math 71: honors algebraclare/m71f15.pdf · math 71: honors algebra ... week 4 problem session 3....
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Math 71: Honors AlgebraSyllabus, problems and solutions
Pierre Clare
Dartmouth College, Fall 2015
1
Textbook
[DF] Abstract Algebra (3rd ed.), by S. Dummit and R. M. Foote.
Effective syllabus
O. Preliminaries - 2 lectures
O.1 Binary relationsO.2 Basic arithmetic
I. Groups - 15 lectures
I.1. GeneralitiesI.2. Examples of groupsI.3. Morphisms and subgroupsI.4. Cyclic groupsI.5. Groups presented by generators and relationsI.6. Quotient groupsI.7. The First Isomorphism TheoremI.8. Lagrange’s TheoremI.9. The alternating group Sn
I.10. Group actionsI.11. Composition series and Holder’s ProgramI.12. Sylow’s TheoremsI.13. The Fundamental Theorem of finitely generated abelian groupsI.14. Direct and semi-direct products
II. Rings - 10 lectures
II.1. GeneralitiesII.2. Properties of idealsII.3. Euclidean domainsII.4. Principal ideal domainsII.5. Unique factorization domainsII.6. Rings of fractionsII.7. Polynomial ringsII.8. Field extensions
III. Introduction to representation theory - 1 lecture
2
Contents
Content of the lectures 4
Problem sets and examinations 9Problem set 1: groups, subgroups and morphisms 9Problem set 2: cyclic groups and quotients 10Midterm 1 12Problem set 3: symmetric groups, group actions 14Midterm 2 - In-class 16Midterm 2 - Take-home 18Problem set 4: rings and ideals 20Problem set 5: polynomial rings 22Final examination 23
Elements of solutions for the homework sets and examinations 25
3
Week 1
Lecture 1. [DF, §0.1-§0.3]Equivalence relations: definition and examples. Equivalence classes, representatives. Cor-respondence between equivalence relations and partitions.Basic arithmetic vocabulary, relatively prime numbers. Congruence modulo n.
Lecture 2. [DF, §0.1-§0.3]Review of Euclidean division, the Euclidean Algorithm.Addition and multiplication are well-defined operations in Z/nZ.Characterization of invertible elements: (Z/nZ)× = {a , a ∧ n = 1}.
Lecture 3. [DF, §1.1-§1.4]Composition laws (binary operationa), associativity, commutativity.Groups: definition, first examples. Abelian groups are denoted additively.General properties: uniqueness of the identity and of the inverse, inverse of a composition.Cancellation laws, conjugation.Fields: definition, examples (Q, R, C, Z/pZ, p prime). Matrix groups: GL(2,F), SL(2, F ).
Week 2
Lecture 4. [DF, §1.2]Dihedral groups: geometric definition, enumeration (#D2n = 2n). Table of a group.Generators and relations: examples of Z/nZ and D2n = 〈r, s | s2 = 1 , rn = 1 , rs = sr−1〉.
Lecture 5. [DF, §1.3-§1.6]Symmetric groups Sn: permutations, cycles. Cycles with disjoint supports commute.Canonical decomposition: every permutation can be uniquely written as a (commuting)product of cycles with disjoint supports. Cycle Decomposition Algorithm. Application tothe determination of the order of a permutation.Group homomorphisms: definition, examples. Morphisms map identity to identity andinverse to inverse. Isomorphism. The isomorphic image of an abelian group is abelian.Isomorphisms preserve the order of elements.
Problem Session 1. A permutation has order 2 if and only if its canonical decompositiononly contains permutations. Presentation of Z/nZ.
Lecture 6. [DF, §2.1-§2.2]Subgroups: definition, examples, simple groups, the Subgroup Criterion.Subgroups of SL(2,R) isomorphic to {z ∈ C , |z| = 1}, R×+, and R.Images and kernels of homomorphims are subgroups (generalizing the case of linear mapsbetween vector spaces). Characterization of injectivity via the kernel.
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Week 3
Lecture 7. [DF, §2.3]Cyclic groups: classification by the cardinality, characterization of generators, subgroups.
Lecture 8. [DF, §2.4 - §3.1]Group 〈A〉 generated by a subset A of a group G: definition as minimal subgroup contain-ing A, characterisation by the intersections of all subgroups containing A, general form ofthe elements.Equivalence relation ∼H on G, with H < G. Left cosets: gH is the class of g ∈ G forthis relation. Composition of cosets g1H ? g2H = g1 · g2H is well defined if and only ifgH = Hg for all g ∈ G. Normal subgroups.
Problem Session 2. A subgroup H of Q such that x ∈ H ⇒ 1x∈ H must be {0} or Q.
Lecture 9. [DF, §3.1 - §3.2 - §3.3]Characterizations of normal subgroups, quotients. Normal subgroups are exactly kernelsof homomorphisms. The First Isomorphism Theorem. Application: R/2πZ ' T.
Week 4
Problem Session 3. Subgroups of µn = {z ∈ C , zn = 1}. Subgroups of D2n generatedby various subsets. Study of GL(n,R)/SL(n,R).
Lecture 10. [DF, §3.2]Lagrange’s Theorem and corollaries: the order of an element divides the order of the group,a group with prime order is cyclic. The index of a subgroup. Finite case: [G : H] = #G
#H,
examples. A subgroup with index 2 is normal. There is no general converse for Lagrange’sTheorem: the tetrahedron group, of order 12, as no subgroup of order 6.
Midterm 1. Left cosets; S3 is not cyclic; order of a permutation; non-isomorphic groupsof order 8; homomorphic image of a cyclic group; multiplicative group generated by theinverses of primes. A finite group generated by two elements of order 2 is a dihedral group.Examples of quotients. Inn(G) ' G/Z(G).
Lecture 11. [DF, §3.5]Cycles and transpositions generate Sn; there is no uniqueness in the way a given permu-tation decomposes into transpositions. Action of σ ∈ Sn on ∆ =
∏1≤i<j≤n
(Xi −Xj). Defi-
nition of the signature: σ ·∆ = ε(σ)∆. This defines a homomorphism ε : Sn −→ {−1, 1},whose kernel is called the alternating group An with cardinality n!
2. Transpositions have
signature −1, notion of even and odd transposition.5
Week 5
Lecture 12. [DF, §1.7,§4.2]Action of a group on a set: definition and examples. The kernel of G y X is the kernelof the homomorphism g 7→ σg, where σg(x) = g · x; faithful actions.The stabilizer StabG(x) of an element x ∈ X is a subgroup of G with index the cardinalityof the orbit of x. Transitive actions.
Lecture 13. [DF, §4.2]Any group acts on itself by left multiplication. Example of Klein’s group V4.More generally, if H < G, the action Gy G/H given by left multiplication is transitive.The stabilizer of 1GH is H, the kernel is N =
⋂g∈G gHg
−1 and N is the largest normalsubgroup of G contained in H.Corollary (Cayley): every finite group is isomorphic to a subgroup of Sn for some n.
Lecture 14. [DF, §4.3]Equivariant maps and isomorphisms in the category of G-sets. Centralizers and normal-izers, action by conjugation. Examples: similar matrices, equivalences classes and rank.The Class Equation and applications: a group of order pα with p prime has non-trivialcenter. A group of order p2 with p prime is abelian, isomorphic to Z/p2Z or Z/pZ×Z/pZ.
Week 6
Lecture 15. [DF, §3.4]Cauchy’s Theorem: if G is a finite group and p prime divides #G, then G containsan element (hence a subgroup) of order p. Proof in the abelian case. Simple groups,composition series, Jordan-Holder Theorem. Holder’s classification program.Feit-Thompson’s Theorem: simple groups of odd order are cyclic.
Lecture 16. [DF, §4.5, §5.2, §5.4]Maximal p-subgroups: Sylow’s theorems. Structure of finitely generated abelian groups.Structure of compactly generated abelian groups. Characterization of direct products.Structure of the group H1H2 when H1 �G, H2 < G and H1 ∩H2 = {1}.
Lecture 17. [DF, §5.5]Semi-direct products: general case. Application: if p and q are prime numbers andp|(q − 1), there exists a non-abelian group of order pq.
Lecture 18. [DF, §7.1]Rings: definition, basic examples. Fields, division rings, example of the quaternions H.Rings of functions with values in a ring. Zero divisors and units.
Week 7
Lecture 19. [DF, §7.1, §7.2]Integral domains. Cancellation laws. A finite integral domain with identity is a field.Wedderburn’s Theorem: a finite division ring is abelian, hence a field.Subgrings and ring morphisms. The kernel of a ring homomorphism is absorbent.
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Ideals. If I is an ideal in A, the quotient group A/I is a ring for the multiplication(a+ I)(b+ I) = ab+ I. Natural projection A −→ A/I, First Isomorphism Theorem.
Lecture 20. [DF, §7.1, §7.4]Finitely supported sequences in a ring: polynomials.Ideal generated by a subset: definition, characterization, commutative case.Principal ideals. b ∈ (a)⇔ a|b⇔ (b) ⊂ (a). If I is an ideal in A, then I = A if and onlyif I contains a unit. A commutative ring with unit is a field if and only if has no otherideals than {0} and A.
Lecture 21[DF, §7.4]Maximal ideals; I is maximal if and only if A/I is a field. Examples: (X) is not maximalin Z[X], the ideal generated by 2 and X is. Prime ideals in a commutative ring.
Week 8
Lecture 22. [DF, §8.1]Euclidean domains, division algorithm. Examples: Z, R[X], fields. In a Euclidean domain,every ideal is principal. Example: Z[X] carries no Euclidean division.Multiples, divisors, notion of g.c.d. Uniqueness up to a unit. The division algorithmallows to compute a g.c.d., Bezout relation.More about polynomial rings: if A is a commutative ring with identity and I is an ideal,then (I) = I[X] and A[X]/I[X] ' A/I[X]. Polynomials in several variables.
Lecture 23. [DF, §8.2]A principal ideal domain (PID) is an integral domain in which every ideal is principal.
Euclidean rings are PIDs, but Z[
1+i√
192
]is a PID that is not Euclidean. For a and b, in
a PID, any generator of (a, b) is a g.c.d. It is unique up to multiplication by a unit andsatisfies a Bezout relation. In a PID, every prime ideal is maximal.Corollary: if A is a commutative ring, then A[X] is a PID ⇒ A is a field. Conversely, ifA is a field, then A[X] is a Euclidean domain hence a PID.
Fun: group rings and convolution.
Lecture 24. [DF, §8.3]Notice that in Z, an alternate way to compute the g.c.d. of two elements is to comparetheir prime factors decompositions. Irreducibles and primes in an integral domain, asso-ciate elements. General fact: prime ⇒ irreducible. The converse does not hold in Z[i
√5]
as 3 is irreducible but not prime: 32 = (2+ i√
5)(2− i√
5). In a PID, prime⇔ irreducible.Unique factorization domains (UFD): definition, examples. In a UFD, prime ⇔ irre-ducible. Computation of g.c.d.’s in a UFD.
Week 9
Lecture 25. [DF, §8.3, §7.5]Every principal ideal domain has the unique factorization property.Rings of fractions: general construction and universal property.Fields of fractions: examples of Q and F (X).
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Problem session 4. Quotients of A[X], nilpotent elements.
Lecture 26. [DF, §9.3]Given a ring A with field of fractions F can results obtained in F [X] be used in A[X]?Gauss’ Lemma: if A is a UFD and P is reducible in F [X], then it is reducible in A[X].Corollary: if A is a UFD and the coefficients of P have g.c.d. 1, then P is irreducible inF [X] if and only if it is irreducible in A[X]. Transfer theorem: A UFD ⇔ A[X] UFD.Corollary: if A is a UFD, so is A[X1, . . . , Xn].
