Lemma 10.36.12 (034K)—The Stacks project

An element of an algebra over a ring is integral over the ring if and only if it is locally integral at every prime ideal of the ring.

Lemma 10.36.12. Let $\varphi : R \to S$ be a ring map. Let $x \in S$. The following are equivalent:

$x$ is integral over $R$, and
for every prime ideal $\mathfrak p \subset R$ the element $x \in S_{\mathfrak p}$ is integral over $R_{\mathfrak p}$.

Proof. It is clear that (1) implies (2). Assume (2). Consider the $R$-algebra $S' \subset S$ generated by $\varphi (R)$ and $x$. Let $\mathfrak p$ be a prime ideal of $R$. Then we know that $x^ d + \sum _{i = 1, \ldots , d} \varphi (a_ i) x^{d - i} = 0$ in $S_{\mathfrak p}$ for some $a_ i \in R_{\mathfrak p}$. Hence we see, by looking at which denominators occur, that for some $f \in R$, $f \not\in \mathfrak p$ we have $a_ i \in R_ f$ and $x^ d + \sum _{i = 1, \ldots , d} \varphi (a_ i) x^{d - i} = 0$ in $S_ f$. This implies that $S'_ f$ is finite over $R_ f$. Since $\mathfrak p$ was arbitrary and $\mathop{\mathrm{Spec}}(R)$ is quasi-compact (Lemma 10.17.8) we can find finitely many elements $f_1, \ldots , f_ n \in R$ which generate the unit ideal of $R$ such that $S'_{f_ i}$ is finite over $R_{f_ i}$. Hence we conclude from Lemma 10.23.2 that $S'$ is finite over $R$. Hence $x$ is integral over $R$ by Lemma 10.36.4. $\square$

Comments (4)

Comment #886 by Konrad Voelkel on August 07, 2014 at 17:39

In the statement, instead of $\mathfrak p \in R$ it should say $\mathfrak p \subset R$ prime ideal. In the proof, instead of Assume (1) it should say Assume (2).

Proposed slogan: An element of an algebra over a ring is integral over the ring if and only if it is locally integral at every prime ideal of the ring.

Comment #903 by Johan on August 10, 2014 at 12:57

Thanks very much! Fixed in this commit.

Comment #8446 by Elías Guisado on June 01, 2023 at 05:18

I think instead of "we have $a_i \in R_f$ and $x^d + \sum_{i = 1, \ldots, d} \varphi(a_i) x^{d - i} = 0$ in $S_f$ " it would be more informative to write "there are $b_i\in R_f$ such that $x^d + \sum_{i = 1, \ldots, d} \varphi(b_i) x^{d - i} = 0$ in $S_f$ ." At least in the calculation I derived I end up with different $a_i$ 's (this doesn't affect the rest of the proof).

Comment #9068 by Stacks project on May 06, 2024 at 14:43

Well, to the extent that $R_f$ is not a subring of $R_\mathfrak p$ I agree that the $a_i$ in the first instance are different from the $a_i$ in the second instance. But if $R$ is a domain, say, then $R_f$ may be viewed as a subring of $R_\mathfrak p$ and in that case you can take those $a_i$ to be the same. Anyway, since the sentence is not claiming these $a_i$ are the same, I think I am going to leave this as is until others complain.

There are also:

2 comment(s) on Section 10.36: Finite and integral ring extensions

Comments (4)

Post a comment