proof of Kummer theory
Proof.
(1) Let ; then is Galois since contains all roots of unity and thus is a splitting field for , which is separable since in . Thus the elements of permute the roots of , which are
and thus for , we have for some . Define a map
We will show that is an injective homomorphism, which proves the result.
Since , each root of unity is fixed by . Then for ,
so that and is a homomorphism. The kernel of the map consists of all elements of which fix , so that is injective and we are done.
(2) Note that since is a root of , so that by Hilbert’s Theorem 90,
But then so that and since it is fixed by a generator of . Then clearly is a splitting field of , and the elements of send into distinct elements of . Thus admits at least automorphisms over , so that . But , so . ∎
References
- 1 Dummit, D., Foote, R.M., Abstract Algebra, Third Edition, Wiley, 2004.
- 2 Kaplansky, I., Fields and Rings, University of Chicago Press, 1969.
Title | proof of Kummer theory |
---|---|
Canonical name | ProofOfKummerTheory |
Date of creation | 2013-03-22 18:42:07 |
Last modified on | 2013-03-22 18:42:07 |
Owner | rm50 (10146) |
Last modified by | rm50 (10146) |
Numerical id | 5 |
Author | rm50 (10146) |
Entry type | Proof |
Classification | msc 12F05 |