Let $k$ be an algebraically closed field. I am looking for an easily quotable description of automorphism groups of $\mathrm{Spec} k[x]/(x^n)$. I could compute explicit matrix representations for several small $n$, but a more intrinsic description would be desirable.

$\begingroup$ One can give a basis for the Lie algebra and describe the Lie bracket, would that be sufficient? $\endgroup$ – Will Sawin Jan 26 '14 at 16:18

1$\begingroup$ if it's just the coefficients of the bracket as a bilinear map, then it's not much different from matrix representations. I am rather curious if there is a description of this group, for example, in terms of simpler groups, as a sequence of extensions. $\endgroup$ – Dima Sustretov Jan 26 '14 at 16:23

5$\begingroup$ It's an extension of one copy of $\mathbb G_m$ by $n2$ copies of $\mathbb G_a$. The tricky bit is describing which extension. Specifically, what's the unipotent subgroup. There is no classification of unipotent groups, and as far as I know, this has no description in terms of other standard unipotent groups. $\endgroup$ – Will Sawin Jan 26 '14 at 17:24

$\begingroup$ I guess I shouldn't have said what I said since I thought about the fact that it is a sequence of extensions of G_m by several copies of G_a, and indeed it's not clear how one would describe in general what kind of extension it is. My hope was that people thought about these automorphism groups before me and either have come up with a description of this extension or maybe something more enlightening. $\endgroup$ – Dima Sustretov Jan 26 '14 at 18:03

