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...

Solution 1:

I'm not so good at drawing but let me at least try to describe the picture here.

Outside of the divisor $\{x=0\} \subset \mathbb{A}^2:= Spec \mathbb{C}[x,y]$ the sheaf corresponding to $M$ looks like functions on $\mathbb{A}^2$. That's just saying that as modules we have the following:

$$\mathbb{C}[x,x^{-1},y,z] /(xz-y) \cong\mathbb{C}[x,x^{-1},y]$$

Consider now $\pi :Spec M \to \mathbb{A}^2$ given by the obvious inclusion of algebras. Geometrically $Spec M$ can be thought of as a family of hyperbolas (parametrized by $y$) which degenerates to a union of lines when $y=0$ (a family of conic sections if you like). As we have established already outside of the divisor $D:=\{x=0\}$ (which is precisely the $y$ axis) this map is a local isomorphism. What happens at the $y$ axis?

Lets consider the preimage $\pi^{-1}(D)$. Calculating we get:

$$\mathbb{C}[x,y,z] /(xz-y,x)\cong \mathbb{C}[z]$$

What happened? $\pi^{-1}(D)$ could equivalently be described as the intersection of the $(x,z)$-plane (which is the pullback of $y$-axis to $\mathbb{A}^3$ the ambient space of $SpecM$) with $SpecM$ which is just the $z$-axis. This $\pi^{-1}(D)$ is now by construction fibered over $Spec \mathbb{C}[y] = \mathbb{A}^1$

Notice a weird thing though. Now $z$ is $y$-torsion! What does this mean?

From the geomeric picture we know that it makes sense because the projection of the $z$-axis to $\mathbb{A}^ {2}$ contains only the origin and so the only point where the fiber should be non-empty is $(x,y)=(0,0)$. So $\pi^{-1}(D) \to D = \mathbb{A}^1$ consists of a line sitting entirely over the origin. And this is already non-flat for obvious reasons (even if you don't know what flatness is precisely it should be close to "continuously varying" and there is no way an entire line sitting over a point in $\mathbb{A}^1$ is "continuously varying").

In general I think it is much less rewarding to ponder the geometric meaning of tensor products of the form $M \otimes I$ where $M$ is a module and $I$ is an ideal and neither are flat. Sometimes these can massaged to give stuff like jet bundles and sheaves over nilpotent thickenings but as of themselves they are rather formal objects.

I will say something about this specific case though.

We may interpret $ M \otimes I \otimes \mathbb{C}[x,y] = M \otimes I/I^2$ as the fiber of $M$ at the origin tensored with the cotangent space there. We've seen that the fiber at the origin is the $z$-axis we may imagine the above vector space as part of the first order thickening of the $z$-axis in $\mathbb{A}^3$.

The element $s = 1 \otimes y - z \otimes x$ at the fiber at the origin has value $\bar{s} = dy - z dx$. This 1-form with values in $M$ is zero in the first order thickening $M/I^2M$. The interpretation I think is that $\bar{s}$ wants to be a function on a first order thickening of the fiber at the origin but it fails to be non-trivial because the nilpotent thickening is not big enough (in a flat family any function which cuts out the fiber can be lifted to a function which cuts a nilpotent thickening of it - this is the intuition behind flat=continuously varying). So perhaps there is a sense in which $\bar{s}$ "points in a direction to which $Spec M$ doesn't deform"