Author: William A. Stein
Compute Environment: Ubuntu 18.04 (Deprecated)
1\documentclass[12pt]{article}
2\author{William A. Stein}
3\title{Algebraic Geometry Homework\\
4II.5, 1,2,3,4,5,7,8}
5
6\font\grmn=eufm10 scaled \magstep 1
7
8\renewcommand{\a}{\mbox{\bfseries A}}
9\newcommand{\gp}{p}
10\newcommand{\ga}{a}
11\newcommand{\so}{\mathcal{O}}
12\newcommand{\sox}{\mathcal{O}_X}
13\newcommand{\soy}{\mathcal{O}_Y}
14\newcommand{\se}{\mathcal{E}}
15\renewcommand{\sf}{\mathcal{F}}
16\newcommand{\sg}{\mathcal{G}}
17\newcommand{\shom}{\mathcal{H}\mbox{\rm om}}
18\renewcommand{\hom}{\mbox{\rm Hom}}
19\newcommand{\tensor}{\otimes}
20\newcommand{\tens}{\otimes}
21\newcommand{\isom}{\cong}
22\renewcommand{\dim}{\mbox{\rm dim}}
23\newcommand{\z}{\mbox{\bfseries Z}}
24\newcommand{\into}{\rightarrow}
25\newcommand{\res}{\mbox{\rm res}}
26\newcommand{\ru}{|_U}
27\newcommand{\ann}{\mbox{\rm Ann}}
28\newcommand{\coker}{\mbox{\rm Coker}}
29
30
31\newtheorem{prob}{Problem}
32\newtheorem{theorem}{Theorem}
33\newtheorem{prop}{Proposition}
34
35\newcommand{\ov}{\overline{\varphi}}
36\renewcommand{\phi}{\varphi}
37\newcommand{\fh}{f^{\#}}
38\newcommand{\proj}{Proj \hspace{.01in}}
39\newcommand{\spec}{Spec \hspace{.01in}}
40\newcommand{\proof}{\mbox{\sc Proof.\hspace{.1in}}}
41
42\begin{document}
43\maketitle
44
45\begin{prob}[II.5.1]
46Let $(X,\sox)$ be a ringed space, and let $\se$ be a locally
47free $\sox$-module of finite rank. We define the dual of
48$\se$, denoted by $\check{\se}$, to be the sheaf
49$\shom_{\sox}(\se,\sox)$.
50
51(a) Show that $(\check{\se})\check{}\cong\se$.
52
53(b) For any $\sox$-module $\sf$, $\shom_{\sox}(\se,\sf) 54\cong \check{\se} \tensor_{\sox} \sf$.
55
56(c) For any $\sox$-modules $\sf, \sg,$
57$\hom_{\sox}(\se\tensor\sf,\sg)\cong 58\hom_{\sox}(\sf,\shom_{\sox}(\se,\sg))$.
59
60(d) (Projection Formula). If $f:(X,\sox)\rightarrow(Y,\soy)$
61is a morphism of ringed spaces, if $\sf$ is an $\sox$-module,
62and if $\se$ is a locally free $\soy$-module of finite
63rank, then there is a natural isomorphism
64$f_{*}(\sf\tensor_{\sox}f^{*}\se)\cong 65f_{*}(\sf)\tensor_{\soy}\se$.
66
67
68\end{prob}
69\proof
70(a)
71A free module of finite rank is {\em canonically}
72isomorphic to its double-dual via
73$\check{m}(\lambda)=\lambda(m)$
74where $m\in M$, $\lambda\in \check{M}$, and
75$\check{m}\in (\check{M})\check{}$.
76Let $U$ be an open set on which $\se|U$ is a
77free $\sox$-module of finite rank. Define a
78map $\phi:\se|_U\into(\check{\se})\check{}|_U$
79by, for all $V\subseteq U$, $\se(V)\into 80(\check{\se})\check{}(V)$ is the isomorphism described
81above.
82Since the isomorphisms are canonical, we can patch on
83intersections and define a global isomorphism.
84
85(b) As above we may assume $\se$ is a free $\sox$-module.
86Let $e_1,\ldots,e_n$ be a basis for $\sf$ and let
87$e_1^{*},\ldots,e_n^{*}$ be the corresponding dual
88basis. Let $U$ be an open subset of $X$.
89Define $\phi_U:\shom_{\sox}(\se,\sf)|_U\into 90(\hom(\se,\sox) \tensor_{\sox}\sf)|_U$ by
91$f\mapsto\sum_{i=1}^{n}e_i^{*}\tens f(e_i).$
92Define $\psi_U: 93(\hom(\se,\sox) \tensor_{\sox}\sf)|_U \into 94\shom_{\sox}(\se,\sf)|_U$ by
95$f\tens a\mapsto(x\mapsto f(x)a)$.
