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As topological spaces, all of $\text{Spec}(k), \text{Spec}(k(x)), \text{Spec}(k[x]/(x^2))$ and $\text{Spec}(k(x_1,\cdots,x_n))$ are all homeomorphic, since they are all one point-spaces.

However, as schemes they are different. We have been told to think of $\text{Spec}(k[x]/(x^2))$ as a point together with a tangent direction, but how do we think of, for example, $\text{Spec}(k(x_1,\cdots,x_n))$?

My guess would be that a good way to think about $\text{Spec}(k(x_1,\cdots,x_n))$ is as a point together with a point in $\mathbb{A}_k^n$.

(inspired by the fact that $\text{hom}(Spec(k),Spec(k(x_1\cdots,x_n))=\mathbb{A}_k^n$ (at least for algebraically closed k))

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Note that Hom$(Spec(k), Spec(k(x_1,\dots, x_n))$ is empty unless $n=0$. You meant Hom$(Spec(k), Spec(k[x_1,\dots, x_n])$ ? –  user18119 Oct 11 '12 at 19:16

3 Answers 3

up vote 2 down vote accepted

One can think $\mathrm{Spec}(k(x_1,\dots,x_n))$ as a point of the affine space $\mathbb A^n_k$ with coordinates which are all variables. It is called the generic point. When you specify a value for each variable, you get a specialization of the generic point, which is then a closed point, a point with coordinates in $k$.

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One-point schemes are of course affine, of the form $X=Spec(A)$ with $A$ a local ring of dimension zero.
The ring $A$ will be artinian if and only if it has finite length $l(A)\lt \infty$ and that length can be used as a measure of the complexity of your one-point scheme $X$.
For example the schemes you mention $Spec (k[x]/\langle x^n\rangle )$ correspond to rings of length $n$ and are in a sense more and more complicated with growing $n$.

Finally beware that there exist local rings of dimension zero (thus yielding one point schemes) which are nevertheless not artinian (nor even noetherian): an example is $k[x_1,x_2,...]/\langle x_ix_j\mid i,j=1,2,...\rangle $

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And it shows up again! :-) –  Mariano Suárez-Alvarez Oct 11 '12 at 22:11
    
I was wondering if I could sneak it in without your noticing, but nothing escapes your hawk-eye: bravo, Mariano! –  Georges Elencwajg Oct 11 '12 at 22:16
    
[The above is a private joke: Mariano and I have used the counterexample in the answer quite a number of times] –  Georges Elencwajg Oct 11 '12 at 22:17

To emphasize that we can think of $\operatorname{Spec}(k(x_1,\ldots,x_n))$ as a generic point, remember that $k(x_1,\ldots,x_n)=k[x_1,\ldots,x_n]_{(0)}$ is the local ring $\mathcal O_{\Bbb A^n,(0)}.$ (This ring, of course is its own residue field $\kappa((0))$ also.) Thus, $\operatorname{Spec}(k(x_1,\ldots,x_n))=\operatorname{Spec}(\mathcal O_{\Bbb A^n,(0)}),$ which we think of as a local scheme corresponding to the generic point $(0)$ of $\Bbb A^n,$ or a single undetermined point of $\Bbb A^n,$ if that helps at all.

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