If $k[S]$ is noetherian, is S finitely generated? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T00:29:02Z http://mathoverflow.net/feeds/question/44833 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44833/if-ks-is-noetherian-is-s-finitely-generated If $k[S]$ is noetherian, is S finitely generated? J.C. Ottem 2010-11-04T15:20:07Z 2010-11-04T22:05:43Z <p>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. </p> <p>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?</p> http://mathoverflow.net/questions/44833/if-ks-is-noetherian-is-s-finitely-generated/44864#44864 Answer by Justin Shih for If $k[S]$ is noetherian, is S finitely generated? Justin Shih 2010-11-04T19:14:38Z 2010-11-04T19:19:50Z <p>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.</p> http://mathoverflow.net/questions/44833/if-ks-is-noetherian-is-s-finitely-generated/44870#44870 Answer by Mariano Suárez-Alvarez for If $k[S]$ is noetherian, is S finitely generated? Mariano Suárez-Alvarez 2010-11-04T20:11:37Z 2010-11-04T20:11:37Z <p>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.</p> <p>This is discussed in chapter 5 of <em>Noetherian semigroup algebras</em> 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.</p>