The Stacks project

Lemma 10.125.9. Let $R$ be a valuation ring with residue field $k$ and field of fractions $K$. Let $S$ be a domain containing $R$ such that $S$ is of finite type over $R$. If $S \otimes _ R k$ is not the zero ring then

\[ \dim (S \otimes _ R k) = \dim (S \otimes _ R K) \]

In fact, $\mathop{\mathrm{Spec}}(S \otimes _ R k)$ is equidimensional.

Proof. It suffices to show that $\dim _{\mathfrak q}(S/k)$ is equal to $\dim (S \otimes _ R K)$ for every prime $\mathfrak q$ of $S$ containing $\mathfrak m_ RS$. Pick such a prime. By Lemma 10.125.6 the inequality $\dim _{\mathfrak q}(S/k) \geq \dim (S \otimes _ R K)$ holds. Set $n = \dim _{\mathfrak q}(S/k)$. By Lemma 10.125.2 after replacing $S$ by $S_ g$ for some $g \in S$, $g \not\in \mathfrak q$ there exists a quasi-finite ring map $R[t_1, \ldots , t_ n] \to S$. If $\dim (S \otimes _ R K) < n$, then $K[t_1, \ldots , t_ n] \to S \otimes _ R K$ has a nonzero kernel. Say $f = \sum a_ I t_1^{i_1}\ldots t_ n^{i_ n}$. After dividing $f$ by a nonzero coefficient of $f$ with minimal valuation, we may assume $f\in R[t_1, \ldots , t_ n]$ and some $a_ I$ does not map to zero in $k$. Hence the ring map $k[t_1, \ldots , t_ n] \to S \otimes _ R k$ has a nonzero kernel which implies that $\dim (S \otimes _ R k) < n$. Contradiction. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00QK. Beware of the difference between the letter 'O' and the digit '0'.