Algebra Qualifying Examination August 5, 2024

Instructions:

  • There are 8 problems worth a total of 100 points. Individual point values are listed next to each problem number.
  • Credit awarded for your answers will be based on the correctness of your answers, as well as the clarity and main steps of your reasoning. "Rough working" will not receive credit: answers must be legible and written in a structured and understandable manner. Do scratch work on a separate page.
  • Read all problems first. Make sure that you understand them, and feel free to ask clarifying questions. Do not interpret a problem in a way that makes it trivial.
  • You may use a calculator to check your computations (but it may not be used as a step in your reasoning).
Notation: Throughout, let Z Z Z\mathbb{Z}Z denote the ring of integers; let Q , R Q , R Q,R\mathbb{Q}, \mathbb{R}Q,R, and C C C\mathbb{C}C denote the fields of rational, real, and complex numbers respectively. For a set S S SSS, we let | S | | S | |S||S||S| denote its cardinality. All non-zero rings are assumed to have a multiplicative identity 1 ≠ 0 1 ≠ 0 1!=01 \neq 01≠0. For a Galois extension of fields L / K L / K L//KL / KL/K, let Gal ( L / K ) Gal ( L / K ) Gal(L//K)\operatorname{Gal}(L / K)Gal(L/K) denote its Galois group.
  1. [12 points] Let G G GGG be a finite group, and let N ⊆ G N ⊆ G N sube GN \subseteq GN⊆G be a normal subgroup.
    (a) Let Aut ( N ) Aut ( N ) Aut(N)\operatorname{Aut}(N)Aut(N) denote the automorphism group of N N NNN. Show that there is a group homomorphism ϕ : G → Aut ( N ) ϕ : G → Aut ( N ) phi:G rarr Aut(N)\phi: G \rightarrow \operatorname{Aut}(N)ϕ:G→Aut(N) whose kernel is the centralizer of N N NNN in G G GGG.
    (b) Suppose | N | = p | N | = p |N|=p|N|=p|N|=p, where p p ppp is the smallest prime divisor of | G | | G | |G||G||G|. Prove that N N NNN is contained in the center of G G GGG.
  2. [12 points] Let G G GGG be a group of order 5 ⋅ 13 ⋅ 23 ⋅ 43 5 ⋅ 13 ⋅ 23 ⋅ 43 5*13*23*435 \cdot 13 \cdot 23 \cdot 435⋅13⋅23⋅43. How many elements of order 5 are contained in G G GGG ?
  3. [12 points] Let R R RRR be a commutative ring that satisfies the descending chain condition. That is, for any descending chain of ideals, I 1 ⊇ I 2 ⊇ ⋯ ⊇ I k ⊇ ⋯ I 1 ⊇ I 2 ⊇ ⋯ ⊇ I k ⊇ ⋯ I_(1)supeI_(2)supe cdots supeI_(k)supe cdotsI_{1} \supseteq I_{2} \supseteq \cdots \supseteq I_{k} \supseteq \cdotsI1⊇I2⊇⋯⊇Ik⊇⋯, there must exist t ≥ 1 t ≥ 1 t >= 1t \geq 1t≥1 so that I k = I t I k = I t I_(k)=I_(t)I_{k}=I_{t}Ik=It for k ≥ t k ≥ t k >= tk \geq tk≥t.
    (a) Let x ∈ R x ∈ R x in Rx \in Rx∈R. Show that there exists t ≥ 1 t ≥ 1 t >= 1t \geq 1t≥1 and r ∈ R r ∈ R r in Rr \in Rr∈R so that x t = r x t + 1 x t = r x t + 1 x^(t)=rx^(t+1)x^{t}=r x^{t+1}xt=rxt+1.
    (b) For a prime ideal P P PPP of R R RRR, use part (a) to prove that R / P R / P R//PR / PR/P is a field.
  4. [12 points] We make C 3 C 3 C^(3)\mathbb{C}^{3}C3 into a C [ x ] C [ x ] C[x]\mathbb{C}[x]C[x]-module by setting f ( x ) ⋅ v = f ( A ) v f ( x ) ⋅ v = f ( A ) v f(x)*v=f(A)vf(x) \cdot v=f(A) vf(x)⋅v=f(A)v, where
A = ( 3 2 0 0 3 0 0 2 5 ) A = 3      2      0 0      3      0 0      2      5 A=([3,2,0],[0,3,0],[0,2,5])A=\left(\begin{array}{lll} 3 & 2 & 0 \\ 0 & 3 & 0 \\ 0 & 2 & 5 \end{array}\right)A=(320030025)
Find polynomials p i ( x ) ∈ C [ x ] p i ( x ) ∈ C [ x ] p_(i)(x)inC[x]p_{i}(x) \in \mathbb{C}[x]pi(x)∈C[x] and exponents e i e i e_(i)e_{i}ei such that C 3 ≅ ⨁ i C [ x ] / ( p i ( x ) e i ) C 3 ≅ ⨁ i   C [ x ] / p i ( x ) e i C^(3)~=bigoplus_(i)C[x]//(p_(i)(x)^(e_(i)))\mathbb{C}^{3} \cong \bigoplus_{i} \mathbb{C}[x] /\left(p_{i}(x)^{e_{i}}\right)C3≅⨁iC[x]/(pi(x)ei) as C [ x ] C [ x ] C[x]\mathbb{C}[x]C[x]-modules. Justify your answer.
