Subring of a finitely generated $R$-algebra is also finitely generated?

Question

Let $R$ be noetherian ring, $A$ a finitely generated $R$-algebra; that means that $A = R[x_1,..., x_n]/I$, $I \subset R[x_1,..., x_n]$ some ideal. Assume $S \subset A$ is an $R$-algebra, which is a subring of $A$.

Is then $S$ also a finitely generated $R$-algebra?

Motivation: proof of Lemma A.16 from David Mumford's Algebraic Geometry 1: Complex projective varieties (p 124):

The claim there is that certain subring $S$ of a finitely generated $\mathbb{C}$-algebra $A$ is finite over certain other $\mathbb{C}$-algebra $B$;
ie finitely generated as $B$-module.
And the proof shows that for every $f \in S$ the $B$-module $B[f]$ is finite.

And my question is if this already sufficient to show that $S$ is finite $B$-module? I know a result which says that an $B$-algebra $S$ is finite if and only if it is finitely generated as $B$-algebra and for every element $f \in S$ the $B$-module $B[f]$ is finite.

Answer 1

The answer is in general no. For example, suppose $R=F$ and let $A=F[x,y]$. Then $A$ is certainly a finitely generated $F$-algebra. But the subring $S=F[x,xy,xy^2,xy^3,\dots]$ of $A$ is not a finitely generated $F$-algebra, since every one of its finitely generated $F$-sub-algebras is contained in some $F[x,xy,\dots,xy^n]$, and $xy^{n+1}\notin F[x,xy,\dots,xy^n]$ for any $n\in\mathbb{N}$.