Let $A=k[x,y]$ be the polynomial ring in two variables over a field $k$,
$\mathbb{A}^2=\mathrm{Spec} A$ the affine plane ,
and $U=\mathbb{A}^2 \setminus \{0\}$ the punctured affine plane.
Then the canonical injection $j : U \to \mathbb{A}^2$ induces
a fully faithful left exact functor $j_{*}: \mathrm{QCoh} (U) \to \mathrm{QCoh} (\mathbb{A}^2)\simeq \mathrm{Mod} (A)$ between the category of quasi-coherent sheaves.
In the page 114 of this book,
the author says
\begin{equation}
\mathrm{QCoh} (U)=\{M\in \mathrm{Mod}(A) \mid \mathrm{Hom}_A(A/\mathfrak{m}, M) =\mathrm{Ext}^1 (A/\mathfrak{m}, M) =0 \}
\end{equation}
as a subcategory of $\mathrm{Mod} (k[x,y])$,
where $\mathfrak{m}=(x,y) \subset A$.
My Question:
- Why can we describe $\mathrm{QCoh} (U)$ as the above?
- For a scheme $X$ with a closed point $x\in X$,
consider the canonical open immersion $j:X\setminus \{x\} \to X.$
Then can we calculate $(j_* \mathcal{F})_{x}$ for a quasi-coherent sheaf on $X\setminus \{x\}$?
Best Answer
Honestly, I can't figure out how the book wants us to explain this fact. Based on the mention of section functors and localizations I assume we are supposed to use the stuff in sections 2.13 and 2.14, so let's give this a shot to answer your question 1.
$\DeclareMathOperator{\QCoh}{QCoh} \newcommand{\QA}{\QCoh(\mathbb A^2)} \newcommand{\m}{\mathfrak m} \newcommand{\QU}{\QCoh(U)} \newcommand{\Id}{\mathrm{Id}} \newcommand{\F}{\mathcal F} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\Ext}{Ext} \DeclareMathOperator{\Spec}{Spec} $ Let $T$ be the full subcategory of $\QA$ consisting of sheaves set-theoretically supported at $\{0\}$. In terms of modules, these are modules such that every element is annihilated by a power of $\m$. It is clearly a Serre subcategory.
Let's just identify $\QA/T$ with $\QU$ from now on.
It remains to prove that if $M\in \QA$ is such that $\Hom(A/\m,M)=\Ext^1(A/\m,M)=0$, then $G$ is in the essential image of $j_*$. For this we use Theorem 14.8. It gives us an exact sequence: $$ 0\to \tau M\to M\xrightarrow{\eta} j_*j^*M\xrightarrow{\pi}M'\to 0. $$ It further tells us that the first and last term are in $T$ and that $\eta$ is an essential map. We are going to heavily use that $A/\m$ generates $T$, in order to show that $\eta$ is an isomorphism.
First, being torsion, if $\tau M\neq 0$ has some submodule isomorphic to $A/\m$. This would mean that $M\supseteq \tau M$ does as well, which contradicts the assumption that $\Hom(A/\m,M)=0$. So $\tau M=0$.
Finally, if $M'\neq 0$ it also has a submodule isomorphic to $A/\m$. We then have a short exact sequence $$ 0 \to M\to \pi^{-1}(A/\m)\to A/\m\to 0 $$ Since $\Ext^1(A/\m,M)=0$ by assumption, this sequence splits, so $A/\m$ embeds into $\pi^{-1}(A/\m)\subseteq j_*j^*M$, into a submodule disjoint from $M$. This contradicts the fact that $\eta$ is essential. So $M'=0$, and $\eta$ is an isomorphism as desired. This answers question 1, modulo some details.
For question 2, I'm not sure what constitutes a satisfactory answer. Since you are looking at the stalk, you can assume that $X$ is affine, say $\Spec R$, and $x = V(f_1,\ldots ,f_n)$ for some $f_i\in R$. If we let $U_i = X\setminus V(f_i) = \Spec R[f_i^{-1}]$ $$ X\setminus x = \bigcup_i U_i $$ So by the property of being a sheaf, for any sheaf $M$ on $X\setminus x$, we have that $\Gamma(X,j_*M)=\Gamma(X\setminus x,M)$ is the kernel of $$ \bigoplus \Gamma(U_i,M) \longrightarrow \bigoplus \Gamma(U_i\cap U_j,M) $$ And I believe that by the exactness of localization, tensoring the above map with the local ring at $x$ will give you the stalk as the kernel. In practice, how hard is this kernel to find? I have no clue. I can show that in the case of $\mathbb A^2$, $j_*\mathcal O_{\mathbb A^2\setminus 0} \cong \mathcal O_{\mathbb A^2}$.