Several times [1, 2, 3, 4] has been questions regarding the following results of Atiyah, Macdonald, Introduction to Commutative Algebra.
Lemma 5.14. Let $A\subseteq B$ be rings and let $\mathfrak{a}\subseteq A$ be an ideal. Let $C$ be 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 $\mathfrak{a}$).
Proof. Clearly $x$ is algebraic over $K$. Let $L$ be an extension field of $K$ which contains all the conjugates $x_1, \ldots, x_n$ of $x$. Each $x_i$ satisfies the same equation of integral dependence as $x$ does, hence each $x_i$ is integral over $a$. The coefficients of the minimal polynomial of $x$ over $K$ are polynomials in the $x_i$, hence by (5.14) are integral over $\mathfrak{a}$. Since $A$ is integrally closed, they must lie in $r(\mathfrak{a})$, by (5.14) again. $\square$
My issue is not addressed in the above linked questions: why do we need minimality of $f=t^n + a_1 t^{n-1} + \cdots+ a_n$? If $f$ was just assumed to be some monic polynomial with $a_i\in K$ and $f(x)=0$, couldn't we still conclude $a_i\in r(\mathfrak{a})$? Basically by same argument above (one takes a field $L/K$ where $f$ fully splits).
