Problem Set 2
II.1
For any open \(U \subseteq X\) show that the functor \begin{align*} {{\Gamma}\qty{U, {-}} }: {\mathsf{Sh}}(X) \to {\mathsf{Ab}}{\mathsf{Grp}} \end{align*} is left-exact, but need not be exact.
We’re given exactness of \begin{align*} \xi: 0 \to {\mathcal{F}}_1 \xrightarrow{f} {\mathcal{F}}_2 \xrightarrow{g} {\mathcal{F}}_3 \to 0 ,\end{align*} which (e.g. ) is data of the form
Applying \(\Gamma(X; {-})\), we want to show exactness of \begin{align*} \xi_X: 0 \to {\mathcal{F}}_1(X) \xrightarrow{f_X} {\mathcal{F}}_2(X) \xrightarrow{g_X} {\mathcal{F}}_3(X) \to \cdots .\end{align*}
Exactness at \(f_X\):
-
Use that \(f\) is exact \(\iff \ker f = \mathbf 0\) as a sheaf, so
\begin{align*}
(\ker f)(U) = (\mathbf 0)(U) = 0 \in \mathsf{CRing}
.\end{align*}
- Use that \begin{align*} (\ker f)(U) &\coloneqq\ker f_U \coloneqq\ker( {\mathcal{F}}_1(U) \xrightarrow{f_U} {\mathcal{F}}_2(U)) \\ \implies \ker f_X &= (\ker f)(X) = (\mathbf 0)(X) = 0 .\end{align*}
- Why this works: the kernel presheaf is already a sheaf, so we can use the presheaf assignment \((\ker f)(U) = \ker f_U\) directly. This won’t work for the cokernel sheaf, since the image presheaf needs to be sheafified.
Alternatively, a direct argument that \(f_X\) is injective:
- A fact we’ll need: \(\xi\) is exact iff locally exact. Sketch of situation, there are commutative squares for all \(p\in X\):
-
Write the kernel out: \begin{align*} \ker f_X \coloneqq\left\{{ s\in {\mathcal{F}}_1(X) {~\mathrel{\Big\vert}~}f_X(s) = 0 \in {\mathcal{F}}_2(X)}\right\} .\end{align*}
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Suppose \(s\in {\mathcal{F}}_1(X)\) and \(f_X(s) = 0\) in \({\mathcal{F}}_2(X)\). Then \begin{align*} ({\mathcal{F}}_2 \mathrel{\Big|}^X_p \circ f_X )(s) &= {\mathcal{F}}_2\mathrel{\Big|}^X_p( 0) = 0 \in ({\mathcal{F}}_2)_p \text{ Ring mor, 0 to 0}\\ \implies (f_p \circ {\mathcal{F}}_1 \mathrel{\Big|}^X_p(s) &= ({\mathcal{F}}_2 \mathrel{\Big|}^X_p \circ f_X)(s) = 0 \text{ by commutativity} \\ \implies {\mathcal{F}}_1 \mathrel{\Big|}^X_p (s) &= 0 \text{ left-cancel $f_p$ since mono} ,\end{align*} which holds for all \(p\).
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Claim: by the sheaf condition on \({\mathcal{F}}_1\), \(s= 0 \in {\mathcal{F}}_1(X)\).
- Fix \(p\). For \(s\in {\mathcal{F}}_1(X)\), write a representative \({\mathcal{F}}_1\mathrel{\Big|}^X_p(s) = [U, \tilde s\in {\mathcal{F}}_1(U)]\).