96For convenience of notation write $\phi=\phi_U$ and
97$\psi=\psi_U$.
98Let $f\tens a\in (\hom(\se,\sox) \tensor_{\sox}\sf)|_U$.
99Then $\phi\circ\psi(f\tens a) 100=\phi(x\mapsto f(x)a)=\sum_{i=1}^{n} e_i^{*}\tens 101f(e_i)a=\sum_{i=1}^{n}e_i^{*}f(e_i)\tens a 102=f\tens a$. Let $f\in\shom_{\sox}(\se,\sf)$, then
103$\psi\circ\phi(f)=\psi(\sum_{i=1}^{n}e_i^{*}\tens f(e_i)) 104=(x\mapsto\sum_{i=1}^{n}e_i^{*}(x)f(e_i)) 105=(x\mapsto f(\sum_{i=1}^{n}e_i^{*}(x)e_i)) 106=(x\mapsto f(x))$.
107Thus $\phi$ and $\psi$ are inverse bijective
108homomorphisms, hence ring isomorphisms, and since they
109respect the restriction maps we see that the
110corresponding sheaves are isomorphic.
111
112(c) On each open set define $\phi|_U: 113\hom_{\sox}(\se\tensor\sf,\sg)\into 114\hom_{\sox}(\sf,\hom_{\sox}(\se,sg))$ by
115$f\mapsto(a\mapsto(e\mapsto f(e\tens a)))$.
116(For notational convenience we omit the sheaf restrictions.)
117If $\phi(f)=0$ then the map
118$(a\mapsto(e\mapsto f(e\tens a)))$ is $0$
119so $f$ is the zero map, hence $\phi$ is injective.
120Let $f\in \hom_{\sox}(\sf,\hom_{\sox}(\se,sg))$. Define
121$g\in\hom_{\sox}(\se\tensor\sf,\sg)$ by
122$g(a\tens b)=(f(b))(a)$.
123Then $\phi(g)=(a\mapsto (e\mapsto g(e\tens a))) 124=(a\mapsto(e\mapsto(f(a))(e))=(a\mapsto f(a))=f$
125so $\phi$ is surjective. Thus $\phi$ is the
126desired isomorphism which, since $\phi$ evidently
127commutes with the restriction maps, induces an
128isomorphism of sheaves.
129
130(d)
131First we consider the case when
132$\se\isom\soy^n$.
133One one hand,
134\begin{eqnarray*}
135f_{*}(\sf)\tensor_{\soy}\se
136 &\isom& f_{*}(\sf)\tensor_{\soy}\soy^n \\
137&\isom& \oplus_{i=1}^{n} (f_{*}(\sf)\tensor_{\soy}\soy)\\
138&\isom& \oplus_{i=1}^{n} f_{*}(\sf).\\
139\end{eqnarray*}
140
141On the other hand,
142\begin{eqnarray*}
143\sf\tensor_{\sox}f^{*}\se &=&\sf\tensor_{\sox}f^{*}(\soy^n)\\
144&\isom&\sf\tensor_{\sox}(f^{-1}(\soy^n)\tensor_{f^{-1}\soy}\sox)\\
145&\isom&\sf\tensor_{\sox}((f^{-1}(\soy)^n\tensor_{f^{-1}\soy}\sox)\\
146&\isom&\sf\tensor_{\sox}(\sum_{i=1}^{n}
147f^{-1}\soy\tensor_{f^{-1}\soy}\sox)\\
148&\isom&\sf\tensor_{\sox}(\sox^n)\\
149&\isom&(\sf\tensor_{\sox}\sox)^n=\sf^n
150\end{eqnarray*}
151where $f^{-1}(\soy^n)\isom f^{-1}(\soy)^n$ since
152$f^{-1}$ is a left adjoint functor hence commutes
153with direct sums (which are a right universal
154construction).
155
156Putting this together we have that
157$$f_{*}(\sf\tensor_{\soy}f^{*}(\se)) 158=f_{*}(\sf^n)=f_{*}(\oplus_{i=1}^{n}\sf) 159=\oplus_{i=1}^{n}f_{*}(\sf).$$
160
161In general, we construct isomorphisms as above
162on an open cover then, since all of the isomorphisms
163are canonical, the isomorphisms match up on
164the intersections so we can glue to obtain an
165isomorphism.
166
167\begin{prob}[II.5.2]
168Let $R$ be a discrete valuation ring with quotient field
169$K$, and let $X=\spec R$.
