Proving a sufficient condition for the separability of a finite extension over a field of non-zero characteristic

Question

I'm taking an intro to Galois theory course, which is rather exciting. We had the following question in a practice paper:

Let $K$ be a field of characteristic $p$, and let $L/K$ be a finite extension with $\left[L:K\right]$ prime to $p$. Show that $L/K$ is separable.

Note that I took "$\left[L:K\right]$ prime to $p$" to mean $p\nmid\left[L:K\right]$.

I managed to reduce the problem to a seemingly simpler one. $L/K$ is separable if and only if $K\left(\alpha\right)$ is separable for all $\alpha\in L$. Also, from the multiplication rule, $p\nmid\left[K\left(\alpha\right):K\right]$. So it's sufficent to prove the proposition for $L=K\left(\alpha\right)$.

I couldn't proceed beyond that. I tried to play a bit with the general form of $\mathrm{irr}\left(\alpha,K\right)$, but that didn't inspire any solution. I had hoped that I might be able to show that $\gcd\left(f,f'\right)=1$, but was unsuccessful. I then tried to understand how an embedding of $L/K$ would "work", but that didn't lead me anywhere either.

I got the sense I'm missing something rather basic. I'm still a newbie when it comes to finite fields and fields with non-zero characteristic. They were mentioned in previous courses I took (say, linear algebra), but we didn't use them as much as fields of zero-characteristic.

I'd love a nudge in the right direction. Thanks!

Answer 1

Let $a\in L$ not in $K$, consider $P$ the minimal polynomial of $a$. Remark that there exists $u_0,...,u_{p-1}\in K$ such that $u_0+u_1a+...+u_{p-1}a^{p-1}+a^p=0$ since $L$ is a $K$-vector space of dimension $P$. Write $Q=u_0+u_1X+...+u_{p-1}X^{p-1}+X^p$, you obtain that $P$ divides $Q$ and $deg(P)\leq p$. You have $[K(a):K]=deg(P)$ and $[L:K]=[L:K(a)][K(a):K]=p$, this implies that $[K(a):K]=p$ since $a$ is not in $K$, you deduce that $P=Q$.

Suppose that the derivative of $P$ is zero, there exists a polynomial $L$ such that $P=L(X^r)$ (see the reference), $deg(P)=rdeg(L)=p$ since $p$ is prime, you deduce that $r=1$. Contradiction.

https://en.wikipedia.org/wiki/Separable_polynomial#Separable_field_extensions

Answer 2

I'm not sure I fully understand Tsemo's approach, but regardless, I believe I found a different approach. I used proof by contradiction, but I wonder if there is a more direct approach.

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As I mentioned in the question, it's sufficient to prove that all $\alpha\in L$ are separable.

Denote $m=\left[K\left(\alpha\right):K\right]$ and $f=\mathrm{irr}\left(\alpha,K\right)$. As explained in the question, $p\nmid m$. Also, $m=\deg\mathrm{irr}\left(\alpha,K\right)$.

Now, assume that $f$ is not separable. That is, $f$ has a multiple root $\beta$. Thus, $f$ and $f'$, the formal derivative of $f$, have a common root, $f\left(\beta\right)=f'\left(\beta\right)=0$. Therefore, $\mathrm{irr}\left(\beta,K \right)\in K\left[X\right]$ divides both $f$ and $f'$.

From this we obtain $\gcd(f,f')\ne 1$ over $K\left[X\right]$. Since $f$ is irreducible, $\gcd(f,f')=f$. This implies $f'=0$, because $\deg f'< \deg f$. Therefore, all the powers of $f$ with non-zero coefficients are divisible by $p$. More formally, there is $g\in K\left[X\right]$ such that $f(X)=g(X^p)$.

But this implies that $p\mid m= \deg f$. Contradiction.