 Solution

This is the Artin--Tate lemma. Choose $A$-algebra generators $x_1,\dots ,x_r$ of $C$ and $B$-module generators $e_1,\dots ,e_s$ of $C$. Writewith coefficients $a_{ij},b_{ij\ell }\in B$, and let $B_0$ be the $A$-subalgebra of $B$ generated by these finitely many coefficients.

The module ${\sum }_j{B}_0{e}_j$ contains the $x_i$, is closed under multiplication, and contains $A$ after including an expression for $1$ among the chosen coefficients. It therefore equals $C$. Thus $C$ is a finite $B_0$-module. The ring $B_0$ is Noetherian by the Hilbert basis theorem, and $B\subseteq C$ is a $B_0$-submodule, so $B$ is a finite $B_0$-module. It follows that $B$ is a finitely generated $A$-algebra.