# $A\otimes \mathbf{Q}\cong B\otimes \mathbf{Q}$ implies $A$ and $B$ are $\mathcal{C}$-isomorphic

I am trying to solve exercise 211 on Davis-Kirk:

Let $$\mathcal{C}$$ be the class of torsion abelian groups. Show that for any abelian groups $$A,~B$$, $$A\otimes \mathbf{Q}\cong B\otimes \mathbf{Q}$$ implies $$A$$ and $$B$$ are $$\mathcal{C}$$-isomorphic, that is,

there exists an abelian group $$C$$ and group homomorphisms $$f:C\to A$$, $$g:C\to B$$ such that $$\ker f,\mathrm{coker} f$$, $$\ker g,\mathrm{coker} g$$ are abelian torsion groups.

My attempt:

Consider the exact sequence $$0\to\mathbf{Z}\to \mathbf{Q}\to \mathbf{Q}/\mathbf{Z} \to 0$$ and the indcued exact sequence $$0\to \mathrm{Tor}(\mathbf{Q}/\mathbf{Z},A)\to A\to \mathbf{Q}\otimes A \to \mathbf{Q}/\mathbf{Z} \otimes A \to 0$$

It is tempting to take the map $$A\to \mathbf{Q}\otimes A$$ whose kernel and cokernel are all torsion groups. But the definition ask us to find a map whose target is $$A$$, not the source. Moreover, we cannot conclude $$\mathrm{Tor}(\mathbf{Q}/\mathbf{Z},A)\cong \mathrm{Tor}(\mathbf{Q}/\mathbf{Z},B)$$.

Besides from this "canonical map", I have no idea how to construct those two maps. Any hints and answer are welcome!

• Related. – Shaun Apr 24 at 10:54
• Are you assuming $A$ and $B$ to be finitely generated abelian groups? – darko Apr 24 at 11:51
• @JyrkiLahtonen They are arbitrary abelian groups. – Aolong Li Apr 24 at 11:53
• @darko abelian groups, not necessarily finitely generated. – Aolong Li Apr 24 at 11:55
• Thanks for the clarification. – Jyrki Lahtonen Apr 28 at 21:02

Let $$D$$ denote both $$A\otimes \mathbb{Q}$$ and $$B\otimes \mathbb{Q}$$, and let $$f: A\to D, g:B\to D$$ denote the canonical morphisms.

Let $$p:C \to A, q:C\to B$$ be the pullback of $$f$$ along $$g$$, that is, concretely $$C=\{(a,b)\in A\times B \mid f(a)=g(b)\}$$.

Now let $$x\in \ker p$$. Then $$x=(0,b)$$ for some $$b$$ such that $$g(b) =0$$. But then $$b$$ is torsion, thus so is $$x$$. Symmetrically, we show that $$\ker q$$ is torsion.

Let's now look at the cokernel : let $$a\in A$$. We wish to find $$b\in B$$ such that $$g(b) = f(a)$$. Of course it isn't always possible, but it's possible up to some torsion.

Indeed $$f(a) = a\otimes 1 = b\otimes \frac{1}{q}$$ for some $$b\in B, q\in \mathbb{N}_{>0}$$. Therefore $$qf(a) = b\otimes 1$$ so that $$(qa,b) \in C$$ and therefore $$qa\in \mathrm{im}\; p$$, so that $$a$$ is torsion in $$\mathrm{coker}\; p$$. Symmetrically, we show that $$\mathrm{coker}\; q$$ is torsion.

Therefore both $$p,q$$ are $$\mathcal{C}$$-isomorphisms, and $$A,B$$ are $$\mathcal{C}$$-isomorphic.

You can have fun by generalizing that and seeing that $$\mathcal{C}$$-isomorphism behaves nicely.

• Perfect answer! Thanks! Could you explain why do you come up with the pull-back construction? I will never think of it before I see your answer... – Aolong Li Apr 25 at 1:10
• @AolongLi : there is probably more theory behind it, but I don't know enough so I'll just tell how I came up with it, not how anyone should : I was looking for a $C$ with arrows to $A, B$, but, as you mentioned we had arrows from $A$ and $B$ to their $\otimes \mathbb Q$;but then I realized these were the same, so instead of a "cocone" going towards $A,B$, I had a cone leaving from them. Now I only know one way to "reverse" this and it's to take the pullback – Max Apr 25 at 6:51
• @AolongLi : I actually just found the "theory" behind it : $A$ is $\mathcal{C}$-isomorphic to $A\otimes \mathbb{Q}$ which is isomorphic (and thus $\mathcal{C}$- isomorphic) to $B\otimes \mathbb{Q}$ which is $\mathcal C$-isomorphic to $B$. Now just apply the lemma that $\mathcal C$-isomorphism is transitive (which, in its proof, uses the pullback construction) – Max Apr 25 at 8:59
• That makes sense!!! Thanks a lot!! – Aolong Li Apr 25 at 11:00