Recall \((U_1, s_1) \sim (U_2, s_2) \in {\mathcal{F}}_p \iff\) they’re both equivalent to \((W, t)\) where \(W \subseteq U_1 \cap U_2\) and \begin{align*} {\mathcal{F}}_1 \mathrel{\Big|}^{U_1}_W (s_1) = t = {\mathcal{F}}_1\mathrel{\Big|}^{U_2}_W (s_2) .\end{align*}
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Then \(s_p \coloneqq{\mathcal{F}}_1 \mathrel{\Big|}^X_p(s) = 0 \sim (W, 0) \in ({\mathcal{F}}_1)_p\) means there exists some \(W_p\) and a lift \(\tilde s(p) = 0 \in {\mathcal{F}}_1(W_p)\) with \({\mathcal{F}}_1\mathrel{\Big|}^{W_p}_p(\tilde s(p)) = s_p\).
-
But this holds for all \(p\), and \(\left\{{W_p}\right\}_{p\in X} \rightrightarrows X\), so by the sheaf gluing axiom for \({\mathcal{F}}_1\), \(\left\{{ \tilde s(p) \in {\mathcal{F}}_1(W_p) {~\mathrel{\Big\vert}~}p\in X}\right\}\) glue to a unique \(\tilde s\in {\mathcal{F}}_1(X)\), and by uniqueness, \(\tilde s = s = 0 \in {\mathcal{F}}_1(X)\).
Exactness at \(g_X\):
-
We want to show \(\ker g_X = \operatorname{im}f_X\). First show \(\operatorname{im}f_X \subseteq \ker g_X\), and let \(s \in \operatorname{im}f_X \subseteq {\mathcal{F}}_2(X)\).
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A small diagram chase:
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Push \(s\) into \(({\mathcal{F}}_2)_p\) and pull back to \(f_X^{-1}(s) \in {\mathcal{F}}_1(X)\), by commutativity, the former is in \(\operatorname{im}f_p\), so \begin{align*} {\mathcal{F}}_1\mathrel{\Big|}^U_p(s) \in \operatorname{im}f_p = \ker g_p .\end{align*}
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Then push \(s \xrightarrow{g_X} \ell\), so \({\mathcal{F}}_3 \mathrel{\Big|}^U_p(\ell) = 0\) by commutativity. Since this is true for all stalks at all \(p\), \(\ell = 0\in {\mathcal{F}}_3(X)\), so \(\ell \in \ker g_X\).
A counterexample:
The exponential SES, assembled from groups:
Here \(\mathop{\mathrm{Hol}}_X({-})^{\times}\) denotes the multiplicatively invertible functions, i.e. nonvanishing functions.
But if the bottom sequence were exact, then every invertible holomorphic function would have a logarithm on all of \({\mathbb{C}}^{\times}\), but for example the identity function does not.
Let \({\mathcal{F}}\in {\mathsf{Sh}}(X)\) and \(s\in {\mathcal{F}}(U)\) be a section, and define \begin{align*} \mathop{\mathrm{supp}}s &\coloneqq\left\{{p\in U {~\mathrel{\Big\vert}~}s_p \neq 0}\right\} \subseteq U \\ \mathop{\mathrm{supp}}{\mathcal{F}}&\coloneqq\left\{{p\in X{~\mathrel{\Big\vert}~}{\mathcal{F}}_p\neq 0}\right\} \subseteq X ,\end{align*} where \(s_p\) denotes the germ of \(s\) in the stalk \({\mathcal{F}}_p\). Show that \(\mathop{\mathrm{supp}}s\) is closed in \(U\) but \(\mathop{\mathrm{supp}}{\mathcal{F}}\) need not be closed in \(X\).
\(\mathop{\mathrm{supp}}(s)\) is closed:
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Write \begin{align*} \mathop{\mathrm{supp}}(s) \coloneqq\left\{{p\in U {~\mathrel{\Big\vert}~}{\mathcal{F}}\mathrel{\Big|}^U_p(s) \neq 0}\right\} \subseteq U \\ \implies \mathop{\mathrm{supp}}(s)^c \coloneqq U\setminus\mathop{\mathrm{supp}}(s) \coloneqq\left\{{p\in U {~\mathrel{\Big\vert}~}{\mathcal{F}}\mathrel{\Big|}^U_p(s) = 0}\right\} \subseteq U \\ .\end{align*}
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Now use that if \(p\in U\) with \(s=0\) in the stalk at \(p\), then \(s=0\) in an open neighborhood \(W_p\) of \(p\) with \(W_p \subseteq U\setminus\mathop{\mathrm{supp}}(s)\), so every such \(p\) is interior.
