The moral of my last post was that automorphism groups of affine varieties over non-reduced bases are huge, and that varieties which we think are different are often actually the same. In particular, I checked that, over , the “punctured curves” and are the same. Perhaps you are all convinced, but I’d like to do a slightly harder example and convince you that the genus one, once punctured, curves and over are isomorphic.

We begin by answering the question at the end of the last post: to construct an automorphism of the affine line taking to .

Very explicitly, it is given by . Of course, it is obvious that this map takes to ; the nonobvious fact is that is is an automorphism! The inverse map is given by

Although it doesn’t look like it, this formula is actually a polynomial of finite degree because is nilpotent, so all of the nested squares are zero after a certain point. Let’s denote this map by . From the analysis in the first post, we see that does not extend to .

Now, take the double covers of branched over and respectively. In equations, they are given by and . Then takes points satisfying the first equation to points satisfying the second, where

.

Here , with nilpotent, is a shorthand for the polynomial . We leave it to the reader to write down the inverse polynomial map.

At this point we have two examples, so it is time for a theorem. In fact **any** deformation of a smooth affine variety, over a -algebra whose reduction is , is trivial. For a precise statement and a proof over the dual numbers (), see Corollary 5.4 of Hartshorne’s notes. The general case also follows from Hartshorne’s notes (combine Theorems 10.1 and 5.3) but it doesn’t seem to be stated in one place.

So infinitesimal deformations of affine schemes are boring. But in mathematics, whenever something is boring, it is well suited to be the building block for an interesting theory about something else. (The great example here is vector spaces. The classification of modules over a field is completely dull — which is why linear algebra is fundamental to every field of mathematics!) Let’s see what happens when you try to deform a smooth, non-affine scheme over an infinitesimal base. Any such deformation can be covered by affine patches, and on each affine patch the deformation is trivial. So all of the interest is conveyed in the transition functions between the charts. This is a very explicit thing, and admirably suited for the tools of sheaf cohomology. In my next post, I’ll use this to work out the deformation theory of curves and state what the general set up is.

Ideally, there will be two more posts after that, exploring two of the things that can go wrong. One will be an example with an obstruction. I’m not sure what the simplest example is, so suggestions are welcome from the experts who I see are reading this. The other will work out a failure of algebraization in the case of surfaces. This means that the behavior over the ring of formal power series is not simply the inverse limit of what happens over the rings . I’ve wanted to work out this example in detail for years, so hopefully writing these posts will force me to do it.

In the statement of your theorem, “In fact any …”, you should specify that your $k$-algebra is Artinian.

I am sure it is more fun for you to work it out yourself, but the K3 example is also worked out in Claim 3.5 of my arXiv preprint, “Artin’s axioms, composition and moduli spaces” (and I’m sure a dozen other places as well). That proof of non-algebraizability goes through the computation of a non-vanishing obstruction: the obstruction to extending an invertible sheaf from a given scheme to a given deformation of the scheme.

That section of that article also contains some other well-known examples: an example using a cardinality argument to prove the stack of affine schemes is not algebraic, and an example of a smooth, projective scheme whose automorphism group is not quasi-compact.