})$ (see [2]). In [17] it is proved that if $\mathcal{V}$ has the Fraser-Horn property, then for $\mathbf{A}\in\mathcal{V}$ , the map

\begin{array}[]{rcl}\lambda:FC(\mathbf{A})&\rightarrow&Z(\mathbf{A})\\ \theta&\rightarrow&\begin{array}[t]{l}\mathrm{unique\ }\vec{e}\in A^{N}\ \mathrm{such\ that}\\ \mathrm{\ \ \ \ \ \ \ \ \ }\vec{e}\equiv\vec{0}(\theta)\ \mathrm{and\ }\vec{e}\equiv\vec{1}(\theta^{\ast})\end{array}\end{array}

(where $\theta^{\ast}$ is the complement of $\theta$ in $FC(\mathbf{A})$ ) is bijective. Thus via the above bijection we can give to $Z(\mathbf{A})$ a Boolean algebra structure. We shall denote by $\mathbf{Z}(\mathbf{A})$ this Boolean algebra. Many of the usual properties of central elements in rings with identity or bounded lattices hold when $\mathcal{V}$ has the FHP. We say that a set of first order formulas $\{\varphi_{r}(\vec{z}):r\in R\}$ defines the property $``\vec{e}\in Z(\mathbf{A})"$ in $\mathcal{V}$ if for every $\mathbf{A}\in\mathcal{K}$ and $\vec{e}\in A^{N}$ , we have that $\vec{e}\in Z(\mathbf{A})$ iff $\mathbf{A}\vDash\varphi_{r}(\vec{e})$ , for every $r\in R$ .

In [17] it is proved

Lemma 4

Let $\mathcal{V}$ be a variety with $\vec{0}$ and $\vec{1}$ with the FHP. Let $\mathcal{L}$ be the language of $\mathcal{V}$ . Then, there is a finite set $\Sigma_{0}$ of $(\forall\exists\bigwedge p=q)$ -formulas in the variables $z_{1},...,z_{N}$ and $(\forall\bigwedge p=q)$ -formulas

F\text{-}PRES(z_{1},...,z_{N})\text{, with }F\in\mathcal{L}

such that $\Sigma_{0}\cup\{F$ - $PRES(\vec{z}):F\in\mathcal{L}\}$ defines the property $``\vec{e}\in Z(\mathbf{A})"$ .

Also in [17] it is proved that there is a $(\exists\bigwedge p=q)$ -formula $\varepsilon(x,y,\vec{z})$ such that for all $\mathbf{A},\mathbf{B}\in$ $\mathcal{V}$ ,

\mathbf{A}\times\mathbf{B}\vDash\varepsilon((a,b),(c,d),[\vec{0},\vec{1}])\text{ if and only if }a=c\text{.}

The formula $\varepsilon(\_,\_,\vec{e})$ defines the factor congruence associated (via the map $\lambda^{-1}$ ) with the central element $\vec{e}$ . We stress that the existence of $\varepsilon(x,y,\vec{z})$ and the set $\Sigma_{0}\cup\{F$ - $PRES(\vec{z}):F\in\mathcal{L}\}$ imply that the central elements (and its Boolean algebra structure) are preserved by surjective homomorphisms and products. That is to say:

Lemma 5

Let $\mathcal{V}$ be a variety with $\vec{0}$ and $\vec{1}$ with the FHP and let $\mathbf{A},\mathbf{B}\in\mathcal{V}$ . If $f:\mathbf{A}\rightarrow\mathbf{B}$ is a surjective homomorphism, then the map $Z(\mathbf{A})\rightarrow Z(\mathbf{B})$ defined by $\vec{e}\mapsto(f(e_{1}),...,f(e_{N}))$ , is a homomorphism from $\mathbf{Z}(\mathbf{A})$ to $\mathbf{Z}(\mathbf{B})$ .

Lemma 6

Let $\mathcal{V}$ be a variety with $\vec{0}$ and $\vec{1}$ with the FHP, $\{\mathbf{A}_{i}\}_{i\in I}$ be a family of members of $\mathcal{V}$ and $\vec{e}\in(\prod\mathbf{A}_{i})^{N}$ . Then, $\vec{e}\in Z(\prod\mathbf{A}_{i})$ if and only if $\vec{e}(i)\in Z(\mathbf{A}_{i})$ for every $i\in I$ . Moreover, $\mathbf{Z}(\prod\mathbf{A}_{i})$ is naturally isomorphic to $\prod\mathbf{Z}(\mathbf{A}_{i})$ .

For the details of the proofs of Lemmas 5 and 6, the reader may consult the proofs of the items $(a)$ and $(b)$ of Lemma 4 in [16].

3.1 Key theorem

Next, we will prove a series of lemmas in order to demonstrate a result (Theorem 12) which will be fundamental in our study of the Pierce stalks.

Lemma 7

Let $\mathcal{V}$ be a variety with $\vec{0}$ and $\vec{1}$ with the FHP such that $\mathbb{P}_{u}(\mathcal{V}_{SI})\subseteq\mathcal{V}_{DI}$ . Then, the property $``\vec{e}\in Z(\mathbf{A})"$ is definable in $\mathcal{V}$ by a single first order formula.

