(v.1.1.1)
3.3: Cohomology of a Noetherian Affine Scheme
3.5: The Cohomology of Projective Space
3.7: The Serre Duality Theorem
3.8: Higher Direct Images of Sheaves
3.11: The Theorem on Formal Functions
3.12: The Semicontinuity Theorem
There are no exercises for this sections. Although I think there should be. Weibel’s, Maclane’s, and Rotman’s books all have excellent exercises on derived functors, there are also some in Vakil’s book as well (Maybe I’ll make a list of them).
Let $X = \A^1$ be the affine line over an infinite field $k$. Let $P,Q$ be distinct closed points of $X$, and let $U = X - P,Q$. Show that $H^1(X,\mathbb{Z}_U)\neq 0$.
proof:
Here $\mathbb{Z}_U:=i_!\mathbb{Z}$ the extension by zero of the constant sheaf outside of $U$. There is exact sequence
\[ 0\to \mathbb{Z}_U\to \mathbb{Z}\to i_P(\mathbb{Z})\oplus i_Q(\mathbb{Z})\to 0 \]
Here $i_P(A)$ stand for the skycraper of $A$ at $P$.
We take the cohomology to get exact sequence.
\[ 0\to H^0(X,\mathbb{Z}_U)\to \mathbb{Z}\xrightarrow{q} H^0(X,i_P(\mathbb{Z})\oplus i_Q(\mathbb{Z}))=\mathbb{Z}\oplus \mathbb{Z}\to H^1(X,\mathbb{Z}_U)\to 0 \] The map $q$ takes $s\mapsto (s|_P, s|_Q)$, which is definitely not surjective.
More generally, let $i:Y \to X:=\A^n_k$ be the union of $n+1$ suitably placed hyperplanes, and let $U = X - Y$. Show that $H^1(X,\mathbb{Z}_U) \neq 0$.
proof:
Once again, there exists exact sequence \[ 0\to \mathbb{Z}_U\to \mathbb{Z}\to \mathbb{Z}_Y=i_*\mathbb{Z}\to 0\]
And there is exact sequence,
\[ 0\to H^0(X,\mathbb{Z}_U)\to \mathbb{Z}\xrightarrow{q} H^0(X, \mathbb{Z}_Y)=H^0(Y, \mathbb{Z})\to H^1(X,\mathbb{Z}_U)\to 0\]
For $Y$ with $d$ connected components, $H^0(Y, \mathbb{Z})\cong \mathbb{Z}^d$. Choose $d\geq 2$, the case would follow.
Let $X = \Proj^1$ be the projective line over an algebraically closed field $k$. Show that the exact sequence \[0\to \Oc \to \mathcal{K}\to \mathcal{K}/\Oc \to 0 \] of (II, Ex. 1.21d) is a flasque resolution of $\mathcal{O}$, here $\mathcal{K}$ is the constant sheaf of the function field $K(X)$. Conclude that $H^i(X, \Oc)=0$ for $i>0$.
proof:
Let $U$ be an open set of $X$, we have that $X$ is irreducible, then $U$ must be irreducible, and hence connected. Therefore, $\mathcal{K}(U)=K(X)$ for all opens, and therefore it is flasque.
By Ex.2.1.21e), $\mathcal{K/O}\cong \sum_{P\in X}i_P(K(X)/\Oc_{X,P})$, therefore it is flasque.
Take the global section gives us an exact sequence \[ 0\to \Gamma(X, \Oc)=k\to \Gamma(X, \mathcal{K})=K(X)\xrightarrow{g}\Gamma(X, \mathcal{K/O})=\bigoplus_{P\in X}K(X)/\Oc_{X,P}\to 0 \]
There is then cochain complex \[ 0\to K(X)\to \bigoplus_{P\in X}K(X)/\Oc_{X,P}\to 0 \] that calculate $H^i$. This immediately tells us $H^i(X, \Oc)=0$ for $i>1$. For $i=1$, we claim that $K(X)\to\bigoplus_{P\in X}K(X)/\Oc_{X,P} $ is surjective, but this directly comes from the previous exact sequence.
Let $X$ be a topological space, let $Y$ be a closed subset, and let $\F$ be a sheaf of abelian groups. Let $\Gamma_Y(X, \F)$ denote the group of sections of $\F$ with support in Y. So $s|_U=0$ for $U\subseteq X-Y$.
