A result in deformation theory states that if every morphism $Y=\operatorname{Spec}(A)\rightarrow X$ where $A$ is a local Artin ring finite over $k$ can be extended to every $Y'\supset Y$ where $Y'$ is an infinitesimal thickening of $Y$, then $X$ is non-singular.

My question is:

if $k$ is algebraically closed, can we say explicitly what every local Artin ring finite over $k$ is?

$A$ can be of the form $k[t]/(t^n)$ or for instance if its maximal ideal isn't principally generated $k[t^2,t^3]/(t^4)$. Are there $A$ which we cannot write in this "adjoin various powers of $t$ and mod out by some power of $t$" form? If there are more exotic $A$ can we say anything non-tautological about the structure of such an $A$?


YuiTo Cheng
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    Well since it's finite, it can be embedded/'found' in $\mathbb{A}^n$, and if my definitions are right, local artin means just one prime/maximal ideal, corresponding to WLOG the origin in $\mathbb{A}^n$. so up to isomorphism you can write it as $k[x_1, ... x_n]/I$ where $\sqrt{I} = (x_1 ... x_n)$. – uncookedfalcon May 07 '13 at 21:12
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    I'll leave it to an expert to make this into an answer, but to the best of my knowledge the answer is a resounding **no**. This is what the Nullstellensatz sweeps under the rug: the structure of non-reduced affine rings is horrendously complicated, even in the zero-dimensional case. I believe that there are for instance difficult open problems in classical algebraic geometry that would be easy to solve if some kind of explicit parameterization of zero-dimensional affine schemes were known: e.g. try looking up the **Harbourne-Hirschowitz Conjecture**. – Pete L. Clark May 07 '13 at 21:13
  • @PeteL.Clark Yes thanks for your intuition. Perhaps another question more to the point--I have a moduli space for which I can prove every morphism from $k[t]/(t^n)$ can be lifted to any infinitesimal thickening of the form $k[t]/(t^m)$. Do I get anything from this? Is there a reasonable example of a singular scheme for which I can lift as above? –  May 09 '13 at 20:27
  • @Ryan: I'm sorry, your followup question is really out of my depth: it sounds like you need a "real" algebraic geometer. I would try making an edit to bump your question, and if that doesn't work, repost to Math Overflow. – Pete L. Clark May 09 '13 at 20:51
  • @PeteL.Clark I don't know if can be said much about the structure of artinian local finite $k$-algebras when $k$ is algebraically closed. For example, in Artin's book, [Lectures on Deformations of Singularities](https://www.math.umass.edu/~hacking/seminarF12/artin.pdf), one can find on page 4 all these algebras for dimension $1, 2, 3$, and $4$. Can we find a hidden pattern in these examples? –  May 11 '13 at 16:46
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    @YACP: I am pretty sure that the answer is no: no one knows such a pattern. The point is that in order for a negative answer like this to be convincing, it has to come from a real expert in the field: someone who is suitably empowered to say not just "I don't know this" but "*we* don't know this". Thus I would be happy for a serious algebraic geometer to essentially repeat my comment as an answer. – Pete L. Clark May 12 '13 at 16:51
  • This is surely a wild problem. Let $V$ a vector space and fix $n_1$, $\dots$, $n_5$ be integers each at most $\dim V$. Consider an ordered tuple $U=(U_1,\dots,U_5)$ of subspaces of $V$ of those dimensions, let $X_U$ be the union of the $U_i$ and let $A_u$ be the quotient of the completion at the origin of $X_U$ modulo the $r$th power of the radical. If $r$ is sufficiently large with respect to $\dim V$, then I would imagine that the algebras $A_U$ corresponding to two $U$'s are isomorphic iff the $U$'s are conjugated in the obvious sense. – Mariano Suárez-Álvarez May 13 '13 at 20:00
  • *(cont.)* Now the classification of such $5$-tuples is a well-known wild problem. – Mariano Suárez-Álvarez May 13 '13 at 20:02
  • @MarianoSuárez-Alvarez Does this well-known problem have a name/do you have a reference to somewhere discussing it? The terms are a bit vague for my googling to turn up something fruitful and my knowledge of "well-known" problems is sometimes lacking. –  May 13 '13 at 20:41
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    It is the "five subspace problem". The corresponding "four subpace problem" was sovled by Gelfand and Ponomarev (it is a tame clasification problem) but I don't know who showed that with 5 subspaces the problem becomes wild. It follows from standard results of the representation theory of quivers, or of posets. See, for example, the book by Gabriel and Roiter on finite dimensional algebras. – Mariano Suárez-Álvarez May 13 '13 at 20:52
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    You might be interested in the paper "Isomorphism types of commutative algebras of finite rank over an algebraically closed field" by Poonen. It seems that up to $n=6,$ there are finitely many isomorphism types, but for $n\ge 7$ there are always infinitely many. – Andrew May 28 '13 at 16:09
  • @user16544: you probably already knew this, but I want to tell you that your criterion for lifting is already enough to say your moduli space (let's say finite type over a field) is smooth. And you can check this by looking at formal neighborhood of each point. – lee Feb 04 '17 at 04:06

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