Can we choose any cyclic group to compute simplicial homology?

Question

Simplicial homology is, as far as I know, usually computed with coefficients from $\mathbb{Z}_2$ or $\mathbb{Z}$. The question is not about whether or not it would be useful or meaningful to compute homology with coefficients from an arbitrary $\mathbb{Z}_n$ but whether or not we can and if we can, how would it be done.

All simplicial homology really amounts to is the computation of the boundary operators and then some row and column operations with the boundary operators. With coefficients from the finite cyclic group $\mathbb{Z}_n$, the $(-1)^k$ in the definition for boundary of simplex with integer coefficients should become $(n-k)\mod \ n$, if I'm thinking about it correctly.

Using the fact that $$ka=k(a\mod\ n)\mod\ n$$ and $$a+b=(a\mod\ n) + (b\mod\ n)\mod\ n$$ It seems like one should be able to do all row and column operations as would usually be done over the integers and then take the appropriate modulus after everything else is complete.

Answer 1

When working with $\mathbb{Z}_n$, the following things change:

• The coefficient group is $\mathbb{Z}_n$ instead of $\mathbb{Z}$. The chain groups are $C_k(X;\mathbb{Z}_n)$ which are $\mathbb{Z}_n$-linear combinations of simplices.
• The boundary maps are defined the same, but now the coefficients involved are in the quotient ring $\mathbb{Z}_n$. When working with $\mathbb{Z}_n$, it is best to think of numbers as being in equivalence classes, and not that there is an infix $\text{mod}$ operator. So, $(-1)^j\equiv (n-1)^j$ in $\mathbb{Z}_n$ if you want, but in practice it is easier to just keep it as $(-1)^j$. By the way, this is why $n=2$ is special: $(-1)^j\equiv 1^j\equiv 1$ in $\mathbb{Z}_2$, hence no need to keep track of signs/orientations.

You are correct that if you do all of the manipulations to the chain maps with $\mathbb{Z}$ coefficients, you can then reduce the matrices modulo $n$ to get $\mathbb{Z}_n$ coefficients. What can happen is that some of the diagonal entries will reduce, and you might have to do some more work to determine the homology groups.

As Lord Shank the Unknown referenced in the comments, there is a Universal Coefficient Theorem. It implies that $H_\bullet(X;\mathbb{Z})$ contains everything there is to know about $H_\bullet(X;\mathbb{Z}_n)$ for all $n$, at least as far as which homology group will be isomorphic to what.