Show that $\Gamma_Y$ is left exact functor $\mathrm{Ab}(X)\to \mathrm{Ab}$ (we denote $H^i_Y$ the right derived functor of $\Gamma_Y$
proof:
Given $f: \F\to \G$ a sheaf map, let $s\in \F(X)$ such that $s|_ {X-Y}=0$, then we have that \[ f(s)|_ {X-Y}=f(s|_ {X-Y})=f(0)=0\] So $\Gamma_Y$ is a functor.
Suppose we have $0\to \F_1\xrightarrow{f} \F_2\xrightarrow{g} \F_3$ exact. Then $f$ is injective and $\ker{g}=\mathrm{im} \ f$.
We know that $\Gamma$ is left exact, so $\Gamma(f)$ is injective and hence $\Gamma_Y(f)$ is injective.
Suppose $\alpha\in\ker(g)$, there is an unique $s\in \Gamma(X, \F_1)$ such that $f(s)=\alpha$. If $s$ is not compactly supported, and since $f$ is injective and thus, $f|_{X-Y}$ is injective. Then $f(s)|_{X-Y}\neq0 =\alpha|_{X-Y}$. So $s\in \Gamma_Y(X,\F_1)$.
If $0\to \F’\xrightarrow{f} \F\xrightarrow{g} \F”\to 0$ is exact with $\F’$ flasque, then \[0\to \Gamma_Y(X, \F’)\to \Gamma_Y(X, \F)\to \Gamma_Y(X, \F”)\to 0 \] is exact.
proof:
From a) we are left with showing $g$ is surjective.
We know from Ex. 2.1.16b) that $g|_U$ is surjective for all open $U$. Suppose we have $\alpha \in \Gamma_Y(X,\F”)$ and a $s\in \Gamma(X,\F)$ such that $\Gamma(g)(s)=\alpha$, is it supported on $Y$?
We can take the restriction of the exact sequence onto $U=X-Y$, which gives us an exact sequence \[ 0\to \F’(U)\xrightarrow{f}\F(U)\xrightarrow{g}\F”(U) \] $\alpha|_U=0$ so $s|_U\in \ker g$, which by exactness, there is $t\in \F’(U)$ such that $f(t)=s|_U$.
Since $\F’$ is flasque, there is $t’\in \F’(X)$ that lifts $t$.
We then have $s$ and $f(t’)$ that both live over the entire $X$ in $\F$.
Let $u=s-f(t’)$, we have that $g(u)=g(s)-g(f(t’))=g(s)=\alpha$, and $u|_{U}=s|_U-f(t’)|_U=s|_U-f(t)=0$. Hence $u\in \Gamma_Y(X,\F)$ and $g(u)=\alpha$.
Show that if $\F$ is flasque, then $H^i_Y(X,\F)=0$ for all $i>0$
proof:
Let $0\to \F\to \F\to 0$ be the exact sequence associated with the identity of $\F$. This is then a flasque resolution. Take $\Gamma_Y$, we then have \[ 0\to \Gamma_Y(X, \F)\to \Gamma_Y(X, \F)\to 0 \] is exact. Hence $H^i_Y=0$ for $i>0$.
If $\F$ is flasque, show that the sequence \[ 0\to \Gamma_Y(X, \F)\xrightarrow{i} \Gamma(X, \F)\xrightarrow{res} \Gamma(X-Y, \F)\to 0\]
is exact.
proof:
Since $i$ is inclusion, it is injective. Since $\F$ is flasque, $res$ is surjective. So we just need $c= \mathrm{im} \ i$
If $s\in \mathrm{im}\ i$, then $res(s)=0$ by definition. If $s\in \ker res$, then $s$ is supported on $Y$, hence $s\in \mathrm{im} \ i$.
Let $U=X-Y$, show that for any $\F$ there is exact sequence \[ 0\to H^0_Y(X, \F)\to H^0(X, \F)\to H^0(U, \F)\to H^1_Y(X, \F)\to H^1(X, \F)\to H^1(U, \F)\to …\]
proof:
Let $\F\to \G_{\bullet}$ be some flasque resolution, $i_{*}\F\to i_{*}\G_{\bullet}$ is then a flasque resolution of $i_{*}\F$.
There is then exact
\[ 0\to \Gamma_{Y}(X, \G_{\bullet})\to \Gamma(X, \G_{\bullet})\to \Gamma(X, i_{*}\G_{\bullet})= \Gamma(U, \G_{\bullet})\to 0\]
Which determines exactly the long exact sequence we want.
