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Given $S=\mathbb{P}^2$ and a locally free $O_S$-module $E$ of rank r and an integer $l\geq 1$. Then it is known that the scheme $Quot(E,l)$ is irreducible, due to Ellingsrud and Lehn. Here $Quot(E,l)$ parametrizes zero dimensional quotients $E\rightarrow T$ of length $l$.

Are there any generalizations of this scheme?

I'm thinking for example: Given a locally free sheaf $R$ of associative $O_S$-algebras, not necessarily commuative, of finite rank and a locally projective $R$-module $E$, which is locally free and of finite rank as an $O_S$-module.

Is there a scheme $Quot_R(E,l)$ parametrizing zero dimensional $R$-quotients $E\rightarrow T$ of $R$-length $l$? Does such a scheme have similar properties, i.e. it is irreducible or connected?

Edit: Thanks to t3suji and Sasha this question is solved. I remoed the rest of the question, so i can accept their answers, and i think the deformation problem deserves its own question anyway :-).

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2 Answers 2

up vote 3 down vote accepted

This can be thought of as a response to your comment to Sasha's answer.

Until you try fixing length, there is no issue. Indeed, consider the scheme $$Quot(E)=\coprod Quot(E,k)$$ of all quotients of $E$. Then, as Sasha says, $Quot_R(E)$ is obviously a closed subscheme.

You can now consider $Quot_R(E,l)$ as subsets of $Quot_R(E)$. It is not hard to see that you get a stratification of $Quot_R(E)$ in this way. This is the main difference between $Quot_R(E)$ and $Quot(E)$: we get a family of locally closed subsets of $Quot_R(E)$, while for $Quot(E)$, the subsets are both open and close.

To me, it seems that now you run into a bit of trouble. Namely, $Quot_R(E,l)$, being a locally closed subset of $Quot_R(E)$, can be equipped with a scheme structure. However, it is not unique (because you do not know the nilpotents in the structure sheaf). You can trace the issue to the definition of $Quot_R(E,l)$: if you want it to parametrize $R$-modules of given length, you have to define what it means to have certain length, not just for $R$-modules, but for their families. The problem is non-existent for $O$-modules because length is invariant in the family (which by the way is only true if the ground field is algebraically closed!)

On the other hand, questions like irreducibility or connectivity do not depend on the scheme structure, so maybe you don't need to worry about this.

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Okay, thanks. This is good. But more and more it seems to me, that asking for irreducibility / connectivity is too much. For my purposes a weaker statement would be enough. I will think about this and update the question. –  TonyS May 17 '11 at 21:55

Of course there is a scheme $Quot_R(E,l)$. Indeed, each $R$-module quotient $E \to T$ is an $O_S$-module quotient, so $Q_R(E,l)$ is a closed subscheme of $Quot(E,l)$ consisting of all surjections $E \to T$ of $O_S$-modules such that the kernel is invariant under the action of $R$.

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Yes, but the problem, from my point of view is that $Q_R(E,l)$ may lie in $Quot(E,k)$ for different k. For example like in my example where the $R_p$ simples have $O_{S_p}$-length 1 and 2. Then an elment of $Q_R(E,1)$ could lie, as an $O_S$-module in $Quot(E,1)$ and $Quot(E,2)$. But in $Quot(E,2)$ there are also e.g. quotients from $Q_R(E,2)$ with support on two points with simple $R_p$-modules of $O_{S,p}$-length. So one needs more conditions to distinguish these quotients. –  TonyS May 17 '11 at 20:32
    
I think at first you should restrict your attention to the quotients $E \to T$ with fixed length of $T$ as an $O_S$-module. This will give you a closed subscheme $Quot_R(E,l) \subset Quot(E,l)$. After that, you can consider the stratification of $Quot_R(E,l)$ by length of $T$ as $R$-module. –  Sasha May 18 '11 at 5:12

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