Some Common Tor and Ext Groups

Abstract

We compute all the groups G ⊗ H, Tor(G, H), Hom(G, H), and Ext(G, H), where G and H can be any of the groups ℤ (the integers), ℤ/n = ℤ/nℤ (the integers mod n), or ℚ (the rationals). All but one are reasonably accessible. Because all these functors are biadditive, these cases suffice to handle any finitely generated groups G and H.

The emphasis here is on computation, not on the abstract definitions (which we don’t give).

From the symmetry of the definition, we obviously have

G ⊗ H ≅ H ⊗ G .  (1)

We write θ: G × H → G ⊗ H for the universal bilinear pairing. Then by the definition of G ⊗ H, any bilinear pairing f: G × H → K of abelian groups factors uniquely through a homomorphism \overline{f}: G ⊗ H → K. For any integer n, we have

f(ng, h) = nf(g, h) = f(g, nh)  (2)

and hence, in G ⊗ H,

(ng) ⊗ h = n(g ⊗ h) = g ⊗ (nh).  (3)

Although the definition of Tor(G, H) is not symmetric, it is true that Tor(G, H) ≅ Tor(H, G). This requires some proof; we shall refrain from using this fact, choosing always to compute Tor(G, H) as the derived functor of −⊗H, by using a free resolution of G.

Computing G ⊗ H and Tor(G, H) We begin with a triviality.

Proposition 4 For any group H, 0 ⊗ H = 0.

Proof From equation (3) with n = 0, we have

0 ⊗ h = (00) ⊗ h = 0(0 ⊗ h) = 0.

Proposition 5 For any H, we have ℤ ⊗ H ≅ H, with the universal pairing θ: ℤ × H → H given by n ⊗ h ↦ nh.

Proof This proof we give in full detail, as a pattern for other proofs.

Given a bilinear pairing f: ℤ × H → K, we need to show there is a unique homomorphism \overline{f}: H → K such that f = \overline{f} ^{∘} θ. This forces \overline{f}h = \overline{f}θ(1, h) = f(1, h); we therefore define \overline{f} by \overline{f}h = f(1, h) for all h. This is a homomorphism, because by bilinearity,

\overline{f}(h+h′) = f(1, h+h′) = f(1, h) + f(1, h′) = \overline{f}h + \overline{f}h′.

It satisfies \overline{f} ^{∘} θ = f, because for any integer n we have

\overline{f}θ(n, h) = \overline{f}(nh) = n\overline{f}h = nf(1, h) = f(n, h).

Proposition 6 We have ℚ ⊗ ℚ ≅ ℚ, with θ: ℚ × ℚ → ℚ given by θ(a × b) = ab.

Proof Given a bilinear pairing f: ℚ × ℚ → K, we are forced to define \overline{f}: ℚ → K by \overline{f}a = f(1, a), and this is a homomorphism. We have to check that it satisfies \overline{f} ^{∘} θ = f. Now \overline{f}θ(a, b) = \overline{f}(ab) = f(1, ab). To see that this agrees with f(a, b), we write a as a fraction m/n, with m and n integers; then by equation (2),

f(1, ab) = f (n\frac{1}{n}, ab) = f (\frac{1}{n}, nab) = f (\frac{1}{n}, mb) = f (\frac{m}{n}, b) = f(a, b).

(We warn that we may not be able to divide by n in the group K.)

This leaves ℤ/n ⊗ H. For this we use the usual free resolution

0 ⟶ ℤ ⟶^{n} ℤ ⟶ ℤ/n ⟶ 0,  (7)

in which the n denotes multiplication by n. We might as well compute Tor(ℤ/n, H) at the same time.

Proposition 8 For any H, we have:

(a) ℤ/n ⊗ H ≅ H/nH, where nH denotes the subgroup of all elements of H that are divisible by n;

(b) Tor(ℤ/n, H) ≅ _{n}H, the subgroup {h ∈ H : nh = 0} of elements of H of order n (or some divisor of n).

Proof When we tensor diagram (7) with H, the resulting long exact sequence simplifies by Proposition 5 to

0 ⟶ Tor(ℤ/n, H) ⟶ H ⟶^{n} H ⟶ ℤ/n ⊗ H ⟶ 0 .  (9)

We read off the kernel and cokernel of n: H → H.

