My book has the following question:
Prove that, for any given element $x$ of an arbitrary module $X$ over $R$, the assignment $\alpha \to \alpha x$ defines a homomorphism
$$h_a: R \to X$$
of the additive group $R$ into the additive group $X$. Hence,
$$0x = 0, (\alpha-\beta)x = \alpha x -\beta x, n(\alpha x) = (n\alpha) x$$
hold for all elements $\alpha, \beta\in R$ and every integer $n$. By means of these, show that $px=0$ holds for all $x\in X$ if $R$ is of characteristic $p$.
Well, since for a module $X$ we must have
$$\alpha(x+y) = \alpha x + \alpha y$$
for all $x,y\in X$
which proves that $h_a(x+y) = h_a(x) + h_a(y)$. But what exactly is the additive group or $R$? Did I do it right?
Since it's an homomorphism, we know that $h_a(0) = 0$, hence $0x = 0$, but what about $(\alpha-\beta)x = \alpha x -\beta x$ and $n(\alpha x) = (n\alpha) x$?
A ring $R$ has two operations defined on its elements, addition ($+$) and multiplication ($\cdot$) with the property that $R$ is an abelian group under addition. Additive group of $R$ refers to this group under addition. Multiplication between elements of $R$ is ignored while considering $R$ as an additive group.
I prefer to denote the map $R\rightarrow X$ defined as $\alpha\mapsto \alpha x$ for given $x\in X$, by $h_x$ instead of $h_a$ (thinking of $h_x$ as a $``$multiplication by $x"$ map). This map is a homomorphism between the additive group of $R$ and $R$-module $X$ (again considered as an additive group) because $$h_x(\alpha+\beta)=(\alpha+\beta)x=\alpha x+\beta x=h_x(\alpha)+h_x(\beta).$$ Now as you noted, $h_x(0)=0\Rightarrow 0x=0$, and $$h_x(\alpha-\beta)=h_x(\alpha)-h_x(\beta)\Rightarrow (\alpha-\beta)x=\alpha x-\beta x,$$ $$h_x(n\alpha)=nh_x(\alpha)\Rightarrow (n\alpha)x=n(\alpha x).$$