Given domain $R$ and $\frac{a}{b}\in frac(R)$, suppose $\frac{a}{b}$ is integral over $R_\frak{m}$ for any maximal ideal ${\frak m}$ of $R$. Then is it right that $\frac{a}{b}$ is integral over $R$?
I know that $R$ is integral closed if and only if each localization w.r.t. a maximal ideal is integral closed. Can I solve my problem by this assertion?
Let $K$ be the field of fractions of $R$, and suppose $x \in K$ is integral over each localization $R_{\mathfrak m}$. Let
$$I = \{ r \in R : rx \textrm{ is integral over } R \}$$
Then $I$ is an ideal of $R$. If we can show that
$$I_{\mathfrak m} = \{r \in R_{\mathfrak m} : rx \textrm{ is integral over } R_{\mathfrak m}\}$$
then we will have $I_{\mathfrak m} = R_{\mathfrak m}$ for all $\mathfrak m$, hence $I = R$.