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I apologize if this is a duplicate. I don't know enough about group cohomology to know if this is just a special case of an earlier post with the same title.

Let $G=\langle\sigma\rangle$ where $\sigma^m=1$. Let $N=1+\sigma+\sigma^2+\cdots+\sigma^{m-1}$. Then it is claimed in Dummit and Foote that $$\cdots\mathbb{Z} G \xrightarrow{\;\sigma -1\;} \mathbb{Z} G \xrightarrow{\;N\;} \mathbb{Z} G \xrightarrow{\;\sigma -1\;} \cdots \xrightarrow{\;N\;} \mathbb{Z} G \xrightarrow{\;\sigma -1\;} \mathbb{Z} G \xrightarrow{\;\text{aug}\;} \mathbb{Z} \longrightarrow 0$$ is a free resolution of the trivial $G$-module $\mathbb{Z}$. Here $\mathbb{Z} G$ is the group ring and $\text{aug}$ is the augmentation map which sums coefficients. It's clear that $N( \sigma -1) = 0$ so that the composition of consecutive maps is zero. But I can't see why the kernel of a map should be contained in the image of the previous map. any suggestions would be greatly appreciated. Thanks for your time.

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for some reason my computer isn't loading the type face so I apologize if there are egregious typos. –  Rory Pulvino May 22 '12 at 21:36
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This follows from a direct computation, really: you pick an element of the kernel, and look at it enough time until you see that it must have a certain form which implies it is in the image. –  Mariano Suárez-Alvarez May 22 '12 at 23:01
    
yeah yeah... I've already slapped my forehead :) –  Rory Pulvino May 22 '12 at 23:30
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2 Answers

up vote 4 down vote accepted

As $(\sigma-1)(c_0+c_1\sigma+\dots c_{n-1}\sigma^{n-1})=(c_n-c_0)+(c_0-c_1)\sigma+\dots (c_{n-2}-c_{n-1})\sigma^{n-1}$, the element $a=c_0+c_1\sigma+\dots c_{n-1}\sigma^{n-1}$ is in the kernel of $\sigma-1$ iff all $c_i$'s are equal, i.e. iff $a=Nc$ for some $c\in\mathbb{Z}$. Similarly, $Na=(\sum c_i)N$, so here the kernel is given by the condition $\sum c_i=0$, but this means $a=(\sigma-1)(-c_0-(c_0+c_1)\sigma-(c_0+c_1+c_2)\sigma^2-\cdots)$.

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Fantastic! Thanks so much. –  Rory Pulvino May 22 '12 at 22:24
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I just wanted to elaborate a bit on minu's excellent answer.

Suppose that $\alpha=\sum_{i=0}^{m-1}a_i\sigma^i$ is in the kernel of $\sigma-1$. Then $$\textstyle(\sigma-1)(\alpha)=\sum\limits_{i=0}^{m-1}a_{i}(\sigma^{i+1}-\sigma^i)=(a_{m-1}-a_0)+(a_0-a_1)\sigma+\cdots+(a_{m-2}-a_{m-1})\sigma^{m-1}=0$$ so that $a_{i-1}-a_i=0$ for all $i$, and so all the $a_i$ are equal. Therefore, $$\alpha=a\left(\textstyle\sum\limits_{i=0}^{m-1}\sigma^i\right)=N(a)$$ where $a=a_0=a_1=\cdots=a_{m-1}$. Thus, $\ker(\sigma-1)\subseteq\operatorname{im}(N)$.


Now suppose that $\alpha=\sum_{i=0}^{m-1}a_i\sigma^i$ is in the kernel of $N$. Then $$N(\alpha)=\textstyle\sum\limits_{i=0}^{m-1}a_i\left(\sum\limits_{j=0}^{m-1}\sigma^{i+j}\right)=\sum\limits_{i=0}^{m-1}a_i\left(\sum\limits_{j=0}^{m-1}\sigma^{j}\right)=\left(\sum\limits_{i=0}^{m-1}a_i\right)\left(\sum\limits_{j=0}^{m-1}\sigma^{j}\right)=0$$ which implies that $\sum_{i=0}^{m-1}a_i=0$, and therefore $$\textstyle(\sigma-1)\left(\sum\limits_{i=0}^{m-1}\left(-\sum\limits_{j=0}^ia_j\right)\sigma^i\right)=\left(a_0-\sum\limits_{j=0}^{m-1}a_j\right)+\sum\limits_{i=1}^{m-1}\left(\sum\limits_{j=0}^{i}a_j-\sum\limits_{j=0}^{i-1}a_j\right)\sigma^i$$ $$=(a_0-0)+\textstyle\sum\limits_{i=1}^{m-1}a_i\sigma^i=\alpha.$$ Thus, $\ker(N)\subseteq\operatorname{im}(\sigma-1)$.

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Thanks for taking the time to spell it out, Zev –  Rory Pulvino May 22 '12 at 23:34
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