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I am a newcomer to schemes. Joe Harris' book suggests we do not get too thrown off by the terminology and machinery.

At this point I can not name any interesting schemes except for $\text{Spec} R$ for a ring $R$. Any other examples? For $\text{Spec} \mathbb{Z}$ the points are the primes. What are the points and stalks for these other schemes?

I wish to understand how $\text{Spec} \mathbb{Z}$ is the terminal object in the category over schemes over $\text{Spec} \mathbb{Z}$. What would be the unique morphism to Today I wish to understand how $\text{Spec} \mathbb{Z}$ is the terminal object in the category over schemes over $\text{Spec} \mathbb{Z}$?

cactus314
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  • similar: http://math.stackexchange.com/questions/1249/motivating-example-for-algebraic-geometry-scheme-theory – cactus314 Mar 13 '17 at 22:32
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    There's already quite a lot of things to say about affine schemes. Arguably the simplest examples of non-affine schemes are the projective spaces $\mathbb{P}^n$, which any textbook on algebraic geometry will get to sooner or later. – Qiaochu Yuan Mar 13 '17 at 22:32
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    Remark: If you have a category $\mathcal{C}$ and an object $C$ of $\mathcal{C}$, then the category of objects of $\mathcal{C}$ over $C$ always has $C$ (or perhaps more precisely $C\xrightarrow{id_C} C$) as a terminal object: every object of this category is a morphism $X\to C$ in $\mathcal{C}$, and a morphism between $X\to C$ and $Y\to C$ is a map $X\to Y$ such that $X\to C$ is the same as $X\to Y\to C$. There is a unique morphism $X\xrightarrow{f} C$ to $C\xrightarrow{id_C}C$: it is $X\xrightarrow{f} C$. Essentially, this category is constructed to have $C$ as a terminal object. – Stahl Mar 13 '17 at 22:44
  • one example that i rather liked was $\mathbb{Q}[x,t]/(x^3-t)$ or possibly with $\mathbb{Q}$ instead of $\mathbb{Z}$. These objects could have no points? Maybe over the various completions $\mathbb{R}$, $\mathbb{Q}_p$ or $\mathbb{Z}_p$ and aren't schemes supposed to organize all of that into one pretty package? Actually profoundly significant package. Hopefully I wrote it correctly. – cactus314 Mar 13 '17 at 22:45
  • The more interesting statement is that the category of schemes (in general, not necessarily schemes over a base) has $\operatorname{Spec}\Bbb Z$ as a final object: this is a consequence of $\Bbb Z$ being initial in the category of rings, although there is now something to show, since not all schemes are $\operatorname{Spec}R$ for some ring $R$. – Stahl Mar 13 '17 at 22:45
  • @stahl all schemes have maps to $\text{ Spec}\mathbb{Q}$? I am rather surprised. I don't even know what that means. OK I remember reading a long time ago about the Hilbert scheme of points over $\mathbb{C}$ it wasn't too complicated. Except it was. Please I really need all the help I can get. – cactus314 Mar 13 '17 at 22:48
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    No, all schemes have a map to $\operatorname{Spec}\Bbb Z$. There are plenty of schemes with no map to $\operatorname{Spec}\Bbb Q$; e.g. $\operatorname{Spec}\Bbb F_p$ for any $p$. The fact that any scheme has a map to $\operatorname{Spec}\Bbb Z$ is a consequence of the fact that any commutative ring with identity $R$ comes with a unique map $\Bbb Z\to R$ determined by $1\mapsto 1$ (then any affine scheme admits a unique map to $\operatorname{Spec}\Bbb Z$ by functoriality), and the fact that schemes are constructed by gluing affine schemes. – Stahl Mar 13 '17 at 22:50

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