An element is integral iff its minimal polynomial has integral coefficients.

Question

This is from Algebraic Number Theory by Neukirch

Let $A$ be an integral domain which is integrally closed, K its field of fractions, $L|K$ a finite field extension, and $B$ the integral closure of $A$ in $L$.

Furthermore, the fact that $A$ is integrally closed has the effect that an element $\beta \in L$ is integral over $A$ if and only if its minimal polynomial $p(x)$ takes its coefficients in $A$. In fact, let $\beta$ be a zero of the monic polynomial $g(x) \in A[x]$.Then $p(x)$ divides $g(x)$ in $K[x]$, so that all zeroes $\beta_1, ..., \beta_n$ of $p(x)$ are integral over A, hence the same holds for all the coefficients, in other words $p(x) \in A[x]$.

Now my question is are $\beta_1, ..., \beta_n$ all the roots of $p(x)$ in $L$ or $\bar K$, the algebraic closure of $K$? I think this arguement will only work when we consider all the roots of $p(x)$ in $\bar K$.

Answer 1

Sorry I misread your question earlier. Yes all the $\beta_i$ have to be in an algebraic closure $\bar{K}$. This is because $L$ may not be a normal extension of $K$. For an example take $K = \Bbb{Q}$ and $L = \Bbb{Q}(\sqrt[3]{2})$. Then the element $\beta = \sqrt[3]{2}$ has minimal polynomial $x^3 - 2$ but the other two roots are $\zeta_3\sqrt[3]{2}$ and $\zeta_3^2\sqrt[3]{2}$ which don't lie in $L$.

So let us recollect the proof of Neukirch. One direction is already clear so we prove that if $\beta \in L$ is integral over $A$ then its minimal polynomial $p(t) \in A[t]$. Now by assumption because $\beta$ is integral over $A$ we can write down the integral dependence relation given by some $g(t) \in A[t]$ which $\beta$ satisfies. Then $p(t) | g(t)$ and thus every root of $p$ is a root of $g$. Thus every root of $\beta_i$ of $p$ lying in some algebraic closure $\overline{K}$ is integral over $A$.

By Vieta's formulas the coefficients of $p(t)$ are linear combinations of products of the roots. Since the product and sum of two integral elements is integral, we conclude that $p(t)$ actually has coefficients in $A$.