5
$\begingroup$

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:

  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 $X\setminus \{x\}$?
$\endgroup$

1 Answer 1

2
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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 $\mathrm{QCoh}(X\setminus \{x\})$? $\endgroup$
    – LOCOAS
    Mar 15, 2021 at 0:19
  • $\begingroup$ 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. $\endgroup$
    – Moisés
    Mar 15, 2021 at 5:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .