Galois theory Exam Questions and Answers 100% Pass

Galois theory Exam Questions and Answers 100% Pass 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[i1i2···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[ij](xi −xj)^2 ∈ Z[x1,...,xn]^Sn; (ii) δ(x1, ..., xn) := Π[ij](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 an