The Stacks project

A composition of integral ring maps is integral

Lemma 10.35.6. Suppose that $R \to S$ and $S \to T$ are integral ring maps. Then $R \to T$ is integral.

Proof. Let $t \in T$. Let $P(x) \in S[x]$ be a monic polynomial such that $P(t) = 0$. Apply Lemma 10.35.4 to the finite set of coefficients of $P$. Hence $t$ is integral over some subalgebra $S' \subset S$ finite over $R$. Apply Lemma 10.35.4 again to find a subalgebra $T' \subset T$ finite over $S'$ and containing $t$. Lemma 10.7.3 applied to $R \to S' \to T'$ shows that $T'$ is finite over $R$. The integrality of $t$ over $R$ now follows from Lemma 10.35.3. $\square$


Comments (1)

Comment #3782 by slogan_bot on

Suggested slogan: A composition of integral ring maps is integral.


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Your email address will not be published.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