Let k be a field, A=k[x,y](y2,xy). I know that the only associated points of Spec A are [(x,y)] and [(y)]. I want to show that 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. Thanks)

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Let k be a field, A=k[x,y](y2,xy). I know that the only associated points of Spec A are [(x,y)] and [(y)]. I want to show that    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. Thanks)

Manraj Horton · Accepted Answer

The problem comes from the the fact that this function should be defined on the open set U=SpecAx,(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&#039;s definition of rational section)

Let k be a field, A=\frac{k[x,y]}{(y^2,xy)}. I know that the only associated points of Spec A

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