The Stacks project

Lemma 10.126.7. Let $R$ be a ring. Let $S$, $S'$ be of finite presentation over $R$. Let $\mathfrak q \subset S$ and $\mathfrak q' \subset S'$ be primes. If $S_{\mathfrak q} \cong S'_{\mathfrak q'}$ as $R$-algebras, then there exist $g \in S$, $g \not\in \mathfrak q$ and $g' \in S'$, $g' \not\in \mathfrak q'$ such that $S_ g \cong S'_{g'}$ as $R$-algebras.

Proof. Let $\psi : S_{\mathfrak q} \to S'_{\mathfrak q'}$ be the isomorphism of the hypothesis of the lemma. Write $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ r)$ and $S' = R[y_1, \ldots , y_ m]/J$. For each $i = 1, \ldots , n$ choose a fraction $h_ i/g_ i$ with $h_ i, g_ i \in R[y_1, \ldots , y_ m]$ and $g_ i \bmod J$ not in $\mathfrak q'$ which represents the image of $x_ i$ under $\psi $. After replacing $S'$ by $S'_{g_1 \ldots g_ n}$ and $R[y_1, \ldots , y_ m, y_{m + 1}]$ (mapping $y_{m + 1}$ to $1/(g_1\ldots g_ n)$) we may assume that $\psi (x_ i)$ is the image of some $h_ i \in R[y_1, \ldots , y_ m]$. Consider the elements $f_ j(h_1, \ldots , h_ n) \in R[y_1, \ldots , y_ m]$. Since $\psi $ kills each $f_ j$ we see that there exists a $g \in R[y_1, \ldots , y_ m]$, $g \bmod J \not\in \mathfrak q'$ such that $g f_ j(h_1, \ldots , h_ n) \in J$ for each $j = 1, \ldots , r$. After replacing $S'$ by $S'_ g$ and $R[y_1, \ldots , y_ m, y_{m + 1}]$ as before we may assume that $f_ j(h_1, \ldots , h_ n) \in J$. Thus we obtain a ring map $S \to S'$, $x_ i \mapsto h_ i$ which induces $\psi $ on local rings. By Lemma 10.6.2 the map $S \to S'$ is of finite presentation. By Lemma 10.126.6 we may assume that $S' = S \times C$. Thus localizing $S'$ at the idempotent corresponding to the factor $C$ we obtain the result. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 10.126: Algebras and modules of finite presentation

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Your email address will not be published.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 00QS. Beware of the difference between the letter 'O' and the digit '0'.