# Math Help - Galois Theory Question

1. ## Galois Theory Question

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

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

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

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

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

My first thought was that we can show $G$ is isomorphic to $S_4$ easily enough anyway - it's not difficult to narrow down the possibilities to either $A_4$ or $S_4$, then we can use the discriminant to rule out the first case. Once we know $G$ is isomorphic to $S_4$ 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 $L\Longrightarrow g(x)=x^3-73$ is an irreducible pol. over $\mathbb{Q}$ and it thus splits in linear factors over $L$ (why?).

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

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