Kernel of a map on tensor product of modules

Question

Let $M,N, P$ be $A$-modules, and let $f:M \otimes N \to P$ be an $A$-homomorphism.

If $m \otimes n \in \ker f$ implies $m\otimes n =0$ for all $m\in M, n\in N$, does it follow that $\ker f=0?$

For instance, does $m_1\otimes n_1+ m_2 \otimes n_2 \in \ker f$ imply $m_1\otimes n_1+ m_2 \otimes n_2=0$?

Answer 1

In general, no. For instance, let $A$ be a field, let $$M = N = \langle v_1, v_2, v_3, v_4 \rangle$$ be a $4$-dimensional vector space over $A$, and let $$ P = (M \otimes N) / \langle v_1 \otimes v_2 + v_3 \otimes v_4\rangle, $$ with $f$ the natural map. No non-zero decomposable tensor lies in the kernel $$\ker(f) = \langle v_1 \otimes v_2 + v_3 \otimes v_4\rangle.$$