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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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Isn't this a completely standard Kunneth theorem calculation? The resolution Sasha gave is the standard one and works fine. –  Daniel Moskovich Aug 26 '10 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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