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 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. $\square$

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## Comments (1)

Comment #6320 by Takumi Murayama on

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