The Stacks project

Lemma 10.126.11. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $S \to S'$ be an $R$-algebra map. Assume that

  1. $I$ is locally nilpotent,

  2. $S/IS \to S'/IS'$ is an isomorphism,

  3. $S$ is of finite type over $R$,

  4. $S'$ of finite presentation over $R$, and

  5. $S'$ is flat over $R$.

Then $S \to S'$ is an isomorphism.

Proof. By Lemma 10.126.9 the map $S \to S'$ is surjective. As $I$ is locally nilpotent, so are the ideals $IS$ and $IS'$ (Lemma 10.32.3). Hence every prime ideal $\mathfrak q$ of $S$ contains $IS$ and (trivially) $S_\mathfrak q/IS_\mathfrak q \cong S'_\mathfrak q/IS'_\mathfrak q$. Thus Lemma 10.126.10 applies and we see that $S_\mathfrak q \to S'_\mathfrak q$ is an isomorphism for every prime $\mathfrak q \subset S$. It follows that $S \to S'$ is injective for example by Lemma 10.23.1. $\square$

Comments (1)

Comment #213 by Rex on

Typo: "is an ismorphism"

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Your email address will not be published.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 07RE. Beware of the difference between the letter 'O' and the digit '0'.