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Let $(A,\mathfrak{m}_A)$ be a local Artinian $k$-algebra with residue field $k$. Then the scheme $\mathrm{Spec}(A)$ can be loosely seen as a "fat point", or an "infinitesimal neighbourhood" of a point $\mathrm{Spec}(k)$. If the maximal ideal satisfies the condition $\mathfrak{m}^k=0$, then I think the spec of $A$ can be seen as a "$(k-1)$-th order infinitesimal neighbourhood" of the closed point.

Recall: a small extension of $A$ is an extension of local Artinian $k$-algebras $0\to I \to B\to A \to 0$ such that $\mathfrak{m}_B\cdot I =0$.

Does the concept of a small extension have any geometric interpretation?

Does it mean that the closed embedding $\mathrm{Spec}(A)\hookrightarrow \mathrm{Spec}(B)$ is such that it only "fattens" already existent "directions" of $\mathrm{Spec}(A)$, making it into a "higher order" fattening of the closed point but without adding new "directions" of fattening? Is my suggestion totally misleading? Edit: as comments point out, my interpretation was not correct.

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I wouldn't call this a soft question! Fat, maybe... – Tom Leinster Jan 27 '11 at 17:23
I think extensions do add "directions", think about the extension $k[\epsilon]\to k$. $k$ is just the point, $k[\epsilon]$ is a point with a tangent vector.. – Mattia Talpo Jan 27 '11 at 19:36
@Mattia: you're definitely right. The usual extension to $k[\epsilon]$ is small and it adds a "direction"... – Qfwfq Jan 27 '11 at 22:57
This is a nice question, I think. Here is a counterexample in the other direction: consider $A = k[x,y]/(x^2,y^2)$ and $B=k[x,y]/(x^3,y^2)$. This is not a small extension since $x^2y$ is a nontrivial element of $\mathfrak{m}_B\cdot I$, but no new "direction" is added. – Dan Petersen Jan 28 '11 at 6:23
up vote 3 down vote accepted

If you think of elements in a local Artin $k$-algebra $R$ as, say functions on the origin of $k^n$ which remember some (finite amount of) higher order information in the various $n$ directions, then a small extension $R'$ of $R$ is just another such ring with functions that remember "at most one order higher".

For example, let $R = k[x,y]/((x,y)^2)$ and consider the small extension $R' = k[x,y]/((x,y)^3)$. The elements of $R$ are functions which remember up to 1st order in the directions $x,y$. The elements of $R'$ are what we get if we take functions from $R$ and stick on some 2nd order terms in the $x,y$.

You can also just extend only in one direction. For example, $R'' = k[x,y]/(x^3,xy,y^2)$ is also a small extension of $R$, but the only 2nd order term we added was $x^2$, not $xy$ nor $y^2$.

However, if you take $A = k[x]/(x^2)$ and $A'' = k[x]/(x^4)$, then this is not a small extension because we went two orders up, from 1st order to 3rd order.

Generally, I interpret a small extension as one that only thickens our fat point by an order of at most 1 in each direction.

Things are more complicated if you add a new direction. If you take $R = k[x,y]/((x,y)^2)$ as before, and consider $R[z]/(z^2)$, then this is not a small extension because of the cross terms $xz$ and $yz$. However, if you get rid of them, then $R[z]/(z^2,xz,yz)$ is a small extension of $R$.

You could say that adding a new direction involves making two small extensions. The first one adds the new variable with no cross terms, and the second adds the cross terms. For instance, let $R = k[x]/(x^2)$, $R' = R[y]/(xy,y^2)$, and $R'' = R[y]/(y^2)$.

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Thank you! That was exactly the kind of answer I was looking for. – Qfwfq Jan 28 '11 at 17:57

The way I see this is the following, which I learnt from Sernesi's book "Deformations of algebraic schemes".

Assume that you have an infinitesimal deformation $\xi$ of a nonsingular scheme $X$ over $\textrm{Spec}(A)$, i.e. a flat map $\mathcal{X} \to \textrm{Spec}(A)$ together with an identification of $X$ with the fibre over $\textrm{Spec}(k)$.

Then, whenever one has a surjection $B \to A$ of local Artinian $k$-algebras, it is natural to ask whether one can lift $\xi$ to a deformation $\xi'$ over $\textrm{Spec}(B)$, i.e. whether one can find $\mathcal{X}' \to \textrm{Spec}(B)$ which extends $\mathcal{X} \to \textrm{Spec}(A)$.

In the case of small extensions, the answer is very simple: in fact, given $\xi$ and any small extension $e$ of $A$, there is associated an element $o_{\xi}(e) \in H^2(X, T_X)$, called "obstruction", which is zero if and only if a lifting $\xi'$ of $\xi$ to $\textrm{Spec}(B)$ exists.

Moreover, if $o_{\xi}(e)=0$ then there is a natural transitive action of $H^1(X, T_X)$ on the set of isomorphism classes of liftings $\xi'$.

Finally, the correspondence $e \to o_{\xi}(e)$ defines a $k$-linear map

$o_{\xi} \colon \textrm{Ex}_k(A,k) \to H^2(X, T_X).$

Summing up, whenever one has a small extension $0 \to I \to B \to A \to 0$ and a smooth scheme $X$, it is possible to lift over $\textrm{Spec}(B)$ exactly those infinitesimal deformations $\xi$ of $X$ over $\textrm{Spec}(A)$ such that the corresponding obstruction $o_{\xi}(e)$ vanishes; moreover, the isomorphism classes of liftings form a homogeneous space under the natural action of the group $H^1(X, T_X)$ of first-order deformations of $X$.

For instance, given the small extension

$0 \to \frac{(t^{n-1})}{(t^n)} \to \frac{k[t]}{(t^n)} \to \frac{k[t]}{(t^{n-1})} \to 0$,

the obstruction map is non-zero exactly for those "directions" in the $n-1$-th infinitesimal neighborhood of $X$ which cannot be "fattened" into the $n$-th infinitesimal neighborhood.

In particular, if $H^2(X, T_X)=0$ then every infinitesimal deformation of $X$ over $\textrm{Spec}(A)$ can be lifted to an infinitesimal deformation over $\textrm{Spec}(B)$.

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So, the smallness condition is just a technical requirement to make the theory work, or does it also have a more direct "visual" interpretaion? – Qfwfq Jan 27 '11 at 22:59
Personally, I am able to have a "visual interpretation" only for small extensions of type $0 \to (t^{n-1})/(t^n) \to K[t]/(t^n) \to k[t]/(t^{n-1}) \to 0$. In this case, the vanishing of the obstruction map characterizes those "infinitesimal tickenings of order n-1" of $X$ which lift to a "tickening of order $n$". Or, if you prefer, the obstruction map is non-zero for those "directions" in the $n-1$-th infinitesimal neighborhood of $X$ that cannot be "fattened" into the $n$-th infinitesimal neighborhood. – Francesco Polizzi Jan 27 '11 at 23:29
In the general case, I lose the visual interpretation but it still makes sense to try lifting infinitesimal deformations from $\textrm{Spec}(A)$ to $\textrm{Spec}(B)$. I added the example above into the answer... – Francesco Polizzi Jan 27 '11 at 23:42
One important point is that you can always reduce to considering small extensions, as long as you're concerned with surjective maps of artinian local $k$-algebras (which correspond to "thickenings"), because any such surjection can always be written as a finite composition of small extensions. – Mattia Talpo Jan 28 '11 at 0:03

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