Are noetherian valuation rings maximal subrings?

Question

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!

Answer 1

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 $K.$

Answer 2

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.