Lecture 27. [DF, §13.1]Two constructions of C: as a subring of M2(R) and as R[X]/(X2 + 1).Fields extensions, degree. If F is a field and P ∈ F [X] is irreducible, then F [X]/(P ) is anextension of degree deg(P ) of F in which P has a root. Conversely, if K is an extensionof F in which P has a root α, then F (α) ' F [X]/(P ).
Week 10
Lecture 28. Introduction to representation theory.Fun: definitions and examples of representations (permutations, left regular).Irreducibles, Schur’s Lemma. Fourier analysis on abelian groups, projective representa-tions of SO(3), representations of SU(2) and spin of particles.
- End of the course -
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Problem set 1: groups, subgroups and morphisms
Solution p.25
(1) Let E be the set of matrices of the form
[a 0b 0
]with a ∈ C× and b ∈ C.
(a) Prove that matrix multiplication is an associative composition law on E.
(b) Does E have a left identity, that is, an element e such that e ·x = x for everyx ∈ E? Is it unique? What about right identities?
(c) Does every element in E have a left inverse? A right inverse? Is it unique?
(2) Let G =
{[a bc d
];a, b, c, d ∈ Zad− bc = 1
}and α =
[0 −11 0
], β =
[0 1−1 1
].
(a) Verify that G is a subgroup of GL(2,R) that contains α and β.
(b) Determine the order of α, β and αβ.
(3) All the sets considered below are equipped with their ordinary additive groupstructures; m and n are positive integers.
(a) Determine all the elements of Hom(Q,Q), Hom(Q,Z) and Hom(Z/nZ,Z)
(b) Show that Γ = Hom(Z/mZ,Z/nZ) is a group for a law to be determined.
(c) Prove that Γ is isomorphic to Z/(m ∧ n)Z.
(4) If A and B are subsets of a group G, we denote by AB the set of products{ab , a ∈ A , b ∈ B}. Let H1 and H2 be subgroups of G.
(a) Find a necessary and sufficient condition for H1H2 to be a subgroup of G.
(b) Assume that H1 and H2 are finite and H1 ∩H2 = {eG}. Prove that
Card(H1H2) = Card(H1) Card(H2).
Hint: construct a bijection between H1 ×H2 and H1H2.
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Problem set 2: cyclic groups and quotients
Solution p.29
(1) Recall that an automorphism of a group is an isomorphism of the group withitself. The set of automorphisms of a group G is denoted by Aut(G).Let G be a group, H a normal subgroup of G and K a subgroup of H.
(a) Assume that ϕ(K) ⊂ K for any ϕ ∈ Aut(H). Is K normal in H?
(b) Under the same hypotheses, prove that K is normal in G.
(c) Find three groups G, H and K such that K �H and H � G but K is notnormal in G.
(2) Recall that R× = R \ {0} and C× = C \ {0} and consider the multiplicativegroups
T = {z ∈ C , |z| = 1}µn = {z ∈ C , zn = 1}
µ∞ = {z ∈ C , ∃n ∈ Z , zn = 1}.
(a) Which of these are cyclic?
(b) Prove the following isomorphisms:
R/Z ' T , C×/R×+ ' T , C×/R× ' T
T/µn ' T , C×/µn ' C× , Q/Z ' µ∞.
(c) Determine all the finite subgroups of µ∞.
(3) The center of a group G is the set Z(G) = {γ ∈ G | ∀g ∈ G, γ · g = g · γ}.
(a) Compute the center of your favorite and least favorite non-abelian groups.
Let G be a group and H a subgroup such that H ⊂ Z(G).
(b) Verify that H �G.
(c) Prove that if G/H is cyclic then G is abelian.10
(4) A commutator in a group G is an element of the form xyx−1y−1 with x, y ∈ G.
(a) Do commutators form a subgroup of G?
Let D(G) be the subgroup of G generated by the commutators.
(b) Prove that D(G) is normal in G.
(c) Prove that G/D(G) is abelian.
(d) Prove that D(G) is the smallest normal subgroup of G with abelian quotient.
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Midterm 1
Solution p.33
This exam consists of 4 independent problems. Treat them in the order of your choosing,starting each problem on a new page.
Every claim you make must be fully justified or quoted as a result studied in class.
Problem 1
1. Let G be a group and H a subgroup. The relation defined on G by
x ∼ y ⇔ x−1y ∈ His an equivalence relation. For g element of G, describe the class of g.
2. Is S3 cyclic?
3. Find the order of
(1 2 3 4 5 6 7 8 9 10 11 1212 4 1 6 8 11 10 5 7 9 2 3
)in S12.
4. Give two non-isomorphic groups of cardinal 8. Explain why they are not isomorphic.
5. Let G, G′ be groups and ϕ ∈ Hom(G,G′). Assume that G is cyclic and ϕ is surjective.Prove that G′ is cyclic.
6. Determine the subgroup of (Q×+,×) generated by A ={
1p, p prime
}.
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Problem 2
Recall that D2n = 〈r, s | rn = 1 , s2 = 1 , rs = sr−1〉.
1. Show that every element of D2n that is not a power of r has order 2.
2. Deduce that every element of D2n is the product of elements of order 2.
3. Let G be a finite group generated by distinct elements a and b, both of order 2.Prove that G is isomorphic to D2n, where n = |ab|.
Hint: prove that 〈a, b〉 = 〈a, ab〉
Problem 3
1. Let G1 and G2 be groups and consider the product G = G1 ×G2, with the group law
(x1, x2) · (y1, y2) = (x1y1, x2y2).
Prove that G1 × {1G2}�G and that the quotient G/G1 × {1G2} is isomorphic to G2.
2. Let F be a field. Consider the following subgroups of SL(2, F ):
P =
{[a t0 a−1
], a ∈ F× , t ∈ F
}, N =
{[1 t0 1
], t ∈ F
}.
Prove that N � P and that P/N is isomorphic to F×.
Problem 4
Recall that Aut(G) denotes the set of isomorphisms of a group G onto itself, and that itis a group under composition. Let
Inn(G) = {cg , g ∈ G} , where cg(x) = gxg−1 for all x ∈ G.
1. Verify that cg ∈ Aut(G) for all g ∈ G.
2. Prove that Inn(G) is a subgroup of Aut(G).
3. Is Inn(G) normal in Aut(G)?
4. Prove that Inn(G) is isomorphic to a quotient of G.13
Problem set 3: symmetric groups, group actions
Solution p.36
(1) Recall that An denotes the alternating group, that is the subgroup of even per-mutations in Sn.
(a) Assume n ≥ 3. Prove that An is generated by 3-cycles.
Let p ≥ 5 be a prime number and H a subgroup of Sp such that [Sp : H] ≤ p−1.
(a) Prove that H contains the p-cycles.
(b) Prove that [Sp : H] ∈ {1, 2}.
(c) Deduce from this another concrete example showing that there is no generalconverse to Lagrange’s Theorem.
(d) Let G be a finite group, and H a subgroup. For g ∈ G, and x ∈ G/H, letσg(x) = gx.
(a) Verify that σg is a well-defined map from G/H to itself.
(b) Prove that the map σ : g 7→ σg is a group homomorphism from G toBij(G/H).
(c) Assume that [G : H] = p, where p is the smallest prime factor of #G.Prove that H is normal in G.
(2) Let G be a finite group acting on a finite set X.
(a) Assume that every orbit contains at least 2 elements, that #G = 15 andthat #X = 17. Determine the number of orbits and the cardinal of each ofthem.
(b) Assume that #G = 33 and #X = 19. Prove that the action has a fixedpoint.
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(3) Let G be a group and H a subgroup.
(a) Recall how G acts on G/H by left multiplication and determine the stabilizerof the coset gH.
If G acts on two sets X and Y , a map f from X to Y is said G-equivariant if itsatisfies
f(g · x) = g · f(x)
for every g ∈ G and x ∈ X.
(b) Prove that the inverse of a G-equivariant bijection is G-equivariant.
(c) Assume that G acts transitively on X, let x ∈ X and H = StabG(x).Construct a G-equivariant bijection between X and G/H.
(d) Let H and K be subgroups of G. Prove that there exists a G-equivariantmap from G/H to G/K if and only if H is contained in a conjugate of K,and that such a map is necessarily surjective.
(e) Prove that there exists a G-equivariant bijection between G/H and G/K ifand only if H and K are conjugate in G.
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Midterm 2 - In-class
Solution p.40
This examination consists of four independent problems. Treat them in the order ofyour choosing, starting each problem on a new page. Every claim you make must befully justified or quoted as a result studied in class.
Problem 1
The questions in this problem are independent.
1. Let G be a group acting on a finite set X and x1, . . . , xn a family of representatives of
all the orbits. Prove that #X =n∑i=1
[G : StabG(xi)].
2. Let A be a ring. Prove that 0× a = 0 for all a ∈ A.
3. Consider the ring A of functions from R to R. For each of the following subsets,determine if it is a subring A. If so, determine if it is an ideal.
(a) A−3 = {f : R −→ R , f(−3) = 0}(b) E = {even functions in A}(c) O = {odd functions in A}
Problem 2
1. Let F be a field. For a ∈ F× let ϕ(a) denote the mapF −→ Ft 7−→ a2t
.
(a) Prove that ϕ(a) is an automorphism of the additive group (F,+).
(b) Prove that ϕ is a group homomorphism from F× to Aut(F ).
2. Prove that FoϕF× is isomorphic to the subgroup P =
{[a b0 a−1
], a ∈ F× , b ∈ F
}of SL(2, F ).
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Problem 3
Let n ≥ 3. Recall that the alternating group An is generated by the 3-cycles.A square in a group G is an element of the form g2 = g · g.
1. Prove that every 3-cycle σ in Sn is a square. Hint: compute σ4.
2. Let H be the subgroup of Sn generated by the squares. Prove that H = An.
3. Let N be a subgroup of index 2 in Sn
(a) Prove that σ2 ∈ N for any σ ∈ Sn.
(b) Deduce that N = An.
Problem 4
Let K be the subgroup of SL(2,R) consisting of matrices of the form
[c −ss c
]with
c2 + s2 = 1 and
g =
{X =
[a bc d
], Tr(X) = a+ d = 0
}.
Note that g is not a subgroup of SL(2,R).
1. Verify that K acts on g by k ·X = kXk−1 where k ∈ K and X ∈ g.
Let a =
{[α 00 −α
], α ∈ R
}⊂ g. Again, a is not a subgroup of SL(2,R).
2. Determine all the elements of the subgroup N = {k ∈ K , k a k−1 ⊂ a} of K.
3. Determine the subgroup C = {k ∈ K , kXk−1 = X for all X ∈ a} of N .
4. Prove that N is isomorphic to Z/4Z and that N/C is isomorphic to Z/2Z.
17
Midterm 2 - Take-home
Solution p.43
Problem 1
The goal of this problem is to determine for which values of n there exists a uniquegroup of order n up to isomorphism. In what follows, n is a positive integer with primedecomposition n = pα1
1 · · · pαss .
1. (a) Assume n prime. Determine, up to isomorphism, all the groups of order n.
(b) Prove that if αi ≥ 2 for some i ∈ {1, . . . , s}, then there are at least two non-isomorphic groups of order n. (Think about Klein’s group V4.)
From now on, we assume that αi = 1 for all i ∈ {1, . . . , s}.
2. Recall (without proof) the expression of the Euler Indicator ϕ(n) in that case.
3. Let p and q be distinct prime numbers such that p|(q − 1).
(a) Prove the existence of a non-abelian group of order pq.
(b) Deduce that if all groups of order n are isomorphic, then n ∧ ϕ(n) = 1.
