2
$\begingroup$

Given a field $\mathbb{F}$ with an irreducible, separable polynomial $f(x),$ let $E$ denote the splitting field of $f$ over $\mathbb{F},$ and assume that $\alpha$ and $\alpha + 1$ are roots of $f(x).$ Prove that the characteristic of $\mathbb{F}$ is not zero. Further, prove that there exists a field $\mathbb{F} \leq L \leq E$ such that $\operatorname{char} (\mathbb{F}) = [E : L].$

Could anyone provide any direction on this problem? Unfortunately, I have been thinking about it for a few hours, and I have not come up with anything close to a solution; however, I have several ingredients with which I have worked. Considering that $f(x)$ is separable, $f(x)$ and $f'(x)$ are relatively prime so that by Bezout's Theorem, there exist polynomials $a(x), b(x) \in \mathbb{F}[x]$ such that $a(x) f(x) + b(x) f'(x) = 1.$ Using the fact that $\alpha$ and $\alpha + 1$ are roots of $f(x)$ but not roots of $f'(x),$ we have that $b(\alpha) f'(\alpha) = 1$ and $b(\alpha + 1) f'(\alpha + 1) = 1.$ We have not really used any facts about the roots yet, so I wonder if that might be a direction worth exploring, but I am stuck at the moment.

$\endgroup$
3
  • 3
    $\begingroup$ Idea: try proving that for every root $\beta$ of $f$, $\beta + 1$ is also a root. $\endgroup$ – Ravi Fernando Jun 10 '18 at 4:55
  • 3
    $\begingroup$ Equivalently, @RaviFernando, show that there’s an automorphism of the splitting field of $f$ with $\alpha\mapsto\alpha+1$. $\endgroup$ – Lubin Jun 10 '18 at 4:57
  • 2
    $\begingroup$ Hint: Show that both $f(x)$ and $f(x-1)$ are minimal polynomials of $\alpha$. Hence ... $\endgroup$ – Jyrki Lahtonen Jun 10 '18 at 5:01
1
$\begingroup$

The Galois group $G$ acts transitively on the roots of $f$ since $\Pi_{s\in G}(X-s(\alpha))\in \mathbb{F}[X]$, there exists $s$ such that $s(\alpha) =\alpha+1$, we deduce that $s(\alpha+1)=s(\alpha)+1=\alpha+2$, and recursively $\alpha+p$ is a root for every integer. Since the number of root is finite, there exists $n, m$ distinct such that $\alpha+n=\alpha+m$, we deduce that $m-n=0$ and the characteristic is finite.

$\endgroup$
1
  • $\begingroup$ Excellent. I appreciate your very elegant solution! $\endgroup$ – Carlo Jun 11 '18 at 4:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.