In a ring extension $R\subseteq S$, in order to check whether an element of $S$ is integral over $R$, I wonder why it is enough to check the minimal polynomial? In a concrete example, I found that the minimal polynomial was not monic (which we require for the element to be integral), how can I conclude from here that there is no monic polynomial with coefficients in $R$ annihilating the considered element of $S$? Couldn’t there be such a polynomial that simply has higher degree than the minimal one? Thanks for your help!
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This is true iff $R$ is integrally closed, as follows by various forms of Gauss's Lemma, e.g. see here. – Bill Dubuque Jan 07 '20 at 17:14
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In the same vein as Bill's comment.... consider $D \subseteq K$ where $D$ is a domain and $K$ its field of fractions. Any element of $a/b \in K$ is a root of the linear polynomial $bx -a$. Meanwhile, an element of $K \setminus D$ is integral over $D$ iff it is the root of a monic non-linear polynomial. – Badam Baplan Jan 07 '20 at 20:14