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Let $k$ be a field of characteristic zero and $I$ be a radical ideal of $k[[x_1,\ldots, x_n]]$. Let $P$ be a minimal prime ideal of the reduced ring $R:=k[[x_1,\ldots, x_n]]/I$. Then, $R_P$ is a field extension of $k$. Let $s:=\dim(R/P)$.

My question is: Is it true that $R_P$ is a finite extension over $k((X_1,\ldots,X_s))$ ?

This would be an "analytic" analogue of the standard fact on $k$-algebras of essentially finite type.

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Clearly $R_P=(R/P)_P$, and so we may replace $R$ by $R/P$. Since its dimension is $s$, one can find a system of parameters $y_1,y_2,\ldots,y_s$ in the maximal ideal of $R/P$. Then the inclusion $k[[y_1,y_2,\ldots,y_n]]\to R/P$ is finite, stronger than what you ask. A proof of the last statement can be found in many places (essentially Weierstrass preparation), for example, Theory of analytic spaces by Raghavan Narasimhan.

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  • $\begingroup$ thanks ... so this is like a power series version of Noether normalization, right? can you please mention where in Narasimhan's book this is mentioned? $\endgroup$
    – Alex
    Commented Jul 19 at 18:31
  • $\begingroup$ @Alex I think in the first ten pages. $\endgroup$
    – Mohan
    Commented Jul 19 at 20:40

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