Let $V$ be an open subset of $X$ containing $Y$. Then there are natural functorial isomorphisms, for all $i$ and $\F$,
\[ H^i_Y(X, \F)\cong H^i_Y(V, \F|_V) \]
proof:
For any flasque resolution $\F\to \G_{\bullet}$, there is flasque resolution $i_{*}\F\to i_{*}\G_{\bullet}$.
There is then natural isomorphism of cochain complex $\Gamma_Y(X, \G_{\bullet})\to \Gamma_Y(V, \G_{\bullet})$ as $Y\subset V$. This then induces isomorphism on the cohomology.
Let $Y_1, Y_2$ be two closed subsets of $X$. Then there is a long exact sequence of cohomology with supports
\[ …\to H^i_{Y_1\cap Y_2}(X, \F)\to H^i_{Y_1}(X, \F)\oplus H^i_{Y_2}(X, \F) \to H^i_{Y_1\cup Y_2}(X, \F)\to H^{i+1}_{Y_1\cap Y_2}(X, \F)\to …\]
proof:
Let $\F\to \G_{\bullet}$ be a flasque resolution.
For each $i$, there is sequence \[ 0\to \Gamma_{Y_1\cap Y_2}(X, \G_i)\xrightarrow{i_1\oplus i_2}\Gamma_{Y_1}(X, \G_i)\oplus\Gamma_{Y_2}(X, \G_i)\xrightarrow{(s,t)\mapsto s-t} \Gamma_{Y_1\cup Y_2}(X, \G_i)\to 0 \]
We want to show that this is exact.
$i_1\oplus i_2$ is injective by construction.
For $\alpha\in \Gamma_{Y_1\cup Y_2}(X, \G_i)$, we have $\alpha|_{X-Y_2}$ and $-\alpha|_{X-Y_1}$. For each $j=1,2$, there is a $s_j\in \Gamma(X, \G_i)$ supported on $Y_j$ that is a lift of local sections defined, similar to that in Ex.3.2.3b). So $\alpha=s_i-s_j$.
Now for the middle. Obviously $\mathrm{im}\ i_1\oplus i_2 \subseteq \ker g$, since $s-s=0$. If $(s,t)\in \ker g$, then $s-t=0$ everywhere, Yet $(s-t)|_{X-Y_1}=-t|_{X-Y_1}$ and $(s-t)|_{X-Y_2}=s|_{X-Y_2}$. So $s$ and $t$ are supported on $Y_1\cap Y_2$.
We are done.
Let $X$ be a noetherian scheme. Show that $X$ is affine if and only if $X_{red}$ is affine.
proof:
If $X$ is affine, then $X_{red}$ is obvious affine.
Now if $X_{red}$ is affine. Denote $i: X_{red}\to X$. Let $\mathcal{N}il$ be the sheaf of nilpotents of $X$. Given a $\F$ quasicoherent over $X$, there is then a filtration $\mathcal{N}il^n \F\subseteq \F$. Denote $\F_j:=\mathcal{N}il^n \F/\mathcal{N}il^{n+1} \F$. This is then quasicoherent as well.
Note that $\mathcal{N}il\F_j=0$, so it is a $X_{red}$-module, and $H^{i}(X,F_j)\cong H^{i}(X_{red},\F_j)$, since they have the same underlying space.
There is then a long exact sequence
\[ …\to H^i(X, \mathcal{N}il^{n+1} \F)\to H^i(X, \mathcal{N}il^{n} \F)\to H^{i}(X_{red},\mathcal{N}il\F_j) \to H^{i+1}(X, \mathcal{N}il^{n+1} \F)\to …\]
Since $X$ is noetherian, $\mathcal{N}il^{n}=0$ for big enough $n$, and since $X_{red}$ is affine, higher cohomology vanishes. Therefore, we can deduce that
$H^i(X, \mathcal{N}il^{n} \F)=0$ for all $n$. Let $n=0$ for our case.
Let $X$ be a reduced noetherian scheme. Show that $X$ is affine if and only if each irreducible component is affine.
proof:
($\implies$) If $X$ is affine, then we have that each irreducible component must be affine.
($\impliedby$) Let $X=X_1\cup X_2\cup…\cup X_n$ be irreducible decomposition. By assumption, $X_i$ are affine $\spec(A_i)$.
We show the statement inductively on $n$.
