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I am having some trouble understanding Proposition 5.15 in Introduction to Commutative Algebra by Atiyah and Macdonald.

Let $A\subset B$ be integral domains, $A$ integrally closed, and $x\in B$ 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 $t^n+a_1t^{n-1} + \cdots + a_n$, then the $a_1, a_2, \ldots , a_n$ all lie in the radical of $\mathfrak{a}$.

The proof of the proposition states that the coefficients of the minimal polynomial of $x$ are polynomials in the $x_i$ (the conjugates of $x$). I don't understand this, and am struggling with the rest of the proof as well. Thanks for any help.

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1 Answer 1

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Since $x\in B$ is integral over $\mathfrak{a}$ it's algebraic over $K$. Since $x_1,\dots, x_n$ are the conjugates of $x$ (including $x$, I am guessing), then the minimal polynomial of $x$ $$f(t):=t^n+a_1t^{n-1}+\cdots+a_n$$ factors as $$f(t)=(t-x_1)(t-x_2)\cdots(t-x_n).$$ Expanding this and equating coefficients we see that \begin{align*}a_n&=(-1)^nx_1x_2\ldots x_n \\ & \vdots \\ a_2&=\sum_{i<j}x_ix_j \\ a_1&=-(x_1+x_2+\ldots+x_n).\end{align*} In particular, the $a_i$ are symmetric polynomials in the $x_i$.

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  • $\begingroup$ Could you clarify, from the last statement, how do we conclude that the $a_i$ are in the radical of $\mathfrak{a}$? $\endgroup$
    – eatfood
    Commented Jul 1, 2020 at 11:57
  • $\begingroup$ @eatfood Using the proposition 5.14 of the same book one can conclude that each of the $x_i's$ are in the radical of the extension of the ideal $\mathfrak{a}$ in its integral closure $C$. Here, $C=A$ and hence it follows. $\endgroup$
    – nkh99
    Commented Jul 23, 2023 at 5:03

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