If $k[S]$ is noetherian, is S finitely generated? - MathOverflow

If $k[S]$ is noetherian, is S finitely generated?

2010-11-04T15:20:07Z

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?

Answer by Justin Shih for If $k[S]$ is noetherian, is S finitely generated?

2010-11-04T19:14:38Z

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.

Answer by Mariano Suárez-Alvarez for If $k[S]$ is noetherian, is S finitely generated?

2010-11-04T20:11:37Z

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.