$\begingroup$ @Dima: Does it matter here what the characteristic of $k$ is? Or is the answer (assuming one can find it) uniform? $\endgroup$ – Jim Humphreys Jan 26 '14 at 18:16
Smooth points are all alike; every fat point is happy in its own way.
"Anna Karenina", Leo Tolstoy
Let $k$ be an algebraically closed field. Denote $A_n:=k[x]/x^{n+2}$, $G_n:=\text{Aut}A_n$, and let $N_n\subset G_n$ stand for the unipotent radical of $G_n$. Since there is no problem in finding a faithful matrix representation of $G_n$, I guess, Dima asks about a coordinatefree description of $G_n$.
We will mainly deal with $N_n$ because $G_n=N_n\rtimes T$ is a semidirect product, where $\text{G}_m\simeq T\subset G_n$ has no canonical choice (though all tori are conjugated in $G_n$).
Answer. We assume for simplicity $k$ of characteristic $0$. For sufficiently big $n$, $N_{n+1}$ is a central extension (with that very centre) of $N_n$. This extension is determined by a certain orbit of the group $\text{Out}N_n$ naturally acting on the projective plane ${\mathbb P}_kH^2(N_n,k)$. The orbits are: a dense one, two $1$dimensional orbits $L'$ and $H'$ dense in the projective lines $L$ and $H$, and a couple of fixed points $f_0,f_1\in L$, where $f_0$ is the intersection $L\cap H$. Our choice determining the extension $0\to k\to N_{n+1}\to N_n\to1$ is $H'$. The action of $T$ can be easily defined on the way.
The interesting stuff appears at the end, when the point becomes really fat.
Boring stuff. Every element $g\in G_n$ is determined by the polynomial $f_g:=gx$, where $f_g(x)=f_0x+f_1x^2+\dots+f_nx^{n+1}$ with $f_i\in k$ and $f_0\ne0$. The composition in $G_n$ is expressed in these terms as $f_{g_1g_2}(x)\equiv f_{g_2}\big(f_{g_1}(x)\big)\mod x^{n+2}$. Note that $g\in N_n$ iff $f_0=1$ and $g\in T$ iff $f_1=\dots=f_n=0$. Sometimes, it is more convenient to use other coordinates: $f_g(x)=x\Big(h_g(x)+\displaystyle\sum_{i=0}^n(p_1x)^i\Big)$, where $h_g(x)=\displaystyle\sum_{i=2}^np_ix^i$. So, $p_1=f_1$ and $p_i=f_if_1^i$ for $i\ge2$.
The Lie algebra $L_n$ of $G_n$ is formed by all the derivations from $A_n$ to $A_n$, i.e., by all the derivations from $k[x]$ to $k[x]$ that preserve $\text{Ideal}(x^{n+2})$. The formula $df(x)=f'(x)dx$ correctly defines a derivation $d\in L_n$ by its value $dx\in A_nx$. Taking $d_ix=x^{i+1}$ for $i=0,1,\dots,n$, we obtain a basis of $L_n$ subject to $[d_i,d_j]=(ji)d_{i+j}$ for $i+j\le n$ and $[d_i,d_j]=0$ for $i+j>n$. The Lie algebra $M_n$ of $N_n$ is simply spanned by $d_1,\dots,d_n$.
0. $N_0=1$ and $G_0=\text{G}_m$.
1. $N_1=\text{G}_a$ and $G_1=N_1\rtimes\text{G}_m$ is the semidirect product with respect to $\text{G}_m=\text{Aut}\text{G}_a$.
2. $N_2=\text{G}_a\oplus\text{G}_a$ (the corresponding coordinates are, for instance, $p_1,p_2$) and $G_1=N_1\rtimes\text{G}_m$, where the action of $c_0\in\text{G}_m$ is given by the rule $c_0\cdot(p_1,p_2):=(c_0p_1,c_0^2p_2)$.
Preliminaries. The homomorphism $\pi:A_n\to A_{n1}$ induces the surjective homomorphisms $\pi:G_n\to G_{n1}$ and $\pi:N_n\to N_{n1}$ whose kernel $\text{G}_a\simeq C_n\subset N_n$, formed by all $c\in N_n$ with $f_c=x(1+fx^n)$, $f\in k$, lies in the centre of $N_n$ and coincides with this centre unless $n=2$.
A few more wellknown and easy facts. Under the assumption that there is a rational section of $\pi$ (which is valid in our case of $X_h=N_{n+1}$ and $X=N_n$), a central extension $0\to k\to X_h\overset\pi\to X\to1$ of algebraic groups is given by an (arbitrary) element $h\in H^2(X,k)$ of rational $2$cohomologies of $X$ (the action of $X$ on $k=\text{G}_a$ is trivial). The groups $\text{Aut}X$ and $\text{G}_m$ act on $X$ and on $k$, hence, on $H^2(X,k)$. Assuming that $k$ coincides with the centre of $X_{h_i}$ for both $i=1,2$, the groups $X_{h_1}$ and $X_{h_2}$ are isomorphic iff $h_1$ and $h_2$ are in a same orbit of $\text{Out}X\times\text{G}_m$ (the inner automorphisms of $X$ act trivially on $H^2(X,k)$).
In our case, the central extension is nontrivial. Denote by $0\ne[h_n]\in H^2(N_n,k)$ an element providing the extension $\pi:N_{n+1}\to N_n$. By the above, the group $N_{n+1}$ is in fact given by the choice of the orbit of $[h_n]$ with respect to the action of $\text{Out}N_n$ on ${\mathbb P}_kH^2(N_n,k)$. Our task is therefore to indicate this orbit and to explain how $T\simeq\text{G}_m$ acts on $N_{n+1}$ (already knowing how does $T$ act on $N_n$).
