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.


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.

  • $\begingroup$ 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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