# Calculating Galois Group of $x^5+2x^2+2x+5$ over $\mathbb{Q}$

First of all i verified that the quintic is irreducible over $\mathbb{Q}$, since it is irreducibile over $\mathbb{Z}_3$ by Eisenstein's Criterion with 2.

I know that the galois group G should be a transitive subgroup of $S_5$.

The polynomial has only 1 real root.

I tried to apply Dedekind's Theorem by considering the polynomial in $\mathbb{Z}_2$ where we have $x^5+2x^2+2x+5=x^5+1=(x+1)*(x^4+x^3+x^2+x+1)$, so (if i have correctly understood the theorem) there is an element $\sigma \in G$ which is a 4-cycle. For this reason i tend to exclude $A_5$ as the solution, but i don't know how to move further.

• For my benefit, could you please explain how you applied Eisenstein in $\mathbb Z_3[x]$? My confusion is that $\mathbb Z_3$ is a field, so the prime ideal $(2)$ is the whole of $\mathbb Z_3$. But then, $5 \in (2)^2$, so it doesn't look like Eisenstein applies. Or am I being dumb? – Kenny Wong Sep 1 '17 at 21:29
• Good point. i did it wrong. anyway after cheking the polynomial on wolfram it is really irreducible, now i try to find a way to prove it – Michele Mascherpa Sep 1 '17 at 21:34
• Can you also find the roots on Wolfram? I wonder if it has a pair of complex conjugate roots? If so, the Galois group must have an involution. – Kenny Wong Sep 1 '17 at 21:36
• it has one real root and two pair of complex conjugate roots. i think irreducibility can be proved in this way: since the quintic hasn't rational roots it can only be a product of a cubic and a quadratic expression, but then it should present a similar factorization also in $\mathbb{Z}_2$ which is false – Michele Mascherpa Sep 1 '17 at 21:42
• It just occurred to me that you get irreducibility of $f(x)$ also as follows. Your modulo two calculation shows that the only way $f(x)$ can factor in $\Bbb{Z}[x]$ is as a product of a linear and quartic polynomial. But, such a factorization implies that $f(x)$ has a rational zero (actually the zero must be an integer). But, you can eliminate the existence of such a zero with the aid of the so called rational root test. Just check that $f(\pm1)$ and $f(\pm5)$ are all non-zero, and you are done. – Jyrki Lahtonen Sep 3 '17 at 6:52

1. As Michele observed, modulo $2$ the polynomial $f(x)$ is a product of a linear and a quartic factor.
2. Modulo three we have $$f(x)=x(x^4-1)+2(x^2+1)=(x^2+1)(x^3-x-1),$$ and neither of those degree $\le3$ factors have any zeros in $\Bbb{F}_3$ so they are irreducible.
3. By the first fact the only way $f(x)$ can factor over $\Bbb{Z}$ is linear $\times$ quartic, and by the second fact the only way is quadratic $\times$ cubic. These are incompatible, so $f(x)$ must be irreducible.
4. Because $f(x)$ is irreducible, the Galois group $G$ is a transitive subgroup of $S_5$. Hence $5\mid |G|$. Dedekind's theorem, together with facts 1 and 2, implies that $|G|$ is divisible by both four and three. We can conclude that $60\mid |G|$ so $G$ is either $A_5$ or $S_5$.
5. Michele already observed that $G$ is not a subgroup of $A_5$, so we have eliminated all the alternatives to $G=S_5$.
• Thank you! The only thing i don't understand is the fact that $|G|$ is divisible by four and three implies that it is divisible by 60 – Michele Mascherpa Sep 2 '17 at 7:49
• @MicheleMascherpa Because $G$ acts transitively on the set of roots its order is automatically also divisible by five. – Jyrki Lahtonen Sep 2 '17 at 7:52
• In order to show that the galois group of an irreducible quintic is $S_5$ it could be enough to obtain that Modulo a prime it can be factorized in the product of a cubic*quadratic? Because with your last statement we have $30 | |G|$ but i think that the only transitive subgroup of $S_5$ with more than 30 elements are $S_5$ and $A_5$, which can be excluded. – Michele Mascherpa Sep 2 '17 at 8:51