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Let $S$ be a semigroup. If $S$ is abelian, then it follows that the semigroup algebra $k[S]$ is finitely generated if and only if $S$ is.

What if we relax the condition on $k[S]$, so that $k[S]$ is only noetherian. Does it in this case follow that $S$ is finitely generated?

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It is an open problem (or was, last time I checked!) whether the noetherianity of $k[S]$ implies finite generation of $S$, when $S$ is not abelian.

This is discussed in chapter 5 of Noetherian semigroup algebras by Eric Jespers and Jan Okniński, along with various cases where we know that $S$ is finitely generated. They prove, for example, that this is so if $k[S]$ satisfies a polynomial identity, and this gives the case in which $S$ is abelian as a corollary.

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  • $\begingroup$ That's interesting, I was actually expecting some strange counterexample. Thanks for the reference! $\endgroup$
    – J.C. Ottem
    Commented Nov 4, 2010 at 23:33
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No, at least in the case that $k$ is infinite, and not prime - for an indeterminate $t$ take $S$ generated by $\{t, t/a, t/a^2, ...\}$ with $0 \ne a \in k$ of infinite (multiplicative) order not in the prime field. Then $k[S] = k[t]$ but $S$ is not finitely generated.

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  • $\begingroup$ Your $k[S]$ is not the group algebra as in it the elements of $S$ are by definition linearly independent. Also $S$ is not a semigroup. $\endgroup$
    – Torsten Ekedahl
    Commented Nov 4, 2010 at 19:19
  • $\begingroup$ Sorry, I guess I mean the semigroup generated by the $t/a^n$. I've edited my post to reflect this change. And I'm missing something, but I don't quite understand what you mean by the elements of $S$ are by definition linearly independent? $\endgroup$
    – mediocrates
    Commented Nov 4, 2010 at 19:25
  • $\begingroup$ Oh, I see now. I guess I'm taking $k[t]$ and then finding a sub-semigroup inside there, instead of taking a semigroup $S$ and then making $k[S]$. $\endgroup$
    – mediocrates
    Commented Nov 4, 2010 at 19:27
  • $\begingroup$ @Justin: By definition, $k[S]$ is the $k$-vector space with basis given by elements of $S$. (Multiplication is induced by the semigroup structure on $S$.) So I'm afraid the edit still doesn't fix this... $\endgroup$
    – Dave Anderson
    Commented Nov 4, 2010 at 19:27
  • $\begingroup$ Right. So then can we make an increasing chain of ideals by picking $s_1 \in S$, setting $I_1 = (s_1)$, and then by induction picking $s_{k+1} \in $S\backslash\I_k$ and setting $I_{k+1} = I_k + (s_{k+1})$? This chain eventually terminates, which I think means that $S$ is finitely generated? $\endgroup$
    – mediocrates
    Commented Nov 4, 2010 at 19:32

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