170
171(a) To give an $\sox$-module $\sf$ is equivalent to giving
172an $R$-module $M$, a $K$-vector space $L$, and a
173homomorphism $\rho:M\tensor_R K\rightarrow L$.
174
175(b) That $\sox$-module is quasi-coherent if and only
176if $\rho$ is an isomorphism.
177\end{prob}
178\proof
179(a) First suppose we are given an $\sox$-module $\sf$. Since $R$
180is a DVR, $X$ has exactly two nonempty open sets,
181$X$ and the set consisting of the generic point,
182$\{\xi\}$. Let $M=\Gamma(\sf,X)$ and let
183$L=\Gamma(\sf,\{\xi\})$.
184Since $\Gamma(\sox,X)=R$ and $\Gamma(\sox,\{\xi\})=K$, $M$ is
185an $R$-module and $L$ is a $K$-vector space. Let
186$g:M\rightarrow L$ be the restriction map. Define
187$\rho$ by $\rho(m\tens\alpha)=\alpha\cdot{}g(m)$.
188Since $g$ is a homomorphism, so is $\rho$.
189
190Now suppose we are given an $R$-module $M$, a $K$-vector
191space $L$, and a homomorphism $\rho:M\tensor_R K\rightarrow L$.
192Define an $\sox$-module $\sf$ as follows. Let
193$\Gamma(\sf,X)=M$ and $\Gamma(\sf,\{\xi\})=L$. Define
194the restriction map $g:M\rightarrow L$ by
195$g(m)=\rho(m\tens 1)$. We just need to check that
196$g$ is a valid restriction map. Let $r\in R, m\in M$,
197then $g$ is a valid restriction map iff
198$r\cdot g(m)=g(rm)$, that is, when
199$r\cdot \rho(m\tens 1)=\rho(rm\tens 1)=\rho(m\tens r)$.
200So we must verify that the given homomorphism $\rho$ is
201$R$-linear. But there is absolutely no reason why this should
202be the case! (For example, let $M=L=K$, then
203$\rho:K\rightarrow K$ and it is easy to construct nontrivial
204homomorphisms of the additive group of a field).
205I think the problem is imprecisely stated. It
206should be assumed throughout that $\rho$ is $K$-linear.
207
208(b) First suppose $\sf$ is quasi-coherent. Then
209$\sf=\tilde{M}$ so
210Proposition 5.1 implies that
211$L=\Gamma(\tilde{M},\{\xi\})=(R-0)^{-1}M\cong 212M\tensor_R K$. Thus $\rho$ must be an isomorphism.
213Conversely, if $\rho$ is an isomorphism, we see that
214$\sf\cong \tilde{M}$ since they are the same on
215each open set and the restriction map is the
216same.
217
218\begin{prob}[II.5.3]
219Let $X=\spec A$ be an affine scheme. Show that the functors
220$\tilde{}$ and $\Gamma$ are adjoint, in the
221following sense: for any $A$-module $M$, and for any sheaf
222of $\sox$-modules $\sf$, there is a natural isomorphism
223$$\hom_A(M,\Gamma(X,\sf))\cong\hom_{\sox}(\tilde{M},\sf).$$
224\end{prob}
225\proof
226
227Define a homomorphism $F:\hom_A(M,\Gamma(X,\sf))\into 228\hom_{\sox}(\tilde{M},\sf)$ as follows. Send
229a ring homomorphism $\phi:M\into\Gamma(X,\sf)$
230to the morphism of sheaves $F(\phi):\tilde{M}\into\sf$.
231It suffices to define $F(\phi)$ on distinguished
232open sets (Eisenbud \& Harris, page 13).
233For $f\in A$ let $F(\phi)_{D(f)}$ be the map
234$${m\over f^n}\mapsto {1\over f^n} 235\cdot\res_{X,D(f)}(\phi(m))$$
236where $\res_{X,D(f)}:\sf(X)\into\sf(D(f))$
237is the restriction map of $\sf$.
238$F(\phi)_{D(f)}$ is a well-defined homomorphism since
239both $\phi$ and $\res_{X,D(f)}$ are homomorphisms and
240$\phi$ is an $A$-module homomorphism. Next note
241that $F(\phi)$ commutes with the
242restriction maps since each $\res_{X,D(f)}$ does.
243To see that $F$ is injective suppose $\phi$ and
244$\psi$ are two homomorphisms $M\into\Gamma(X,\sf)$.
245If $F(\phi)=F(\psi)$ then, in particular,
246$\phi=F(\phi)_{X}=F(\psi)_{X}=\psi$.
