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

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.

• Irreducible over what? $\mathbb{Q}$? Mar 15 at 12:48
• Is the rational root theorem not an algebraic method? Mar 15 at 12:50
• @Cameron Williams Yes Mar 15 at 13:11
• @aschepler It was used to conclude there is no linear factor, but Rouche’s theorem is not an algebraic method. Mar 15 at 13:12
• If $a$ has at least one prime factor that has an exponent of $1$, then the polynomial is irreducible by Eisenstein's criterion. Mar 15 at 13:13

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 $$a=(5c)^5$$.

• See here math.stackexchange.com/questions/193201/… to complete the proof. I assume the same method can be employed without using p-adics (since it is a qual problem); namely one can reduce to a degree $2$ and a degree $3$ factor and try to match coefficients. Mar 15 at 15:22
• Indeed I was trying to see why it didn't have a linear factor. Mar 15 at 15:32
• @reuns, your polynomial $g(x)=x^5+5b^{16}x+1$ clearly doesn't have a linear factor if $b\not=0$. (The rational root theorem says the only possible roots to check are $1$ and $-1$.) Mar 15 at 15:37