For $n=2$, $X=X_1\cup X_2$. Since $X_1\cap X_2$ is an irreducible of affines, it is itself affine. For any quasicoherent $\F$, there is exact sequence \[ 0\to \F\to i_1^{*}\F\oplus i_2^{*}\F\to i_{12}^{*}\F\to 0 \] There is then a long exact sequence
\[…\to H^i(X, \F)\to H^i(X_1,\F)\oplus H^i(X_2,\F)\to H^i(X_1\cap X_2,\F)\to H^{i+1}(X, \F)\to …\] Since $X_i$ and $X_1\cap X_2$ are all affine and noetherian, $H^i=0$. Therefore, $H^i(X,\F)=0$ for $i>0$. So affine.
Let $X=X_1\cup…\cup X_{n-1}\cup X_n$, since $X_1\cup…\cup X_{n-1}$ is affine, and $(X_1\cup…\cup X_{n-1})\cap X_n$ is irreducible of affine, so affine as well, our argument follows.
Let $A$ be a noetherian ring, and let $a$ be an ideal of $A$.
Show that $\Gamma_a(-)$ is a left exact functor from the category of $A$-modules to itself. We denote its right derived functors, calculated in $\mathrm{Mod}(A)$ by $H^i_a(-)$
proof:
By definition $\Gamma_a(M)={m\in M: \text{there is } n, a^nm=0 }$. For $f: M\to N$, then we have that if $m\in M$ such that $a^nm=0$ for some $n$, then $a^nf(m)=f(a^nm)=0$.
Given exact sequence \[ 0\to M’\xrightarrow{f} M \xrightarrow{g} M” \]
Injectivity is trivial. If $m\in \ker g$, then there is an $m’ \in M’$ such that $f(m’)=m$. Since $a^nm=0$, then $f(a^nm’)=0$. But $f$ is injective, so $a^nm’=0$.
Now let $X = \spec(A)$, $Y = V(a)$. Show that for any $A$-module $M$, $H^i_a(M)\cong H^i_Y(X, \tilde{M})$
proof:
Since $A$ is noetherian, we can find an injective resolution $M\to I_{\bullet}$, which determines a flasque resolution $\tilde{M}\to \tilde{I_{\bullet}}$. Therefore, as long as we show that $\Gamma_a(N)\cong \Gamma_Y(X, \tilde{N})$, then the statement follows. Note that there is isomorphism $N\cong \Gamma(X, \tilde{N})$, we just need to show that a section is annihilated by power of $a$ iff $s$ is supported on $V(a)$.
Since $A$ is noetherian, $a$ is finitely generated $(f_1,…,f_n)$, so $V(a)=V(f_1)\cap …\cap V(f_n)$.
Suppose, $s\in \Gamma_Y(X, \tilde{N})$, then $s|_{D(f_i)}=0$ for all $i$. Then by Lemma 1.5.3a), $f_i^{n_i}s=0\in \Gamma(X, \tilde{N})$. Therefore, $s\in \Gamma_a(N)$.
On the other hand, if $s\in\Gamma_a(N)$, then $s|_{D(f_i)}=b/f_{i}^n$ for some $b \in N$. But $f_{i}^n s=0$ so $s\in \Gamma_Y(X, \tilde{N})$.
For any $i$, show that $\Gamma_a(H^i_a(M))=H^i_a(M)$.
proof:
Let $M\to I_{\bullet}$ be an injective resolution. Apply $\Gamma_a(-)$ we have cochain complex \[ 0\to \Gamma_a(I_1)\to \Gamma_a(I_2)\to… \]
Since every element in the cochain complex are annihilated by powers of $a$, then the cohomology are annihilated as well.
If $A$ is a ring, $a$ an ideal, and $M$ an $A$- module, then $depth_a M$ is the maximum length of an $M$ -regular sequence $x_1, … ,x_n$ with all $x_i\in a$.
Assume that $A$ is noetherian. Show that if $depth_a M \geq 1$, then $\Gamma_a(M)=0$, and the converse is true if $M$ is finitely generated.
proof:
Since $depth_a M\geq 1$, there exists $x\in a$ such that $x$ is not a zero divisor. If $m\in\Gamma_a(M)$, then there is $n$ such that $a^nm=0$. But then $x^nm \in a^nm$ which is not zero for any $n$ unless $m=0$.
On the other hand, suppose $\Gamma_a(M)=0$ and $M$ is finitedly generated.
If $depth_a M=0$, then for every $x\in a$, there is $m\in M$ such that $xm=0$. In other words, all $x\in a$ are zero divisors.
For noetherian ring $A$, we have that the zero divisors are $\cup_{p \in \mathrm{Ass}_A(M)}p$, here $\mathrm{Ass}_A(M)$ is the set of associated prime ideals of $M$. That is $\mathrm{Ann}_A(m)$ for some $m$ that is maximal.
