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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.

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    $\begingroup$ Do you agree that $k$ is initial in the category of $k$-algebras? $\endgroup$
    – Daniel McLaury
    Mar 24, 2015 at 20:54
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    $\begingroup$ Yes, there is a unique $k$-algebra homomorphism. That's the important thing! $\endgroup$
    – Zhen Lin
    Mar 24, 2015 at 21:11
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    $\begingroup$ @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.) $\endgroup$
    – Daniel McLaury
    Mar 24, 2015 at 21:13
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    $\begingroup$ 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). $\endgroup$
    – Qiaochu Yuan
    Mar 24, 2015 at 21:18
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    $\begingroup$ You are using the wrong definition for a morphism of $k$-schemes. $\endgroup$
    – Daniel McLaury
    Mar 24, 2015 at 21:35

1 Answer 1

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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.)

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  • $\begingroup$ 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?) $\endgroup$
    – user211392
    Mar 24, 2015 at 21:41
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    $\begingroup$ @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. $\endgroup$
    – Qiaochu Yuan
    Mar 24, 2015 at 21:43
  • $\begingroup$ I see. Thank you very much! $\endgroup$
    – user211392
    Mar 24, 2015 at 21:43

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