# Show that a distributive lattice can be embedded into product of two-element lattices

I have seen this exercise in Bergman: Universal Algebra: Fundamentals and Selected Topics.

Let $$L$$ be a distributive lattice and let $$2$$ be a 2-element lattice. Show that there is a set $$J$$ and lattice embedding $$h: L \rightarrow 2^J$$ such that for every $$j \in J$$, $$\pi_j \circ h$$ is surjective.

My progress: The two-element lattice is defined as $$2 = ({0,1}, \land, \lor)$$. I understand the $$\pi_j$$ as a projection $$\pi_j: 2^J \rightarrow 2$$.

There is a theorem that may be used:

If L is a distributive lattice and $$Prm(L)$$ is the set of prime ideals in $$L$$, then there is a monomorphism $$L \rightarrow (P(Prm(L), \cap, \cup)$$.

Also, by the definitions it looks like $$h(L)$$ could be a subdirect product of $$2^J$$, but I am not sure how that could be useful.

Do you have any advice on how to proceed? Thank you.

• It seems essentially you already have the result, or at least its ingredients. Indeed, given that there is a homomorphism from $L$ into $\wp(X)$, where $X$ is the set of prime ideals of $L$, then compose it with an isomorphism between $\wp(X)$ and $2^X$. Of course you still have to prove that the first homomorphism is injective, and for that I think you need (DPI), the Distributive Prime Ideal principle, according to which, if $J$ is an ideal and $G$ is a filter and they're disjoint, then there is a prime ideal $I$ whose complement $F$ is a prime filter and $J\subseteq I$ and $G\subseteq F$. Jan 22 at 10:58
• I think you want to show that any two elements of a distributive lattice $L$ canbe separated by a homomorphism $f:L\to2$. I. E., if $a,b\in L$ and $a\not\ge b$, then there is a homomorphism $f:L\to2$ such that $f(a)=0$ and $f(b)=1$, something like that. Should be straightforward.
– bof
Jan 22 at 10:59
• Another (more Universal Algebraic) approach is to show that $\mathbf 2$ (the two-element lattice) is the only s.i. distributive lattice. Then, since lattices (distributive or not) are congruence-distributive, you can apply a result from Jónsson from which, in particular, $V(\mathbf 2)=IP_S(HS(\mathbf 2))$, whence $V(\mathbf 2)$, the variety of distributive lattices, is $IP_S(\mathbf 2) \subseteq SP(\mathbf 2)$. Jan 22 at 14:14

I'll give you two different ways of proving that.

The first makes use of the Distributive Prime Ideal principle:

(DPI) Given a distributive lattice $$\mathbf L$$ and an ideal $$J$$ and a filter $$G$$ of $$\mathbf L$$ such that $$J \cap G = \varnothing$$, there exist a prime ideal $$I$$ and a prime filter $$F=L\setminus I$$ such that $$J\subseteq I$$ and $$G \subseteq I$$.

The (DPI) principle is a weak form of the Axiom of choice (AC). In fact, it is equivalent to (AC)$$_F$$, which states that every family of non-empty finite sets has a choice function (see [Davey&Priestley], page 237).

Now, you already have a result stating that a distributive lattice $$\mathbf L$$ can be embedded in $$\wp(X)$$, where $$X$$ is the set of its prime ideals (this is Lemma 10.20 (page 238) in [Davey&Priestley]).
Since $$\wp(Y) \cong 2^Y$$ for any set $$Y$$ (just take a subset $$A$$ to the function $$\chi_A:Y\to 2=\{0,1\}$$ that makes $$\chi_A(y)=1$$ iff $$y\in A$$), the result is proven. Again, this result is also in [Davey&Priestley], Theorem 10.21.

(Added. Actually we don't need (DPI) in the above proof (indeed, I didn't use it!) because its use must already be incorporated in the proof of the result mentioned by the OP. I misread it as "there is a homomorphism..." when the OP claims "there is a monomorphism..."; (DPI) is useful to prove the homomorphism is one-to-one.)

A different approach would be to show that $$\mathbf2$$, the two-element lattice is the only (up to isomorphism) distributive lattice which is subdirectory irreducible (and therefore, the variety of distributive lattices is generated by $$\mathbf2$$) and the fact that lattices are congruence-distributive.
Then, you can apply this result from [Burris&Sankappanavar]:

Corollary 6.10 (Jónsson). If $$\mathcal K$$ is a finite set of finite algebras and $$V(\mathcal K)$$, the variety generated by $$\mathcal K$$ is congruence-distributive, then the subdirectory irreducible algebras of $$V(\mathcal K)$$ are in $$HS(\mathcal K)$$ and $$V(\mathcal K)=IP_S(HS(\mathcal K)).$$

It follows that $$V(\mathbf2)=IP_S(HS(\mathbf2)) \subseteq SP(\mathbf2)$$, and so every distributive lattice is a sub-lattice of a power of $$\mathbf 2$$, i.e, a sub-lattice of $$\mathbf2^J$$, for some $$J$$.

(Let me know if you have difficulty with some of the auxiliary results I'm using.)

[Burris&Sankappanavar] S. Burris and H.P. Sankappanavar, A course in Universal Algebra, The Millennium Edition

[Davey&Priestley] B.A. Davey and H.A. Priestley, Introduction to lattices and order, Cambridge University Press, 2nd Edition