Let $M=\mathbf{Z}/2$, $N=\mathbf{Z}/3$, the free-module $\mathbf{Z}^{M \times N}$ and his sub-module $H$ generated by elements of the following forms $(m_1+m_2,n)-(m_1,n)-(m_2,n), (m, n_1+n_2)-(m,n_1)-(m,n_2), (am,n)-a(m,n), (m,an)-a(m,n), m, m_1, m_2 \in M, n ,n_1, n_2 \in N, a \in \mathbf{Z}$.
And then consider the quotient-module $\mathbf{Z}^{M \times N}/H=\mathbf{Z}/2 \otimes_{\mathbf{Z}} \mathbf{Z}/3$. Then a bilinear map $\tau$ from $N \times M$ to the quotient module vanishes on $(1,1) \in M \times N$ and I would like to know if we can find a generator of $H$ that verify the following equality \begin{equation} (1,1)=(3.1,1)=(1,3.1)+\text{a generator of} H. \end{equation} Many thanks for any suggestions.
I'd prefer a different notation, namely $[x]_2$ and $[x]_3$ for the residue classes modulo $2$ and $3$.
First some general fact: if $M=Rx$ and $N=Ry$ are cyclic modules over the commutative ring $R$, then also $M\otimes_RN$ is cyclic, generated by $x\otimes y$.
Indeed, for a generic element of $M\otimes N$ has the form $$ \sum_{i=1}^n r_ix\otimes s_iy=\Bigl(\,\sum_{i=1}^n r_is_i\Bigr)(x\otimes y) $$ because you can move scalars past the tensor product. With more details $$ \sum_{i=1}^n r_ix\otimes s_iy= \sum_{i=1}^n r_is_ix\otimes y= \Bigl(\,\sum_{i=1}^n r_is_ix\Bigr)\otimes y= \Bigl(\,\sum_{i=1}^n r_is_i\Bigr)(x\otimes y) $$ In your case $W$ is cyclic, generated by $[1]_2\otimes[1]_3$. On the other hand $$ 3[1]_2=[1]_2 $$ so you have $$ [1]_2\otimes[1]_3=3[1]_2\otimes[1]_3=[1]_2\otimes 3[1]_3=[1]_2\otimes [0]_3 $$ so the generator is zero and consequently $W$ is the zero module.