Hartshorne (Chapter 2, 4.8), why isn't a open immersion written to be a proper morphism?

(I currently possess only the Japanese Edition of Hartshorne's book. Therefore, I kindly request that any answers provided do not rely on page numbers (of the English Edition) for reference. Thank you.)

In Corollary 4.8 in Chapter 2 of the book, it is stated that only closed immersions (and not open immersions) are considered proper. However, I believe the open immersions are proper under the assumption of the corollary. The following is the proof I wrote.

This corollary assumes that all schemes are noetherian. Since $$X$$ is noetherian, an open immersion $$f: X \to Y$$ should also be quasi-compact. Thus $$f$$ is of finite type, by Exercise 3.13.$$\require{AMScd}$$

$$\begin{CD} \mathrm{Spec}\ K @>>> B \\ @V{\text{i}}VV @VV{\text{f}}V \\ \mathrm{Spec}\ R @>{\text{\phi}}>> D \end{CD}$$

Now we can apply valuative criterion (Theorem 4.7). $$f$$ is a open immersion, so $$f$$ is a split mono and there is $$g : Y \to X$$ such that $$gf = 1_X$$. Then $$g\phi : T \to X$$ is the only morphism that remains the diagram above commutative. (The commutativity is evident. The uniqueness is followed by the fact that $$f$$ is separated or the fact that $$f$$ is a mono in the meaning of category theory.) By the criterion, it follows that $$f$$ is proper. $$\blacksquare$$

Is the proof above is wrong? Or was the fact that open immersions are proper under the assumption accidentally not written in the corollary?

1 Answer

The proof is wrong: open immersions are not in general split monos. For instance, over a field $$k$$, the open immersion of $$\Bbb A^1\to\Bbb P^1$$ as $$D(t_1)$$ cannot be split, because every map from $$\Bbb P^1$$ to $$\Bbb A^1$$ has image a point.

(Another reason this is wrong because a proper map is defined to be universally closed, and in particular closed. So an open immersion which does not have closed image can't be proper.)