\documentclass[12pt]{article}1\author{William A. Stein}2\title{Algebraic Geometry Homework}34\font\grmn=eufm10 scaled \magstep 156\newcommand{\gp}{p}7\newcommand{\ga}{a}8\newcommand{\so}{\mathcal{O}}910\newtheorem{prob}{Problem}11\newtheorem{theorem}{Theorem}1213\newcommand{\ov}{\overline{\varphi}}14\renewcommand{\phi}{\varphi}15\newcommand{\fh}{f^{\#}}16\newcommand{\proj}{Proj \hspace{.01in}}17\newcommand{\spec}{Spec \hspace{.01in}}18\newcommand{\proof}{\mbox{\sc Proof.\hspace{.1in}}}192021\begin{document}22\maketitle2324\section{Solutions}2526\begin{prob}[II.2.14(a)]27Let $S$ be a graded ring. Show that $\proj S=\emptyset$28iff every element of $S_{+}$ is nilpotent.29\end{prob}30\proof This is equivalent to showing that the nilradical of $S$ is equal31to the intersection of all homogenous primes of $S$. Indeed, if every32element of $S_{+}$ is nilpotent then every homogenous prime contains33$S_{+}$ so $\proj S=\emptyset$. Conversely, if $\proj S=\emptyset$, then34every homogenous prime contains $S_{+}$ so the nilradical of $S$ contains35$S_{+}$ so that every element of $S_{+}$ is nilpotent.3637It remains to show that the intersection of all homogenous primes of $S$38is the nilradical of $S$. Using Zorn's lemma we can show that if $I$ is39a proper homogenous ideal then there is at least one maximal homogenous40ideal containing $I$. (The proof preceeds just as on page 2 of Matsumura41except one notes that the union of a chain of {\em homogenous} ideals42is homogeneous. If $I_1 \subset I_2 \subset \cdots$ is a chain of43homogeneous ideals, then $\cup_{n=1}^{\infty} I_n$ is an ideal and44if $x\in\cup_{n=1}^{\infty} I_n$ then $x\in I_n$ for some $n$, so the45homogenous components of $x$ are in $I_n$, so they are in46$\cup_{n=1}^{\infty} I_n$, so $\cup_{n=1}^{\infty} I_n$ is47homogeneous.)4849\begin{theorem}50Let $T$ be a multiplicative set and $I$ a homogeneous ideal disjoint51from $T$; then there exists a homogeneous prime ideal containing52$I$ and disjoint from $T$.53\end{theorem}54\proof55Using Zorn's lemma we see that the set of homogeneous ideals disjoint56from $T$ and containing $I$ contains a maximal element, say $P$. Then57$P$ is prime. For if $x\notin P$, $y\notin P$ are homogenous, then58$P+(x)$ and $P+(y)$ are both homogenous and so they meet $T$, so their59product also meets $T$. However,60$$(P+(x))(P+(y))\subset P+(xy)$$61so $xy\notin P$ since $P$ does not meet $T$.6263\begin{theorem}64The nilradical of $S$ is the intersection of the homogeneous primes65of $S$.66\end{theorem}67\proof68Suppose $x$ is in the nilradical of $S$ so that $x$ is nilpotent, say69$x^n=0$. If $I$ is a homogenous prime then $x^n=0\in I$ so, by induction,70$x\in I$. Conversely, suppose $x$ is not nilpotent. Then71$T=\{1,x,x^2,\ldots\}$ is a multiplicative set disjoint from $(0)$. So,72by the above theorem, there is a homogeneous prime ideal containing73$(0)$ disjoint from $T$. Thus $x$ is not in the intersection of all74homogeneous prime ideals of $S$.7576\begin{prob}[II.2.14(b)] Let $\varphi:S\rightarrow T$ be a graded homomorphism77of graded rings. Let78$U=\{\gp \in \proj T : \gp \not\supseteq \varphi(S_{+})\}.