Exercise 3.3.1 By definition, ℤ_{p^{∞}} = ℤ[\frac{1}{p}]/ℤ (example 2.3.3). So we have an exact sequence,

0 ⟶ ℤ ⟶ ℤ[\frac{1}{p}] ⟶ ℤ_{p^{∞}} ⟶ 0

This induces a long exact sequence: 0 ⟶ Ext^{0}(ℤ_{p^{∞}}, ℤ) ⟶ Ext^{0}(ℤ[\frac{1}{p}], ℤ) ⟶ Ext^{0}(ℤ, ℤ) ⟶ Ext^{1}(ℤ_{p^{∞}}, ℤ) ⟶

Ext^{1}(ℤ[\frac{1}{p}], ℤ) ⟶ Ext^{1}(ℤ, ℤ) ⟶ … But Ext^{1}(ℤ, ℤ) = 0, since ℤ is a projective (in fact free) ℤ-module. Also, Ext^{0}(ℤ[\frac{1}{p}], ℤ) = Hom_{ℤ}(ℤ[\frac{1}{p}], ℤ). I claim that this is 0. Let ϕ ∈ Hom_{ℤ}(ℤ[\frac{1}{p}], ℤ). Suppose that ϕ(1) =

a ∈ ℤ. Then ϕ(\frac{1}{p^{n}}) = \frac{a}{p^{n}} ∈ ℤ for all n.

Thus, a = 0. So ϕ = 0. So we have a short exact sequence:

0 ⟶ Ext^{0}(ℤ, ℤ) ⟶ Ext^{1}(ℤ_{p^{∞}}, ℤ) ⟶ Ext^{1}(ℤ[\frac{1}{p}], ℤ) ⟶ 0.

But Ext^{0}(ℤ, ℤ) = Hom_{ℤ}(ℤ, ℤ) ≃ ℤ, and Ext^{1}(ℤ_{p^{∞}}, ℤ) = ℤ̂_{p}, from Example 3.3.3. Thus, Ext^{1}(ℤ[\frac{1}{p}], ℤ) ≃ ℤ̂_{p}/ℤ, as required.