As a follow-up to this Galois theory course and the one in commutative algebra, could you do some lectures on algebraic number theory and local fields?
@@tiop52 Thanks, I'll check those out. Incidentally I'll be doing the virtual Arizona Winter School, organized by Professor Kedlaya in just a couple of weeks.
~2:10 Frobenius endomorphism maps x to x^p where x is an element of finite field of characteristic p. I am not sure why “ p= 0” appeared at 1:15 ….. In case of finite field, Frobenius endomorphism is automorphisim, but it is not in general.For example, let k be a field of characteristic p, then Phi( (f(x) ) = f(x^p) for any polynomial f over k. Thus, Phi( k(x) ) is contained in k(x^p) which is strictly contained in k(x). Thus for a filed F = k(x), the surjectivity Phi(k(x)) = k(x) is not true. 2:40 Suppose we have a Galois extension M/Q for rationals Q, where M = Q[x]/(f(x)) and polynomial f(x) is in Z[x] and f is monic. For example, f(x) = x^2+1 and M = Q[i]. M is a splitting field of f over Z. Because Q contains 1/p for any prime, M/pZ = 0. To consider Frobenius map, we replace Q by Z and consider R = Z[x]/(f(x)) instead of M = Q[x]/(f(x)). For example, f=x^2+1, them R is the Gaussian integers R = Z[i]. 5:16 Assume that ASSUMPTION(1) R is a splitting field of f over Z (that is all roots of f are in R), ASSUMPTION(2) f(x) is separable on Z/pZ[x]. For example, 7:42 if f(x) = x^2+1, which is irreducible in Z[x], but reducible in (Z/5Z)[x]. In fact, (x + 2)(x + 3) = x^2 +5x + 6 = x^2 + 0x + 1 = f(x) in (Z/5Z)[x]. So, if we take f(x) = x^2 +1, on Z/5Z f splits into linear factors (x + 2)(x + 3) and the roots 2, and 3 are distinct in Z/5Z, we can say that x^2 +1 is separable on Z/5Z/ Separability of f over Z/pZ is equivalent that the discriminant of f does not vanish in Z/pZ. Note that discriminant is the product of the squared differences of all roots of f and this is symmetric by permutations of roots of f, so discriminant can be written by coefficients of f and hence discriminant is Z-valued function because now f(x) is in Z[x]. 7:20 Consider R/(p) = (Z/pZ)[x]/(f(x)). Recall that Q[x]/(f(x)) is Galois extension of Q, so Q[x]/(f(x)) is a field and hence we implicitly assumed that f(x) is an irreducible polynomial over Z. However this irreducibility of f over Z does not means irreducibility of f over Z/pZ. Because ASSUMPTION(1) and (2), f factors into distinct coprime factors in (Z/pZ)[x], i.e., f (x)= f1(x)*f2(x)*….. in (Z/pZ)[x]. So, by Chinese remainder theorem, (Z/pZ)[x]/(f) = (Z/pZ)[x]/(f1) @ (Z/pZ)[x]/(f2) @ …. , where @ denotes the product of fields. For example, f(x) = x^2 +1 and p =5, then (Z/5Z)[x]/(x^2+1) = (Z/5Z)[x]/(x+2) @ (Z/5Z)[x]/(x+3) ( = (Z/5Z) @ (Z/5Z) ).. 9:10 If alpha_1,…,alpha_n are roots of f(x) in (Z/pZ)[x]/(f), then they are distinct because we pick a prime p so that f is separable in (Z/pZ)[x]/(f), namely, f has distinct roots….. (Z/pZ)[x]/(f) = products of (Z/pZ)[x]/(f_i) = products of (Z/pZ)(alpha_i) (= n copies of Z/pZ.?..NO, but a vector space over Z/pZ of dimension = degree of f_i) I cannot prove all about slide 11:53 Count the number of automorphisms of Galois extension Q[x]/(f(x)) of Q via Chinese remainder theorem.. I do not understand. 15:19 Main Thorem,… now, I give up. This theorem connects a Frobenius map for fields of nonzero characteristic and a Galois groups for fields of zero characteristic. 18:51 If p = 1 mod 4, then there is some integer so that x^2 + 1 = 0 mod p. If p = 3 mod 4, then there are no integers so that x^2 + 1 = 0 mod p.
As a follow-up to this Galois theory course and the one in commutative algebra, could you do some lectures on algebraic number theory and local fields?
Not by Richard but see UCSD math 204A and 204B by Kiran Kedlaya for lectures on algebraic number theory, global and local fields.
