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From "A course in Universal Algebra" of Burris and Sankappanavar, exercise 4 page 16.

If $L$ is a finite lattice let $J(L)$ be the poset of join irreducible elements of $L$, where $a \le b$ in $J(L)$ means $a \le b$ in $L$. Show that if L is a finite distributive lattice then $L$ is isomorphic to $L(J(L))$, the lattice of nonempty lower segments of $J(L)$.

What I have tried

I have tried to use the isomorphism of $L$ with the lattice of its principal ideals. I would prove that:

$\{x\in J(L)| x \le a \} \cup \{x\in J(L)| x \le b \} = \{x\in J(L)| x \le a\lor b \}$

But i need to show that the meet of two join irreducible elements is join irreducible, is this true in distributive lattices? There are minimum conditions for this to be true?

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Welcome to MSE!

Unfortunately, it is not true that the meet of two join-irreducible elements is always join-irreducible. For a simple example, consider:

a lattice

here $d$ and $e$ are both join irreducible, but their meet is not.


The "standard" proof (that is, the proof that I am most familiar with) of Birkhoff's theorem goes as follows:

First, show that join-irreducible elements act like primes, in the following sense:

If $p$ is join-irreducible, and $p \leq a_1 \lor a_2 \lor \cdots \lor a_n$, then actually $p \leq a_i$ for some $i$.

This is where we (heavily) use distributivity of our lattice. It is analogous to the statement "If $p$ is prime, and $p \mid a_1 \times a_2 \times \cdots \times a_n$ then actually $p \mid a_i$ for some $i$".

Next, we show that every element of our lattice can be "factored" into join-irreducibles. This is where we heavily use finiteness of our lattice (a chain condition would work for this lemma too). You will need the first lemma while proving it.

Each $x$ can be written uniquely as a irredundant join of join-irreducible elements.

This is analogous to the unique factorization of some integer into primes.

Finally, we consider the map $\varphi : L \to L(J(L))$ given by

$$\varphi(x) = \{p \in J(L) ~|~ p \leq x\}$$

Can you show that this map is an isomorphism?


Edit:

It is worth noting that we can go the other way as well. Instead of sending $L \to L(J(L))$, we can send a poset $P$ to $J(L(P))$ by $\psi(y) = \langle y \rangle$. This is also an isomorphism!

This information together shows that $\varphi$ and $\psi$ actually form an Equivalence of Categories between the category of finite posets (with monotone maps) and the category of finite distributive lattices (with bounded homomorphisms). For more information, see this wikipedia page.


I hope this helps! ^_^

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    $\begingroup$ Apparently you're using the notation $J(P)$ to denote the lattice of order-ideals of the poset $P$. In the question, that would be $L(P)$. This can raise some confusion, I think. $J(L)$ is used to denote join-irreducibles of $L$ (instead of $Irr(L)$). And in what sense are $\varphi$ and $\psi$ inverse of each other? $\endgroup$ – amrsa Mar 20 at 19:59
  • $\begingroup$ oooooh, you're so right. Good catch! I saw J and just assumed it was what it meant when I learned this, haha. I'll update the answer. $\endgroup$ – HallaSurvivor Mar 20 at 20:00
  • $\begingroup$ As for the sense in which $\varphi$ and $\psi$ are inverses, I perhaps spoke too quickly. I meant to say that they relate the category of finite distributive lattices to the category of posets, where, for any lattice $L$, $L \cong J(\text{Irr}(L))$ and for any poset $P$, $P \cong \text{Irr}(J(P))$ (using the notation in my answer). I will also clarify this ^_^ $\endgroup$ – HallaSurvivor Mar 20 at 20:03
  • $\begingroup$ Good point. I'll include it either way as it remains good to know. Thanks for pointing this out ^_^ $\endgroup$ – HallaSurvivor Mar 20 at 20:11
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    $\begingroup$ Thanks for your detailed explanation. I can see that your first counterexample is also distributive. Using the lemma you reported, which are easy enough to prove for me, 𝜑 can be shown to be an invertible order preserving bijection hence a lattice isomorphism. $\endgroup$ – Giovanni Barbarani Mar 21 at 7:36

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