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Let a (rational) Puiseux monoid be a non-trivial submonoid of the non-negative rational numbers under (the usual operation of) addition. It is not difficult to show that, if $f \colon H \to K$ is a (monoid) homomorphism between Puiseux monoids, then there exists a rational number $r \ge 0$ such that $f(x) = r x$ for every $x \in H$.

Proof. Set $H^+ := H \setminus \{0\}$ and pick $p, q \in H^+$ (since $f$ maps $0$ to $0$, we can focus on the positive elements). There then exist positive integers $a$, $b$, $c$, and $d$ such that $p = a/b$ and $q = m/n$. It follows that $$ mb f(p) = f(mbp) = f(ma) = f(naq) = na f(q), $$ and hence $f(p)/p = f(q)/q$. That is, the function $H^+ \to \mathbb Q_{\ge 0} \colon x \mapsto f(x)/x$ is constant. []

Question. The result, along with a slighly more complicated proof, appears as Proposition 3.2(1) in [F. Gotti, Puiseux monoids and transfer homomorphisms, J. Algebra 516 (2018), 95–114], but I feel it must be older. Do you know of any reference that supports my feeling?

Note that the case of numerical monoids (that is, submonoids of $(\mathbb N, +)$ with finite complement in $\mathbb N$) is covered by Theorem 3.2 in [J.C. Higgins, Representing $N$-semigroups, Bull. Austral. Math. Soc. 1 (1969), 115–125], as I learned from Benjamin Steinberg's answer in another thread some time ago.

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This seems to be Corollary 4 of the paper Morio Sasaki, Takayuki Tamura, Positive rational semigroups and commutative power joined cancellative semigroups without idempotent Czechoslovak Mathematical Journal, Vol. 21 (1971), No. 4, 567–576 that I mentioned in the comments to your previous question.

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  • $\begingroup$ Thanks! Corollary 4 in Sasaki and Tamura's paper does indeed imply the result in the OP, after noticing that a monoid homomorphism between Puiseux monoids must either be identically zero or map positive rationals to positive rationals (and hence restrict, in the terminology of Sasaki and Tamura, to a semigroup homomorphism of positive rational semigroups). Before asking here, I had the term "rational semigroup" in mind but couldn't remember why... $\endgroup$
    – Salvo Tringali
    Commented Feb 13 at 3:29
  • $\begingroup$ I am confused. I think what they mean by positive rational semigroup is exactly puiseux monoids (except without the identity) and Cor 4 says every homomorphism between them is given by multiplication by a rational $\endgroup$
    – Benjamin Steinberg
    Commented Feb 13 at 3:41
  • $\begingroup$ I think that, by the term "positive rational semigroup", Sasaki and Tamura mean a (non-empty) subsemigroup of the positive rational numbers under addition. Corollary 4 in their paper says that any homomorphism between positive rational semigroups is in fact an isomorphism (of the form $x \mapsto rx$, where $r$ is a positive rational number). This is not the case with Puiseux monoids, but only because it is missing the zero homomorphism. $\endgroup$
    – Salvo Tringali
    Commented Feb 13 at 3:47
  • $\begingroup$ So, to recover the result in the OP from Sasaki and Tamura's corollary, I think one should first argue (as trivial as it may be) that a monoid hom $f \colon H \to K$ between Puiseux monoids restricts to a sgrp hom between the "positive cones" of $H$ and $K$, unless $f$ is identically zero. $\endgroup$
    – Salvo Tringali
    Commented Feb 13 at 3:56
  • $\begingroup$ @SalvoTringali, yes. This is easy to see though since any two elements have a common multiple and so if one element goes to 0 they all do. $\endgroup$
    – Benjamin Steinberg
    Commented Feb 13 at 4:15

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