Finite type/finite morphism - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T20:45:23Z http://mathoverflow.net/feeds/question/1634 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/1634/finite-type-finite-morphism Finite type/finite morphism Ho Chung Siu 2009-10-21T11:45:33Z 2009-11-26T03:10:42Z <p>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?</p> http://mathoverflow.net/questions/1634/finite-type-finite-morphism/1643#1643 Answer by Peter Arndt for Finite type/finite morphism Peter Arndt 2009-10-21T12:25:56Z 2009-10-21T12:25:56Z <p>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. </p> <p>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)).</p> <p>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).</p> http://mathoverflow.net/questions/1634/finite-type-finite-morphism/1648#1648 Answer by David Speyer for Finite type/finite morphism David Speyer 2009-10-21T13:10:03Z 2009-10-21T13:10:03Z <p><a href="http://mathoverflow.net/questions/1634/finite-type-finite-morphism/1643#1643" rel="nofollow">Peter-arndt</a> 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.</p> <p>I believe that a precise statement is that finite fibers + proper implies finite.</p> http://mathoverflow.net/questions/1634/finite-type-finite-morphism/1839#1839 Answer by Andrew Critch for Finite type/finite morphism Andrew Critch 2009-10-22T07:34:49Z 2009-11-25T10:15:30Z <p>I definitely agree with Peter's general intuitive description.</p> <p>In response to some of the subsequent comments, here are some implications to keep in mind: </p> <p><strong>Finite</strong> ==> <strong>finite fibres</strong> (1971 EGA I 6.11.1) and <strong>projective</strong> (EGA II 6.1.11), hence <strong>proper</strong> (EGA II 5.5.3), but <em>not conversely</em>, contrary to popular belief ;)</p> <p><strong>Proper</strong> + <strong>locally finite presentation</strong> + <strong>finite fibres</strong> ==> <strong>finite</strong> (EGA IV (part 3) 8.11.1)</p> <p>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" $\mathbb{P}^n_Y$.</p> <p>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 </p> <p><strong>finite</strong> &lt;==> <strong>proper</strong> + <strong>finite fibres</strong></p> <p>Regarding "locally finite type", its does <em>not</em> 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.</p>