## Problem Set 2 ## II.1 ::: {.exercise .proofenv .proofenv .proofenv .proofenv .proofenv title="II.1.8"} 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*} `{=html} is left-exact, but need not be exact. ::: ::: {.solution .proofenv .proofenv .proofenv .proofenv .proofenv} 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*} `{=html} which (e.g. ) is data of the form `

`{=html} > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMiwwLCJcXG1jZl8xKFUpIl0sWzAsMCwiVSJdLFswLDIsIlYiXSxbMiwyLCJcXG1jZl8xKFYpIl0sWzQsMiwiXFxtY2ZfMihWKSJdLFs0LDAsIlxcbWNmXzIoVSkiXSxbMiwxLCJcXGlvdGFfe1VWfSIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzAsMywiXFxSZXNfMShVLCBWKSJdLFs1LDQsIlxcUmVzXzIoVSwgVikiXSxbMCw1LCJmX1YiXSxbMyw0LCJmX1UiLDJdLFs2LDcsImYiLDAseyJzaG9ydGVuIjp7InNvdXJjZSI6MjAsInRhcmdldCI6MjB9fV1d) 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*} `{=html} **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*} `{=html} - 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*} `{=html} - 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 $p\in X$: `

`{=html} > [Link to Diagram](https://q.uiver.app/?q=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) - 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*} `{=html} - 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*} `{=html} which holds for all $p$. - 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*} `{=html} - 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 $s \in \operatorname{im}f_X \subseteq {\mathcal{F}}_2(X)$. - A small diagram chase: `

`{=html} > [Link to Diagram](https://q.uiver.app/?q=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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*} `{=html} - 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: `

`{=html} > [Link to Diagram](https://q.uiver.app/?q=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) 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. ::: ::: {.exercise .proofenv .proofenv .proofenv .proofenv .proofenv title="II.1.14"} 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*} `{=html} 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 $X$. ::: ::: {.solution .proofenv .proofenv .proofenv .proofenv .proofenv} **$\mathop{\mathrm{supp}}(s)$ is closed**: - 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*} `{=html} - 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 $q$: `

`{=html} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwyLCJxIl0sWzIsMiwiWCJdLFswLDAsIlxcdWx7QX0iXSxbMiwwLCJxXiogXFx1bHtBfSJdLFswLDEsInEiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFsyLDMsInFeKiJdXQ==) - 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*} `{=html} - 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*} `{=html} - $\not q^*\underline{A}$ 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*} `{=html} - 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. - Now concretely take \$X `\coloneqq{\mathbb{A}}`{=tex}\^1\_{/ {k}} \$ and $q=0$, then $\left\{{0}\right\} = V(x)$ for $x\in k[x]$ is closed, so ${ \operatorname{cl}} _X(\left\{{0}\right\}) = \left\{{0}\right\}$. - 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*} `{=html} is open and not closed when $k$ is infinite. ::: ::: {.exercise .proofenv .proofenv .proofenv .proofenv .proofenv title="II.1.17"} 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*} `{=html} 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. ::: ::: {.solution .proofenv .proofenv .proofenv .proofenv .proofenv} 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*} `{=html} - 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 ::: {.exercise .proofenv .proofenv .proofenv .proofenv .proofenv title="II.2.1"} 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*} `{=html} ::: ::: {.strategy .proofenv .proofenv .proofenv .proofenv .proofenv} ```{=tex} \envlist ``` - Take $\iota: A\to A_f$ and induced maps $\iota^*: \operatorname{Spec}A_f \to \operatorname{Spec}A$. - 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 $\psi^# = \operatorname{id}$. ::: ::: {.solution .proofenv .proofenv .proofenv .proofenv .proofenv} 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*} `{=html} 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*} `{=html} 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*} `{=html} - 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*} `{=html} Here $i^*(q) = i^{-1}(q)$ is $i_*(p) = \left\langle{i(p)}\right\rangle$. - So take $\psi: \operatorname{Spec}A_f \to D(f)$ to be $i^*$, which is a bijection. - 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*} `{=html} - 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*} `{=html} using that ${\mathcal{O}}_X(D(h)) = k[X]_{h}$, so the coordinate ring of $D(f)$ is $A_f$. - Similarly, ` \begin{align*} {\mathcal{O}}_{\operatorname{Spec}A_f}(D(g)) = (A_f)_g ,\end{align*} `{=html} and these are equal. - 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. ::: ::: {.exercise .proofenv .proofenv .proofenv .proofenv .proofenv title="II.2.3"} 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. a. Show that $X$ is reduced iff for every $p\in X$, the local ring ${\mathcal{O}}_{X, p}$ has no nilpotents. b. 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. c. 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: `

`{=html} > [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJYIl0sWzIsMCwiWSJdLFsyLDIsIllfXFxyZWQiXSxbMCwxLCJmIl0sWzIsMSwiXFxyZWQiLDJdLFswLDIsIlxcZXhpc3RzICFnIiwyXV0=) ::: ::: {.strategy .proofenv .proofenv .proofenv .proofenv .proofenv} 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. ::: ::: {.solution .proofenv .proofenv .proofenv .proofenv .proofenv title="1"} $\implies$: 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. ::: ::: {.solution .proofenv .proofenv .proofenv .proofenv .proofenv title="2"} ::: ::: {.exercise .proofenv .proofenv .proofenv .proofenv .proofenv title="II.2.5"} Describe $\operatorname{Spec}{\mathbb{Z}}$ and show it is terminal in ${\mathsf{Sch}}$, i.e. each $X\in {\mathsf{Sch}}$ admits a unique morphism $X\to \operatorname{Spec}{\mathbb{Z}}$. ::: ::: {.strategy .proofenv .proofenv .proofenv .proofenv .proofenv} ```{=tex} \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*} `{=html} ::: ::: {.exercise .proofenv .proofenv .proofenv .proofenv .proofenv title="II.2.7"} 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*} `{=html} ` \begin{align*} x^2 - y^q = 1 && x^p - y^2 = 1 .\end{align*} `{=html} ::: ::: {.strategy .proofenv .proofenv .proofenv .proofenv .proofenv} ? ::: ::: {.warnings .proofenv .proofenv .proofenv .proofenv .proofenv} - 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 $\mathsf{C}$). - 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. :::