Susan Marie Cooper

Monday November 13, 2000

Exercise B23 - Weibel, page 74 - Exercise 3.3.2: When R = ℤ/m and B = ℤ/p with p | m, show that

0 → ℤ/p ↪^{i} ℤ/m →^{p} ℤ/m →^{m/p} ℤ/m →^{p} ℤ/m →^{m/p} ⋯

is an infinite periodic injective resolution of B.

Then compute the groups Ext_{ℤ/m}^{n}(A, ℤ/p) in terms of A^{∗} =Hom(A, ℤ/m). In particular, show that if p^{2} | m, then Ext_{ℤ/m}^{n}(ℤ/p, ℤ/p) ≅ ℤ/p for all n.

Proof. Let us prove that the given sequence is an infinite periodic injective resolution of B. Clearly the sequence is infinite and periodic. Thus, we need to show that ℤ/m is an injective R-module and that the sequence is exact. The fact that ℤ/m is an injective R-module is Exercise 2.3.1 of Weibel (page 39).

We now show that the sequence is exact. For this we let p^{r}c = m, where r ≥ 1 and gcd(c, p) = 1. To begin we show that i is an inclusion. To see this it is best to think of the map i as multiplication by m/p. Then take two different cosets a + pℤ, b + pℤ. Since a + pℤ ≠ b + pℤ, a − b ∉ pℤ. Now i(a + pℤ) = (m/p)(a + pℤ) = ma/p + mℤ = p^{r−1}ca + mℤ, and similarly i(b + pℤ) = mb/p + mℤ = p^{r−1}cb + mℤ. If we can show that ma/p + mℤ ≠ mb/p + mℤ we will have shown that i is an inclusion map. Now ma/p + mℤ = mb/p + mℤ ⇔ p^{r−1}ca − p^{r−1}cb ∈ mℤ ⇔ p^{r−1}c(a − b) ∈ mℤ ⇔ p | a − b. This contradicts the assumption that a+pℤ ≠ b+pℤ. Thus, ma/p+mℤ ≠ mb/p+mℤ, and so i is an inclusion.

We now show exactness at ℤ/p →^{i} ℤ/m →^{p} ℤ/m. To see this let b+mℤ ∈ im(i). Let a ∈ ℤ be such that b+mℤ = i(a+pℤ) = (m/p)(a+pℤ) = ma/p+mℤ = p^{r−1}ca+mℤ. Thus (p)(b + mℤ) = (p)(p^{r−1}ca + mℤ) = p^{r}ca + mℤ = ma + mℤ = 0 + mℤ. Thus, b + mℤ ∈ ker(p). Now note that the composition p ∘ i is the zero map, and that if c + mℤ ∈ ker(p), then pc would be a multiple of m. Thus, m/p divides c, and hence, c + mℤ ∈ im(i). Therefore, ker(p) = im(i).

Now consider a portion

ℤ/m →^{p} ℤ/m →^{m/p} ℤ/m.

Then take b + mℤ ∈ im(p). Let a ∈ ℤ be such that p(a + mℤ) = ap + mℤ = b + mℤ. Then (m/p)(b + mℤ) = (m/p)(ap + mℤ) = ma + mℤ = 0 + mℤ. Therefore, b + mℤ ∈ ker(m/p). Now note that the composition (m/p) ∘ p is the zero map and that if c + mℤ ∈ ker(m/p), then (m/p)(c) would be a multiple of m. Thus, p | c, and hence, c + mℤ ∈ im(p). Therefore, ker(m/p) = im(p).

Finally, consider the portion

ℤ/m →^{m/p} ℤ/m →^{p} ℤ/m.

Take c + mℤ ∈ im(m/p). Take a ∈ ℤ such that (m/p)(a + mℤ) = (ma/p) + mℤ = c + mℤ. Then p(c + mℤ) = p((ma/p) + mℤ) = ma + mℤ = 0 ∈ ℤ/mℤ. Thus, c + mℤ ∈ ker(p). Now note that the composition p ∘ (m/p) is the zero map and that if b + mℤ ∈ ker(p), then pb is a multiple of m. Thus, (m/p) | b, and hence b + mℤ ∈ im(m/p). This finishes the proof that the given sequence is exact.

