When $N \to M \otimes_R N$ is not an embedding.

Question

Can someone please provide an example of the following (or tell me why such an example doesn't exist): Let $R$ be a (not necessarily commutative) ring, $M$ an $R$-$R$-bimodule containing a copy of $R$, and $N$ a left submodule of $M$, such that the map $$ N \to M \otimes_R N, ~~~~~~ n \mapsto 1 \otimes n, $$ is not an embedding.

Answer 1

Take $R=\mathbb{Z}$, $M=\mathbb{Q}\oplus \mathbb{Z}/2\mathbb{Z}$ and $N= \mathbb{Z}/2\mathbb{Z}$, with inclusion $R\to M:n\mapsto (n,\overline{0})$. Then, in $M\otimes_\mathbb{Z} N$, we have that $$ (1,\overline{0})\otimes \overline{1} = (\frac{1}{2},\overline{0})\otimes 2\cdot\overline{1} = (\frac{1}{2},\overline{0})\otimes \overline{0} = 0.$$

Therefore the map $N\to M\otimes_\mathbb{Z} N$ described in the question is the zero map, and is not injective.