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Let $f: X \to Y$ a finite morphism of schemes (therefore $f$ affine and the quasi koherent $\mathcal{O}_Y$-algebra $\mathcal{A}= f_*(\mathcal{O}_X)$ such that $X = Spec(\mathcal{A})$ is a finite $\mathcal{O}_Y$-algebra)

Therefore practically that means that for every affine open set $V = Spec(S) \subset Y$ the $S=\Gamma(V,\mathcal{O}_Y)$-algebra $A= \Gamma(V,\mathcal{A})$ is finite

I want to show that then $f$ is proper.

Here my attempts:

I want to use following valuative criterion for properness:

$f$ is proper (so separated, of finite type, and universally closed) if and only if in every diagram

$$ \require{AMScd} \begin{CD} Spec(F) @>{g} >> X \\ @VViV @VVfV \\ Spec(R) @>{t}>> Y \end{CD} $$

for every discrete valuation ring $R$ and $F = Frac(R)$ there exist $l:Spec(R) \to X$ such that $a = l \circ i$ and $t = f \circ l$ holds.

The spectrum of DVR $R$ has the structure $Spec(R) = \{\sigma, \eta\} = \{\sigma\} \cup Spec(F)$ with generic point $ \eta$ and unique maximal ideal $\sigma$.

Since $Spec(F)$ is a sigleton, we can reduce the problem to affine case $X= Spec(A), Y = Spec(S)$ with $A$ finite $S$-algebra.

For underlying topological spaces it's obviously how to define $l: Spec(R) \to X = Spec(A)$:

Let $s_0:= t(\eta), s_1 := t(\sigma)$. Obviously $s_0 \subset s_1$ as prime ideals. Because $= s_0= t \circ i (\eta) = f \circ g(\eta)$, we conclude that $a_0:= g(\eta)$ is lying over $s_0$. Since $f$ is finite morphism the going up theorem holds, therefore there exist a $a_1$ lying over $s_1$.

Define $l$ set theoretically by setting $l(\eta)= a_0, l(\sigma) := a_1$.

Now to my problem: How to see that $l$ defined in this way provides a morphism of schemes, not only a map which commutates with the commutative square?

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    $\begingroup$ You should show that a finite map is separated (every affine map trivially is), of finite type (trivial) and universally closed (finite maps are closed by going-up and they are universally closed since finite is preservered under base change). $\endgroup$
    – MooS
    Commented Mar 4, 2018 at 20:38
  • $\begingroup$ To put it more concretely: Showing that a finite morphism is proper by using the valuative criterion will basically just reproof the direction 'proper --> valuative criterion holds' of the valuative criterion. $\endgroup$
    – MooS
    Commented Mar 4, 2018 at 21:01

1 Answer 1

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As you have remarked, it is sufficient to look at the case where $X = \operatorname{Spec} A$ and $Y = \operatorname{Spec} S$ are both affine. In general, when defining morphisms between affine schemes, it is easier to define the corresponding map between the rings. So instead of trying to define construct directly a scheme map from $\operatorname{Spec} R$ to $X$, it is easier to construct a ring map from $A$ to $R$. This is the same thing, since we have a canonical isomorphism $\operatorname{Hom}_{Schemes}(\operatorname{Spec} R, X) = \operatorname{Hom}_{Rings}(A, R)$.

Turning the valuative criterion that you quoted into commutative algebra, we see that what you want to show is equivalent to the following proposition.

Proposition Let $A$ be a finite $S$-algebra. Then for any discrete valuation ring $R$ with fraction field $F$ and any a commutative diagram $$ \require{AMScd} \begin{CD} S @>{t} >> R \\ @VVfV @VViV \\ A @>{g}>> F \end{CD} $$ there is a ring map $A \to R$ that fits in the diagram.

Of course, we usually think of $R$ as a subring of $F$, so saying that there is a morphism $A \to R$ as in the diagram really just means that the map $g$ has image contained in $R \subset F$.

To prove the above proposition, we just note that the image $g(A)$ of $A$ will be finite and therefore integral over $R$. So $g(A)$ is contained in the integral closure of $R$ in $F$. But $R$ is a discrete valuation ring, and such rings are integrally closed in their fraction field. This gives us $g(A) \subset R$ as needed.

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