We shall prove the converse by contradiction. Let n be the smallest integer for whichn ∧ ϕ(n) = 1 and there exists a group G of order n that is not isomorphic to Z/nZ(assuming the existence of such integers).
4. (a) Prove that m ∧ ϕ(m) = 1 for any divisor m of n.
(b) Prove that every proper subgroup and every non-trivial quotient group of G iscyclic.
(c) Deduce that the center of G is trivial. (Hint: consider G/Z(G) and use the Fundamental
Theorem.)
A maximal subgroup of a group Γ is a proper subgroup H such that the only subgroupsof Γ containing H are H and Γ.
5. Let U be a maximal subgroup of G and x 6= 1 in U .
(a) Prove that U = CG(x).
(b) Deduce that any two distinct maximal subgroups of G have trivial intersection.
18
We admit the following result: every maximal subgroup of G is equal to its own normalizer:
U = NG(U).
6. Let U be a maximal subgroup, u its order and U the union of all conjugates of U in G.
(a) Determine the number of conjugates of U and the order of each such conjugate.
(b) Verify that the conjugates of U are maximal and deduce that U contains n − nu
elements different from the identity.
7. Let x ∈ G \ U. Consider V a maximal subgroup of G containing x. Denote by v itsorder and by V the union of all conjugates of V .
(a) Prove that U ∪V contains 2n− nu− n
velements different from 1.
(b) Compare to the cardinality of G \ {1} and deduce a contradiction.
8. Conclude.
Problem 2
Let F be a field and consider the groups G = SL(2, F ) and N =
{[1 t0 1
], t ∈ F
}.
Let X2 denote the set F 2 \ {(0, 0)}, with elements written as column matrices.
1. Is N is normal in G?
2. Prove that G acts on X2 via left matrix multiplication.
For g ∈ G, let c1(g) denote the first column of g.
3. Prove that the map ϕ :G/N −→ X2
gN 7−→ c1(g)is well-defined.
4. Prove that ϕ is a G-equivariant bijection.
From now on, assume n ≥ 1 and let G = SL(n + 1, F ), while Yn+1 denotes the set ofF -valued matrices with n+ 1 rows and n columns.
5. Verify that G acts on Yn+1 via left matrix multiplication.
6. Let x0 =
In
0 . . . 0
. Determine the group N = StabG(x0).
7. Describe the map b :G −→ Yn+1
g 7−→ g · x0.
8. Prove that G/N is in G-equivariant bijection with the subset Xn+1 of Yn+1 consistingof the elements of rank n.
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Problem set 4: rings and ideals
Solution p.48
(1) Let A be a commutative ring with identity and A[X] the ring of polynomials withcoefficients in A.
(a) Prove that (X) is a prime ideal of A[X] if and only if A is an integral domain.
(b) Prove that (X) is maximal if and only if A is a field.
(c) Let I and J be ideals in A and P a prime ideal such that IJ ⊂ P . Provethat either I or J is contained in P .
(2) Let A be a commutative ring with identity and consider a polynomial P ∈ A[X]of degree n ≥ 1 with leading coefficient 1.
(a) Verify that every class in A[X]/(P ) has a representative in A[X] of degreeat most n− 1.
(b) Let U, V ∈ An−1[X] be distinct polynomials. Prove that their images inA[X]/(P ) are distinct.
(c) Assume that P can be factored in A[X]. Prove that A[X]/(P ) has zerodivisors.
(d) Assume that P = Xn − a where a ∈ A is nilpotent1. Prove that the imageof X in A[X]/(P ) is nilpotent.
(3) Consider the ring Z[i] = {a+ ib , a, b ∈ Z}.
(a) Prove that every point in the complex plane is at distance strictly less than1 from an element in Z[i].
(b) Prove that Z[i] is a Euclidean domain and determine Z[i]×.
1An element a in a ring is said nilpotent if am = 0 for some positive integer m.20
(4) Let A be a ring with identity and consider the map ϕ : N −→ A defined byϕ(n) = 1 + . . .+ 1︸ ︷︷ ︸
n times
.
(a) Prove that ϕ extends uniquely to a ring homomorphism from Z to A.
(b) Prove the existence of a non-negative integer κ such that kerϕ = κZ.
The integer κ of the previous question is called the characteristic of A.
(c) Prove that the characteristic of an integral domain is either 0 or a primenumber.
(d) Determine the characteristics of Q, Z[X] and Z/nZ[X].
(e) Let p be a prime number and A a commutative ring of characteristic p. Provethat (a+ b)p = ap + bp in A2.
2Be warned: mentioning this result to calculus students may constitute a violation of the Honor Code.21
Problem set 5: polynomial rings
Solution p.51
(1) If A is a ring, denote by A[X, Y ] the ring of polynomials in two variables withcoefficients in A.
(a) Is the ideal (X, Y ) of Q[X, Y ] principal?
(b) Are the ideals (X) and (X, Y ) prime in Q[X, Y ]? Are they maximal?
(c) Are the ideals (X, Y ) and (2, X, Y ) prime in Z[X, Y ]? Are they maximal?
(2) Let F be a field and A = F [X,X2Y,X3Y 2, . . . , XnY n−1, . . .] ⊂ F [X, Y ].
(a) Prove that A consists of all polynomials of the form a0 +∑ak,`X
kY ` withak,` = 0 if 1 ≤ k ≤ `.
(b) Prove that A and F [X, Y ] have the same field of fractions.
(c) Prove that the ideal (X,X2Y,X3Y 2, . . .) of A is not finitely generated.
(3) Let A = Z +XQ[X].
(a) Prove that A is an integral domain and determine A×.
(b) Show that the irreducibles of in A are prime numbers, opposites of primenumbers and irreducible polynomials in Q[X] with constant term ±1.
(c) Are the irreducibles prime?
(d) Show that X cannot be written as the product of irreducibles. Is A a UFD?
(4) Let F be field and P a polynomial of degree n ≥ 1.
(a) Prove that F [X]/(P ) is a vector space of dimension n over F .
(b) Assume that F is finite, of order q. What is the cardinality of F [X]/(P )?
(c) Prove that F [X]/(P ) is a field if and only if P is irreducible.22
Final examination
Solution p.54
1. Cyclotomic polynomials
For n ≥ 2, let µn denote the multiplicative group of nth roots of 1: µn = {z ∈ C , zn = 1}and Πn the set of generators of µn. The cyclotomic polynomial of order n is
Φn =∏ξ∈Πn
(X − ξ).
We recall that the Euler indicator ϕ satisfies the formula∑d|nϕ(d) = n.
1. Consider the polynomial Pn =∏ξ∈µn
(X − ξ).
(a) Prove that Pn = Xn − 1.
(b) Determine Φp for p prime.
2. Let ω = e2iπn and k an integer such that 0 ≤ k ≤ n− 1.
(a) Let d be the order of ωk in µn. Prove that ωk ∈ Πd.
(b) Deduce that Xn − 1 divides∏d|n
Φd.
(c) Prove that∏d|n
Φd = Xn − 1.
3. We will prove by induction that Φn has integer coefficients.
(a) Verify the result for n = 1.
(b) Assuming the result true up to n− 1, find a monic polynomial P ∈ Z[X] such that
Xn − 1 = P Φn.
(c) Prove the existence of polynomials Q and R in Z[X] with deg(R) < deg(P ), such that
Xn − 1 = PQ+R.
(d) Prove that the couple (Q,R) is unique and conclude that Φn ∈ Z[X].
23
2. Application: proof of Wedderburn’s Theorem
We shall prove that every finite division ring is commutative. Let K be a finite divisionring. We argue by induction on the cardinality of K.
0. Prove the following result.
Lemma. If A is a finite division ring and F a subring of A that is a field, then A is afinite dimensional vector space over F .
1. Prove that a division ring of cardinality 2 is commutative.
From now on, we assume that every division ring of cardinality < #K is commutativeand that K is noncommutative.
2. Let Z = {x ∈ K | xy = yx for all y ∈ K} be the center of K and q = #Z.
(a) Prove that Z is a subring of K.
(b) Prove the existence of an integer n ≥ 2 such that #K = qn.
3. For x ∈ K, let Kx = {y ∈ K | xy = yx}.
(a) Verify that either Kx = K or Kx is a field extension of Z and a subring of K.
(b) Deduce the existence of a divisor d of n such that #Kx = qd.
4. Recall that the multiplicative group K× = K \ {0} acts on itself by conjugation.
(a) Prove that every stabilizer has a cardinality of the form qd − 1 with d a divisor of n.
(b) Using the class equation, prove the existence of integers λd such that
#K× = q − 1 +∑
d|n , d 6=n
λdqn − 1
qd − 1.
5. Assume that d|n and d 6= n.
(a) Prove that Φn dividesXn − 1
Xd − 1in Z[X].
(b) Prove that Φn divides (Xn − 1)−∑
d|n , d 6=n
λdXn − 1
Xd − 1in Z[X].
(c) Deduce that Φn(q) divides q − 1.
6. Prove that |Φn(q)| >ϕ(n)∏i=1
|q − 1| ≥ |q − 1| and conclude.
24
Problem set 1: groups, subgroups and morphisms - Elements of solution
(1) Let E be the set of matrices of the form
[a 0b 0
]with a ∈ C× and b ∈ C.
(a) Prove that matrix multiplication is an associative composition lawon E.
For (a, b) (x, y) in C× × C, observe that[a 0b 0
]·[a′ 0b′ 0
]=
[aa′ 0ba′ 0
]which is an element of E since ax 6= 0. Matrix multiplication is associativein general hence in particular when restricted to E.
(b) Does E have a left identity, that is, an element e such that e ·x = xfor every x ∈ E? Is it unique? What about right identities?
For
[a 0b 0
]in E, the equation
[x 0y 0
]·[a 0b 0
]=
[a 0b 0
]is equivalent
to
{xa = aya = b
, which cannot be satisfied for every value of (a, b) so there is
no left identity in E. On the other hand, the equation
[a 0b 0
]·[x 0y 0
]=[
a 0b 0
]is equivalent to
{ax = abx = b
hence will be satisfied if and only if
x = 1. In other words, the right identities of E are all the matrices
[1 0y 0
]with y ∈ C.
(c) Does every element in E have a left inverse? A right inverse? Isit unique?
For
[a 0b 0
]in E, the equation
[x 0y 0
]·[a 0b 0
]=
[1 0z 0
]always has
a unique solution, namely x = 1a, y = z
a. On the other hand, the equation[
a 0b 0
]·[x 0y 0
]=
[1 0z 0
]can be solved if and only if b
a= z, in which
case every matrix of the form
[1a
0y 0
]is a solution (for any choice of y ∈ C).
25
(2) Let G =
{[a bc d
];a, b, c, d ∈ Zad− bc = 1
}and α =
[0 −11 0
], β =
[0 1−1 1
].
(a) Verify that G is a subgroup of GL(2,R) that contains α and β.
The inclusion G ⊂ GL(2,R) is obvious, as is the fact that G is stable undermatrix multiplication. To see that it is stable under inverses, we need toverify that for g ∈ G with inverse g−1 in GL(2,R), the coefficients of g−1
are integers and the determinant of g−1 is 1. Both follow from the formulagiving g−1.
(b) Determine the order of α, β and αβ.
Direct computations show that |α| = 4, |β| = 6 and (αβ)n =
[1 −n0 1
]so
that αβ has infinite order (although it is the product of torsion elements ofG).
(3) All the sets considered below are equipped with their ordinary additivegroup structures; m and n are positive integers.
(a) Determine all the elements of Hom(Q,Q), Hom(Q,Z) and Hom(Z/nZ,Z).
