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I was reading Mumford again and I noticed a comment in the beginning of the book: "Finally, in any study of general preschemes, the varieties are bound, for many reasons which I will not discuss here, to play a unique and central role."

My question is: is there a particular theorem (or series of theorems) Mumford might have been hinting at?

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4 Answers 4

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Here is a really cool illustration of the principle which Emerton was outlining. We know that the Picard group of projective $(n-1)$-space over a field $k$ is $\mathbf{Z}$ ($n \ge 2$), generated by $\mathcal{O}(1)$. This underlies the proof that the automorphism group of such a projective space is ${\operatorname{PGL}}_n(k)$. But what is the automorphism group of $\mathbf{P}^{n-1}_A$ for a general ring $A$? Is it ${\operatorname{PGL}}_n(A)$? That is, there is a natural map $${\operatorname{PGL}} _n (A) \rightarrow {\operatorname{Aut}} _A (\mathbf{P}^{n-1} _A)$$

(see my answer in the posting called What is the difference between PSL_2 and PGL_2?) and we want to know if it is an isomorphism. It's a really important fact that the answer is yes. But how to prove it? It's a shame that this isn't done in Hartshorne.

By an elementary localization (as in the last step of Emerton's answer), we may assume $A$ is local. In this case we claim that ${\operatorname{Pic}}(\mathbf{P}^{n-1}_A)$ is infinite cyclic generated by $\mathcal{O}(1)$. Since this line bundle has the known $A$-module of global sections, it would give the desired result if true (since ${\operatorname{PGL}}_n(A) = {\operatorname{GL}}_n(A)/A^{\times}$ for local $A$) by the same argument as in the field case. And since we know the Picard group over the residue field, we can twist to get to the case when the line bundle $\mathcal{L}$ is trivial on the special fiber and then we can formulate the problem in two equivalent ways: (I) "lift" the generating section there to a generating section over $A$, or (II) prove that $f _{\ast}(\mathcal{L})$ is invertible in $A$ with the natural map $f^{\ast}(f _{\ast} \mathcal{L}) \rightarrow \mathcal{L}$ an isomorphism. How to do it?

Step 0: The case when $A$ is a field. Done.

Step 1: The case when $A$ is artin local, via (I): this goes via induction on the length, the case of length 0 being Step 0 and the induction resting on cohomological results for projective space over the residue field.

Step 2: The case when $A$ is complete local noetherian ring. This goes via (I) using Step 1 and the theorem on formal functions (formal schemes in disguise).

Step 3: The case when $A$ is local noetherian. This is faithfully flat descent for (II) from Step 2 applied over $\widehat{A}$.

Step 4: The case when $A$ is local: descent from the noetherian local case in Step 3 via direct limit arguments.

"QED"

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  • $\begingroup$ Brian: you'll know the answer to this instantly but I'm not at work and can't look it up. In Katz' Antwerp paper I am pretty sure there's another example of this. He makes an assertion and then says "WLOG base is local; WLOG it's complete local; WLOG it's artinian; done". If you don't answer I'll look it up Thurs lunchtime UK time. $\endgroup$ Commented Feb 24, 2010 at 23:49
  • $\begingroup$ @Kevin Buzzard: It's possible that by local, he meant "Noetherian + unique maximal ideal", which is not uncommon. For the weaker condition that most people call local, commutative algebraists use the term "quasilocal". It's similar to compact/quasicompact. $\endgroup$ Commented Feb 25, 2010 at 0:00
  • $\begingroup$ @Lasergun: oh yeah I'm sure there was probably a WLOG before the WLOGs I remembered, saying "WLOG base is Noetherian". $\endgroup$ Commented Feb 25, 2010 at 0:44
  • $\begingroup$ @Lasergun: usually only commutative algebraists have an implicit "noetherian" when saying "local" (in my experience). I am 100% certain that Katz or anyone "raised on EGA" does not use the phrase "quasi-local". In fact, Grothendieck was quite opposed to noetherian hypotheses (after being convinced by Serre), so he always made them explicit in his terminology. So I think Kevin is right. $\endgroup$
    – BCnrd
    Commented Feb 25, 2010 at 1:42
  • $\begingroup$ I've never heard of "quasi-local" as an adjective for what I call "local", and so I concur with Brian. Do commutative algebraists really incorporate Noetherian as a condition in local? $\endgroup$
    – Emerton
    Commented Feb 25, 2010 at 3:43
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The fibres of any morphism of schemes are schemes over a field. If the morphism if of finite type, then they are finite type schemes over a field, i.e. are (essentially) varieties.

Many arguments proceed from the case of a variety over a field, to the case of a scheme over an Artinian local ring, to a scheme over a complete local ring (a projective limit of Artinian local rings), to a scheme over a local ring (say by descent), to the general case (by passing from stalks to neighbourhoods). For example, many results about abelian schemes are proved in this (or some closely related) way. (See e.g. Kevin Buzzard's answer, and see also Brian Conrad's answer for an example of this kind of argument in a different context.)

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I'm never sure what exactly the word variety means, but finite-type is always included, I think, and it hasn't been brought up yet. Every ring is a direct limit of finitely generated rings. Even further, the category of rings is equivalent to the ind-category of finitely generated rings. So, in some sense, finitely generated rings determine everything. This also schemeifies: the category of quasi-compact quasi-separated schemes is equivalent to the pro-category of schemes of finite type where all the transition maps are affine. (I hope I got that right.) It's very standard to reduce theorems about schemes which are possibly not of finite type to the case where they are.

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The kernel of an isogeny between abelian schemes is flat. The reason is that it's true over a field, and then you use the fibrewise criterion for flatness. I don't know of any other proof. So there's an example where as far as I know you need to study varieties even if you're really interested in schemes...

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