Lemma 31.13.5. Let $S$ be a scheme. Let $D \subset S$ be an effective Cartier divisor. Let $s \in D$. If $\dim _ s(S) < \infty$, then $\dim _ s(D) < \dim _ s(S)$.

Proof. Assume $\dim _ s(S) < \infty$. Let $U = \mathop{\mathrm{Spec}}(A) \subset S$ be an affine open neighbourhood of $s$ such that $\dim (U) = \dim _ s(S)$ and such that $D = V(f)$ for some nonzerodivisor $f \in A$ (see Lemma 31.13.2). Recall that $\dim (U)$ is the Krull dimension of the ring $A$ and that $\dim (U \cap D)$ is the Krull dimension of the ring $A/(f)$. Then $f$ is not contained in any minimal prime of $A$. Hence any maximal chain of primes in $A/(f)$, viewed as a chain of primes in $A$, can be extended by adding a minimal prime. $\square$

## Comments (1)

Comment #736 by Keenan Kidwell on

The $X$ in the second line of the proof should be $s$.

There are also:

• 3 comment(s) on Section 31.13: Effective Cartier divisors

## 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Your email address will not be published.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 056N. Beware of the difference between the letter 'O' and the digit '0'.