let $A$ and $B$ be rings. Any $A\times B$ module is of the form $M\oplus N$ for some $A$-module $M$ and $B$-module $N$ with $(a,b)\cdot (m,n)=(am,bn)$

Question

Let $A$ and $B$ be two rings, and let $L$ be an $A\times B$-module. Prove that $L$ is of the form $M\oplus N$ for some $A$-module $M$ and $B$-module $N$, with $(a,b)\cdot (m,n) = (am,bn)$.

Let $M := \{ (a_1,0)l_1 + \cdots + (a_n,0)l_n \mid n \in \mathbb N, a_i \in A, \text{ and } l_i \in L \text{ for all } 1 \leq i \leq n \}$ and similarly let $N := \{ (0, b_1)r_1 + \cdots + (0, b_n)l_m \mid m \in \mathbb N, b_j \in A, \text{ and } r_j \in L \text{ for all } 1 \leq j \leq m \}$. Then we may check that $M$ is an $A$-module and $N$ is a $B$-module when we define $a \cdot m := (a,0)m$ and $b \cdot n := (0,b)n$. I claim that $L = M \oplus N$. Indeed, for any $l \in L$ we have $l = (1,1)l = (1,0)l + (0,1)l$.

However, I am not sure whether $M$ and $N$ are share only $0$ in their intersection, i.e. whether $L = M + N$ is a direct sum or not. Could anyone help?

Answer 1

For $m\in M$ we have $(1,0)m=m$.

For $n \in N$ we have $(1,0)n=0$.

Thus if $x\in M\cap N$, we have $$x=(1,0)x=0.$$