# Every irreducible polynomial over a finite field is separable

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?

• I don't have Dummit and Foote's book, but I suppose the last sentence either refers to the case where the field is not necessarily finite, or the proposition is not yet proved. – Bernard Feb 21 '18 at 16:44
• It says "over a field of characteristic p" – ponchan Feb 21 '18 at 16:47
• @ponchan There are infinite fields of characteristic $p$. For instance, $\mathbb{F}_p(t)$, the field of rational functions. – Viktor Vaughn Feb 21 '18 at 16:53

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 $$K$$.

• I can't understand how the first sentence matches the reasoning that follows. The reasoning basically says that if $f'(x)=0$ then $f$ is reducible, and yet "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 irreducible."? Did you mean "reducible" maybe? – polettix Apr 14 '19 at 17:37
• @polettix Yes. There was a typo.Thanks. – Jyrki Lahtonen Apr 14 '19 at 18:40