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.

  • 2
    $\begingroup$ 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

1 Answer 1


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.

  • $\begingroup$ Whoops, that Hartshorne reference should be III.9. $\endgroup$
    – Andrew
    Oct 19, 2012 at 2:44
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.