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Let $k$ be a field of characteristic $p$ and $K$ the field of fractions of the formal power series ring $k[[X_1,\dots,X_n]]$. Let $L$ be a finite purely inseparable field extension of $K$, then there exists an integer $m$ such that $L\subseteq K^{{1/q}}$ with $q=p^m$. Set $k'=k^{1/q}$ and $Y_i=X_i^{1/q}$, let $K'$ and $k'((Y_1,\dots,Y_n))$ denote the field of fractions of $k[[Y_1,\dots,Y_n]]$ and $k'[[Y_1,\dots,Y_n]]$ respectively, then $K'\subseteq L(Y_1,\dots,Y_n) \subseteq k'((Y_1,\dots,Y_n))$. The integral closure $A$ of $k[[Y_1,\dots,Y_n]]$ in $L(Y_1,\dots,Y_n)$ is just $k'[[Y_1,\dots,Y_n]] \cap L(Y_1,\dots,Y_n)$. Now the question is:

Why the maximal ideal $\mathfrak{m}$ of $A$ is generated by $Y_1,\dots, Y_n$?

From $k[[Y_1,\dots,Y_n]] \subseteq A \subseteq k'[[Y_1,\dots,Y_n]]$ and $A$ is integral over $k[[Y_1,\dots,Y_n]]$, I can only know $(Y_1,\dots,Y_n)$ is $\mathfrak{m}$-primary.

nick
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  • If every element in the complement of a proper ideal is invertible then the ideal is maximal. – OR. Aug 18 '13 at 10:03
  • @ABC, thank you. But I still do not know how to show the contraction of the ideal $(Y_1,\cdots,Y_n)$ of $k^{'}[[Y_1,\cdots,Y_n]]$ in $A$ is still $(Y_1,\cdots,Y_n)$. – nick Aug 18 '13 at 15:47

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