Lemma 10.112.6. Let $R \to S$ be a homomorphism of Noetherian rings. Let $\mathfrak q \subset S$ be a prime lying over the prime $\mathfrak p$. Then
Proof. We use the characterization of dimension of Proposition 10.60.9. Let $x_1, \ldots , x_ d$ be elements of $\mathfrak p$ generating an ideal of definition of $R_{\mathfrak p}$ with $d = \dim (R_{\mathfrak p})$. Let $y_1, \ldots , y_ e$ be elements of $\mathfrak q$ generating an ideal of definition of $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ with $e = \dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q})$. It is clear that $S_{\mathfrak q}/(x_1, \ldots , x_ d, y_1, \ldots , y_ e)$ has a nilpotent maximal ideal. Hence $x_1, \ldots , x_ d, y_1, \ldots , y_ e$ generate an ideal of definition of $S_{\mathfrak q}$. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #4320 by Zhiyu Zhang on
Comment #4478 by Johan on