Corollary 10 ℤ/n ⊗ ℚ = 0 and Tor(ℤ/n, ℚ) = 0.

In view of the symmetry (1), the only tensor product left is ℤ/n ⊗ ℤ/m.

Lemma 11 As subgroups of ℤ, we have nℤ+mℤ = dℤ, where d denotes the greatest common divisor gcd(n, m) of n and m.

Proof This is almost the definition of gcd(n, m).

Proposition 12 Let d = gcd(n, m). Then (a) ℤ/n ⊗ ℤ/m ≅ ℤ/d;

(b) Tor(ℤ/n, ℤ/m) ≅ ℤ/d.

Proof We apply Proposition 8. For (a) we get ℤ/(nℤ + mℤ), which Lemma 11 identifies. For (b) we need {i ∈ ℤ/m : ni = 0}. Write n = n′d and m = m′d, so that m′ and n′ are coprime; we need ni to be divisible by m, which reduces to having i divisible by m′. Thus the answer is m′ℤ/mℤ ≅ ℤ/d.

Now to finish off the remaining Tor groups. Because ℤ is projective, we have immediately

Tor(ℤ, H) = 0.  (13)

On the other hand, because − ⊗ ℤ is essentially the identity functor by Proposition 5 and therefore exact, we have

Tor(G, ℤ) = 0.  (14)

As a companion to Corollary 10 we have

Tor(ℚ, ℤ/n) = 0.  (15)

This can be done by a simple trick, by writing multiplication by n in Tor(ℚ, ℤ/n) in two different ways. First, we write it as Tor(n, ℤ/n), which is invertible with obvious inverse Tor(1/n, ℤ/n). Second, we write it as Tor(ℚ, n), which is plainly zero. (More generally, F (ℚ, ℤ/n) = 0 for any biadditive functor F, as in Corollary 10.)

This leaves only Tor(ℚ, ℚ). To handle this properly, we need a better description of G ⊗ ℚ.

Lemma 16 Let G be an abelian group.

(a) If G is a torsion group, G ⊗ ℚ = 0.

(b) If G is a torsion-free group, every element of G ⊗ ℚ has the form g ⊗ (1/n) for some g ∈ G and integer n ≠ 0, and g ⊗ (1/n) = g′ ⊗ (1/n′) if and only if n′g = ng′ in G.

Proof For (a), the only bilinear pairing f: G × ℚ → K is zero, because if mg = 0,

f(g, b) = f (g, m\frac{b}{m}) = f (mg, \frac{b}{m}) = f (0, \frac{b}{m}) = 0.

For (b), we take the set of all formal symbols g/n, where g ∈ G and n ∈ ℤ is nonzero, and impose the relation that g/n = g′/n′ if and only if n′g = ng′. This is an equivalence relation, because n′g = ng′ and n′′g′ = n′g′′ imply n′n′′g = n′ng′′, and hence n′′g = ng′′ (this is where we use the hypothesis that G is torsion-free).

We then define addition on the set E of equivalence classes by the usual rule for fractions,

\frac{g}{n} + \frac{g′}{n′} = \frac{n′g + ng′}{nn′} ,

and show it is well defined and makes E an abelian group. We define the bilinear pairing θ: G×ℚ → E by θ(g, m/n) = (mg)/n. Given a bilinear pairing f: G×ℚ → K, we must define \overline{f}: E → K by \overline{f}(g/n) = f(g, 1/n). This is a homomorphism because

\begin{aligned}\overline{f} (\frac{n′g + ng′}{nn′} ) & = f (n′g + ng′, \frac{1}{nn′} ) \\ & = f (n′g, \frac{1}{nn′} ) + f (ng′, \frac{1}{nn′} ) = f (g, \frac{1}{n}) + f (g′, \frac{1}{n′} ) . \end{aligned}

It satisfies \overline{f} ^{∘} θ = f because

\overline{f}θ (g, \frac{m}{n}) = \overline{f} (\frac{mg}{n} ) = f (mg, \frac{1}{n}) = f (g, \frac{m}{n}) .

Corollary 17 For any group G, we have Tor(G, ℚ) = 0.

