Algebraic Elements are Integral, if their Minimal Polynomial is. In an upcoming exercise class in commutative algebra I would like to discuss how to detect, whether an algebraic element $\alpha$ over $\Bbb Q$ is integral over $\Bbb Z$. The claim is that it is precisely the case if $\operatorname{Mipo}_{\Bbb Q}(\alpha) \in \Bbb Z[X]$. This is indeed true, see for example this question.
The usual proof of this fact goes along the lines of showing that the minimal polynomial has integral coefficients, so since $\Bbb Z$ is integrally closed the minimal polynomial lies in $\Bbb Z[X]$. However, I think I was told another proof some while ago. I tried to remember it, but it seems to circumvent the integral closedness of $\Bbb Z$, which confuses me. So I would really appreciate a second opinion.

Let $\alpha$ be algebraic over $\Bbb Q$. Then $\alpha$ is integral over $\Bbb Z$ if and only if $\operatorname{Mipo}_\Bbb{Q}(\alpha) \in \Bbb Z[X]$.

One direction is immediate since minimal polynomials are assumed to be monic. So let $\alpha$ be integral over $\Bbb Z$. This means that $\Bbb Z[\alpha]$ is a finitely generated $\Bbb Z$-module. Since $\Bbb Z[\alpha]$ is torsion-free and $\Bbb Z$ is a principal ideal domain, $\Bbb Z[\alpha]$ is a free $\Bbb Z$-module of rank $k \in \Bbb N$. Tensoring with $\Bbb Q$ shows that $\Bbb Z[\alpha] \otimes \Bbb Q \cong \Bbb Q[\alpha] = \Bbb Q(\alpha)$ is a free $\Bbb Q$-module of rank $k$. By IBN $k = \deg \operatorname{Mipo}_\Bbb{Q}(\alpha)$.
Now multiplication with $\alpha$ gives a $\Bbb Z$-linear endomorphism $\cdot\alpha:\Bbb Z[\alpha] \rightarrow \Bbb Z[\alpha]$. It gives rise to a characteristic polynomial $\operatorname{CP}(\cdot\alpha) \in \Bbb Z[X]$, which is a monic polynomial of degree $k$. Cayley-Hamilton implies that evaluating it at $\cdot\alpha$ yields the zero-endomorphism, ie. that $\operatorname{CP}(\cdot\alpha)(\cdot\alpha) = 0$. In particular $\operatorname{CP}(\cdot\alpha)(\cdot\alpha)(1) = \operatorname{CP}(\cdot\alpha)(\alpha) = 0$.
We have found a monic polynomial in $\Bbb Z[X]\subseteq \Bbb Q[X]$, which annihilates $\alpha$ and has the same degree as $\operatorname{Mipo}_\Bbb{Q}(\alpha)$, so both polynomials have to coincide.

Does this proof fail, and if so where?

Thank you for your help.
 A: Your proof is correct, but it uses the fact that $\mathbb{Z}$ is a PID, a property stronger than integral closedness.
If $R$ is not integrally closed, there exists an $\alpha \in Q(R)-R$ that is integral over $R$. Its minimum polynomial over $Q(R)$, however, is $X-\alpha$, which is not in $R[X]$. (@ChrisH: $R[\alpha]$ is finitely generated as an $R$-module, but not free.)
Edit:
To see that $R[\alpha]$ is not free over $R$, consider Chris' example $R = k[x,y]/(x^2-y^3)$, and take $\alpha = x/y$. Then $\alpha^2 = y$, so $\alpha$ is integral over $R$. But the minimum polynomial over $Q(R)$ is $X-\alpha$ $\notin$ $R[X]$. Assuming $S$ = $R[\alpha]$ is free, it is certainly flat over $R$, so that $(I\cap J)S$ = $IS \cap JS$ for every pair of ideals $I,J$ of $R$ (cf. Matsumura, Commutative Ring Theory, Th.7.4.ii). Taking $I=xR$ and $J=yR$, we have $IS$ = $xR$ + $y^2R$ and $JS$ = $xR$ + $yR$, with intersection $xR+y^2R$. But $I \cap J$ = $x^2R$ + $xyR$, which is already an ideal of $S$, but does not contain the element $x$, or $y^2$, for that matter. Contradiction.
End of insert.
When $R$ is an integrally closed domain, $K = Q(R)$ its quotient field, $L$ a field extension of $K$, and $\alpha \in L$ an element that is integral over $R$, the minimum polynomial $f$ of $\alpha$ over $K$ is in $R[X]$. For $f$ factorizes as $\prod_{1 \leq i \leq n}(X-\alpha_i)$ in $E[X]$, where $E \supseteq K(\alpha)$ is a splitting field for $f$ over $K$, $n = \text{deg}(f)$, and $\alpha = \alpha_1$. Let $g \in R[X]$ be a monic polynomial such that $g(\alpha) = 0$. Then $f$ divides $g$ in $K[X]$. So every $\alpha_i$ is a root of $g$, and therefore integral over $R$. The coefficients of $f$ are (elementary symmetric) polynomials in the $\alpha_i$ (with coefficients in $\mathbb{Z}$), hence they are also integral over $R$. But the coefficients are in $K$, and because $R$ is integrally closed in $K$, they must be in $R$.
Compare Corollary 2.6.a in Fröhlich & Taylor, Algebraic Number Theory. There, the additional assumption is made that $\alpha$ is separable over $K$, which is not necessary for the result at hand.
A: To add to the correct answer above, here is an example where the result breaks down when $R$ is not integrally closed.
Let $R=k[x,y]/(x^2-y^3)$, and let $z:=\sqrt \frac{x}{y}$ be contained in an algebraic extension of $Q(R)$. Then the minimal polynomial of $z$ over $Q(R)$ is $X^2-\frac{x}{y}$, which isn’t contained in $R[X]$, but $R[z]$ is finite, since it’s spanned by $1,z,z^2,z^3$, which one can check since $z^4=y$ is in $R$, so $z$ is integral over $R$.
(EDIT: The following was my incorrect original answer) This proof works, and it proves the more general statement that if $R$ is a domain, and $\alpha$ is algebraic over $Q(R)$, (say contained in some extension $L$ of $Q(R)$) then being integral over $R$ is equivalent to having minimal $Q(R)$ polynomial having coefficients in $R$. Here $Q(R)$ is the fraction field of $R$.
The proof of the nontrivial direction is exactly as you gave, but one can note that for an integral element $\alpha$ over $R$, the $R$ module $R[\alpha]$ is free over $R$, since by assumption this lies in the ambient field extension of $Q(R)$.
The lemma used for the previous statement is that for any ring, any monic polynomial $f$ in $R[x]$, then $R[x]/(f)$ is free as an $R$ module of rank $deg(f)$, coupled with the isomorphism theorem.