Let ϕ ∈ Hom(Q,Q). Then the additivity of ϕ implies that ϕ(n) = nϕ(1) forany n ∈ Z. It follows that, for (a, b) ∈ Z× Z \ {0},
ϕ(a) = ϕ(b · a
b
)= b · ϕ
(ab
)hence a · ϕ(1) = b · ϕ
(ab
). Therefore, Hom(Q,Q) consists of all the maps
x 7−→ α · x, with α ∈ Q.Observe that Hom(Q,Z) ⊂ Hom(Q,Q). For a map of the form x 7−→ α · x,the condition α·x ∈ Z for all x ∈ Q implies that α = 0: there is no non-trivialhomomorphism between Q and Z.Finally, let ϕ ∈ Hom(Z/nZ,Z). Observe that the natural surjection p :Z 7−→ Z/nZ is a morphism hence ϕ◦p : x 7−→ ϕ(x) is group morphism fromZ to itself. It must therefore be of the form x 7−→ α · x. However, since ϕmust have finite range hence ϕ ◦ p too, which can only happen if α = 0. Wehave proved that Hom(Z/nZ,Z) = {0}.
(b) Show that Γ = Hom(Z/mZ,Z/nZ) is a group.
It is a general fact that functions from a set X to a group (G, ?) form agroup under the law ∗ defined by pointwise composition, that is f ∗ g : x 7→f(x) ? g(x). The identity is the constant function x 7→ eG and the inverseis defined by f−1(x) = (f(x))−1. Letting X = Z/mZ and G = Z/nZ, it iseasy to see that Γ is a subgroup.
26
(c) Prove that Γ is isomorphic to Z/(m ∧ n)Z.
For a ∈ Z, we will denote by a and a the classes of a modulo m and nrespectively, and by m′ and n′ the integers m
m∧n and nm∧n .
First observe that, since Z/mZ is generated by 1, any element ϕ of Γ isdetermined by ϕ(1) and that the map T : ϕ 7→ ϕ(1) is a morphism from Γto Z/nZ.
We will prove that the image T (Γ) is isomorphic to Z/(m ∧ n)Z.Note that m · ϕ(1) = ϕ(m) = ϕ(0) = 0.
Lemma. The elements x ∈ Z/nZ such that mx = 0 are the multiples of n′.
Proof. If x = n′a, then mx = mn′a = m′na = 0. Conversely, if mb = 0then n divides mb so n′ divides m′b. Since m′ and n′ are relatively prime, itfollows that n′ divides b �It follows that T (Γ) ⊂ {0, n′, 2n′, . . . , (m∧ n− 1)n′}, which is a subgroup ofZ/nZ isomorphic to Z/(m∧n)Z. To verify the converse inclusion, note thatthe map ϕc : k 7→ ck with c multiple of n′ is a well-defined morphism withT (ϕc) = c. Finally, Γ ' T (Γ) ' Z/(m ∧ n)Z.
(4) If A and B are subsets of a group G, we denote by AB the set ofproducts {ab , a ∈ A , b ∈ B}. Let H1 and H2 be subgroups of G.
(a) Find a necessary and sufficient condition to have H1H2 < G.
Assume that H1H2 is a subgroup of G and let h1 ∈ H1 and h2 ∈ H2. Theng = h−1
1 h−12 ∈ H1H2, subgroup, so g−1 ∈ H1H2. Since g−1 = h2h1, we
have proved that H2H1 ⊂ H1H2. Since it is a subgroup the inversion mapx 7−→ x−1 is a bijection of H1H2 to itself. Since it also maps H1H2 to H2H1,we get H1H2 = H2H1.Conversely, assume that H1H2 = H2H1 and let h1, h′1 ∈ H1 and h2, h′2 ∈ H2.Then
h1h2(h′1h′2)−1 = h1h2h
′2−1h′1−1 ∈ H1H2H1 = H1H1H2 = H1H2
so H1H2 is a subgroup of G.
(b) Assume that H1 and H2 are finite and H1 ∩H2 = {eG}. Prove that
Card(H1H2) = Card(H1) · Card(H2).
The multiplication map from H1 ×H2 to H1H2 is surjective by definition ofH1H2. Now assume that h1h2 = h′1h
′2 with hi, h
′i ∈ Hi. Then
h′1−1h1︸ ︷︷ ︸
∈H1
= h′2h−12︸ ︷︷ ︸
∈H2
27
so h′1−1h1 = h′2h
−12 = eG, which means that h1 = h′1 and h2 = h′2 so the map
is also injective. It follows that H1H2 and H1×H2 have the same cardinality,namely Card(H1) · Card(H2).
28
Problem set 2: cyclic groups and quotients - Elements of solution
(1) Recall that an automorphism of a group is an isomorphism of thegroup with itself. The set of automorphisms of a group G is denotedby Aut(G).Let G be a group, H a normal subgroup of G and K a subgroup of H.
(a) Assume that ϕ(K) ⊂ K for any ϕ ∈ Aut(H). Is K normal in H?
Yes: among the automorphisms of H are the ones, called inner automor-phisms, given by conjugations, i.e. the maps of the form h 7→ h0hh
−10 . If
K is stable under every automorphism of H, it is in particular stable underthese, which exactly means that it is normal.
(b) Under the same hypotheses, prove that K is normal in G.
It suffices to prove that for every k ∈ K and g ∈ G, the element gkg−1 belongsto K. Observe that the fact that H is normal in G is equivalent to sayingthat H is stable under every inner automorphism x 7→ gxg−1. By restriction,this map defines an automorphism of H, which therefore preserves K by theworking assumption. The same argument as in (a) shows that K �G.
(c) Find G, H and K with K �H and H �G but K is not normal in G.
Consider
G = D8 = {1, r, r2, r3, s, sr, sr2, sr3},H = {1, s, r2, sr2}
K = {1, s}.Normality of K in H is automatic since H is abelian. Moreover, H is normalin G. Although direct verification is a reasonable approach, another possi-bility is to observe that the map ϕ defined on G by ϕ(sirj) = j mod 2 is awell-defined homomorphism from G to Z/2Z with kerϕ = H.Finally, K is not normal in G since rKr−1 contains sr2 /∈ K.
29
(2) Recall that R× = R\{0} and C× = C\{0} and consider the multiplicativegroups
T = {z ∈ C , |z| = 1}µn = {z ∈ C , zn = 1}
µ∞ = {z ∈ C , ∃n ∈ Z , zn = 1}.
(a) Which of these are cyclic?
We have seen in class that µn ' Z/nZ is cyclic, generated by e2iπn . If the
infinite group T was cyclic, it would be isomorphic to Z, hence in bijectionwith a countable set, which it is not possible, as it is in bijection with [0, 2π),uncountable. For a more group-theoretic argument, assume to the contrarythat T = 〈z〉 for some z = eiθ. If θ ∈ 2πQ, then z must have finite order,which is impossible since T is infinite. Therefore θ /∈ 2πQ, so that no powerof z is equal to −1. This contradicts the assumption that z generates T.Finally, every element of µ∞ has finite order hence generates a finite group,so µ∞ is not cyclic.
(b) Prove the following isomorphisms:
R/Z ' T , C×/R×+ ' T , C×/R× ' TT/µn ' T , C×/µn ' C× , Q/Z ' µ∞.
All these isomorphisms follow from the First Isomorphism Theorem appliedto the following homomorphisms.
R −→ Tt 7−→ e2iπt ,
C× −→ Tz 7−→ z
|z|,
C× −→ Tz 7−→ z2
|z|2
T −→ Tz 7−→ zn
,C× −→ C×z 7−→ zn
,Q −→ µ∞q 7−→ e2iπq .
(c) Determine all the finite subgroups of µ∞.
For every n ∈ N, the finite group µn is a subgroup of µ∞ by definition. Toprove that there are no other finite subgroups, it suffices to show that everyfinite subgroup H < µ∞ is cyclic. Every element h of H different from 1 has
finite order nh. It is therefore a power of e2iπnh and H is a subgroup of µN ,
with N = l. c.m(nh , h ∈ H), so H is cyclic.
30
(3) The center of a group G is the set Z(G) = {γ ∈ G | ∀g ∈ G, γ · g = g · γ}.
(a) Compute the center of your favorite and least favorite non-abeliangroups.
Let F be a field and G = GL(n, F ). For x ∈ F×, and k ∈ {1, . . . , n} letDk(x) be the diagonal matrix with all diagonal terms equal to one, exceptfor the one in position (k, k), equal to x. A direct computation shows thata matrix A satisfying Dk(x)A = ADk(x) must be diagonal. Furthermore,if a diagonal matrix commutes to the matrices Tτ defined by Tτ ij = δi,τ(j)
with τ transposition, all its coefficients must be equal. Therefore, Z(G) ⊂F×.In. The converse inclusion is easy, so Z(GL(n, F )) consists of the non-zero multiples of the identity.
Let G = Sn. If n = 1 or 2, then G is abelian. Assume n ≥ 3, and let σ bean element of G different from the identity. Then, there exists an elementi ∈ {1, . . . , n} such that σ(i) = j 6= i. Then if k is different from i and j, onedirectly verifies that σ does not commute to the transposition (j, k) (checkthe image of i).It follows that the center of G is trivial: Z(Sn) = {1} for n 6= 2.
Let G be a group and H a subgroup such that H ⊂ Z(G).
(b) Verify that H �G.
For g ∈ G and h ∈ H, one has ghg−1 = gg−1h = h ∈ H so gHg−1 ⊂ H.
(c) Prove that if G/H is cyclic then G is abelian.
Assuming G/H is cyclic, there exists g0 ∈ G such that G/H = 〈g0H〉 andevery element of G/H is of the form gn0H for some n ∈ Z. The classesmodulo H form a partition of G so every element in G is of the form gn0hwith n ∈ Z and h ∈ H. Since H < Z(G), it follows that G is abelian.
31
(4) A commutator in a group G is an element of the form xyx−1y−1 withx, y ∈ G. Let D(G) be the subgroup of G generated by the commutators.
We will use the following (rather common) notation: for x, y in G,
[x, y] = xyx−1y−1.
(a) Prove that D(G) is normal in G.
Every element of D(G) is a product of commutators and conjugations arehomomorphisms, so it suffices to prove that
g[x, y]g−1 ∈ D(G)
for x, y ∈ G. This actually follows from the general an easily verified factthat
ϕ([x, y]) = [ϕ(x), ϕ(y)]
for any endomorphism ϕ of G.
(b) Prove that G/D(G) is abelian.
A group Γ is abelian if and only if [γ1, γ2] = 1 for all γ1, γ2 in Γ.For i = 1, 2, let γi = giD(G) be an element of Γ = G/D(G). Since thenatural surjection π : G −→ G/D(G) is a group homomorphism,
[γ1, γ2] = [π(g1), π(g2)] = π ([g1, g2]) = [g1, g2]︸ ︷︷ ︸∈D(G)
D(G) = D(G),
that is [γ1, γ2] = 1G/D(G), so G/D(G) is abelian.
(c) Prove that D(G) is the smallest normal subgroup of G with abelianquotient.
Let H �G be a normal subgroup such that G/H is abelian. We shall provethat H contains D(G). To do so, it suffices to prove that H contains allcommutators. Suppose to the contrary the existence of x and y in G suchthat [x, y] /∈ H. Then [x, y]H 6⊂ H, which means that [xH, yH] 6= H. Inother words,
[xH, yH] 6= 1G/H ,
thus contradicting the assumption that G/H is abelian.
32
Midterm 1 - Elements of solution
Problem 1
1. Let G be a group and H < G. The relation defined on G by x ∼ y ⇔ x−1y ∈ His an equivalence relation. For g element of G, describe the class of g.
By definition g ∼ g′ is equivalent to the existence of h ∈ H such that g−1g′ = h, that isg′ = gh. Therefore, the class of g for this relation is the set gH = {gh , h ∈ H}.