5. [12 points] Let R R RRR be a ring with 1 ≠ 0 1 ≠ 0 1!=01 \neq 01≠0. Let M M MMM be a left R R RRR-module, and N ⊆ M N ⊆ M N sube MN \subseteq MN⊆M a submodule. Suppose that M M MMM and M / N M / N M//NM / NM/N are projective R R RRR-modules. Prove that N N NNN is also projective.
6. [14 points] Let K K KKK be a field, and let R = K [ x , y ] R = K [ x , y ] R=K[x,y]R=K[x, y]R=K[x,y] be the polynomial ring in variables x x xxx and y y yyy over K K KKK. Let M = ( x , y ) ⊆ R M = ( x , y ) ⊆ R M=(x,y)sube RM=(x, y) \subseteq RM=(x,y)⊆R (i.e., M = R x + R y M = R x + R y M=Rx+RyM=R x+R yM=Rx+Ry, the ideal generated by x x xxx and y y yyy ).
(a) We make K K KKK into an R R RRR-module by setting f ( x , y ) ⋅ α = f ( 0 , 0 ) α f ( x , y ) ⋅ α = f ( 0 , 0 ) α f(x,y)*alpha=f(0,0)alphaf(x, y) \cdot \alpha=f(0,0) \alphaf(x,y)⋅α=f(0,0)α, for f ∈ R f ∈ R f in Rf \in Rf∈R and α ∈ K α ∈ K alpha in K\alpha \in Kα∈K. Show that there is an R R RRR-module homomorphism ϕ : M ⊗ R M → K ϕ : M ⊗ R M → K phi:Mox_(R)M rarr K\phi: M \otimes_{R} M \rightarrow Kϕ:M⊗RM→K such that on pure tensors
ϕ ( f ⊗ g ) = ∂ f ∂ x ( 0 , 0 ) ⋅ ∂ g ∂ y ( 0 , 0 ) ϕ ( f ⊗ g ) = ∂ f ∂ x ( 0 , 0 ) ⋅ ∂ g ∂ y ( 0 , 0 ) phi(f ox g)=(del f)/(del x)(0,0)*(del g)/(del y)(0,0)\phi(f \otimes g)=\frac{\partial f}{\partial x}(0,0) \cdot \frac{\partial g}{\partial y}(0,0)ϕ(f⊗g)=∂f∂x(0,0)⋅∂g∂y(0,0)
Here partial derivatives are defined formally in the usual way.
(b) Show that x ⊗ y ≠ y ⊗ x x ⊗ y ≠ y ⊗ x x ox y!=y ox xx \otimes y \neq y \otimes xx⊗y≠y⊗x in M ⊗ R M M ⊗ R M Mox_(R)MM \otimes_{R} MM⊗RM.
(c) Show that x ⊗ y − y ⊗ x x ⊗ y − y ⊗ x x ox y-y ox xx \otimes y-y \otimes xx⊗y−y⊗x is non-zero and torsion in M ⊗ R M M ⊗ R M Mox_(R)MM \otimes_{R} MM⊗RM. That is, there exists r ∈ R , r ≠ 0 r ∈ R , r ≠ 0 r in R,r!=0r \in R, r \neq 0r∈R,r≠0, such that r ( x ⊗ y − y ⊗ x ) = 0 r ( x ⊗ y − y ⊗ x ) = 0 r(x ox y-y ox x)=0r(x \otimes y-y \otimes x)=0r(x⊗y−y⊗x)=0.
7. [14 points] Let f = x 4 − 2 ∈ Q [ x ] f = x 4 − 2 ∈ Q [ x ] f=x^(4)-2inQ[x]f=x^{4}-2 \in \mathbb{Q}[x]f=x4−2∈Q[x], and let E ⊆ C E ⊆ C E subeCE \subseteq \mathbb{C}E⊆C be the splitting field of f f fff.
(a) Show that E = Q ( 2 4 , i ) E = Q ( 2 4 , i ) E=Q(root(4)(2),i)E=\mathbb{Q}(\sqrt[4]{2}, i)E=Q(24,i) and determine [ E : Q ] [ E : Q ] [E:Q][E: \mathbb{Q}][E:Q].
