Given two finite posets $P$ and $Q$, we can form the direct product poset $P \times Q$ whose elements are pairs $(p,q) \in P \times Q$ with $(p,q) \leq (p',q')$ if $p \leq p'$ and $q \leq q'$. Let us say a finite poset $P$ on $\geq 2$ elements is *indecomposable* if $P=P\_1 \times P\_2$ implies that either $P\_1$ or $P\_2$ is equal to $P$.


Is it true that a finite poset $P$ on $\geq 2$ elements can be written as a product $P = P\_1 \times \cdots \times P\_n$ of indecomposable posets in a unique way up to permutation of the factors?


Surely this is a classical question/result. I am apparently having trouble coming up with the right terms to google for, however. I would definitely appreciate any pointer to the literature. If the general result fails, I would also be interested in what mild conditions we can put on $P$ (e.g. gradedness) to guarantee a positive result.



Unique factorization of posets
Basically repeating what I said in the comments:


* In full generality, the answer to your question is "No". For a counterexample, see: [Tadasi Nakayama and Junji Hashimoto, *On a problem of G. Birkhoff*, Proc. Amer. Math. Soc. 1 (1950), pp. 141--142](http://www.ams.org/journals/proc/1950-001-02/S0002-9939-1950-0035279-X/).
* However, if you assume that $P$ is connected (i.e., the undirected graph obtained by forgetting the directions of the arcs of the Hasse diagram of $P$ is connected), then the answer is "Yes". For a proof, see [Junji Hashimoto, *On Direct Product Decomposition of Partially Ordered Sets*, Annals of Mathematics, Second Series, Vol. 54, No. 2 (Sep., 1951), pp. 315--318](http://sci-hub.io/10.2307/1969532).


(Note: I have verified the counterexample, but I haven't looked at the proof.)


