4

I am working on an exercise trying to show that $Spec \ k$ is final in category of $k$ schemes. I am stuck and I would appreciate any assistance. Thank you!

PS The definition I have for $k$ scheme is that it is a morphism of the form $X \rightarrow \operatorname{Spec} \ k$. And then I know from the exercise I did that $X \rightarrow \operatorname{Spec} \ A$ are in natural bijection with ring morphisms $A \rightarrow \Gamma (X, O_X)$.

So I figured if I have a $k$ scheme, then it follows that there exists a corresponding ring morphism $k \rightarrow \Gamma (X, O_X)$. I guess I was wondering how this is unique.

Sha Vuklia
  • 3,960
  • 4
  • 19
  • 37
user211392
  • 767
  • 6
    Do you agree that $k$ is initial in the category of $k$-algebras? – Daniel McLaury Mar 24 '15 at 20:54
  • @DanielMcLaury I guess that is what I was confused about. Is it obvious that the map from $k$ to the global section of some $k$ scheme is unique? – user211392 Mar 24 '15 at 21:03
  • 3
    Yes, there is a unique $k$-algebra homomorphism. That's the important thing! – Zhen Lin Mar 24 '15 at 21:11
  • 1
    @user211392: remember, it's not just a homomorphism of rings. It's a homomorphism of $k$-algebras. (If you don't see the difference, think about the $k$-algebra structure on $k$ itself and carefully unpack the definition.) – Daniel McLaury Mar 24 '15 at 21:13
  • 5
    What is your definition of the category of schemes over $k$? The right definition makes this obvious (it's the category of schemes equipped with a map to $\text{Spec } k$; this is a very general construction called taking the overcategory, and the object you're taking the overcategory of is always terminal). – Qiaochu Yuan Mar 24 '15 at 21:18
  • I have added PS to make my specific question more clear. Thank you! – user211392 Mar 24 '15 at 21:33
  • @DanielMcLaury Does it follow from construction of the correspondence between $X \rightarrow Spec \ k$ and $k \rightarrow \Gamma (X, O_X)$ that it is in fact a $k$ morphism and not just a ring morphism? (The exercise I did only mentioned it was a ring morphism...) – user211392 Mar 24 '15 at 21:35
  • 1
    You are using the wrong definition for a morphism of $k$-schemes. – Daniel McLaury Mar 24 '15 at 21:35
  • Thank you for all the help. This whole thing makes a lot more sense to me now! – user211392 Mar 24 '15 at 21:44

1 Answers1

3

A $k$-scheme is a scheme $X$ together with a morphism $X \to \operatorname{Spec} k$. A morphism of $k$-schemes is a morphism $\varphi : X \to Y$ of schemes such that the diagram $$\begin{array}{c} X & \xrightarrow{\varphi} & Y \\ \downarrow & & \downarrow \\ \operatorname{Spec} k & = & \operatorname{Spec} k \end{array}$$ commutes. In particular, not every morphism $\varphi : X \to Y$ of schemes is a morphism of $k$-schemes. This appears to be the sticking point for you.

(This is just an unpacking of what Qiaochu Yuan mentioned in the comments.)

Daniel McLaury
  • 24,656
  • I see. Thank you. So then in the exercise I did which states "show that morphism $X \rightarrow Spec \ A$ are in natural bijection with ring morphisms $A \rightarrow \Gamma (X, O_X)$" are they referring to just morphisms of schemes or morphisms of $k$ schemes (or are they equivalent in this case?) – user211392 Mar 24 '15 at 21:41
  • 4
    @user211395: morphisms of schemes. If you want morphisms of schemes over $k$ on the left then that corresponds to $k$-algebra morphisms on the right. – Qiaochu Yuan Mar 24 '15 at 21:43
  • I see. Thank you very much! – user211392 Mar 24 '15 at 21:43