# Galois group of $x^5-5x+10$

I was illustrating the theorem on solvability by radicals through some examples of degree $5$ polynomials. One I chose was $x^5-5x+10$. I was (perhaps wrongly) going to prove that the Galos group is $S_5$, with expectation that it has exactly three real roots.

However, it has exactly one real root (which I saw by looking graph of this polynomial in $\mathbb{R}^2$ using some online software). Then I get stucked, and couldn't completely find the Galois group of this polynomial.

How do we proceed to determine the Galois group of this polynomial over $\mathbb{Q}$?

• have you caculat the discriminant of this – m.idaya Jul 2 '16 at 10:56
• – Watson Dec 26 '16 at 13:31

Observing discriminant of polynomial is very helpful to find its Galois group. It is known that if $D$ is discriminant of polynomial $f(x)$, then Galois group $G=G_{f}$ of $f(x)$ is contained in $A_{5}$ iff $D\in \mathbb{Q}^{2}$. (This holds for general fields with $char\neq 2$, not only $\mathbb{Q}$. Proof is not hard and you can find proof in "Abstract Algebra", Chap 14, Proposition 33 of Dummit-Foote.) As Wolfram says, its discriminant is 30450000 which is not a square. So $G$ is not contained in $A_{5}$.
Also, by Eisenstein criterion with $p=5$, $f(x)$ is irreducible and $[\mathbb{Q}(\alpha):\mathbb{Q}]=5$ for any $f(\alpha)=0$. So order of $|G|$ has to be divided by $5$.
Then if you see here, the only candidates are $S_{5}$ and general Affine group $GA(1,5)$. If we reduce the polynomial modulo $3$, then $\overline{f}(x)=x^{5}+x+1=(x-1)^{2}(x^{3}-x^{2}+1)$ in $\mathbb{F}_{3}[x]$. Since $x^{3}-x^{2}+1$ is irreducible over $\mathbb{F}_{3}$, order of $G_{\overline{f}}$ has to be divided by $3$. Since $G_{\overline{f}}\leq G_{f}$ (this is non-trivial result, and see here for the Tate's proof), $|G_{f}|$ is also divided by $3$ and contains a $3$-cycle. We can show that $5$-cycle and $3$-cycle in $S_{5}$ generates $A_{5}$, so the answer is $G_{f}=S_{5}$.
I think it will be possible to find a prime $p$ s.t. $\overline{f}(x)\in\mathbb{F}_{p}[x]$ has an irreducible factor of degree $2$, and then you don't have to calculate discriminant of $f$.
• Useful points. Just to make sure I will remark that finding a prime such tha $\overline{f}$ has an irreducible quadratic factor does not quite suffice. If $\overline{f}$ splits as a product of a linear factor and two distinct quadratics that only guarantees the existence of a permutation of cycle type $(2,2,1)$ in $G_f$. But such a permutation is still even. You need cycle types $(2,1,1,1)$ or $(4,1)$ or $(3,2)$ to be sure of finding odd permutations. Also, if $\overline{f}$ has a repeated factor, then you cannot use that prime. For example, your modulo $3$ factorization doesn't help... – Jyrki Lahtonen Jul 4 '16 at 9:06
• (cont'd) because the factor $(x-1)$ is repeated. Theorem A in your link does require that $\overline{f}$ should not have roots of multiplicity $>1$. So, modulo $13$ factorization gives a product of two disjoint 2-cycles, modulo $17$ gives a 3-cycle, and modulo $41$ a 4-cycle. At that point we can conclude that $G_f=S_5$. – Jyrki Lahtonen Jul 4 '16 at 9:11
• @JyrkiLahtonen Thank you for your comment! Actually I find the theorem $G_{\overline{f}}\leq G_{f}$ right before answer the question, so I missed about some important details. I will modify this answer later. – Seewoo Lee Jul 6 '16 at 16:08