# First-Order Deformations

Today we actually get to some deformations. Let ${X_0}$ be a scheme of finite type over ${k}$. First, we’ll be working with the “ring of dual numbers” a lot, so we’ll just define it to be ${R=k[\epsilon]/(\epsilon^2)}$. Let’s recall a few useful properties first.

To give a map in ${k}$-schemes: Spec ${R\rightarrow X_0}$ is equivalent specifying a ${k}$-point and an element of the Zariski tangent space at that point.

Flatness is another important concept. An ${R}$-module, ${M}$, is flat if and only if the map ${M/\epsilon M\rightarrow M}$ by multiplication by ${\epsilon}$ is an injective. The proof is quite straightforward: Consider the exact sequence ${0\rightarrow k\stackrel{\epsilon}{\rightarrow} R\rightarrow k \rightarrow 0}$. If ${M}$ flat, tensor this with ${M}$ and you get ${0\rightarrow M/\epsilon M\stackrel{\epsilon}{\rightarrow} M\rightarrow M/\epsilon M\rightarrow 0}$, and hence injectivity. If the map is injective, then ${Tor^1(M,k)=0}$, so ${M}$ is flat.

Let’s move on now that those are out of the way. What exactly should these deformations be? Let’s say we have a nice family of schemes. This means that there is a map ${f:X\rightarrow T}$, and nice in our case means flat. ${T}$ parametrizes this family, since the fiber over any point gives a scheme ${X_t}$. There is a special fiber ${X_0}$, and deformations of ${X_0}$ are the schemes that occur in this family in a neighborhood of ${X_0}$. (Recall, this is still a “touchy-feely” idea of what a deformation is, don’t take this to be the definition).

For instance, you could parametrize some curves over ${\mathbb{A}^1_k}$ by ${Spec\left(k[x,y,t]/(xy-t)\right)\rightarrow Spec(k[t])}$. We could think about this family as hyperbolas $X_t=\{(x,y): xy=t\}$. As $t$ approaches 0, the hyperbolas degenerate into the coordinate axes. Deformations in this family of the special fiber ${X_0}$ are all irreducible, yet the special fiber is reducible.

Now for what we really care about in this post. A first-order deformation of ${X_0}$ is a scheme ${X'}$, flat over ${R}$ such that ${X'\otimes_R k\simeq X_0}$. Note the terminology comes from the fact that the family is over Spec ${R}$ which remembers “tangent” information. A second-order deformation would keep track of information via a family over Spec ${k[\epsilon]/(\epsilon^3)}$, etc (anyone see a completion happening in the near future?).

A first-order deformation ${X'}$ is something that completes the fiber diagram:

${\begin{matrix} X_0 & \rightarrow & X'\\ \downarrow & & \downarrow f\\ Spec k & \rightarrow & Spec R \end{matrix}}$

where ${f}$ is flat. Maybe a notation will be useful later: Def(${X_0/k}$).

Let’s use the last couple of posts to classify the first-order deformations of a non-singular scheme over an algebraically closed field. The claim is that Def(${X/k}$) (where these are up to isomorphism) is in bijective correspondence with ${H^1(X, \mathcal{T}_X)}$. So by the last post it is enough to show bijective correspondence with the infinitesimal extensions of ${X}$ by ${\mathcal{O}_X}$.

The proof is just adapting what we said at the start of the post. If ${X'\in Def(X/k)}$, then by flatness we can tensor with ${0\rightarrow k\rightarrow R\rightarrow k\rightarrow 0}$ and it remains exact: ${0\rightarrow \mathcal{O}_X\rightarrow \mathcal{O}_{X'}\rightarrow \mathcal{O}_X}$. Thus ${X'}$ is an infinitesimal extension of ${X}$ by ${\mathcal{O}_X}$. Conversely, any such extension is flat and hence a first-order deformation.

## 5 thoughts on “First-Order Deformations”

1. Is Hartshorne your source for this material? I vaguely remember seeing it in the exercises some time back, but I wasn’t able to solve it then.

Also, do we need $M$ to be finitely generated for the statement about flatness over the ring of dual numbers to hold? The proof I know of the statement that $Tor_1(M,k)=0$ implies flatness requires finite generation (via Nakayama’s lemma), but it is possible things are different for this particular ring (or that the result I know is weaker than necessary).

2. hilbertthm90 says:

The past two posts were a set of three exercises that I solved. This post was extrapolated from the flatness section of III in Hartshorne. Overall, I’ve been glancing through Hartshorne’s GTM on Deformation Theory and Ravi Vakil’s notes to get some more of an overview of the subject.

The more general statement of the Tor result is: If M is any module over a Noetherian ring, then it is flat if and only if for every prime $p\subset A$, $Tor_1^A(M, A/p)=0$.

A quick sketch: $Tor_1(M, N)=0$ for all N if and only if the functor $- \otimes_A M$ is exact and hence $M$ flat. But Tor commutes with limits, so $Tor_1(M, N)=0$ for all N if and only if $Tor_1(M,N)=0$ for all finitely generated N. But finitely generated modules have a filtration with quotients of the form $A/p_i$ for some primes $p_i$. So take the long exact sequence associated to $0\to p_i \to A \to A/p_i \to 0$ to see that if $Tor_1(M, p)=0$ for all primes, then $Tor_1(M, N)=0$ for all finitely generated modules.

3. That’s a nice argument; thanks for explaining.