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I am reading the book An Introduction of Homological Algebra by Rotman. I'm disturbed by Proposition 7.33 on page 427, which claim that

If $F$ is a torsion-free abelian group and $T$ is a group of bounded order (i.e. $nT=\{0\}$ for some positive integer $n$), then $Ext^1(F,T)=\{0\}$.

I can't understand the proof given by Rotman. Here is the skretch of Rotman.

Since $F$ is torsion-free, then there is a short exact sequence $$0\to F\to V \to V/F\to0,$$ where $V$ is a vector space over $\mathbb{Q}$.

We can get an exact sequence $$Ext^1（V,T）\to Ext^1（F,T）\to Ext^2（V/F,T）.$$ The last term is zero, we can obtain $Ext^1（V,T）$ is divisible.

There is no problem for me at present. But then, Rotman wrote that the multiplication $\mu_n:\mathbb{Q}\to\mathbb{Q}$ is an isomorphism, and so the induced map $\mu_n^*:Ext^1（F,T）\to Ext^1（F,T）$is an isomorphism. This sentence make me puzzled. I think $\mu_n$ should be $\mu_n:F\to F$, but this new $\mu_n$ may be not an isomorphim.

Could someone give me a explanation of this proposition?

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liwolf
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1 Answers1

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I have not here a copy of Rotman's, but in the diagram (the vertical maps induced by multiplication by $n$) $$\require{AMScd} \begin{CD} Ext^1（V,T） @>>> Ext^1（F,T） @>>> Ext^2（V/F,T）=0\\ @VVV @VVV \\ Ext^1（V,T） @>>> Ext^1（F,T） @>>> Ext^2（V/F,T）=0 \end{CD}$$ the vertical map on the left is an isomorphism, so the middle vertical map is surjective but this map is zero since $nT=0$.

SlavaM
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