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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.

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    $\begingroup$ If the posets in question all have a global minimum and a global maximum, then this follows from Lemma 6.1 in: William R. Schmitt, Incidence Hopf algebras, home.gwu.edu/~wschmitt/papers/iha.pdf . $\endgroup$
    – darij grinberg
    Commented Apr 28, 2016 at 0:16
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    $\begingroup$ Actually, a positive answer to your question in the case where $P$ is connected is claimed in: Junji Hashimoto, On Direct Product Decomposition of Partially Ordered Sets, Annals of Mathematics, Second Series, Vol. 54, No. 2 (Sep., 1951), pp. 315--318, sci-hub.io/10.2307/1969532# . $\endgroup$
    – darij grinberg
    Commented Apr 28, 2016 at 0:58
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    $\begingroup$ In full generality, your question has a negative answer: See Tadasi Nakayama and Junji Hashimoto, On a problem of G. Birkhoff, Proc. Amer. Math. Soc. 1 (1950), pp. 141--142, ams.org/journals/proc/1950-001-02/S0002-9939-1950-0035279-X . $\endgroup$
    – darij grinberg
    Commented Apr 28, 2016 at 0:59
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    $\begingroup$ Darij, those last two references answer my question perfectly: false in general, true when $P$ is connected. You could post it as an answer. $\endgroup$
    – Sam Hopkins
    Commented Apr 28, 2016 at 1:10

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Basically repeating what I said in the comments:

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

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    $\begingroup$ For some recent work on arithmetic of finite ordered sets, Ralph McKenzie has a couple of papers from 2000 showing that one can cancel exponents for certain bases and exponents. Gerhard "Exponents Aren't Just For Numbers" Paseman, 2016.04.27. $\endgroup$
    – Gerhard Paseman
    Commented Apr 28, 2016 at 2:44

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