# Are noetherian valuation rings maximal subrings?

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!

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 $$K.$$

• 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 $n\in \mathbb Z$, $k\in R'$. – Ivon Mar 2 '20 at 20:32
• @Ivon Exactly; the point is that $K = R[1/\pi]$ already! – Stahl Mar 2 '20 at 20:34

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.