# What geometrical obstructions to $M$ being flat do elements which map to 0 in $M \otimes I$ represent?

I'm trying to get geometric intuition for the notion of a flat module over a ring, and am running into some problems with my intuition. I am comfortable with flat modules and tensor products from the commutative side, so when I ask what an object "is", I mean how can I translate the commutative algebra into geometric reasoning.

Consider $M=\mathbb{C}[x,y,z]/(xz-y)$ as an $R=\mathbb{C}[x,y]$ module.

As was explained very well by many users here: Why isn't $\mathbb{C}[x,y,z]/(xz-y)$ a flat $\mathbb{C}[x,y]$-module, the reason that $M$ is not flat as an $R$-module is because if we consider the ideal $I=(x,y)$, then $M \otimes I$ contains the non-zero element $1 \otimes y-z \otimes x$, which is 0 if considered as an element of $M$ (it happens to have x-torsion).

We also know that the geometric reason for the failure of flatness is that a whole line is mapped to $(0,0)$ when we map the variety corresponding to $M$ to the plane (the variety corresponding to $R$).

My question is what geometric "object" exactly is $1 \otimes y-z \otimes x$, which maps to $0$ when we consider multiplication but happens to be non-zero when we consider some other random bilinear form, and how does this object correspond to a non-constant dimensional fiber if we consider it geometrically?

If the object is in fact just something we can apply bilinear forms to, then I would like to know if the general bilinear forms on $M \times I$ are possibly "geometrical" in some sense (i.e. products of nth derivatives of functions, which could correspond to nth order behavior around a specific point).

On a related note, I'm not sure what the "functions" in $M \otimes \kappa(P)$ actually correspond to (here $\kappa$ mean residue class field); I would really appreciate if anyone could clear this up (specifically, what exactly do these functions "act on", and where do they "live").

Thank you,

Rofler

Edit: I thought understanding $M \otimes \kappa(P)$ would shed some light on $M \otimes P$, but actually the former is actually easy to understand, since none of the elements in the tensor products are non-elementary, and we've really just inverted a few functions, and set a few others to be $0$. $M \otimes P$ on the other hand...

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