# Galois Theory Question

• Jan 15th 2010, 01:20 AM
Boysilver
Galois Theory Question
Let $\displaystyle f(x)=x^4+6x^2+3x-3$ and let $\displaystyle L$ be a splitting field of $\displaystyle f$ over $\displaystyle \mathbb{Q}$.

I can show that 73 has a cube root in $\displaystyle L$. Now I'm asked to "deduce" that the Galois group $\displaystyle G$ of $\displaystyle f$ over $\displaystyle \mathbb{Q}$ has a normal subgroup $\displaystyle H$ such that the quotient group $\displaystyle G/H$ is isomorphic to $\displaystyle S_3$, and then deduce that $\displaystyle G$ is isomorphic to $\displaystyle S_4$.

My first thought was that we can show $\displaystyle G$ is isomorphic to $\displaystyle S_4$ easily enough anyway - it's not difficult to narrow down the possibilities to either $\displaystyle A_4$ or $\displaystyle S_4$, then we can use the discriminant to rule out the first case. Once we know $\displaystyle G$ is isomorphic to $\displaystyle S_4$ then we can show the "normal subgroup" part.

But this is obviously not what the question is getting at...the idea is to presumably use the fact that 73 has a cube root. Can anyone help out? Thanks.
• Jan 15th 2010, 02:36 AM
tonio
Quote:

Originally Posted by Boysilver
Let $\displaystyle f(x)=x^4+6x^2+3x-3$ and let $\displaystyle L$ be a splitting field of $\displaystyle f$ over $\displaystyle \mathbb{Q}$.

I can show that 73 has a cube root in $\displaystyle L$. Now I'm asked to "deduce" that the Galois group $\displaystyle G$ of $\displaystyle f$ over $\displaystyle \mathbb{Q}$ has a normal subgroup $\displaystyle H$ such that the quotient group $\displaystyle G/H$ is isomorphic to $\displaystyle S_3$, and then deduce that $\displaystyle G$ is isomorphic to $\displaystyle S_4$.

My first thought was that we can show $\displaystyle G$ is isomorphic to $\displaystyle S_4$ easily enough anyway - it's not difficult to narrow down the possibilities to either $\displaystyle A_4$ or $\displaystyle S_4$, then we can use the discriminant to rule out the first case. Once we know $\displaystyle G$ is isomorphic to $\displaystyle S_4$ then we can show the "normal subgroup" part.

But this is obviously not what the question is getting at...the idea is to presumably use the fact that 73 has a cube root. Can anyone help out? Thanks.

73 has a root in $\displaystyle L\Longrightarrow g(x)=x^3-73$ is an irreducible pol. over $\displaystyle \mathbb{Q}$ and it thus splits in linear factors over $\displaystyle L$ (why?).

We thus have the fields tower $\displaystyle \mathbb{Q}\le \mathbb{Q}(\alpha)\le L$ , with $\displaystyle \alpha=\sqrt[3]{73}\in L$.

Use now the Fund. Theorem of Galois Theory with the group of $\displaystyle \mathbb{Q}-$automorphisms of $\displaystyle L$ that leave $\displaystyle \mathbb{Q}(\alpha)$ fixed...

Of course, you STILL have to prove that the resulting group is the non-abelian group of order 6, but if you succeed with the above I think this already will be easier.

Tonio