In the proposition above, the assumption that can not be dropped. If fact, quadratic extensions of do not exist, for if , then .
For the rest of the discussion, we assume that .
Pick any element in . Then and . So is a root of the irreducible polynomial in . If we define to be , then is the other root of , clearly also in . This implies that the minimal polynomial of every element in has degree at most 2, and splits into linear factors in .
Since , are two distinct roots of . This shows that is separable over .
Now, let be any irreducible polynomial over which has a root in . Then the minimal polynomial of in must divide . But because is irreducible, . This shows that is normal over . Since is both separable and normal over , it is a Galois extension over .
|Date of creation||2013-03-22 15:42:34|
|Last modified on||2013-03-22 15:42:34|
|Last modified by||CWoo (3771)|