Let $K/F$ be a finite separable extension of number fields of finite degree over $\mathbb{Q}$. Let $A = \mathcal{O}_F$ and $B$ the integral closure of $A$ in $K$. Is $B = \mathcal{O}_K$?
Let us recall that given the rings $R_1\subseteq R_2$, an element $\alpha\in R_2$ is integral over $R_1$ if $\alpha$ is a root of a monic polynomial $P\in R_1[X]$. With that definition, the integral closure of $R_1$ in $R_2$ is the set of elements in $R_2$ that are integral over $R_1$. IF $K$ is a field then we denote by $\mathcal{O}_K$ the integral closure of $\mathbb{Z}$ in the field $K$.
Here's my attempt:
First of all, since $\mathbb{Z}\subseteq \mathcal{O}_F$ then any element $\beta\in K$ integral over $\mathbb{Z}$ is trivially integral over $A$, so $\mathcal{O}_K \subseteq B$. I suspect that the converse inclusion is also true.
If $\alpha$ in $B$ then it is a root of some monic polynomial
$$X^n + a_{n-1}X^{n-1} + \ldots + a_1X + a_0,$$
where each $a_i\in \mathcal{O}_F$ is integral over $\mathbb{Z}$, i.e. $a_i$ is the root of a monic polynomial of degree $d_i$ in $\mathbb{Z}[X_i]$:
$$X_i^{d_i} + b_{i,d_i-1}X_i^{d_i-1} + \ldots + b_{i,1}X_i + b_{i,0}$$
where each $b_{i,j}\in \mathbb{Z}$ for all $0 \leq i < n$, $0\leq j < d_i$
We can construct a polynomial in $\mathbb{Z}[X,X_{n-1},\ldots,X_1,X_0]$ that has $(\alpha,a_{n-1},a_{n-2},\ldots,a_1,a_0)$ as a zero:
$$X^n + \sum_{i=0}^{n-1}\left( X^iX_i^{d_i} + \sum_{j=0}^{d_i-1} b_{i,j}X^iX_i^j\right).$$
I'm trying to find a clever way to unify all this variables into a single one without compromising the coefficients, but so far I've been unable to do so.
Don't try to find a specific polynomial with coefficients in $A$ for $\alpha\in B$. Simply observe that, since $\mathbf Q\subset A \subset B$, $B$ is a finitely generated $A$-module. On the other hand, $A$ is a finitely generated $\mathbf Z$-module, so $B$ is also a finitely generated $\mathbf Z$-module, therefore it is integral over $\mathbf Z$. This proves $B\subset \mathcal O_K$.
Conversely, any element of $O_K$, being integral over $\mathbf Z$ is a fortiori integral over A, so $O_K\subset B$.