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.

  • 2
    $\begingroup$ 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 $G\subseteq F$. $\endgroup$
    – amrsa
    Jan 22 at 10:58
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    $\begingroup$ 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. $\endgroup$
    – bof
    Jan 22 at 10:59
  • 1
    $\begingroup$ 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 $IP_S(\mathbf 2) \subseteq SP(\mathbf 2)$. $\endgroup$
    – amrsa
    Jan 22 at 14:14

1 Answer 1


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


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