Proof Take any free resolution

0 ⟶ F_{1} ⟶^{∂} F_{0} ⟶^{ϵ} G ⟶ 0  (18)

of G. The explicit description furnished by the Lemma shows that

∂ ⊗ ℚ: F_{1} ⊗ ℚ ⟶^{∂⊗ℚ} F_{0} ⊗ ℚ

is a monomorphism, because if x ∈ F_{1} is nonzero, so is (∂ ⊗ ℚ)x/n = ∂x/n. Computing Hom(G, H) and Ext(G, H) We start with the analogues of Proposition 5 and equation (13).

Proposition 19 For any group H we have Hom(ℤ, H) ≅ H and Ext(ℤ, H) = 0.

Proof Homomorphisms ω: ℤ → H correspond 1–1 to elements h ∈ H by h = ω1. Conversely, given h, we define ωn = nh. (This is almost one definition of ℤ.) The second statement is immediate because ℤ is projective.

Next we deal with G = ℤ/n, using the same free resolution (7) as before.

Proposition 20 For any group H we have:

(a) Hom(ℤ/n, H) ≅ _{n}H, the subgroup of H as in Proposition 8;

(b) Ext(ℤ/n, H) ≅ H/nH.

Proof When we apply the functor Hom(−, H) to equation (7) and use Proposition 19, we obtain the exact sequence

0 ⟶ Hom(ℤ/n, H) ⟶ H ⟶^{n} H ⟶ Ext(ℤ/n, H) ⟶ 0,

which has to be isomorphic to diagram (9). (Part (a) was obvious directly.)

This result allows us to read off all the groups Hom(ℤ/n, H) and Ext(ℤ/n, H) as in Proposition 12 etc.; several of them are obvious anyway.

Corollary 21 We have the following groups:

(a) Hom(ℤ/n, ℤ) = 0 and Ext(ℤ/n, ℤ) ≅ ℤ/n;

(b) Hom(ℤ/n, ℤ/m) ≅ ℤ/d and Ext(ℤ/n, ℤ/m) ≅ ℤ/d, where d = gcd(n, m);

(c) Hom(ℤ/n, ℚ) = 0 and Ext(ℤ/n, ℚ) = 0.

This leaves only the case G = ℚ.

Proposition 22 For the group ℚ we have:

(a) Hom(ℚ, ℚ) ≅ ℚ;

(b) Ext(G, ℚ) = 0 for any group G.

Proof Part (a) is easy enough: homomorphisms ω: ℚ → ℚ correspond 1–1 to elements h ∈ ℚ by h = ω1 and ωa = ah. These correspondences are inverse by the identities 1h = h and ωa = a(ω1). To see the second, we must write a = m/n and use

n(ωa) = nω (\frac{m}{n}) = ωm = m(ω1) = na(ω1) .

Because ℚ is torsion-free, we deduce ωa = a(ω1).

Part (b) is equivalent to the injectivity of ℚ. If we use the free resolution (18) of G, we have to show that Hom(F_{0}, ℚ) → Hom(F_{1}, ℚ) is surjective. (The fact that any divisible group is injective is standard, but not trivial.)

By the same trick as for equation (15), we have

Hom(ℚ, ℤ/n) = 0 and  Ext(ℚ, ℤ/n) = 0,  (23)

except that this time, direct proof of the second equation from the definitions is not so easy.

It is obvious that

Hom(ℚ, ℤ) = 0,  (24)

because no nonzero element of ℤ is divisible by n for all n. This leaves only one group to determine.

The group Ext(ℚ, ℤ) This subsection is strictly optional. The group Ext(ℚ, ℤ) is much more difficult to determine. It is easy to see that it is a rational vector space, simply from the presence of ℚ, but harder to see what its dimension is. This group is not as mysterious as is sometimes claimed, but is related to adèle groups familiar to number theorists. [The result is surely not new, but I don’t have a reference.]

Denote by ℚ_{p} the field of p-adic numbers, for each prime p, and by ℤ_{p} ⊂ ℚ_{p} the subring of p-adic integers (not to be confused with ℤ/p).

