Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question is based on the following proposition (EGA IV, 2.7.1)

Let $f: X \rightarrow Y$ be a $S$-morphism of $S$-schemes, $g: S'\rightarrow S$ a faithfully flat and quasi-compact morphism. Denote $X \times_S S'$ by $X'$, and denote $Y \times_S S'$ by $Y.$ We have a natural morphism $f': X\rightarrow Y.$ Consider the following properties of morphisms:

(i) separated

(ii) quasiseparated

(iii) locally of finite type

(iv) locally of finite presentation

(v) finite type

(vi) finite presentation

(vii) proper

(viii) isomorphism

(ix) monomorphism

(x) open immersion

(xi) quasi-compact immersion

(xii) closed immersion

(xiii) affine

(xiv) quasi-affine

(xv) finite

(xvi) quasi-finite

(xvii) entire (I'm not sure exactly what a "morphisme entier" is, but some reading of the french wikipedia gave me the impression that it's an integral morphism)

If $P$ is one of the preceding properties, then $f$ has property $P$ if and only if $f'$ has property $P.$

My impression of the proof is that there are quite a few ingredients needed to prove this proposition, where different ingredients are needed for different properties, and it surprises me that there seems to be no "unifying principle" that covers all the proofs.

What I would like is something like this theorem: Let $P$ be a property closed under composition and base change. Then, if $f:X \rightarrow Y$ is a map of $Z$-schemes, such that the structure morphism $X \rightarrow Z$ is in $P$ and the diagonal morphism of the structure morphism $Y \rightarrow Z$ is in $P$, then $f$ is in $P$. I like this theorem because it explicitly states the conditions the property needs to satisfy, and the conditions are fairly loose. Then, you have essentially the same proof of this theorem for every such property $P$.

Is there any such unification of the proofs for this result? To me, it seems like this result is some incredibly mysterious miracle. I would appreciate any intuition behind this result.

share|improve this question

2 Answers 2

up vote 7 down vote accepted

I think that a "faithfully flat" morphism in algebraic geometry can be thought of as analogous to a surjective map of sets (maybe more accurately as something like a fibration with surjective image). A lot of elementary properties one might formulate about a map of sets can be tested after base-change by a surjection. For instance, if $f: X \to Y$ is a function, then $f$ is surjective (resp. injective) if and only if the base change of $f$ along a surjection $Y' \to Y$ is. There is more to this analogy: one can recover the function $f$ from its base-change along $Y' \to Y$ (under the condition that the two base-changes to $Y' \times_Y Y'$ agree); the analog of this for schemes is (a special case of) faithfully flat descent, explained for instance in Vistoli's article.

I don't know of a general categorical set up relevant here. However, to demystify this result, you might try proving the analogs in commutative algebra, for instance, the following: given an $A$-module $M$ (for $A$ a ring) and a faithfully flat extension $A \to B$, then $M$ is finitely generated if and only if $M \otimes_A B$ is finitely generated.

share|improve this answer
    
Incidentally, a faithfully flat morphism is "conservative": pull-back along it reflects isomorphisms. This is at least an explanation for faithfully flat descent (from the point of view of the Barr-Beck theorem, conservativity is the crucial part of the explanation for descent-type results), and maybe it's "morally" why pull-back along a faithfully flat, quasi-compact morphism "reflects" a whole set of additional properties. –  Akhil Mathew Aug 27 '12 at 22:16

Just to add a little to Akhil's nice answer:

Akhil's answer explains that a faithfully flat morphism is analogous to a surjective map of sets. This shouldn't be a complete surprise, because the faithful part of faithful flatness is precisely that the morphism be surjective.

There is another analogy that is also intuitively useful, in which we make a comparison not with the category of sets, but with a slightly more geometric category, the category of topological spaces:

Namely, a faithfully flat morphism in algebraic geometry is roughly analogous to a quotient morphism of topogical spaces (i.e. a surjective morphism in which the target is given the quotient topology induced by the topology on the source). Again, this should seem quite natural once you recall that a faithfully flat morphism of schemes is a quotient map when we pass to underlying topological spaces. (See the wikipedia article for the citation to EGA IV.)

So the idea is that the target of a faithfully flat morphism can be obtained as the quotient of its source just by imposing an appropriate equivalence relation, and hence many properties downstairs can instead be checked upstairs.

share|improve this answer

Your Answer

 
discard

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.