$79Show that $U$ is an open subset of $\proj T$, and show that80$\varphi$ determines a natural morphism $f:U\rightarrow\proj S$.81\end{prob}82\proof83$U$ is open because84$\proj T - U = \{\gp \in \proj T:\gp \supseteq \varphi(S_{+})\}85=\{\gp \in \proj T : \gp \supseteq T\varphi(S_{+})\}86=V(T\varphi(S_{+}))$87and the ideal $T\varphi(S_{+})$ is homogenous because it is generated88by the homogeneous elements $\{\varphi(f):f\in S_{+} \mbox{ is homogeneous}\}$.8990The natural morphism $f:U-\proj S$ is defined as follows. As a map on91topological spaces we specify that, for $x\in U$, $f(x)=\varphi^{-1}(x)$.92Because of the way $U$ was chosen and since $\varphi$ is homogeneous93$f$ maps $U$ into $\proj S$, so $f$ is well-defined. If $V(\ga)$ is94a closed subset of $\proj S$ with $\ga$ homogeneous, then95$f^{-1}(V(\ga))=\{\gp\in U:\varphi^{-1}(\gp)\supseteq \ga\}96=\{\gp\in U:\gp \supseteq\varphi(\ga)\}=V(T\varphi(\ga)).$97As above, $T\varphi(\ga)$ is homogeneous so $f^{-1}(V(\ga))$ is98closed so $f$ is continuous.99100To define the associated map $f^{\#}:\so_{\proj S} \rightarrow101f_{*}\so_{U}$ of sheaves let $V\subset \proj S$ be open. Then an102element of $\so_{\proj S}(V)$ is a map103$s:V\rightarrow\bigsqcup_{p \in V}S_{(p)}$ such that $s$ is104locally a quotient of elements of $S$. We specify that105$f^{\#}(s)=\varphi \circ s\circ f:f^{-1}(V)\rightarrow106\bigsqcup_{p\in f^{-1}(V)} T_{(p)}$. To see that107$f^{\#}(s)\in f_{*}\so(V)$ we must check that it is locally108a quotient. So suppose $\gp\in f^{-1}(V).$ Let $W\subset V$ be109an open neighborhood of $f(\gp)$ on which $s$ is represented110as a quotient. Then $f^{\#}(s)$ is represented as a quotient on111the open set $f^{-1}(W)$, as required. Since $f^{\#}$ respects112the restriction maps we see that $f$ is a morphism.113114\begin{prob}[II.2.14(c)]115$f$ can be an isomorphism even when $\varphi$ is not. For example,116suppose that $\varphi_d:S_d\rightarrow T_d$ is an isomorphism117for all $d\geq d_0$. Show that $U=\proj T$ and the morphism118$f:\proj T\rightarrow \proj S$ is an isomorphism.119\end{prob}120\proof121To see that $U=\proj T$ note that if $\gp \in \proj T$ but122$\gp \notin U$ then $\gp \supseteq \varphi(S_{+})$. In particular,123$\gp \supseteq \bigoplus_{d\geq d_0} T_d$, so124$\gp \supseteq (T_{+})^d$ so, since $\gp$ is prime,125$\gp \supseteq T_{+}$, a contradiction.126127128Let $\{g_{\alpha}\}$ be a set of generators of $T_{+}$. Then129$\cup_{\alpha}D_T(g_{\alpha})=130\cup_{\alpha}\{x\in\proj T : g_{\alpha} \notin x\}=\proj T$131since every prime in $\proj T$ must