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Galois theory Exam Questions and Answers 100% Pass

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  • Galois theory
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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]...

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  • May 5, 2024
  • 14
  • 2023/2024
  • Exam (elaborations)
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  • Galois theory
  • Galois theory
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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[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), . . . , x σ(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 AutRi ngs(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)

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