Let be a Kummer -extension. Write . If we define and then it can be shown that . Therefore, the degree of the extension, , is . Define the embedding by . Then, . It turns out that (this notation means 'group generated by'). By the embedding it means . And this gives us a way to find the degree of a Kummer extension. What I do not see is why . I tried out some examples with and convinced myself that the only way to get an element when squared is when the element is a product of the -th roots. But I cannot find a nice proof to this.
No, I am not familar with this theorem. I do not even understand the notation you are using.
However, I did think of an elementary way of proving . We note that every element in has to be expressed in terms of powers and linear combinations of 's. For example, consider (here ) then is basis for and is basis for . Thus, basis for is . Therefore, any element can be written as . And this is true in general, any element in a Kummer extension can be written out in terms of linear combinations of the products of the -th roots. Let be a basis for where 's are products of the 's by above. If then we can write where is a polynomial over with variables. Assume that has the property that . We wish to argue that all terms of must be zero except for possibly one of them, and this will show that the only elements in the Kummer group are the products of the -th roots. Suppose this was not the case and there were two coefficients for which were non-zero. Then in the expansion of we would have a non-zero "cross term" . But then this means by the uniqueness of the representation of an element in terms of a basis.
this proof is not complete! you can't just extend things from a simple extension such as to the general case! ok, let's see:
i think you chose your basis to be all possible products of powers of then you need to show that this set is linearly independent
over F. then when you choose an element of (which by the way would be just a simple linear combination of and you don't need
to consider it as a polynomial ), then in computing the charactersitic of F will also get involved!
Consider this. Let . Then will spam . It is not necessarily a basis so is a basis - say for simplicity sake. Now will spam . But it is not necessarily a basis. Say is this basis. Thus, is basis for . And so on. The same idea for . Thus, any element is a linear combination of the extraction of roots in general.
But the charachteristic of should not come into play, should it? Because since has all -roots of unity it means the characheristic of does not divide . Thus, we do not get multiple of when we expand . Do you buy that?to consider it as a polynomial ), then in computing the charactersitic of F will also get involved!
then you need to prove that in general K has a basis chosen from the set of products of powers of which is
closed under multiplication ... and proving the linear independence in general is not an easy task!
i know that the char(F) doesn't divide n, but char(F) can still divide some of the coefficients in the expansion ofBut the charachteristic of should not come into play, should it? Because since has all -roots of unity it means the characheristic of does not divide . Thus, we do not get multiple of when we expand . Do you buy that?
Let be an algebraic extension and let . It is a known result that if then is a basis for . Furthermore, if are fields with having basis and having basis then is basis for . Now let us apply these facts to . Let . Now define . We have the tower of fields . Let for . Because has as a root it means . This means is a basis for . Thus, is a basis for . Let be all the elements of . This means if then we can write .
If then all 's are zero expect for possibly one of them. We will assume . If this was not the case then there are (by relabeling if necessary) in the linear expansion of such that the coefficients of are non-zero. Then in the expansion of we will have a 'cross term' which a non-zero coefficient. Let and where . Then the cross term . Now realize that . For this to be an element of we need that are multiples of . Note that if then we can write with . Because of the constrainst it is impossible for all to be multiples of unless which is not the case. It seems that the problem is that that we do not know if there is a triple cross term, for example, whose basic reduced form (of the exponents) is not the same as . And therefore it is possible we get cancellation from the cross terms.
let and then both and are subgroups of which contain
Claim:
Proof: since and we must have and now if then
thus for some n-th root of unity hence so finally if then for some
which gives us for some n-th root of unity so and thus
so by the theorem now it's obvious that
I would restate the theorem as: is a bijection from the set of all subgroups of which contain onto the set of all Kummer n-extension of which are contained in (the algebraic closure of ). Is this okay (this does not damage the theorem in any way) ?
That was stupid of me. When I looked at your proof I used .
I think I understand your proof if I do it in the following manner: Given and . Construct so that . Define and . So this is basically what you are doing except I picked a big extension field to work under - this makes me feel safe. And now I just carry through the exact same argument. I think that is okay to do.
My last question is how do you prove this bijection theorem between subgroups and Kummer n-extensions. In fact, I do not even see why is a Kummer n-extension over . The definition I am using is that is a Kummer n-extension when is Galois and has primitive -th root of unity and is a finite abelian group whose exponent divides . I can see that is a splitting field over and each polynomial is seperable over . However, I do not see why is a finite abelian group whose exponent divides .
Note, I am familar with the theorem that defined by is a non-degenerate pairing. And so . Maybe your bijection theorem is somehow related to it, but I do not see the connection.
Congratulations on your 1th post!
Here is another approach that my professor showed me without using that theorem.
We know that (the Kummer group) is isomorphic to . Let and be the -th roots of unity. Define by . This is a bilinear pairing, since it is non-degenerate it means . But since is a finite abelian group it is isomorphic to its charachter group. And this means that is isomorphic to the Kummer group, since is a subgroup of the Kummer group it means that in fact the Kummer group.