Another question in the proof of A&M Thm 9.5 I have a question in the proof of Thm 9.5. In the middle of the proof, it shows that the ring of integers (integral closure of $\Bbb Z$ in $K$ where $K/\Bbb Q$ is a finite field extension). Then it says that $A$ is integrally closed by (5.5)

(5.5) : Let $A\subset B$ be rings and let $C$ be the integral closure of $A$ in $B$. Then $C$ is integrally closed in $B$.

From this, we can conclude $A$ is integrally closed in $K$ not in its field of fraction. Why can one say $A$ is integrally closed?
 A: Since $A$ is a subring of $K$, every nonzero element of $A$ is invertible in $K$, so $F := \mathrm{Frac}(A)$ injects into $K$ by, say, the universal property of localization. (In plainer but less precise terms, for any nonzero element $a \in A$, $1/a$ belongs to $K$.) Hence, $F$ is a subfield of $K$. This means being integrally closed in $K$ is, in principle, a stronger condition than being integrally closed in $F$, so we could be done here.
However, you can in fact show that $F = K$. I claim that for every nonzero element $u \in K$, there exists an integer $a \in \mathbb{Z}$ such that $au \in A$. This will prove the claim because if $au = b$ for $b \in A$, then $u = b/a \in F$.
Indeed, $u$ satisfies a polynomial relation with coefficients in $\mathbb{Q}$, since $K$ is a finite extension of $\mathbb{Q}$. Clearing denominators, there exist $a_{0}, \ldots, a_{n} \in \mathbb{Z}$ such that
$$a_{n}u^{n} + a_{n-1}u^{n-1} + \cdots + a_{1}u + a_{0} = 0.$$
By multiplying each side by $a_{n}^{n-1}$, we then see that $a_{n}u$ is a root of the monic polynomial
$$X^{n} + a_{n-1}X^{n-1} + \cdots + a_{1}a_{n}^{n-2}X + a_{0}a_{n}^{n-1} \in \mathbb{Z}[X],$$
and therefore is integral over $\mathbb{Z}$. Hence, $a_{n}u \in A$.
