Thread: Galois Theory Question

Let and let be a splitting field of over .

I can show that 73 has a cube root in . Now I'm asked to "deduce" that the Galois group of over has a normal subgroup such that the quotient group is isomorphic to , and then deduce that is isomorphic to .

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

Originally Posted by Boysilver

Let and let be a splitting field of over .

I can show that 73 has a cube root in . Now I'm asked to "deduce" that the Galois group of over has a normal subgroup such that the quotient group is isomorphic to , and then deduce that is isomorphic to .

My first thought was that we can show is isomorphic to easily enough anyway - it's not difficult to narrow down the possibilities to either or , then we can use the discriminant to rule out the first case. Once we know is isomorphic to 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 is an irreducible pol. over and it thus splits in linear factors over (why?).

We thus have the fields tower , with .

Use now the Fund. Theorem of Galois Theory with the group of automorphisms of that leave 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