direct sum of modules is a module itself

Question

I am trying to learn modules for the first time. I have read that direct sum of two modules $M, N$ is a module itself, but I am not sure how to prove it. Can someone please explain it to me?

Thanks a lot in advance!

Answer 1

Let $M$ and $N$ be left modules over a ring $R$. Define (as a set) $$M \oplus N = \{(x,y) : x \in M, y \in N\}.$$ Define $$ + : M \oplus N \times M \oplus N \rightarrow M \oplus N \qquad +((x,y), (x',y')) = (x,y) + (x',y') = (x+x', y+y').$$ Check this makes sense (i.e. well-defined).

Define $$ \cdot : R \times M \oplus N \rightarrow M \oplus N, \qquad \cdot(\alpha, (x,y)) = \alpha \cdot (x,y) = (\alpha x, \alpha y).$$ Check this makes sense (i.e. well-defined).

Things left to verify:

• $r \cdot ((x,y) + (x',y')) = r \cdot (x,y) + r \cdot (x',y').$

• $(r+s) \cdot (x,y) = r \cdot (x,y) + s \cdot (x,y).$

• $(rs) \cdot (x,y) = r \cdot ( s \cdot (x,y)).$

• $1_R \cdot (x,y) = (x,y)$.

Can you take it from here?