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

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 $$S$$?

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 $$K$$
• 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 $S$. – Cornelius Dec 11 '18 at 10:28
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 $$S = K$$.
• 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 $R\subseteq S \subsetneq K$ – Cornelius Dec 8 '18 at 13:28
• 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 ? – Cornelius Dec 8 '18 at 17:11
• 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 $n-Nn_0 \geq 0$. – punctured dusk Dec 8 '18 at 17:15