Galois Group

**ZetaX** · May 17th 2009, 12:50 PM

Find the galois group of $f(x) =1+x+x^2/2+x^3/6+x^4/24.$

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

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

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

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

**ThePerfectHacker** · May 17th 2009, 01:33 PM

Originally Posted by ZetaX

Find the galois group of $f(x) =1+x+x^2/2+x^3/6+x^4/24.$

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

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

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

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

Let $a\in \mathbb{Q}^{\times}$ then $f(x)$ and $af(x)$ have the same Galois group. Because the splitting field of $f(x)$ and $af(x)$ 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.

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