Proof. By Lemma 4, there is a finite set $\Sigma_{0}$ of $(\forall\exists\bigwedge p=q)$ -formulas in the variables $z_{1},...,z_{N}$ and $(\forall\bigwedge p=q)$ -formulas

F\text{-}PRES(z_{1},...,z_{N})\text{, with }F\in\mathcal{L}

such that $\Sigma_{0}\cup\{F$ - $PRES(\vec{z}):F\in\mathcal{L}\}$ defines the property $``\vec{e}\in Z(\mathbf{A})"$ . Since $\mathbb{P}_{u}(\mathcal{V}_{SI})\subseteq\mathcal{V}_{DI}$ we have that

\mathbb{P}_{u}(\mathcal{V}_{SI})\models\left(\bigwedge_{\varphi\in\Sigma_{0}}\varphi(\vec{z})\wedge\bigwedge_{F\in\mathcal{L}}F\text{-}PRES(\vec{z})\right)\rightarrow\left((\vec{z}=\vec{0})\vee(\vec{z}=\vec{1})\right)

So, by compactness there exists a finite subset $\mathcal{L}_{0}\subseteq\mathcal{L}$ such that:

\mathbb{P}_{u}(\mathcal{V}_{SI})\models\left(\bigwedge_{\varphi\in\Sigma_{0}}\varphi(\vec{z})\wedge\bigwedge_{F\in\mathcal{L}_{0}}F\text{-}PRES(\vec{z})\right)\rightarrow\left((\vec{z}=\vec{0})\vee(\vec{z}=\vec{1})\right)

(1)

We will see that the formula $\psi(\vec{z})=\bigwedge_{\varphi\in\Sigma_{0}}\varphi(\vec{z})\wedge\bigwedge_{F\in\mathcal{L}_{0}}F$ - $PRES(\vec{z})$ defines the property $``\vec{e}\in Z(\mathbf{A})"$ in $\mathcal{V}$ . To do so, it is enough to check that if $\mathbf{A}\in\mathcal{V}$ and $\mathbf{A}\models\psi(\vec{e})$ , then $\mathbf{A}\models F\text{-}PRES(\vec{e})$ , for every $F\in\mathcal{L}-\mathcal{L}_{0}$ . From Birkhoff’s subdirect representation theorem, we can assume that $\mathbf{A}\leq\prod\mathbf{A}_{i}$ is a subdirect product with subdirectly irreducible factors. Since the formula $\psi$ is positive and the $\mathbf{A}_{i}$ are quotients of $\mathbf{A}$ , we obtain that $\mathbf{A}_{i}\models\psi(\vec{e}(i))$ , for every $i\in I$ . But every $\mathbf{A}_{i}$ belongs to $\mathbb{P}_{u}(\mathcal{V}_{SI})$ so, from (1) we have that $\vec{e}(i)\in\{\vec{0}^{\mathbf{A}_{i}},\vec{1}^{\mathbf{A}_{i}}\}$ , for every $i\in I$ . Since $\vec{0}^{\mathbf{A}_{i}},\vec{1}^{\mathbf{A}_{i}}\in Z(\mathbf{A}_{i})$ we have that

\mathbf{A}_{i}\models F\text{-}PRES(\vec{0}^{\mathbf{A}_{i}})\wedge F\text{-}PRES(\vec{1}^{\mathbf{A}_{i}})\text{, for every }F\in\mathcal{L}

therefore, we get that

\mathbf{A}_{i}\models F\text{-}PRES(\vec{e}(i))\text{, for every }F\in\mathcal{L}-\mathcal{L}_{0}

for every $i\in I$ . But the formulas $F\text{-}PRES(z_{1},...,z_{N})$ are $(\forall\bigwedge p=q)$ -formulas, so they are preserved by subdirect products. Therefore, we can conclude that

\mathbf{A}\models F\text{-}PRES(\vec{e})\text{, for every }F\in\mathcal{L}-\mathcal{L}_{0}.

Lemma 8

Let $\mathcal{V}$ be a variety with $\vec{0}$ and $\vec{1}$ with the FHP, such that $\mathbb{P}_{u}(\mathcal{V}_{SI})\subseteq\mathcal{V}_{DI}$ . Then, the following are equivalent:

1.

The property $``\vec{e}\in Z(\mathbf{A})"$ is definable in $\mathcal{V}$ by an existential positive formula.
2.

The homomorphisms in $\mathcal{V}$ preserve central elements. I.e. if $\sigma:\mathbf{A}\rightarrow\mathbf{B}$ is a homomorphism between elements of $\mathcal{V}$ and $\vec{e}\in Z(\mathbf{A})$ , then $\sigma(\vec{e})\in Z(\mathbf{B})$ .
3.

The property $``\vec{e}\in Z(\mathbf{A})"$ is definable in $\mathcal{V}$ by a $(\exists\bigwedge p=q)$ -formula.

