Is there a DVR with quotient field $k(X)$ that doesn't contain $k$? What if $k$ is algebraically closed? What if $k=\mathbb{Q}$?

Question

Some background - I've been working on Fulton's Algebraic Curves, problem 2.27, and it is raising a lot of questions. I've tried to prove one general theorem so far here though my proof may be wrong, and is definitely cumbersome. In my third reference there it is assumed that if $k$ is algebraically closed and $R$ is a DVR with quotient field $k(X)$ then $k\subseteq R$. It was also implied that this wouldn't be true if $k$ wasn't algebraically closed. Can anyone give a formal proof of the first and an example of the second? Preferably with $\mathbb{Q}$ as the ring, but any other ring would also be appreciated.

Answer 1

Definition. If $A,B$ are local rings contained in a field $K$, then $B$ dominates $A$ if $A\subset B$ and $\mathfrak{m}_A = A\cap \mathfrak{m}_B$.

Theorem. (Bourbaki's Algebra, Ch. VI, Section 1, 3) Let $K$ be a field. A local ring $R$ contained in $K$ is a valuation ring of $K$ iff it is a maximal element of the set of local rings contained in $K$ with respect to the relation of domination. Every local ring contained in $K$ is dominated by some valuation ring of $K$.

This theorem will give us the tools to show that some pretty wild stuff happens here in response to your question. Suppose $k$ is a field which contains a local ring $R_0$ with $r\in R_0$ a nonzero nonunit. Then by the above theorem, for any field extension $k\subset K$, there is a DVR $R\subset K$ which contains $R_0$, has field of fractions $K$, and $r\in R$ is again a nonzero nonunit. This implies that $k\not\subset R$, so plentiful counterexamples to your statement abound. Unfortunately, they're hard to explicitly describe - if $k=\Bbb Q$ or $k=\Bbb C$, we can take any affine $k$-variety $X$, take $R_0=\Bbb Z_{(p)}$, and then let $R$ be some valuation ring dominating $R_0$ inside $K=k(X)$. This will have $p$ non-invertible, which means the ring certainly can't contain $k$.

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When studying a $k$-variety $X$, the reason we frequently restrict ourselves to considering DVRs $R$ satisfying $k\subset R \subset k(X)$ is because we're interested in looking at interactions with other things defined over $k$. There's all sorts of interesting behavior when you don't add this assumption - sometimes that's what you want, sometimes it's not.