247To see that $F$ is surjective, let $\phi\in\hom_{\sox} 248(\tilde{M},\sf)$. Define $\psi:M\into\Gamma(X,\sf)$
249by letting $\psi=\phi_X$, that is, by taking the
250induced map on global sections.
251Then, for $f\in A$,
252$F(\psi)_{D(f)}:\tilde{M}(D(f))\into\sf(D(f))$ is
253the map
254$({m\over f^n}\mapsto {1\over f^n}\res_{X,D(f)}\circ\phi_X(m))$
255which, since $\phi$ is a morphism of sheaves, equals
256$=({m\over f^n}\mapsto {1\over f^n}\phi_{D(f)}(m) 257=\phi_{D(f)}({m\over f^n})=\phi_{D(f)}$.
258(We are just using the fact that $\phi$ commutes with
259the appropriate restriction maps.)
260Thus $F(\psi)$ agrees with $\phi$ on a basis for $X$
261hence $F(\psi)=\phi$. This shows that $F$ is surjective.
262So $F$ is an isomorphism, as required.
263
264\begin{prob}[II.5.4]
265Show that a sheaf of $\sox$-modules $\sf$ on a scheme $X$
266is quasi-coherent
267if and only if every point of $X$ has a neighborhood $U$, such
268that $\sf|_{U}$ is isomorphic to a cokernel of a morphism of
269free sheaves on $U$. If $X$ is noetherian, then $\sf$ is coherent
270iff it is locally a cokernel of a morphism of free sheaves
271of finite rank.
272\end{prob}
273\proof
274Suppose first that $\sf$ is quasi-coherent. Let $x\in X$.
275Then there is an affine open neighborhood
276$U=\spec A$ of $x$ such that $\sf|_U\isom \tilde{M}$,
277$M$ an $A$-module. It suffices to show that $M$ is
278isomorphic to a cokernel of a morphism of finitely
279generated $A$-algebras. Indeed, if
280$\phi:A^{(I)}\into A^{(J)}$ then
281$\coker(\tilde{\phi})=\tilde{A^{(J)}}/\phi(A^{(I)})\tilde{} 282=(A^{(J)}/\phi(A^{(I)}))\tilde{}$
283since, for all $f\in A$,
284$(A^{(J)})_f/\phi(A^{(I)})_f 285=(A^{(J)}/\phi(A^{(I)}))_f$
286so that they agree on a basis.
287
288Let $A^{|M|}$ be the free $A$-module on the elements
289of $M$.
290Let $\phi:A^{|M|}\into M$ be the natural map.
291Similiary, let $\psi:A^{|\ker(\phi)|}\into\ker(\phi) 292\subseteq A^{|M|}$ be the natural map.
293Then $\coker(\psi)\isom A^{|M|}/\ker(\phi)\isom M$,
294as required.
295
296Now assume the $\sf$ is coherent. We proceed as above
297but now $M$ is a finitely generated $A$-module,
298generated by $e_1,\ldots,e_n$, say, and we must
299show that $M$ is the cokernel of a morphism of
300free modules of finite rank. Let $\phi:A^n\into M$
301be the map which takes the $i$th generator
302$(0,\ldots,1,\ldots,0)$ of $A^n$ to $e_i\in M$.
303Then $\ker(\phi)$ is a submodule of $M$ so, since
304$M$ is neotherian (any finitely generated module
305over a noetherian ring is noetherian), $\ker(\phi)$
306is finitely generated. Let $f_1,\ldots,f_m$ be
307a generating set. Let $\psi:A^m\into\ker(\phi)$ be
308the surjection defined by sending the $i$th
309basis element of $A^m$ to $f_i$. Then
310$\coker(\psi)\isom A^n/\psi(A^m)\isom A^n/\ker(\phi)\isom M$.
311Thus $M$ is isomorphic to a cokernel of a morphism
312of free sheaves of finite rank.
313
314
315
316
317
318\begin{prob}[II.5.5]
319Let $f:X\rightarrow Y$ be a morphism of schemes.
320
321(a) Show by example that if $\sf$ is coherent on $X$, then
322$f_{*}\sf$ need not be coherent on $Y$, even if $X$ and $Y$
323are varieties over a field $k$.
324
325(b) Show that a closed immersion is a finite morphism.
326
327(c) If $f$ is a finite morphism of neotherian schemes, and if
328$\sf$ is coherent on $X$, then $f_{*}\sf$ is coherent on $Y$.
329\end{prob}
330\proof
331(a) Lex $k$ be a field, let $X=\spec (k[x]_x)$,
332$Y=\spec (k[x])$ and $\sf=\sox$. Then
333$f_{*}(\sf)(U)=\sf(f^{-1}(U))$, so
334$f_{*}(\sf)=(k[x]_x)^{\tilde{}_Y}$.
