Proof that field extensions of degree 3 over $K$ with $\mathrm{char}(K) \neq 3$ are separable

Question

My question is how one can prove, for all field extensions $K \subset L$ with $[L:K]=3, $ char($K$) $\not=3$, that $L$ is separable over $K$.

I understand this proof with 2 in stead of 3. I imagine a similar proof, but with the Cardano formula instead of the $abc$ formula. If so, isn't there a nicer proof?

I'm also wondering whether this holds for all primes $p$.

Thanks.

Answer 1

Suppose $[L:K]=p$ and ${\rm char}\,K\ne p$ with $p$ prime. Show the following:

• For all $\alpha\in L$ either $\alpha\in K$ or $K[\alpha]=K(\alpha)=L$.
• If $\alpha\in L\setminus K$ then $p(x):={\rm minpoly}_{\alpha,K}(x)\in K[x]$ has degree $p$.
• If $p(x)$ has repeated factors then $\gcd(p(x),p'(x))\ne1$ in $K[x]$.
• By irreducibility, the only monic divisors of $p(x)$ are $1$ and $p(x)$.
• For all $f(x)\in K[x]$, either $f(x)=0$ or $\deg f'(x)<\deg f(x)$.
• $\gcd(p(x),p'(x))=p(x)$ is not possible unless $p'(x)=0$.
• For all $f(x)\in K[x]$, $f'(x)=0$ implies $({\rm char}\,K)\mid \deg f$.

Answer 2

Fix an arbitrary minimal polynomial of $L/K$. It is of the form $x^3 + bx^2 + cx +d$. The derivative of this poly is $3x^2 + 2bx + c$. The extension is separable iff this derivative $\neq 0$, which is the case since ${\rm char}(K)\ne3$. Same method for the general case of $[L:K] = p \neq {\rm char}(K)$. QED