Exercise B1 (b): Find a free resolution for ℤ/2 as a ℤ/8-module.

Solution: We construct the desired resolution. It is not difficult to see that

ℤ/8 →^{π_{1}} ℤ/2 → 0

where π_{1} is the natural projection \overline{x} ↦ \overline{x} is exact at ℤ/2. The kernel of this map is the set \{\overline{0}, \overline{2}, \overline{4}, \overline{6}\}, or equivalently, ker(π_{1}) = ⟨\overline{2}⟩ ⊆ ℤ/8. Thus, we have a short exact sequence

0 → ⟨\overline{2}⟩ → ℤ/8 →^{π_{1}} ℤ/2 → 0. (1)

We observe that ⟨\overline{2}⟩ ≅ ℤ/4, and since ℤ/4 is not a free ℤ/8 module, we need to extend the resolution. Consider the map

ℤ/8 →^{×\overline{2}} ℤ/8

where ×\overline{2} is the map \overline{x} ↦ \overline{2x}. The image of this map is ⟨\overline{2}⟩ = \{\overline{0}, \overline{2}, \overline{4}, \overline{6}\}, which is just the kernel of π_{1}. So we get an exact sequence

ℤ/8 →^{×\overline{2}} ℤ/8 →^{π_{1}} ℤ/2 → 0

The kernel of ×\overline{2} is the ideal ⟨\overline{4}⟩ ⊆ ℤ/8, and ⟨\overline{4}⟩ ≅ ℤ/2 which is not a free ℤ/8 module. Thus we again need to extend the resolution

ℤ/8 →^{×\overline{4}} ℤ/8 →^{×\overline{2}} ℤ/8 →^{π_{1}} ℤ/2 → 0 (2)

The kernel of ×\overline{4} is the same as ker π_{1} so we must repeat the above constructions, obtaining an infinite free resolution of ℤ/2 as a ℤ/8-module given by

⋯ → ℤ/8 →^{×\overline{2}} ℤ/8 →^{×\overline{4}} ℤ/8 →^{×\overline{2}} ℤ/8 →^{π_{1}} ℤ/2 → 0.

(with ×\overline{4} and ×\overline{2} alternating).