Galois theory: Kummer extensions

This lecture is part of an online graduate course on Galois theory.

We describe Galois extensions with cyclic Galois group of order n in the case when the base field contains all n'th roots of unity and has characteristic not dividing n. We show that all such extensions are radical. As an example we express cos 2pi/7 explicitly using cube roots. Finally we mention the Kummer pairing.

[k-theory]

Thank you! I think this is my favorite video of yours so far. I had read some basic stuff about Kummer theory in the intro sections of class field theory texts before, but it never really "clicked" until this video.

[hausdorffm]

For an extension of field M/K of Gal(M/K) = Z/nZ,
Show that M = K(n-th root of something in K) under the assumptions
K contains n-th roots of 1 and char K dose not divide n.
I guess, from the example of this video, this extension is a Galois extension.... I am not sure... Galois is redundant?

We regard M as a vector space over K and Gal(M/K) = K-linear maps.

Around 8:00, in case of n=3, concrete eigenspace decomposition and projections to each eigen space are constructed.
The eigen-values of a linear map, denoted by sigma are, one, omega and omega squared which are distinct. So, if dimension of M over K is three, the map sigma is diagonalizable…..I am not sure why dimension of M over K is three.
In his lecture of Galois extensions, he said if an extension of field M/K is Galois extension, then dimension of M over K as vector space is equal to the order of G, i.e., [M:K] = |Gal(M/K)|. Now, if we assume that M/K is a Galois extension, then dimension of M over K is three, because now Galois group is Z/nZ and n = 3, so, [M:K] = |Z/nZ| = 3.

At 10:15, If eigenvalue is not 1, then corresponding eigenvector moves by sigma, this means that this vector is not in K. So, the eigen vector, denoted by t at 10:15 is not in K but its n-th power is in K. I am not sure why K(t) = M….K(t) strictly contains K, because t is not in K. How can we prove K(t) is whole M???



16:31 ~ 22:22 For a field K, there is a map,
K*/(K*^n) -> { Isomorphism class of K(n-th root of x); x is in K}
a -> K(n-th roof of a)

10:12 Typo. M = K(sqrt(T)) will be M = K(n-th root of T)


5:41 Let K be a field of characteristic p. If z^n = 1 in K and n and p are coprime, then order of z is n, that is, z, z^2, z^3, …, z^n are distinct.
PROOF: for the sake of contradiction, suppose that order of z is n and p divides n, i.e., n = pq for some q of Z. Then (z^n) - 1 = z^(pq) -1 = (z^q -1 )^p in K. Because any field is integral domain, z^q – 1 = 0. This contradicts the assumption that z is of order n.

[maxwibert]

Is there an exhaustive technique for determining eigenvalues and such from non-matrix linear transformations of M/K like sigma @8:34? My first guess is to find sigma's minimal polynomial in the space of linear transformations to get at the eigenvalues. Even if the minimal poly of this particular sigma is clear, should we always be able to find the minimal polynomial given the order of the automorphism?

[f-th]

Seems related to representation theory

[diribigal]

"7 is a completely different prime from 3". To be fair, 7 is 3!+1, but as shown that's not really relevant.