\(\mathop{\mathrm{supp}}({\mathcal{F}})\) is not closed:
- Take the skyscraper sheaf: take the constant sheaf on a point \(q\in X\), then push it forward along the inclusion of \(q\):
-
Then check \begin{align*} q^* \underline{A}(U) = \begin{cases} A & q\in U \\ 0 & \text{else}. \end{cases} ,\quad (q^* \underline{A})_p = \begin{cases} A & p=q \\ 0 & \text{else}. \end{cases} \end{align*}
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Now invert this construction by taking the sheaf \begin{align*} {\mathcal{F}}\coloneqq \left( \not q^* \underline{A} \right)^{\scriptscriptstyle \mathrm{sh}} : \left( U \mapsto \begin{cases} A & q\not\in U \\ 0 & q\in U. \end{cases} \right)^{\scriptscriptstyle \mathrm{sh}} \end{align*}
-
\(\not q^*\underline{A}\) and \({\mathcal{F}}\) have the same stalks, and a computation shows \begin{align*} (\not q^*\underline{A})_p = \begin{cases} A & p\neq q \\ 0 & p=q. \end{cases} .\end{align*}
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Then for a fixed \(q\in X\), we have \(\mathop{\mathrm{supp}}{\mathcal{F}}= X \setminus{ \operatorname{cl}} _X(\left\{{ q}\right\})\).
- Why: if there is some neighborhood \(U\ni p\) that doesn’t meet \(q\), then \({\mathcal{F}}\mathrel{\Big|}^X_U(V) = A\) for every \(V \subseteq U\), so the colimit stabilizes and is equal to \(A\).
- Conversely, if every neighborhood of \(p\) meets \(q\), then \({\mathcal{F}}\mathrel{\Big|}^X_U(V) = 0\) for every \(V \subseteq U\) (including \(U\)), so the colimit stabilizes to zero.
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Now concretely take $X \coloneqq{\mathbb{A}}^1_{/ {k}} $ and \(q=0\), then \(\left\{{0}\right\} = V(x)\) for \(x\in k[x]\) is closed, so \({ \operatorname{cl}} _X(\left\{{0}\right\}) = \left\{{0}\right\}\).
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Thus \begin{align*} \mathop{\mathrm{supp}}{\mathcal{F}}= {\mathbb{A}}^1 \setminus{ \operatorname{cl}} _X(\left\{{0}\right\}) = {\mathbb{A}}^1\setminus\left\{{0}\right\}= D(x) \end{align*} is open and not closed when \(k\) is infinite.
Let \(X\in {\mathsf{Top}}, A\in {\mathsf{Ab}}{\mathsf{Grp}}, p\in X\) and define the skyscraper sheaf as \begin{align*} \iota_p(A)(U) \coloneqq \begin{cases} A & p\in U \\ 0 & \text{else}. \end{cases} \end{align*} Show that the stalk \(\iota_p(A)_q = A\) when \(q\in { \operatorname{cl}} _X(\left\{{p}\right\})\) and 0 otherwise, and that there is an equality of sheaves \(\iota_p(A) = \iota_*(\underline{A})\) where \(\iota: { \operatorname{cl}} _X(\left\{{p}\right\}) \hookrightarrow X\) is the inclusion.
Computation of stalks: see previous problem.
