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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
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1 Answer 1

This is not true in this form:

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.
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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
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