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$is not finitely generated. 1 No, at least in the case that$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.