Since $A$ is noetherian, $ \mathrm{Ass}_A(M)$ is finite. We have $a\subseteq \cup_{p \in \mathrm{Ass}_A(M)}p$, so by prime avoidance, $a\subseteq p=\mathrm{Ann}_A(m)$ for some $m\in M$.
Therefore, $am=0$ so $m\in\Gamma_a (M)$, a contradiction.
Show inductively, for $M$ finitely generated, that for any $n \geq 0$, the following conditions are equivalent:
i) $depth_a M\geq n$ ii) $H^i_a(M)=0$ for $i <n$
proof:
For $n=1$, this is a).
Suppose this is true for $n-1$. ($\implies$) Let $M$ be a module of $depth_a M\geq n$, so there is a $x_1,…,x_n\in a$ that is a $M$-regular sequence, so $x_2,…,x_n$ is a $M/x_1$-regular sequence, and $depth _a M/x_1\geq n-1$. Therefore, $H^i_a( M/x_1)=0$ for $i <n-1$
We have that \[ 0\to M\xrightarrow{x_1}M \to M/x_1\to 0 \] is short exact, since $x_1$ is not a zero divisor.
There is then long exact sequence \[ 0\to H^{n-1}_a(M)\xrightarrow{x_1}H^{n-1}_a(M)\to H^{n-1}_a(M/x_1) \] So $x_1$ is not a zero divisor in $H^{n-1}_a(M)$.
Therefore $depth_a H^{n-1}_a(M)\geq 1$, hence $\Gamma_a(H^{n-1}_a(M))=0$ by a). However, from Ex 3.3.3c), $\Gamma_a(H^{n-1}_a(M))=H^{n-1}_a(M)$, so $H^{n-1}_a(M)=0$.
($\impliedby$) Once again, induction on $n$. Suppose true for $n-1$. If $H^i_a(M)=0$ for $i <n$. We know that $M$ has at least depth $n-1$. So there is a $M$-regular sequence $x_1,…,x_{n-1}\in a$. Then $M/(x_1)$ has depth $\geq n-2$. If $H^{n-2}_a(M/(x_1))=0$, then we have that $M/(x_1)$ has depth $\geq n-1$, and $M$ has depth $\geq n$.
We have that $H^{n-1}_a(M)=0$ and $H^{n-2}_a(M)=0$, so there is \[ H^{n-2}_a(M)=0\to H^{n-2}_a(M/(x_1))\to H^{n-1}_a(M)=0 \] exact. This shows our argument.
Let $X$ be a noetherian scheme.
Show that the sheaf $\G$ constructed in the proof of (3.6) is an injective object in the category $\mathrm{Qcoh}(X)$ of quasi-coherent sheaves on $X$. Thus $\mathrm{Qcoh}(X)$ has enough injectives.
proof:
Let’s recall the construction of $\G$.
Let $\F$ be a quasicoherent $X$-module. Find a (finite, since $X$ is noetherian) affine cover $U_i=\spec(A_i)$ of $X$ such that $\F|_{U_i}=\tilde{M_i}$ for some $A_i$-module $M_i$. This always exists by construction. Find an injective $A_i$-module $I_i$ and embedding $M_i\to I_i$ for each $i$. There is then a sheaf map $\F|_{U_i}\to \tilde{I_i}$. Define $\G:=\bigoplus_i f_{*}\tilde{I_i}$ (or simply the product as $i$ is finite). There is an morphsm $\F\to \G$, which is clearly an injection.
The key is to show that $\Hom_{\Oc_X}(-, \G)$ is exact. First of all, we have that \[\Hom_{\Oc_X}(-, \G)=\Hom_{\Oc_X}(-, \bigoplus_i f_{*}\tilde{I_i}) \cong \bigoplus_i \Hom_{\Oc_X}(-, f_{*}\tilde{I_i})\cong\bigoplus_i \Hom_{\Oc_{U_i}}(f^{*}(-), \tilde{I_i}) \]
Since $f:U_i\to X$ is an open immersion, $f^{-1}\mathcal{N}\cong \mathcal{N}|_{U_i}$. Therefore, $f^{-1}$ is exact. Moreover, since $f^{-1}\Oc_{X}\cong \Oc_{U_i}$ and $f^*=\Oc_{U_i}\otimes_{f^{-1}\Oc_{X}}f^{-1}$, so $f^{*}$ is also exact.
