A note on homomorphisms between products of algebras

Let $${\mathcal {K}}$$K be a congruence distributive variety and call an algebra hereditarily directly irreducible (HDI) if every of its subalgebras is directly irreducible. It is shown that every homomorphism from a finite direct product of arbitrary algebras from $${\mathcal {K}}$$K to an HDI algebra from $${\mathcal {K}}$$K is essentially unary. Hence, every homomorphism from a finite direct product of algebras $${\mathbf {A}}_i$$Ai ($$i\in I$$i∈I) from $${\mathcal {K}}$$K to an arbitrary direct product of HDI algebras $${\mathbf {C}}_j$$Cj ($$j\in J$$j∈J) from $${\mathcal {K}}$$K can be expressed as a product of homomorphisms from $${\mathbf {A}}_{\sigma (j)}$$Aσ(j) to $${\mathbf {C}}_j$$Cj for a certain mapping $$\sigma $$σ from J to I. A homomorphism from an infinite direct product of elements of $${\mathcal {K}}$$K to an HDI algebra will in general not be essentially unary, but will always factor through a suitable ultraproduct.


Introduction
Let A i , i ∈ I, and B j , j ∈ J, be algebras of the same type and f a homomorphism from i∈I A i to j∈J B j . For every k ∈ J let p k denote the projection from j∈J B j onto B k and f k := p k • f . More generally, for any J 0 ⊆ J we let p J0 : j∈J B j → j∈J0 B j be the canonical projection map. It is evident that f = (f j : j ∈ J). Hence, the task of describing f is reduced to the task of describing the homomorphisms f k from i∈I A i to B k .
In [3] the authors solve this problem for the case that the algebras A i and B j are conservative median algebras and the index sets are finite. More generally, Couceiro et al. [2] considers the case that the A i are median algebras and the B j are tree-median algebras. It turns out that the method developed in [2,3] can be further generalized to lattices. Let us note that every distributive lattice is a median algebra (but not conversely). We are even able to extend this result to arbitrary lattices A i provided the B j are chains. For lattice concepts used in the rest of the paper the reader is referred to the monographs [1,5].
We call a mapping f : i∈I A i → C essentially unary if there exists an i 0 ∈ I and a mapping g : A i0 → C with g • p i0 = f . In this case we say that "f depends only on the i 0 -th coordinate", or that "f factors through p i0 ".
From f = g • p i0 it easily follows that g is a homomorphism if and only if f is.

Fraser-Horn property and HDI algebras
The following lemma is known from [4].

Lemma 2.2. Let K be a congruence distributive (CD) variety. Then K has the Fraser-Horn property.
For the rest of the paper we fix a variety K with the Fraser-Horn property. We call an algebra non-trivial if its universe contains at least two elements. Definition 2.3. We call an algebra A hereditarily directly irreducible (HDI) if every subalgebra B ≤ A is directly irreducible, i.e., is not isomorphic to a direct product of two non-trivial factors. (2) A lattice is HDI if and only if it is a chain.
Theorem 2.5. Let K be a variety with the Fraser-Horn property. If n is a positive integer, A 1 , . . . , A n are in K and C ∈ K is HDI, then every homomorphism f from A 1 × · · · × A n to C is essentially unary, i.e., factors through one of the projections p i .
Proof. Let θ = ker(f ). By a straightforward generalization of the Fraser-Horn property we know that θ = θ 1 × · · · × θ n , where each θ i is a congruence on A i . The homomorphism theorem tells us that B := f (A 1 × · · · × A n ) is isomorphic to the direct product (A 1 /θ 1 ) × · · · × (A n /θ n ). By our assumption, B is directly irreducible, so at most one of these factors can be non-trivial, Vol. 79 (2018) A note on homomorphisms between products of algebras Page 3 of 7 25 so there is at most one i such that A i /θ i has more than one element. So f depends only on the i-th coordinate.
As a consequence of the above theorem we obtain the following statement.
Theorem 2.7. If n is a positive integer, A 1 , . . . , A n ∈ K, where K has the Fraser-Horn property, (C j ; j ∈ J) is a non-empty family of HDI algebras in K, and f is a homomorphism from A 1 × · · · × A n to j∈J C j then there exists a mapping σ : J → {1, . . . , n} and for every j ∈ J a homomorphism g j from A σ(j) to C j such that Proof. Apply Theorem 2.5 to the mappings f j := p j • f , j ∈ J.