omit some $g_\alpha$.132Since $g_{\alpha}\notin{}x$ iff $g_{\alpha}^{d_0}\notin{}x$ for133$x$ prime, we may replace the $g_{\alpha}$ by elements of134$T_{\geq d_0}$ and still have a cover of $\proj T$ by distinguished135open sets.136137Our strategy is as follows. We first show that138$f|_{D_T(g_{\alpha})}:D_T(g_{\alpha})\rightarrow{}D_S(\varphi^{-1}139(g_{\alpha})$ is an isomorphism for each $\alpha$ and then show that140the open sets $D_S(\varphi^{-1}(g_{\alpha}))$ cover $\proj S$.141Then showing that $f$ is injective completes the proof.142143Let $g=g_{\alpha}$ be one of our $g_{\alpha}$.144By Proposition 2.5, $D_T(g)\cong\spec T_{(g)}$145and so $f'=f|_{D_T(g)}$ is a morphism of affine schemes146$f':\spec T_g\rightarrow\spec S_{(\varphi^{-1}(g))}$.147This map is induced by $\ov:S_{(\varphi^{-1}(g))}148\rightarrow{}T_{(g)}$ where $\ov$ is149the localization of the ring homomorphism150$\varphi:S\rightarrow T$.151So we just need to verify that $\ov$ is an isomorphism.152Suppose $\ov(a/b)=0$. Then $\ov(a\phi^{-1}(g)/b\phi^{-1}(g))=\ov(a/b)=0$153so $\phi(a\phi^{-1}(g))/\phi(b\phi^{-1}(g))=0$ in $T_{(g)}$ so154there is $n$ such that $g^n\phi(a\phi^{-1}(g))=0$ in $T$ so155$\phi(a\phi^{-1}(g^{n+1})=0$. Thus $a\phi^{-1}(g)^{n+1}=0$ since156$\phi$ is an isomorphism in high enough degree. Thus157$a=0$ in $S_{(\phi^{-1}(g))}$, so $a/b=0$ in $S_{(\phi^{-1}(g))}$.158This shows that $\ov$ is injective. To see that $\ov$ is surjective159let $a/g^n\in T_{(g)}$. Then $\phi^{-1}(ag)/\phi^{-1}(g^{n+1})$ is160a well-defined element of $S_{(\phi^{-1}(g))}$ and161$\ov(\phi^{-1}(ag)/\phi^{-1}(g^{n+1}))=ag/g^{n+1}=a/g^n$,162which shows that $\ov$ is surjective.163164Next we verify that $\cup_{g_{\alpha}}D_S(\phi^{-1}(g_{\alpha})=\proj S$.165Suppose $x\notin D_S(\phi^{-1}(g_{\alpha})$ for all $\alpha$. Then166$\phi^{-1}(g_{\alpha})\in x$ for each $\alpha$, so, since we may assume167that the $g_{\alpha}$ generate $T_{\geq d_0}$, and $x$ is prime,168$\phi^{-1}(T_{\geq d_0}) \subseteq x$, a contradiction since169$S_{+}\not\subseteq x$.170171Next we show that the induced map $f:\proj T \rightarrow \proj S$172is injective. Let $p, q \in \proj T$ and suppose $f(p)=f(q)$.173Then $\varphi^{-1}(p)=\varphi^{-1}(q)$ so, since $\varphi$ is an174isomorphism, for $d\geq d_0$ we see that $p\cap T_d=q\cap T_d$.175So if $a\in p$ is homogeneous then $a^n \in p\cap T_d$ for176some $n$ and $d\geq d_0$. So $a^n\in q\cap T_d$ so $a^n\in q$177so $a\in q$. Likewise $a\in q$ implies $a\in p$. Thus178$p=q$ so $f$ is injective.179180Finally, no solution is complete without an actual example of a181map $\varphi:S\rightarrow{}T$ which satisfies the hypothesis182of the theorem. Let $T=k[x,y]$ and let $S=T_0 + T_2 + \cdots$ and183let $\varphi:S\hookrightarrow T$ be the inclusion map. Then184$\varphi$ is graded and an isomorphism for $d\geq 2$. But $\varphi$185is not