proof of fundamental theorem of Galois theory
Let be an extension field of contained in . Then is Galois over , and is a subgroup of .
Note that is normal and separable because it is a Galois extension; it remains to prove that is also normal and separable. Since is normal and finite over , it is the splitting field of a polynomial over . Now is also the splitting field of over (because ), so is normal.
To see that is also separable, suppose , and let be its minimal polynomial over . Then the minimal polynomial of over divides , which has no double roots in its splitting field by the separability of . Therefore has no double roots in its splitting field for any , so is separable over .
The assertion that is a subgroup of is clear from the fact that . ∎
The definition of makes sense because of Lemma 1. The
for all subgroups and all fields with follow from the properties of the Galois group. The fixed field of is precisely ; on the other hand, since is the fixed field of in , is the Galois group of .
For extensions and of with , we have
so . This shows that is inclusion-reversing. ∎
The following lemmas show that normal subextensions of are Galois extensions and that their Galois groups are quotient groups of .
: Since for all and , is a zero of the minimal polynomial of over , we have by the of .
: For all the equality
holds for all (from the assumption it follows that , which is fixed by ). This implies that
for all .
: Let , and let be the minimal polynomial of over . Since is normal, splits into linear factors in . Suppose is another zero of , and let be such that (such a always exists). By assumption, for all we have , so that
This shows that lies in as well, so splits in . We conclude that is normal over . ∎
Let be a normal subgroup of . Then is a Galois extension of , and the homomorphism
induces a natural identification
The map is well-defined by the implication from Lemma 3. It is surjective since every automorphism of that fixes can be extended to an automorphism of (if , for example, we can choose an such that using the primitive element theorem, and we can extend to by putting ). The kernel of is clearly equal to , so the first isomorphism theorem gives the claimed identification. ∎
|Title||proof of fundamental theorem of Galois theory|
|Date of creation||2013-03-22 14:26:38|
|Last modified on||2013-03-22 14:26:38|
|Last modified by||pbruin (1001)|