Exercise C7

Let p and q are two arbitrary integers in ℕ.

(a) For the ℤ-module ℤ we have the injective resolution

0 ⟶ ℤ ⟶ ℚ ⟶ ℚ/ℤ ⟶ 0

since ℚ and ℚ/ℤ are injective ℤ-modules (because divisible). For the ℤmodule ℤ/pℤ we have the injective resolution

0 ⟶ ℤ/pℤ ⟶ ℚ/pℤ ⟶ ℚ/ℤ ⟶ 0

with ℚ/pℤ injective since divisible, and the quotient (ℚ/pℤ)/(ℤ/pℤ) ≅ ℚ/ℤ by the second isomorphism theorem.

(b) This injective resolution gives us the chain complex

C_{•} 0 ⟶ ℚ/pℤ ⟶^{d_{1}} ℚ/ℤ ⟶ 0 .

Applying the functor Hom_{ℤ}(ℤ/qℤ, −) to this complex gives the chain complex

D_{•} 0 ⟶ Hom_{ℤ}(ℤ/qℤ, ℚ/pℤ) ⟶^{d_{1}^{∗}} Hom_{ℤ}(ℤ/qℤ, ℚ/ℤ) ⟶ 0 .

We then have

Ext_{ℤ}^{1}(ℤ/qℤ, ℤ/pℤ) = H^{1}(D_{•}) = Hom_{ℤ}(ℤ/qℤ, ℚ/ℤ)/Imd_{1}^{∗}
Ext_{ℤ}^{i}(ℤ/qℤ, ℤ/pℤ) = H^{i}(D_{•}) = 0 for i ≥ 2.

First we claim that we have an isomorphism Hom_{ℤ}(ℤ/qℤ, ℚ/ℤ) ≅ ℤ/qℤ. Proof. Given the short exact sequence 0 ⟶ ℤ ⟶^{q} ℤ ⟶ ℤ/qℤ ⟶ 0, the contravariant functor Hom_{ℤ}(−, ℚ/ℤ) gives a short exact sequence

0 ⟶ Hom_{ℤ}(ℤ/qℤ, ℚ/ℤ) ⟶ \underbrace{Hom_{ℤ}(ℤ, ℚ/ℤ)}_{≅ℚ/ℤ} ⟶^{q} \underbrace{Hom_{ℤ}(ℤ, ℚ/ℤ)}_{≅ℚ/ℤ}

hence Hom_{ℤ}(ℤ/qℤ, ℚ/ℤ) ≅ ker(ℚ/ℤ ⟶^{q} ℚ/ℤ) = ker(q) and we have ker(q) ≅ ℤ/qℤ via the homomorphism ℤ/qℤ ⟶^{f} ker(q), f(1) = [\frac{1}{q}]. Note that f is well defined because x ≡ y mod q gives f(y) = [\frac{x+kq}{q}] = [\frac{x}{q} +k] = [\frac{x}{q}] = f(x). f is injective because [\frac{x}{q}] = 0 ⇒ \frac{x}{q} ∈ ℤ ⇒ q|x ⇒ x ≡ 0 mod q. f is clearly surjective. □ Now we shall show that Imd_{1}^{∗} = p(ℤ/qℤ). Let f ∈ Hom_{ℤ}(ℤ/qℤ, ℚ/pℤ), then we have [0] = f(\overline{q}) = qf(\overline{1}) ⇒ qf(1) ∈ pℤ ⇒ f(\overline{1}) = [\frac{p}{q}] = p[\frac{1}{q}] in ℚ/pℤ. Therefore d_{1}f(\overline{1}) = p[\frac{1}{q}] in ℚ/ℤ. But by the previous discussion, [\frac{1}{q}] corresponds to the generator 1 of ℤ/qℤ in the isomorphism Hom_{ℤ}(ℤ/qℤ, ℚ/ℤ) ≅ ℤ/qℤ, so we have Imd_{1}^{∗} = p(ℤ/qℤ). Finally we have

Ext_{ℤ}^{1}(ℤ/qℤ, ℤ/pℤ) ≅ ℤ/qℤ/p(ℤ/qℤ) ≅ ℤ/(p, q)ℤ

where (p, q) is the gcd of p and q. The result in Weibel (p.73) corresponds to this one since it says that for every ℤ-module B, we have Ext_{ℤ}^{1}(ℤ/qℤ, B) ≅ B/qB, taking B = ℤ/pℤ gives our result.