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Can you please give some examples of computation of the derived functors $\operatorname{Tor}_1$ and $\operatorname{Ext}^1$ for some simple cases, say $R=\mathbb{Z}$ or $R=\mathbb{Z}[G]$ for some finite group $G$?

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math.wichita.edu/~pparker/classes/handout/torext.pdf lists basic properties and some examples. –  M.B. Feb 12 '12 at 18:01
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All textbooks about homological algebra that I know of include examples. Which have you looked? –  Mariano Suárez-Alvarez Feb 12 '12 at 21:17

2 Answers 2

Here's a simple example of $\mathrm{Tor}_n^R$ for $R = \mathbb Z$:

Example 1: $\mathrm{Tor}_n^{\mathbb Z} (\mathbb Z / 2 \mathbb Z, \mathbb Z / 2 \mathbb Z)$

(i) Choose a projective (in this case free) resolution of $\mathbb Z / 2 \mathbb Z$:

$$ \dots \to 0 \to \mathbb Z \xrightarrow{\cdot 2} \mathbb Z \xrightarrow{\pi} \mathbb Z / 2 \mathbb Z \to 0$$

(ii) Remove $M = \mathbb Z / 2 \mathbb Z$ and apply $- \otimes_{\mathbb Z} \mathbb Z / 2 \mathbb Z$ to get

$$ 0 \to \mathbb Z \otimes_{\mathbb Z} \mathbb Z/ 2 \mathbb Z \xrightarrow{id \otimes (\cdot 2)} \mathbb Z \otimes_{\mathbb Z} \mathbb Z / 2 \mathbb Z \to 0$$

Then simplify using $R \otimes_R M \cong M$ to get $$ 0 \to \mathbb Z / 2 \mathbb Z \xrightarrow{f} \mathbb Z / 2 \mathbb Z \to 0$$

Find $f$ for example, by using the isomorphism $R \otimes_R M \cong M , r \otimes m \mapsto rm $:

$$ \mathbb Z / 2 \mathbb Z \to \mathbb Z \otimes_{\mathbb Z} \mathbb Z / 2 \mathbb Z \xrightarrow{id \otimes (\cdot 2)} \mathbb Z \otimes_{\mathbb Z} \mathbb Z / 2 \mathbb Z \to \mathbb Z / 2 \mathbb Z$$

where we have the maps $m \mapsto 1 \otimes m, 1 \otimes m \mapsto 1 \otimes 2m , 1 \otimes 2m \mapsto 2m$, in this order.

Hence we see that $f \equiv 0$.

(iii) Hence we get

$$ \mathrm{Tor}_0^{\mathbb Z} (\mathbb Z/2 \mathbb Z , \mathbb Z/ 2 \mathbb Z ) = \mathbb Z / 2 \mathbb Z\otimes_{\mathbb Z} \mathbb Z / 2 \mathbb Z$$

$$ \mathrm{Tor}_n^{\mathbb Z} (\mathbb Z / 2 \mathbb Z, \mathbb Z / 2 \mathbb Z) = 0 \text{ for } n \geq 2$$

$$ \mathrm{Tor}_1^{\mathbb Z}(\mathbb Z / 2 \mathbb Z, \mathbb Z/ 2 \mathbb Z ) = \mathrm{ker}0 / \mathrm{im}f = \mathbb Z / 2 \mathbb Z$$

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Is the $\Bbb{Z}$ in $\textrm{Tor}$ written with a subscript or superscript? –  user38268 Oct 27 '12 at 9:08
    
@BenjaLim If you are referring to the $R$ in $\mathrm{Tor}^R$ then yes, the $R$ is written with a superscript. See Wikipedia. –  Matt N. Oct 27 '12 at 9:14
    
Nice answer :D +1 –  user38268 Oct 27 '12 at 10:15
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@BenjaLim: The index of Tor is a subscript because it is a left derived functor (otherwise you'd write $\mathrm{Tor}^{-n}$). Thus, the only free spot for the $\mathbb{Z}$ is the superscript. Likewise, the index of Ext is a superscript because it is a right derived functor, and therefore you write the $\mathbb{Z}$ as a subscript in Ext. –  commenter Oct 27 '12 at 10:32
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@BenjaLim: yes that's right. The "algebraic geometer's" convention is to restrict attention to cochain complexes $(C^{n})_{n \in \mathbb{Z}}$. A chain complex is then a complex $(C^n)_{n \in\mathbb{Z}}$ with $C^n = 0$ for $n \gt 0$ and to obtain the usual indexing convention for chain complexes one simply puts $C_n = C^{-n}$ an $H_n(C) = H^{-n}(C)$. That's why one writes $\operatorname{Ext}^n$ and $\operatorname{Tor}_n = \operatorname{Tor}^{-n}$. –  commenter Oct 27 '12 at 13:09

Here is a slightly more complicated example than just free abelian groups. We will compute $\textrm{Ext}_{\Bbb{Z}/4}^1(\Bbb{Z}/2,\Bbb{Z}/2)$.

