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> [!proposition] Proposition. ([[characterization of integrality over an ideal (in an integrally closed domain)]])
>
Let $A\subset B$ be an extension of [[integral domain|integral domains]], where $A$ is [[integral closure|integrally closed]][^1]. Let $\mathfrak{a} \subset A$ be an [[ideal]], and take $b \in B$. Consider the [[field extension]] $\text{Frac }A \subset \text{Frac }B$. Then the following are equivalent:
>1. $b$ is [[integrality over an ideal|integral over]] $\mathfrak{a}$;
>2. $b$ is [[integrality over an ideal|integral over]] the [[radical of an ideal|radical]] $\sqrt{ \mathfrak{a} }$ ;
>3. $\underbrace{\frac{b}{1}}_{ \in \text{Frac }B}$ is [[algebraic element|algebraic]] over $\text{Frac }A$ with [[classification of simple field extensions|minimal polynomial]] over $\text{Frac }A$ of the form $T^{n}+\frac{a_{1}}{1}T^{n-1}+\dots+ \frac{a_{n}}{1}T^{0}$, $n \geq 1$, $a_{i} \in \sqrt{ \mathfrak{a} }$.
^proposition
> [!note] Remark.
> This is useful for (at least) two reasons:
> 1. Working with [[field|fields]] rather than [[ring|rings]] is nice, since we can divide;
> 2. $\mathfrak{a}$-integrality of $b$ in general just says that there is *some* [[polynomial 4|polynomial]] with coefficients in $\mathfrak{a}$ producing an [[integral element of an algebra|integrality equation]]. But in this case we are saying that we're concerned with a *very specific* [[polynomial 4|polynomial]] — if its coefficients are not in the [[radical of an ideal|radical]] $\sqrt{ \mathfrak{a} }$, then we can rule out $\mathfrak{a}$-integrality of $b$.
>
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####
[^1]: Recall that, for an [[integral domain]], "$A$ is integrally closed" automatically means "$A$ is integrally closed in its [[field of fractions]] $\text{Frac }A
quot;.
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#### References
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> ```
> [!frontlink]
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> ```