Looks like he's going to...ua-cam.com/play/PL8yHsr3EFj52Qf7lc3HHvHRdIysxEcj1H.html I am HYPED for when this drops
@@TheAcer4666 Sweet! Me too
@@tiop52 Thanks, I'll check those out. Incidentally I'll be doing the virtual Arizona Winter School, organized by Professor Kedlaya in just a couple of weeks.
I can't thank you enough professor!😀
19:35 note: If x^2+1 is irreducible mod p, then Fp[x]/(x^2+1)=F_{p^2}, on which Frobenius automorphism acts non trivially. Hence, x^2+1 is reducible.
Minor correction, at 12:45 it should be Number
~2:10
Frobenius endomorphism maps x to x^p where x is an element of finite field of characteristic p. I am not sure why “ p= 0” appeared at 1:15 …..
In case of finite field, Frobenius endomorphism is automorphisim, but it is not in general.For example, let k be a field of characteristic p, then Phi( (f(x) ) = f(x^p) for any polynomial f over k. Thus, Phi( k(x) ) is contained in k(x^p) which is strictly contained in k(x). Thus for a filed F = k(x), the surjectivity Phi(k(x)) = k(x) is not true.
2:40 Suppose we have a Galois extension M/Q for rationals Q, where M = Q[x]/(f(x)) and polynomial f(x) is in Z[x] and f is monic. For example, f(x) = x^2+1 and M = Q[i]. M is a splitting field of f over Z.
Because Q contains 1/p for any prime, M/pZ = 0. To consider Frobenius map, we replace Q by Z and consider R = Z[x]/(f(x)) instead of M = Q[x]/(f(x)). For example, f=x^2+1, them R is the Gaussian integers R = Z[i].
5:16 Assume that
ASSUMPTION(1) R is a splitting field of f over Z (that is all roots of f are in R),
ASSUMPTION(2) f(x) is separable on Z/pZ[x]. For example, 7:42 if f(x) = x^2+1, which is irreducible in Z[x], but reducible in (Z/5Z)[x]. In fact, (x + 2)(x + 3) = x^2 +5x + 6 = x^2 + 0x + 1 = f(x) in (Z/5Z)[x]. So, if we take f(x) = x^2 +1, on Z/5Z f splits into linear factors (x + 2)(x + 3) and the roots 2, and 3 are distinct in Z/5Z, we can say that x^2 +1 is separable on Z/5Z/
Separability of f over Z/pZ is equivalent that the discriminant of f does not vanish in Z/pZ. Note that discriminant is the product of the squared differences of all roots of f and this is symmetric by permutations of roots of f, so discriminant can be written by coefficients of f and hence discriminant is Z-valued function because now f(x) is in Z[x].
7:20 Consider R/(p) = (Z/pZ)[x]/(f(x)). Recall that Q[x]/(f(x)) is Galois extension of Q, so Q[x]/(f(x)) is a field and hence we implicitly assumed that f(x) is an irreducible polynomial over Z. However this irreducibility of f over Z does not means irreducibility of f over Z/pZ.
Because ASSUMPTION(1) and (2), f factors into distinct coprime factors in (Z/pZ)[x], i.e., f (x)= f1(x)*f2(x)*….. in (Z/pZ)[x]. So, by Chinese remainder theorem,
(Z/pZ)[x]/(f) = (Z/pZ)[x]/(f1) @ (Z/pZ)[x]/(f2) @ …. ,
where @ denotes the product of fields. For example, f(x) = x^2 +1 and p =5, then
(Z/5Z)[x]/(x^2+1) = (Z/5Z)[x]/(x+2) @ (Z/5Z)[x]/(x+3) ( = (Z/5Z) @ (Z/5Z) )..
9:10
If alpha_1,…,alpha_n are roots of f(x) in (Z/pZ)[x]/(f), then they are distinct because we pick a prime p so that f is separable in (Z/pZ)[x]/(f), namely, f has distinct roots…..
(Z/pZ)[x]/(f) = products of (Z/pZ)[x]/(f_i) = products of (Z/pZ)(alpha_i) (= n copies of Z/pZ.?..NO, but a vector space over Z/pZ of dimension = degree of f_i)
I cannot prove all about slide 11:53
Count the number of automorphisms of Galois extension Q[x]/(f(x)) of Q via Chinese remainder theorem.. I do not understand.
15:19 Main Thorem,… now, I give up. This theorem connects a Frobenius map for fields of nonzero characteristic and a Galois groups for fields of zero characteristic.
18:51
If p = 1 mod 4, then there is some integer so that x^2 + 1 = 0 mod p.
If p = 3 mod 4, then there are no integers so that x^2 + 1 = 0 mod p.
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