an isomorphism.186187\begin{prob}[II.2.14(d)]188Let $V$ be a projective variety with homogeneous coordinate189ring $S$. Show that $t(V)\cong \proj S$.190\end{prob}191\proof192Define $f:t(V)\rightarrow\proj S$ by $f$ takes a point193$x\in t(V)$ to its homogeneous ideal $I(x) \in \proj S$.194By exercise I.2.4 $f$ is injective and surjective hence a195bijection. Furthermore, $x\supseteq{}y$ iff $f(x)\subseteq{}f(y)$196so $f$ and $f^{-1}$ send closed sets to closed sets hence197$f$ is a homeomorphism.198199To define $\fh$ let $U$ be an open subset of $\proj S$. Then we200must define $\fh$ so that (notation as in the proof of201proposition 2.6)202$\fh:\so_{\proj S}(U)\rightarrow f_{*}\so_{t(V)}(U)203=\so_{t(V)}(f^{-1}(U))=\so_V(\alpha^{-1}(f^{-1}(U)))$.204Let $s\in\proj_S(U)$. Then $s$ can locally be represented205in the form $g/h$ where $g,h \in S$ have the same degree and206$h$ is nonzero on the appropriate subset of $V$. Thus207$s$ naturally defines an element of $\so_V(\alpha^{-1}(f^{-1}(U)))$208and any element of $\so_V(\alpha^{-1}(f^{-1}(U)))$ defines an209element of $\so_{\proj S}(U).$ Thus $\fh$ is an isomorphism.210[I'm glossing over a lot of details!]211212\begin{prob}[II.2.16(a)]213Let $X$ be a scheme, let $f\in\Gamma(X,\so_X)$, and define $X_f$ to214be the subset of points $x\in{}X$ such that the stalk $f_x$ of $f$215at $x$ is not contained in the maximal ideal $m_x$ of the local216ring $\so_x$. (a) If $U=\spec B$ and217$\overline{f}\in{}B=\Gamma(U,\so_{X|U})$ is the restriction of $f$,218show that $U\cap{}X_f=D(\overline{f})$ so $X_f$ is an open subset219of $X$.220\end{prob}221\proof222Note that $D(\overline{f})=\{x\in U:\overline{f}\notin x\}223=\{x\in U:\overline{f}_x\notin m_x\}224=\{x\in U:f_x\notin m_x\}225=U\cap{}X_f$.226Thus $U\cap{}X_f$ is an open subset of $U$. Now let227$\{U_{\alpha}\}$ be an affine open over of $X$. Then228$X_f=\cup_{\alpha}(U_\alpha \cap X_f)$ is the union229of open sets, hence open.230231\begin{prob}[II.2.16(b)]232Assume that $X$ is quasi-compact. Let $A=\Gamma(X,\so_X)$, and let233$a\in A$ be an element whose restriction to $X_f$ is $0$. Show that234for some $n>0$, $f^na=0$.235\end{prob}236\proof237Using the fact that $X$ is quasi-compact we can find a finite cover238$\{U_i=\spec B_i\}_{i=1}^m$ of $X$ by affine open sets. Since $a|_{X_f}$239is $0$, the image of $a$ in $\so_{X_f\cap{}U_i}=(B_i)_{\overline{f}}$240is 0. Thus there exists $n_i$ such that $\overline{f}^{n_i}\overline{a}=0$241in $B_i$. That is, $f^{n_i}a|_{X_f\cap{}U_i}$ is $0$. Letting242$n=\mbox{max}\{n_1,\ldots,n_m\}$ we see that $f^n a|_{X_f\cap{}U_i}$ is $0$ for243each $i$ whence $f^n a=0$ in $A=\Gamma(X,\so_X)$.244245\begin{prob}[II.2.16(c)]246Assume $X$ has a finite cover by open affines $U_{i}$ such that247each intersection $U_i\cap{}U_j$ is quasi-compact. Let248$b\in\Gamma(X_f,\so_{X_f})$. Show that for some $n>0$,249$f^n b$ is the restriction of an element of $A$.250\end{prob}251\proof252Write $U_i=\spec B_i$. Then $b|_{U_i}\in{}(B_i)_f$ so,253from the definition of $(B_i)_f$, there exists $n_i$254such that $f^{n_i}b|_{U_i}\in{}B_i=\so_X(U_i)$.255Let $N=\mbox{max}\{n_i\}$ and let $g_i=f^{N}b|_{U_i}\in\so_X(U_i).