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 $G$?
Computation of $\operatorname{Tor}_1$ and $\operatorname{Ext}^1$
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2http://www.math.wichita.edu/~pparker/classes/handout/torext.pdf lists basic properties and some examples. – 2012-02-12
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7All textbooks about homological algebra that I know of include examples. Which have you looked? – 2012-02-12
2 Answers
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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1Das ist ausgezeichnet! – 2012-12-08
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0@awllower Thank you :D I don't understand german though. – 2012-12-17
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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0Is the $\Bbb{Z}$ in $\textrm{Tor}$ written with a subscript or superscript? – 2012-10-27
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0@BenjaLim If you are referring to the $R$ in $\mathrm{Tor}^R$ then yes, the $R$ is written with a superscript. See [Wikipedia](http://en.wikipedia.org/wiki/Tor_functor). – 2012-10-27
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0Nice answer :D +1 – 2012-10-27
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1@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. – 2012-10-27
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0@commenter So I am getting it right then in writing $\textrm{Ext}$ with a superscript. I get that it it's kinda why we write cohomology as $H^n$ and homology as $H_n$. – 2012-10-27
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1@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 $\operatorname{Tor}_n = \operatorname{Tor}^{-n}$. – 2012-10-27
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0A little question: why does one get o when $Z/2Z$ is tensored with $Z/2Z$? I guess it should still be $Z/2Z$, right? Because every element tensored with 0 gives 0 again, so there are two elements in the tensor product: 0 and $1\otimes_{\mathbb Z} 1$? Thanks for the attention. – 2012-12-08
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0@awllower You have $Z/nZ \otimes_Z Z/mZ \cong Z / \gcd(n,m) Z$. If $n=m$ then, as you say, $Z/nZ \otimes_Z Z/nZ \cong Z / n Z$. I am not sure why you think it should be zero, I don't claim so in my answer, so I assume you are referring to something else. – 2012-12-08
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0Sorry for not specifying what I referred to . I was alluding to the statement that "Remove $M=Z/2Z$". Maybe I am not understanding what you talked about. Thanks for the explanations. – 2012-12-09
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0@awllower The $M$ in the sequence is removed so that there is no instance of tensoring $Z/2Z$ with itself. – 2012-12-09
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0But how can one remove it? The removed sequence should not be exact, right? – 2012-12-09
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0@awllower Exactly. If it was it would be boring! : ) – 2012-12-09
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0But you stated the exactness of the tensored sequence!This is exactly my point.Thanks for the clarification. – 2012-12-09
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0@awllower Exact for $n\geq 2$, only. Can you point me to where I stated it was exact? – 2012-12-09
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0let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6672/discussion-between-awllower-and-matt-n) – 2012-12-09