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When are Noetherian conditions on a scheme genuinely essential in algebraic geometry?

I am under the impression that most of the time these conditions are imposed for expositional clarity and to simplify the commutative algebra involved (for example throughout Hartshorne and FGA), and that they can be removed by paying more attention to finiteness conditions on the morphisms involved.

So what are the things that genuinely require Noetherian conditions? Even better, is there a "slogan" that tells you when they will be needed?

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    $\begingroup$ This is best learned by experience with techniques of argument than by seeking an all-purpose general "slogan"; i.e., one really should look under the hood at the engine. It is akin to the fact that technique is far more important than general theorems in the study of PDE. The methods are developed and used in a masterful way in EGA, so poke around in there to learn. Read EGA I and II (skip the final section 8 there), then section 8 of IV$_4$, and then explore. $\endgroup$
    – nfdc23
    Commented Dec 8, 2016 at 20:56
  • $\begingroup$ Dear @nfdc23, Thank you for the advice - I will admit I have been avoiding EGA IV through fear, but perhaps it is time to brave it. $\endgroup$
    – Ben Morley
    Commented Dec 9, 2016 at 19:05

2 Answers 2

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Here are some examples illustrating the genuine necessity of noetherian assumptions:

1) Every scheme with just one point is the spectrum of a local artinian ring?
This is true for every noetherian one point scheme and false for every non-noetherian one point scheme.
A non-noetherian (and thus non-Artinian) example : $\operatorname {Spec}(\mathbb Q[T_1,T_2,T_3,\cdots]/\langle T_1^2,T_2^2,T_3^2 \rangle )$

2) A scheme has only finitely many irreducible components ?
A noetherian scheme has only finitely many irreducible components, but a non-noetherian scheme may have infinitely many irreducible components: this is the case for any disjoint union of infinitely many non-empty schemes.

3) Every scheme has a closed point?
This is true for every noetherian scheme (actually for any quasi-compact scheme), but there exist schemes without any closed point: Qing Liu, Chapter 3, Exercise 3.27, page 114.

4) Injective modules give injective sheaves?
If $I$ is an injective module over the ring $A$, then the associated quasi-coherent sheaf $\tilde I$ on $X=\operatorname {Spec}(A)$ is an injective sheaf of $\mathcal O_X$- Modules if $A$ is noetherian but is not necessarily injective for $A$ non-noetherian: SGA6, Exposé II, Appendice I Un contre-exemple de Verdier, page 195.

5) A finitely presented sheaf is coherent?
Given on a scheme $X$ a sheaf of $\mathcal F$ of $\mathcal O_X$-Modules, does the existence of an open covering $(U_i)$ of $X$ for which one has exact sequences $\mathcal O_{U_i}^{n_i}\to \mathcal O_{U_i}^{m_i}\to \mathcal F\vert _{U_i}\to 0$ imply that $\mathcal F$ is coherent?
The answer is yes if $X$ is noetherian (or even locally noetherian) but no in general: there exist non-noetherian rings $A$ such that the structural sheaf $\mathcal O_X$ on $X=\operatorname {Spec}(A)$ is not coherent!

6) A scheme is affine if all its quasi-coherent sheaves are acyclic ?
Serre's criterion is that a noetherian scheme $X$ is affine if and only if $H^p(X, \mathcal F)=0$ for all quasi-coherent sheaves $\mathcal F$ on $X$ and all $p\gt 0$.
This no longer holds if $X$ is not assumed noetherian: given a field $k$, any infinite disjoint sum $X=\coprod \operatorname {Spec}k$ satisfies the cohomology condition but is not affine since it is not quasi-compact.

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    $\begingroup$ Except in 1), noetherianness is not necessary (but, of course, cannot just be omitted). $\endgroup$ Commented Dec 9, 2016 at 9:43
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    $\begingroup$ Regarding 5), many people do not even know the definition of "coherent" in the non-noetherian case. Thanks to a certain popular textbook in algebraic geometry. $\endgroup$
    – HeinrichD
    Commented Dec 9, 2016 at 9:54
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    $\begingroup$ Dear @Fred, oh yes I agree. Often quasi-compactness suffices, but maybe the philosophy might be to tread carefully when applying "well-known theorems" to non-noetherian schemes... . $\endgroup$ Commented Dec 9, 2016 at 12:07
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    $\begingroup$ Dear @Georges, thank you for the many examples. I agree with Fred though that while the first is an example of the necessity noetherian conditions, the later examples 2,3,6 showcase more why we need conditions such as quasicompactness or that the underlying topological space is noetherian (and I certainly agree that the resulting noetherian induction arguments are powerful and interesting). Apologies if this sounds like I am quibbling over details - all your examples are certainly very useful for understanding what can go wrong in the non-noetherian case. $\endgroup$
    – Ben Morley
    Commented Dec 9, 2016 at 19:02
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    $\begingroup$ @FredRohrer: the confusion about your first comment comes from the word 'except', which you use to mean in all examples except (1). I (and probably other people too) read it as stating your objection: everything you say is correct, except that in (1), .... $\endgroup$ Commented Dec 9, 2016 at 19:18
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Just to give one example: There are certain questions about the derived category $D(R)$ of a commutative ring $R$ where noetherian assumptions can really result in different behaviour. For example, Neeman proved that if $R$ is noetherian then the Bousfield classes of $D(R)$ correspond to arbitrary subsets of the spectrum $Spec(R)$. On the other hand, we do not have an understanding of the Bousfield classes of $D(R)$ when $R$ is not noetherian --- and there can be a huge number of them. For example, Dwyer & Palmieri wrote a paper in which they considered a truncated polynomial algebra in countably many variables $R=k[x_1,x_2,\ldots,x_n]/(x_i^{n_i} \text{for all } i)$ and proved that in this case the Bousfield lattice of $D(R)$ has cardinality at least $2^{2^{\aleph_0}}$.

Such questions are particularly interesting when one considers the analogy between derived categories $D(R)$ and the stable homotopy category $SH$ (which morally is the derived category of the sphere spectrum). The stable homotopy groups of spheres are highly non-noetherian and much of our intuition about derived categories of noetherian rings does not carry over to $SH$.

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    $\begingroup$ I do not really understand this, but it does not seem to be of the kind "requires noetherian hypothesis". Rather: In the noetherian case certain nice things happen, and there are non-noetherian examples were these nice things do not happen. $\endgroup$ Commented Dec 8, 2016 at 19:38

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