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In chapter 5 of Atiyah Macdonald

Lemma 5.14: Let $C$ is the integral closure of $A$ in $B$. Then the integral closure of ideal $\mathfrak{a}$ in $B$ is the $r(\mathfrak{a}^e)$ where $\mathfrak{a}^e$ is the extension of $\mathfrak{a}$ in $C$

Proposition 5.15: Let $A \subseteq B$ be integral domains, $A$ integrally closed, and let $x \in B$ be integral over an ideal $\mathfrak{a}$ of $A$. Then $x$ is algebraic over the field of fractions $K$ of $A$, and if its minimal polynomial over $K$ is $t^n + a_1 t^{n-1} + \cdots+ a_n$, then $a_1, \dots, a_n$ lie in $r(\mathfrak{a})$ (the radical of $\mathfrak{a}$).

Now in the proof of proposition lemma $x$ is algebraic over $K$. Let $L$ contain all the roots $x_1,\dots,x_n$ of the minimal polynomial of $x$. Each $x_i$ satisfies the same polynomial of integral dependence as $x$ does. So each $x_i$ is integral over $\mathfrak{a}$. The coefficients of the minimal polynomial are polynomials of $x_i$ over $K$.

Then how does Lemma 5.14 implies that the coefficients are integral over $\mathfrak{a}$

Soham Chatterjee
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  • Each $x_i$ is integral of $\mathfrak{a}$.Lemma 5.14 says that the integral closure of $\mathfrak{a}$ in $B$ is $r(\mathfrak{a}^e)$ - it is an ideal.Ideals are closed under multiplication and addition and hence any polynomial in $x_i$'s must be in $r(\mathfrak{a}^e)$ and integral over $\mathfrak{a}$. – user631874 Jul 10 '23 at 08:56
  • How any polynomial in $x_i$'s in $r(\mathfrak{a}^e)$ implies the coefficients are integral over $\mathfrak{a}$ – Soham Chatterjee Jul 11 '23 at 05:29
  • Lemma 5.14 says that exactly.The integral closure of ideal $\mathfrak{a}$ in $B$ is precisely $r(\mathfrak{a}^{e})$,i.e. if $x$ is integral over $\mathfrak{a}$ then $x\in r(\mathfrak{a}^{e})$ and vice versa.Thus if $x_1$ and $x_2$ are integral over $\mathfrak{a}$ then $x_{1}x_{2}$ is too. – user631874 Jul 11 '23 at 11:23
  • If $x_1$ and $x_2$ are integral over $\mathfrak{a}$ then they must be in $r(\mathfrak{a}^{e})$.Hence $x_{1}x_{2}\in r(\mathfrak{a}^{e})\implies x_{1}x_{2}$ is integral over $\mathfrak{a}$ and so any polynomial in $x_{i}$'s too. – user631874 Jul 11 '23 at 11:30

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