The Stacks project

Lemma 10.37.4. Let $R$ be a domain with fraction field $K$. If $u, v \in K$ are almost integral over $R$, then so are $u + v$ and $uv$. Any element $g \in K$ which is integral over $R$ is almost integral over $R$. If $R$ is Noetherian then the converse holds as well.

Proof. If $ru^ n \in R$ for all $n \geq 0$ and $v^ nr' \in R$ for all $n \geq 0$, then $(uv)^ nrr'$ and $(u + v)^ nrr'$ are in $R$ for all $n \geq 0$. Hence the first assertion. Suppose $g \in K$ is integral over $R$. In this case there exists an $d > 0$ such that the ring $R[g]$ is generated by $1, g, \ldots , g^ d$ as an $R$-module. Let $r \in R$ be a common denominator of the elements $1, g, \ldots , g^ d \in K$. It is follows that $rR[g] \subset R$, and hence $g$ is almost integral over $R$.

Suppose $R$ is Noetherian and $g \in K$ is almost integral over $R$. Let $r \in R$, $r\not= 0$ be as in the definition. Then $R[g] \subset \frac{1}{r}R$ as an $R$-module. Since $R$ is Noetherian this implies that $R[g]$ is finite over $R$. Hence $g$ is integral over $R$, see Lemma 10.36.3. $\square$

Comments (1)

There are also:

  • 3 comment(s) on Section 10.37: Normal rings

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Your email address will not be published.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