Lemma 10.35.12. 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 $R_{\mathfrak p}$.

** 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 10.35.12. 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 10.16.10) 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 10.22.2 that $S'$ is finite over $R$. Hence $x$ is integral over $R$ by Lemma 10.35.4.
$\square$

Your email address will not be published.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.

## Comments (2)

Comment #886 by Konrad Voelkel on

Comment #903 by Johan on