Bland - Rings and Their Modules (2011)
Exercise 2 of Problem Set 4.1
$\Leftarrow)$ $M$ and $N$ are right $R$-modules.
For each nonzero R-linear mapping $f:N \to N’$ (for each $R$-module $N'$) there is an R-linear mapping $h:M \to N$ such that $fh \neq 0$. Prove that $M$ generates $N$.
According to definition 4.1.2. we have to find an epimorpism $\phi:M^{(\Delta)} \to N$ for some index-set $\Delta$.
$M^{(\Delta)} = \bigoplus_\Delta M_\alpha$, where $M_\alpha = M$, for all $\alpha \in \Delta$.
I suggested to take $\Delta = \text{Hom }(M,N)$ as our index-set. Why? Our hypothesis says that we can get an $R$-map $h:M \to N$ such that $fh \neq 0$, for every nonzero R-map $f:N \to N’$, that map $h$ has to come from somewhere. Sometimes you have to except something and see at the end why.
Every set can be taken as an index-set, see post #45. At this moment I could not find more information on index-sets, maybe you can find something.
We have to find an $R$-map $\phi:\bigoplus_\Delta M_\alpha \to N$. I suggested to use proposition 2.1.5. to construct $\phi$. I think it is the only tool we have to do that. (After we have found $\phi$, we will prove that it is surjective)Proposition 2.1.5
If $\{M_\alpha | \alpha \in \Delta\}$ is a family of R-modules, then the direct sum $(\bigoplus_\Delta M_\alpha, i_\alpha)$,
has the property that for every $R$-module $N$ and every family $\{\phi_\alpha:M_\alpha \to N | \alpha \in \Delta\}$
there is a unique R-map $\phi:\bigoplus_\Delta M_\alpha \to N$
such that $\phi \circ i_\alpha = \phi_\alpha$, for all $\alpha \in \Delta$,
where the $i_\alpha$ are the canonical injections $i_\alpha:M_\alpha \to \bigoplus_\Delta M_\alpha$ for $\alpha \in \Delta$,
The $R$-map $\phi$ is defined on $\bigoplus_\Delta M_\alpha$ in this way:
$\phi((x_\alpha)_\Delta) = \Sigma_\Delta \text{ } \phi_\alpha (x_\alpha)$ for $((x_\alpha)_\Delta) \in \bigoplus_\Delta M_\alpha$Now I want you to write down the premises or conditions of this theorem and translate them into the premises in our case. For instance: the $\Delta$ in the theorem, becomes $\text{Hom }(M,N)$ in our case, and so on. (see post #35). Use a better index-variable than $\alpha$ ...
Good luck.
