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

Let $(A,\mathfrak{m})=k[[x,y]]$ with $char(k)=0$ and $K=Quot(A)$. Set $X=Spec(A)$, $U=Spec(A)\backslash \lbrace \mathfrak{m} \rbrace$ the pointed spectrum. Furthermore given an $A$-algebra $B$, which can be embedded in $C=M_n(A)$, where $B$ is free $A$-module of rank $n^2$. One can see the algebra as a sheaf on $X$ resp. $U$ and $B^{\times}$ denotes its group of units.

Now we get an exact sequence $0\rightarrow B^{\times} \rightarrow C^{\times}=Gl_n(A) \rightarrow F \rightarrow 0$ and $F$ is supported on $Y=\lbrace x=0 \rbrace$, where it can be identified with some flag space.

Now in the article i'm reading, there are 3 facts i don't quite understand (now it is just one):

a) $H^0(X,C^{\times})=H^0(U,C^{\times})$ because $C$ is a free $A$-module

b) $H^0(X,C^{\times}) \rightarrow H^0(X,F)$ is surjective, because $X$ is local (see vytas comment)

c) $H^1(U,C^{\times})=0$ because $C$ and $A$ are Morita equivalent and this holds for $A$ (here one has to use "reflexive" Morita equivalence, then this follows from the fact that every reflexive A-module is free)

My question is: a) why does this follow from the freeness. If it was just $C$ i would believe this, because sections extend uniquely to codimension 2 points, which works here because $dim(A)=2$. But why should this be true for the group/sheaf of units? More generally this should be true if we replace $A$ by a free $A$-algbera $D$, i.e $H^0(X,Gl_n(D))=H^0(U,Gl_n(D))$. I tried using the Cech complex for $U$, as suggested in the comments, but it didn't help.

The article/text i'm referring to is $\href{http://www.math.lsa.umich.edu/courses/711/ordpages60-85.ps}{this}$. It is called "Stable orders and the Riemann-Roch Theorem". It is Lemma 3.1.9 on page 62 ( Page 3 in this document ).

share|improve this question
    
Would you mind giving the title of the article you are reading? –  Hailong Dao Jul 4 '11 at 18:31
1  
I think i understand b). Since X is local the only open subset containing $\mathfrak m$ is $X$ (all non-empty closed subsets contain $\mathfrak m$), hence if F is a sheaf on X, the stalk at $\mathfrak m$ is equal to F(X). One obtains b) by looking at the exact sequence of stalks at $\mathfrak m$. –  vytas Jul 4 '11 at 20:08
    
If you write down the Check complex for the covering $\{x\neq 0\}$ and $\{y\neq 0\}$ of $U$ for the sheaves $C^{\times}$ and $C$, you seem to get a). Not sure, whether this the intended proof though. –  vytas Jul 4 '11 at 20:54
    
Thanks for your answers. I added the name of the text i was referring to. I will try to understand your tips tomorrow vytas and will maybe ask additional questeions :-). –  TonyS Jul 4 '11 at 21:17

1 Answer 1

up vote 2 down vote accepted

a): An element of $C^\times$ can be thought of as a pair $(a,b)$ of elements of $C$ with $ab=1$. This gives a) by applying of existence extension to $a$ and $b$ and unicity to $ab$ and $1$.

b): The relation $F=C^\times/B^\times$ is a relation of a sheaf quotient for the flat topology (even if one defines $F$ as schematic quotient as $B^\times$ is flat. Hence by general nonsensen, the obstruction to lifting a section of $F$ to one of $C^\times$ is a $B^\times$-torsor. This can also be interpreted as as a locally free $B$-module of rank $1$ which is trivial as $B$ is finite over a local base. Hence the obstruiction vanishes and the section lifts.

c): As you say this follows from Morita equivalence (it would actually be enough that $C$ be an Azumaya algebra).

share|improve this answer
    
Thanks for your answer. There is just one question: I don't quite get the idea behind a). I take an element $s$ in $C^{\times}(U)$. Then i can think of $s$ as $(a,b)$ with $ab=1$ (why this identification? $a=s$, $b=s^{-1}$?). Then $a$ and $b$ are also in $C(U)$ and satisfy $ab=1$. Now i can extend them to $C(X)$ but also the relation can be extend, so the pair is even in $C^{\times}(X)$. Why/where do i need unicity? –  TonyS Jul 5 '11 at 18:34
1  
Yes, projection on the first factor of $(a,b)$ gives the identification. Unicity is needed because we have $ab=1$ on $U$ and then we get $ab=1$ on $X$ by unicity. –  Torsten Ekedahl Jul 6 '11 at 4:40
    
Okay. Thanks a lot for your help. –  TonyS Jul 6 '11 at 7:13

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.