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:

1. $x$ is integral over $R$, and

2. 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 10.17.10) 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. $\square$

Comment #886 by on

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.

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