# Examples of computing Ext and Tor functors?

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.

• Use a projective resolution. – Mariano Suárez-Álvarez May 17 '10 at 21:52
• of the other one, of course! – Homology May 17 '10 at 21:59
• 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. – Jack Schmidt May 17 '10 at 22:55
• @adeel: have you consulted an arbitrary introduction to homological algebra? – Martin Brandenburg May 18 '10 at 0:53
• Exceptionally, I don't think the downvote is right here... – user717 May 18 '10 at 8:51

$\mathbb{Q}/\mathbb{Z}$ 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.