2. Is S3 cyclic?
No: beside the identity, S3 contains only transpositions ((1 2), (1 3), (2 3)) of order 2 and3-cycles ((1 2 3), (1 3 2)) of order 3. None of this element can generate the full group.
3. Find the order of
(1 2 3 4 5 6 7 8 9 10 11 1212 4 1 6 8 11 10 5 7 9 2 3
)in S12.
The cycle decomposition of this permutation is (1 12 3)(2 4 6 11)(5 8)(7 10 9). Its orderis the lowest common multiple of the orders of the cycles, namely lcm(3, 4, 2, 3) = 12
4. Give two non-isomorphic groups of cardinal 8.
The dihedral group D8 and Z/8Z both have 8 elements, yet they are not isomorphic asZ/8Z is abelian while D8 is not.
5. Let ϕ ∈ Hom(G,G′) with G is cyclic and ϕ surjective. Prove that G′ is cyclic.
Let x be a generator of G. Then every element of G is of the form xa with a ∈ Z. Bysurjectivity of ϕ, every element of G′ is of the form ϕ(xa) = ϕ(x)a, wich means that ϕ(x)generates G′.
6. Determine the subgroup of (Q×+,×) generated by A ={
1p, p prime
}.
Since 〈A〉 contains all the inverses of elements of A, it contains all prime numbers hence,considering products all the fractions of the form p1
p2with p1, p2 prime or equal to 1. By
the Fundamental Theorem of Arithmetic, every element of Q×+ is a product of finitelymany such fractions so 〈A〉 = Q×+.
33
Problem 2
Recall that D2n = 〈r, s | rn = 1 , s2 = 1 , rs = sr−1〉.
1. Show that every element of D2n that is not a power of r has order 2.
Every such element is of the form g = sri with 0 ≤ i ≤ n− 1. Therefore,
g2 = srisri = s2r−iri = 1.
2. Deduce that every element of D2n is the product of elements of order 2.
Since s · sr = s2r = r, D2n is generated by s and sr, both of which have order 2.
3. Let G be a finite group generated by distinct elements a and b, both oforder 2. Prove that G is isomorphic to D2n, where n = |ab|.
Since a and b have order 2, the elements ρ = ab and σ = a satisfy the relations
σ2 = 1 and ρσ = aba = ab−1a−1 = a(ab)−1 = σρ−1.
In addition, G is assumed finite with |ab| = n so ρn = 1. By the same argument as in theprevious question, ρ and σ satisfy the same relations as r and s in D2n. Therefore themorphism defined by s 7→ a and r 7→ ab is well-defined and an isomorphism between D2n
and G.
Problem 3
1. Let G1 and G2 be groups and consider the product G = G1×G2. Prove thatG1 × {1G2}�G and that the quotient G/G1 × {1G2} is isomorphic to G2.
The result follows from the First Isomorphism Theorem, after observing that the map(g1, g2) 7→ g2 is a surjective homomorphism from G to G2 with kernel G1 × {1G2}.
2. Let F be a field. Consider the following subgroups of SL(2, F ):
P =
{[a t0 a−1
], a ∈ F× , t ∈ F
}, N =
{[1 t0 1
], t ∈ F
}.
Prove that N � P and that P/N is isomorphic to F×.
The result follows from the First Isomorphism Theorem, after observing that the map[a t0 a−1
]7→ a is a surjective homomorphism from P to F× with kernel N .
34
Problem 4
Recall that Aut(G) denotes the set of isomorphisms of a group G onto itself,and that it is a group under composition. Let
Inn(G) = {cg , g ∈ G} , where cg(x) = gxg−1 for all x ∈ G.
1. Verify that cg ∈ Aut(G) for all g ∈ G.
Let g ∈ G. For x, y ∈ G, cg(xy) = gxyg−1 = gx g−1g︸︷︷︸=1
xg−1 = cg(x)cg(y). Moreover cg is
bijective, with inverse cg−1 , so cg ∈ Aut(G).
2. Prove that Inn(G) is a subgroup of Aut(G).
A direct computation shows that cg1 ◦ cg2 = cg1g2 and the Subgroup Criterion applies.
3. Is Inn(G) normal in Aut(G)?
Yes: another direct computation shows that ϕ ◦ cg ◦ ϕ−1 = cϕ(g) for any ϕ ∈ Aut(G).
4. Prove that Inn(G) is isomorphic to a quotient of G.
The relation cg1 ◦ cg2 = cg1g2 shows that the map g 7→ cg is a group homomorphism fromG to Aut(G). Its image is Inn(G) and its kernel is the center Z(G) of G, so the FirstIsomorphism Theorem implies that Inn(G) ' G/Z(G).
35
Problem set 3: symmetric groups, group actions - Elements of solution
(1) Recall that An denotes the alternating subgroup of Sn.
(a) Assume n ≥ 3. Prove that An is generated by 3-cycles.
Since transpositions generate Sn, it suffices to prove that the product of an evennumber of transposition is equal to a product of 3-cycles. This reduces to thefollowing observations, where i, j, k and l are all different:
(i j)(k l) = (i j k)(j k l),
(i j)(i k) = (i k j),
(i j)(j i) = Id.
Let p ≥ 5 be a prime number and H < Sp such that [Sp : H] ≤ p− 1.
(b) Prove that H contains the p-cycles.
Let c be a p-cycle and consider the cosets H, cH, ..., cp−1H. All of themhave the same cardinality asH so they cannot be mutually disjoint, otherwisetheir union would have cardinality p ·#H which contradicts the assumptionthat [Sp : H] ≤ p − 1. Therefore, there exist i 6= j such that ciH = cjH,so that cj−i ∈ H. Since c generates a group of prime order, 〈cj−i〉 = 〈c〉 soc ∈ H.
(c) Prove that [Sp : H] ∈ {1, 2}.
Observe that 3-cycles are products of p cycles:
(i j k) = (k j i a1 . . . ap−3)(i k j ap−3 . . . a1).
Therefore, since H contains all the p-cycles by the previous result, it alsocontains the 3-cycles hence, by (a), the alternating group Ap.Therefore [Sp : H] ≤ [Sp : Ap] = 2.
(d) Deduce from this another concrete example showing that there isno general converse to Lagrange’s Theorem.
Let p = 5. Then S5 does not admit any subgroup of index 3 or 4, that is ofcardinal 40 or 30, although these numbers divide #S5 = 120.
36
(2) Let H < G finite group. For g ∈ G, and x ∈ G/H, let σg(x) = gx.
(a) Verify that σg is a well-defined map from G/H to itself.
Let x, x′ ∈ G be representatives of x, that is x′−1x ∈ H. Then,
σg(x′) = gx′ = gx′H = gx′ x′−1x︸ ︷︷ ︸∈H
H = gxH = gx = σg(x)
so the definition is independent of a choice of representative.
(b) Prove that σ : g 7→ σg is a homomorphism from G to Bij(G/H).
It follows directly from the associativity of the group law in G.
(c) Assume that [G : H] = p, where p is the smallest prime factor of#G.Prove that H is normal in G.
We have proved in class that kerσ is a subgroup of H. Moreover, the FirstIsomorphism Theorem implies that [G : ker σ] = # Imσ hence divides# Bij(G/H).Since [G : H] = #G/H = p and Bij(G/H) has cardinality p!, it follows that
[G : H] =#G
# ker σdivides p!.
Now p is prime and the smallest prime factor of #G so #G# kerσ
divides p.
Therefore, #G# kerσ
≤ p, hence
#H =#G
p≤ # ker σ.
Since kerσ ⊂ H, it follows that kerσ = H so that H �G.
(3) Let G be a finite group acting on a finite set X.
(a) Assume that every orbit contains at least 2 elements, that #G = 15and #X = 17. Determine the number and the cardinals of theorbits.
Each orbit as cardinality [G : H] for H an appropriate stabilizer subgroupof G. By Lagrange’s Theorem, the possible cardinalities for subgroups are1, 3, 5 and 15 (excluded since we assume that no orbit consists of a singlepoint). Therefore the possible cardinalities for orbits are 15, 5 and 3. Theorbits form a partition of X and the only way of writing 17 as a sum of orbitcardinalities is 17=3+3+3+3+5. It follows that this action has exactly fourorbits of cardinality 3 and one orbit of cardinality 5.
37
(b) Assume that #G = 33 and #X = 19. Prove that the action has afixed point.
Following the same line of reasoning, the possible cardinalities for orbits are1, 3 and 11. The equation 3a+11b = 19 has no integer solution with a, b ≥ 0so there is at least one orbit that consists of a single point.
(4) Let G be a group and H a subgroup.
(a) Recall how G acts on G/H and determine the stabilizer of g0H.
The action is defined by g · g0H = (gg0)H. An element g ∈ G stabilizes g0Hif and only if gg0 and g are in the same equivalence class modulo H, that is ifg−1
0 gg0 ∈ H. It follows that StabG(g0H) = g0Hg−10 .
If G acts on two sets X and Y , a map f from X to Y is said G-equivariantif it satisfies f(g · x) = g · f(x) for every g ∈ G and x ∈ X.
(b) Prove that the inverse of a G-equivariant bijection is G-equivariant.
Let f : X −→ Y be a G-equivariant bijection. Let y ∈ Y and g ∈ G. Then,
f−1(g · y) = f−1(g · f(f−1(y)))
= f−1(f(g · f−1(y))) by G-equivariance of f
= g · f−1(y).
(c) Assume that G acts transitively on X, let x ∈ X and H = StabG(x).Construct a G-equivariant bijection between X and G/H.
The mapG/H −→ XgH 7−→ g · x is well-defined, G-equivariant by associativity
of the group law and surjective by transitivity of the action. To prove in-jectivity, note that g1 · x = g2 · x if and only if g−1
2 g1 · x = x, which exactlymeans that g1 and g2 are equal modulo H = StabG(x).
(d) Let H and K be subgroups of G. Prove that there exists a G-equivariant map from G/H to G/K if and only if H is contained ina conjugate of K, and that such a map is necessarily surjective.
Assume given an equivariant map f : G/H −→ G/K and let g0K denotethe image of H by f . By equivariance,
g0K = f(H) = f(hH) = h · f(H) = hg0K
for every h ∈ H. This means that H ⊂ StabG(g0K)(a)= g0Kg
−10 . Fur-
thermore, let gK be an arbitrary element of G/K. Since G acts tran-sitively, there exists g′ ∈ G such that gK = g′g0K. By equivariance,gK = g′f(H) = f(g′H), which shows that f is surjective.
38
Conversely, assume that H ⊂ g0Kg−10 for some g0 ∈ G. Then the map
G/H −→ G/Kg0H 7−→ gK
is well-defined since any representative of gH is of the
form gh with h ∈ H, hence mapped to ghg0K ⊂ gg0Kg−10 g0K = gg0K.
Once again, associativity of the group law entails G-equivariance.
(e) Assume G finite. Prove that there exists a G-equivariant bijectionbetween G/H and G/K if and only if H and K are conjugate in G.
By the previous results, G/H and G/K are isomorphic as G-sets if and onlyif there exists g ∈ G such that H ⊂ gKg−1 and #G/H = #G/K. The latterimplies that H and K have same cardinality so H = gKg−1.
39
Midterm 2 - In-class - Elements of solution
Problem 1
1. Let G be a group acting on a finite set X and x1, . . . , xn a family of repre-
sentatives of all the orbits. Prove that #X =n∑i=1
[G : StabG(xi)].
The set X is the disjoint union of the Orbits Ox1 , . . . ,Oxn . For each i, the orbit Oxi is inbijection with the quotient G/ StabG(xi)], with cardinality [G : StabG(xi)].
