Finitely Generated Algebra over Commutative Ring

Question

Let $R$ be a commutative noetherian ring, A a finitely generated $R$-algebra (not necessarily commutative). Let $B$ be a subalgebra of the center $Z(A)$. Assume $A$ is finitely generated $B$-module. Can we show $B$ is finitely generated $R$-algebra?

Answer 1

Yes. When $A$ is commutative, this is known as the Artin-Tate lemma, and the same proof works in general. In detail, let $x_1,\dots,x_n$ generate $A$ as an $R$-algebra and let $y_1,\dots,y_m$ generate $A$ as a $B$-module. We can then choose scalars $b_{ij},b_{ijk}\in B$ such that $$x_i=\sum_j b_{ij}y_j$$ and $$y_iy_j=\sum_k b_{ijk} y_k.$$ Let $B_0\subseteq B$ be the $R$-subalgebra generated by the $b_{ij}$ and $b_{ijk}$, and note that the equations above imply $A$ is still generated by $y_1,\dots,y_m$ as a $B_0$-module (here we use the fact that $B$ is central in $A$, to write any finite product of the $x_i$ as a $B_0$-linear combination of the $y_j$ using the equations above since we can freely pull out the coefficients). Moreover, since $B_0$ is a finitely generated commutative $R$-algebra, it is a Noetherian ring by the Hilbert basis theorem, and hence $B$ is also finitely generated as a $B_0$-module. Since $B_0$ is a finitely generated $R$-algebra, this implies $B$ is also a finitely generated $R$-algebra.