No, at least in the case that $k$ is infinite, and not $\mathbb{Q}$ prime - for an indeterminate $t$ take $S = {t, S$ generated by ${t, t/a, t/a^2, ...}$ with $0 \ne a \in k\backslash\mathbb{Q}$ k$ of infinite (multiplicative) order not in the prime field. Then $k[S] = k[t]$ but $S$ is not finitely generated.
|
2 | deleted 4 characters in body; added 13 characters in body | ||
|
|
||||
|
|
1 |
|
||
|
No, at least in the case that $k$ is infinite, and not $\mathbb{Q}$ - for an indeterminate $t$ take $S = {t, t/a, t/a^2, ...}$ with $0 \ne a \in k\backslash\mathbb{Q}$ of infinite (multiplicative) order. Then $k[S] = k[t]$ but $S$ is not finitely generated. |
||||
