I am not too certain what these two properties mean geometrically. It sounds very vaguely to me that finite type corresponds to some sort of "finite dimensionality", while finite corresponds to "ramified cover". Is there any way to make this precise? Or can anyone elaborate on the geometric meaning of it?
3 Answers
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I definitely agree with Peter's general intuitive description.
In response to some of the subsequent comments, here are some implications to keep in mind:
Finite ==> finite fibres (1971 EGA I 6.11.1) and projective (EGA II 6.1.11), hence proper (EGA II 5.5.3), but not conversely, contrary to popular belief ;)
Proper + locally finite presentation + finite fibres ==> finite (EGA IV (part 3) 8.11.1)
When reading about these, you'll need to know that "quasi-finite" means "finite type with finite fibres." Also be warned that in EGA (II.5.5.2) projective means $X$ is a closed subscheme of a "finite type projective bundle" $\mathbb{P}_Y(\mathcal{E})$, which gives a nice description via relative Proj, whereas "Hartshorne-projective" more restrictively means that $X$ is closed subscheme of "projective n-space" $\mathbb{P}^n_Y$.
When the target (or "base" scheme) is locally Noetherian, like pretty much anything that comes up in "geometry", a proper morphism is automatically of locally finite presentation, so in that case we do have
finite <==> proper + finite fibres
Regarding "locally finite type", its does not imply finite dimensionality of the fibres; rather, it's about finite dimensionality of small neighborhoods of the source of the map. For example, you can cover a scheme by some super-duper-uncountably-infinite disjoint union of copies of itself that is LFT but not FT, since it has gigantic fibres.
answered Oct 22, 2009 at 7:34
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$\begingroup$ Sure! Also, I just clarified that the locally Noetherian hypothesis was on the target scheme. $\endgroup$ Commented Oct 22, 2009 at 7:56
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1$\begingroup$ BTW, just out of curiosity, how do you manage to be so familiar with EGA? Quoting section by section is really impressive. $\endgroup$– user709Commented Oct 22, 2009 at 12:23
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5$\begingroup$ My main advice is to just start reading through it. I found it a lot easier to read Hartshorne after a (very) cursory reading of EGA I. Get a copy printed off, so you can annotate it for yourself. For my own sanity, I also keep a few notes of my own with references to the theorems I think are important, like the ones I mentioned above. (By the way, my EGA I reference is for the 1971 edition.) $\endgroup$ Commented Oct 22, 2009 at 13:38
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If you have a morphism X-->Y of schemes, finite type means that the fibers are finite dimensional and finite, that the fibers are zero-dimensional.
Take for a finite type example K[x]-->K[x,y]. This corresponds to the projection A^2-->A^1, the fibers are 1-dimensional, which is reflected by K[x,y] being a K[x]-algebra of rank one (or K(x,y) having transcendence degree 1 over K(1)).
For a finite type example take K[x]-->K[x,y]/(y^2-x). For every prime ideal P in K[x] you find two prime ideals in K[x,y]/(y^2-x) whose preimage is P, so the fibers of the corresponding scheme map have cardinality 2 (and dimension zero).
answered Oct 21, 2009 at 12:25
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$\begingroup$ Does finite dimensionality of fibers imply finite type? $\endgroup$– user709Commented Oct 21, 2009 at 13:47
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$\begingroup$ One more thing: what about locally of finite type? Does it mean finite dimensional fibers as well? $\endgroup$– user709Commented Oct 21, 2009 at 16:25
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Peter-arndt gives exactly the right intuition. I will add a cautionary example: It is not literally true that finite fibers implies finite. For example, consider Spec k[x, y]/(xy-1) ---> Spec k[x]. The way I think about this intuitively is that xy=1 has a vertical asymptote at 0 so, even though the fiber over 0 is empty, the map acts like there is an infinite fiber over 0 and is not a finite map. So a more precise intuition is that finite maps have finite fibers, and none of the preimages go off to infinity.
I believe that a precise statement is that finite fibers + proper implies finite.
answered Oct 21, 2009 at 13:10
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$\begingroup$ So is this the "right" intuition? Proper, acting like compactness, prevents the preimage to go off to infinity thus finite fibers + proper should imply finite. Anyway, where can I see a proof of "finite fiber + proper => finite?" I don't think I've seen this in Hartshorne. $\endgroup$– user709Commented Oct 21, 2009 at 13:31
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1$\begingroup$ Its kind of in Hartshorne, but actually kind of hard (but also pretty). Its an exercise in the section on Formal Functions, which is III.11. If you assume Stein factorization its not so hard. $\endgroup$ Commented Oct 21, 2009 at 16:08
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$\begingroup$ It's not quite true that proper + finite fibres ==> finite; see below. (It does work on a locally Noetherian base.) $\endgroup$ Commented Oct 22, 2009 at 7:38
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