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.
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.
The reason that K(t)=M is that [K(t):K]=n since the minimal polynomial of t over K is f=X^n-t^n: Note that f=\prod (X-t\zeta^i) (note that this works because n is prime, I’ll add in more details if n isn’t prime while of course assuming that char(K) mid n). Let m denote the minimal polynomial. Since m\mid f, then m=(X-t)\prod_i=0^k-1 (X-t\zeta^(n_i)) for k\leq n, where n_i are distinct integers (the tzeta^n_i are distinct because the derivative of f is nX^n-1 which only has 0 as a root, notice that Chark(K) mid n, which means that f’ eq 0. So f has no multiple roots, which means that neither can m. In particular, this means that tzeta^n_i eq tzeta^n_j for i eq j, and that means that n_i eq n_j) with 0\leq n_i\leq n-1 and labelled such that n_i\leq n_j whenever i\leq j and such that n_0=0 (since X-t has to be a factor of m since m is the minimal polynomial of t over K) Since m\in K[X], then the constant term is in K. This implies that t^k is in K since K contains all roots of unity. Hence sigma(t^k)=t^k. That means that zeta^k=1. Hence, n|k, but since k\leq n, this implies that k=n. Since the n_i are distinct and they lie between 0 and n-1 and n_i
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?
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.
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.
The reason that K(t)=M is that [K(t):K]=n since the minimal polynomial of t over K is f=X^n-t^n: Note that f=\prod (X-t\zeta^i) (note that this works because n is prime, I’ll add in more details if n isn’t prime while of course assuming that char(K)
mid n). Let m denote the minimal polynomial. Since m\mid f, then m=(X-t)\prod_i=0^k-1 (X-t\zeta^(n_i)) for k\leq n, where n_i are distinct integers (the tzeta^n_i are distinct because the derivative of f is nX^n-1 which only has 0 as a root, notice that Chark(K)
mid n, which means that f’
eq 0. So f has no multiple roots, which means that neither can m. In particular, this means that tzeta^n_i
eq tzeta^n_j for i
eq j, and that means that n_i
eq n_j) with 0\leq n_i\leq n-1 and labelled such that n_i\leq n_j whenever i\leq j and such that n_0=0 (since X-t has to be a factor of m since m is the minimal polynomial of t over K)
Since m\in K[X], then the constant term is in K. This implies that t^k is in K since K contains all roots of unity. Hence sigma(t^k)=t^k. That means that zeta^k=1. Hence, n|k, but since k\leq n, this implies that k=n. Since the n_i are distinct and they lie between 0 and n-1 and n_i
Love your videos!
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?
Here there are n distinct eigenvalues so sigma is diagonalisable...
excellent!
Can't thank you enough!
Seems related to representation theory
yeeeee
"7 is a completely different prime from 3". To be fair, 7 is 3!+1, but as shown that's not really relevant.