Example of a nonsmoothable scheme

Question

I try to understand Iarrobinos example of a nonsmoothable 0-dimensional scheme with the help of Artins notes on it:

But I have some difficulties with this topic. So here are my questions:

Very supid question: Are "smooth" and "nonsingular" schemes the same?
Just to be sure: If we are talking about deforming a scheme into a nonsingular one, it means that the total scheme of the deformation is nonsingular, right?
In Artins notes, there are no restrictions stated about the parameter space of the deformation. Can it be any scheme? Or does it have to be $Spec (k)$, since the term "smooth" is only defined for schemes over a field $k$?
In Artins notes, we are only looking for affine schemes $X=Spec (\mathcal{O})\hookrightarrow \mathbb{A}^n$ of krull dimension 0, so $\mathcal{O}$ is a finite-dimensional $k$-algebra of dimension $d$. Artin writes: "In our particular case the question is whether $X$ can be deformed into $d$ distinct points of $\mathbb{A}^n$." This sentence confueses me very much, and I have absolutely no idea what it means.

Thank you for your help.

2. Probably it means that there is a non-singular fiber in your family, I don't think that it refers to the total space. — user141498, Jun 7, 2019 at 16:35
3. $\mathrm{Spec}\:k$ has only one point so not much space for deformations there. I am not sure what are the parameter spaces allowed to be (I have seen DVRs, Artinian local rings, affine spaces, projective spaces used, but it would be hard to write down a complete list as essentially any morphism could be considered as a family, depending on your needs). — user141498, Jun 7, 2019 at 16:37

Mohan · Accepted Answer · 2019-06-07 17:34:05Z

1

In the situation at hand, non-singular and smooth are interchangeable.
As @Asura Path mentioned, the total space need not be smooth.
Parameter space can be any scheme (over $k$), since one is looking at deformations of a finite dimensional $k$-algebra. If such an algebra can be smoothened, then you may assume that the parameter space is a smooth irreducible curve for the smoothening.
If $k$ is algebraically closed, the only smooth finite dimensional algebra over it is the ring of functions on a finite set of points, since any other would have nilpotents and thus not smooth.

answered Jun 7, 2019 at 17:34

Mohan

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$\begingroup$ Thank you for your answer! The first three questions are now clear to me, but I am still confused about 4. You are talking about smooth algebras, but in which way does this algebra occur in the deformation diagram, is it the ring of global sections of the total space? $\endgroup$
– flitwick
Jun 7, 2019 at 17:58
$\begingroup$ @flitwick If $S$ is the parameter space, then total space $X$ maps (flat and in this case finite) to $S$ and a special fiber is your finite dimensional $R$ you started with and general fiber is the ring of functions on finitely many points. If no such $X,S$ exist, then $R$ can not be smoothened as in Iarrobino's example. $\endgroup$
– Mohan
Jun 7, 2019 at 18:45
$\begingroup$ I think my problem was that I didnt realize that $X\rightarrow S$ is finite - why does this hold? But ok, if this map is finite, then the fibers constist of finitely many points. I have one fiber which is isomorphic to $R$ and if $R$ is smoothable, there is at least one other fiber which is smooth over $k$. You said that this other fiber is finite the (spectrum of a) ring of functions on finitely many points, because it is smooth and finite dimensional, but why is it finite dimensional? $\endgroup$
– flitwick
Jun 7, 2019 at 21:00
$\begingroup$ @flitwick Finite and flat implies all fibers have the same dimension over $k$, so they all have the same dimension as that of $R$. $\endgroup$
– Mohan
Jun 7, 2019 at 21:37

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