# Galois group of a polynomial of degree seven

Let $K$ be the splitting field over $\mathbb{Q}$ w.r.t. the polynomial $x^7 - 10x ^5+15x+5$. I think its Galois group is the symmetric group $S_7$. I tried to prove it using a theorem which says: "If the degree of the polynomial is a prime $p$, the polynomial is irreducible and it has exactly two non real roots, then its Galois group is $S_p$." In this case, I know that $x^7 - 10x ^5+15x+5$ is irreducible (by the Eisenstein criterion). However, I could not study its roots... I tried to study its derivative... My methods were effective in the remaining questions... Someone, please, know how to solve this problem?

Thank you.

• Also since the degree is odd, you know it must have at least one real root. – Laars Helenius Nov 25 '15 at 18:39
• @Kaj: your first claim is only true for quadratics and cubics. It's possible for quartics and up to have positive discriminant and non-real roots. – Qiaochu Yuan Nov 25 '15 at 18:40
• Oops, you are correct @QiaochuYuan. Fortunately, for this polynomial, the discriminant appears to be negative. – Kaj Hansen Nov 25 '15 at 18:43

Maybe you'll find this cheating but check out this table:

$$\begin{array}{c|c}i&f(i)\\ \hline-4 & -6199 \\ -3 & 203 \\ -2 & 167 \\ -1 & -1 \\ 0 & 5 \\ 1 & 11 \\ 2 & -157 \\ 3 & -193 \\ 4 & 6209 \end{array}$$

By the intermediate value theorem, this implies that $f$ has at least $5$ real zeroes (in the intervals [-4,-3], [-2,-1], [-1,0], [1,2] and [3,4].

(In fact you don't really have to compute $f(-4)$ and $f(4)$ if you note instead that $\lim_{x\to\pm\infty}f(x)=\pm\infty$).

Its discriminant $\prod_{i<j} (x_i-x_j)^2$ turns out to be negative, so the polynomial cannot have only real roots. So there must be at least $2$ non-real roots $z$ and $\bar z$.

(In fact the discriminant is equal to $-576043678484375$, to the best of my knowledge the easiest way to compute it with pen and paper is to compute the determinant of the resultant $R(f,f')$ as explained on wikipedia.)

So together there must be exactly $5$ real roots and you can apply the theorem.

• Not cheating, exact the way to calculate it by hand! (Or did you calculate the discriminant with a computer ? ) – Peter Nov 25 '15 at 19:06
• I computed it in my head while translating James Joyce's Ulysses to Sumerian :-) – Myself Nov 25 '15 at 19:10
• that was hilarious – Bogdan Simeonov Nov 25 '15 at 21:56

Myself's answer is very good, but I want to show how you can check the maximum number of roots with Descartes's sign rule :

$$f(x)=x^7-10x^5+15x+5$$ has two sign changes, so at most $2$ positive roots.

$$-f(-x)=x^7-10x^5+15x-5$$ has three sign changes, so at most $3$ positive roots, so $f(x)$ has at most $3$ negative roots.

With Myself's answer you can deduce that there are $5$ real roots.

• Nice, saves the headache of computing a discriminant! – Myself Nov 25 '15 at 19:50