Since $U_i$ are affine, and ${I_i}$ is an injective $A_i$-module, $\Hom_{\Oc_{U_i}}(-, \tilde{I_i})$ is exact.
Now, $\Hom_{\Oc_{U_i}}(f^{*}(-), \tilde{I_i})$ is the composition of exact functors, so it is exact as well. Lastly, $\Hom_{\Oc_X}(-, \G)$ is the product of exact functors, it is itself exact.
Therefore, $\G$ is injective.
Show that any injective object of $\mathrm{Qcoh}(X)$ is flasque.
proof:
We first show that for any open subset $i:U\to X$, if $\F$ is injective over $X$ then $\F|_{U}$ is injective $\Oc_U$. Suppose we have $\G\to \G’$ injection over $U$, and a map $\G\to \F|_{U}$, we wish to extend to a $\G’\to\F|_{U}$.
Suppose $\G$ and $\G’$ are coherent, then by Ex 2.5.15, there exists coherent module and injection $\G_X\to \G’_X$ that restrict to $\G\to \G’$, and there is a map $\G_X\to \F$ that restrict to $\G\to \F|_{U}$. Thus a lift exists.
By Ex 2.5.15e), any quasicoherent modules are union of coherent modules, so we are good.
Let $\F\in \mathrm{Qcoh}(X)$ be an injective module. Choose an affine cover $U_i=\spec(A_i)$ such that $\F|_{U_i}=\tilde{I_i}$. We have that $I_i$ must be injective from the above. Then by Lemma 3.3.3, the map between any injective module $I$ to its localization over a noetherian ring is surjective. Hence, $\F$ is flasque.
Conclude that one can compute cohomology as the derived functors of $\Gamma(X, -)$ considered as a functor from $\mathrm{Qcoh}(X)$ to $\mathrm{Ab}$.
proof:
Ah Doiii.
Let $f:X \to Y$ be an affine morphism of noetherian separated schemes. Show that for any quasi-coherent sheaf $\F$ on $X$, there are natural isomorphisms for all $i\geq 0$ \[ H^{i}(X, \F)\cong H^{i}(Y, f_{*}\F) \] proof:
Since $Y$ is noetherian and separated, theorem 3.4.5 says that for an ordered open cover $\mathfrak{U}$, we have
\[ {H}^i(\mathfrak{U}, \G)\xrightarrow{\cong} H^i(Y, \G) \] The same would follow for $X$.
Choose an ordered affine cover $\mathfrak{U}$ of $Y$, then $f^{-1}(\mathfrak{U})$ is then an affine cover of $X$, as $f$ is affine.
The Cech cochain of $f_{*}$ over $\mathfrak{U}$ is \[ {C}^n(\mathfrak{U}, f_{*}\F)=\prod_{i_0\leq i_1\leq …\leq i_n}f_{*}\F(U_{i_0,i_1,…,i_n}) \cong \prod_{i_0\leq i_1\leq …\leq i_n}\F(f^{-}(U_{i_0,i_1,…,i_n})) ={C}^n(f^{-1}\mathfrak{U}, \F) \] Therefore, the cohomology are the same.
Let $X =\A^{2}_k = \spec( k[x,y])$, and let $U = X - (0,0)$. Using a suitable cover of $U$ by open affine subsets, show that $H^1(U, \Oc_U)$ is isomorphic to the $k$-vector space spanned by $x^iy^j$, $i, j<0$.
proof:
There are $U_0:=D(x)=\spec(k[x^{\pm}, y])$ and $U_1:=D(y)=\spec(k[x, y^{\pm}])$, which forms an open affine cover $\mathfrak{U}$ of $U$. Then we have \[ d_1:C^0(\mathfrak{U}, \Oc_U)=k[x^{\pm}, y]\times k[x, y^{\pm}]\to C^1(\mathfrak{U}, \Oc_U)=k[x^{\pm}, y^{\pm}] \] $d_1(a, b)=a-b$, and $C^2=0$,
Define linear transformation, $f:k[x^{\pm}, y^{\pm}]\to k(x^iy^j | i, j<0)$, such that $f(x^iy^j)=0$ if $i$ or $j\geq 0$.
This is obviously a surjection. The kernel is the spanned by $x^iy^j$ for $i$ or $j\geq 0$.
Suppose, $i\geq 0$, then $x^iy^j \in k[x, y^{\pm}]$, hence lies in the image of $d_1$, similarly for $j\geq 0$. On the other hand, image of $d_1$ is spanned by $x^iy^j$, and we are done