That \(\iota_p A \coloneqq(U\mapsto A \chi_{p\in U})\) is equal to \(\iota_* \underline{A}\) the pushforward sheaf:
-
Note that \(\underline{A}\) on \(\left\{{p}\right\}\) is given by ` \begin{align*} \underline{A}(U) \coloneqq {\mathsf{Top}}(U, a)
\begin{cases} A & U = \left{{p}\right}
\ 0 & U = \emptyset. \end{cases} .\end{align*} `{=html} -
Now check \begin{align*} \iota_* \underline{A}(U) \coloneqq\underline{A} (\iota^{-1}(U)) &= \begin{cases} A & \iota^{-1}(U) = p \\ 0 & \iota^{-1}(U) = \emptyset. \end{cases} \\ &= \begin{cases} A & U \ni \iota(p) = p \\ 0 & U\not\ni \iota(p) = p \end{cases} \end{align*}
-
Now take the identity maps as the components of a morphism \(\iota_p A\to \iota_* \underline{A}\), which induces the identity on stalks, making these sheaves equal.
II.2
Let \(A\in \mathsf{Ring}\) and \(X\coloneqq\operatorname{Spec}(A)\), and for \(f\in A\) let \(D(f) \coloneqq V(\left\langle{f}\right\rangle)^c\). Show that there is an isomorphism of ringed spaces \begin{align*} (D(f), { \left.{{{\mathcal{O}}_X}} \right|_{{D(f)}} }) \xrightarrow{\sim} \operatorname{Spec}(A_f) .\end{align*}
\envlist
- Take \(\iota: A\to A_f\) and induced maps \(\iota^*: \operatorname{Spec}A_f \to \operatorname{Spec}A\).
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Use \(\operatorname{Spec}A \left[ { \scriptstyle { {S}^{-1}} } \right] \cong \left\{{p\in \operatorname{Spec}A {~\mathrel{\Big\vert}~}p \cap S = \emptyset }\right\}\)
- So \(\operatorname{Spec}A_f = \left\{{p\in \operatorname{Spec}A {~\mathrel{\Big\vert}~}p \not\supseteq\left\langle{f}\right\rangle}\right\}\).
- Construct \(\psi \coloneqq\iota^*\), check \(D(g/f^k) \xrightarrow{\psi} D(gf)\) and \(D(g) \xrightarrow{\psi^{-1}} D(g/1)\).
- Use \({\mathcal{O}}_{\operatorname{Spec}A}{ \left.{{}} \right|_{{D(f)}} }(D(g)) = (A_f)_g\) and define \(\psi^# = \operatorname{id}\).
Recall: for \(I{~\trianglelefteq~}A\) any ideal, \begin{align*} V(I) = \left\{{p\in \operatorname{Spec}A {~\mathrel{\Big\vert}~}p\supseteq I}\right\} \\ D(I) \coloneqq V(I)^c = \left\{{p\in \operatorname{Spec}A {~\mathrel{\Big\vert}~}p\not\supseteq I}\right\} ,\end{align*} and there is a correspondence \begin{align*} i: A&\mapsto A_f \\ a &\mapsto {\left[ {a\over 1} \right]} \\ \\ \adjunction{i_*}{i^*} { \left\{{p\in \operatorname{Spec}A {~\mathrel{\Big\vert}~}p \cap\left\{{f^n}\right\} = \emptyset}\right\}} {\operatorname{Spec}A_f} \\ p \mapsto \left\langle{i(p)}\right\rangle = \left\{{p'/s {~\mathrel{\Big\vert}~}p'\in p, s\in \left\{{f^n}\right\} }\right\}\\ i^{-1}(q) \mapsfrom q = \left\{{g/f^n}\right\}{~\trianglelefteq~}A_f ,\end{align*} i.e. prime ideals of \(A_S\) are prime ideals of \(A\) not meeting \(S\).
