[proofplan] The lattice $J(P)$ has bottom element $\varnothing$ and joins given by unions, so join-irreducibility can be tested by decompositions of an ideal as a union of two ideals. First we show that every principal ideal $\downarrow p$ is join-irreducible, because any union of ideals equal to $\downarrow p$ must place the top element $p$ in one of the two ideals. Conversely, a nonempty non-principal ideal has at least two maximal elements, and finiteness lets us decompose it as the union of the principal ideals generated by those maximal elements; splitting the maximal elements into two nonempty parts gives a nontrivial join decomposition. Finally, inclusion of principal ideals is exactly the original order relation on $P$. [/proofplan] [step:Identify joins and the bottom element in the ideal lattice] An order ideal in $P$ is a subset $I\subset P$ such that, whenever $x\in I$ and $y\le x$, one has $y\in I$. The empty set $\varnothing$ is an order ideal and is the least element of $J(P)$ under inclusion. If $I,K\in J(P)$, then $I\cup K$ is again an order ideal: if $x\in I\cup K$ and $y\le x$, then $x\in I$ or $x\in K$, hence $y\in I$ or $y\in K$, so $y\in I\cup K$. Since $I\subset I\cup K$ and $K\subset I\cup K$, and every order ideal containing both $I$ and $K$ contains $I\cup K$, the join in $J(P)$ is \begin{align*} I\vee K=I\cup K. \end{align*} [/step] [step:Show every principal ideal is join-irreducible] Fix $p\in P$. The principal ideal $\downarrow p$ is nonempty, since $p\in\downarrow p$, so it is not the bottom element of $J(P)$. Suppose $I,K\in J(P)$ satisfy \begin{align*} \downarrow p=I\vee K=I\cup K. \end{align*} Since $p\in\downarrow p$, either $p\in I$ or $p\in K$. If $p\in I$, then the order-ideal property gives $\downarrow p\subset I$, while $I\subset \downarrow p$ follows from $I\cup K=\downarrow p$; hence $I=\downarrow p$. If $p\in K$, the same argument gives $K=\downarrow p$. Thus every decomposition of $\downarrow p$ as a join of two elements of $J(P)$ has one factor equal to $\downarrow p$, so $\downarrow p$ is join-irreducible. [/step] [step:Decompose every finite ideal using its maximal elements] Let $I\in J(P)$ be nonempty. Define \begin{align*} \max(I)=\{m\in I:\text{ there is no }x\in I\text{ with }m<x\}. \end{align*} Because $P$ is finite, every element $x\in I$ lies below some element of $\max(I)$. Indeed, if $x$ is not maximal in $I$, choose $x_1\in I$ with $x<x_1$; if $x_1$ is not maximal, continue. Since $I$ is finite, this strictly increasing process must stop at an element $m\in\max(I)$ with $x\le m$. It follows that \begin{align*} I=\bigcup_{m\in\max(I)}\downarrow m. \end{align*} The inclusion from right to left holds because each $m\in I$ and $I$ is an order ideal. The inclusion from left to right holds because every $x\in I$ satisfies $x\le m$ for some $m\in\max(I)$. [guided] The point of passing to maximal elements is that a finite ideal is completely determined by its maximal elements. Let \begin{align*} \max(I)=\{m\in I:\text{ there is no }x\in I\text{ with }m<x\}. \end{align*} We first verify that every $x\in I$ lies below at least one element of $\max(I)$. If $x$ is already maximal in $I$, take $m=x$. Otherwise there is some $x_1\in I$ with $x<x_1$. If $x_1$ is not maximal, there is $x_2\in I$ with $x_1<x_2$. This cannot continue indefinitely, because $I$ is a finite set and the sequence is strictly increasing. Therefore the process terminates at some $m\in I$ such that no element of $I$ is strictly above $m$, and then $m\in\max(I)$ with $x\le m$. Now we prove the decomposition. If $m\in\max(I)$, then $m\in I$, and because $I$ is an order ideal, every element below $m$ also belongs to $I$. Hence $\downarrow m\subset I$, so \begin{align*} \bigcup_{m\in\max(I)}\downarrow m\subset I. \end{align*} Conversely, for each $x\in I$, the preceding paragraph gives $m\in\max(I)$ with $x\le m$. This says exactly that $x\in\downarrow m$, and therefore \begin{align*} I\subset \bigcup_{m\in\max(I)}\downarrow m. \end{align*} Combining the two inclusions gives \begin{align*} I=\bigcup_{m\in\max(I)}\downarrow m. \end{align*} This is the structural fact that will let us turn a non-principal ideal into a nontrivial join decomposition. [/guided] [/step] [step:Show every non-principal ideal is not join-irreducible] Let $I\in J(P)$ be nonempty and not principal. We first show that $\max(I)$ has at least two elements. If $\max(I)=\{m\}$, then the decomposition from the previous step gives \begin{align*} I=\downarrow m, \end{align*} contradicting that $I$ is not principal. Hence $\max(I)$ contains at least two distinct elements. Choose $m_0\in\max(I)$, and define two subsets of $\max(I)$ by \begin{align*} A=\{m_0\},\qquad B=\max(I)\setminus\{m_0\}. \end{align*} Both $A$ and $B$ are nonempty. Define ideals \begin{align*} I_A=\bigcup_{a\in A}\downarrow a,\qquad I_B=\bigcup_{b\in B}\downarrow b. \end{align*} Each is an order ideal, being a union of order ideals, and the maximal-element decomposition gives \begin{align*} I=I_A\cup I_B=I_A\vee I_B. \end{align*} The inclusion $I_A\subset I$ is proper: choose $b\in B$. If $b\in I_A$, then $b\le m_0$, and since both $b$ and $m_0$ are maximal elements of $I$, this forces $b=m_0$, contradicting $b\in B$. Thus $b\in I\setminus I_A$. Similarly, $m_0\in I\setminus I_B$, so $I_B\subsetneq I$. Therefore $I$ is a join of two strictly smaller ideals and is not join-irreducible. The empty ideal is the bottom element of $J(P)$, so it is not join-irreducible by the standard convention that join-irreducible elements are non-bottom elements. Consequently, the only join-irreducible elements of $J(P)$ are the principal ideals $\downarrow p$. [/step] [step:Verify that principal ideals preserve and reflect the order] Define \begin{align*} \Phi:P&\to J_{\mathrm{irr}}(J(P)) \end{align*} \begin{align*} p&\mapsto \downarrow p. \end{align*} The previous steps show that $\Phi$ is well-defined and surjective. For $p,q\in P$, if $p\le q$, then every $r\in\downarrow p$ satisfies $r\le p\le q$, so $r\in\downarrow q$ and hence \begin{align*} \downarrow p\subset \downarrow q. \end{align*} Conversely, if $\downarrow p\subset \downarrow q$, then $p\in\downarrow p\subset\downarrow q$, so $p\le q$. Therefore \begin{align*} p\le q\quad\Longleftrightarrow\quad \downarrow p\subset \downarrow q. \end{align*} This equivalence shows that $\Phi$ preserves and reflects the order. It also implies injectivity, since $\downarrow p=\downarrow q$ gives $p\le q$ and $q\le p$, hence $p=q$ by antisymmetry of the poset order. Thus $\Phi$ is an isomorphism from $P$ onto the poset of join-irreducible elements of $J(P)$. [/step]