# Galois group of a polynomial

• Apr 19th 2009, 03:26 AM
georgel
Galois group of a polynomial
The problem is to identify Galois group of the polynomial $f=x^4 + 2x^2 + x +3$.
We have done a couple of examples in class, but the I am unable to solve this. :(

Thank you.
• Apr 19th 2009, 04:27 AM
peteryellow
There are many ways to do it. Which method have you learned so far?
You can use reduction modulo $p$, or you can use the method where you find cubic resolvent and then determine the Galois group. Have you learned this?

Anyways, It is easily seen that the polynomial is irreducible over $\mathbb Q$. You can try to show this as following.

It is easily seen by Rational root theorem that the polynomial has no rational root, furthermore assume that the polynomial can be written as
$x^4+2x^2+x+3 = (x^2+bx+c)(x^2+dx+e)$ and then try to get a contradiction. Therefore you can conlude that the polynomial is irreducible over $\mathbb Q$.

By reducing the polynomial modulo $3$ we have $(x^3+2x+1)(x)$ and modulo 5 we have
$(x+2)(x+1)(x^2+2x+4)$. Since The Galois group contains a trasoposition and a 3-cycle we conclude that the Galois group is $S_4.$
• Apr 19th 2009, 04:40 AM
georgel
Quote:

Originally Posted by peteryellow
There are many ways to do it. Which method have you learned so far?
You can use reduction modulo p, or you can use the method where you find cubic resolvent and then determine teh Galois group. Have you learned this?

Yes, we've learned the method with cubic resolvent, but I'm unable to solve it by myself.
I would be really grateful for your help and time.
• Apr 19th 2009, 05:14 AM
peteryellow
The cubic resolvent is
$g(y) = y^3-2y^2-12y-25$ (If I have solved it correctly.)You can recheck it. then it is easily seen that the Galois group if g is $S_3$.
Then by using some Theorems you can conclude that $S_4$ is the Galois group.

Ask if something is not very clear.
• Apr 19th 2009, 05:26 AM
georgel
Thank you very very very much, I will read more about Galois groups and than ask if there are any questions left.