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Let $ R\subset S $ be an integral extension of domains with $R$ normal, $ K $ the quotient field of $R$, $ \alpha\in S,\, x^n + a_1x^{n-1} + \cdots + a_n $ the minimal polynomial of $ \alpha $ over $K$. Show that for any $R$-ideal $I$, $ \alpha\in\sqrt{IS} $ if and only if $ a_i\in\sqrt{I} $ for $ 1\le i\le n. $

I believe I have one direction.

$(\Leftarrow):\,$ Suppose $ a_i\in\sqrt{I} $ for $ 1\le i\le n. $ Observe that $$ \alpha^n = -(a_1\alpha^{n-1} + \cdots + a_n), $$ where the $ a_i $ are in $\sqrt{I}$. Then since $ \sqrt{I}S = \sqrt{IS} $ is an $S$-ideal, we see that $ \alpha^n\in\sqrt{IS}. $ Hence there exists some $ m $ such that $ \alpha^m\in IS $ so that $ \alpha\in\sqrt{IS}. $

$(\Rightarrow):\,$ Suppose $\alpha\in\sqrt{IS}$. I'm not sure how to start getting that the $a_i\in\sqrt{I}$.

user26857
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user346096
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  • Although $\sqrt IS\subseteq\sqrt{IS}$ (and this is enough for you), equality is unlikely. – user26857 Dec 07 '17 at 20:52
  • If $\alpha^k\in IS$ then $\alpha^k$ is integral over $I$ (in the sense that there is an integral dependence relation with all coefficients in $I$), so $\alpha$ is integral over $I$. Now use Proposition 5.15 from Atiyah and Macdonald; https://math.stackexchange.com/questions/892340/proposition-5-15-from-atiyah-and-macdonald-integral-closure-and-minimal-polynom/892355. – user26857 Dec 09 '17 at 14:51

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