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Hartshorne, Algebraic Geometry, Exercise III.9.1 asks one to prove

A flat morphism $f : X \to Y$ of finite type of Noetherian schemes is open, i.e., for every open subset $U \subseteq X$, $f(U)$ is open in $Y$.

So far as I can tell this is essentially equivalent to the going down theorem, which only needs the hypothesis of flatness. Are the Noetherian and finite-type conditions actually needed here?

Daniel McLaury
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3 Answers3

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Let me just say the finite type hypothesis is 100% necessary; otherwise the result is false. Consider the inclusion $k[t] \to k(t)$. A module over a PID is flat iff it is torsion-free, so the induced map on spec here is flat. But $\operatorname{Spec} k(t)\to \operatorname{Spec} k[t]$ is not open!

darij grinberg
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  • I am a little bit confused by your example. It seems the map you are describing is induced by localization i.e. inverting t. But these maps are basis of the Zariski topology so it should be open. – user127776 Feb 08 '21 at 18:49
  • @user127776 Inverting $t$ is not enough: for instance, it does not make $t-1$ invertible. You have to invert all irreducible monic polynomials, and as a result you get an intersection of infinitely many basic opens. – Geva Yashfe Feb 12 '21 at 14:57
  • I see! I was confused about the notation. – user127776 Feb 12 '21 at 14:59
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Just to comment, the finite type and Noetherian hypotheses are probably so one can give a quick proof using Chevalley. The idea of this proof goes as follows:

  1. (General Topology) One can show that a constructible subset of a scheme is open if and only if it's closed under generization.

  2. (General Topology) One can also show that $f:X\to Y$ is such that $f(X)$ is closed under generization if and only if $\text{Spec}(\mathcal{O}_{X,x})\to\text{Spec}(\mathcal{O}_{Y,f(x)})$ is surjective for all $x\in X$.

  3. (Algebra) Then one proves the following easy fact: Let $(A,\mathfrak{m})\to (B,\mathfrak{n})$ be a local homomorphism, and $M$ a finitely generated $B$-module. Then, $M$ is faithfully flat over $A$ if and only if it is flat over $A$ and non-zero.

  4. (Algebra) One also checks that if $A\to B$ is faithfully flat, then $\text{Spec}(B)\to\text{Spec}(A)$ is surjective.
  5. Finally, to prove that $f$ is open (if it's flat, finite type and $X,Y$ Noetherian) we just put the pieces together. Clearly it suffices to show that $f(X)$ is open (because then apply this case to the map $U\hookrightarrow X\xrightarrow{f} Y$). By Chevalley's theorem we know that $f(X)$ is constructible. Thus, by 1. it suffices to prove that $f(X)$ is closed under generization. But, by 2. this is equivalent to checking that $\text{Spec}(\mathcal{O}_{X,x})\to\text{Spec}(\mathcal{O}_{Y,f(x)})$ is surjective for all $x\in X$. But, using 3. (with $A=\mathcal{O}_{Y,f(x)}$, $B=M=\mathcal{O}_{X,x}$) we see that $\mathcal{O}_{Y,f(x)}\to\mathcal{O}_{X,x}$ is faithfully flat, and so by 4. the induced map $\text{Spec}(\mathcal{O}_{X,x})\to\text{Spec}(\mathcal{O}_{Y,f(x)})$ is indeed surjective.

This proof can actually be adapted to the general case (of just finite presentation) by using Grothendieck's "passing to the limit" technique. Namely, we can reduce to the affine case $X=\text{Spec}(B)$ and $Y=\text{Spec}(A)$. We can then write $B=\varinjlim B_\lambda$ and $A=\varinjlim A_\lambda$ where $A_\lambda$ is a finitely generated algebra over $\mathbb{Z}$ (and so, in particular, Noetherian) and $B_\lambda$ is a finitely presented $A_\lambda$-algebra. We can then use the fact that since we're in the finitely presented case, there exists some $\lambda$ such that

$$\begin{matrix}\text{Spec} (B) & \to & \text{Spec}(B_\lambda)\\ \downarrow & & \downarrow\\ \text{Spec}(A) & \to & \text{Spec}(A_\lambda)\end{matrix}$$

is fibered.

For a reference for this limit technique, I would suggest looking at section 1.10 of Lei Fu's Etale Cohomology.

darij grinberg
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Alex Youcis
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  • Yeah, that's essentially what Hartshorne outlines in the hint for that problem. – Daniel McLaury Feb 14 '14 at 21:22
  • @DanielMcLaury Ah, I'm sorry for the empty verbiage then :) I was never really a fan of Hartshorne, and so didn't know this is what he suggested. – Alex Youcis Feb 14 '14 at 23:06
  • What do you recommend instead? – Daniel McLaury Feb 14 '14 at 23:14
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    @DanielMcLaury It depends upon your style. I personally believe that if you want to be led through a long, epic, arduous (but extremely rewarding journey) you should read Vakil, if you want someone to look quickly at proofs for then Qing Liu, and if you want middle ground then Gortz and Wedhorn. Said differently, Vakil is a conversation, Qing Liu is technical excellence, and Gortz and Wedhorn is, as the porridge of the littlest bear, just right. – Alex Youcis Feb 14 '14 at 23:20
  • Sorry for the question 9 years later (exactly 9 years later, funny enough) but how does the limit technique reduce this problem to one about Noetherian schemes? Seems like you would need to know that the openness of $\text{Spec} B_{\lambda}\rightarrow \text{Spec} A_{\lambda}$ is preserved by base change, which is not obvious to me – Nathan Lowry Feb 14 '23 at 14:41
  • @NathanLowry Hey Nathan. I can't remember what I was thinking then, but it suffices to show that the image of quasi-compact opens are open. By considering what I said for arbitrarily large $\lambda$ you win from the fact that any quasi-compact open descends. Compare this with the proof of openess in Proposition 2.3.8 of Lei Fu's book on etale cohomology. – Alex Youcis Feb 19 '23 at 12:15
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Flat and locally of finite presentation is sufficient, see (Stacks Project, 01UA).

  • As is going-down and of finite presentation. Is a condition of this sort known which is an "if and only if"? – Daniel McLaury Feb 14 '14 at 17:16
  • Sure, that is 01U1 if you want, and doesn't require flatness. As for a necessary condition for a flat morphism to be open, I doubt there is one. –  Feb 14 '14 at 17:24