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# join irreducibility

An element $a$ in a lattice $L$ is said to be *join irreducible* iff $a$ is not a bottom element, and, whenever $a=b\vee c$, then $a=b$ or $a=c$. Dually, $a\in L$ is *meet irreducible* iff $a$ is not a top element, and, whenever $a=b\wedge c$, then $a=b$ or $a=c$. If $a$ is both join and meet irreducible, then $a$ is said to be *irreducible*. Any atom in a lattice is join irreducible.

Example. In the lattice diagram (Hasse diagram) below,

$\xymatrix{&1\ar@{-}[d]&\\ &a\ar@{-}[ld]\ar@{-}[rd]\\ b\ar@{-}[rd]&&c\ar@{-}[ld]\\ &d\ar@{-}[d]&\\ &0}$ |

$a$ is meet irreducible but not join irreducible, $d$ is join irreducible but not meet irreducible, while $b,c$ are irreducible.

From this, we make the observations that in any chain, all the elements except the bottom one are join irreducible. Dually, all the elements except the top one are meet irreducible. An element is join irreducible iff it covers at most one other element. An element is meet irreducible iff it is covered by at most one other element.

Remark. If a lattice satisfies the descending chain condition, then every element can be expressed as a join of join irreducible elements. This statement can be dualized: if a lattice satisfies the ascending chain condition, then every element is the meet of meet irreducible elements.

# References

- 1 B. A. Davey, H. A. Priestley, Introduction to Lattices and Order, 2nd Edition, Cambridge (2003)

## Mathematics Subject Classification

06B99*no label found*

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## Comments

## Confusing example

In example, not only a,b,c,d are join-irreducible, but f,g are also irreducible. Or am I wrong? (and a,b are meet-irreducible and therefore a,b,d,f,g are irreducible)

## Re: Confusing example

I changed the example so it's less confusing now. Thanks

## Re: Confusing example

I thank you. I'm preparing for exams and I think that example is "didactic enought" now :-).