335But $(k[x]_x)^{\tilde{}_Y}$ is not a coherent sheaf
336of $\soy$-modules. Indeed, if it were, there would
337be a distinguished neighborhood $D(f)$ of $0$ so that
338$(k[x]_x)^{\tilde{}_Y}|_{D(f)}$
339is a finitely generated module over $k[x]$ (here I'm
340using the fact that over open set is distinguished
341in the topology of $\spec(k[x])$). But,
342$k[x]_f$ is never a finitely generated module over
343$k[x]$ for any $f$ of degree at least $1$. For $k[x]_f$
344contains elements of arbitrarily small degree whereas
345the degrees of elements of $k[x]\{\alpha_1,\ldots,\alpha_n\}$
346are bounded below. ($k[x]\{\alpha_1,\ldots,\alpha_n\}$
347is the $k[x]$-module generated by $\alpha_1,\ldots,\alpha_n$.)
348
349(b) Let $f:Y\into X$ be a closed immersion. Let $U=\spec(A) 350\subseteq X$ and let $W=f^{-1}(U)\subseteq Y$. Then
351$W$ is a a scheme (give it the induced scheme structure as
352an open subset of $Y$). Furthermore,
353$f(W)=U\cap f(Y)$ is a relatively closed subset of $U$,
354that is, a closed subset of $\spec(A)$. The
355map $\fh:\so_{U}\into\so_{W}$ is surjective since
356$\fh:\sox\into\soy$ is surjective and surjectivety
357is a local property. Thus $W$ is a closed subscheme
358of $\spec(A)$ so, by Corollary 5.10, $W\cong\spec(A/I)$
359for some ideal $I$ of $A$. Since $A/I$ is a finitely
360generated $A$ module (generated by $1+I$), $f$
361is a finite morphism.
362
363(c) Let $U=\spec(A)\subseteq Y$ be an affine open subset
364of $Y$. Then, since $f$ is finite, $f^{-1}(U)=\spec(B)$
365is affine with $B$ is a finitely generated $A$-module.
366By Proposition 5.4 it suffices to show that
367$f_{*}(\sf)|_U$ is $\tilde{M}$ for some finitely
368generated $A$-module $M$. Now by Proposition 5.4,
369since $f^{-1}(U)$ is affine and $B$ is noetherian,
370$\sf|_{f^{-1}(U)}=\tilde{M}$ for some finitely
371generated $B$ module $M$.
372But
373$f_{*}(\sf)|_U=(f|_{f^{-1}(U)})_{*}\sf|_{f^{-1}(U)} 374=(f|_{f^{-1}(U)})_{*}(\tilde{M})=\tilde{(A^M)}$
375where the last equality follows from Proposition 5.2(d).
376Since $B$ is a finite module over $A$ and $M$
377is a finite module over $B$, it follows that $M$
378is a finite module over $A$ which completes the proof.
379
380\begin{prob}[II.5.7]
381Let $X$ be a noetherian scheme, and let $sf$ be a coherent sheaf.
382
383(a) If the stalk $\sf_x$ is a free $\sox$-module for some point
384$x\in X$, then there is a neighborhood $U$ of $x$ such that $\sf|_U$
385is free.
386
387(b) $\sf$ is locally free iff its stalks $\sf_x$ are free $\sox$-modules
388for all $x\in X$.
389
390(c) $\sf$ is invertible iff there is a coherent sheaf $\sg$ such
391that $\sf\tensor\sg\isom\sox$.
392\end{prob}
393\proof
394(a) Let $U=\spec(A)$ be a neighborhood of $x$ so that
395$\sf\ru=\tilde{M}$, $M$ a finitely generated $A$-module.
396Then $\sf_x=M_x$ so we have reduced the problem to
397the following purely algebraic result.
398\begin{prop}
399If $M$ is a finitely generated free $A$ module and there
400is a prime $\wp$ of $A$ such that $M_{\wp}$ is a free
401$A_{\wp}$-module, then there exists $f\in A$ such that
402$f\not\in\wp$ and $M_f$ is a free $A_f$-module.
403\end{prop}
404
405Once we have proven this we will know that $\sf|_{D_U(f)}$
406is free since Proposition 5.1 asserts that
407$\tilde{M}(D(f))\cong M_f$.