2. Let A be a ring. Prove that 0× a = 0 for all a ∈ A.
Observe that 0× a = (a− a)× a = a× a− a× a = 0.
3. Consider the ring A of functions from R to R. For each of the followingsubsets, determine if it is a subring A. If so, determine if it is an ideal.
(a) A−3 = {f : R −→ R , f(−3) = 0}: ideal.
(b) E = {even functions in A}: subring but not an ideal, since the product of x 7→ xwith a non-zero even function is not even.
(c) O = {odd functions in A}: not a ring, since a product of non-zero odd functionsis never odd.
Problem 2
1. Let F be a field. For a ∈ F× let ϕ(a) denote the mapF −→ Ft 7−→ a2t
.
(a) Prove that ϕ(a) is an automorphism of the additive group (F,+).
The map ϕ(a) is linear hence a group morphism. It is bijective with inverseϕ(a−1).
(b) Prove that ϕ is a group homomorphism from F× to Aut(F ).
A direct computation shows that ϕ(ab) = ϕ(a) ◦ ϕ(b) for a, b ∈ F×.
2. Prove that FoϕF× is isomorphic to the subgroup P =
{[a b0 a−1
], a ∈ F× , b ∈ F
}.
Note that
[a 00 a−1
] [1 t0 1
]=
[a at0 a−1
]. This suggests to consider the map
π :F × F× −→ P
(t, a) 7−→[a a−1t0 a−1
].
40
To see that π is a group homomorphism, note that for (t, a) and (s, b) in F oϕF×, one has
(t, a)(s, b) = (t + a2s, ab) so that π((t, a)(s, b)) =
[ab (ab)−1t+ ab−1s0 (ab)−1
]. On the other
hand,
π(t, a)π(s, b) =
[a a−1t0 a−1
] [b b−1s0 b−1
]=
[ab (ab)−1t+ ab−1s0 (ab)−1
].
The kernel of π is reduced to the singleton (0, 1) so π is injective. It is also surjective,hence an isomorphism of groups.
Problem 3
1. Let n ≥ 3. Prove that every 3-cycle σ in Sn is a square.
Any 3-cycle σ has order 3 so σ = σ4 = (σ2)2 is a square.
2. Let H be the subgroup of Sn generated by the squares. Prove that H = An.
3-cycles generate An so it follows from the previous result that An ⊂ H. Conversely, thesignature map being a group homomorphism from Sn to {−1, 1}, products of squares inSn must have signature 1, so that H ⊂ An.
3. Let N be a subgroup of index 2 in Sn.
(a) Prove that σ2 ∈ N for any σ ∈ Sn.
A subgroup of index 2 is necessarily normal so Sn/N is a a group, isomorphic toZ/2Z. Therefore, for every σ ∈ Sn, the coset σN has order 2 in σ ∈ Sn, whichmeans that σ2N = 1Sn/N = N or, equivalently, that σ2 ∈ N .
(b) Deduce that N = An.
By (a), a subgroup N of index 2 in Sn contains all the squares, hence containsH = An. Since [Sn : An] = 2, it follows that N = An: the alternating group isthe only subgroup of index 2 in Sn.
41
Problem 4
Let K be the subgroup of SL(2,R) consisting of matrices of the form
[c −ss c
]with c2 + s2 = 1 and
g =
{X =
[a bc d
], Tr(X) = a+ d = 0
}.
1. Verify that K acts on g by k ·X = kXk−1 where k ∈ K and X ∈ g.
The trace property Tr(AB) = Tr(BA) implies that Tr(kXk−1) = Tr(Xk−1k) = Tr(X),so k · X ∈ g for all k ∈ K and X ∈ g. The identity of K is the matrix I2 and satisfiesI2 ·X = X for all X. Finally, associativity holds for matrix multiplication in M2(R).
Let a =
{[α 00 −α
], α ∈ R
}⊂ g.
2. Determine all the elements of the subgroup N = {k ∈ K , k a k−1 ⊂ a} of K.
Notice that , for k =
[c −ss c
]and X =
[α 00 −α
],
k ·X =
[c −ss c
] [α 00 −α
] [c s−s c
]=
[αc2 − αs2 2αcs
2αcs αs2 − αc2
]Therefore k ·X ∈ a for all X ∈ a if and only if cs = 0. Considering the relation c2 +s2 = 1,the set N consists of the matrices
I − 2 =
[1 00 1
], −I2 =
[−1 00 −1
], w =
[0 1−1 0
], −w =
[0 −11 0
].
To verify that it is a group, one can proceed directly or reproduce the proof that anormalizer is always a subgroup since K acts by conjugation (although not on a subgroupof SL(2,R)).
3. Determine the subgroup C = {k ∈ K , kXk−1 = X for all X ∈ a} of N .
Observe that C ⊂ N by definition and verify that C = {−I2, I2}.
4. Prove that N is isomorphic to Z/4Z and that N/C is isomorphic to Z/2Z.Since N is a group of order 4, it is necessarily isomorphic to Z/4Z or to V4. Notice that|w| = 4 so that N ' Z/4Z. It is abelian so C is normal and N/C = {±I2,±w} ' Z/2Z.
42
Midterm 2 - Take-home - Elements of solution
Problem 1
The goal of this problem is to determine for which values of n there exists aunique group of order n up to isomorphism. In what follows, n is a positiveinteger with prime decomposition n = pα1
1 · · · pαss .
1. (a) For n prime, determine, up to isomorphism, all the groups of order n.
In a group of order n prime, every element different from 1 has order n so the group iscyclic, hence isomorphic to Z/nZ.
(b) Prove that if αi ≥ 2 for some i ∈ {1, . . . , s}, then there are at least twonon-isomorphic groups of order n.
The assumption entails that n is of the form pαm with α ≥ 2 and p prime not dividingm. An element of the group Z/pZ× . . .× Z/pZ︸ ︷︷ ︸
α times
×Z/mZ has order at most pm whereas
Z/nZ has elements of order pαm so they cannot be isomorphic.
From now on, we assume that αi = 1 for all i ∈ {1, . . . , s}.
2. Recall the expression of the Euler Indicator ϕ(n) in that case.
For n =∏s
i=1 pi the value of the Euler indicator is ϕ(n) =∏s
i=1(pi − 1).
3. Let p and q be distinct prime numbers such that p|(q − 1).
(a) Prove the existence of a non-abelian group of order pq.
Let P = Z/pZ and Q = Z/qZ and recall that Aut(Q) is cyclic of order p. Sincep|(q − 1), there exist non-trivial morphisms from P to Aut(Q). Let ϕ be such amorphism and consider QoϕP . This group is non-abelian since ϕ is not constantand it has order pq by construction.
(b) Deduce that if all groups of order n are isomorphic, then n ∧ ϕ(n) = 1.
Assume that n and ϕ(n) are not relatively prime. Then in view of the formulafor ϕ(n) given in 2., there exist primes factors p and q of n such that p|(q − 1).Let H be non-abelian group of order pq, whose existence is guaranteed by 3.(a)and let n′ = n
pq. Then H × Z/n′Z is a non-abelian group of order pqn′ = n.
43
We shall prove the converse by contradiction. Let n be the smallest integer forwhich n∧ϕ(n) = 1 and there exists a group G of order n that is not isomorphicto Z/nZ (assuming the existence of such integers).
4. (a) Prove that m ∧ ϕ(m) = 1 for any divisor m of n.
For k ∈ Z, let Dk denote the set of positive divisors of k. Since ϕ is multiplicative, m|nimplies Dϕ(m) ⊂ Dϕ(n) so n ∧ ϕ(n) = 1 implies that m ∧ ϕ(m) = 1 if m divides n.
(b) Prove that proper subgroups and quotients of G are cyclic.
Lagrange’s theorem implies that every such group H has a divisor of n for its order. Bythe previous question #H ∧ ϕ(#H) = 1 with #H < n so by the minimality assumptionon n, the group H must be cyclic.
(c) Deduce that the center of G is trivial.
Assume that the center Z(G) of G is not reduced to the identity. Then G/Z(G) is cyclicby the previous result. It follows that G is abelian (see Problem (3) in Homework #2).By the structure theorem of finitely generated abelian groups, G is isomorphic to a directproduct Z/n1Z× . . .×Z/nsZ with ni+1|ni for all i. In particular, n = n1 · . . . ·ns and anyprime factor of n divides n1. Since n is square-free, this implies that n = n1 and G mustbe cyclic, which contradicts our assumption.
A maximal subgroup of a group Γ is a proper subgroup H such that the onlysubgroups of Γ containing H are H and Γ.
5. Let U be a maximal subgroup of G and x 6= 1 in U .
(a) Prove that U = CG(x).
As a proper subgroup of G, the group U is cyclic hence abelian so U ⊂ CG(x).The inclusion cannot be strict, because the maximality of U would then implythat CG(x) = G and x would be a non-trivial element in the center of G.
(b) Deduce that distinct maximal subgroups of G have trivial intersection.
Assume that U and V are maximal subgroups of G and x 6= 1 is in the intersectionU ∩ V . Then U = CG(x) = V . This proves that two maximal subgroups areidentical or have trivial intersection.
We admit the following result: every maximal subgroup of G is equal to itsown normalizer:
U = NG(U).
6. Let U be a maximal subgroup, u its order and U the union of all conjugatesof U in G.
44
(a) Determine the number of conjugates of U and the order of each suchconjugate.
The class equation indicates that the number of conjugates is the index of NG(U)in G, that is, according to the admitted result, n
u. Conjugations are bijections so
each conjugate has the same cardinality u as U .
(b) Verify that the conjugates of U are maximal and deduce that U containsn− n
uelements different from the identity.
Let g ∈ G and H a subgroup of G such that gUg−1 ⊂ H. Then U ⊂ g−1Hgso either U = g−1Hg (in which case H = gUg−1) or g−1Hg = G, implying thatH = G. Since maximal subgroups have trivial intersection, it follows from theprevious question that U \ {1} is the disjoint union of n
usets of cardinality u− 1,
hence the result.
7. Let x ∈ G \ U. Consider V a maximal subgroup of G containing x. Denoteby v its order and by V the union of all conjugates of V .
(a) Prove that U ∪V contains 2n− nu− n
velements different from 1.
By the similar arguments, U \ {1} ∪ V \ {1} is the disjoint union of nu
sets ofcardinality u− 1 and n
vsets of cardinality v − 1.
(b) Compare to the cardinality of G \ {1} and deduce a contradiction.
The inclusion of U∪V \ {1} in G \ {1} implies that 2n− nu− n
v≤ n− 1, leading
to 1u
+ 1v> 1, which is impossible since u and v are the cardinalities of maximal
subgroups of G.
8. Conclude.
Observe that n ∧ ϕ(n) = 1 ensures that n is square-free, so the result proved in 3.(a)shows that the condition is necessary. Questions 4. to 7. show that the assumption thatthere exists a (smallest) non-cyclic group of order n satisfying the condition is absurd, sothe condition is sufficient too.pNotice that the proof of the second implication is done by induction: assuming that allgroups of order m < n with ϕ(m) ∧m = 1 are cyclic, we showed that it must also be thecase for a group of order n satisfying the same condition.References for this result are When is Zn the only group of order n?, by J. Gallian, andD. Moulton, Elem. Math. 48 (1993), no. 3, 117–119 and On the uniqueness of the cyclicgroup of order n, by D. Jungnickel, Monthly, Vol. 99 (1992), pp. 545–547.
45
Problem 2
Let F be a field and consider the groups G = SL(2, F ) and N =
{[1 t0 1
], t ∈ F
}.
Let X2 denote the set F 2 \ {(0, 0)}, with elements written as column matrices.
