# galois group and intermediate fields

• January 9th 2010, 07:17 PM
sporkmonkey
galois group and intermediate fields
Here is the problem I have:
Let E be a splitting field of f = x^4 - 4 in Q[x]

1. determine the galois group of E/Q (up to isomorphism)

2. Determine all intermediate fields between E and Q. Which of them are Galois over Q?

so far I have this:
letting a = 2^(1/2) then the roots are +/-a and +/- ai

I know Q[a] < E and |Q[a]:Q| = 2 (minimal polynomal is x^2 - 2) and ai is not in Q[a].

Letting E = Q[a, i] we have all roots of f in E and |Q[a, i]: Q[a]| = 2 since minimal polynomial is x^2 + 1.

thus |E| = |Q[a, i]: Q[a]||Q[a]:Q| = 4

from here I start drawing a big blank. Any help is appreciated.
• January 9th 2010, 07:56 PM
NonCommAlg
Quote:

Originally Posted by sporkmonkey
Here is the problem I have:
Let E be a splitting field of f = x^4 - 4 in Q[x]

1. determine the galois group of E/Q (up to isomorphism)

2. Determine all intermediate fields between E and Q. Which of them are Galois over Q?

so far I have this:
letting a = 2^(1/2) then the roots are +/-a and +/- ai

I know Q[a] < E and |Q[a]:Q| = 2 (minimal polynomal is x^2 - 2) and ai is not in Q[a].

Letting E = Q[a, i] we have all roots of f in E and |Q[a, i]: Q[a]| = 2 since minimal polynomial is x^2 + 1.

thus |E| = |Q[a, i]: Q[a]||Q[a]:Q| = 4

from here I start drawing a big blank. Any help is appreciated.

well, if $\sigma \in \text{Gal}(E/\mathbb{Q}),$ then $\sigma(\sqrt{2})$ has to be a root of $x^4 - 4.$ clearly we cannot have $\sigma(\sqrt{2})=\pm i \sqrt{2}$ because then $2=\sigma(2)=(\sigma(\sqrt{2}))^2=(\pm i\sqrt{2})^2=-2,$ which is not possible.

similarly we cannot have $\sigma(i \sqrt{2})=\pm \sqrt{2}.$ so every non-identity element of $\text{Gal}(E/\mathbb{Q})$ has order two and thus $\text{Gal}(E/\mathbb{Q}) \cong C_2 \times C_2.$ (here $C_2$ is the cyclic group of order two)