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By definition $\operatorname{Spec}k$ is a point for any field $k$. So $\operatorname{Spec}\mathbb{Q}, \operatorname{Spec}\mathbb{R}, \operatorname{Spec}\mathbb{C}$ etc are all the same as topological spaces. But according to the natural inclusion map $$ \mathbb{Q} \rightarrow \mathbb{R} \rightarrow \mathbb{C} $$ there exist natural morphisms $$ \operatorname{Spec}\mathbb{Q} \leftarrow \operatorname{Spec}\mathbb{R} \leftarrow \operatorname{Spec}\mathbb{C}, $$ but not the other direction. So $\{\operatorname{Spec}k\}_k$ must carries more information than merely one point topological space.

I would appreciate it if someone could kindly explain what is going on.

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    $\begingroup$ The ring of globally defined regular functions on $\text{Spec R}$ is $R$. Thus the difference between these three schemes are the functions on them. $\endgroup$ Oct 19, 2012 at 1:59

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The extra information that's carried along is the scheme structure. I.e., these are all locally ringed spaces with a single closed point, but with different sheaves of regular functions corresponding to the rings $\Bbb Q,\Bbb R,\Bbb C.$ The functions you describe carry along this sheaf information via pushforward along a trivial (set-theoretic/topological) map.

Note that $\operatorname{Spec}(\Bbb C[t]/t^n)$ is another one-pointed space with a different scheme structure from the rest. And there are many more examples, in fact you can take the spectrum of any local artinian ring.

PS - Don't worry, this makes algebraic geometry very rich! In a certain sense, the scheme structure "remembers" information that the topological space forgets, for example in degenerating families. Eisenbud-Harris and Hartshorne have nice examples, in chapter II and chapter II.9 respectively, if I remember correctly.

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  • $\begingroup$ Whoops, that Hartshorne reference should be III.9. $\endgroup$
    – Andrew
    Oct 19, 2012 at 2:44
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    $\begingroup$ I see. I should think of $Spec k$ as a pair of a topological space and the ring of functions on it. Your answer makes thing clearer. $\endgroup$
    – M. K.
    Oct 19, 2012 at 3:26
  • $\begingroup$ Dear @M.K., that's right. Glad to help! $\endgroup$
    – Andrew
    Oct 19, 2012 at 3:27

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