# Nonzerodivisor of a module under localization

If $$x_1,x_2,...,x_r\in R$$ is an $$M$$-sequence (i.e. a regular sequence on $$R$$-module $$M$$), then $$x_1^t,...,x_r^t$$ is an $$M$$-sequence.

This is corollary 17.8 from Eisenbud's commutative algebra. The proof is doing induction on $$r$$. Thus, it suffices to show $$x_r$$ is a nonzerodivisor on $$M/(x_1^t,...,x_{r-1}^t)M$$. Then he says we may localize at a prime $$P$$ containing $$x_1,x_2,...,x_r$$, and reduce the case to that $$R$$ is a local ring.

I am confused here. If we do localization, we view $$x_r\in R_P$$ and then show it is a nonzerodivisor on $$M_P/(x_1^t,...,x_{r-1}^t)M_P$$. But it seems to me this doesn't imply $$x_r\in R$$ must be a nonzerodivisor on $$M/(x_1^t,...,x_{r-1}^t)M$$, because in general, if $$x\in R_P$$ is a nonzerodivisor on $$M_P$$ and $$xm/s=0\in M_P$$, we can only conclude that there exists some $$s'\in R -P$$ such that $$s'm=0$$. However, $$m$$ need not to be zero and not every element in $$M$$ can be written as the product of some $$s$$ and $$m$$. Did I misunderstand his argument?

The author wants to show that the multiplication by $$x_r$$ on $$M/(x_1^t,...,x_{r-1}^t)M$$ is injective and tests this locally.