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Consider the family $S_n:=\operatorname{Spec} A[x]/(x^n)$ of schemes, $A$ denoting any ring (which in our subject always means commutative and with identity). Is there some intuitive picture for $S_n$ available?

For instance I heard somewhere that $S_2$ (aka dual numbers) can be thought of as being two points coming together (well here $A$ was a algebraically closed field), is this correct?

Also as for the ring theoretic properties of $A[x]/(x^n)$, what can be said generally about this ring, in particular its Krull dimension, is it a local ring, etc?

Feel free to make some assumptions on $A$ since obviously without some assumptions hardly anything can be said about this ring (or family of rings whatever, I am interested in all $n$'s for I guess they should be all alike somehow). I guess that if $A$ is a field then it is a local ring of dimension $0$, and $S_n$ is a one point space. I am interested in these rings since they should be somehow understandable and are rather fundamental.

Also feel free to give references! Thanks in advance.

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These are the infinitesimal neighborhoods of the "origin" in $\mathbb{A}^1_A$. They have the same topological space as $\mathrm{Spec}(A)$ (in particular the same dimension). But the structure sheaf differs of course. Notice that the limit of the rings $A[x]/x^n$ is $A[[x]]$, the ring of formal power series. You can think of $A[x]/x^n$ as the ring of $n$th order approximations of formal power series.

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  • $\begingroup$ Thanks! is there a concise way to see that $S_n$ and $\mathrm{Spec} \ A$ are homeomorphic? $\endgroup$ – Zlatan der Zechpreller Jun 24 '14 at 10:33
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    $\begingroup$ Yes. If $I$ is any ideal, then $V(I)=V(I^n)$ as topological spaces. Now apply this to the ring $A[x]$ and the ideal $I=(x)$. Notice that $V(x)=\mathrm{Spec}(A)$ as schemes. $\endgroup$ – Martin Brandenburg Jun 24 '14 at 12:24

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