Given a semisimplicial set $X_*$ we define it's chain complex to be $C_n=\mathbb {Z}[X_n] $ the free group over the integers generated by $X_n $. Why do we choose is over the integers and not over the reals or complex numbers?
Indeed, we would lose the beautiful intuition given by the homology group of the projective plane, but are there any deeper justifications? Is the fact we are over the integers used in the homology theory?
On
It is true that $\mathbb Z$ is the most natural example, and captures some important cases. However, there's a deeper reason for this choice.
The reason that coefficients are generally taken in $\mathbb Z$ is due to the Universal Coefficient Theorem. Essentially, it says that homology with coefficients in $\mathbb Z$ completely determines homology with coefficients in $R$ for any other ring $R$. Hence $\mathbb Z$ is "universal."
This is essentially because $\mathbb Z$ is the initial object in the category of rings so there's a natural homomorphism $\mathbb Z \to R$ for any $R$.
In many cases, it also yields a straightforward formula (exact sequence) to compute the homology with coefficients in $R$ given the homology with coefficients in $\mathbb Z$.
This what Lord Shark the Unknown's is saying in his or her comment: homology with coefficients in $R$ is the same as homology of the complex $C_* \otimes R$.
As Eric Wofsey points out (and as you may know), this is merely a convention motivated by the Universal Coefficient Theorem; it is possible to define homology in other coefficient rings.
You don't have to use the integers! For any abelian group $G$, you can take $C_n$ to be the abelian group of formal linear combinations of elements of $X_n$ with coefficients in $G$. (If $G$ is a ring, then this will be not just an abelian group but a $G$-module.) The homology of this chain complex is called the homology of $X_*$ with coefficients in $G$. The choice $G=\mathbb{Z}$ is just the "default" choice.
As for why $G=\mathbb{Z}$ is the default, it is in a certain sense the most basic example. Taking formal $\mathbb{Z}$-linear combinations just means that we are taking the free abelian group on $X_n$. That is, we are formally allowing ourselves to add and subtract elements of $X_n$, but doing nothing else. Every other choice of coefficient group can be obtained from this basic construction: if we want to take coefficients in $G$, that's equivalent to first constructing the chain group $C_n$ with coefficients in $\mathbb{Z}$ and then taking a tensor product with $G$.