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I encounter the following result in a lecture notes about algebraic curves:

Let $R,S$ be noetherian valuation rings over their common field of fractions $K$, and $R,S$ are not fields. Let $M=R\setminus R^*$, $N=S\setminus S^*$, then:

  • $M,N$ are principal ideals;
  • $R,S$ are maximal subrings of $K$;
  • $M\subseteq N\Leftrightarrow M=N\Leftrightarrow R=S\Leftrightarrow R\subseteq S$.

For the 2nd statement, I am unsure how to prove. It is not hard to show that $R$, $S$ are maximal valuation rings of $K$, but I am not sure why $K$ cannot have any bigger subrings that does not admit a valuation.

Could anyone confirm me that the original statement is correct by giving a clue of proof, or give a counterexample that says it is not true?

Thank you very much!

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In this situation, $R$ and $S$ are in fact discrete valuation rings. Suppose that $M = (\pi).$ Then all nonzero elements of $R$ can be written uniquely as $u\pi^n,$ where $n\geq 0$ and $u\in R^\times,$ which implies that you can write any nonzero element of $K$ uniquely as $u\pi^n,$ where $u\in R^\times$ and $n\in\Bbb{Z}$.

Try proving the claims I've made above, and use the presentation of elements given to show that if you have a ring $R'$ such that $R\subsetneq R'\subseteq K,$ then it necessarily must be $K.$

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    $\begingroup$ Thanks for the hint! I proceed like this: Take $u\cdot \pi^{-n}\in R'\setminus R$ where $u$ is a unit of $R$ and $n\ge 1$, then $\pi^{-1}=u\cdot\pi^{-n}\cdot u^{-1}\pi^{n-1}\in R'R=R'$, moreover, $\pi\in R\subsetneq R'$, so for any $k=v\pi^{n}\in K$ where $n\in \mathbb Z$, $k\in R'$. $\endgroup$ – Ivon Mar 2 '20 at 20:32
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    $\begingroup$ @Ivon Exactly; the point is that $K = R[1/\pi]$ already! $\endgroup$ – Stahl Mar 2 '20 at 20:34
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I have just noticed that, besides @Stahl's elegant answer using DVRs, the statement itself is in fact equivalent to what I have shown before that they are just maximal valuation rings. This is because any intermediate ring between a valuation ring and its field of fractions is still a valuation ring.

To show $R$ is a maximal valuation ring, assume $R\subseteq T\subsetneq K$ for another valuation ring $T$ that is not a field. If $Q$ is the nonzero maximal ideal of $T$, then $Q\subseteq M=(m)$. Now there exists $k$ such that $m^k\in Q$ which implies $m\in Q$ as $Q$ is maximal, hence prime. This gives $M=Q$, so $R=T$.

We have used the fact that any such ring $R$ has principal maximal ideal. In fact, since $R$ is noetherian, $M=(m_1,\cdots,m_k)$. Then as a valuation ring, $R$ is in particular uniserial, i.e. any two ideals of $R$ are comparable. Thus we can choose $1\le k\le n$ such that $(m_k)\supseteq (m_i)$, $\forall 1\le i\le n$, and $M=(m_k)$.

In fact, this proof uses some known results about valuation rings, which is equivalent to the fact that $R,S$ are indeed discrete valuation rings.

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