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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
Group action - ANSWER: Let S be a set and let G be a group. Write Aut[Sets](S) for
the group of bijective maps a : S → S (where the group law is given by the
composition of maps). An action of G on S is a group homomorphism φ : G →
Aut[Sets](S)
S^G - ANSWER: S^G := {s ∈ S : γ(s) = s ∀γ ∈ G} (set of invariants of S under the
action of G)
Orbits of s under G - ANSWER: Orb(G, s) := {γ(s) : γ ∈ G}
Stabiliser of s - ANSWER: Stab(G, s) := {γ ∈ G : γ(s) = s}
Action compatible with ring structure - ANSWER: We shall say that the action of G on
R is compatible with the ring structure of R, or that G acts on the ring R, if the image
of φ lies in the subgroup Aut[Rings](R) ⊆ Aut[Sets](R) of Aut[Sets](R)
Properties of the set of invariants for rings - ANSWER: Let G act on the ring R. (i) R^G
is a subring of R. (ii) If R is a field, then R^G is a field.
Symmetric polynomial - ANSWER: A symmetric polynomial with coefficients in R is an
element of R[x1, . . . , xn]^Sn
Elementary symmetric function - ANSWER: For any k ∈ {1, ..., n}, the polynomial k
sk := Σ[i1<i2<···<ik]Π[j=1 to k]xij ∈ Z[x1,...,xn]
is symmetric. It is called the k-th elementary symmetric function (in n variables)
Fundamental theorem of the theory of symmetric functions - ANSWER:
R[x1,...,xn]^Sn = R[s1,...,sn]. More precisely: Let φ : R[x1,...,xn] → R[x1,...,xn] be the
map of rings, which sends xk to sk and which sends constant polynomials to
themselves. Then (i) the ring R[x1,...,xn]^Sn is the image of φ; (ii) φ is injective.
Some useful polynomials - ANSWER: (i) ∆(x1, ..., xn) := Π[i<j](xi −xj)^2 ∈
Z[x1,...,xn]^Sn; (ii) δ(x1, ..., xn) := Π[i<j](xi −xj) ∈ Z[x1, ..., xn]^An; (iii) If σ ∈ Sn, then
δ(xσ(1), . . . , xσ(n)) = sign(σ)·δ(x1, ..., xn)
Gauss's content function - ANSWER: For r∈Q s.t. |r| = p1^m1 . . . p1^mk , where m
∈ Z. We define ordp(r) := mi if p=pi and 0 otherwise. For f(x)=∑cₙxⁿ, we define
c(f)=Πp^min{ordp(ci)} i.e. product of p to their smallest power s.t. they feature in the
prime factorisations of all the coefficients
, Field extension - ANSWER: Let K be a field. A field extension of K, or K-extension, is
an injection K → M of fields. This injection endows M with the structure of a K-vector
space. Alternate notation: M − K, M|K, M : K. We shall mostly use the notation M|K.
Maps between extensions - ANSWER: A map from the K-extension M|K to the K-
extension M′|K is a ring map M → M′ (which is necessarily injective), which is
compatible with the injections K → M and K → M′.
Automorphisms of extensions - ANSWER: If M|K is a field extension, we shall write
AutK(M) for the group of bijective maps of K-extensions from M to M (where the
group law is the composition of maps). In other words, the group AutK(M) is the
subgroup of AutRings(M), consisting of ring automorphisms, which are compatible
with the K-extension structure of M.
Degree of the extension M|K - ANSWER: We shall write [M : K] for dimK(M)
Annihilator - ANSWER: Let M|K be a field extension and let a ∈ M. We define Ann(a)
:= {P(x) ∈ K[x]|P(a) = 0}. The set Ann(a) ⊆ K[x] is called the annihilator of x. It is an
ideal of K[x]
Transcendental element - ANSWER: We say that a is transcendental over K if Ann(a)
= (0)
Algebraic element - ANSWER: We say that a is algebraic over K if Ann(a) ≠ (0)
Minimal polynomial - ANSWER: If a is algebraic over K, then the minimal polynomial
ma is by definition the unique monic polynomial, which generates Ann(a)
Algebraic extensions - ANSWER: We say that a field extension M|K is algebraic if for
all m ∈ M, the element m is algebraic over K.
Transcendental extensions - ANSWER: We say that a field extension M|K is
transcendental if it is not algebraic over K.
Finite extensions transcendental vs algebraic - ANSWER: If M|K is finite, then M|K is
algebraic
Formal derivative - ANSWER: Let K be a field. Let P(x) ∈ K[x]. Suppose that P(x) =
adx^d +a[d−1]x^(d−1) + ··· + a0. We define P′(x)= d/dxP(x) := d*ad*x^(d−1) +
(d−1)*a[d−1]x^(d−2) + ··· + a1
Multiple roots - ANSWER: We say that P(x) has no multiple roots if (P(x),P′(x)) = (1).
Otherwise, we say that P(x) has multiple roots. Equivalently, P(x) has multiple roots
iff gcd(P(x), P'(x)) ≠ 1
Multiple roots terminology origins - ANSWER: If P(x) = (x − ρ1 )(x − ρ2 ) · · · (x − ρd )
then P(x) has multiple roots iff there are i,j ∈ {1,...,d} such that i ≠ j and ρi = ρj.
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