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 ).

  • $\begingroup$ Would you mind giving the title of the article you are reading? $\endgroup$ – Hailong Dao Jul 4 '11 at 18:31
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    $\begingroup$ 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 $\mathfrak m$. $\endgroup$ – vytas Jul 4 '11 at 20:08
  • $\begingroup$ 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. $\endgroup$ – vytas Jul 4 '11 at 20:54
  • $\begingroup$ 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 :-). $\endgroup$ – TonyS Jul 4 '11 at 21:17

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).

  • $\begingroup$ 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? $\endgroup$ – TonyS Jul 5 '11 at 18:34
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    $\begingroup$ 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. $\endgroup$ – Torsten Ekedahl Jul 6 '11 at 4:40

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