Another question about proposition 5.15 of Atiyah-Macdonald

Question

I have a trouble in understanding the proposition 5.15 in the book from Atiyah and Macdonald. I see that some time ago another user asked a similar question (Proposition 5.15 from Atiyah and Macdonald: Integral Closure and Minimal Polynomial) but unfortunately my problem is different.

The proposition says:

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 $\mathfrak{a}$).

And here is the proof suggested by the book:

Clearly $x$ is algebraic over $K$. Let $L$ be an extension field of $K$ witch contains all the conjugates $x_1,...,x_n$ of $x$.

Here I am ok, but now starts the problem:

Each $x_i$ satisfies the same equation of integral dependence as $x$ does...

My answer is: why? I think I am probably missing some property of conjugates elements in an extension field :) However, assuming that it is true, the end of the proof sounds clear (I'll write it for sake of completeness):

...hence each $x_i$ is integral over $\mathfrak{a}$. The coefficients of the minimal polynomial of $x$ over $K$ are polynomials in the $x_i$, hence they are integral over $\mathfrak{a}$ (by a previous Lemma). Since $A$ is integrally closed, they must lie in $r(\mathfrak{a})$ (again thanks to the same a previous Lemma).

I think my answer can be reformulated in the following manner. We know that - being integral over the ideal $\mathfrak{a} \subseteq A$ - $x$ satisfies a monic polynomial with coefficients in $\mathfrak{a}$, let's call it $f \in A[t]$. Then $x$ is algebraic in $K$ and so it has a minimal polynomial $g \in K[t]$ (and in general $f$ and $g$ are different). Now we can extend the field $K$ to a field $L$ that contains all the roots $x_i$ of $g$ (and we call them the associated to $x$). Of course, by definition, $g(x_i)=0$, but the proof seems to claim that $f(x_i)=0$ too. Is it correct? If yes, could you explain me the reason?

Important idea (is it a solution?): $f$ can be seen as a polynomial in $K[t]$. So $f,g \in K[t]$ are both polynomial such that $f(x)=g(x)=0$. But $g$ is the minimal polynomial of $x$, then $g | f$. Do I remember this fact correctly? Because in that case, seeing $f, g \in L[t]$, since $g | f$ we will have that all the conjugates $x_i$ - being roots of $g$ - are actually roots of $f$ too - and we should have done.

Thanks for your help!!! Cheers

Answer 1

Yes, your argument is correct. You don't need to see the polynomials as elements of $L[t]$, though: by minimality of $g$ you know that $\deg(f) \geq \deg g$, so by Euclidean division you can find $q,r \in K[t]$ with $q \neq 0$ and either $r \equiv 0$ or $0 \leq \deg(r) < \deg(g)$ such that $$ f = qg + r $$ On the other hand, $f(x) = g(x) = 0$ implies that $r(x) = f(x) - q(x) g(x) = 0$, which by minimality of $g$ gives $\deg(r) \geq \deg(g)$. Thus you can conclude that $r \equiv 0$, so $f = qg$ and every root of $g$ (over $K$) is also a root of $f$ (over $K$).