Artin–Tate lemma

Let $A\subseteq B\subseteq C$ be commutative rings. If $A$ is Noetherian, $C$ is a finitely generated algebra over $A$, and $C$ is a finitely generated module over $B$, then $B$ is a finitely generated algebra over $A$.^[ref]

Choose algebra generators $x_1,\dots ,x_r$ for $C$ over $A$ and module generators $c_1=1,c_2,\dots ,c_m$ for $C$ over $B$. Express every $x_i{c}_j$ as a $B$-linear combination of the $c_k$. Let $B_0$ be the $A$-subalgebra generated by the finitely many coefficients in these expressions. The $B_0$-module $M={\sum }_k{B}_0{c}_k$ contains one and is stable under multiplication by every $x_i$, so $M=C$. Thus $C$ is finite as a $B_0$-module.

The Hilbert basis theorem makes $B_0$ Noetherian. Hence its submodule $B\subseteq C$ is finitely generated over $B_0$. Combining those module generators with the finite algebra generators of $B_0$ gives finitely many algebra generators for $B$ over $A$.