The Stacks project

Lemma 16.13.4. Let $R$ be a Noetherian ring. Let $\mathfrak p \subset R$ be a prime ideal. Let $f_1, \ldots , f_ m \in R[x_1, \ldots , x_ n]$. Suppose that $(a_1, \ldots , a_ n) \in ((R_\mathfrak p)^\wedge )^ n$ is a solution. If $R_\mathfrak p$ is a G-ring, then for every integer $N$ there exist

  1. an étale ring map $R \to R'$,

  2. a prime ideal $\mathfrak p' \subset R'$ lying over $\mathfrak p$

  3. a solution $(b_1, \ldots , b_ n) \in (R')^ n$ in $R'$

such that $\kappa (\mathfrak p) = \kappa (\mathfrak p')$ and $a_ i - b_ i \in (\mathfrak p')^ N(R'_{\mathfrak p'})^\wedge $.

Proof. By Theorem 16.13.2 we can find a solution $(b'_1, \ldots , b'_ n)$ in some ring $R''$ étale over $R_\mathfrak p$ which comes with a prime ideal $\mathfrak p''$ lying over $\mathfrak p$ such that $\kappa (\mathfrak p) = \kappa (\mathfrak p'')$ and $a_ i - b'_ i \in (\mathfrak p'')^ N(R''_{\mathfrak p''})^\wedge $. We can write $R'' = R' \otimes _ R R_\mathfrak p$ for some étale $R$-algebra $R'$ (see Algebra, Lemma 10.143.3). After replacing $R'$ by a principal localization if necessary we may assume $(b'_1, \ldots , b'_ n)$ come from a solution $(b_1, \ldots , b_ n)$ in $R'$. Setting $\mathfrak p' = R' \cap \mathfrak p''$ we see that $R''_{\mathfrak p''} = R'_{\mathfrak p'}$ which finishes the proof. $\square$

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Your email address will not be published.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