# Galois Group

• May 17th 2009, 12:50 PM
ZetaX
Galois Group
Find the galois group of $\displaystyle f(x) =1+x+x^2/2+x^3/6+x^4/24.$

I multiply the polynomial with 24 then get
$\displaystyle 24f(x) =24+24x+12x^2+4x^3+x^4.$

Then I have that the cubic resolvent of $\displaystyle 24f(x)$ is

$\displaystyle h(x) = z^3-12z^2+192.$
Therefore, I get that the galois group of $\displaystyle 24f(x)$ is $\displaystyle A_4.$

My question is: Can I so conclude that the Galois group of $\displaystyle f(x)$
is $\displaystyle A_4.$
• May 17th 2009, 01:33 PM
ThePerfectHacker
Quote:

Originally Posted by ZetaX
Find the galois group of $\displaystyle f(x) =1+x+x^2/2+x^3/6+x^4/24.$

I multiply the polynomial with 24 then get
$\displaystyle 24f(x) =24+24x+12x^2+4x^3+x^4.$

Then I have that the cubic resolvent of $\displaystyle 24f(x)$ is

$\displaystyle h(x) = z^3-12z^2+192.$
Therefore, I get that the galois group of $\displaystyle 24f(x)$ is $\displaystyle A_4.$

My question is: Can I so conclude that the Galois group of $\displaystyle f(x)$
is $\displaystyle A_4.$

Let $\displaystyle a\in \mathbb{Q}^{\times}$ then $\displaystyle f(x)$ and $\displaystyle af(x)$ have the same Galois group. Because the splitting field of $\displaystyle f(x)$ and $\displaystyle af(x)$ are identical. The Galois group of a polynomial is defined to be the Galois group of the splitting field over the base field, since the splitting fields are the same it follows that the Galois groups are the same.