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Where can I find a clear exposé of the so called "standard reduction to the local artinian (with algebraically closed residue field", a sentence I read everywhere but that is never completely unfold?


EDIT: Here, was a badly posed question.

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Dear Workitout: In order for people to even try giving a reasonable answer, you probably need to be a lot more specific. –  Hailong Dao Aug 23 '10 at 20:32
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Read pp. 8--9 in Introduction to EGA I (IHES version) and then Introduction to EGA IV_1 (which is embedded near the end of that volume, after the long Chapter 0). Then think about Krull intersection theorem and fact that every artin local ring admits a faithfully flat extension which is another artin local ring having whatever desired residue field you want (such as alg. closed) -- this fact (which is also valid for complete local noetherian rings more generally) is EGA 0$_{\rm{III}}$, 10.3.1. That's it. If still unclear and you know someone who can generate EGA-style arguments, ask them. –  BCnrd Aug 23 '10 at 22:56
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Hailong, this question is formulated in exactly the spirit of how the technique being asked about is generally used, which is to say that almost no information is provided. The truth is that if someone doesn't know what is being asking about from the little bit that is said, there's no way they'd know the answer (since what is said is exactly what is actually written in papers using this powerful method). So in that sense, it is a well-formulated question (from the perspective of getting an answer, which is ultimately what Workitout wants, and has now received). –  BCnrd Aug 23 '10 at 22:59
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Dear Workitout: One other point you should keep in mind is that for a proper scheme over a local ring, the only open set around the special fiber is the entire space. So when you have enough openness theorems at your disposal either for many local properties upstairs (flatness, etc.) or for fibral properties over the base (e.g., relative ampleness of a line bundle on top), it is sufficient to work infinitesimally around the special fiber (once you have exploited finite presentation to reduce to the case of a noetherian base scheme). –  BCnrd Aug 23 '10 at 23:37
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Dear Hailong: there are at least 5 or 6 others here (perhaps more) who are entirely familiar with this kind of reduction argument and so would know the answer too. So no luck necessary. :) –  BCnrd Aug 24 '10 at 2:58

2 Answers 2

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Dear Workitout: The list of comments above is getting unwieldy, so let me post an answer here, now that you have finally identified 1.10.1 in Katz-Mazur as (at least one) source of the question. As I predicted, you'll see that the basic technique to be used adapts to many other settings, and that it is very hard to formulate a "meta-theorem" to cover all cases. The only sure-fire method I know is to read all of sections 8, 9, 11, 12 in EGA IV$_3$ and sections 17 and 18 in EGA IV$_4$, and then it becomes really routine. Maybe there's a better method (well I can think of one, but I won't say it here). I will use the terminology and notation around 1.10.1 from Katz-Mazur without explanation below.

Being a "full set of sections" of $Z/S$ is something which is sufficient to check using the constituents of a single open affine cover of $S$, and also a finite generating set of the coordinate ring of $Z$ over each such open. Thus, by working Zariski-locally on $S$ we may assume $S = {\rm{Spec}}(R)$ and that both sides of (1) in KM 1.10.1 have $R$-free coordinate rings (when viewed as finite $R$-schemes), and likewise for their sum (as effective Cartier divisors). Now the assertions (1) and (2) in KM 1.10.1 are identities among finitely many elements of some finite free $R$-modules. For instance, (1) asserts that certain elements in a finite free $R$-module have vanishing image in a finite free $R$-module quotient. This is all now a bunch of identities among finitely many elements of $R$.

OK, finally we come to the part with a real idea. Consider the subring $R_0$ of $R$ generated over $\mathbf{Z}$ generated by those finitely many elements. It is noetherian. Now unfortunately your initial algebro-geometric setup over $R$ (the smooth separated finite-type curve $C$, the various effective relative Cartier divisors, etc.) probably does not arise via base change from the exact same setup (including flatness properties!) over $R_0$. But that doesn't matter: what would really be swell is if some noetherian subring of $R$ which contains $R_0$ permits such a descent of the situation. Now express $R$ as the direct limit of its finitely generated $R_0$-subalgebras (all of which are noetherian). Does the entire situation descend to one of those? If it did, we'd be in great shape, since it would then suffice to solve the problem in the case of a noetherian base ring (as that would imply the result over $R$ by suitable base change of the descent from the noetherian subring back up to $R$).

