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 $R$.

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