Proof. Let $\mathcal{L}$ be the language of $\mathcal{V}$ and let $\mathcal{L}^{\prime}=\mathcal{L}\cup\{R\}$ where $R$ is a $N$ -ary relation symbol. Given an algebra $\mathbf{A}\in\mathcal{V}$ we define:

R^{\mathbf{A}}=Z(\mathbf{A})

Let $\mathcal{K}$ be the following class of $\mathcal{L}^{\prime}$ -structures

\mathcal{K}=\{(\mathbf{A},R^{\mathbf{A}}):\mathbf{A}\in\mathcal{V}\}

From Lemma 7, there exists a first order formula $\sigma(z_{1},...,z_{N})$ , in the language of $\mathcal{V}$ , such that for every $\mathbf{A}\in\mathcal{V}$ , we have that $\vec{e}\in Z(\mathbf{A})$ if and only if $\mathbf{A}\models\sigma(\vec{e})$ . That is to say, for every $\mathbf{A}\in\mathcal{V}$ , we have that $(e_{1},...,e_{N})\in R^{\mathbf{A}}$ if and only if $\mathbf{A}\models\sigma(\vec{e})$ . It is easy to see that the class $\mathcal{K}$ is axiomatizable by the set of sentences

\Sigma\cup\{\forall z_{1}...z_{N}\ \ \left(R(z_{1},...,z_{N})\leftrightarrow\sigma(z_{1},...,z_{N})\right)\}

where $\Sigma$ is any set of axioms defining $\mathcal{V}$ . Observe that since $\mathcal{K}$ is a first order class, it is closed by ultraproducts; and furthermore, from Lemma 6, $\mathcal{K}$ is closed under the formation of direct products. Hence, from Lemma 3 we obtain that the following are equivalent:

(a)

There is an existential positive $\mathcal{L}$ -formula which defines $R$ in $\mathcal{K}$ .
(b)

If $(\mathbf{A},R^{\mathbf{A}}),(\mathbf{B},R^{\mathbf{B}})\in\mathcal{K}$ and $\sigma:\mathbf{A}\rightarrow\mathbf{B}$ is a homomorphism, then $\sigma:(\mathbf{A},R^{\mathbf{A}})\rightarrow(\mathbf{B},R^{\mathbf{B}})$ is a homomorphism
(c)

There is a $(\exists\bigwedge p=q)$ -formula which defines $R$ in $\mathcal{K}$ .

But $1.$ , $2.$ and $3.$ are restatements of $(a)$ , $(b)$ and $(c)$ , respectively. That is to say, $1.$ , $2.$ and $3.$ are equivalent as required.

Lemma 9

Let $\mathcal{V}$ be a variety with $\vec{0}$ and $\vec{1}$ with the FHP and $\mathbf{A}\in\mathcal{V}$ . Let $\vec{e}\in A^{N}$ . Then, $\vec{e}\in Z(\mathbf{A})$ if and only if $\theta^{\mathbf{A}}(\vec{0}^{\mathbf{A}},\vec{e})$ and $\theta^{\mathbf{A}}(\vec{1}^{\mathbf{A}},\vec{e})$ are a pair of complementary factor congruences of $\mathbf{A}$ .

Proof. For details of the proof, the reader may consult Corollary 4 of [17].

Lemma 10

Let $\mathcal{V}$ be a variety with $\vec{0}$ and $\vec{1}$ with the FHP and let $\{\mathbf{A}_{i}\}_{i\in I}$ be a family of non-trivial members of $\mathcal{V}$ . Suppose $\mathbf{A}\leq\prod\mathbf{A}_{i}$ is a global subdirect product. If $\vec{e}\in A^{N}$ is such that $\vec{e}(i)\in\{\vec{0}^{\mathbf{A}_{i}},\vec{1}^{\mathbf{A}_{i}}\}$ , for every $i\in I$ , then $\vec{e}\in Z(\mathbf{A})$ .

Proof. Let $\mathbf{A}\in\mathcal{V}$ and $\vec{e}\in A^{N}$ satisfying the hypothesis of the statement. Since $\mathcal{V}$ has FHP, in order to prove that $\vec{e}\in Z(\mathbf{A})$ , from Lemma 9, we must verify that $\theta^{\mathbf{A}}(\vec{0}^{\mathbf{A}},\vec{e})$ and $\theta^{\mathbf{A}}(\vec{1}^{\mathbf{A}},\vec{e})$ are a pair complementary factor congruences of $\mathbf{A}$ . So, let $J=\{i\in I:\vec{e}(i)=\vec{0}^{\mathbf{A}_{i}}\}$ . Notice that, since $\mathbf{A}_{i}$ is not trivial, for every $i\in I$ , then we obtain that $I-J=\{i\in I:\vec{e}(i)=\vec{1}^{\mathbf{A}_{i}}\}$ . Let $i\in I$ . Then, it follows that $\theta_{i}^{\mathbf{A}}=\{(a,b)\in A\times A:i\in E(a,b)\}$ . Moreover, for every $F\subseteq I$ , it is clear that $\bigcap_{i\in F}\theta_{i}^{\mathbf{A}}=\{(a,b)\in A\times A:F\subseteq E(a,b)\}$ . Now let us to consider $\theta=\bigcap_{i\in J}\theta_{i}^{\mathbf{A}}$ and $\delta=\bigcap_{i\in I-J}\theta_{i}^{\mathbf{A}}$ . We will prove that $\theta$ and $\delta$ are a pair of complementary factor congruences of $\mathbf{A}$ . If $(a,b)\in\theta\cap\delta$ , then $J\subseteq E(a,b)$ and $I-J\subseteq E(a,b)$ , thus $E(a,b)=I$ . Hence, $\theta\cap\delta=\Delta^{\mathbf{A}}$ . In order to show that $\theta\circ\delta=\nabla^{\mathbf{A}}$ , let $(a,b)\in\nabla^{\mathbf{A}}$ . By assumption, $\mathbf{A}$ is a global subdirect product of $\{\mathbf{A}_{i}\}_{i\in I}$ . So, since $J=\bigcap_{k=1}^{N}E(e_{k},0_{k}^{\mathbf{A}})$ and $I-J=\bigcap_{k=1}^{N}E(e_{k},1_{k}^{\mathbf{A}})$ , then $J$ and $I-J$ are open sets of the topology over $I$ which contains all the equalizers of elements of $A$ . Therefore, because $\emptyset=J\cap(I-J)\subseteq E(a,b)$ , from the Patchwork Property it follows that there is a $z\in A$ , such that $J\subseteq E(a,z)$ and $I-J\subseteq E(z,b)$ . I.e., $(a,z)\in\theta$ and $(z,b)\in\delta$ . In consequence, $(a,b)\in\theta\circ\delta$ .

