# Why is $\operatorname{Spec} \ k$ final in category of $k$ schemes?

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.

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

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.
• 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?) Commented Mar 24, 2015 at 21:41
• @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. Commented Mar 24, 2015 at 21:43