I am studying modules $M$ on rings with units $A$ and it is given as an exercise to prove the following from basic properties of the definition: $a\in A, x,y \in M$
(1) $a \cdot 0_M = \cdot 0_M $
(2) $a \cdot (x-y) = a \cdot x - a \cdot y $
(3) $0_A \cdot x = 0_M $
(4) $(a-b) \cdot x = a \cdot x - b \cdot x$
It is quite clear to me how to prove them if I consider the ring homomorphism $\alpha : A \to End(M) $ and I re-define all the properties. For instance, (1) comes from
$ a \cdot 0_M = \alpha(a)(0_M) = f(0_M) = 0_M$ where $f = \alpha(a)\in End(M)$
What I would like to do is to prove them only by using the definition. Is there any book in which this simple exercise is solved or explained?
For example, $$a \cdot 0_M = a \cdot (0_M + 0_M) = a\cdot 0_M + a\cdot 0_M,$$ whence $a\cdot 0_M=0_M$.
To prove the second assertion note that both $A$ and $M$ are groups, where $x-y$ is, by definition, $x+(-y)$.
Thus, $$a \cdot (x - y) = a \cdot (x + (-y)) = a\cdot x + a \cdot (-y),$$ and as $$a \cdot (-y) = a \cdot (-1\cdot y) = (a \cdot (-1)) \cdot y = -a\cdot y,$$ we get $$a\cdot (x-y) = a\cdot x - a\cdot y.$$
The third is similar to the first, and the fourth to the second.