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

I came across the notion as follows:

Let $X$ be a projective, smooth scheme. And let $$ 0\to M\to N\to \mathcal{O}_{X}\to0 $$ be an exact sequence of coherent $\mathcal{O}_X$-modules. What is meant by “the above exact seqence defines an $M$-torsor on $X$”? I think it may be a standard use of terminology. I just lack knowledge. Thank you!

share|improve this question

1 Answer 1

up vote 9 down vote accepted

From the associated long exact sequence you obtain a map $H^0(X,\mathcal O_X) \to H^1(X, M)$. $H^1(X,M)$ is just the group paramaterizing $M$-torsors. Since $H^0(X,\mathcal O_X)$ is free of rank one as a module over itself, this map is equivalent to a single element of $H^1(X,M)$, the image of $1$ - in other words, a single $M$-torsor.

Explicitly one is just viewing the inverse image of $1$ in $N$ as an $M$-torsor.

share|improve this answer
    
It helps . Thank you very much! –  TOM Jun 10 '13 at 3:22
1  
The torsor is the inverse image of 1 in $N$. –  Angelo Jun 10 '13 at 5:08

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.