Let $k$ be a field, $A = k[x,y] / (y^2, xy)$. I know that the only associated points of Spec $A$ are $[(x,y)]$ and $[(y)]$. I want to show that $$ \frac{x-2}{(x-1)x} $$ is not a rational function.
Intuitively it makes sense to me that this function is not defined at the origin $[(x,y)]$, hence not a rational function. But I am having trouble arguing it precisely. I would appreciate any comments. Thank you!
ps I apologize as this question is similar to my previous question to show something is regular. A basic question regarding a rational function on a locally Noetherian scheme
The problem comes from the the fact that this function should be defined on the open set $U=\operatorname{Spec}A_{x,(x-1)}$ (inverting $x$ and $x-1$) or on a smaller open set so that $x$ and $x-1$ can be invertible. However such $U$ could not contain all the associated points of the scheme (which is necessary in Vakil's definition of rational section). For instance the origin is an associated point (by exercise 5.5A in Vakil's notes) and $U$ does not contain the origin, so indeed $\frac{x-2}{(x-1)x}$ cannot be a rational function