Application of Kulikov's theorem

Question

I follow the Rotman's textbook "Introduction to homological algebra". This is the exercise 3.43 in the book.

Recall Kulikov's theorem: if $G$ is a $p$-primary abelian group, then there exists a pure exact sequence $0\rightarrow B\rightarrow G\rightarrow D\rightarrow 0$ with $B$ a direct sum of cyclic groups and $D$ divisible.

By using this theorem, I want to prove that if $H$ and $K$ are torsion abelian groups, prove that $H\otimes_{\mathbb{Z}}K$ is a direct sum of cyclic groups.

However I don't know how to use this theorem.

Answer 1

Let me try to provide a sketch of the general answer. First of all, assuming that $G, G'$ are two torsion abelian groups you know that they decompose as direct sums of their $p$-primary components. Since tensor products satisfy a general property of distributivity with respect to direct sums, in order to prove your general claim it thus suffices to prove it in the special case when $G, G'$ are both primary.

Thus your claim is reduced to the particular case when $G$ is $p$-primary and $G'$ is $q$-primary for two primes $p, q$. Then, by Kulikov's theorem you can infer the existence of abelian groups $H, K, D, D'$ such that:

• there exist sequences $$\{0\} \to H \to G \to D \to \{0\} \tag{1}$$ $$\{0\} \to H' \to G' \to D' \to \{0\} \tag{2}$$ which are pure exact
• $H, H'$ are direct sums of cyclic groups (and are hence torsion groups)
• $D, D'$ are divisible groups (and since they are quotients of $G$ and respectively $G'$ they are furthermore torsion groups as well).

By tensoring the pure sequence (1) with $G'$ you obtain a new short exact sequence

$$ \{0\} \to H \otimes_{\mathbb{Z}} G' \to G \otimes_{\mathbb{Z}} G' \to D \otimes_{\mathbb{Z}} G' \to \{0\} \tag{3}$$

Bearing in mind that the tensor product between a divisible group and a torsion group is null, from sequence (3) you can infer that $H \otimes_{\mathbb{Z}} G' \approx G \otimes_{\mathbb{Z}} G'$. In order to analyze this tensor product we now tensor the pure exact sequence (2) with $H$, obtaining:

$$\{0\} \to H \otimes_{\mathbb{Z}} H' \to H \otimes_{\mathbb{Z}} G' \to H \otimes_{\mathbb{Z}} D' \to \{0\} \tag{4}$$

By the same token as above, $H \otimes_{\mathbb{Z}} D'$ vanishes leading you to the conclusion that $$H \otimes_{\mathbb{Z}} H' \approx H \otimes_{\mathbb{Z}} G' \approx G \otimes_{\mathbb{Z}} G' \tag{5}$$

By virtue of the universal distributivity of the tensor product with respect to direct sums, $H \otimes_{\mathbb{Z}} H'$ can be developed into the direct sum of tensor products of pairs of cyclic groups, and it is well-known that these latter objects are themselves cyclic groups (in general one has $\mathbb{Z}_m \otimes_{\mathbb{Z}} \mathbb{Z}_n \approx \mathbb{Z}_{(m,n)}$ for any $m, n \in \mathbb{N}$). Hence your claim.