In the following, $R$ is a subring of a ring $S,$ and $v\in S.$ All of my rings are commutative with $1.$
I like to use the following terminology.
Definitions. I say that $S$ is module-finite over $R$ if there is a surjective $R$-module homomorphism $R^{\oplus n}\to S$ for some $n$, and I say that $S$ is ring-finite over $R$ if there is a surjective ring homomorphism $R[x_{1},\ldots,x_{n}]\to S$ for some $n$.
It seems that most people prefer other terminology. To make sure everyone is on the same page:
- module-finite over $R$ implies ring-finite over $R,$ but the converse does not hold.
- "integral + ring-finite = module-finite", i.e., $S$ is integral over $R$ and ring-finite over $R$ if and only if $S$ is module-finite over $R.$
I feel that the answer to the following question is "no", but I'm not sure how to go about looking for a counterexample.
Question. Suppose every subring $R^{\prime}$ of $S$ with $R[v]\subseteq R^{\prime}$ is ring-finite over $R.$ Does this imply that $v$ is integral over $R$?
Equivalent question. Suppose every subring $R^{\prime}$ of $S$ with $R[v]\subseteq R^{\prime}$ is ring-finite over $R.$ Does this imply that $R[v]$ is module-finite over $R$?
To be clear: the answer is "no" if there exists a triple $(R,S,v)$ as above such that $v$ is not integral over $R$ but whenever $R[v]\subseteq R^{\prime}\subseteq S$ is an intermediate ring then $R^{\prime}$ is ring-finite over $R.$
No. Indeed, take any example of a non-integral element $v$ over $R$ in some $R$-algebra and let $S$ be the subalgebra generated by $v$. Then the only choice of $R'$ is $S$ itself, which is ring-finite over $R$.