1. Is N is normal in G? No.
2. Prove that G acts on X2 via left matrix multiplication.
Multiplication by an invertible matrix is injective so x 6= (0, 0) implies g · x 6= (0, 0) forg ∈ G. The matrix I2 acts trivially; associativity follows from that of products in M2(F ).
For g ∈ G, let c1(g) denote the first column of g.
3. Prove that the map ϕ :G/N −→ X2
gN 7−→ c1(g)is well-defined.
A direct computation shows that c1(gn) = c1(g) for all g ∈ G and n ∈ N .
4. Prove that ϕ is a G-equivariant bijection.
Equivariance follows from the associativity of matrix products. To prove that ϕ is in-
jective, assume that c1(g1) = c1(g2), that is, g1 =
[a bc d
]and g2 =
[a βc δ
]with
ad − bc = 1 = aδ − βc. To prove that g1 and g2 are equal modulo N , it suffices to find
t ∈ F such that g1nt = g2 with nt =
[1 t0 1
]. This amounts to solving the system{
at+ b = βct+ d = δ
. Note that the determinants equality implies that a(d − δ) = c(b − β). If
a 6= 0 and c 6= 0, it follows that t = β−ba
= δ−dc
is a solution. If a = 0, then c 6= 0, so b = β
and t = δ−dc
is solution. The case c = 0 is similar. Finally, surjectivity follows from the
observation that
[a −b
a2+b2
b aa2+b2
]is an inverse image of (a, b).
From now on, assume n ≥ 1 and let G = SL(n+ 1, F ), while Yn+1 denotes the setof F -valued matrices with n+ 1 rows and n columns.
5. Verify that G acts on Yn+1 via left matrix multiplication.
The proof is essentially the same as in 2., using the fact that multiplication by an invertiblematrix turns an independent family into another independent family.
46
6. Let x0 =
In
0 . . . 0
. Determine the group N = StabG(x0).
A direct computation shows that N consists of all the matrices of the form1 0 . . . 0 x1
0 1 . . . 0 x2... 0
. . ....
......
... . . . 1 xn0 0 . . . 0 1
.Observe that the notation N is compatible with that of question 1. when n = 1.
7. Describe the map b :G −→ Yn+1
g 7−→ g · x0.
This map extracts the block consisting of the first n columns of g.
8. Prove that G/N is in G-equivariant bijection with the subset Xn+1 of Yn+1
consisting of the elements of rank n.
A direct computation shows that b(gn) = b(g) for any g ∈ G and n ∈ N so we defineϕ(gN) = b(g). This map is the natural bijection between G/ StabG(x0) and the orbitof x0 so injectivity is automatic. Equivariance is, once again, a consequence of the asso-ciativity of matrix multiplication. To verify surjectivity, consider a family (x1, . . . , xn) oflinearly independent vectors in F n+1. Then there exists a vector y 6= 0 in F n+1 such that(x1, . . . , xn, y) is a basis of F n+1, so that det(x1, . . . , xn, y) = δ 6= 0. Then the matrix offamily (x1, . . . , xn,
yδ) has determinant 1 and is a preimage for (x1, . . . , xn).
47
Problem set 4: rings and ideals - Elements of solution
(1) Let A be a commutative ring with identity and A[X] the ring of poly-nomials with coefficients in A.
(a) Prove that (X) is a prime ideal of A[X] ⇔ A is an integral domain.
The ideal (X) is prime if and only if the quotient A[X]/(X) is an integraldomain. The evaluation map P 7→ P (0) is surjective with kernel (X). There-fore, by the First Isomorphism Theorem, it induces an isomorphism betweenA[X]/(X) and A hence the result.
(b) Prove that (X) is maximal if and only if A is a field.
Recall that an ideal is maximal if and only if the corresponding quotient isa field. The result therefore follows from the isomorphism A[X]/(X) ' A.
(c) Let I and J be ideals in A and P a prime ideal such that IJ ⊂ P .Prove that either I or J is contained in P .
Assume for instance that I is not contained in P and fix j0 in P \ J . Thehypothesis IJ ⊂ P implies that ij0 ∈ P for all i ∈ I. Since P is prime, itfollows that I is contained in P .
(2) Let A be a commutative ring with identity and consider a polynomialP ∈ A[X] of degree n ≥ 1 with leading coefficient 1.
(a) Verify that every class in A[X]/(P ) has a representative in A[X] ofdegree at most n− 1.
Two polynomials are in the same class modulo (P ) if they differ by a multipleof P . Let U =
∑aiX
i be a polynomial of degree m in A[X]. If m ≤ n− 1,there is nothing to do. Otherwise consider U1 = U − amX
m−nP . Sincethe leading coefficient of P is 1, U1 has degree < m. Define inductivelypolynomials
Uk+1 = Uk − akXdeg(Uk)−nP
as long as deg(Uk) ≥ n. All the polynomials Uk are in the same class as Uand since deg(Uk+1) < deg(Uk), one of them finally has degree < n.
48
(b) Let U, V ∈ An−1[X] be distinct polynomials. Prove that their imagesin A[X]/(P ) are distinct.
Assume that U and V in An−1[X] are in the same class modulo (P ). ThenU − V is a multiple of P with degree < n. Therefore U − V = 0.
(c) Assume that P can be factored in A[X]. Prove that A[X]/(P ) haszero divisors.
Assume that P = AB with deg(A), deg(B) ≥ 1. Then in A[X]/(P ) we have
(A+ (P ))(B + (P )) = AB + (P ) = P + (P ) = (P ) = 0A[X]/(X)
showing that A+ (P ) and B + (P ) are zero divisors in A[X]/(X).
(d) Assume that P = Xn − a where a ∈ A is nilpotent. Prove that theimage of X in A[X]/(P ) is nilpotent.
Since
P︷ ︸︸ ︷Xn − a+(P ) = (P ), (the images of) Xn and a are equal in A[X]/(P ).
The natural projection is a ring homomorphism so (Xn)k = ak in A[X]/(P )for every k. Since a is nilpotent in A, it follows that Xn and therefore X isnilpotent in A[X]/(P ).
(3) Consider the ring Z[i] = {a+ ib , a, b ∈ Z}.
(a) Prove that every point in the complex plane is at distance strictlyless than 1 from an element in Z[i].
Let z = x + iy be an arbitrary complex number. There are integers a andb such that |x − a| ≤ 1
2and |y − b| ≤ 1
2so that u = a + bi ∈ Z[i] and
d(z, u) ≤ 12< 1.
(b) Prove that Z[i] is a Euclidean domain and determine Z[i]×.
Let (u, v) ∈ Z[i] × Z[i] \ {0}. By the previous result, there exists q in Z[i]
such that∣∣∣q − u
v
∣∣∣ < 1. Let r = u− vq. Then u = vq + r and
|r| = |v|∣∣∣uv− q∣∣∣ < |v|.
Therefore, the norm u 7→ |u|2 turns Z[i] into a Euclidean domain.
To determine the invertible elements, observe that Z[i] is a subring of Cwhich is a field. If u = a + ib ∈ Z[i] \ {0}, then u−1 exists in C and equalsa−iba2+b2
. It follows that u−1 is in Z[i] if and only if a2 + b2 divides a and b,
which can only happen if a2 + b2 = 1. Therefore, Z[i]× = {1,−1, i,−i}.
49
(4) Let A be a ring with identity and consider the map ϕ : N −→ A definedby ϕ(n) = 1 + . . .+ 1︸ ︷︷ ︸
n times
.
(a) Prove that ϕ extends uniquely to a ring morphism from Z to A.
Any Φ ∈ Hom(Z, A) is determined by its values on N since any negativeinteger is the opposite of a positive integer and Φ(−n) = −Φ(n). The onlyextension of ϕ to a ring homomorphism is therefore defined by
Φ(n) =
{ϕ(n) if n ≥ 0−ϕ(−n) if n ≤ 0
.
(b) Prove the existence of a non-negative integer κ such that kerϕ = κZ.
The kernel of Φ is an ideal in the PID Z so it is generated by a single elementk ∈ Z. Note that (k) = (−k) and let κ = |k|.
The integer κ is called the characteristic of A.
(c) Prove that the characteristic of an integral domain is either 0 or aprime number.
Assume that κ = pq with p, q > 1. Then p < κ and q < κ so Φ(p) 6= 0 andΦ(q) 6= 0. However, Φ(p)Φ(p) = Φ(pq) = Φ(κ) = 0 so Φ(p) is a zero divisor.
(d) Determine the characteristics of Q, Z[X] and Z/nZ[X].
The characteristic of Q and Z[X] is 0, the characteristic of Z/nZ[X] is n.
(e) Let p be a prime number and A a commutative ring of characteristicp. Prove that (a+ b)p = ap + bp in A.
Using the commutativity of A, one can prove (for instance by induction,using Pascal’s relation
(n+1k+1
)=(n+1k
)+(n+1k
)) that the binomial expansion
holds:
(a+ b)p =
p∑k=0
(p
k
)akbp−k = ap +
p−1∑k=1
(p
k
)akbp−k + bp.
Observe that p divides(pk
)= p!
k!(n−k)!for all k ∈ {1, . . . , p − 1} since p is
prime and p > k and p > p − k for all these values of k, so that all thecorresponding terms in the sum vanish in A.
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Problem set 5: polynomial rings - Elements of solution
(1) If A is a ring, denote by A[X, Y ] the ring of polynomials in two variableswith coefficients in A.
(a) Is the ideal (X, Y ) of Q[X, Y ] principal?
If the ideal (X, Y ) was generated by a single polynomial T 6= 0, then Twould divide both X and Y hence would be of degree 0. Then T would bea unit in Q hence in Q[X, Y ], so that (X, Y ) = Q[X, Y ] which is absurd.
(b) Are the ideals (X) and (X, Y ) prime in Q[X, Y ]? Are they maximal?
Recall that Q[X, Y ] = Q[Y ][X] so that
Q[X, Y ]/(X) = Q[Y ][X]/(X) ' Q[Y ].
Since Q[Y ] is an integral domain, (X) is a prime ideal. However, Q[X] isnot a field so (X) is not maximal.Similarly, the First Isomorphism Theorem applied to the evaluation mapQ[X, Y ] 3 P 7→ P (0, 0) ∈ Q shows that
Q[X, Y ]/(X, Y ) ' Q.Since Q is a field, (X, Y ) is maximal hence prime.
(c) Are the ideals (X, Y ) and (2, X, Y ) prime in Z[X, Y ]? Maximal?
As in the previous question, Z[X, Y ]/(X, Y ) is isomorphic to Z integral do-main but not a field so (X, Y ) is prime but not maximal.Let p denote the natural projection from Z to Z/2Z. The composition map
Z[X, Y ]eval(0,0)−→ Z p−→ Z/2Z
P 7−→ P (0, 0) 7−→ [P (0, 0)]
is a surjective ring homomorphism with kernel (2, X, Y ) so Z[X, Y ]/(2, X, Y )is a field and (2, X, Y ) is maximal, hence prime.
(2) Let F be a field and A = F [X,X2Y,X3Y 2, . . . , XnY n−1, . . .] ⊂ F [X, Y ].
(a) Prove that A consists of all polynomials of the form a0 +∑ak,`X
kY `
with ak,` = 0 if 1 ≤ k ≤ `.
All F -combinations of monomials Xk+1Xk are of this form. Moreover,
(?) Xa+1Xa ·Xb+1Xb = Xa+b+2Xa+b
so products also share this property. Conversely, all polynomials of the givenform belong to A: observe that XkY ` = Xk−`−1︸ ︷︷ ︸
∈A
X`+1Y `︸ ︷︷ ︸∈A
.
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(b) Prove that A and F [X, Y ] have the same field of fractions.
