discriminant and irreducibility of $x^p - (p+1)x - 1$

Question

I proved that if the discriminant of a monic polynomial with $p > 5$ an odd prime is nonzero it has at least 3 roots in the reals. I'm trying to apply that result to this polynomial $f(x) = x^p - (p+1)x - 1$, how can I show that it has a nonzero discriminant?

Also, how do I show that $f(x)$ mod $p$ is irreducible in $\mathbb{F}_p[x]$? My approach was to suppose that $\alpha \in \mathbb{F}_p$ is a root, then $f(\alpha) = \alpha^p - (p+1)\alpha - 1$. But the field has characteristic $p$, so $\alpha^p = \alpha$ and $p\alpha = 0$, which means $f(\alpha) = -1$, and $\alpha$ can't be a root. How can I proceed from here?

Here please find several nice proofs for the irreducibility of this polynomial in $\Bbb{F}_p[x]$. — Jyrki Lahtonen, Jan 15 '18 at 22:13
Can you use the formula for the discriminant of trinomials of this form? (I swear to the deity of your choice that I didn't remember having answered that latter question as well. I just knew that we had handled it here. May be I'm getting better with the search engine?) — Jyrki Lahtonen, Jan 15 '18 at 22:20

score 1 · Answer 1 · answered Jan 15 '18 at 16:55

1

Hint for the second question:

If $$g(X)h(X)\equiv X^p-X-1\pmod{p}$$ then $$g'(X)h(X)+g(X)h'(X)\equiv -1\pmod p$$

answered Jan 15 '18 at 16:55

ajotatxe

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discriminant and irreducibility of $x^p - (p+1)x - 1$

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