Algebra 111C Assignment 6
Hello, dear friend, you can consult us at any time if you have any questions, add WeChat: daixieit
Algebra 111C
Assignment 6
SOLUTIONS TO SELECTED PROBLEMS
1. Prove that S4 is solvable. Hint: start with S4 p A4, then the only thing needed is some/any G a A4 of order 4 .
Solution. Want Ga A4 with |G| = 4. Then at once
S4 B A4 B G B {1}
with S4 /A4 Z2 abelian, A4 /Ghas order 12/4 = 3 今 abelian, G/{1} = G abelian since |G| = 4.
Group A4 consists of all even permutations of {1, 2, 3, 4}. Look at disjoint
transpositions τ1 τ2 E A4 . All such τ1 τ2 are
(12)(34), (13)(24), (14)(23).
Denote by a,b,c. Easy: a2 = b2 = c2 = 1. Longer computations:
ab = ba = c, ac = ca = b, bc = cb = a.
Hence G < A4 and G = Z2 根 Z2 , the Klein group!
Need normality. But actually G a S4 , not just A4 . Indeed, Aσ E S4 and any disjoint τ1 τ2
στ1 τ2 σ __ 1 = ╱στ1 σ __ 1、╱ στ2 σ __ 1、.
For any τ = (ij) the Sn-calculus formula gives
╱σ(ij)σ __ 1 、= ╱σ(i)σ(j)、.
Therefore
στ1 τ2 σ __ 1 = ˜(τ)1 ˜(τ)2
with ˜(τ)1 , ˜(τ)2 disjoint because τ1 , τ2 are. Hence στ1 τ2 σ __ 1 is some element of
G. □
2. You need the full discriminant problem from HW4 for this one. You may use ∆ = a2 b2 − 4b3 − 4a3 c + 18abc − 27c2 for the discriminant of x3 + ax2 + bx + c. Let f(x) E F [x] be irreducible over F of degree 3, F be a number field. Prove that GalF (f) = S3 or A3 .
Compute Galois group over Q of the following polynomials. You can using any method/short-cut you prefer.
(a) x3 + 2x + 5
(b) x3 一 1
(c) x3 - 5x2 - 5x + 10
3. Let extension E/F be Galois. Prove that Gal(E/F) is cyclic if and only ifF = E φ (fixed field of φ) for some φ e Gal(E/F).
Solution. Assume G = Gal(E/F) is cyclic with the generator φ, |φ| = n.
Then F = EG , and x e F 今 φj (x) = x Aj = 1, . . . , n. But
φj (x) = x Aj = 1, . . . , n 今 φ(x) = x.
Indeed, 今 is trivial. To prove — just observe φ(x) = x 今 φ2 (x) = φ(φ(x)) = φ(x) = x, and soon.
Assume F = E φ . Take subgroup〈φ) < G generated by φ . We have
{1} <〈φ) < Gal(E/F)
and the corresponding
E >〈φ)# > F,
where〈φ)# = E (φ; is the fixed field. By the Fundamental Theorem
┌E (φ; : F┐ = [Gal(E/F) :〈φ)].
Since Eφ = F conclude Gal(E/F) =〈φ) . □
2023-06-15