21:57 I have a question about how to construct the desired extension. Let L=Q[x]/( phi_{\prod p_i} (x)). If we denote by f the map from (Z/p_1)* \times ... \times (Z/p_k)* onto G, then ker f\subset (Z/p_1)* \times ... \times (Z/p_k)* and (Z/p_1)* \times ... \times (Z/p_k)*/ ker f \equiv G. There exists an extension L/K whose Galois group is ker f. This K is the one?
18:40 Inverse Problem Let G be any given finite abelian group. then find extension K/Q such that Gal(K/Q) = G. We can solve this in some case, key point is the irreducibility of cyclotomic polynomials and Dirichlet theorem. By the fundamental theorem of Abelian groups, G is the following form. G = products of cyclic groups of order n_1,n_2,n_3,… By the Dirichlet theorem, we can pick distinct primes p1,p2,p3.. so that p_i= 1 mod n_i. Note that we can choose p_1, p_2,… are distinct, because for any n the set {x in Z; x =1 mod n} has infinitely many x by Dirichlet theorem. Because p_i = 1 mod n_i, it follows that p_i-1 divides n_i. Because the order of the group (Z/p_iZ)^* = {x in Z; gcd(x,p_p) =1} is p_i-1, which is divided by n_i = order of the group Z/n_iZ. So, there is an surjective map (Z/p_iZ)^* to Z/n_iZ. ( I do not know why ..). Define an extended field K/Q by adjoining roots of unity of order n := p_1*p_”2*p_3*, namely, K = Q[x]/(phi_n(x)) we get a field. Note that because phi_n is irreducible K is a field. Then I am not sure but Gal(K/Q) = products of (Z/p_iZ)^* among all i. ??? What we want is not Gal(K/Q) = products of (Z/p_iZ)^* over all i but some K such that Gal(K/Q) = products of (Z/n_iZ) over all i…. Humm I do not understand. 16:39 Here, phi_n(0) = 1 is used, but in general, phi_n(0) = +1 or “-1”., How to show phi_n(0) =1 in this context….??? 13:02 ~ 17:20 Proof of Dirichlet theorem in special case, a =1. THEOREM: If a,n are coprime, then there are infinitely many primes so that it equals to a modulo n. That is #{x is prime in Z| x = a mod n} is infinite. To prove this, we use the following lemma. LEMMA: If prime p and integers k and n satisfy p | phi_n(k) and gcd(p, n) = 1, then p = 1 mod n. PROOF OF LEMMA: ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- Because p | phi_n(k) and gcd(p, n) = 1, it follows that {k, k^2,….,,k^n=1 mod p} are all distinct in Z/pZ,.. I do not understand this …. Or I am wrong. ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- Because {k, k^2,….,,k^n=1 mod p} is a subgroup of the group (Z/pZ)* = {x is in Z/pZ| gcd(x, p)=1}, we can say that #{k, k^2,….,,k^n=1 mod p} divides # (Z/pZ)* = p-1, namely, n | p-1. So, p = 1 mod n. We get back to the proof of theorem. PROOF of THEOREM: For simplicity a =1, then n is any positive integer. Suppose we have found finite number of primes p1, p2,…., =1 mod n. Pick a prime p dividing phi_n(k’), where k’= n*p1*p2*….that is, p|phi_n(k’). Because in modular n*p1*p2*…, we get phi_n(n*p1*p2*….) = phi_n(0) = 1 mod (n*p1*p2*… ) (or -1 mod (n*p1*p2*… )??? I guess). So, p is not equal to p1,p2,… and p does not divide n, that is p is a new prime number. Because, p|phi_n(k’), we apply the lemma, so p =1 mod n. 6:40 ~ 12:40 Proof of irreducibility of cyclotomic polynomial Phi_n(x). If n is prime (or n = p^k for some prime p), then by Eisenstein criterion, Phi_n(x+1) is irreducible, so Phi(x) is. (Because all coefficients of Phi(x) is 1, we cannot apply Eisenstein criterion for Phi_n(x).). 2:50 the factorization of x^8 -1 = 0 is awesome. I do not understand the forumalae for general n, that is, x^n -1 = 0. 4:40 To calculate phi_12(x), we use the formula at 2:00. Using the formula, we get X^{12} - 1 = phi_1 *phi_2 * phi_3 * phi_4 * phi_6 *phi_12. Hence Phi_12 = (x^12 -1)/ ( phi_1 *phi_2 * phi_3 * phi_4 * phi_6 ) = (x^12 -1)/ ( ( phi_1 *phi_2 * phi_3 * phi_6 )* phi_4 ) = (x^12 -1)* phi_1 *phi_2/ ( ( phi_1 *phi_2 * phi_3 * phi_6 )*( phi_1 *phi_2* phi_4) ). Using the formulae, X^6 -1 = phi_1 * phi_2 * phi_3 * phi_6, X^4 -1 = phi_1 *phi_2 * phi_4, X^2 -1 = phi_1 *phi_2,
We get phi_12 = (x^12 -1)*(x^2-1) / ( (x^6-1)*( x^4-1) ) . About the exercise at 5:29 it seems difficult for me…. In the exercise at 5:29, I examine the first coefficients of phi(x). ( I know this is a miss direction) Let p,q,r be distinct primes. At x = 0, the values of phi_p(x), phi_{pq}(x), (d/dx) phi_1(x), (d/dx) phi_{p}(x) are all 1 but note that phi_1(0) = (d/dx) phi_{pq}(0) = -1. Using these, it is easy to calculate (d/dx)phi_{pqr}(x) at x = 0 and it is 1. So,…, the first coefficient of phi_{pqr}(x) is in the set {1,-1,0}. Derivations are easy to calculate and is this scheme is available to solve this?? 8:37 (Z/nZ)^* := { x is in Z/nZ; gcd(x,n) = 1 }, so #(Z/nZ)^* = phi(n) := Euler’s phi. If zeta is a primitive root of x^n-1 =0, then for any k with gcd(k,n) = 1, zeta^k is also primitive, which is easy to prove. I cannot prove the inverse. That is, for any primitive root, say r , there is some k of gcd(k,n) =1 so that r = zeta^k.
