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I would like to write down explicitly the generating cocycles of this second cohomology group, $H^2(Z_n \times Z_n, k^*)$. Here $k$ is an algebraically closed field of characteristic zero and $Z_n$ is the cyclic group with $n$ elements.

I need to know what resolution to use and how to get the formulas!

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    $\begingroup$ Isn't this a completely standard Kunneth theorem calculation? The resolution Sasha gave is the standard one and works fine. $\endgroup$ Commented Aug 26, 2010 at 12:17

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I would take the standard cyclic resolution of $G = Z/nZ$: $$ \dots \stackrel{1-t}\to Z[G] \stackrel{\sum t^i}\to Z[G] \stackrel{1-t}\to Z[G] \to Z \to 0, $$ where $t$ is the generator of $G$, and then take the tensor square of two such --- this would give a resolution $$ \dots \to Z[G_1\times G_2]^3 \stackrel{d_2}\to Z[G_1\times G_2]^2 \stackrel{d_1}\to Z[G_1\times G_2] \to Z \to 0, $$ where $G_1 = G_2 = Z/nZ$ and the maps are given by $$ d_1 = (1-t_1,1-t_2), \qquad d_2 = \left(\begin{array}{ccc} \sum t_1^i & 1-t_2 & 0 \cr 0 & 1-t_1 & \sum t_2^i \end{array}\right) $$ ($t_1$ and $t_2$ are the generators of $G_1$ and $G_2$ respectively). I think you can use this for the calculations.

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  • $\begingroup$ Sorry for such a late comment. Yet I'm a little bit confused at the term "tensor square" of two resolutions. Is there an operation called "tensor square of resolutions" and here is a particular example? Or this is not actually a mathematical notion? Thank you! $\endgroup$
    – Hetong Xu
    Commented Sep 12, 2022 at 8:51
  • $\begingroup$ If you have two rings $R_1$ and $R_2$ and two modules $M_1$ over $R_1$ and $M_2$ over $R_2$, then $M_1 \otimes M_2$ is naturally a module over $R_1 \otimes R_2$ (tensor products are over the integers). Similarly, if you have two complexes $M_1^\bullet$ and $M_2^\bullet$ their tensor product is the complex with $n$-th term $\oplus_{i+j = n} M_1^i \otimes M_2^j$. This is what I mean by tensor square of resolutions. $\endgroup$
    – Sasha
    Commented Sep 13, 2022 at 9:17
  • $\begingroup$ Thank you! So we can also regard it as the total complex of a double complex $M_1^{\bullet} \otimes_{\mathbb{Z}} M_2^{\bullet}$? (So we can get the description of the differential map in the new complex.) $\endgroup$
    – Hetong Xu
    Commented Sep 13, 2022 at 10:20
  • $\begingroup$ Yes, precisely. $\endgroup$
    – Sasha
    Commented Sep 13, 2022 at 11:49
  • $\begingroup$ Thank you! (Since when I'm checking $d_1 \circ d_2=0$ in your notation, there is a quite subtle sign issue. Maybe the second column in $d_2$ should be $(1-t_1, -(1-t_2))^{t}$ or conversely?) $\endgroup$
    – Hetong Xu
    Commented Sep 13, 2022 at 12:06

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