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


<http://www.math.tifr.res.in/~publ/ln/tifr54.pdf> (pages 4-6)


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


1. Very supid question: Are "smooth" and "nonsingular" schemes the same?
2. 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?
3. 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 $k$?
4. 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.



Example of a nonsmoothable scheme
1. In the situation at hand, non-singular and smooth are interchangeable.
2. As @Asura Path mentioned, the total space need not be smooth.
3. 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.
4. 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.