Let \(Y\coloneqq\operatorname{Spec}A \left[ { \scriptstyle { {f}^{-1}} } \right]\). Need \begin{align*} f&\in {\mathsf{Top}}(D(f), Y)\\ f^# &\in \mathop{\mathrm{Mor}}_{{\mathsf{Sh}}_{/ {X}} }({\mathcal{O}}_{Y}, f^* { \left.{{{\mathcal{O}}_X}} \right|_{{D(f)}} }) .\end{align*}
-
Use the commutative algebra fact: primes of localizations lift to primes not intersecting the local set. Let \(i: A\to A_f\) be the ring morphism \(a\mapsto a/1\), this induces \begin{align*} \operatorname{Spec}A \left[ { \scriptstyle { {f}^{-1}} } \right] &\xrightarrow{i^*} \left\{{p\in \operatorname{Spec}A{~\mathrel{\Big\vert}~}p \cap\left\{{f^n}\right\}_{n\geq 1} = \emptyset }\right\} \\ &= \left\{{p\in \operatorname{Spec}A {~\mathrel{\Big\vert}~}p \not\supseteq\left\langle{f}\right\rangle}\right\} \\ &\coloneqq D(\left\langle{f}\right\rangle) = D(f) .\end{align*}
Here \(i^*(q) = i^{-1}(q)\) is \(i_*(p) = \left\langle{i(p)}\right\rangle\).
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So take \(\psi: \operatorname{Spec}A_f \to D(f)\) to be \(i^*\), which is a bijection.
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Check this is a homeomorphism: it’s an open map, since \begin{align*} D(g/f^k ) &\xrightarrow{\psi} D(gf) \\ D(g) &\xrightarrow{\psi^{-1}} D(g/1) ,\end{align*}
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Then \(\psi^#\coloneqq\operatorname{id}\) induces an isomorphism of sheaves: check that on distinguished opens,
\begin{align*} D(g) \subseteq Y \implies \psi^* { \left.{{{\mathcal{O}}_X}} \right|_{{D(f)}} }(D(g)) &\coloneqq { \left.{{{\mathcal{O}}_X}} \right|_{{D(f)}} }(\psi^{-1}( D(g)) ) \\ &= { \left.{{{\mathcal{O}}_X}} \right|_{{D(f)}} }( D(g/1) ) \\ &= (A_f)_{g/1} ,\end{align*} using that \({\mathcal{O}}_X(D(h)) = k[X]_{h}\), so the coordinate ring of \(D(f)\) is \(A_f\).
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Similarly, \begin{align*} {\mathcal{O}}_{\operatorname{Spec}A_f}(D(g)) = (A_f)_g ,\end{align*} and these are equal.
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Now for any \(U \subseteq \operatorname{Spec}A_f\), taken an open cover by distinguished opens \(D(g_k)\rightrightarrows U\); then the sections of each sheaf agree on each \(D(g_k)\) and using the sheaf axioms they glue to agreeing sections on \(U\), so this induces an isomorphism of sheaves.
Note that \((X, {\mathcal{O}}_X)\in {\mathsf{Sch}}\) is reduced iff \({\mathcal{O}}_X(U)\) has no nilpotents, and for \(A\in \mathsf{Ring}\) define \(A^{{ \text{red} }}\coloneqq A/\sqrt{0}\) to be \(A\) modulo its ideal of nilpotents.
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Show that \(X\) is reduced iff for every \(p\in X\), the local ring \({\mathcal{O}}_{X, p}\) has no nilpotents.
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Let \({\mathcal{O}}_X^{{ \text{red} }}\) be the sheafification of \(U \mapsto {\mathcal{O}}_X(U)^{ \text{red} }\). Show that \(X_{ \text{red} }\coloneqq(X, {\mathcal{O}}_X^{ \text{red} })\) is a scheme, and there is a morphism of schemes \(X_{ \text{red} }\xrightarrow{{ \text{red} }} X\) which induces a homeomorphism \({\left\lvert {X_{ \text{red} }} \right\rvert}\to {\left\lvert {X} \right\rvert}\) on underlying topological spaces.
