# Join-Irreducibles and Direct Products

I'm trying to characterize the join-irreducible elements of the direct product of two lattices $L$ and $K$.

If we denote by $\mathscr{J}(L)$ the set of all join-irreducible elements of $L$, then my guess so far is that $(a,b) \in \mathscr{J}(L \times K)$ iff $a \in \mathscr{J}(L)$ or (exclusive) $b \in \mathscr{J}(K)$. I've tried to prove this, but I'm getting stuck with some technical details. Moreover, my guess looks right when you look at diagrams of finite lattices.

Can someone give me a hand with this conjecture?

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What "technical details" do you get stuck at? –  Fredrik Meyer Oct 23 '12 at 7:13

Consider for example $K=L=\{0,1\}$ the 2 element lattice. Then both $0$ and $1$ are joint-irreducibles in $L$, but
• $(1,1)\in K\times L$ can be written as $(1,0)\lor(0,1)$.
• $(1,0)$ is join-irred., though both $0$ and $1$ are, so neither the 'exlusive or' condition works.
As far as I know, $0$ is not considered as a join-irreducible element. On the other hand, any idea on what could the characterization be? –  ragrigg Oct 23 '12 at 14:55
Let's say you're defining an element $j$ of a lattice $L$ to be "join-irreducible" if it is not $0_L$, the bottom element of $L$ (should it exist), and for all $a,b\in L$, if $a\vee b=j$ then $a=j$ or $b=j$. Let $K$ also be a lattice. If $0_K$ and $0_L$ exist, then $\mathscr{J}(L \times K)=\{(a,0_K): a\in\mathscr{J}(L)\}\cup\{(0_L,b): b\in\mathscr{J}(K)\}$. Proof: $0_{L \times K}=(0_L,0_K)$, so if $a\in\mathscr{J}(L)$, then $(a,0_K)\ne0_{L\times K}$; also, if $(c,d)\vee(e,f)=(a,0_K)$, then $d=0_K=f$ and $c\vee e=a$ in $L$, so $c=a$ or $e=a$. Conversely, let $(a,b)\in \mathscr{J}(L \times K)$. Since $(a,b)=(a,0_K)\vee(0_L,b)$, then without loss of generality $(a,b)=(a,0_K)$. Clearly, $a>0_L$. Also, if $a=c\vee e$ in $L$, then $(a,0_K)=(c,0_K)\vee(e,0_K)$ and hence $(a,0_K)=(c,0_K)$ without loss of generality, or $a=c$, so $a\in\mathscr{J}(L)$. If $L$ or $K$ lacks a bottom element, you should now be able to answer your question.