Why $0\rightarrow B\otimes I\rightarrow B\otimes R$ is exact?

Question

Let $B$ is a right $R$-module. Show that $$0\rightarrow B\otimes I\rightarrow B\otimes R$$ is exact for every finitely generated left ideal $I$.

Any help is appreciated, thanks a lot.

Answer 1

A counterexample: take the ideal $n\mathbb{Z} \subset \mathbb{Z}$ and tensor the inclusion with $\mathbb{Z}/n\mathbb{Z}$.

Tensor product is right exact, not left exact in general. If $R$ is a commutative ring and $M$ is an $R$-module, then the following are equivalent:

1. $M$ is flat (i.e. $M\otimes_R -$ is exact).
2. For every finitely generated ideal $I \subset R$ the morphism $M\otimes_R I \to M\otimes_R R$ is mono.

—see any textbook on commutative algebra, e.g. Matsumura, "Commutative ring theory", Theorem 7.7.