Why is $\mathbb Z$ a $\mathbb Z[\pi]-$module?

Question

Let $X$ be a topological space with fundamental group $\pi_1X:=\pi$. I read that to recover the cellular homology of $X$ with coefficients in $\mathbb Z$ from the homology of $X$ with local coefficients in a $\mathbb Z[\pi]-$module $A$ we just take $A$ to be $\mathbb Z$. My problem is how to see $\mathbb Z$ as a $\mathbb Z[\pi]-$module ? Thank you for your help ?

Answer 1

Hint: More generally, for any ring $R$ with (two-sided) ideal $I$, there's a natural left and a natural right $R$-module structure on $R/I$.

Hint 2: More generally, if there is a homomorphism of rings $R\rightarrow S$, such that the image of $R$ is contained in the centre of $S$, every $S$-module can be given the structure of an $R$-module via that homomorphism.