- Let \(X \xrightarrow{f} Y\in {\mathsf{Sch}}\) with \(X\) reduced. Show that there is a unique morphism \(X \xrightarrow{g} Y_{ \text{red} }\) such that \(f\) is the composition:
Part a:
- Zero in every stalk implies zero by the sheaf axiom.
Part b:
- Cover by affines.
- \({\sqrt{0_{R}} } \leq {\mathfrak{p}}\) for every \({\mathfrak{p}}\in \operatorname{Spec}R\).
- \(R_{ \text{red} }\coloneqq R/{\sqrt{0_{R}} }\) is a quotient, and localization commutes with quotients.
- Maps \(R\to S\) with \(S\) reduced factor through \(R_{ \text{red} }\).
- Sheafification has the same stalks, and isomorphism on stalks iff isomorphism of sheaves.
- Pushforwards of reduces sheaves are reduced.
\(\implies\): If \({\mathcal{O}}_{X, p}\) has nilpotents, pick \(s\) with \(s^n\in 0 \in {\mathcal{O}}_{X, p}\). This lifts to some \(s^n = 0 \in {\mathcal{O}}_X(U)\), so \(s\) is nilpotent in \({\mathcal{O}}_X(U)\), a contradiction.
\(\impliedby\): If \(s\in {\mathcal{O}}_X(U)\) has nilpotents, then \({\mathcal{O}}_X\mathrel{\Big|}^X_p(s^n) = 0\) in the stalk, making \(s\) nilpotent in the stalk.
Describe \(\operatorname{Spec}{\mathbb{Z}}\) and show it is terminal in \({\mathsf{Sch}}\), i.e. each \(X\in {\mathsf{Sch}}\) admits a unique morphism \(X\to \operatorname{Spec}{\mathbb{Z}}\).
\envlist
- An adjunction inducing an equivalence: \begin{align*} \adjunction{{{\Gamma}\qty{{-}} }}{\operatorname{Spec}({-})}{{\mathsf{Sch}}^{\operatorname{op}}}{\mathsf{CRing}}\hspace{6em} \\ \\ \mathop{\mathrm{Mor}}_{{{\mathsf{Sch}}}}(X, \text{Spec}(R)) \cong \mathop{\mathrm{Mor}}_{{\mathsf{CRing}}}(R, \Gamma(X;\mathcal{O}_X)) .\end{align*}
Let \(X\in {\mathsf{Sch}}\) and for \(x\in X\) let \({\mathcal{O}}_x\) be the local ring at \(x\) and \({\mathfrak{m}}_x\) its maximal ideal. Let \(\kappa(x) \coloneqq{\mathcal{O}}_x/{\mathfrak{m}}_x\) be the residue field at \(x\).
Then for \(k\) any field, show that giving a morphism \(\operatorname{Spec}(k) \to X \in {\mathsf{Sch}}\) is equivalent to giving a point \(x\in X\) and an inclusion \(\kappa(x) \hookrightarrow k\).
\begin{align*} x^2 - y^q = 1 && x^p - y^2 = 1 .\end{align*} \begin{align*} x^2 - y^q = 1 && x^p - y^2 = 1 .\end{align*}
?
- The image presheaf isn’t necessarily a sheaf.
- Filtered direct limits are exact.
- Finite limits commute with filtered colimits in most categories
- Right adjoints preserve colimits
- \(\mathop{\mathrm{Hom}}({-}, {-}): \mathsf{C}^{\operatorname{op}}\times \mathsf{C}\to \mathsf{C}\) is continuous (for regular limits, noting limits in \(\mathsf{C}^{\operatorname{op}}\) are colimits in \(\mathsf{C}\)).
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Right adjoints preserve limits, left adjoints preserve colimits. Sheafification is left adjoint to the forgetful functor, so right exact, so preserves colimits
- So exact in presheaves implies exact in sheaves, but exact in shaves only implies left-exact in presheaves. Sheaf cohomology measures how much the presheaf fails exactness.