408
409\proof (of Proposition) Let $a_1,\ldots,a_n\in M$ be a free
410basis of $M_{\wp}$ over $A_{\wp}$ (we can clear denominators
411so that we may assume all $a_i$ lie in $M$.) Let
412$b_1,\ldots,b_m\in M$ be a generating set for $M$
413over $A$. For each $i$ we can write $b_i$ as
414an $A_{\wp}$-linear combination of the $a_i$. Clearing
415denominators we see that $d_i b_i\in A\{a_1,\ldots,a_n\}$
416for some $d_i\not\in\wp$. Let $f=\prod d_i$, then
417$f\not\in\wp$ and $a_1,\ldots,a_n$ have
418$A_f$-span including all of the $b_i$,  and thus
419including $M$, and thus including $M_f$. But
420$a_1,\ldots,a_n$ is free over $A_{\wp}$ hence over
421$A_f$ since $A_f\subseteq A_{\wp}$. [This proposition
422is in Bourbaki, {\em Commutative Algebra},
423II.5.1, although the proof is more abstract
424than mine.]
425
426(b) ($\Longrightarrow$) Let $x\in X$ and let $U=\spec(A)$ be an open
427neighborhood of $x$ such that $\sf\ru=\tilde{M}$
428with $M$ a free $A$ module. Suppose $\wp$ is the
429prime of $A$ corresponding to $x$. Then $\sf_x=M_\wp$
430which is a free $A_{\wp}$-module. Indeed,
431if $e_1,\ldots,e_n$ is a free $A$-basis for $M$,
432then it is also a free $A_{\wp}$-basis for
433$M_{\wp}$. For if
434${a_1\over b_1}e_1 + \cdots + {a_n\over b_n}e_n=0, 435b_i\not\in \wp$,
436then
437$a_1{b\over b_1}e_1+\cdots+a_n{b\over b_n}e_n=0, b=\prod b_i$
438so $b{a_i\over b_i}=0$ for each $i$,
439so ${a_i\over b_i}=0$ in the localization $A_{\wp}$
440since $b\not\in\wp$.
441
442($\Longleftarrow$) By part (a) every point has a neighborhood
443on which $\sf$ is free. Therefore $X$ can be covered by
444open affines on which $\sf$ is free so $\sf$ is locally free.
445
446(c) ($\Longrightarrow$)
447Let $\sg=\check{\sf}=\shom(\sf,\sox)$, we will show that
448$\sf\tensor\sg\isom\sox$. To define a morphism
449$\phi:\sf\tensor\shom(\sf,\sox)\rightarrow\sox$
450it is enough to define $\phi$ on the presheaf
451$(U\mapsto \sf(U)\tensor_{\sox(U)}\shom(\sf(U),\sox(U))$.
452Define $\phi$ by $a\tens f\mapsto f(a)$. Thus $\phi$ commutes
453with the restrictions so $\phi$ defines a valid morphism
454of sheaves.
455
456Let $U$ be an open affine subset of $X$ such that $\sf(U)$
457is a free $\sox(U)$-module of rank 1 with basis $e_0$.
458Then $\shom(\sf(U),\sox(U))$ has basis $e_0^{*}$ where
459$e_0^{*}(e_0)=1$. Thus $\shom(\sf(U),\sox(U))$ is a free
460$\sox(U)$-module of rank 1. Thus every element of
461$\sf(U)\tensor \hom(\sf(U),\sox(U))$ can be written in the
462form $a\tens f$ (as opposed to as a sum of such products).
463Now $\phi_U(a\tens f)=0$ implies $f(a)=0$ which implies
464$a=0$ or $f=0$ so $a\tens f=0$, so $\phi_U$ is injective.
465Since $\phi_U(ae_0\tens e_0^{*})=a$, $\phi_U$ is surjective.
466Thus $\phi_U$ is an isomorphism.
467
468Use the definition of locally free of rank 1 to cover
469$X$ by affine open sets $U$ such that $\sf\ru$ is
470a free $\sox\ru$ module of rank 1. Then, by Proposition
4715.1 (c) and an argument like that for (b) above, any
472distinguished open subset of $U$ has a $\sf$-sections
473free of rank 1. Since $\phi$ is an isomorphism on each
474of these distinguished open sets (use the argument in the
475above paragraph) and these distinguished open sets form
476a basis for the topology on $X$ it follows that $\phi$
477must be an isomorphism.
478
479($\Longleftarrow$) Because of part (b) above it suffices
480to show that $\sf_x$ is free of rank one for each $x\in X$.
481Since $\sf_x\tensor_{\so_x}\sg_x 482=(\sf\tensor\sg)_x\isom\so_{X,x}$
483the problem is reduced to the following purely
484algebraic statement.
485
486\begin{prop} Let $M$ and $N$ be finitely generated modules
487over a local ring $(A,m)$ and suppose that
488$M\tensor_A N\isom A$. Then $M$ is free of rank 1.