Now we will see that $\theta=\theta^{\mathbf{A}}(\vec{0}^{\mathbf{A}},\vec{e})$ . Since $\theta=\{(a,b)\in A\times A:J\subseteq E(a,b)\}$ and $\vec{e}(i)\in\{\vec{0}^{\mathbf{A}_{i}},\vec{1}^{\mathbf{A}_{i}}\}$ , then $(0_{k}^{\mathbf{A}},e_{k})\in\theta_{i}^{\mathbf{A}}$ for every $1\leq k\leq N$ , so $\theta^{\mathbf{A}}(\vec{0}^{\mathbf{A}},\vec{e})\subseteq\theta$ . In order to prove the other inclusion, notice that $\theta^{\mathbf{A}}(1_{k}^{\mathbf{A}},e_{k})\subseteq\delta$ for every $k$ . Therefore $\theta\cap\theta^{\mathbf{A}}(\vec{1}^{\mathbf{A}},\vec{e})=\Delta^{\mathbf{A}}$ . Because $\mathcal{V}$ has the FHP, thus factor congruences distribute with any other (c.f. [2]) and since $\nabla^{\mathbf{A}}=\theta^{\mathbf{A}}(\vec{0}^{\mathbf{A}},\vec{e})\vee\theta^{\mathbf{A}}(\vec{1}^{\mathbf{A}},\vec{e})$ , we obtain that

	$\displaystyle\theta$	$\displaystyle=\theta\cap\nabla^{\mathbf{A}}$
		$\displaystyle=\theta\cap(\theta^{\mathbf{A}}(\vec{0}^{\mathbf{A}},\vec{e})\vee\theta^{\mathbf{A}}(\vec{1}^{\mathbf{A}},\vec{e}))$
		$\displaystyle=(\theta\cap\theta^{\mathbf{A}}(\vec{0}^{\mathbf{A}},\vec{e}))\vee(\theta\cap\theta^{\mathbf{A}}(\vec{1}^{\mathbf{A}},\vec{e}))$
		$\displaystyle=\theta^{\mathbf{A}}(\vec{0}^{\mathbf{A}},\vec{e})\vee\Delta^{\mathbf{A}}$
		$\displaystyle=\theta^{\mathbf{A}}(\vec{0}^{\mathbf{A}},\vec{e}).$

The proof for $\delta=\theta^{\mathbf{A}}(\vec{1}^{\mathbf{A}},\vec{e})$ is similar. This completes the proof.

Lemma 11

Let $\mathcal{V}$ be a variety with $\vec{0}$ and $\vec{1}$ with the FHP and let $\{\mathbf{B}_{i}\}_{i\in I}$ be a family of members of $\mathcal{V}$ such that every subalgebra of each $\mathbf{B}_{i}$ is directly indecomposable. Suppose $\mathbf{B}\leq\prod\mathbf{B}_{i}$ is a global subdirect product, and let $\mathbf{A}\leq\mathbf{B}$ . Then $Z(\mathbf{A})\subseteq Z(\mathbf{B})$ .

Proof. Let us assume that $\vec{e}\in Z(\mathbf{A})$ . Then $\vec{e}\in B^{N}$ . Let $A_{i}=\{a(i):a\in A\}$ and let $\mathbf{A}_{i}$ denotes the algebra whose universe is $A_{i}$ . Since $\mathbf{A}_{i}$ is subalgebra of $\mathbf{B}_{i}$ for every $i\in I$ , then it follows that $\mathbf{A}_{i}\in\mathcal{V}_{DI}$ . Consider the canonical projection $\pi_{i}^{\mathbf{A}}:\mathbf{A}\rightarrow\mathbf{A}_{i}$ . Because $\pi_{i}^{\mathbf{A}}$ is onto $\mathbf{A}_{i}$ and $\mathcal{V}$ has the FHP, from Lemma 5 we obtain that $\pi_{i}^{\mathbf{A}}(\vec{e})\in Z(\mathbf{A}_{i})$ , for every $\vec{e}\in Z(\mathbf{A})$ . But $Z(\mathbf{A}_{i})=\{\vec{0}^{\mathbf{A}_{i}},\vec{1}^{\mathbf{A}_{i}}\}$ , therefore $\vec{e}(i)\in\{\vec{0}^{\mathbf{A}_{i}},\vec{1}^{\mathbf{A}_{i}}\}=\{\vec{0}^{\mathbf{B}_{i}},\vec{1}^{\mathbf{B}_{i}}\}$ . Since $\mathbf{B}$ is a global subdirect product of non trivial algebras, from Lemma 10 we conclude that $\vec{e}\in Z(\mathbf{B})$ .

