proof that the compositum of a Galois extension and another extension is Galois
Proof.
The diagram of the situation of the theorem is:
To see that is Galois, note that since is Galois, is a splitting field of a set of polynomials over ; clearly is a splitting field of the same set of polynomials over . Also, if is separable over , then also is separable over . Thus is normal and separable over , so is Galois. is obviously Galois over since .
Let be the restriction map
is clearly a group homomorphism, and since is normal over , is well-defined.
Claim is injective. For suppose and is the identity. Then is fixed on (since it is in and on (since its restriction to is the identity), so is fixed on and thus is itself the identity.
Now, the image of is a subgroup of with fixed field , and thus the image of is . Claim . is obvious: any element is fixed by for each since fixes . Thus . To see the reverse inclusion, choose ; then is fixed by each for . But , so that (as an element of ), is fixed by each . Thus so that .
Thus , and is then an isomorphism . ∎
References
- 1 Morandi, P., Field and Galois Theory, Springer, 1996.
Title | proof that the compositum of a Galois extension and another extension is Galois |
---|---|
Canonical name | ProofThatTheCompositumOfAGaloisExtensionAndAnotherExtensionIsGalois |
Date of creation | 2013-03-22 18:41:58 |
Last modified on | 2013-03-22 18:41:58 |
Owner | rm50 (10146) |
Last modified by | rm50 (10146) |
Numerical id | 6 |
Author | rm50 (10146) |
Entry type | Proof |
Classification | msc 11R32 |
Classification | msc 12F99 |