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

lattice-ordersmaximal-and-prime-idealsuniversal-algebra

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.

Best Answer

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 (JΓ³nsson). 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

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