Let $R$ be a <valuation ring> with <fraction field> $K$. For every commutative square
$$
\begin{array}{ccc}
\operatorname{Spec}K&\longrightarrow&X\\
\downarrow&&\downarrow f\\
\operatorname{Spec}R&\longrightarrow&Y,
\end{array}
$$
the <valuative criterion for separatedness> says that a <finite type morphism> $f$ between <Noetherian scheme>[Noetherian schemes] is <separated morphism>[separated] exactly when there is at most one dotted lift $\operatorname{Spec}R\to X$ completing the diagram.

Under the same finiteness hypotheses, the <valuative criterion for properness> says that $f$ is <proper morphism>[proper] exactly when every such square has a unique lift. Thus separatedness supplies uniqueness, while properness supplies existence as well.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 113 / 1 / a / Solution

{wiki=Morphism_of_finite_type}

A morphism $f:X\to Y$ is of finite type when every point of $Y$ has an affine neighborhood $V=\operatorname{Spec}A$ for which $f^{-1}V$ has a finite affine cover $\operatorname{Spec}B_i$ with each $B_i$ a finitely generated $A$-algebra.


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 Incoming links: Finite type morphism