So we work with the complex

0 → ℤ/m →^{p} ℤ/m →^{m/p} ℤ/m →^{p} ℤ/m →^{m/p} ⋯

Applying Hom_{ℤ/m}(A, −) to this we get the following complex

0 → Hom_{ℤ/m}(A, ℤ/m) →^{p_{∗}} Hom_{ℤ/m}(A, ℤ/m) →^{(m/p)_{∗}} Hom_{ℤ/m}(A, ℤ/m) →^{p_{∗}}

Hom_{ℤ/m}(A, ℤ/m) →^{(m/p)_{∗}} ⋯

Or, equivalently,

0 → A^{∗} →^{p_{∗}} A^{∗} →^{(m/p)_{∗}} A^{∗} →^{p_{∗}} A^{∗} →^{(m/p)_{∗}} ⋯

By definition, Ext_{ℤ/m}^{n}(A, ℤ/p) =ker(m/p)_{∗}/im(p_{∗}) if n > 0 is odd, if n = 0, then Ext_{ℤ/m}^{n}(A, ℤ/p) =ker(p_{∗}), and Ext_{ℤ/m}^{n}(A, ℤ/p) =ker(p_{∗})/im((m/p)_{∗}) if n > 0 is even.

Now suppose p^{2} | m and let A = ℤ/p. If p^{2} | m, then we can write m = p^{a}n where a ≥ 2 and gcd(p, n) = 1. Thus, ℤ/m ≅ ℤ/p^{a} ⊕ ℤ/n. Hence, Hom_{ℤ/m}(ℤ/p, ℤ/m) ≅ Hom_{ℤ/m}(ℤ/p, ℤ/p^{a} ⊕ ℤ/n) ≅ Hom_{ℤ/m}(ℤ/p, ℤ/p^{a})⊕ Hom_{ℤ/m}(ℤ/p, ℤ/n).

Take Γ ∈ Hom_{ℤ/m}(ℤ/p, ℤ/n). Then pΓ(1 + pℤ) = Γ(p + pℤ) = Γ(0 + pℤ) = 0 + nℤ. Thus, if Γ(1 + pℤ) = a + nℤ, then pa ∈ nℤ. So let pa = nr, where r ∈ ℤ. Since gcd(n, p) = 1 this implies that n | a. But then Γ(1) = 0 + nℤ. Therefore, Hom_{ℤ/m}(ℤ/p, ℤ/n) ≅ 0. Hence, Hom_{ℤ/m}(ℤ/p, ℤ/m) ≅ Hom_{ℤ/m}(ℤ/p, ℤ/p^{a}). But f ∈ Hom_{ℤ/m}(ℤ/p, ℤ/p^{a}) is identified with its value at 1 ∈ ℤ/p and in order for f(1) to be of order p, f(1) must be a multiple of p^{a−1}. Thus, Hom_{ℤ/m}(ℤ/p, ℤ/p^{a}) ≅ ℤ/p, and so Hom_{ℤ/m}(ℤ/p, ℤ/m) ≅ Hom_{ℤ/m}(ℤ/p, ℤ/p^{a}) ≅ ℤ/p.

Hence,

0 → A^{∗} →^{p_{∗}} A^{∗} →^{(m/p)_{∗}} A^{∗} →^{p_{∗}} A^{∗} →^{(m/p)_{∗}} ⋯

is equivalent to

0 → ℤ/p →^{p} ℤ/p →^{p^{a−1}n} ℤ/p →^{p} ℤ/p →^{p^{a−1}n} ⋯

Thus, Ext_{ℤ/m}^{0}(ℤ/p, ℤ/p) ≅ker(p) = ℤ/p.

If n > 0 is even, then Ext_{ℤ/m}^{n}(A, ℤ/p) ≅ker(p)/im(p^{a−1}n) = (ℤ/p)/(0) = ℤ/p. If n > 0 is odd, then Ext_{ℤ/m}^{n}(ℤ/p, ℤ/p) =ker(p^{a−1}n)/im(p) = (ℤ/p)/(0) = ℤ/p. (Note: all of these are easily seen with the observation that a ≥ 2, and hence

a − 1 ≥ 1.)