Now we can prove the key result.

Theorem 12

Let $\mathcal{L}$ be a language of algebras with at least a constant symbol. Let $\mathcal{V}$ be a variety of $\mathcal{L}$ -algebras with the FHP. Suppose that there is a universal class $\mathcal{F}\subseteq\mathcal{V}_{DI}$ such that every member of $\mathcal{V}$ is isomorphic to a global subdirect product with factors in $\mathcal{F}$ . Then there exists a $(M+2)$ -ary term $U(x,y,\vec{z})$ and $0$ -ary terms $0_{1},\ldots,0_{M},1_{1},\ldots,1_{M}$ such that

\mathcal{V}\vDash U(x,y,\vec{0})=x\wedge U(x,y,\vec{1})=y

Proof. First notice that, since no algebra of $\mathcal{F}$ has a trivial subalgebra and every member of $\mathcal{V}$ is a subdirect product with factors in $\mathcal{F}$ , then no non-trivial algebra of $\mathcal{V}$ has a trivial subalgebra. So, from Proposition 2.3 of [4] we get that there are unary terms $0_{1}(w),...,0_{N}(w),1_{1}(w),...,1_{N}(w)$ such that

\mathcal{V}\vDash\vec{0}(w)=\vec{1}(w)\rightarrow x=y\text{,}

where $w,x,y$ are distinct variables. Let $c\in\mathcal{L}$ be a constant symbol. Since $\mathcal{V}\vDash\vec{0}(c)=\vec{1}(c)\rightarrow x=y$ , we can redefine $\vec{0}=\vec{0}(c)$ and $\vec{1}=\vec{1}(c)$ to get that $\mathcal{V}$ is a variety with $\vec{0}$ and $\vec{1}$ .

Next, we will prove that

(1)

$\mathbb{P}_{u}(\mathcal{V}_{SI})\subseteq\mathcal{V}_{DI}$ .

Since every algebra of $\mathcal{V}$ is a subdirect product with factors in $\mathcal{F}$ , we have that $\mathcal{V}_{SI}\subseteq\mathcal{F}$ . So we have that $\mathbb{P}_{u}(\mathcal{V}_{SI})\subseteq\mathbb{P}_{u}(\mathcal{F})\subseteq\mathcal{F}\subseteq\mathcal{V}_{DI}$ , which proves (1).

Since by hypothesis we have that every member of $\mathcal{V}$ is isomorphic to a global subdirect product with factors in $\mathcal{F}$ and $\mathbb{S}(\mathcal{F})\subseteq\mathcal{F}\subseteq\mathcal{V}_{DI}$ , Lemma 11 says that

(2)

If $\mathbf{A}\leq\mathbf{B}\in$ $\mathcal{V}$ , then $Z(\mathbf{A})\subseteq Z(\mathbf{B})$ .

But (2) and Lemma 5 say that 2. of Lemma 8 holds, which implies that the property $"\vec{e}\in Z(\mathbf{A})"$ is definable in $\mathcal{V}$ by a $(\exists\bigwedge p=q)$ -formula. Let

\phi(\vec{z})=\exists\vec{w}\bigwedge_{k=1}^{n}p_{k}(\vec{z},\vec{w})=q_{k}(\vec{z},\vec{w})

define the property $``\vec{e}\in Z(\mathbf{A})"$ in $\mathcal{V}$ .

Due to in $\mathcal{L}$ there is at least a constant symbol, there exists $\mathbf{F}_{\mathcal{V}}\mathbf{(\emptyset)}$ . Since $\vec{0},\vec{1}\in Z(\mathbf{F}_{\mathcal{V}}\mathbf{(\emptyset)})$ , there are $0$ -ary terms $c_{1},...,c_{N},d_{1},...d_{N}$ , such that $\mathbf{F}_{\mathcal{V}}\mathbf{(\emptyset)}\models\bigwedge_{k=1}^{n}p_{k}(\vec{0},\vec{c})=q_{k}(\vec{0},\vec{c})$ and $\mathbf{F}_{\mathcal{V}}\mathbf{(\emptyset)}\models\bigwedge_{k=1}^{n}p_{k}(\vec{1},\vec{d})=q_{k}(\vec{1},\vec{d})$ . This says that

(3)

For every $\mathbf{A}\in\mathcal{V}$ ,

	$\displaystyle p_{k}(\vec{0}^{\mathbf{A}},\vec{c}^{\mathbf{A}})$	$\displaystyle=$	$\displaystyle q_{k}(\vec{0}^{\mathbf{A}},\vec{c}^{\mathbf{A}})\text{, }k=1,...,n$
	$\displaystyle p_{k}(\vec{1}^{\mathbf{A}},\vec{d}^{\mathbf{A}})$	$\displaystyle=$	$\displaystyle q_{k}(\vec{1}^{\mathbf{A}},\vec{d}^{\mathbf{A}})\text{, }k=1,...,n$