How to implement this strategy of reduction to the noetherian case (after which we'll need to deal with the passage to artin local base with algebraically closed residue field)? OK, so that's where we are all very fortunate that Grothendieck actually wrote out the entire formalism in total detail to handle basically every such situation one could ever want to handle. So it becomes kind of a game in finding the references in EGA (which I admit is hard to do if one doesn't know where to look, but is really easy if one has read the right parts). For your particular situation with some smooth separated curves and some relative effective Cartier divisors, etc., the results you need are: EGA IV$_3$ 8.3.4, 8.9.1, 8.10.5 (e.g., (v)), 9.2.6.1, 11.2.6(ii), and IV$_4$ 17.7.9.

I'm not going to say more about how you combine those references here; that is where I again remind you of your pseudonym.

OK, now $R$ is noetherian (even finite type over $\mathbf{Z}$, which is very useful extra stuff to have for other arguments with excellence later in life), and you're trying to prove some finite collection of identities among elements of $R$. To do that it suffices to check in the local rings of $R$, so you can assume $R$ is local. Now a pair of elements of a local noetherian ring are equal if and only if they have equal images in each artinian quotient (Krull intersection theorem). So it suffices to prove the general result over arbitrary artin local rings (really just the artinian quotients of finitely generated $\mathbf{Z}$-algebra, so there's no set-theoretic quantification nonsense going on). Now $R$ is artin local. To check an identity in a ring it suffices to do so in a faithfully flat extension ring. So finally we haul out EGA 0$_{\rm{III}}$ 10.3.1 to find a faithfully flat artin local extension with an algebraically closed residue field. That's it!

I presume you can now see why anyone who knows how to fill in such arguments never actually writes them out in papers: it is much simpler to say "by standard limit methods from EGA IV$_3$, sections 8,9, etc." (maybe even to be a bit more specific, as K-M are at the beginning of their 1.8.1), and to trust that the reader will pick up the clue that they should read those parts of EGA if they want to understand what is going in such arguments for themself. It is bad when people don't at least mention the relevance of sections 8, 9, etc., but things could be worse (e.g., Grothendieck could have not written EGA, leaving stuff in a complete mess reference-wise).

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I should stress that it is a terrible idea to do "math via literature search". If you really want to understand what makes these reduction steps work, learn the methods that make them work so you can generate whatever arguments you need on your own (apart from some really subtle things, like Raynaud's amazing descent of flatness in 11.2) without having to go back to the source for each bit. Later in life maybe you'll work with algebraic spaces or who-knows-what, and you then want to easily figure out by yourself how to get what you need in your new setup, etc. This is all just a technique. –  BCnrd Sep 2 '10 at 3:44
    
Damn brilliant answer! –  Daniel Larsson Sep 2 '10 at 8:26

Since I was somewhat critical of the question, I would offer here some reasons why the phrase: "reduction to the artinian local case" may mean quite differently to an algebraist (I was also vaguely aware of the geometric use BCnrd explained, that's why I asked the OP to clarify).

To prove some statements about a commutative ring $R$, a very often used technique is to pass to the quotient of $R$ by a non-zerodivisor. This is a local analogue of "taking the hyperplane section" in geometry. For examples of this technique, see the proof of the Auslander-Buchsbaum formula in Bruns-Herzog, or this answer.

Now, if the ring you started with is local and Cohen-Macaulay, then after killing enough regular elements you ended up in a local Artinian ring $(A,m,k)$. There are a lot of nice things about such rings: all f.g. modules have finite lengths, the Grothendieck group is $\mathbb Z$, local duality is simple as $A$ is already complete, the representation theory is well-studied, blah blah...

Sometimes you also wish that the residue field $k$ is algebraically closed. This can be obtained by making a faithfully flat extension by the EGA result BCnrd mentioned. Here are a few examples where such extension is useful:

1) The Auslander-Reiten quiver is easier to write down (all arrows has trivial valuations).

2) If $A$ is a complete intersection, then one can attach to each pair of $A$-modules $M,N$ a $k$-variety, whose dimension equals to the rate of (polynomial) growth of the lengths of $\text{Ext}^i(M,N)$. This has a non-trivial consequence that $\text{Ext}^i(M,N)$ and $\text{Ext}^i(N,M)$ grow at the same rate! (see this paper).

3) To prove that there are only finitely many semi-dualizing modules over Cohen-Macaulay algebras, the only known proof is to pass to the Artinian case with alg. closed field and then use relevant results from representation theory.

My hope is although this post did not answer what the OP wanted to ask, perhaps it can be of help to someone who googles "reduction to artinian case" with different motivation and find this MO question.

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