Can you tell me, how to prove that every proüper and affine morphism between locally Noetherian schemes is finite? Every help will be appreciated.
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9$\begingroup$ This is in EGA2, 6.7.1. $\endgroup$– Matthieu RomagnyCommented Mar 27, 2013 at 17:02
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1$\begingroup$ Is this one of your homework problems? $\endgroup$– Jason StarrCommented Mar 27, 2013 at 17:43
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13$\begingroup$ Dear nicolas, Do you understand intuitively why this should be true (or might be true)? If not, why not first consider the case when the base is a point; then you are asking why an affine variety that is proper is necessarily zero-dimensional. Hopefully you can prove this. The general problem for affine morphisms doesn't directly reduce to this particular case, but it should at least provide you with some intuition. Regards, $\endgroup$– EmertonCommented Mar 27, 2013 at 19:13
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3 Answers
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There is an elementary proof of the result "universally closed + affine $\Rightarrow$ integral" that I learnt from Olivier's paper "Going up along absolutely flat morphisms." In fact, it's so simple, I can present it here.
Observation 1: Say $\phi:A \to B$ is an injective ring map that is closed on $\mathrm{Spec}$. Then $\phi^{-1}(B^\ast) = A^\ast$.
(This proof was edited and corrected to reflect xuhan's comment.)
Proof: Fix $a \in A$ with $\phi(a) \in B^\ast$. We must show $a \in A^\ast$ or, equivalently, $a$ is non-zero in the residue field $\kappa(\mathfrak{p})$ of $A$ at any prime $\mathfrak{p} \in \mathrm{Spec}(A)$. Note that the last statement is clearly true if $\mathfrak{p}$ lies in the image $Z$ of $\mathrm{Spec}(\phi)$. So it suffices to show $Z = \mathrm{Spec}(A)$. By closedness, $Z = V(I)$ for some ideal $I \subset A$ (set-theoretically). Localizing at any prime $\mathfrak{p}$ shows $I \subset \mathfrak{p}$ by the injectivity hypothesis. Then $I$ is contained in all primes of $A$, so it contains only nilpotents, and hence $Z = \mathrm{Spec}(A)$.
Observation 2: Say $\phi:A \to B$ is an injective ring map, and $\phi[T]:A[T] \to B[T]$ is closed on $\mathrm{Spec}$. Then $\phi$ is integral.
Proof: Fix some $f \in B$, and consider the surjective map $B[T] \to B[\frac{1}{f}]$ given by $T \mapsto \frac{1}{f}$. If we write $C \subset B[\frac{1}{f}]$ for the image of the composite $A[T] \to B[T] \to B[\frac{1}{f}]$, then $C \to B[\frac{1}{f}]$ is an injective ring map that is closed on $\mathrm{Spec}$. The image of $T$ in $C$ becomes a unit in $B[\frac{1}{f}]$, and hence must be a unit on $C$ by Observation 1, so we can write $f = \sum_{i=0}^n a_i \big(\frac{1}{f}\big)^i$ in $B[\frac{1}{f}]$ for $a_i \in A$. Clearing denominators shows that $f \in B$ satisfies a monic polynomial over $A$.
Observation 2 + killing the kernel shows:
Theorem: If $\phi:A \to B$ is a ring map that is universally closed on $\mathrm{Spec}$, then it is integral.
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1$\begingroup$ The end of the proof of Observation 1 seems incorrect when $B$ does not have only finitely many minimal primes, since for a collection of ideals $J_i$ of $A$ and an $a \in A$ it isn't generally true (when the collection is infinite) that if $a$ is a unit in every $A/J_i$ then $a$ is a unit in $A/(\cap J_i)$. For example, let $A = \mathbf{Z}$, $a = 10$, and consider the collection of prime ideals distinct from $(2)$ and $(5)$. (If the collection of $J_i$'s is finite then all is OK, since then $A/(\prod J_i)$ makes sense and has $A/(\cap J_i)$ as a quotient. So Obs. 1 is OK for noetherian $B$.) $\endgroup$ Commented Mar 28, 2013 at 9:52
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2$\begingroup$ Here is a corrected proof of Observation 1. Since being a unit amounts to being nonzero at all points of Spec, it suffices to show that the map ${\rm{Spec}}(B) \rightarrow {\rm{Spec}}(A)$ is surjective. The image of this map is closed and contains all generic points of Spec($A$) (due to the injectivity hypothesis, combined with localization at minimal primes of $A$), so it suffices to show that if $Z = {\rm{Spec}}(A/I)$ contains all generic points of Spec($A$) then the elements of $I$ are nilpotent. This in turn follows from the fact that the intersection of all minimal primes is Nil($A$).QED $\endgroup$ Commented Mar 28, 2013 at 9:56
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$\begingroup$ In my first comment, for the counterexample with $A = \mathbf{Z}$ and $a = 10$ I should have said "collection of nonzero prime ideals distinct from (2) and (5)". $\endgroup$ Commented Mar 28, 2013 at 9:58
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$\begingroup$ Thanks! I corrected the proof. (This mistake was mine, not Olivier's.) $\endgroup$– anonCommented Mar 28, 2013 at 15:19
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Have a look at Ravi Vakil's notes on Algebraic geometry, 18.1.8, 18.9.A (and possibly tracing through the results used in these sections) http://math.stanford.edu/~vakil/216blog/. The point is that $f:X\rightarrow Y$ is affine, then $X\cong \underline{\mathrm{Spec}}(f_*\mathcal{O}_X)$, and if $f$ is proper, then $f_*\mathcal{O}_X$ is coherent. Hence $X$ is the relative spectrum of a coherent $\mathcal{O}_Y$-algebra, and hence finite.
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See also Liu, Lemma 3.3.17 (for a proof without using the coherence theorem).