I have a very basic question that's been vexing me. Let $a\in A$ be a nonzero element of a Noetherian ring. Then $a$ lies in some maximal ideal $M$. Consider the localization $A_M$ at that maximal ideal. Is it true then that the element $\frac{a}{1}$ is nonzero in $A_M$? The text I'm reading (Milne's Algebraic Geometry) seems to say so (Proposition 1.16, page 18).
I was thinking that if $\frac{a}{1}=\frac{0}{1}$, then this implies that $ta=0$ for some $t\in A\setminus M$.
As I pointed out in the comments, this is not true, with an explicit counterexample given by $A= k\times k$ for some field $k$, and $M= k\times\{0\}$.
However, this is not what Milne claims : in his text, $M$ is a maximal ideal containing $\{b \mid ba = 0\}$, but in particular, $a$ is rarely in there (sometimes it is). But because of this, any $t$ as you describe will have to be in $M$.