Exercise 3A, Page 16 of Tony’s notes

i) Prove that ℤ/nℤ is not a projective ℤ-module

Proof: There is an obvious map, f, from ℤ to ℤ/nℤ, given by sending a ∈ ℤ to ā ∈ ℤ/nℤ. This map is certainly onto, so

ℤ →^{f} ℤ/nℤ → 0

is exact.

If ℤ/nℤ is a projective ℤ module, then there exists a homomorphism g : ℤ/nℤ → ℤ such that f ∘ g = 1_{ℤ/nℤ}.

It suffices for us to show that g must be the zero map, since this forces the composition f ∘ g to be zero as well, and the identity map on ℤ/nℤ is certainly not the zero map.

Let g be any homomorphism from ℤ/nℤ to ℤ. Then g(¯0) = 0. However, n̄ = ¯0 in ℤ/nℤ. Thus, we have 0 = g(¯0) = g(n̄) = g(n · ¯1) = ng(¯1). Since n ≠ 0, g(¯1) = 0. g is a homomorphism, so g(ā) = ag(¯1) = a · 0 = 0. Thus, g is the zero map, and ℤ/nℤ is not a projective ℤ-module. ⊓ ⊔

ii) Show that finitely generated projective ℤ-modules are free.

Proof: Let M be a finitely generated projective ℤ-module. By the structure theorem,

M ≅ ℤ ⊕ … ⊕ ℤ ⊕ ℤ/p_{1}^{a_{1}}ℤ ⊕ … ⊕ ℤ/p_{k}^{a_{k}}ℤ.

Since M is projective, all of the summands above must be projective. However, in part (i) we showed that ℤ/nℤ is not projective. Thus,

M ≅ ⊕_{l}ℤ

and hence is free, by the proposition on page 12 of Tony’s notes. ⊓ ⊔

iii) Show that there exist finitely generated projective ℤ/nℤ-modules that are not free, if n is not a prime power.

Proof: If n is not a prime power, then n = n_{1}n_{2}, where gcd(n_{1}, n_{2}) = 1. Then, by applying the Chinese Remainder theorem, it is easy to show we have the following isomorphism:

ℤ/nℤ ≅ ℤ/n_{1}ℤ ⊕ ℤ/n_{2}ℤ.

Certainly ℤ/nℤ is a free ℤ/nℤ-module, so ℤ/n_{1}ℤ is a projective ℤ/nℤmodule. It is finitely generated (by the element ¯1), but it is not free, since ¯0 = n_{1} · ¯1, where n_{1} is an element of ℤ/nℤ. ⊓ ⊔