The homomorphism $\pi$ induces a $k$linear map $\pi^*:H^2(N_{n1},k)\to H^2(N_n,k)$ and, unless $n=2$, a homomorphism $\pi:\text{Out}N_n\to\text{Out}N_{n1}$. Another elementary remark: an element $\alpha\in\text{Out}N_{n1}$ belongs to the image $\pi\text{Out}N_n$ (i.e., is liftable) iff $[h_{n1}]\in{\mathbb P}_kH^2(N_{n1},k)$ is a fixed point of $\alpha$.
Using the ChevalleyEilengebrg complex of $M_n$, one can calculate $H^2(M_n,k)\simeq H^2(N_n,k)$ for $n\ge2$ (the action of $M_n$ on $k$ is trivial), getting the following result. For any $n\ge4$, we have the cocycle $a:=d_2^*\wedge d_3^*$; for any $n\ge6$, we have the cocycle $b:=d_2^*\wedge d_5^*3d_3^*\wedge d_4^*$; for any $n\ge2$, we have the cocycle $h_n:=\displaystyle\sum_{1\le i<j\le n\atop i+j=n+1}(ji)d_i^*\wedge d_j^*$. Now, $1$dimensional $H^2(M_2,k)$ and $H^2(M_3,k)$ are spanned respectively by $[h_2]$ and $[h_3]$, $2$dimensional $H^2(M_4,k)$ and $H^2(M_5,k)$ are spanned respectively by $[a],[h_4]$ and $[a],[h_5]$, and a $3$dimensional $H^2(M_n,k)$ is spanned by $[a],[b],[h_n]$ when $n\ge6$. Note also that (when it makes sense) $\pi^*:[a]\mapsto[a]$, $\pi^*:[b]\mapsto[b]$, and $\pi^*:[h_n]\mapsto0$.
First steps. I claim (without proof as it is bulky) that $\text{Out}N_n$ is $4$dimensional.
Since $N_2$ is in fact a $2$dimensional $k$linear space, we conclude that $\text{Out}N_2=\text{GL}_2k$. In view of $\dim_kH^2(N_2,k)=1$, we have a unique possible extension, thus obtaining the famous Heisenberg group $N_3$. The group $\text{Out}N_2$ is entirely liftable, and $\text{Aut}N_3$ is formed by $3\times 3$matrices $G:=\left[\begin{smallmatrix}g&0\\{*}&\det g\end{smallmatrix}\right]$, where $g\in\text{GL}_2k$. Taking $g=1$, we obtain the normal subgroup $N_2\subset\text{Aut}N_3$ of inner automorphisms of $N_3$, hence, $\text{Out}N_3=\text{GL}_2k$ is $4$dimensional, as claimed. The action of $T$ on $N_3$ will be defined later.
Since $\dim_kH^2(N_3,k)=1$, we again get a unique possible extension $N_4$, the entirely liftable $\text{Out}N_3$, and $\text{Out}N_4=\text{GL}_2k$.
When $\dim_kH^2(N_4,k)=2$, we are stuck.
Lemma. For $n\ge5$, there is a $1$dimensional subgroup $\text{G}_a\simeq U_n\subset\text{Out}N_n$ such that $\text{Out}N_n=U_n\cdot\pi\text{Out}N_{n+1}$, $U_n\cap\pi\text{Out}N_{n+1}=1$, $\pi^3U_n=1$, and $\pi^2U_n\ne1$.
In place of a proof, we only define how $u\in k$ acts on $f(x)=x\Big(1+\displaystyle\sum_{i=1}^nf_ix^i\Big)\in N_n$. (Let me skip some reasons allowing to get rid of an unpleasant straightforward verification.) We use the coordinates $p_1:=f_1$ and $p_i:=f_if_1^i$ for $i\ge2$. Consider the "quadratic" polynomial $q_f(x):=q_{f,0}+q_{f,1}x+q_{f,2}x^2$, where $$q_{f,0}:=p_2,\qquad q_{f,1}:=(n3)p_32(n4)p_2p_1,$$ $$q_{f,2}:=\frac{(n2)(n3)}2p_4(n3)(n4)p_3p_1\frac{(3n14)(n1)}4p_2^2+\frac{(n4)(n9)}2p_2p_1^2.$$ The action of $u\in k$ on $f$ is given by the rule $u\cdot f:=f+ux^{n1}q_f$.
Fat point fun. Using Lemma and omitting the $\pi$'s, we can write $\text{Out}N_n=U_nU_{n+1}U_{n+2}T$. As $U_{n+1},U_{n+2},T$ are liftable, they fix the point $[h_n]\in H^2(N_n,k)$. As $U_n$ is not liftable, the orbit $(\text{Out}N_n)[h_n]=U_n[h_n]$ is $1$dimensional. The action of $U_n,U_{n+1},U_{n+2}$ on $[a]$ and $[b]$ is trivial because it can be performed at the level of $H^2(N_{n3},k)$, where $U_n,U_{n+1},U_{n+2}$ vanish. It is easy to see that $[a]$ and $[b]$ are fixed points of $T$ of different weights. Finally, we can get the picture of $\text{Out}N_n$orbits in ${\mathbb P}_kH^2(N_n,k)$. There is a dense orbit which is the complement of two lines $L$ and $H$. The intersection $L\cap H$ is a fixed point $f_0$ (either $[a]$ or $[b]$; no idea which one). The line $L$ spanned by $[a]$ and $[b]$ contains therefore one more fixed point $f_1$. Our choice is the $1$dimensional orbit $H'$ such that the compliment $H\setminus H'$ in its closure $H$ is a single point $f_0$.
I forgot to define the action of $T$ on $N_n$. For big $n$, it is easy: just take any torus $T\subset\text{Aut}N_n$. In order to get the action of $T$ for small $n$, simply apply a number of $\pi$'s.
Conclusion. When we killed $[h_n]$ by inserting it as a centre, a new $[h_{n+1}]$ was born, and the other cohomologies $[a]$ and $[b]$ survived. What if we kill $[a]$ or $[b]$, once or several times. How probable is that we obtain a group of automorphisms of some fat point? (The algebras $A_n$ provide possibly the simplest series of artinian local algebras. Of course, there is no hope to classify all local artinian algebras. Nevertheless, my question may have a tame nature.)