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Galois Theory
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Exam study book Galois' Theory of Algebraic Equations of Jean-Pierre Tignol - ISBN: 9789810245412 (Galois Theory)
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Summary Galois' Theory of Algebraic Equations - Mathematics
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SOLUTION MANUAL For Galois Theory 5th Edition byIan Stewart
Degrees and simple extensions - ANSWER: Every finite degree extension of Q is
simple.
Finite degree extensions cannot contain transcendental elements.
Splitting field - ANSWER: Given a field K, non-zero f(x) in K[x], a field extension L/K is
called a splitting field for f(x) over K if:
i) f(x) = c(x-θ1)...(x-θn), θi in L, c a unit in K
ii) L=K(θ1,...,θn)
Splitting fields and roots - ANSWER: A splitting field of f(x) over K contains all the
roots of f(x) and is no bigger than it needs to be for this to happen
They are unique up to isomorphism
Splitting field of x^3-2 over Q(cube root 2) - ANSWER: Q(cube root 2, w) where w = (-
1+sqrt(-3))/2
Degree of splitting field - ANSWER: Given a field K and non-zero f(x) in K[x],
degf(x)=n, then there is a splitting field L of f(x) over K with [L:K]<=n!
phi*f(x) - ANSWER: Given a field homomorphism phi:K--<K' and f(x)=sum(aix^i),
phi*f(x)=sum(phi(ai)x^i)
Unique K(θ) isomorphism - ANSWER: Given a field isomorphism sigma:K-->K' and a
simple extension K(θ)/K where θ has minimal polynomial m(x) in K[x] and θ' a root of
sigma*m(x) in K[x].
There exists a unique isomorphism tau:K(θ)-->K'(θ') where tau(θ)=θ' and
tau(a)=sigma(a) for a in K.
Minimal polynomial of θ' over K - ANSWER: m(x), the same as θ over K
Splitting field isomorphism - ANSWER: Given an isomorphism sigma:K-->K' and non-
zero f(x) in K[x], let L be a splitting field of f(x) over K and let L' be a splitting field of
sigmaxf(x) over K'.
Then there is an isomorphism tau:L-->L' where tau(a)=sigma(a) for a in K.
Uniqueness of splitting fields - ANSWER: Setting K=K' and sigma=id implies that
splitting fields are unique.
Extending sigma to tau - ANSWER: An isomorphism K--K' can be extended to L-->L' in
at most [L:K] ways
, Normal field extension - ANSWER: A field extension L/K is normal if whenever f(x) in
K[x] is irreducible and L contains one root of f(x), L contains all roots of f(x).
In particular, f(x) splits completely as a product of linear factors in L[x].
Normal field extensions and minimal polynomials - ANSWER: If θ is in L, the minimal
polynomial of θ over K splits in K[x], ie it can be written as a product of linear factors
in L[x] and m(x) has all its roots in L
Normal field extensions: deg1 - ANSWER: If [L:K]=1, then L=K so L/K is normal
Normal field extensions: deg2 - ANSWER: If [L:K]=2, L/K is normal
Normality criterion - ANSWER: L/K is finite and normal if and only if L is the splitting
field of some non-zero f(x) in K[x] over K.
Normality of Q(sqrt(2), sqrt(3), sqrt(5)) - ANSWER: Normal as it's a splitting field of
(x^2-2)(x^2-3)(x^2-5)
Field extensions that are not normal - ANSWER: Q(cube root 2)/Q is not normal but
Q(cube root 2, w)/Q is
Q(fourth root 2)/Q is not normal but Q(fourth root 2, i)/Q is
Normality of Fp(θ)/Fp(t) where t is indeterminate, θ^p=t - ANSWER: Normal as
splitting field of f(x)=x^p-t=x^p-θ^p=(x-θ)^p
L(θ)=L - ANSWER: L(θ)=L when θ is in L
Adjoining elements to non-normal extensions - ANSWER: Elements can be adjoined
to make non-normal extensions normal
Normal closure - ANSWER: A field N is a normal closure of L/K if K c L c N for N/K
normal and if K c L c N' c N with N/K normal then N=N'
Normal closures of finite extensions - ANSWER: A finite extension L/K has a normal
closure unique up to K-isomorphism.
Normality of F3(cube root 2)/F3 - ANSWER: Normal as (x^3-2)=(x-cube root 2)^3
K-automorphism - ANSWER: For a field extension L/K, a K-automorphism of L is a
field isomorphism sigma:L-->L with sigma |K=idK
Ie sigma(a)=a, sigma(a+b)=sigma(a)+sigma(b), sigma(1)=1
Sigma is a bijection
Aut(L/K) - ANSWER: Group of K-automorphisms of L/K with composition as the group
operation
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