You are correct to worry whether k really has order n (in proof of Dirichlet's The); however, the proof of this is a simple induction. So, assume that for all cyclotomic polynomials phi_d(x) have phi(d) distinct roots mod p with order d, where d is a divisor of n and less than n. (Since x^n - 1 is separable mod p, so are all these polynomials mod p). Recall from Euler's identity for the totient function phi, that n = sum_{d|n} phi_d. so the sum over all such d
The fact that there is a surjective map follows from the fundamental theorem for finite abelian groups. To show this, show first that it is true for cyclic groups, then use the fundamental theorem to show that it is true for any abelian group. Since there exists a surjective map out of \prod (Z/p_iZ)* onto \prod Z/n_iZ, then there exists a surjective map from Gal(L/Q) onto \prod Z/n_iZ since Gal(L/Q) and \prod (Z/p_iZ)* are isomorphic. If we let H denote the kernel of this map, then there exists a subextension K/Q such that Gal(L/K)=H, since L/Q is a galois extension (L is a splitting field of Q and therefore normal over Q, and since char(Q)=0, then it is both normal and separable). By the first isomorphism theorem Gal(K/Q)\congGal(L/Q)/Gal(L/K)\cong \prod Z/n_iZ\cong G.
This is far more advanced then I am ready for, currently entering Calc 2 Integrals; will return later -Chemist that likes Math a Lot
How are the integrals going?
n = 105 = 3*5*7, the first number with three distinct odd prime factors
Definitely an odd situation...
21:57 I have a question about how to construct the desired extension. Let L=Q[x]/( phi_{\prod p_i} (x)). If we denote by f the map from (Z/p_1)* \times ... \times (Z/p_k)* onto G, then ker f\subset (Z/p_1)* \times ... \times (Z/p_k)* and (Z/p_1)* \times ... \times (Z/p_k)*/ ker f \equiv G. There exists an extension L/K whose Galois group is ker f. This K is the one?
18:40 Inverse Problem
Let G be any given finite abelian group. then find extension K/Q such that Gal(K/Q) = G.
We can solve this in some case, key point is the irreducibility of cyclotomic polynomials and Dirichlet theorem.
By the fundamental theorem of Abelian groups, G is the following form.
G = products of cyclic groups of order n_1,n_2,n_3,…
By the Dirichlet theorem, we can pick distinct primes p1,p2,p3.. so that p_i= 1 mod n_i. Note that we can choose p_1, p_2,… are distinct, because for any n the set {x in Z; x =1 mod n} has infinitely many x by Dirichlet theorem.
Because p_i = 1 mod n_i, it follows that p_i-1 divides n_i. Because the order of the group (Z/p_iZ)^* = {x in Z; gcd(x,p_p) =1} is p_i-1, which is divided by n_i = order of the group Z/n_iZ. So, there is an surjective map (Z/p_iZ)^* to Z/n_iZ. ( I do not know why ..).
Define an extended field K/Q by adjoining roots of unity of order n := p_1*p_”2*p_3*, namely, K = Q[x]/(phi_n(x)) we get a field. Note that because phi_n is irreducible K is a field. Then I am not sure but Gal(K/Q) = products of (Z/p_iZ)^* among all i. ???
What we want is not Gal(K/Q) = products of (Z/p_iZ)^* over all i but some K such that
Gal(K/Q) = products of (Z/n_iZ) over all i…. Humm I do not understand.
16:39 Here, phi_n(0) = 1 is used, but in general, phi_n(0) = +1 or “-1”., How to show phi_n(0) =1 in this context….???
13:02 ~ 17:20 Proof of Dirichlet theorem in special case, a =1.
THEOREM: If a,n are coprime, then there are infinitely many primes so that it equals to a modulo n. That is #{x is prime in Z| x = a mod n} is infinite.
To prove this, we use the following lemma.
LEMMA: If prime p and integers k and n satisfy p | phi_n(k) and gcd(p, n) = 1, then p = 1 mod n.
