Let $M$ and $N$ be $R$- modules and $M'$ be a submodule of $M$. Let $v\in M $ but $v\notin M'$. My question is -- Can we find a module homomorphism $\phi$ from $M$ to $N$ such that $$\phi(M')=\{0\}$$ but $$\phi(v)\neq0.$$ We already know that this is true for vector space but is it true for general module? Is there any result regarding this?
Thank you!
A necessary criterion for the existence of such a map is that $$Hom_R(M/M', N) \neq \{0\}$$ which depends on $M$, $M'$ and $N$ in a highly non-trivial way for general rings $R$. However, this is not even sufficient in general. E.g. if $R=\mathbb{Z}, M = \mathbb{Z} \oplus \mathbb{Z}/4, M' = 0 \oplus 2\mathbb{Z}/4$ (so $M/M' \simeq \mathbb{Z} \oplus \mathbb{Z}/2$), then both for $N_1 = \mathbb{Z}$ and $N_2 =\mathbb{Z}/2$ we have $Hom_R(M/M', N_l) \neq \{0\}$, but for a homomorphism as desired to $N_1$ you additionally need your $v$ to have non-zero first coordinate ($v \notin 0 \oplus \mathbb{Z}/4$), whereas for a homomorphism as desired to $N_2$, you need $v$ to have non-zero second coordinate ($v \notin \mathbb{Z} \oplus 0$).
If your ring $R$ is semisimple, then your criterion is equivalent to: there is at least one isomorphism class of simple modules $\tau$ such that $(Rv+M')/M'$ and $N$ both have non-zero $\tau$-isotypic components. (In other words, simple modules of that isoclass occur in both.)
If $R$ additionally is simple (that is, it has only one isotypic component; that is, it is isomorphic to $M_n(D)$ for some skew field $D$), then this is always true as soon as $N \neq \{0\}$. This includes the case of vector spaces you quote, and is sort of the furthest one can generalize the result without further restriction.
For more general rings, it becomes highly dependent on $v, M, M'$ and $N$. E.g. if $R$ is a PID and $M$ is a finitely generated module, so that (with $p_i$ primes or $0$) $$M \simeq \bigoplus_{i=1}^n (R/p_i^{k_i})$$ $$M' \simeq \bigoplus_{i=1}^n p_i^{l_i}(R/p_i^{k_i})$$ with $0 \le l_i \le k_i$, at least one $l_i \ge 1$, and $$N \simeq \bigoplus_{j \in J} (R/p_j^{k_j}),$$ then a homomorphism as desired to exists if and only if: for at least one coordinate $i$ such that $v_i \notin p_i^{l_i}(R/p_i)^{k_i}$ (where $v = (v_i)_{i=1, ..., n}$), there exists a $j \in J$ such that $(p_j) = (p_i)$ and $k_j \le l_i$. Compare the example in the first paragraph.
