Galois theory: Frobenius automorphism

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

We show that the Frobenius automorphism of a finite field an sometimes be lifted to characteristic 0. As an example we use the Frobenius automorphisms of Q[i] to prove that -1 i a square mod an odd prime p if and only if p is 1 mod 4.

[theflaggeddragon]

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?

[shubhmishra]

I can't thank you enough professor!😀

[vs-cwwc]

Minor correction, at 12:45 it should be Number <= Sum over i: # of maps from F_p(x)/(f_1) to F_p(x)/(f_i)

[SG-kjuy]

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.

[hausdorffm]

~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.

[migarsormrapophis]

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