Splitting a polynomial with one root

2011-07-11T07:30:45Z

Suppose we have an irreducible polynomial $f\in K[x]$. Is there some way to sometimes tell whether $f$ splits completely after adjoining just one root of $f$ to $K$?

I am mostly interested in the case where $K$ is a function field $\mathbb{F}_{q}(t_{1},\ldots,t_{m})$ over some finite field, so it might not be feasible to explicitly compute roots.

Answer by Georges Elencwajg for Splitting a polynomial with one root

2011-07-11T11:17:16Z

Here is the best I can come up with. Consider an algebraic closure $\bar K$ of $K$ and a root $\alpha \in \bar K$.

The number of roots of $f$ in $K(\alpha)$ doesn't depend on $\alpha$: call it $ r(f)$
Moreover call $s(f)$ the number of the different subfields $K(\alpha)\subset \bar K$ obtained by adjoining roots of $f$ to $K$. Then you have the pleasant equality $$deg(f)=r(f).s(f)$$ This shows in particular that the number of roots that you get by just adjoining one root divides the degree $deg(f)$ of your polynomial.
For example if $K=\mathbb Q$ and $f(x)=X^8-2$ you have $r(f)=2$ and $s(f)=4$, since the fields you get by adjoining roots of $f$ to $\mathbb Q$ are [with $\omega =\frac{1}{\sqrt 2}(1+i)$]:
$\mathbb Q(\sqrt[4]2)$
$\mathbb Q(\pm \omega \sqrt[4]2)$
$\mathbb Q(\pm \bar{\omega} \sqrt[4]2)$
$\mathbb Q(\pm i \sqrt[4]2)$

These results are due to Perlis , and although not difficult have found their way in exactly zero books, as far as I am aware.

Splitting a polynomial with one root - MathOverflow

Splitting a polynomial with one root

Answer by Georges Elencwajg for Splitting a polynomial with one root