$256Then $g_i|_{X_f\cap{}U_i\cap{}U_j}=257f^{N}b|_{U_i\cap{}U_j\cap{}X_f}=258g_j|_{X_f\cap{}U_i\cap{}U_j}$259so $(g_i-g_j)|_{(U_i\cap{}U_j)_f}=0$.260By part (b) since $U_i\cap{}U_j$ is quasi-compact there is an integer261$n_{ij}$ such that $f^{n_{ij}}(g_i-g_j)=0$ in262$\so_{U_i\cap{}U_j}$. Let $M=\mbox{max}\{n_{ij}\}$, and let263$h_i=f^{M}g_i.$ Then $h_i\in\so(U_i)$ for each $i$ and264$h_i|_{U_i\cap{}U_j}=h_j|_{U_i\cap{}U_j}$265so we can find $h\in\Gamma(X,\so_X)$ such that266$h|_{U_i}=h_i$. But, for each $i$,267$h|_{X_f\cap{}U_i}=f^{M}g_i|_{X_f\cap{}U_i}=f^{M}f^{N}b|_{X_f\cap{}U_i}268=f^{M+N}b|_{X_f\cap{}U_i}$ so by uniqueness (sheaf axiom iii),269$h|_{X_f}=f^{M+N}b$, as desired.270271\begin{prob}[II.2.16(d)]272With the hypothesis of (c), conclude that273$\Gamma(X_f,\so_{X_f})\cong{}A_f$.274\end{prob}275\proof276Define a homomorphism $\phi:A_f\rightarrow\Gamma(X_f,\so_{X_f})$ by277$\phi(a/f^n)=a|_{X_f}/f^n|_{X_f}$. This is well-defined because the278stalk $f_x$ is invertible in each local ring $\so_{x}$ for each279$x\in{}X_{f}$. $\phi$ is a homomorphism since restriction is280a homomorphism. Suppose $\phi(a/f^n)=0$, then281$a|_{X_f}/f|_{X_f}^n=0$ so $a|_{X_f}=0$. By282part (b) there exists $m$ such that $f^{m}a=0$, so283$a$ is $0$ in $A_f$, whence $\phi$ is injective.284Suppose $b\in\Gamma(X_f,\so_{X_f})$, then by285part (c) there exists $n$ such that $f^n{}b$ is the restriction286of some $a\in\Gamma(X,\so_X)$ to $X_f$. Then287$\phi(a/f^n)=a|_{X_f}/f|_{X_f}^n=f^n{}b/f^n|_{X_f}=b$ so288$\phi$ is surjective, which establishes the desired289isomorphism.290291\begin{prob}[II.2.17(a)]292A Criterion for Affineness.293Let $f:X\rightarrow{}Y$ be a morphism of schemes, and assume that294$Y$ can be covered by open sets $U_i$, such that for each $i$, the295induced map $f^{-1}(U_i)\rightarrow{}U_i$ is an isomorphism.296Then $f$ is an isomorphism.297\end{prob}298\proof299Define a morphism $g:Y\rightarrow X$ as follows. On each open set300$U_i$ let $g_i$ be the morphism inverse to301$f|_{f^{-1}(U_i)}:f^{-1}(U_i)\rightarrow{}U_i$.302Then303$g_i|_{U_i\cap{}U_j}=f|_{f^{-1}(U_i)\cap{}f^{-1}(U_j)}304=g_j|_{U_i\cap{}U_j}$305so, as in step 3 of the proof of Theorem 3.3, we can glue the morphisms306$g_i$ to obtain a morphism $g:Y\rightarrow X$ such that307$g|_{U_i}=g_i$. As a map of spaces $g$ is clearly inverse to $f$.308Since each $g_i$ is an isomorphism, the induced maps $g^{\#}$ on stalks309are isomorphisms so the induced map on sheaves is an isomorphism.310Thus $g$ is an isomorphism.311312313314315316\end{document}317318