MATH 612 Due Wednesday April 21, 1993

Lemma. Let L be a left R-module and M a right R-module . Let ℓ_{1}, …, ℓ_{n} ∈ L and m_{1}, …, m_{n} ∈ M be such that ∑ ℓ_{i} ⊗ m_{i} = 0 ∈ L ⊗_{R} M. Then there exist finitely generated submodules L_{1} ⊆ L and M_{1} ⊆ M such that

ℓ_{1}, …, ℓ_{n} ∈ L_{1}
m_{1}, …, m_{n} ∈ M and
∑ ℓ_{i} ⊗ m_{i} = 0 ∈ L_{1} ⊗_{R} M_{1}.

(The fact that the tensor product is not left exact makes this lemma non-trivial.)

Definition. Let {M_{i}}_{i∈I} be a family of submodules of an R-module M. We say that this family is directed if (∀i, j ∈ I) (∃k ∈ I) M_{i} ∪ M_{j} ⊆ M_{k}. In this case, ⋃_{I} M_{i} is called a direct union.

1. a) Prove that a direct union of submodules of M is a submodule. b) Prove that a direct union of flat submodules of M is flat.

c) Prove that every R-module is a direct union of finitely generated submodules.

d) Prove that a torsion free module over a principal ideal domain is flat.

Definition. If N ⊆ M are modules over a PID R then N is called a pure submodule of M if and only if for every r ∈ R, N ∩ rM = rN. (note: This is not the correct definition over a general ring R.)

2. Let N ⊆ M be modules over a PID R.

a) Prove that N is pure in M if and only if for every cyclic R-module L the induced map L ⊗_{R} N → L ⊗_{R} M is monic.

b) Prove that N is a pure submodule of M if and only if every coset x ∈ M/N has a representative m′ such that ann m′ = ann x. (For abelian groups, we can translate this condition as “Every coset in M/N has a representative in M whose order is the same as the order of the coset.”)

c) Prove that if M/N is torsion free then N is pure in M.

Definition. The conjugacy class of an element g in a group G is {h^{−1}gh | h ∈ G}. If G is infinite, then a conjugacy class may be either finite or infinite.

3. Prove that the center of a group ring R[G] is free as an R-module with a basis consisting of those elements ∑_{C} g, where C ranges over the finite conjugacy classes of G.

For any group algebra R[G] there exists a unique R-algebra morphism ε: R[G] → R such that ε(g) = 1 for all g ∈ G. This is called the augmentation map. Ker ε is called the augmentation ideal of R[G].

4. a) Prove that the augmentation ideal is the ideal of R[G] generated by all elements g − 1 for g ∈ G.

b) Let I be the augmentation ideal of R[G]. Prove that for g_{1}, g_{2} ∈ G, g_{1}g_{2} − 1 ≡ (g_{1} − 1) + (g_{2} − 1) (mod I^{2}).

c) Let [G, G] be the commutator subgroup of G, i. e. the subgroup generated by all elements g_{1}g_{2}g_{1}^{−1}g_{2}^{−1} and let I be the augmentation ideal of ℤ[G]. Prove that G/[G, G] ≈ I/I^{2} .

d) Prove that if G and G′ are (not necessarily finite) abelian groups and ℤ[G] ≈ ℤ[G′] then G ≈ G′.

April 21 Answers

Lemma. Let G_{1} and G_{2} be groups and R a commutative ring. Let ε_{i}: R[G_{i}] → R be the augmentation ideals for the group rings and let I_{i} = Ker ε_{i}. If R[G_{1}] ≈ R[G_{2}] as R-algebras then there exists an isomorphism θ: R[G_{1}] → R[G_{2}] such that θ(I_{1}) = I_{2} . proof: Let ζ: R[G_{1}] → R[G_{2}] be an isomorphism of R-algebras and let ρ = ε_{2}ζ. Then ρ is a ring morphism from R[G_{1}] to R, so for all g ∈ G_{1} , ρ(g) is a unit in R. Now since R[G_{1}] is a free R-module with basis G_{1} , there is a unique R-linear map θ: R[G_{1}] → R[G_{2}] such that θ(g) = ρ(g)^{−1}ζ(g). Notice that for g, h ∈ G, θ(gh) = θ(g)θ(h). It follows easily that θ is an isomorphism of R-algebras. Furthermore, for g ∈ G_{1} , ε_{2}θ(g) = ρ(g)^{−1}ε_{1}ζ(g) = ρ(g)^{−1}ρ(g) = 1. Thus θ(g − 1) ⊆ Ker ε_{2} = I_{2} so θ(I_{1}) ⊆ I_{2} . Thus θ induces an R-algebra morphism θ̄: R[G_{1}]/I_{1} → R[G_{2}]/I_{2} . Now if x ∈ R then x ≡ ε_{1}(x) (mod I_{1}) and since ε_{1}(x) ∈ R it follows that θ̄(x + I_{1}) = θ(ε_{1}(x) + I_{1}) = ε_{1}(x) + I_{2} . Thus θ̄ is an isomorphism, so

θ(I_{1}) = I_{2} . ✓

4. a) In fact, one sees that I is the free abelian group with basis {g − 1 | g ∈ G}. b) Clearly I^{2} is generated by those elements of the form

(g_{1} − 1)(g_{2} − 1) = (g_{1}g_{2} − 1) − (g_{1} − 1) − (g_{2} − 1).

c) Let φ: G → I/I^{2} be defined by φ(g) = g − 1 + I^{2} . By b) this is a homomorphism from G into the additive group I/I^{2} . Since the latter is abelian, it induces a homomorphism φ̄: G/[G, G]: → I/I^{2} . On the other hand, since I is the free abelian group with basis consisting of those elements g − 1 for g ∈ G and since G/[G, G] is an abelian group, there is a unique homomorphism ψ: I → G/[G, G] with ψ(g − 1) = g[G, G] (and taking addition to multiplication). Since clearly ψ( (g_{1}g_{2} − 1) − (g_{1} − 1) − (g_{2} − 1) ) = 0, ψ induces a homomorphism ψ̄: I/I^{2} → G/[G, G]. It is readily apparent that φ̄ and ψ̄ are inverse to each other. ✓

Note. The trick above which changes multiplication into addition is worth remembering. The homomorphism φ: g ↦ g − 1 could almost be thought of here as the algebraist’s logarithm It has other applications beyond the context of group rings.