Why are cubics separable over fields that are not characteristic 2 or 3

Why is it that cubics are separable over fields that are not of characteristic $2$ or $3$?

This is the starting point for some a discussion of the Galois group of a cubic, but I seem to be stuck right off the bat.

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A polynomial has multiple roots if and only if it is not relatively prime to its formal derivative.

Suppose that $f(x) = x^n + a_{n-1}x^{n-1}+\cdots+a_0$ is an irreducible polynomial of degree $n\gt 0$ over a field $\mathbf{F}$. Then $f'(x) = nx^{n-1}+(n-1)a_{n-1}x^{n-1}+\cdots + a_1$. Because $f(x)$ does not have any proper factors in $\mathbf{F}[x]$, then the only possible common factors of $f(x)$ and $f'(x)$ are, up to units, $1$ and $f(x)$. But if $f'(x)\neq 0$, then it is of degree strictly smaller than $f(x)$, so it cannot be divisible by $f(x)$. So either $f'(x)=0$, or else $\gcd(f,f')=1$. But the only way in which we can have $f'(x)=0$ is if $n=0$ in $\mathbf{F}$; that is, the characteristic of $\mathbf{F}$ divides $n$. (In fact, more is true: $f(x)$ must be expresssible as $g(x^p)$, where $g(x)$ is a polynomial and $p$ is the characteristic of $\mathbf{F}$, but that is not needed here).

Now, a polynomial is separable if and only if its irreducible factors are separable. Take a cubic polynomial $f(x) = x^3 + ax^2+bx+c$.

• If $f(x)$ is irreducible, then the only way it can fail to be separable is if $\mathrm{char}(\mathbf{F})|3$, that is, if the characteristic of $\mathbf{F}$ is $3$.

• If $f(x)$ is the product of an irreducible quadratic and a linear polynomial, $f(x) = \ell(x)q(x)$, then $\ell(x)$ is certainly separable, so the only way for $f(x)$ to fail to be separable is for $q(x)$ to not be separable; this is a monic polynomial of degree $2$, so for $q(x)$ to not be separable we need $\mathrm{char}(\mathbf{F})$ to divide $2$; that is, $\mathbf{F}$ to be of characteristic $2$.

• If $f(x)$ is a product of three linear polynomials (possibly repeated), then $f(x)$ is separable, because each irreducible factor is linear and hence separable.

So, a necessary condition for a cubic polynomial to not be separable is for the characteristic of the field of definition to be either $2$ or $3$. In consequence, if $f(x)$ is a cubic over a field of characteristic not equal to $2$ or to $3$, then it must be separable.

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Thanks for the detailed explanation. Turns out that the point I was stuck on was in the general considerations about irreducible polynomials being separable, so it's good that you gave a thorough answer. –  Lost In Math Nov 25 '10 at 19:43
@Lost In Math: You're welcome. Remember to formally "accept" the answer you found most helpful/useful (if there is one that actually answers your question); click on the check-mark next to the question. You can only accept one answer. –  Arturo Magidin Nov 25 '10 at 19:46
You can actually do this "by hand" too. Our goal is to show that a monic irreducible polynomial $f(x)$ of degree $2$ or $3$ over $F$ splits into distinct linear factors in an algebraic closure $E$, as long as the characteristic of the $F$ is not $2$ or $3$. Suppose the irreducible $f(x)$ does not split into linear factors in $E$; we will get a contradiction.
First, if the polynomial is of the form $(x - a)^3 = x^3 - 3ax^2 + 3a^2x + a^3$, then the $x^2$ coefficient $-3a$ must be in $F$. So $a$ is in $F$ too, using that $F$ is not of characteristic $3$.
The remaining case is that the polynomial is of the form $(x - a)^2(x - b)$. For this we use the fact that for any polynomial $p(x)$, ${p'(x) \over p(x)} = \sum {1 \over x - r}$, where the sum is taken over the roots of $p(x)$. Applying this to $f(x)$ gives that ${f'(x) \over f(x)} = {2 \over (x - a)} + {1 \over x - b}$. Note that $f(x) = x^3 - (b + 2a)x^2 + ...$, so that $b + 2a \in F$. Hence ${b + 2a \over 3} \in F$ too. Substituting this value of $x$ into ${f'(x) \over f(x)}$ gives gives ${9 \over 2(b - a)}$, which must be in $F$. Thus $b - a\in F$; note we use the characteristic is not $2$ or $3$ here. So $b = {2 \over 3}(b - a) + {1 \over 3}(b + 2a)$ is in $F$ as well, again contradicting the irreducibility of $f(x)$.