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This question is related to 3-cocycle in $\mathbb{Z}/n\mathbb{Z}$

Let $a,b,c,d\in\mathbb{Z}/n\mathbb{Z}$.

Is $a(b-d+[c+d]-[b+c])$ equal to $0\mod n^2$?

Here the $[b+c]$ notation means $b+c\mod n$.

I cannot think of any counterexamples, yet I do not see why it is true.

Frime Kemic
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  • @MichaelBurr My original question was how to show that the given formula is a cocycle, this question is about modular arithmetic – Frime Kemic Jun 07 '17 at 13:15

1 Answers1

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The notation seems a little off because while there is a nice map from $\mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Z}/n^2\mathbb{Z}$, it corresponds to multiplying by $n$, which might not be what you want. I think that I know what you mean (so here's how I interpret your question).

Setup: \begin{align} x_1&=a+k_1n\\ x_2&=b+k_2n\\ x_3&=c+k_3n\\ x_4&=d+k_4n \end{align} where $0\leq a,b,c,d<n$ and the $k_i$'s are integers.

Is $$ x_1(x_2-x_4+[x_3+x_4]-[x_2+x_3])\equiv 0\pmod{n^2}? $$ Now, we replace \begin{align} [x_3+x_4]&=x_3+x_4+k_5n\\ [x_2+x_3]&=x_2+x_3+k_6n\\ \end{align} since reducing modulo $n$ means subtracting some multiple of $n$ (here $k_i$ is also an integer). Therefore, we get that the original equation reduces to:

\begin{align} x_1(x_2-x_4+[x_3+x_4]-[x_2+x_3])&= x_1(x_2-x_4+x_3+x_4+k_5n-x_2-x_3-k_6n)\\ &=x_1(k_5n-k_6n)\\ &=a(k_5-k_6)n+k_1(k_5-k_6)n^2\equiv a(k_5-k_6)n\pmod{n^2}. \end{align}

This suggests that you claim is not true. If we can choose $a$ not a multiple of $n$, say $a=1=x_1$, $x_2=n-1$, $x_3=1$, and $x_4=1$. Suppose that $n>2$. Then $[x_3+x_4]=[2]=2$, $[x_2+x_3]=n=0$. Then, $$ x_1(x_2-x_4+[x_3+x_4]-[x_2+x_3])=1((n-1)-1+2-0)=n $$ which is not a multiple of $n^2$.

Michael Burr
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  • Thank you for your reply. Could you take a look at the link in my question? Because from what I explain there, it should be $0\mod n^2$. What is going wrong? – Frime Kemic Jun 07 '17 at 12:46