Is $\_ \otimes_R R / m$, for some maximal ideal $m$ of $R$ left exact?

Question

Is $\_ \otimes _R R / m$, for some maximal ideal $m$ of $R$ left exact?

I don't think so (vaguely, because given an injection $f : M \to N$, it is conceivable that $mN \cap M$ is larger than $mM$), but I can't think of a counter example.

Thanks!

Answer 1

If $R$ is a von Neumann regular ring (e.g. $R=\mathbb F_2^{\mathbb N}$), then $R/m$ is $R$-flat for every maximal ideal $m$ of $R$.

Answer 2

Take $R = \mathbb{Z}$ and $m = (2)$. Then the inclusion $\mathbb{Z} \to \mathbb{Q}$ becomes the map $\mathbb{Z} / 2 \mathbb{Z} \to 0$, which is clearly not injective. Exactly what you suspected could happen in fact does, since $(2) \mathbb{Q} \cap \mathbb{Z} = \mathbb{Z}$.