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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.

The geegeezgeegee
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  • Well, this is just general algebra of matrices. – egreg Feb 15 '18 at 15:02
  • @egreg does it matter which letter, n, r or s i choose to be my matrix? – The geegeezgeegee Feb 15 '18 at 15:04
  • The only issue is that the matrix sizes combine for doing multiplication. If you look at the proofs of associativity of matrix multiplication and of distributivity with respect to addition, you'll see that they apply to any ring, even noncommutative. – egreg Feb 15 '18 at 15:11
  • What difficulty do you have doing these the exact same way you did the first one? Just explicitly compute both sides to see they are the same... – Eric Wofsey Feb 15 '18 at 17:43

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