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

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.

• 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. Oct 19, 2012 at 1:59

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.
• 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. Oct 19, 2012 at 3:26