proof of Galois group of the compositum of two Galois extensions
Consider the diagram
(1): Let with a root . Then since (resp. ) is Galois over , all the roots of lie in (resp. ) and thus in . The result follows.
(2): We first show that is Galois over . Choose separable polynomials so that (resp. ) is a splitting field for (resp. ). Then is a splitting field for the squarefree part of , which is separable since it is squarefree and since are separable.
This map is a group homomorphism; its kernel is precisely those elements that leave both and fixed. Any such element must thus leave fixed, so that is injective. The image obviously lies in
by construction: . We will show that is precisely the image of by showing that the order of is the same as the index of the field extension .
For each , there are precisely elements of whose restrictions to are . Thus directly from the definition of ,
By the corollary to the theorem regarding the compositum of a Galois extension and another extension (http://planetmath.org/CorollaryToTheCompositumOfAGaloisExtensionAndAnotherExtensionIsGalois), we have
- 1 Dummit, D., Foote, R.M., Abstract Algebra, Third Edition, Wiley, 2004.
|Title||proof of Galois group of the compositum of two Galois extensions|
|Date of creation||2013-03-22 18:42:01|
|Last modified on||2013-03-22 18:42:01|
|Last modified by||rm50 (10146)|