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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!

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

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  • $\begingroup$ 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... $\endgroup$ – Aolong Li Apr 25 at 1:10
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    $\begingroup$ @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 $\endgroup$ – Max Apr 25 at 6:51
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    $\begingroup$ @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) $\endgroup$ – Max Apr 25 at 8:59
  • $\begingroup$ That makes sense!!! Thanks a lot!! $\endgroup$ – Aolong Li Apr 25 at 11:00

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