Trying to understand why $\gamma(x) = \langle p : p \in Irr(L), p \leq x \rangle$ is a lattice isomorphism A lattice $L$ is called distributive if the join and meet operations distribute, i.e., for all $x,y,z \in L$ we have that
$$x \vee (y \wedge z) = (x \wedge y) \vee (x \wedge z)$$
$$x \wedge (y \vee z) = (x \vee y) \wedge (x \vee z)$$
An element $x \in L$ is called join-irreducible if it can't be written as the join of two other elements. That is, if $x = y \vee z$ then either $x = y$ or $x=z$. The subposet (not sublattice!) of $L$ consisting of all join-irreducible elements is denoted by $Irr(L)$.
Now, let $J(Irr(L))$ denote the set of order ideals of $Irr(L)$.
For a finite distributive lattice $L$, i'm trying to show that the map $\gamma: L \rightarrow J(Irr(L))$ given by $$\gamma(x) = \langle p : p \in Irr(L), p \leq x \rangle$$ is a lattice isomorphism
Can somebody help me to see what is happening here? Thank you for your time
 A: This is called the Birkhoff's representation theorem for finite distributive lattices.
I'll use the following lemma, which I'll state without proof:

Lemma. Let $L$ be a finite lattice.
Then, for all $a \in L$,
$$a = \bigvee \{ p \in Irr(L) : p \leq a \}.$$

If $a \in L$, then let us see that $\gamma(a) \in J(Irr(L))$.
Now
$$\gamma(a) = \{p \in Irr(L) : p \leq a\},$$
and if $q \in Irr(L)$ is such that $q \leq p_1$, for some $p_1 \in \gamma(a)$, then, by transitivity (since $p_1 \leq a$), we have $q \leq a$;
thus, $q \in \gamma(a)$.
We conclude that $\gamma$ is an application.
If $a \leq b$ in $L$, then, for any $p \in Irr(L)$, if $p \leq a$ then $p \leq b$, by transitivity.
Hence $\gamma(a) \subseteq \gamma(b)$.
Conversely, if $\gamma(a) \subseteq \gamma(b)$, then, by the lemma
$$a = \bigvee \gamma(a) \leq \bigvee \gamma(b) = b.$$
So we have
$$a \leq b \quad\text{ iff }\quad \gamma(a) \leq \gamma(b),$$
that is, $\gamma$ is an order-embedding.
If $\gamma$ is also onto, then it's an isomorphism.
So let $A \in J(Irr(L))$, and let us see that there exists $a \in L$ with $\gamma(a) = A$.
If $A = \varnothing$, then $A = \gamma(0)$, as it is clear.
So suppose that $A \neq \varnothing$.
$A$ is certainly finite, so we can take $A = \{p_1, \ldots, p_n\}$.
Let $a = \bigvee_{i=1}^n p_i$.
We'll see that $\gamma(a) = A$.
Indeed, if $p \in \gamma(a)$, then $p \leq p_1 \vee \cdots \vee p_n$, and, since these are join-irreducible, $p \leq p_i$, for some $i$.
Since $A$ is an order-ideal, it follows that $p \in A$.
So $\gamma(a) \subseteq A$.
Conversely, if $p \in A$, then $p \in \gamma(a)$ because $p \in Irr(L)$ and $p \leq a$.
