Let's see what happens when the ground ring consists of the integers $Z$.

Review point: an $R$-module is a commutative group ("abelian group") $G$, along with a ring homomorphism from $R$ into the ring of endomorphisms of $G$. $G$ comprises the "vectors," along with their commutative addition, and the homomorphism tells us how to scale a "vector" in $G$ by a scalar in $R$. It is assumed that $1$ in $R$ maps to the identity mapping on $G$ -- so that scaling a vector by 1 always leaves it unchanged.

So $Z$-module is just a commutative group, where the scalars are integers. The full structure of the scaling operation is already determined by the structure of $G$. For example, $3 * v = (1 + 1 + 1) * v = 1 * v + 1 * v + 1 * v = v + v + v$. So a $Z$-module is just a commutative group with the added perspective of being able to scale an element by an integer.

Review point: an $R$-module is a commutative group ("abelian group") $G$, along with a ring homomorphism from $R$ into the ring of endomorphisms of $G$. $G$ comprises the "vectors," along with their commutative addition, and the homomorphism tells us how to scale a "vector" in $G$ by a scalar in $R$. It is assumed that $1$ in $R$ maps to the identity mapping on $G$ -- so that scaling a vector by 1 always leaves it unchanged.

So $Z$-module is just a commutative group, where the scalars are integers. The full structure of the scaling operation is already determined by the structure of $G$. For example, $3 * v = (1 + 1 + 1) * v = 1 * v + 1 * v + 1 * v = v + v + v$. So a $Z$-module is just a commutative group with the added perspective of being able to scale an element by an integer.