Let $R \subseteq S$ be two local PIDs with the same field of fractions, then $R=S$.

Question

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 $S$?

Answer 1

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 $K$

Answer 2

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 $S = K$.

Answer 3

Here is my attempt.

R and S be defined as given in the statement. Let $\displaystyle M_{R}$ and $\displaystyle M_{S}$ be the unique maximal ideals of R and S respectively.

Claim: $\displaystyle M_{R} \subseteq M_{S}$

Let $\displaystyle x\in M_{R}$ then this implies that $\displaystyle x$ is a non unit and $\displaystyle x\in R\subseteq S$. And hence $\displaystyle x\in S$ and it is also a non unit in S. Since $\displaystyle M_{S}$ is maximal ideal of non units in S, hence $\displaystyle x\in M_{S}$. Claim is proved . Now by the Lemma given in the question we conclude that $\displaystyle R=S$.