Let $X=\{x,y,\vec{z},\vec{w}\}$ and $\theta=\bigvee_{k=1}^{n}\theta^{\mathbf{\mathbf{F}_{\mathcal{V}}}(X)}(p_{k}(\vec{z},\vec{w}),q_{k}(\vec{z},\vec{w}))$ . Let $\mathbf{H}=\mathbf{F}_{\mathcal{V}}(X)/\theta$ . Since $p_{k}^{\mathbf{H}}(\vec{z}/\theta,\vec{w}/\theta)=q_{k}^{\mathbf{H}}(\vec{z}/\theta,\vec{w}/\theta)$ , for $k=1,...,n$ , we have that $\mathbf{H}\models\phi(\vec{z}/\theta)$ . That is to say, $\vec{z}/\theta\in Z(\mathbf{H})$ .

Thereby, since $(x/\theta,y/\theta)\in\nabla^{\mathbf{H}}=\theta^{\mathbf{H}}(\vec{z}/\theta,\vec{0}/\theta)\circ\theta^{\mathbf{H}}(\vec{z}/\theta,\vec{1}/\theta)$ , there is a term $t(x,y,\vec{z},\vec{w})$ , such that

(x/\theta,t^{\mathbf{\mathbf{F}_{\mathcal{V}}}(X)}(x,y,\vec{z},\vec{w})/\theta)\in\theta^{\mathbf{H}}(\vec{z}/\theta,\vec{0}/\theta)

and

(t^{\mathbf{\mathbf{F}_{\mathcal{V}}}(X)}(x,y,\vec{z},\vec{w})/\theta,y/\theta)\in\theta^{\mathbf{H}}(\vec{z}/\theta,\vec{1}/\theta).

Hence we have

(4)

$(x/\theta,t^{\mathbf{H}}(x/\theta,y/\theta,\vec{z}/\theta,\vec{w}/\theta))\in\theta^{\mathbf{H}}(\vec{z}/\theta,\vec{0}/\theta)$
(5)

$(t^{\mathbf{H}}(x/\theta,y/\theta,\vec{z}/\theta,\vec{w}/\theta),y/\theta)\in\theta^{\mathbf{H}}(\vec{z}/\theta,\vec{1}/\theta)$

We will prove that

(6)

$\mathcal{V}\vDash t(x,y,\vec{0},\vec{c})=x\wedge t(x,y,\vec{1},\vec{d})=y$

Let $\mathbf{A}\in\mathcal{V}$ and $a,b\in A$ . From (3) and Lemma 2, there exists a unique $\Omega:\mathbf{H}\rightarrow\mathbf{A}$ such that $\Omega(x/\theta)=a$ , $\Omega(y/\theta)=b$ , $\Omega(\vec{z}/\theta)=\vec{0}^{\mathbf{A}}$ and $\Omega(\vec{w}/\theta)=\vec{c}^{\mathbf{A}}$ . Hence, from (4) and Lemma 1, we obtain that $t^{\mathbf{A}}(a,b,\vec{0}^{\mathbf{A}},\vec{c}^{\mathbf{A}})=a$ . In a similar fashion, again from (3) and Lemma 2 applied to (5) together with Lemma 1, we obtain that $t^{\mathbf{A}}(a,b,\vec{1}^{\mathbf{A}},\vec{d}^{\mathbf{A}})=b$ and so (6) is proved.

To conclude the proof we can redefine $\vec{0}=(0_{1},...,0_{N},c_{1},...,c_{N})$ and $\vec{1}=(1_{1},...,1_{N},d_{1},...,d_{N})$ and take $U(x,y,z_{1},...,z_{2N})=t(x,y,z_{1},...,z_{2N})$ .

4 Pierce stalks

We shall use the above theorem and some results of [16] to give important information on the Pierce stalks for preprimal varieties corresponding to Rosemberg types 6. and 7. of the introduction. An $h$ -ary relation $\sigma$ on a finite set $P$ is central if

-

it is totally symmetric, that is, for all $\vec{a}\in\sigma$ , if $\pi$ is a permutation of $\{1,\ldots,h\}$ , then $(a_{\pi(1)},\ldots,a_{\pi(h)})\in\sigma$ .
-

it is totally reflexive, that is, for all $\vec{a}\in P^{h}$ with at least two of the $a_{i}$ equal, we have that $\vec{a}\in\sigma$ .
-

there is an $a_{1}$ such that for all $a_{2},\ldots,a_{h}$ in $P$ we have $\vec{a}\in\sigma$ ; and
-

$\sigma\neq P^{h}$ .

If $\sigma$ is a central relation on a set $P$ , let $\mathbf{P}_{\sigma}$ be the preprimal algebra whose universe is $P$ and whose fundamental operations are all the operations preserving $\sigma$ . We use $\mathbb{V}(\mathbf{P}_{\sigma})$ to denote the variety generated by $\mathbf{P}_{\sigma}$ .

Proposition 13

Let $\sigma$ be a $h$ -ary central relation on $P$ , with $h\geq 3$ . There is no universal class $\mathcal{F}\subseteq\mathbb{V}(\mathbf{P}_{\sigma})_{DI}$ such that every member of $\mathbb{V}(\mathbf{P}_{\sigma})$ is isomorphic to a global subdirect product with factors in $\mathcal{F}$ . There are Pierce stalks in $\mathbb{V}(\mathbf{P}_{\sigma})$ which are not directly indecomposable.

