Polynomial over a field whose power has coefficients in the ring of algebraic integers.

Question

I have proven the result that given a number field $K$, and a monic polynomial $f \in \mathcal O_K\left[x\right]$ where $\mathcal O_K$ is the ring of algebraic integers in $K$, then any root of $f$ must also be an algebraic integer (in some field containing the root and $K$)

Now I am asked to deduce that if $h \in K\left[x\right]$ is monic and $h^n \in \mathcal O_K\left[x\right]$, then $h \in \mathcal O_K\left[x\right]$.

The only thing I can think to do is take $h^n$ to be our $f$ from earlier. This then tells us that all the roots of $f$ and hence $h$ are algebraic integers.

Therefore, if we suppose $h$ has roots $a_1, \dots,a_k$ then $\exists, f_1, \dots, f_k \in \mathbb Z\left[x\right]$ such that $f_i(a_i) = 0 \;\forall i$ with the $f_i$ all monic.

Now if it happens that $a_1, \dots, a_k \in \mathcal O_K$ then we would be done, so it would suffice to show that $a_1, \dots, a_k \in K$.

Why though would this be the case? I can't see why this would be true. This leads me to believe then that maybe it is not the case that the roots all land in $\mathcal O_K$ but rather that the combinations of them that make up the coefficients of $h$ are in $\mathcal O_K$, but again I am struggling to show that also.

Any help would be greatly appreciated, thank you!

Answer 1

Sorry about my previous flawed attempt.

But it's actually easy . . .

Since the roots of $h$ are algebraic integers, and since $h$ is monic, the coefficients of $h$ are also algebraic integers (by Vieta's formulas).

Since $h\in K[x]$, the coefficients of $h$ are in $K$, hence, since they're algebraic integers which are in $K$, they're in $O_K$.

Therefore $h\in O_K[x]$.

Answer 2

Perhaps it is way easier than it appears at first sight: observe that the roots of $\;h\;$ and of $\;h^n\;$ , in some algebraic extension of $\;K\;$ , are the same (without regard to multiplicity) . Since $\;h^n\in\mathcal O_K[x]\;$ and it obviously is monic just as $\;h\;$ is, the first part of the question means those roots are algebraic integers.

Thus, the roots of $\;h\;$ ( which are the same, up to multiplicity, to those of $\;h^n\;$), are elements in the ring $\;\mathcal O_K[x]\;$, and so thus are the coefficients of $\;h\;$ by Viete's formulas, which means $\;h\in\mathcal O_K[x]\;$