Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The Serre's condition $(S_n)$, especially $(S_2)$, has been mentioned in a few MO answers: see here and here for example. I am pretty sure I have seen it in other questions as well, but could not remember exactly.

I have always been confused by this condition, especially for a sheaf $F$ on a locally Noetherian scheme $X$, mainly because as far as I know, there are at least three different definitions in the literature. Let us look at them, $F$ is said to satisfy condition $(S_n)$ if:

$$ depth_x (F_x) \ge \min (\text{dim} \mathcal O_{X,x},n) \ \forall x\in X \ (1) $$

$$ depth_x (F_x) \ge \min (\dim F_x,n) \ \forall x\in X \ (2) $$

$$ depth_x (F_x) \ge \min (\dim F_x,n) \ \forall x\in \text{Supp}(F) \ (3) $$

Definition (1) can be found in Evans-Grifffith book "Syzygies", definition (2) is given in EGA IV (definition 5.7.2) or Bruns-Herzog book "Cohen-Macaulay modules". Definition (3) is what VA used in his answer to the second question quoted above (and I certainly have seen it in papers or books, but can't locate one right now, so references would be greatly appreciated).

When $F$ is the structure sheaf or a vector bundle (of constant positive rank), then they all agree. However, they can differ when $E$ is a sheaf. For example, (1) allows us to say that if $X$ is normal, then $E$ is reflexive if and only if it is $(S_2)$. But according to (2) or (3), if $X=Spec(R)$ for $(R,m)$ local, then $k=R/m$ would satisfy $(S_n)$ for all $n$. (2) and (3) are equivalent if we assume that the depth of the $0$ module is infinity, but I have seen papers which do not use that convention, adding to the confusion.

Since a result surely depends on which definition we use (I have certainly made mistakes because of this confusion, and I think I am not alone), I would like to ask:

Question: Is there an agreement on what exactly is condition $(S_n)$ for sheaves? If not, what are the advantages and disadvantage for each of the different definition?

Some precise references:

Bruns-Herzog "Cohen-Macaulay modules" : after Theorem 2.1.15, version (2)

Evans-Griffith "Syzygies": part B of Chapter 0, version (1)

Kollar-Mori "Birational geometry of algebraic varieties": definition 5.2, version (1)

EGA, Chapter 4, definition 5.7.2, version (2).

share|improve this question
1  
This isn't really even a question about sheaves; I've seen both (1) and (2) used for modules (though never (3), as far as I can remember). As you point out, they aren't equivalent even then, since every module of finite length has ($S_n$) under (2). For me, that's enough to say that we should be using (1), since I want "maximal Cohen-Macaulay" to mean "($S_n$) for all $n$", and I don't want the residue field to be maximal Cohen-Macaulay. –  Graham Leuschke Apr 22 '10 at 20:14
    
The Kollar-Mori reference was pointed out in an answer (now deleted) by VA. –  Hailong Dao Apr 23 '10 at 4:45
    
In this writeup (www-personal.umich.edu/~kschwede/GeneralizedDivisors.pdf), Karl Schwede gives definition (1) (Definition 1.12), and asserts that it is "highly non-standard". He says "I am largely following an outline of Hartshorne where he uses this non-standard definition". The two Hartshorne articles he might be referring to are ams.org/mathscinet-getitem?mr=1291023 and ams.org/mathscinet-getitem?mr=597077 . I don't have access to either one to check right now. It's worth pointing out that Hartshorne's book only defines $(S_2)$ for rings, not modules. –  Graham Leuschke Apr 23 '10 at 16:06
1  
Dear Hailong, sorry, I had missed that. Well, you called me Algelo, we'll call it even :) –  Angelo Oct 24 '10 at 8:59
1  
No offence taken of course, I mistype all the time. It was meant as a joke, in case it wasn't obvious. –  Angelo Oct 24 '10 at 16:09
show 4 more comments

1 Answer

I would add one more observation to the other comments. Let me not worry about (2) vs (3) as the difference is only about the zero module so this is more of a philosophical question than a mathematical one.

I would just like to point out that there is a very useful characterization of depth and dimension of a module, namely Grothendieck's vanishing theorem which says that at any $x\in X$, the local cohomology of $M$ vanishes for $i$ strictly between the depth and the dimension of the module and does not vanish for $i$ equal either the depth or the dimension.

In my eyes this suggest that one should use the dimension of the module in the definition, i.e., use (2).

Another argument to support the use of (2) is that we like to say that CM is equivalent to "$S_n$ for all $n$". Now if you use definition (1) then only modules supported on the entire $X$ have even a chance to be CM, but I don't see how one would gain from assuming that. More specifically, a module could never satisfy $S_n$ for any $n$ that's larger than the dimension of the module, but not larger than $\dim X$.

Kind of along the same lines, let $A\to B$ be a surjective morphism of rings (commutative with an identity) and $M$ a $B$-module. I.e., $\operatorname{Spec}B$ is a closed subset of $\operatorname{Spec}A$. Now both $\operatorname{depth}M$ and $\operatorname{supp}M$ are independent of the fact whether one views $M$ as a $B$-module or an $A$-module. It is reasonable that then whether it is $S_n$ would be also independent.

The main difference between (1) and (2) is whether one wants to compare to the support of the module (i.e., view it over ring/annihilator) or the whole ring. To me, the second seems more natural. This way a sheaf/module that is $S_n$ on a subscheme remains $S_n$ when viewed on an ambient scheme. The definition (1) seems to prefer to compare to the fixed ring. One way some people try to bridge the gap between the two definitions is to say "$M$ is $S_n$ over its support", meaning that one should mod out by annihilator first before applying (either of the) definition(s). Then the two definitions are equivalent. As for (3), some people go the distance to say "a non-zero module is $S_n$ if..."

share|improve this answer
    
Dear Sandor, thank you for your insight. –  Hailong Dao Oct 23 '10 at 14:34
add comment

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.