Quasi-coherent sheaves on punctured affine plane

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 $$$$\mathrm{QCoh} (U)=\{M\in \mathrm{Mod}(A) \mid \mathrm{Hom}_A(A/\mathfrak{m}, M) =\mathrm{Ext}^1 (A/\mathfrak{m}, M) =0 \}$$$$ as a subcategory of $$\mathrm{Mod} (k[x,y])$$, where $$\mathfrak{m}=(x,y) \subset A$$.

My Question:

1. Why can we describe $$\mathrm{QCoh} (U)$$ as the above?
2. 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 $$X\setminus \{x\}$$?

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.

• We have $$j^*\colon \QA\to \QU$$. It sends $$T$$ to $$0$$, so by the universal property (Theorem 13.9), it factors through $$\QA\to \QA/T\to \QU$$. We claim the second functor is an equivalence. The inverse is the composition $$\QU\xrightarrow{j_*} \QA\to \QA/T$$. Going one way, $$j^*j_*\cong \Id_{\QU}$$. Going the other way, we have a natural transformation on $$\QA$$: $$\Id_{\QA}\to j_*j^*.$$ I believe that all we need to see is that it becomes an isomorphism after modding out by $$T$$.

Let's just identify $$\QA/T$$ with $$\QU$$ from now on.

• $$T$$ is localizing, since $$j^*\colon \QA\to \QU$$ has a right adjoint, $$j_*$$. In terms of Section 2.14, $$j_*$$ is the "section functor". Proposition 14.7(6) tells us already that $$j_*$$ is fully faithful. Proposition 14.7(3) tells us that if $$\F\in \QU$$ and $$M\in T$$, $$\Hom_{\QA}(M,j_*\F)\cong \Hom_{\QU}(j^*M,j^*j_*\F) \cong \Hom_{\QU}(0,\F)=0.$$ Proposition 14.7(3) tells us that $$\mathrm{Ext}^1_{\QA}(M,\F)=0.$$So this gives us one inclusion.

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 $$\mathbb A^2$$, $$j_*\mathcal O_{\mathbb A^2\setminus 0} \cong \mathcal O_{\mathbb A^2}$$.

• Thank you, that is precisely what I want. It seems that $\mathrm{QCoh}(U)$ is a localization of $\mathrm{QCoh}(X)$ for an open subscheme of a scheme $X$. Is this true? Also, is the description of Question 1 true for $\mathrm{QCoh}(X\setminus \{x\})$? Mar 15, 2021 at 0:19
• Yes, I think so. Replacing $\mathfrak m$ by the ideal that cuts out $X\setminus U$, I believe everything should go through. If $X$ is not affine, you don't get to think of $\operatorname{QCoh}(X)$ as modules over a ring, but you can still work with sheaves and open coverings. Mar 15, 2021 at 5:22