Existence of module homomorphism $\phi : M \rightarrow N$ such that $\Phi =\phi^2$

Question

This is a homework problem and I have solved part (a), but I am not sure how to approach part (b). Should I approach it the way how the universal property of free modules are defined? Any hint(s) would be highly appreciated.

Answer 1

Hint: for all matrices $A \in M_2(R)$ and all elements of $x \in M^2$, we have $$ \Phi(Ax) = A\Phi(x) $$ Assuming $R$ is a ring with unity, consider $x = (m,0)^T \in M^2$ and $A = \pmatrix{0&1\\1&0}$.

Answer 2

Hint: Write $\Phi=(\Phi_1,\Phi_2)$ for $\Phi_1$ and $\Phi_2$ functions from $M$ to $N$. Choose appropriate elements of $M_2(R)$ and use the fact that $\Phi$ commutes with the action of $M_2(R)$ to show that:

(1) $\Phi_i:M \rightarrow N$ is an $R$-module homomorphism.

(2) $\Phi_1=\Phi_2$.