Showing that $x^5 - 5 a^4 x + a$ is irreducible.

Question

A problem of finding a Galois group of $x^5 - 5 a^4 x + a$ appeared in a previous prelim at my school for $a \in \mathbb N$, and the only hard part seems like showing that the polynomial is irreducible, since it has 2 complex roots.

However, I do not know about any algebraic way to show that above is irreducible—other person from my school showed it by using rational root theorem and Rouche’s theorem.

So I was wondering if there is any way to show that above polynomial is irreducible in a purely algebraic way.

Answer 1

Too long for a comment. For $a\in \Bbb{Z}$

• If $a=0$ then it is reducible, if $a=\pm 1$ then it is irreducible $\bmod 3$.

• If $a$ is not a 5th power then $f(x)=x^5+5a^4x+a$ is irreducible over $\Bbb{Q}_p$ whenever $5\nmid v_p(a)$, because $f(\gamma)=0$ gives that $v_5(\gamma)=v_5(a)/5$ so that $[\Bbb{Q}_p(\gamma):\Bbb{Q}_p]\ge 5$.

• Otherwise $a= b^5$. Let $g(x)=a^{-1} f(xb)=x^5+5b^{16}x+1$, $h(x)=g(x-1)=x^5 - 5x^4 + 10x^3 - 10x^2 + 5xb^{16} + 5x- 5b^{16}$

  If $5\nmid b$ then it is Eisenstein at $5$.

• It remains to check the case $a=(5c)^5$.