# How are $Spec \mathbb{Q}, Spec \mathbb{R}, Spec \mathbb{C}$ etc different?

By definition $Spec k$ is a point for any field $k$. So $Spec \mathbb{Q}, Spec \mathbb{R}, Spec \mathbb{C}$ 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 $$Spec \mathbb{Q} \leftarrow Spec \mathbb{R} \leftarrow Spec \mathbb{C},$$ but not the other direction. So $\{Spec k\}_k$ 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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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. –  Michael Joyce Oct 19 '12 at 1:59

## 1 Answer

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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Whoops, that Hartshorne reference should be III.9. –  Andrew Oct 19 '12 at 2:44
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. –  M. K. Oct 19 '12 at 3:26
Dear @M.K., that's right. Glad to help! –  Andrew Oct 19 '12 at 3:27
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