Understanding Hartshorne's proof that every projective morphism is proper.

Question

In chapter $II$ of Hartshorne, theorem $4.9$ shows that every projective morphism is proper, using the valuative criterion for properness. I understand how the required morphism is constructed, but I don't understand why it is unique. I would greatly appreciate an explanation explaining why "The uniqueness of this morphism follows from the construction and the way the $V_i$ patch together."

I can see why any morphism with image lying inside $V_k$, the affine open piece $D_+(X_k)$, is unique but I don't see why it should be the case that the image of any morphism of the correct form should lie inside $V_k$.

Answer 1

The key fact seems to be this:

Let $R$ be a local ring. Then any open cover of $\operatorname{Spec} R$ must contain $\operatorname{Spec} R$ itself.

Hence, for any local ring $R$, any scheme $X$, any open cover $\{ U_\alpha \}$ of $X$, and any morphism $\operatorname{Spec} R \to X$, there is some $U_\alpha$ through which the morphism factors.

To apply this to the situation at hand, simply observe that valuation rings are local rings.

Answer 2

Just to suggest a different approach: the uniqueness is equivalent to the fact that a projective morphism is separated (the valuative criterion for separatedness), and this can be shown by directly showing the diagonal morphism $\mathbb P^n \to \mathbb P^n \times \mathbb P^n$ is a closed immersion. (In fact, since formation of diagonals and closed immersions are both compatible with base-change, it suffices to check this over $\mathbb Z$.) Now you could try to check directly that the diagonal is closed, say by computing how it interacts with the standard cover of $\mathbb P^n$ by $n+1$ affine spaces.