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Let $R\subseteq S$ be a finite ring extension such that $R$ and $S$ are both complete noetherian local domains. Let $\mathfrak{p}$ be a prime ideal of $R$ and denote by $\mathfrak{q}_1,\ldots, \mathfrak{q}_n$ all the prime ideals of $S$ that lie over $\mathfrak{p}$.

Assume that $R$ is regular and the local rings $S_{\mathfrak{q_1}}, \ldots, S_{\mathfrak{q}_n}$ are all of finite global dimension, can we deduce that $S_\mathfrak{p}$ is of finite global dimension?

Any comments are welcome. Thanks a lot!

G.-S. Zhou
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    Let $A$ be a Noetherian ring. If global dimension of $A_P$ is at most $n$ for all maximal ideals $P$ of $A$, then global dimension of $A$ is at most $n$. In your case, take $A=S_{\mathfrak{p}}$ and $n$ to be the maximum global dimensions of $S_{\mathfrak{q}_i}$. – Mohan Mar 10 '18 at 15:47
  • @Mohan, Thanks a lot. – G.-S. Zhou Mar 11 '18 at 10:01

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