489\end{prop}
490\proof
491Let $k=A/m$. Let $\nu:M\tensor_A N\into A$ be
492the given isomorphism. Then, taking the product
493with the identity, we get an isomorphism
494$\nu\tensor 1_k:(M\tensor_A N)\tensor_A k\into 495A\tensor_A k\cong k$ (it is obvious that
496$\nu\tensor 1_k$ is surjective, but it is
497not at all obvious that it is injective,
498for this see the Bourbaki reference below.)
499Thus $k\cong M\tensor_A N\tensor_A k 500=M\tensor_A(k\tensor_A k)\tensor_A N 501=(M\tensor_A k)\tensor_A(N\tensor_A k) 502=(M\tensor_A k)\tensor_k(N\tensor_A k) 503=(M/mM)\tensor_k (N/mN)$
504so, since the $k$-rank of
505$(M/mM)\tensor_k (N/mN)$
506is $1$ and is the product of
507the ranks of $M/mM$ and $N/mN$, each has
508rank 1. In particular, $M/mM$ is monogeneous (generated
509by one element) as an $A/m$-module and hence as
510an $A$-module so,
511by Nakayama's lemma, $M$ is monogeneous
512as an $A$-module. Since
513$\ann_A(M)$ anihilates $M\tensor_A N=A$ as well,
514it is $0$ (any element of $\ann_A(M)$ would have
515to annihilate the identity of $A$ and hence be $0$).
516Thus $M$ is a free module of rank one over $A$.
517[See Bourbaki, {\em Commutative Algebra}, II.5.4, for a
518more general theorem.]
519
520\begin{prob}[II.5.8]
521Again let $X$ be a noetherian scheme, and $\sf$ a coherent sheaf
522on $X$. We will consider the function
523$$\phi(x)=\dim_{k(x)}\sf_x\tensor_{\sox}k(x)$$
524where $k(x)=\sox/m_x$ is the residue field at the point $x$. Use
525Nakayama's lemma to prove the following results.
526
527(a) The function $\phi$ is upper semi-continuous, i.e., for any
528$n\in \z$, the set $\{x\in X:\phi(x)\geq n\}$ is closed.
529
530(b) If $\sf$ is locally free, and $X$ is connected, then
531$\phi$ is a constant function.
532
533(c) Conversely, if $X$ is reduced, and $\phi$ is constant,
534then $\sf$ is locally free.
535\end{prob}
536\proof
537
538(a) We must show that $\{x:\phi(x)\geq n\}$ is closed.
539A good way to do this is by showing that
540$\{x:\phi(x)<n\}$ is open. To do this we show that
541if $\phi(x)=m$ then there is an open neighborhood $U$ of
542$x$ so that, for all $y\in U$, $\phi(y)\leq m$.
543Since we need only look locally, we can assume that
544$X=\spec A$, $\sf=\tilde{M}$, $M$ a finitely generated
545$A$-module. Note that $\sf_x\tensor_{\sox}k(x)= 546M_{\wp}\tensor_{A_{\wp}}A_{\wp}/{\wp A_{\wp}}= 547M_{\wp}/{\wp M_\wp}$.
548Let $s_1,\ldots,s_m\in M$ be elements whose images
549form a basis for the vector space $M_{\wp}/\wp M_{\wp}$
550over $A_{\wp}/\wp A_{\wp}$ (to do this choose a basis
551for $M_{\wp}/\wp M_{\wp}$ then clear fractions). Note
552that the images of the $s_i$ in fact generate
553$M_{\wp}/{\wp M_{\wp}}$ as an $A_{\wp}$-module.
554By Nakayama's lemma the $s_i$ generate
555$M_{\wp}$ as an $A_{\wp}$-module.
556Let $m_1,\ldots,m_k$ be a generating set for
557$M$ over $A$. Write $m_j=\sum {a_i\over b_i}s_i$,
558$b_i\not\in\wp$, then, if $c_j=\prod b_i$,
559$c_jm_j$ is in the $A$-span of the $s_i$. Let
560$f=\prod c_j$. Then $\wp\in D(f)$ and if
561$q\in D(f)$, then $m_1,\ldots,m_k$ all lie
562in the $A_q$-span of $s_1,\ldots,s_m$ (since
563$c_jm_j$ is in the $A$-span of the $s_i$
564and $c_j$ is inverted in $A_q$. Thus $M$ is spanned
565by the $s_i$ over $A_q$, so $M_q$ is spanned by
566the $s_i$ over $A_q$. It follows that
567$\phi(q)=\dim M_q/qM_q\leq m$ since the images
568of the $s_i$ generate $M_q/qM_q$ as a vector
569space over $A_q/qM_q=k(q)$. Taking $D(f)$ as
570our open neighborhood completes the proof.
