Artin–Tate lemma

Remarks

Let $x_1, \ldots, x_m$ generate $C$ as an $A$-algebra and $y_1, \ldots, y_n$ generate $C$ as a $B$-module. Write $x_i = \sum_j b_{ij} y_j$ and $y_i y_j = \sum_l b_{ijl} y_l$ with $b_{ij}, b_{ijl} \in B$. Let $B_0$ be the $A$-algebra generated by all $b_{ij}$ and $b_{ijl}$; it is Noetherian by Hilbert's basis theorem, and $C$ is a finitely generated $B_0$-module. Then $B$ is a $B_0$-submodule of $C$, hence finitely generated as a module over the Noetherian ring $B_0$, so $B$ is of finite type over $A$.

Example: let $A = k$ be a field, $C = k[x]$ and $B = k[x^2]$; then $C$ is generated by $1$ and $x$ as a $B$-module, and indeed $B$ is generated by $x^2$ as a $k$-algebra.

Going deeper

For a student

The lemma gives a short proof of Zariski's lemma. Suppose a field $K$ is a finitely generated $k$-algebra but not finite over $k$, and choose a transcendence basis $t_1, \ldots, t_r$ with $r \ge 1$. Then $K$ is finite over $k(t_1, \ldots, t_r)$, and the lemma would make $k(t_1, \ldots, t_r)$ a finitely generated $k$-algebra, which is impossible. The same idea shows that rings of invariants of finite groups acting on finitely generated algebras over a field are finitely generated.