(b) Show that there is an automorphism σ : E → E σ : E → E sigma:E rarr E\sigma: E \rightarrow Eσ:E→E such that σ ( 2 4 ) = i 2 4 σ ( 2 4 ) = i 2 4 sigma(root(4)(2))=iroot(4)(2)\sigma(\sqrt[4]{2})=i \sqrt[4]{2}σ(24)=i24 and σ ( i ) = i σ ( i ) = i sigma(i)=i\sigma(i)=iσ(i)=i.
(c) Let τ : E → E τ : E → E tau:E rarr E\tau: E \rightarrow Eτ:E→E be the restriction of complex conjugation. As elements of Gal ( E / Q ) Gal ( E / Q ) Gal(E//Q)\operatorname{Gal}(E / \mathbb{Q})Gal(E/Q), what are the orders of τ , σ τ , σ tau,sigma\tau, \sigmaτ,σ, and σ 2 τ σ 2 τ sigma^(2)tau\sigma^{2} \tauσ2τ ?
(d) Let H = ⟨ σ 2 τ ⟩ ⊆ Gal ( E / Q ) H = σ 2 τ ⊆ Gal ( E / Q ) H=(:sigma^(2)tau:)sube Gal(E//Q)H=\left\langle\sigma^{2} \tau\right\rangle \subseteq \operatorname{Gal}(E / \mathbb{Q})H=⟨σ2τ⟩⊆Gal(E/Q). What is the fixed field of H H HHH ?
8. [12 points] Fix a field K K KKK, and let K [ t ] K [ t ] K[t]K[t]K[t] be the polynomial ring in a variable t t ttt over K K KKK. Let K ( t ) K ( t ) K(t)K(t)K(t) denote the field of fractions of K [ t ] K [ t ] K[t]K[t]K[t]. That is, K ( t ) K ( t ) K(t)K(t)K(t) is the field of rational functions over K K KKK. In a natural way, K ⊆ K ( t ) K ⊆ K ( t ) K sube K(t)K \subseteq K(t)K⊆K(t) as a subfield. Let σ : K ( t ) → K ( t ) σ : K ( t ) → K ( t ) sigma:K(t)rarr K(t)\sigma: K(t) \rightarrow K(t)σ:K(t)→K(t) be defined by setting for polynomials f , g ∈ K [ t ] , g ≠ 0 f , g ∈ K [ t ] , g ≠ 0 f,g in K[t],g!=0f, g \in K[t], g \neq 0f,g∈K[t],g≠0,
σ ( f ( t ) g ( t ) ) = f ( t + 1 ) g ( t + 1 ) σ f ( t ) g ( t ) = f ( t + 1 ) g ( t + 1 ) sigma((f(t))/(g(t)))=(f(t+1))/(g(t+1))\sigma\left(\frac{f(t)}{g(t)}\right)=\frac{f(t+1)}{g(t+1)}σ(f(t)g(t))=f(t+1)g(t+1)
(a) Show that σ σ sigma\sigmaσ is a well-defined field automorphism of K ( t ) K ( t ) K(t)K(t)K(t) that restricts to the identity on K K KKK.
(b) Depending on the characteristic of K K KKK, what is the order of σ σ sigma\sigmaσ as an element of the group Aut ( K ( t ) / K ) Aut ( K ( t ) / K ) Aut(K(t)//K)\operatorname{Aut}(K(t) / K)Aut(K(t)/K) ?
(c) Let K = F 2 K = F 2 K=F_(2)K=\mathbb{F}_{2}K=F2, the field with 2 elements. Let G = ⟨ σ ⟩ ⊆ Aut ( F 2 ( t ) / F 2 ) G = ⟨ σ ⟩ ⊆ Aut F 2 ( t ) / F 2 G=(:sigma:)sube Aut(F_(2)(t)//F_(2))G=\langle\sigma\rangle \subseteq \operatorname{Aut}\left(\mathbb{F}_{2}(t) / \mathbb{F}_{2}\right)G=⟨σ⟩⊆Aut(F2(t)/F2), and let E ⊆ F 2 ( t ) E ⊆ F 2 ( t ) E subeF_(2)(t)E \subseteq \mathbb{F}_{2}(t)E⊆F2(t) be the fixed field of G G GGG. Show that E = F 2 ( t 2 + t ) E = F 2 t 2 + t E=F_(2)(t^(2)+t)E=\mathbb{F}_{2}\left(t^{2}+t\right)E=F2(t2+t). (Hint: What is [ F 2 ( t ) : F 2 ( t 2 + t ) ] ? ) F 2 ( t ) : F 2 t 2 + t ? {:[F_(2)(t):F_(2)(t^(2)+t)]?)\left.\left[\mathbb{F}_{2}(t): \mathbb{F}_{2}\left(t^{2}+t\right)\right] ?\right)[F2(t):F2(t2+t)]?)