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Let $(R,\mathfrak m)$ be a ramified complete regular local ring of dimension $n$ with $\operatorname{char} R/\mathfrak m=p$. Then $\operatorname{ht}(pR)=1$ and the question is:

How to choose a regular system of parameters $x_1,\dots,x_n$ of $R$ such that $p,x_2,\dots,x_n$ is a system of parameters of $R$, that is, $\sqrt{(p,x_2,\dots,x_n)}=\mathfrak m$?

nick
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  • @YACP, the definition of the ramification of $R$ is that $R$ is of unequal characteristic and $p\in m^2$. Since $R$ is a regular local ring, $R$ is a domain, so ramification of $R$ means $Char R=0$, $Char R/m=p$ and $p\in m^2$; also $p\in m^2$ means $p$ is not in any regular system of parameters of $R$ – nick Jun 01 '13 at 09:24
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    Ok, I've read unramified at first sight. (Btw, why don't include your comment into the body of the question?) Anyway, nice question inspired by Matsumura's proof of Theorem 29.8 who skip this elegantly by saying: "by a skilful choice of a regular system of parameters" –  Jun 01 '13 at 22:43

1 Answers1

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Let $x_1,\dots,x_n$ be a regular system of parameters, that is, $\mathfrak m=(x_1,\dots,x_n)$. Then $\overline R=R/(x_1,\dots,x_{n-1})$ is a regular local ring of dimension $1$ and its maximal ideal $\overline{\mathfrak m}=\mathfrak m/(x_1,\dots,x_{n-1})$ is generated by $\overline x_n$. We have $\overline p\in\overline{\mathfrak m}^2$, and therefore there exists an invertible element $\overline a\in\overline R$ and $t\geq 2$ such that $\overline p=\overline a\overline x_n^t$. This shows that $x_n^t\in(p,x_1,\dots,x_{n-1})$ and thus we get $\mathfrak m^t\subseteq (p,x_1,\dots,x_{n-1})$ which means that $p,x_1,\dots,x_{n-1}$ is a system of parameters.

However a problem can occur in the proof above: we could have $\overline p=\overline 0$, that is, $p\in(x_1,\dots,x_{n-1})$.

Now our task is to choose a regular system of parameters $x_1,\dots,x_n$ such that $p\notin(x_1,\dots,x_{n-1})$. We shall do this by induction on $n\ge 1$. If $n=1$ there is nothing to prove. In the general case take $(y_1,\dots,y_n)=\mathfrak m$ an arbitrary regular system of parameters. We know that $\operatorname{ht}(p)=1$ and consider $\mathfrak p_1,\dots,\mathfrak p_r$ all height $1$ prime ideals of $R$ containing $(p)$. None of these prime ideals contain all $y_i$. We can assume $y_n\notin\mathfrak p_1,\dots,\mathfrak p_s$ and $y_n\in\mathfrak p_{s+1},\dots,\mathfrak p_r$ and define $x_n=y_n+\sum_{j=s+1}^ra_jy_{k_j}$, where $y_{k_j}\notin\mathfrak p_j$ and $a_i\in\bigcap_{j\neq i}\mathfrak p_j-\mathfrak p_i$. Then $x_n\notin\bigcup_{i=1}^r\mathfrak p_i$. The elements $y_1,\dots,y_{n-1},x_n$ are also a regular system of parameters and moreover $p\notin (x_n)$, otherwise $(p)\subseteq (x_n)$ and $(x_n)$ is a prime ideal of height $1$, so $(x_n)$ must coincide with some $\mathfrak p_i$, a contradiction. All we have to do now is to reduce the problem to $R/(x_n)$ and apply the induction hypothesis.