Consider $\Bbb{Z}/2$ as a $\Bbb{Z}/4$ - module; the universal property of quotients gives us a map (call it $g_0$) from $\Bbb{Z}/4 \to \Bbb{Z}/2$. Then we get by continuing a similar process a free resolution of $\Bbb{Z}/2$ by free $\Bbb{Z}/4$ modules

$$\ldots \stackrel{g_3}{\longrightarrow} (\Bbb{Z}/4)^8 \stackrel{g_2}{ \longrightarrow} (\Bbb{Z}/4)^2 \stackrel{g_1}{ \longrightarrow } \Bbb{Z}/4 \stackrel{g_0}{\longrightarrow} \Bbb{Z}/2 \longrightarrow 0$$

where the map $g_1$ sends $(1,0)$ to $0$ and $(0,1)$ to $2$, $g_2$ sends each canonical generator to an element of the kernel of $g_1$ and so on. Now we take $\textrm{Hom}(-,\Bbb{Z}/2)$ (where our homs are now $\Bbb{Z}/4$ - homs) of this exact sequence to get the chain complex

$$\ldots \stackrel{g_3^\ast}{\longleftarrow} \textrm{Hom}((\Bbb{Z}/4)^8 ,\Bbb{Z}/2) \stackrel{g_2^\ast}{ \longleftarrow} \textrm{Hom}( (\Bbb{Z}/4)^2 ,\Bbb{Z}/2) \stackrel{g_1^\ast} { \longleftarrow } \textrm{Hom}(\Bbb{Z}/4,\Bbb{Z}/2) \stackrel{g_0^\ast}{\longleftarrow} \textrm{Hom}(\Bbb{Z}/2,\Bbb{Z}/2) \\ \hspace{5.5in}\longleftarrow 0$$

Now firstly we have $\textrm{Ext}_{\Bbb{Z}/4}^0(\Bbb{Z}/2,\Bbb{Z}/2)$ being isomorphic to $\textrm{Hom}(\Bbb{Z}/2,\Bbb{Z}/2) \cong \Bbb{Z}/2.$

To compute $\textrm{Ext}^1$, first notice that $g_1^\ast$ is the zero map. For if $f : \Bbb{Z}/4 \to \Bbb{Z}/2$, precomposing it with $g_1$ gives that

$$f \circ g_1= 0$$

because the image of $g_1$ is the two point set $\{0,2\}$. Any $f :\Bbb{Z}/4 \to \Bbb{Z}/2$ evaluated on this set is zero. It remains to determine the kernel of $g_2^\ast$.

Now notice that each of the homs is isomorphic to a direct sum of $\Bbb{Z}/2$'s. In the case of computing the kernel of $g_2^\ast$, we get that $g_2^\ast$ is a map from $\textrm{Hom}((\Bbb{Z}/4)^2,\Bbb{Z}/2) \cong \textrm{Hom}((\Bbb{Z}/4),\Bbb{Z}/2)^2$ to

$$\textrm{Hom}((\Bbb{Z}/4)^8,\Bbb{Z}/2) \cong \textrm{Hom}((\Bbb{Z}/4),\Bbb{Z}/2)^8.$$

We now just need to compute the kernel of the associated map $$h : \textrm{Hom}((\Bbb{Z}/4),\Bbb{Z}/2)^2 \to \textrm{Hom}((\Bbb{Z}/4),\Bbb{Z}/2)^8.$$

The kernel of this map consists of the 0 tuple and the tuple $(\varphi,\varphi)$ where $\varphi : \Bbb{Z}/4 \to \Bbb{Z}/2$ that sends $1$ to $1$.

It follows that

$$\textrm{Ext}_{\Bbb{Z}/4}^1(\Bbb{Z}/2,\Bbb{Z}/2) \cong \Bbb{Z}/2.$$

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Das ist ausgezeichnet! –  awllower Dec 8 '12 at 13:51
    
@awllower Thank you :D I don't understand german though. –  user38268 Dec 17 '12 at 10:18

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