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Let $R$ and $S$ be two local principal ideal domains with the same field of fractions $K$. I want to show that if $R\subseteq S$ then $R=S$.

I will denote as $\mathfrak{m}_R=(m_R)$ and $\mathfrak{m}_S=(m_S)$ the unique maximal ideal of the local rings $R$ and $S$ respectively. Then every non-unit of $R$ is of the form $x=m_R^nu$ for some $n\in\mathbb{N} $ and some unit $u\in R$. Same goes for $S$. I though of using the following Lemma to attack this problem.

$\textbf{Lemma}$ $\colon$ Let $R$ be a local PID with field of fractions $K$. Let $S$ be any local domain with $R\subseteq S\subseteq K$. If $\mathfrak{m}_R \subseteq \mathfrak{m}_S$, then $R=S$

I know that for local $R$ it is true that its maximal ideal consists of all the non-units. Also, if $R$ is a PID then, $x\in K$ implies that $x\in R$ or $x^{-1} \in R$ (or both). Let $x\in \mathfrak{m}_R$, thus $x^{-1} \notin R$. Now $x^{-1}$ cannot be in $\mathfrak{m}_S$ because then, since $x\in \mathfrak{m}_R\Rightarrow x \in R \Rightarrow x\in S$ we would have that $1=xx^{-1} \in \mathfrak{m}_S$ which is a contradiction. How can i prove that in fact $x^{-1}$ can't be in $S$?

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First show that $S=T^{-1}R$ where $T$ is a multiplicative subset of $R$. This is true whenever $R$ is a PID. Now use the fact that R is local PID to show that $T^{-1}R = R $ or $K$

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  • $\begingroup$ Thanks! Your answer was enlightening. The intuition i thought of is that if an element $x\in \mathfrak{m}_R$ ie $x=m_R^nu$ is a unit in $S$ then, $m_R$ is invertible in $S$ and hence, every element of $R$ is invertible in $S$. $\endgroup$ – Cornelius Dec 11 '18 at 10:28
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This is not true: Take $S = K$ the field of fractions of $R$.

This is the only counter-example: if $R \subseteq S \subseteq K$ and $S$ contains an element of the form $u m_{R}^{n}$ with $u \in R^{\times}$ and $n < 0$, it follows that $S = K$. Thus either $S = R$ or $S = K$.

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    $\begingroup$ The exercise as i wrote it, is Exercise 2.4 in Lorenzini's book An invitation in Arithmetic Geometry I see why you say it is not true. Can it be true if we assume that $R\subseteq S \subsetneq K$ $\endgroup$ – Cornelius Dec 8 '18 at 13:28
  • $\begingroup$ I can't see why an element of the form $um_R^n$ with $u \in R^{\times}$ and $n>0$ is in $S$ implies that $S=K$. Can you be more specific ? $\endgroup$ – Cornelius Dec 8 '18 at 17:11
  • $\begingroup$ Every element of $K$ is of the form $u m_R^n$ with $n \in \mathbb Z$. If $m_R^{n_0} \in S$ with $n_0 < 0$, then for each $n < 0$, $um_R^{n} = um_R^{n-N n_0} \cdot (m_R^{n_0})^{N} \in R \cdot S = S$, by choosing $N$ large enough so that $n-Nn_0 \geq 0$. $\endgroup$ – punctured dusk Dec 8 '18 at 17:15

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