The inclusion of the rings A ⊂ F [X, Y ] automatically implies that of their
fields of fractions: A ⊂ F [X, Y ]. Conversely, if PQ
is a fraction with P,Q ∈F [X, Y ], then by the previous question, there exists a positive integer n largeenough so that XnP, xnQ ∈ A and
P
Q=Xn
Xn
P
Q∈ A.
(c) Prove that the ideal (X,X2Y,X3Y 2, . . .) of A is not finitely generated.
If the ideal was finitely generated, then it would be generated by a finitefamily {X,X2Y, . . . , XnY n−1} for some n. Using (?), one can show that thisimplies that Xn+1Y n does not belong to the ideal, which is a contradiction.
(3) Let A = Z +XQ[X].
(a) Prove that A is an integral domain and determine A×.
The set A consists of all polynomials in Q[X] with integer constant term.It is an integral domain as a subring of Q[X]. The units of Q[X] are thenon-zero constants so the units of A are 1 and -1.
(b) Show that the irreducibles of in A are prime numbers, oppositesof prime numbers and irreducibles in Q[X] with constant term ±1.
Let P be an irreducible element of A. If deg(P ) = 0, then P must be a primenumber or the opposite of a prime number, otherwise it is reducible in A.Now assume deg(P ) ≥ 1. If P is reducible in Q[X], say P = QR, there existsan integer d such that dP ∈ Z[X] and by Gauss’ Lemma dP is reducible inZ[X], that is dP = ST with S, T ∈ Z[X] non-units. It follows that d dividesthe constant term S(0)T (0), so we can write d = st with s and t integerssuch that s|S(0) and t|T (0). Then P = 1
sS 1tT is a non-trivial factorization
of P in A, which contradicts the irreducibility assumption. Finally, if theconstant term is a0 6= ±1, then a0( 1
a0P ) is a non-trivial factorization of P in
A (note that a0 = 0 would also imply P reducible).Conversely, if P = P (0)+XQ(X) is in A and n|P (0) with n 6= ±1, then P =n(
1nP)
is a non-trivial decomposition of P which is therefore not irreducible.
(c) Are the irreducibles prime?
Yes. Let P be irreducible and assume that P |ST with S, T ∈ A. If P hasdegree 0, then it is a prime number or the opposite of a prime number so itdivides S(0)T (0) hence S(0) or T (0). Assume P |S(0). Then P |S since theother coefficients are in Q. If deg(P ) ≥ 1, then P divides ST in the UFDQ[X] so it divides S or T in Q[X]. Since P is irreducible, the previous resultimplies that P (0) = ±1 so P divides S or T in A.
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(d) Can X be written as the product of irreducibles? Is A a UFD?
No: the equation X = PQ implies that either P or Q has degree 0, so anydecomposition of X into irreducibles must be of the form X = p(aX + b)with a ∈ Q and b = ±1, which is impossible. Note that X is not irreducible
either since X = 2X
2is a non-trivial decomposition. It follows that A is not
a UFD.
(4) Let F be field and P a polynomial of degree n ≥ 1.
(a) Prove that F [X]/(P ) is a vector space of dimension n over F .
Observe that F [X]/(P ) is an abelian additive group and that F acts onF [X]/(P ) by λ·Q = λQ, defining a structure of F -vector space on F [X]/(P ).Furthermore, the division algorithm in the Euclidean domain F [X] showsthat every element of F [X]/(P ) has a unique representative of degree at
most deg(P ) − 1. It follows that {1, X, . . . , Xn−1} is a basis of F [X]/(P ),which has therefore dimension n over F .
(b) Assume that F has cardinality q. What is the cardinality of F [X]/(P )?
It follows from the previous result that F [X]/(P ) is isomorphic to F n as avector space, hence has cardinality qn.
(c) Prove that F [X]/(P ) is a field if and only if P is irreducible.
In the Euclidean domain F [X], irreducible ⇔ prime ⇔ maximal.
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Final examination - Elements of solution
1. Cyclotomic polynomials
For n ≥ 2, let µn denote the multiplicative group of nth roots of 1: µn = {z ∈C , zn = 1} and Πn the set of generators of µn. The cyclotomic polynomial oforder n is
Φn =∏ξ∈Πn
(X − ξ).
We recall that the Euler indicator ϕ satisfies the formula∑d|nϕ(d) = n.
1. Consider the polynomial Pn =∏ξ∈µn
(X − ξ).
(a) Prove that Pn = Xn − 1.
All the elements of µn are roots of Xn − 1 so it is divisible by Pn. Since both poly-nomials have the same degree, they differ by a multiplicative constant and since theyare both monic, they must be equal.
(b) Determine Φp for p prime.
If p is prime, then Πp = µp \ {1} so Φp = PpX−1
. By the previous result,
Φp =Xp − 1
X − 1= 1 +X +X2 + . . .+Xp−1.
2. Let ω = e2iπn and k an integer such that 0 ≤ k ≤ n− 1.
(a) Let d be the order of ωk in µn. Prove that ωk ∈ Πd.
This is immediate form the definition of d.
(b) Deduce that Xn − 1 divides∏d|n
Φd.
By 1.(a), it suffices to prove that (X − ξ) divides∏
d|n Φd for all ξ in µn. Since
µn = 〈ω〉, every ξ ∈ µn is of the form ωk for some 0 ≤ k ≤ n − 1, hence belongs tosome Πd where d|n by Lagrange’s Theorem. It follows that (X − ξ) divides Φd, hencethe result.
(c) Prove that∏d|n
Φd = Xn − 1.
Note that both polynomials are monic. Moreover, the relation∑
d|n ϕ(d) = n impliesthat they have the same degree. Since one divides the other, they are equal.
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3. We will prove by induction that Φn has integer coefficients.
(a) Verify the result for n = 1.
By definition, Φ1 = X − 1.
(b) Assuming the result true up to n − 1, find a monic polynomial P ∈ Z[X]such that Xn − 1 = P Φn.
It follows from 2.(c) that Xn − 1 = Φn ·
∈Z[X]︷ ︸︸ ︷∏d|n , d 6=n
Φd.
(c) Prove the existence of polynomials Q and R in Z[X] with deg(R) < deg(P ),such that Xn − 1 = PQ+R.
The division algorithm in the Euclidean ring Q[X] provides Q and R with rationalcoefficients such that Xn−1 = PQ+R. Moreover, since P is monic, and both Xn−1and P have integer coefficients, the polynomials occurring at every step of the divisionalgorithm all have integer coefficients too. It follows that Q,R ∈ Z[X].
(d) Prove that the couple (Q,R) is unique and conclude that Φn ∈ Z[X].
The couple (Q,R) is unique in C[X] so it is in Z[X] too. Since Xn − 1 = P Φn inC[X], we deduce that R = 0 and Φn = Q belongs to Z[X].
2. Application: proof of Wedderburn’s Theorem
We shall prove that every finite division ring is commutative. Let K be a finitedivision ring. We argue by induction on the cardinality of K.
0. Prove the following result.
Lemma. If A is a finite division ring and F a subring of A that is a field, then A is afinite dimensional vector space over F .
Observe that (A,+) is an abelian group. Moreover, the map F × A −→ A definedby (λ, x) 7−→ λ · x, where · denotes the multiplication in A satisfies all the properties(distributivity etc) making A a vector space over the field F . The family of all non-zerovectors in A spans A so it contains a basis over F . Such a basis must be finite since A is.
1. Prove that a division ring of cardinality 2 is commutative.
All possible products in such a ring are 0 · 0 = 0, 1 · 1 = 1, 1 · 0 = 0 = 0 · 1.55
From now on, we assume that every division ring of cardinality < #K iscommutative and that K is noncommutative.
2. Let Z = {x ∈ K | xy = yx for all y ∈ K} be the center of K and q = #Z.
(a) Prove that Z is a subring of K. This is a straightforward computation.
(b) Prove the existence of an integer n ≥ 2 such that #K = qn.
Observe that Z is commutative. Moreover, for x ∈ Z \ {0} and y ∈ K arbitrary,
x−1y − yx−1 = x−1(y − xyx−1) = x−1(y − yxx−1) = x−1(y − y) = 0
where the second equality holds because x ∈ Z. It follows that Z is a field. Then, bythe lemma proved above, K is a finite-dimensional vector space over Z. Therefore,there is a (linear) bijection K
∼−→ Zn where n = dimZ K. Note that n ≥ 2 by theassumption that K is noncommutative. The result follows.
3. For x ∈ K, let Kx = {y ∈ K | xy = yx}.(a) Verify that if Kx ( K, then Kx is a field extension of Z and a subring of K.
Note that Z is trivially contained in Kx. Moreover, the same argument as in 2.(b)shows that Kx is a division algebra. If it is strictly contained in K, it is a field by theinduction hypothesis.
(b) Deduce the existence of a divisor d of n such that #Kx = qd.
If Kx = K, there is nothing to prove. Otherwise, the lemma implies that Kx isa finite-dimensional vector space over Z hence of cardinality q′ = qd for some d.Moreover, K is a finite vector field over Kx so there exists an integer m such that#K = q′m = (qd)m. Since we proved that #K = qn, we conclude that n = md.
4. Recall that the group K× = K \ {0} acts on itself by conjugation.
(a) Prove that every stabilizer has a cardinality qd − 1 with d a divisor of n.
It follows immediately from the observation that StabK×(x) = Kx \ {0} and theprevious result.
(b) Using the class equation, prove the existence of integers λd such that
#K× = q − 1 +∑
d|n , d 6=n
λdqn − 1
qd − 1.
The class equation implies that
#K× = #Z(K×) +∑
x∈K×/∼
[K× : StabK×(x)].
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Note that Z(K×) = Z \ {0} so that #Z(K×) = q − 1. The expected formula isobtained by letting λd denote the number of distinct K×-orbits corresponding to astabilizer with cardinality qd − 1, for every given divisor d of n.
5. Assume that d|n and d 6= n.
(a) Prove that Φn dividesXn − 1
Xd − 1in Z[X].
It follows from the result of 2.(c) is the first part, that Xn − 1 =∏
δ|n Φδ. Therefore,
Xn − 1 = Φn ·∏
δ|n , δ 6=n
Φδ = Φn ·∏
δ|n , δ 6=n , δ|d
Φδ ·∏
δ|n , δ 6=n , δ-d
Φδ
= Φn ·∏δ|d
Φδ︸ ︷︷ ︸=Xd−1
·∏
δ|n , δ 6=n , δ-d
Φδ︸ ︷︷ ︸=Q
Therefore,Xn − 1
Xd − 1= Φn · Q and the result follows from the fact, proved in the first
part, that cyclotomic polynomials have integer coefficients.
(b) Prove that Φn divides (Xn − 1)−∑
d|n , d 6=n
λdXn − 1
Xd − 1in Z[X].
The constants λd are integers by definition and previous results show that Φn dividesall the terms in the sum.
(c) Deduce that Φn(q) divides q − 1.
It follows from the result proved in 4.(b) that
q − 1 = qn − 1−∑
d|n , d 6=n
λdqn − 1
qd − 1.
The previous result implies that Φn(q) divides the right-hand side, hence the result.
6. Prove that |Φn(q)| >ϕ(n)∏i=1
(q − 1) ≥ q − 1 and conclude.
Note that if q ≥ 1 is an integer, the closest point to q on the unit circle is 1. In otherwords, if |ξ| = 1 and ξ 6= 1, then |q − ξ| > |q − 1|. By definition of Φn and the fact thatdeg(Φn) = ϕ(n), we get
|Φn(q)| >ϕ(n)∏i=1
(q − 1).
With q ≥ 2, it follows that |Φn(q)| > q − 1 which contradicts the result of 5.(c).
The assumption that K is noncommutative is therefore absurd, which concludes the proof.57