571
572(b) Choose $n$ so that some section of $\sf$
573has rank $n$. Let $U$ be the union of all open sets $W$
574such that $\sf|_W\isom\sox^n|_W$. Then $U$ is nonempty.
575Let $V$ be the
576union of all open sets $W$ such that
577$\sf|_W\isom\sox^m|_W$, $m\not=n$. Since $\sf$ is
578locally free, $U\cup V=X$. Suppose $x\in U\cap V$,
579then $\sf_x$ has rank $n$ and rank $m\not=n$ (since
580rank is preserved under localization), a
581contradiction. Thus $U\cap V=\emptyset$. Since $U$
582is nonempty and open, $X-U=V$ is open and $X$ is
583connected, thus we conclude that $V=X-U=\emptyset$. Thus
584every point is contained in an open set $W$ such
585that $\sf|_W\isom\sox^n|_W$.
586
587Let $x\in X$ and let $U=\spec(A)$ be an affine open set
588containing $x$ such that $\sf|_U\isom\tilde{M}$. By the above
589argument, $M$ is a free $A$-module of rank $n$.
590Thus $\phi(x)=\dim_k M_x\tensor_{A_x}A_x/m_x 591=\dim_k A_x^n\tensor_k A_x/m_x=\dim_k (A_x/m_x)^n 592=\dim_k k^n=n$, as desired.
593
594(c) Let $x\in X$. By exercise 5.7b it suffices to
595show that the stalk $\sf_x$ is free. Since $\sf$
596is coherent we can find an affine open set
597$U=\spec(A)$ such that $\sf\ru=\tilde{M}$ for
598some finitely generated $A$-module, $x\in U$,
599and $A_f$ is reduced for each $f\in A$. Let
600$\wp$ be the prime of $A$ corresponding to
601$x$. We must show that $M_{\wp}$ is free over
602$A_{\wp}$. Let $s_1,\ldots,s_n\in M$ be preimages
603of a basis of $M_{\wp}/\wp M_{\wp}$ over
604$k(x)=A_{\wp}/\wp A_{\wp}$ (find these
605as in part (a)). Then, by Nakayama's lemma, the
606$s_i$ generate $M_{\wp}$ over $A_{\wp}$.
607
608We must show that the $s_i$ are linearly
609independent over $A_{\wp}$. It will then follow
610that $M_{\wp}$ is free of rank $n$ over $A_{\wp}$.
611So suppose
612$${a_1\over b_1}s_1+\cdots+{a_n\over b_n}s_n=0$$
613in $M_{\wp}$ with ${a_i\over b_i}\in A_{\wp}$.
614Then for each $i$, $b_i\not\in\wp$ and
615$a_i\in A$. Since the $s_i$ are linearly independent
616over $A_{\wp}/\wp A_{\wp}$, for each $i$ there
617exists $c_i\not\in\wp$ such that ${c_i a_i\over b_i}\in\wp$.
618Thus $c_i a_i\in\wp$ so $a_i\in\wp$.
619Let $r$ be as in part (a) so that $q\in D(r)$ implies
620the $s_i$ generate at least $M$ over $A_q$.
621By definition there exists $c\in A$ such that
622$$c(b_2\cdots b_n a_1 s_1+\cdots+b_1\cdots b_{n-1} a_n s_n)=0$$
623in $A$.
624Let $f=rc\prod b_i$, then if $q\in D(f)$, then
625$s_1,\ldots,s_n$ generate $M_q/qM_q$ over $A_q$ and,
626since $M_q/qM_q$ has dimension $n$
627(since $\phi$ is constant),
628the $s_i$ are actually a basis for $M_q/qM_q$ over
629$A_q/qA_q$.
630
631Since $c|f$, $c\not\in q$ so, as above,
632${a_1\over b_1}s_1+\cdots+{a_n\over b_n}s_n=0$ in
633$M_q$, so, as above, $a_i\in q$ for each $i$.
634Thus, for all $q\in D(f)$, $a_i\in q$, so
635$a_i$ lies in the nilradical of $A_f$ which, since
636$A_f$ is reduced, means that $a_i=0$ in $A_f$.
637So $a_i$ maps to $0$ under the map $A_f\into A_{\wp}$.
638Thus $s_1,\ldots,s_n$ are linearly independent
639over $A_{\wp}$ so $M_{\wp}$ is free of rank $n$
640over $A_{\wp}$. Applying exercise (5.7b) then completes
641the proof.
642
643
644
645
646\end{document}
647