Theorem 25 We have Ext(ℚ, ℤ) ≅ A/ℚ, where A denotes the adèle group consisting of all sequences (x_{2}, x_{3}, x_{5}, x_{7}, …) of p-adic numbers x_{p} ∈ ℚ_{p} such that x_{p} ∈ ℤ_{p} for all except finitely many p, and ℚ ⊂ A denotes the subgroup of all sequences (x, x, x, …) with x ∈ ℚ. (Note to number theorists: A has no coordinate indexed by the reals ℝ.) It is thus an uncountable rational vector space.

We begin by applying the functor Hom(ℚ, −) to the short exact sequence

0 ⟶ ℤ ⟶ ℚ ⟶ ℚ/ℤ ⟶ 0,

to obtain the long exact sequence

Hom(ℚ, ℤ) ⟶ Hom(ℚ, ℚ) ⟶ A ⟶ Ext(ℚ, ℤ) ⟶ Ext(ℚ, ℚ),  (26)

where we write A = Hom(ℚ, ℚ/ℤ) and both end groups vanish. Since Hom(ℚ, ℚ) ≅ ℚ, we have only to identify A to establish the Theorem.

The torsion group ℚ/ℤ decomposes as ⨁_{p} ℤ/p^{∞}, where ℤ/p^{∞} is the well-known divisible group defined as ℤ[p^{−1}]/ℤ. Here, ℤ[p^{−1}] ⊂ ℚ denotes the subring of all rationals of the form m/p^{n}, with integers n ≥ 0 and m. We therefore study Hom(ℚ, ℤ/p^{∞}), which first requires Hom(ℤ/p^{∞}, ℤ/p^{∞}).

Lemma 27 We have the following descriptions of the p-adic numbers:

(a) The endomorphism ring End(ℤ/p^{∞}) = Hom(ℤ/p^{∞}, ℤ/p^{∞}) of ℤ/p^{∞} may be identified with the ring of p-adic integers ℤ_{p};

(b) There are isomorphisms of groups

Hom(ℚ, ℤ/p^{∞}) ≅ Hom(ℤ[p^{−1}], ℤ/p^{∞}) ≅ ℚ_{p},

where ω: ℚ → ℤ/p^{∞} corresponds to an element of ℤ_{p} ⊂ ℚ_{p} if and only if ω1 = 0.

Proof In (a), the group ℤ/p^{∞} is the union of the cyclic subgroups ℤ/p^{n} generated by 1/p^{n}, for n > 0. Any homomorphism ω: ℤ/p^{∞} → ℤ/p^{∞} must map ℤ/p^{n} into itself; thus the endomorphism ring End(ℤ/p^{∞}) is the limit lim_{n} End(ℤ/p^{n}) of the endomorphism rings End(ℤ/p^{n}). By Corollary 21, End(ℤ/p^{n}) ≅ ℤ/p^{n} as a ring, and we may therefore identify the limit with ℤ_{p}.

Because ℤ/p^{∞} has unique division by any integer m prime to p, every homomorphism ℤ[p^{−1}] → ℤ/p^{∞} extends uniquely to a homomorphism ℚ → ℤ/p^{∞}; hence the first isomorphism in (b). Any ω: ℤ[p^{−1}] → ℤ/p^{∞} must satisfy p^{n}(ω1) = 0 for some n, so that (ωp^{n})ℤ = 0; then ω^{∘} p^{n} factors through ℤ/p^{∞} = ℤ[p^{−1}]/ℤ. This gives enough information to identify the inclusion End(ℤ/p^{∞}) ⊂ Hom(ℤ[p^{−1}], ℤ/p^{∞}) with ℤ_{p} ⊂ ℚ_{p}.

Proof of Theorem 25 Instead of dealing with Hom(ℚ, ℚ/ℤ) directly, we first embed

ℚ/ℤ = ⨁_{p} ℤ/p^{∞} ⊂ ∏_{p} ℤ/p^{∞},

and write

A = Hom(ℚ, ℚ/ℤ) ⊂ Hom(ℚ, ∏_{p} ℤ/p^{∞}) ≅ ∏_{p} ℚ_{p},

with the help of Lemma 27. Then A consists of those homomorphisms ω: ℚ → ℚ/ℤ whose coordinates ω_{p}: ℚ → ℤ/p^{∞} satisfy ω_{p}1 = 0 for all except finitely many p. Thus A is as described. Finally, we note that the subgroup Hom(ℚ, ℚ) ≅ ℚ of A given by diagram (26) is as indicated.