MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can you tell me, how to prove that every proüper and affine morphism between locally Noetherian schemes is finite? Every help will be appreciated.

share|cite|improve this question
This is in EGA2, 6.7.1. – Matthieu Romagny Mar 27 '13 at 17:02
Is this one of your homework problems? – Jason Starr Mar 27 '13 at 17:43
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, – Emerton Mar 27 '13 at 19:13
up vote 7 down vote accepted

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.

share|cite|improve this answer
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 $B$.) – user29283 Mar 28 '13 at 9:52
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 – user29283 Mar 28 '13 at 9:56
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)". – user29283 Mar 28 '13 at 9:58
Thanks! I corrected the proof. (This mistake was mine, not Olivier's.) – anon Mar 28 '13 at 15:19

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) 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.

share|cite|improve this answer

See also Liu, Lemma 3.3.17 (for a proof without using the coherence theorem).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.