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. $\square$

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).