Processing math: 4%

The Stacks project

Lemma 31.22.4. Let X \to S be a morphism of schemes. Let Z \to X be an immersion. Assume

  1. X \to S is flat and locally of finite presentation,

  2. Z \to X is a relative quasi-regular immersion.

Then Z \to X is a regular immersion and the same remains true after any base change.

Proof. Pick x \in Z with image s \in S. To prove this it suffices to find an affine neighbourhood of x contained in U such that the result holds on that affine open. Hence we may assume that X is affine and there exist a quasi-regular sequence f_1, \ldots , f_ r \in \Gamma (X, \mathcal{O}_ X) such that Z = V(f_1, \ldots , f_ r). By More on Algebra, Lemma 15.31.4 the sequence f_1|_{X_ s}, \ldots , f_ r|_{X_ s} is a quasi-regular sequence in \Gamma (X_ s, \mathcal{O}_{X_ s}). Since X_ s is Noetherian, this implies, possibly after shrinking X a bit, that f_1|_{X_ s}, \ldots , f_ r|_{X_ s} is a regular sequence, see Algebra, Lemmas 10.69.6 and 10.68.6. By Lemma 31.18.9 it follows that Z_1 = V(f_1) \subset X is a relative effective Cartier divisor, again after possibly shrinking X a bit. Applying the same lemma again, but now to Z_2 = V(f_1, f_2) \subset Z_1 we see that Z_2 \subset Z_1 is a relative effective Cartier divisor. And so on until on reaches Z = Z_ n = V(f_1, \ldots , f_ n). Since being a relative effective Cartier divisor is preserved under arbitrary base change, see Lemma 31.18.1, we also see that the final statement of the lemma holds. \square


Comments (0)


Your email address will not be published.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.