# Proving that polynomials belonging to a certain family are reducible

In an article, I've found the following result. Unfortunately, it was derived from a general, somewhat complicated theory, that would be cumbersome for this result alone.

Assume that $$\mathbb F_p$$ is the field with $$p$$ elements, and that we choose an integer $$a$$ with $$1< a < 4p$$, such that $$a$$ be coprime to $$p(p-1)$$ (for example, $$a = 2p - 1$$ is suitable). Let $$t$$ be transcendental over $$\mathbb F_p$$, and set $$K = \mathbb F_p(t)$$. Consider the polynomial $$P(X) = X^{4p} +t^a$$ in $$K[X]$$.

Then for every $$f\in \mathbb F_p[t]$$, $$P(f)$$ is reducible in $$\mathbb F_p[t]$$.

I would like a direct proof of this result. Any idea?

• There appears to be a typo in the setup; what role does $b$ play? Dec 22, 2021 at 11:50
• thx. question edited. Dec 22, 2021 at 13:26
It follows from this result of Swan: If $$t$$ divides $$f(t)$$ there is nothing to do. So assume that this is not he case. Then with $$F(t)=P(f(t))=f(t)^{4p}+t^a$$, $$F'(t)=at^{a-1}$$ and $$F(t)$$ is separable. From this one easily gets that the discriminant of $$F(t)$$ is a square in $$\mathbb F_p$$ (see the alternative formula for the discriminant at the beginning of the article of Swan). By Swan's result, the number of irreducible factors of $$F(t)$$ is congruent to $$\deg F(t)=4p\deg f$$ mod $$2$$, hence it is even and therefore $$>1$$.
Note added later: In fact the special case needed here is already due to Dickson, as pointed out by Swan in his paper. The argument is easy: If $$F(t)$$ is separable of degree $$n$$ and irreducible, then its Galois group contains an $$n$$-cycle. If $$n$$ is even, then this $$n$$-cycle is an odd permutation, so the discriminant of $$F(t)$$ is not a square. (Assuming that we are working over a finite field of odd characteristic.)