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So I understand in theory the definition of Ext and Tor, but when it comes to actually computing them, I'm stuck. For example, could someone show me how to compute $\text{Ext}(\mathbb{Z}/m\mathbb{Z}, \mathbb{Z}/n\mathbb{Z})$? I tried this by taking an injective resolution ($0 \to \mathbb{Z}/m\mathbb{Z} \to \mathbb{Q}/\mathbb{Z} \to \mathbb{Q}/\mathbb{Z} \to 0$?) and I got an exact sequence with $\text{Ext}^1(\mathbb{Z}/m\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) = 0$, but I don't see what to do next?

P.S. No, this is not a homework question.

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    $\begingroup$ Use a projective resolution. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented May 17, 2010 at 21:52
  • $\begingroup$ of the other one, of course! $\endgroup$
    – Homology
    Commented May 17, 2010 at 21:59
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    $\begingroup$ Ext(Z/mZ,Z/nZ) is the quotient of Hom(Z/nZ,Q/Z) by the image of multiplication by m from Hom(Z/nZ,Q/Z). Hom(Z/nZ,Q/Z) is equal to Hom(Z/nZ,Z/nZ) since an element of order n must go to an element of order n in Q/Z. If you'd like, Z/nZ can be rewritten as (1/n)Z / Z ≤ Q/Z. Hom(Z/nZ,Z/nZ) is the ring Z/nZ and multiplication of m has image (mZ+nZ)/(nZ) = gcd(m,n)Z/nZ, so the quotient is Z/gcd(m,n)Z. In other words, Ext(Z/mZ,Z/nZ) = Z/gcd(m,n)Z. If m,n are coprime, this is Z/Z = 0. Hom, tensor, Ext and Tor are somewhat silly on finitely generated abelian groups, so you might try other groups. $\endgroup$
    – Jack Schmidt
    Commented May 17, 2010 at 22:55
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    $\begingroup$ @adeel: have you consulted an arbitrary introduction to homological algebra? $\endgroup$
    – Martin Brandenburg
    Commented May 18, 2010 at 0:53
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    $\begingroup$ Exceptionally, I don't think the downvote is right here... $\endgroup$
    – user717
    Commented May 18, 2010 at 8:51

1 Answer 1

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$\mathbb{Q}/\mathbb{Z}$ is a pretty terrible abelian group, or a rather hard one, there may be better injective resolutions to work with. It would certainly be easier to do the projective resolution, use $0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/n \to 0$. this will surely be easier to work through than the one involving $\mathbb{Q}/\mathbb{Z}$. Then compute the appropriate tensor product or hom group.

I started learning this stuff on more interesting modules as Schmidt suggests. For example, modules over the group ring of some cyclic group, or maybe an exterior algebra on two generators (if you make the generators of different gradings, in particular 1 and 3). This happens to be the category of modules you need to understand in order to compute complex connective k-theory!

this should help get you going. These computations are very fun!

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