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I am studying Algebraic Geometry with this book.

Through the Serre's criterion for affinity, we can know that $H^1(X,\mathcal{F})=0$ then $X$ is affine with some conditions on $X$ and $\mathcal{F}$. I have a question about the proof.

In the book, the proof starts with choosing a closed point $x\in X$ and an affine open neighborhood $U$ of X. Then let $\mathcal{M}$ be the sheaf of ideals of $\mathcal{O}_X$ which defines the reduced closed subscheme of $X$ whose underlying space is $\{x\}$, and $\mathcal{J}$ be the sheaf of ideals on $X\setminus U$.

With above, the author says, $0 \to \mathcal{MJ} \to \mathcal{J} \to \mathcal{J}/ \mathcal{MJ} \to 0$ is an exact sequence of sheaves.

  1. What is the sheaf $\mathcal{MJ}$ means? Actually, I have never seen the notation $\mathcal{FG}$ for two sheaves $\mathcal{F}$, $\mathcal{G}$.
  2. It seems like that the underlying topological spaces of each sheaves in the sequence are different, but then how can we think about their sequence?
User0829
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    Dear User, Probably $\mathcal I$ is intended to be the sheaf of ideal on $X$ which defines $ X \setminus U$, rather than a sheaf on $X \setminus U$. Then both $\mathcal I$ and $\mathcal M$ will be sheaves on $X$, and your second question will be resolved. Regards, – Matt E Jun 06 '13 at 19:16

1 Answers1

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The point is that both $\mathcal M$ and $\mathcal J$ are not only sheaves of $\mathcal O_X$-modules, but they are also sheaves of ideals of $\mathcal O_X.$ Thus, locally it makes sense to multiply sections of the two sheaves.

For example, if there exists some open affine subscheme $V$ which intersects $X\setminus U$ and $x$ nontrivially, then the above sequence would locally become $0\to\mathfrak m_x\cdot J\to J\to J/\mathfrak m_x\cdot J\to 0$ where $\mathfrak m_x$ is the maximal ideal of $x$, and $(X\setminus U)\cap V$ has ideal $J,$ in $\Gamma(V,\mathcal O_X).$

The underlying topological space is $X$ for all three sheaves (they are $\mathcal O_X$-modules), though the support may vary.

Andrew
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