A question about the ring of integers and the coefficients Let $F,F(x)$ be number fields, and the ring of integers of $F$ be $A$, the ring of integers of $F(x)$ be $B$. $f(y)$ is the minimal polynomial of $x$ over $F$, with $f(y)=(y-x)h(y)$, s.t. $h(y)=a_0+a_1y+...a_{n-1}y^{n-1}$. If $x\in B$, then how to see the $a_i$ are in $B$? 
 A: Since $x$ is an algebraic integer, there exists a monic, irreducible polynomial $p$ in $\mathbb{Z}[y]$ such that $p(x) = 0$.

Since $p \in \mathbb{Z}[y]$, $p$ is also in $F[y]$.

Then, since $f(y)$ is the minimal polynomial for $x$ over $F$, it follows that $f$ divides $p$ in $F[y]$, hence $f$ also divides $p$ in $\mathbb{C}[y]$.

Let $f(y) = (y - x_1)\cdots (y-x_n)$ be the complete factorization of $f$ in $\mathbb{C}[y]$.

Then $x_1,...,x_n$ are also roots of $p$, hence $x_1,...,x_n$ are algebraic integers.

It follows that all coefficients of $f(y)$ are algebraic integers. Then since


*

*$F$ is the field of fractions of $A$

*$A$ is integrally closed

*$f \in F[y]$


it follows that $f \in A[y]$, hence, since $A \subseteq B$, we also have $f \in B[y]$.

Let $K = F(x)$.
Then since


*

*$f \in B[y]$

*$y-x \in B[y]$, with $y - x$ monic

*$f(y) = (y-x)h(y)$


it follows, by polynomial long division in $K[y]$, that $h(y) \in B[y]$.

Thus, all coefficients of $h$ are in $B$, as was to be shown.