Proof. First, we note that by [7] the variety $\mathbb{V}(\mathbf{P}_{\sigma})$ is congruence distributive, hence it has the FHP (see [8]). Let $a_{1},\ldots,a_{N},b_{1},\ldots,b_{N}\in P$ be such that $(a_{1},\ldots,a_{N})\neq(b_{1},\ldots,b_{N})$ . Let $U:P^{N+2}\rightarrow P$ be such that

\begin{array}[]{ccc}U(x,y,a_{1},\ldots,a_{N})&=&x\vskip 6.0pt plus 2.0pt minus 2.0pt\\ U(x,y,b_{1},\ldots,b_{N})&=&y\end{array}

for any $x,y\in P$ . It is easy to check that $U$ does not preserve $\sigma$ . For example, if $h$ is odd, we can take $(c_{1},\ldots,c_{h})\notin\sigma$ and note that

\begin{array}[]{rcl}U(c_{1},c_{2},a_{1},\ldots,a_{N})&=&c_{1}\vskip 6.0pt plus 2.0pt minus 2.0pt\\ U(c_{1},c_{2},b_{1},\ldots,b_{N})&=&c_{2}\vskip 6.0pt plus 2.0pt minus 2.0pt\\ U(c_{3},c_{4},a_{1},\ldots,a_{N})&=&c_{3}\vskip 6.0pt plus 2.0pt minus 2.0pt\\ U(c_{3},c_{4},b_{1},\ldots,b_{N})&=&c_{4}\vskip 6.0pt plus 2.0pt minus 2.0pt\\ \vdots&&\ \vdots\vskip 6.0pt plus 2.0pt minus 2.0pt\\ U(c_{h-1},c_{h},a_{1},\ldots,a_{N})&=&c_{h-1}\vskip 6.0pt plus 2.0pt minus 2.0pt\\ U(c_{h-1},c_{h},b_{1},\ldots,b_{N})&=&c_{h}\end{array}

which, since $\sigma$ is totally reflexive, says that $U$ does not preserve $\sigma$ . Thus the conclusion of Theorem 12 does not hold and hence there is no universal class $\mathcal{F}\subseteq\mathbb{V}(\mathbf{P}_{\sigma})_{DI}$ such that every member of $\mathbb{V}(\mathbf{P}_{\sigma})$ is isomorphic to a global subdirect product with factors in $\mathcal{F}$ . In [2] it is proved that if every Pierce stalk is in $\mathbb{V}(\mathbf{P}_{\sigma})_{DI}$ , then the class $\mathbb{V}(\mathbf{P}_{\sigma})_{DI}$ is universal. Since the Pierce representation is always a global subdirect representation, there are Pierce stalks in $\mathbb{V}(\mathbf{P}_{\sigma})$ which are not directly indecomposable.

In order to analyze the case of a $2$ -ary central relation we need the following lemma from [15].

Lemma 14

Let $\mathcal{V}$ be a congruence distributive variety. If each member of $\mathcal{V}$ is isomorphic to a global subdirect product with factors in $\mathcal{V}_{SI}$ , then $\mathcal{V}$ is congruence permutable.

Proof. It follows from (1) $\Rightarrow$ (2) of [15, Corollary. 1 of Thm. 3.4].

Proposition 15

Let $\sigma$ be a $2$ -ary central relation on a set $P$ . Every Pierce stalk in $\mathbb{V}(\mathbf{P}_{\sigma})$ is directly indecomposable. There are Pierce stalks in $\mathbb{V}(\mathbf{P}_{\sigma})$ which are not subdirectly irreducible.

Proof. It is an exercise [10, X.5.4] to check that there are term-operations $+$ and $\times$ on $\mathbf{P}_{\sigma}$ , and elements $0,1\in P$ such that $\mathbf{P}_{\sigma}$ satisfies the following identities

\begin{array}[]{r}x\times 0=0\times x=0\vskip 6.0pt plus 2.0pt minus 2.0pt\\ x\times 1=1\times x=x\vskip 6.0pt plus 2.0pt minus 2.0pt\\ x+0=0+x=x\vskip 6.0pt plus 2.0pt minus 2.0pt\end{array}

Since $\sigma$ is reflexive, all constant functions on $P$ preserve $\sigma$ , which says that there are terms $0(w)$ and $1(w)$ such that $0^{\mathbf{P}_{\sigma}}(a)=0$ and $1^{\mathbf{P}_{\sigma}}(a)=1$ , for every $a\in P$ . Let $U(x,y,z,w)$ be the term $(x\times w)+(y\times z)$ . Note that

\begin{array}[]{ccc}U(x,y,0(w),1(w))&=&x\vskip 6.0pt plus 2.0pt minus 2.0pt\\ U(x,y,1(w),0(w))&=&y\end{array}

are identities of $\mathbb{V}(\mathbf{P}_{\sigma})$ . Thus, in the terminology of [16], $\mathbb{V}(\mathbf{P}_{\sigma})$ is a Pierce variety. Also we note that, since all constant functions on $P$ are term-operations, we have that $\mathbb{S(}\mathbf{P}_{\sigma})=\{\mathbf{P}_{\sigma}\}$ . By [7] $\mathbf{P}_{\sigma}$ is the only subdirectly irreducible algebra in $\mathbb{V}(\mathbf{P}_{\sigma})$ . Now, (2) $\Rightarrow$ (1) of [16, Theorem 8] says that every Pierce stalk is directly indecomposable.

