Susan Marie Cooper

Wednesday October 11, 2000

Exercise B4 - Tony’s Notes, Exercise 4C, page 20:

1) No finitely generated abelian group is a divisible ℤ-module.

Proof. Let M be a finitely generated abelian group. Since the finitely generated group 0 is divisible, we assume that M ≠ 0. Recall that

M ≅ ℤ ⊕ ⋯ ⊕ ℤ ⊕ ℤ/(p_{1}^{a_{1}}) ⊕ ℤ/(p_{2}^{a_{2}}) ⊕ ⋯ ⊕ ℤ/(p_{k}^{a_{k}}),

where the p_{i} are primes (not necessarily distinct), a_{i} ≥ 1, and the number of summands of just ℤ = r ≥ 0. Moreover, this decomposition is unique up to the order of the summands.

First, I claim that ℤ is not a divisible ℤ-module. To see this take r = 6 ≠ 0 in ℤ and d = 3 ∈ ℤ. Note that there is no element c in ℤ such that rc = 6c = 3 = d.

Second, I claim that for any prime p ∈ ℤ, the ℤ-module ℤ/(p^{a}), for a ≥ 1, is not a divisible ℤ-module. To see this take r = p^{a} ≠ 0 in ℤ and \overline{b} ∈ ℤ/(p^{a}) such that \overline{b} ≠ \overline{0}. Then note that multiplying anything in ℤ/(p^{a}) by r gives the zero coset. Thus, there is no element \overline{c} ∈ ℤ/(p^{a}) such that r(\overline{c}) = \overline{b}.

To prove the claim of the exercise recall that, by Exercise B5 (Tony’s Notes, Exercise 4D, page 22), every quotient of a divisible ℤ-module is again a divisible ℤmodule. Suppose that M is a divisible ℤ-module. We now first suppose that M has at least one summand ℤ, say the first summand. Consider the quotient M/S, where S = {0} ⊕ ℤ ⊕ ℤ ⊕ ⋯ ⊕ ℤ ⊕ ℤ/(p_{1}^{a_{1}}) ⊕ ℤ/(p_{2}^{a_{2}}) ⊕ ⋯ ⊕ ℤ/(p_{k}^{a_{k}}). Then M/S is clearly isomorphic to ℤ. But we have already seen that ℤ is not a divisible ℤ-module. Thus, M/S cannot be a divisible ℤ-module. However, this contradicts Exercise B5. Hence, if M has at least one summand of ℤ, then M cannot be a divisible ℤ-module.

So now suppose that M does not have any summands equal to ℤ and that M is a divisible ℤ-module. That is, suppose M = ℤ/(p_{1}^{a_{1}}) ⊕ ℤ/(p_{2}^{a_{2}}) ⊕ ⋯ ⊕ ℤ/(p_{k}^{a_{k}}). Consider the quotient M/P, where P = \{\overline{0}\} ⊕ ℤ/(p_{2}^{a_{2}}) ⊕ ⋯ ⊕ ℤ/(p_{k}^{a_{k}}). Then M/P is clearly isomorphic to ℤ/(p_{1}^{a_{1}}). But we have already seen that ℤ/(p_{1}^{a_{1}}) is not a divisible ℤ-module. Thus, M/P cannot be a divisible ℤ-module. However, this contradicts Exercise B5. Hence, if M has no ℤ summands, then M cannot be a divisible ℤmodule.

Therefore, no finitely generated non-zero abelian group is a divisible ℤ-module.

2) The direct product of divisible R-modules is again divisible.

Proof. Let R be an integral domain. Consider the R-module M = ∏_{i∈I} M_{i}, where I is some index set and each M_{i} is a divisible R-module. We want to show that M is a divisible R-module. Since each M_{i} is a divisble R-module we have, by definition, that for all n_{i} ∈ M_{i} and for all s ≠ 0, s ∈ R, there is a d_{i} ∈ M_{i} such that sd_{i} = n_{i}, for each M_{i}. Now take r ≠ 0, r ∈ R to be arbitrary. Also let m = (m_{i})_{i∈I} ∈ M be arbitrary. Define c = (c_{i})_{i∈I} ∈ M to be the element in M such that rc_{i} = m_{i} for all i (we know c exists by the above observation). Then we have that rc = (rc_{i})_{i∈I} = (m_{i})_{i∈I} = m. Since r ∈ R and m ∈ M were chosen arbitrarily, we see that M is a divisible Rmodule. Therefore, the direct product of divisible R-modules is again divisible.