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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?

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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.)

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  • $\begingroup$ Very nice. Thank you. I would like to insert this example in an article I am writing. Could I cite you? $\endgroup$
    – MikeTeX
    Dec 24, 2021 at 7:43
  • $\begingroup$ @MikeTeX Of course, you may write that I pointed to Swan's paper. $\endgroup$ Dec 24, 2021 at 9:35
  • $\begingroup$ Thx. The article is here: arxiv.org/abs/2112.14711. You are cited at the beginning of Sec. 7. $\endgroup$
    – MikeTeX
    Jan 3 at 9:53

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