[Math] Every irreducible polynomial over a finite field is separable

abstract-algebrafield-theoryfinite-fields

On page 549 in Dummit and Foote is a proposition which states "Every irreducible polynomial over a finite field is separable. A polynomial in $\mathbb{F}_p[x]$ is separable if and only if it is the product of distinct irreducible polynomials in $\mathbb{F}_p[x]$." However, we know that if the derivative of a polynomial $f(x)$ is $0$, then every root is a multiple root, and so $f(x)$ is not separable. Does this mean, then, that any polynomial in $\mathbb{F}_p$ with derivative $0$ is reducible? The reason I ask is because on the next page, the authors state "If $p(x)$ is an irreducible polynomial which is not separable…" But doesn't the proposition above preclude this?

Best Answer

Over $K=\Bbb{F}_p$ it does hold that if, for some $f(x)\in K[x]$, we have $f'(x)=0$, then $f$ is reducible. The reason is the following.

If $f'(x)=0$ this means that all the terms in $f(x)$ have degrees that are multiples of $p$. In other words, $$ f(x)=\sum_{i=0}^na_ix^{pi} $$ for some natural number $n$ and some coefficients $a_i\in\Bbb{F}_p$. Two key results then come to the fore:

  • In a commutative ring $R$ of characteristic $p$ we have the formula (also known as Freshman's dream) for all $a,b\in R$: $$(a+b)^p=a^p+b^p.$$
  • For all $a\in\Bbb{F}_p$ we have $a^p=a$ (Little Fermat).

Put together these imply that the above polynomial $f(x)$ $$ f(x)=\sum_{i=0}^na_ix^{pi}=\sum_{i=0}^n(a_ix^{i})^p=\left(\sum_{i=0}^na_ix^i\right)^p $$ is actually the $p$th power of a lower degree polynomial, hence reducible.

The same result holds for other fields $K$ of characteristic $p$ as long as all the elements of $K$ are $p$th powers of some element of $K$ (above it would have been enough to have $a_i=b_i^p$ for some $b_i\in K$). Such fields are called perfect, and whenever $K$ is a perfect field we see that irreducible polynomials over $K$ are necessarily separable.

Therefore we can use as the field $K$ any finite field. This is because the Frobenius automorphism $z\mapsto z^p$ is an injective endomorphism of $K$ (trivial kernel). When $K$ is finite "injectivity $\implies$ surjectivity" and we are done.

It doesn't work as nicely for all fields of characteristic $p$. The textbook counterexample is $K=\Bbb{F}_p(t)$. The polynomial $m(T)=T^p-t$ is irreducible (Eisenstein), but it is not separable. It has a single zero $t^{1/p}$ of multiplicity $p$ in an extension field of $K$.

Related Question