Algebra 111C Assignment 5
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Algebra 111C
Assignment 5
SOLUTIONS TO SELECTED PROBLEMS
1. Let p be prime.
Prove that if G < Sp contains a cycle of order p anda transposition then
G = Sp.
Let f (x) e Q[x] be irreducible over Q, degf (x) = p, having exactly two roots α1 ,2 with Imα1 ,2 0. Use the previous claim to prove Gal。f = Sp.
Hint: for both parts follow the steps for x5 - px + p we did in class. Use the Cauchy theorem about abstract group G with p │ │G│ to reduce the second part to the first.
Solution.
Part 1 . Use the theorem (have proved in class, see the notes): p prime, G < Sp acts transitively, G 3 a transposition ÷ G = Sp.
Just need to verify G in the problem acts transitively. Element c e Sp is ap-cycle means c = (1 . . . p) after a possible renumbering. Then Vi,j can find k such that ck (i) = j. Say, cp−1(1) = p.
Part 2. For number fields irreducible ÷ separable. Thus f has distinct
roots ρ1 , . . . , ρp−2 e 1 ,2. Coefficients off are rl ÷ α 1 ,2 = a土ib,
Always know one element φ e Aut(C/R) < Gal。(f), namely φ(z) = ¯(z)
complex conjugation. It is identified with transposition╱(p __ 1) p、e G < Sp.
Existence of ap-cycle. Since f irreducible conclude f (x) = mρ1 ,。(x) and [Sf : Q] = p, where Sf is the splitting field over Q of f (x). Extension Sf /Q is Galois, hence lGal。(f)l = [Sf : Q] = p. Cauchy theorem says lGl = p prime ÷ G has element of order p. Let c e G < Sp has order lcl = p. Then c is a cycle of length p. Indeed, write c = c1 . . . cN , the
unique disjoint cycles decomposition, ci id. Know that c1(p) = id. If
length of c1 is k then lc1 l = k and then k│p. Thus k = p and c1 contains
all elements 1, . . . , p. But then c = c1 , N = 1. □
2. Suppose f, g e F [x]. Prove that GalF (fg) H , where
H < GalF (f) x GalF (g).
You may assume F isa number field.
Solution. All fields are number fields, so (non-)separability off, g isirrel- evant.
Let α1 , . . . , αn be roots off , β1 , . . . , βm be roots of g ,
E = F (α1 , . . . , αn , β1 , . . . , βm ),
E1 = F (α1 , . . . , αn ), E2 = F (β1 , . . . , βm ). Then
GalF (fg) = Gal(E/F), GalF (f) = Gal(E1 /F), GalF (g) = Gal(E2 /F).
Take any φ e Gal(E/F). Since coefficients of f(x) are in F conclude φ(αi ) = (some αj ). Therefore φ(E1 ) c E1 by constructive definition of E1 . Set
φ 1 = φl E1 .
Then φ1 : E1 二 E1 is a fieldisomorphism such that φ1 l F = φl F = idF , so φ 1 e Gal(E1 /F).
Define φ2 e Gal(E2 /F) similarly. Map GalF (fg) 二 GalF (f) x GalF (g) given by φ 二1 (φ1 , φ2 ) is a group homomorphism because the restriction maps φ 1二 φi are. If φ 1二 (idE1 , idE2 ) then φ(αi ) = αi , φ(βi ) = βi , and
φlF = idF . But then φ = idE by the constructive definition of E. □
3. The number
__1 +,17 + ′2(17 __ ,17) + 2 ′17 + 3,17 __ ′2(17 __ ,17)
belongs to a radical extension of Q. Write explicitly a radical tower prov- ing this statement.
2023-06-15