Let me start with the formal definition of Rees algebra. If $A$ is a commutative ring over some field $k$, $I \subset A$ is an ideal, then Rees algebra is by definition $$ R=\oplus_{i \in \mathbb{Z}} I^{i} t^{-i} \subset A[t, t^{-1}], $$ where $I^{i}=A$ for negative $i$. Basic fact about this algebra is that obvious map $k[t] \to R$ is flat and we get a flat family over an affine line, where fiber over $0$ is spec of associated graded algebra and all other fiber are isomorphic to the $\operatorname{spec}(A)$.

For example, if I take smooth hypersurface $S \subset \mathbb{A}^n$ then normal bundle is trivial and I get a family where fiber over $0$ is $S \times \mathbb{A}^1$, where this affine line "corresponds to normal bundle direction" and all other fibers are isomorphic to $\mathbb{A}^n$. This looks to me like an extremely weird family.

Intuitively, this construction suppose to be an analog of a tubular neighborhood theorem for differentiable manifolds, so in the example above is should be something like small neighborhood of $S$ in the total space of normal bundle is isomorphic to a small neighborhood of $S$ in $\mathbb{A}^n$. But in algebraic geometry we can't talk about "small neighborhoods" and naive version would be $S \times \mathbb{A}^1 \cong \mathbb{A}^n$ that fails in general, so instead this two spaces are fibers of a family.

Suppose that I don't know such algebraic construction and a want to construct geometrically such flat family over affine line. What train of geometric thoughts leads to such algebraic answer?

  • $\begingroup$ A dumb question, what is the obvious map $k[t] \to R$? In particular, if $I \neq A$, where does $t$ get sent? Do you mean $A \to R$ being flat? (But that's not a family over the affine line of course). $\endgroup$ – Karl Schwede Oct 2 '13 at 3:16
  • $\begingroup$ @Karl I assume that $A$ is a $k$-algebra (in the first line), then $R$ is a $k[t]$ algebra. $\endgroup$ – Sasha Pavlov Oct 2 '13 at 10:36
  • $\begingroup$ You misunderstood my confusion. $t$ is not an element of $R$ unless $I = R$. Thus I don't see a natural map $k[t] \to R$. Am I missing something obvious here? $\endgroup$ – Karl Schwede Oct 2 '13 at 12:12
  • $\begingroup$ Just to chime in: the correct definition is to take the sum of $A[t^{-1}]$ and the algebra $R$ as $A$-submodules of $A[t,t^{-1}]$. Then, it is $k[t^{-1}]$ that naturally sits in this algebra. This also gives the correct algebra upon taking the quotient by the ideal $\langle t^{-1} \rangle \subset k[t^{-1}]$. $\endgroup$ – Jason Starr Oct 2 '13 at 12:32
  • $\begingroup$ @Jason Thanks, you are absolutely right! I should correct my definition. By the way this shows again that I don't really feel this construction in such pure commutative algebra terms. $\endgroup$ – Sasha Pavlov Oct 2 '13 at 13:02

Let me try to trace a path from the picture in differential geometry to the algebraic formula you gave.

Starting in differential geometry, let $X$ be a manifold and $Z$ a submanifold with normal bundle $N$. Consider $X\times \Bbb R \to \Bbb R$. To construct the deformation to the normal bundle, the idea is that we want to replace the fiber at $0$ with a copy of $N$. The trick is in specifying the topology, i.e., given a sequence of points $(x,t)$ tending to the fiber at $0$ (so $t\to 0$), which ones do we declare to have a limit in $N$, and how do we describe that limit?

If we give ourselves a tubular neighborhood, this is easily done: we say that such a sequence $(x,t)$ as $t\to 0$ has a limit in $N$ if $x$ tends to a point in the tubular neighborhood, in which case the limit is that point, viewed as a point of $N$. (In simpler terms, the deformation to the normal bundle is just obtained by removing from $X\times \Bbb R$ the complement to the tubular neighborhood in the fiber at $0$.)

But we can accomplish the same thing more canonically without a tubular neighborhood. The idea is that we consider only sequences of points $(x,t)$ with $t\to 0$ for which $x$ actually tends to $Z$, and the limit in $N$ will be gotten by remembering the direction along which we've approached $Z\times 0$.

More precisely, consider the possible directions along which a point of $X\times \Bbb R$ can approach $Z\times 0$. This is the projective bundle of the normal bundle of $Z\times 0$ in $X\times \Bbb R$, or in other words the projective bundle of the direct sum of $N$ with a copy of $\Bbb R$. If we remove the section at infinity, we see just a copy of $N$. The section at infinity corresponds to points approaching $Z\times 0$ vertically, i.e. asymptotically along the fiber at $0$, $X\times 0$. So, truly, possible directions of points $(x,t)$ approaching $Z\times 0$ non-vertically correspond to points of the normal bundle $N$. This gives another description of the deformation to the normal bundle.

This description is also easy to translate into algebraic geometry. What we did is to take $X\times \Bbb A^1$, blow up $Z\times 0$ inside it, then remove the closure of the preimage of $(X-Z)\times 0$ in this blowup. So, we are more-or-less reduced to seeing how to describe blowups in terms of coordinate algebras. Maybe that's no less transparent than the original situation... but maybe I'll stop here for now and see if this was at all helpful. If so maybe we can try to figure out why the algebraic construction of the blow-up is what it is.

Response to Dmitri's comments (this is dustin again, i just don't know how to log on anymore): yes, the part about removing the closure of the preimage of $(X-Z)\times 0$ might seem opaque from an algebraic perspective. But actually it's quite the opposite. Recall that, in general, one gets affine charts for a blowup by choosing generators for the ideal. If $J$ is the ideal in the ring $R$ and $f$ is an element of $J$, then the coordinate algebra of such a chart is $R[J/f]$. In our case one need only take $f$ to be the coordinate function $t$. In other words, the deformation to the normal cone is just the most obvious affine piece of the blowup $\operatorname{Bl}_{Z\times 0}(X\times \Bbb A^1)$. One can check chart-by-chart that this description agrees with the more geometric one, in terms of the preimage of $(X-Z)\times 0$ (this is also a specialization of a general fact about blowups).

  • $\begingroup$ I found this explanation very helpful, thank you. $\endgroup$ – Sasha Pavlov Oct 2 '13 at 15:20
  • $\begingroup$ I dont have any problems understanding blow ups algebraicaly or geometrically, but if you can expand part $Bl_I A[t]$-closure of the preimage of $(X-Z) \times 0$ =$R$ that would be helpful. $\endgroup$ – Sasha Pavlov Oct 2 '13 at 15:29

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