Theorem 2.8. Let K be a variety with the Fraser-Horn property. Let n, k be positive integers and let
The injectivity of f implies k ≥ n and the injectivity of f −1 implies n ≥ k. This shows n = k. Moreover, again since f is injective we have σ ∈ S n . Finally, the injectivity of f implies the injectivity of g 1 , . . . , g n and the surjectivity of f implies the surjectivity of g 1 , . . . , g n . This shows that g 1 , . . . , g n are isomorphisms, i.e. C i ∼ = A σ(i) for all i = 1, . . . , n.
Corollary 2.9. If an algebra in K is isomorphic to a finite product of nontrivial HDI algebras, then these factors are uniquely determined up to order and isomorphisms.
Proof. This follows from Theorem 2.8.
We can generalize this to infinite direct products as follows. Recall that an ultrafilter on a set I is a family U of subsets of I which is upwards closed and also closed under intersections such that for all I 0 ⊆ I exactly one of I 0 , I\I 0 is in U . For any family (A i : i ∈ I) of sets and any ultrafilter U on I we define the equivalence relation ∼ U on i A i by and we write i A i /U for the set of equivalence classes, the "ultraproduct of the A i modulo U ". The canonical map from i A i to A i /U is denoted by κ U . If (A i ) i∈I is a family of algebras of the same type, then the relation ∼ U is a congruence relation on the product i A i .
It is clear that U is upwards closed, and from Theorem 2.5 and Remark 2.6 we get: If M 1 ⊆ I and M 2 := I\M 1 , then M 1 ∈ U and M 2 / ∈ U or conversely. As h is not constant, we have ∅ / ∈ U . We now show that U is closed under intersections: Given M 1 , M 2 ∈ U , then we can write i∈I A i as the direct product of four factors: with corresponding projections p 11 , p 10 , p 01 , p 00 .
Since none of the sets M 1 \M 2 , M 2 \M 1 , and I\(M 1 ∪ M 2 ) are in U , h cannot factor through any of p 10 , p 01 , or p 00 . Hence (by Theorem 2.5), h must factor through p 11 , so M 1 ∩ M 2 ∈ U . So we have shown that U is a filter, and even an ultrafilter.
We now check that h factors through the canonical map κ U : Finally, we show that U is unique. So let U be an ultrafilter such that h factors through κ U . It is enough to show U ⊆ U : Let

Lattices
Theorem 2.10 is in some sense best possible, in the sense that homomorphisms from an infinite product i A i into an HDI algebra will in general not factor through any single projection p j , as the following example shows. Example 3.1. Let U be an ultrafilter on the infinite index set I, and for each i ∈ I let A i be the 2-element lattice {0, 1}. Then the ultraproduct i∈I A i /U is again the 2-element lattice.
Identifying i A i with the power set lattice (P (I), ∪, ∩), the canonical map κ U : P (I) → {0, 1} maps each element of U to 1 and everything else to 0. If U is a non-principal ultrafilter, then h U does not factor through any projection.
This example can be generalized to any Fraser-Horn variety where the class of HDI algebras is described by a set of first order formulas: If i A i is a product of algebras, and (h i : i ∈ I) is a family of homomorphisms If U is an ultrafilter on I, then the algebra C := i C i /U is again HDI (as C satisfies all first order statements that are true in each C i ). Let By the uniqueness claim in Theorem 2.10, we see that U is the set of all M ⊆ I such thath factors through p M . So if U is non-principal, thenh does not factor through any p i .

Fact 3.2. Let
A be a lattice. Then the following are equivalent: • There is a non-constant homomorphism from A into a chain.
• There is a non-constant homomorphism from A into the 2-element chain.
• The lattice A has a prime ideal.
The following corollary can be seen as a weak version of Theorem 2.5.

Corollary 3.3. The class of lattices without a prime ideal is closed under finite direct products.
The following example shows that even this weak version cannot be generalized to infinite products, not even if all factors are equal. If M = i∈I M i is an arbitrary direct product with factors M i ∈ M, and h : M → C is a homomorphism into a chain, then h factors through an ultraproduct i∈I M i → i∈I M I /U → C, h = h • κ U . The map h and therefore also h must be constant.
Proof of (b). Let A be the lattice obtained from N = {0, 1, 2, . . . } by replacing each odd number 2k + 1 by a 3-element antichain a k , b k , c k , and each even number 2k by a new element d k . It is easy to see that A has no prime ideal.
We will show that every infinite power A I contains a prime ideal. Clearly it is enough to show this for the case of countable I, say I = N.
For any ultrafilter U on N the following set J U is an ideal on i∈I A i : We now show that J U is a prime ideal. If x = (x i : i ∈ N),ȳ = (y i : i ∈ N),z = (z i : i ∈ N),x ∧ȳ =z ∈ J U , then there is some set C ∈ U and some natural number k ∈ N such that z n ≤ d k holds for all n ∈ C. Now the two sets {n : x n ≥ d k+1 }, {n : y n ≥ d k+1 } cannot both belong to U , as their intersection D is disjoint to C. (Since n ∈ D implies x n ∧ y n ≥ d k+1 .) Without loss of generality we have {n : x n ≤ d k+1 } ∈ U , sox ∈ J U .