Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given $F$, a field with characteristic $p > 0$, and a finite purely inseparable field extension $E$. Then prove that the minimal polynomial $f(x)$ of any $\alpha \in E\backslash F$ will be of the form $X^{p^n} - \alpha^{p^n}$, with $n \geq 0$ being an integer.

Now, I've tried to prove this, using Isaacs' "Algebra: A Graduate Course" as a guide. However, after a series of corollaries and theorems referring to at least one previous theorem or corollary every time, I think I have it. I'll try to put the proof here (more or less):

Isaacs proves that $f(x)$ must be of the form $X^{p^n} - \alpha^{p^n}$, because $f(x)$ can be written as $g(x^{p^n})$ with $g \in F[X]$ irreducible and separable over $F$. This is proven because we can write $f(x) = h(x^p)$ for some irreducible $h \in F[x]$, and then using induction on the degree.

That $f(x)$ can be written as $h(x^p)$ is in turn proven by using that $f' = 0$, then deriving $f$ and showing that coefficients $a_i$ must equal $0$ if $p\nmid i$.

Again, in turn is proven that $f' = 0$ because, since $f$ does not have distinct roots, there must be an $a$ in a certain field $K$ such that $(x-a)^2|f(x)$ in $K[x]$. Writing $f(x) = (x-a)^2h(x)$ gives that $f'(x) = 2(x-a)h(x) + (x-a)^2h'(x)$, which shows that $f'(a)=0$. Then, since $f$ is irreducible, it divides every polynomial in $F[x]$ that has $a$ as a root. Thus $f|f'$ but since $\deg f > \deg f'$, $f' = 0$.

I'm wondering if I made any mistakes/inconsistencies/... ?

And if there is really no easier or more straightforward way to prove this?

share|cite|improve this question
What is $A$? Of course an arbitrary $\alpha \in E$ works. – Martin Brandenburg Apr 12 '13 at 15:43
Ah, my apologies, that should've been F of course. – user49719 Apr 12 '13 at 16:28
Dear @Koeneuze, What is your definition of a purely inseparable extension? One in which the minimal polynomial of each element has only one root? The reasoning in the box seems fine to me. Do you see why it implies the result you're interested in? – Keenan Kidwell Apr 12 '13 at 17:23
The definition I'm working with is the following: "An algebraic field extension $E$ of $K$ is called 'purely inseparable' over $K$ if no element of $E\backslash K$ is separable over $K$". Yes, it is quite clear to me why this proof should be conclusive, but I was hoping to find something more... elegant, perhaps? – user49719 Apr 12 '13 at 17:58
I see. I can't think of anything else. It is a general fact that, for an algebraic extension $E/K$, if $\alpha\in E$, then there is an integer $n\geq 0$ so that $\alpha^{p^n}$ is separable over $K$. The argument in the box is proving this. That's the only argument I've ever seen for the fact that the minimal polynomials all look like $X^{p^n}-a$ for some $n$ and some $a\in K$. – Keenan Kidwell Apr 12 '13 at 18:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.