I need to prove that $\mathbb{Z}^{2}$ is a right module over the ring $M_{2}(\mathbb{Z})$, with scalar multiplication $$(a,b)\cdot\begin{bmatrix}e & f \\g & h \end{bmatrix}=(ae+bg,af+bh)$$ I have already proved the following axiom holds:
For any r ∈ R and m, n ∈ N, (m + n) · r = m · r + n · r.
I have done this by letting m=(a,b), n=(c,d) and letting $$r = \begin{bmatrix}e & f \\g & h \end{bmatrix}$$.
I have then done $(m+n)\cdot r =((a,b)+(c,d)) \cdot \begin{bmatrix}e & f \\g & h \end{bmatrix}$ e.t.c.
and ended up with (ae + ce + bg + dg, af + cf + bh + dh).
I have then looked on the other hand at: $m \cdot r + n \cdot r$.
However, I am unsure as how to prove the following two axioms in the same case:
For any r,s in R, $n \in N$, $(n \cdot r) \cdot s = n \cdot (rs)$
and for any $r,s \in R$, $n \in N$, $n \cdot (r+s)=(n \cdot r)+ (n \cdot s)$
Could someone help please.