PROOF OF LEMMA:
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Because p | phi_n(k) and gcd(p, n) = 1, it follows that {k, k^2,….,,k^n=1 mod p} are all distinct in Z/pZ,.. I do not understand this …. Or I am wrong.
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Because {k, k^2,….,,k^n=1 mod p} is a subgroup of the group (Z/pZ)* = {x is in Z/pZ| gcd(x, p)=1}, we can say that #{k, k^2,….,,k^n=1 mod p} divides # (Z/pZ)* = p-1, namely, n | p-1. So, p = 1 mod n.
We get back to the proof of theorem.
PROOF of THEOREM: For simplicity a =1, then n is any positive integer. Suppose we have found finite number of primes p1, p2,…., =1 mod n. Pick a prime p dividing phi_n(k’), where k’= n*p1*p2*….that is, p|phi_n(k’). Because in modular n*p1*p2*…, we get phi_n(n*p1*p2*….) = phi_n(0) = 1 mod (n*p1*p2*… ) (or -1 mod (n*p1*p2*… )??? I guess). So, p is not equal to p1,p2,… and p does not divide n, that is p is a new prime number. Because, p|phi_n(k’), we apply the lemma, so p =1 mod n.
6:40 ~ 12:40 Proof of irreducibility of cyclotomic polynomial Phi_n(x). If n is prime (or n = p^k for some prime p), then by Eisenstein criterion, Phi_n(x+1) is irreducible, so Phi(x) is. (Because all coefficients of Phi(x) is 1, we cannot apply Eisenstein criterion for Phi_n(x).).
2:50 the factorization of x^8 -1 = 0 is awesome. I do not understand the forumalae for general n, that is, x^n -1 = 0.
4:40 To calculate phi_12(x), we use the formula at 2:00. Using the formula, we get
X^{12} - 1 = phi_1 *phi_2 * phi_3 * phi_4 * phi_6 *phi_12.
Hence
Phi_12 = (x^12 -1)/ ( phi_1 *phi_2 * phi_3 * phi_4 * phi_6 )
= (x^12 -1)/ ( ( phi_1 *phi_2 * phi_3 * phi_6 )* phi_4 )
= (x^12 -1)* phi_1 *phi_2/ ( ( phi_1 *phi_2 * phi_3 * phi_6 )*( phi_1 *phi_2* phi_4) ).
Using the formulae,
X^6 -1 = phi_1 * phi_2 * phi_3 * phi_6,
X^4 -1 = phi_1 *phi_2 * phi_4,
X^2 -1 = phi_1 *phi_2,
We get phi_12 = (x^12 -1)*(x^2-1) / ( (x^6-1)*( x^4-1) ) .
About the exercise at 5:29 it seems difficult for me….
In the exercise at 5:29, I examine the first coefficients of phi(x). ( I know this is a miss direction)
Let p,q,r be distinct primes.
At x = 0, the values of phi_p(x), phi_{pq}(x), (d/dx) phi_1(x), (d/dx) phi_{p}(x) are all 1 but note that phi_1(0) = (d/dx) phi_{pq}(0) = -1. Using these, it is easy to calculate (d/dx)phi_{pqr}(x) at x = 0 and it is 1. So,…, the first coefficient of phi_{pqr}(x) is in the set {1,-1,0}. Derivations are easy to calculate and is this scheme is available to solve this??
8:37 (Z/nZ)^* := { x is in Z/nZ; gcd(x,n) = 1 }, so #(Z/nZ)^* = phi(n) := Euler’s phi.
If zeta is a primitive root of x^n-1 =0, then for any k with gcd(k,n) = 1, zeta^k is also primitive, which is easy to prove. I cannot prove the inverse. That is, for any primitive root, say r , there is some k of gcd(k,n) =1 so that r = zeta^k.
You are correct to worry whether k really has order n (in proof of Dirichlet's The); however, the proof of this is a simple induction. So, assume that for all cyclotomic polynomials phi_d(x) have phi(d) distinct roots mod p with order d, where d is a divisor of n and less than n. (Since x^n - 1 is separable mod p, so are all these polynomials mod p). Recall from Euler's identity for the totient function phi, that n = sum_{d|n} phi_d. so the sum over all such d
The fact that there is a surjective map follows from the fundamental theorem for finite abelian groups. To show this, show first that it is true for cyclic groups, then use the fundamental theorem to show that it is true for any abelian group.
Since there exists a surjective map out of \prod (Z/p_iZ)* onto \prod Z/n_iZ, then there exists a surjective map from Gal(L/Q) onto \prod Z/n_iZ since Gal(L/Q) and \prod (Z/p_iZ)* are isomorphic. If we let H denote the kernel of this map, then there exists a subextension K/Q such that Gal(L/K)=H, since L/Q is a galois extension (L is a splitting field of Q and therefore normal over Q, and since char(Q)=0, then it is both normal and separable). By the first isomorphism theorem Gal(K/Q)\congGal(L/Q)/Gal(L/K)\cong \prod Z/n_iZ\cong G.
Cyclotomic? More like "Cool and complex; I like it!" 👍
yeeeeeeeee
e
Any amount of thanks is less
Oh stop acting like you won the Fields Medal or something.