Suppose every Pierce stalk of $\mathbb{V}(\mathbf{P}_{\sigma})$ is subdirectly irreducible. We will arrive to a contradiction. Since $\mathbf{P}_{\sigma}$ is the only subdirectly irreducible algebra in $\mathbb{V}(\mathbf{P}_{\sigma})$ , we have that $\mathbf{P}_{\sigma}$ is hereditarily simple. By [7] the variety $\mathbb{V}(\mathbf{P}_{\sigma})$ is congruence distributive. By Lemma 14 we have that $\mathbb{V}(\mathbf{P}_{\sigma})$ is arithmetical. Pixley theorem [3, IV.10.7] says that $\mathbf{P}_{\sigma}$ is quasiprimal, i.e. the ternary discriminator is a term-function of $\mathbf{P}_{\sigma}$ . It is easy to check that the ternary discriminator does not preserve $\sigma$ . Thus we have arrived to a contradiction and hence we have proved that there are Pierce stalks in $\mathbb{V}(\mathbf{P}_{\sigma})$ which are not subdirectly irreducible.

Next, we analyze the preprimal variety given by a non-trivial proper equivalence relation $\sigma$ on a finite set $P$ . Let $\mathbf{P}_{\sigma}$ be the preprimal algebra whose universe is $P$ and whose fundamental operations are all the operations preserving $\sigma$ .

Proposition 16

Let $\sigma$ be a non-trivial proper equivalence relation on a finite set $P$ . Every Pierce stalk in $\mathbb{V}(\mathbf{P}_{\sigma})$ is directly indecomposable. There are Pierce stalks in $\mathbb{V}(\mathbf{P}_{\sigma})$ which are not subdirectly irreducible.

Proof. Since $\sigma$ is reflexive, we have that the constant operations on $P$ are term-operations of $\mathbf{P}_{\sigma}$ . Take $0,1\in P$ such that $0/\sigma\neq 1/\sigma$ . Let $0(w)$ and $1(w)$ be terms such that $0^{\mathbf{P}_{\sigma}}(a)=0$ and $1^{\mathbf{P}_{\sigma}}(a)=1$ , for every $a\in P$ . Define $f:P^{4}\rightarrow P$ as follows

f(x,y,z,w)=\left\{\begin{array}[]{ccl}x&&\text{if }(z,0)\in\sigma\text{ and }(w,1)\in\sigma\\ y&&\text{if }(z,1)\in\sigma\text{ and }(w,0)\in\sigma\\ 0&&\text{otherwise}\end{array}\right.

Since $f$ preserves $\sigma$ we have that there is a term $U(x,y,z,w)$ such that $U^{\mathbf{P}_{\sigma}}=f$ . Note that

\begin{array}[]{ccc}U(x,y,0(w),1(w))&=&x\vskip 6.0pt plus 2.0pt minus 2.0pt\\ U(x,y,1(w),0(w))&=&y\end{array}

are identities of $\mathbb{V}(\mathbf{P}_{\sigma})$ . Thus, in the terminology of [16], $\mathbb{V}(\mathbf{P}_{\sigma})$ is a Pierce variety. Note that $\mathrm{Con}(\mathbf{P}_{\sigma})=\{\Delta^{\mathbf{P}_{\sigma}},\sigma,\nabla^{\mathbf{P}_{\sigma}}\}$ . Since $\mathbb{V}(\mathbf{P}_{\sigma})$ is congruence distributive [9] and $\mathbb{S(}\mathbf{P}_{\sigma})=\{\mathbf{P}_{\sigma}\}$ , the subdirectly irreducibles of $\mathbb{V}(\mathbf{P}_{\sigma})$ are $\mathbf{P}_{\sigma}$ and $\mathbf{P}_{\sigma}/\sigma$ . Note that $\mathbf{P}_{\sigma}/\sigma$ is primal. Thus $\mathbb{P}_{u}\mathbb{SV}(\mathbf{P}_{\sigma})_{SI}\subseteq\mathbb{V}(\mathbf{P}_{\sigma})_{DI}$ and hence (2) $\Rightarrow$ (1) of [16, Theorem 8] says that every Pierce stalk is directly indecomposable.

Suppose every Pierce stalk of $\mathbb{V}(\mathbf{P}_{\sigma})$ is subdirectly irreducible. We will arrive to a contradiction. Since every directly indecomposable algebra of $\mathbb{V}(\mathbf{P}_{\sigma})$ is a Pierce stalk of itself we have that every directly indecomposable member of $\mathbb{V}(\mathbf{P}_{\sigma})$ is subdirectly irreducible. By [3, IV.12.5] we have that $\mathbb{V}(\mathbf{P}_{\sigma})$ is semisimple which is impossible since $\sigma\in\mathrm{Con}(\mathbf{P}_{\sigma})$ . Thus we have arrived to an absurd and hence we have proved that there are Pierce stalks which are not subdirectly irreducible.

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