Lemma 10.51.7 (Artin-Tate). Let $R$ be a Noetherian ring. Let $S$ be a finitely generated $R$-algebra. If $T \subset S$ is an $R$-subalgebra such that $S$ is finitely generated as a $T$-module, then $T$ is of finite type over $R$.

Proof. Choose elements $x_1, \ldots , x_ n \in S$ which generate $S$ as an $R$-algebra. Choose $y_1, \ldots , y_ m$ in $S$ which generate $S$ as a $T$-module. Thus there exist $a_{ij} \in T$ such that $x_ i = \sum a_{ij} y_ j$. There also exist $b_{ijk} \in T$ such that $y_ i y_ j = \sum b_{ijk} y_ k$. Let $T' \subset T$ be the sub $R$-algebra generated by $a_{ij}$ and $b_{ijk}$. This is a finitely generated $R$-algebra, hence Noetherian. Consider the algebra

$S' = T'[Y_1, \ldots , Y_ m]/(Y_ i Y_ j - \sum b_{ijk} Y_ k).$

Note that $S'$ is finite over $T'$, namely as a $T'$-module it is generated by the classes of $1, Y_1, \ldots , Y_ m$. Consider the $T'$-algebra homomorphism $S' \to S$ which maps $Y_ i$ to $y_ i$. Because $a_{ij} \in T'$ we see that $x_ j$ is in the image of this map. Thus $S' \to S$ is surjective. Therefore $S$ is finite over $T'$ as well. Since $T'$ is Noetherian and we conclude that $T \subset S$ is finite over $T'$ and we win. $\square$

Comment #4264 by Manuel Hoff on

Typo in the last sentence of the statement. ...then $T$ is of finite type over $R$.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00IS. Beware of the difference between the letter 'O' and the digit '0'.