Sharedwww / hartsoln2.texOpen in CoCalc
Author: William A. Stein
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\documentclass[12pt]{article}
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\author{William A. Stein}
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\title{Algebraic Geometry Homework\\
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II.3 \#'s 6,7,8,12,14\\ II.4 \#'s 2, 4}
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\font\grmn=eufm10 scaled \magstep 1
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\newcommand{\gp}{p}
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\newcommand{\ga}{a}
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\newcommand{\so}{\mathcal{O}}
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\newcommand{\id}{\mbox{\rm id}}
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\newtheorem{prob}{Problem}
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\newtheorem{prop}{Proposition}
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\newtheorem{theorem}{Theorem}
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\newcommand{\ov}{\overline{\varphi}}
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\newcommand{\ol}{\overline}
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\renewcommand{\phi}{\varphi}
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\newcommand{\fh}{f^{\#}}
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\newcommand{\proj}{\mbox{\rm Proj \hspace{.01in}}}
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\newcommand{\spec}{\mbox{\rm Spec \hspace{.01in}}}
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\newcommand{\proof}{\mbox{\sc Proof.\hspace{.1in}}}
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\newcommand{\diagram}{$$\vspace{.75in}$$} %% space for diagrams
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\begin{document}
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\maketitle
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\begin{prob}[3.6]
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Let $X$ be an integral scheme. Show that the local ring $\so_{\xi}$ of
33
the generic point $\xi$ of $X$ is a field. Call it $K(X)$. Show also
34
that if $U=\spec A$ is any open affine subset of $X$, then $K(X)$ is
35
isomorphic to the quotient field of $A$.
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\end{prob}
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\proof
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Let $U=\spec A$ be any nonempty open affine subset of $X$. Then since
39
the closure of a generic point of $X$ is all of $X$, every open set
40
must contain a generic point. Thus if $\xi$ is a generic point, then
41
$\xi\in U$. But $A$ is an integral domain so $(0)$ is the unique
42
generic point of $U$, whence $\xi=(0)$. This shows the generic point is
43
unique if it exists. Since $X$ is integral it is irreducible so every
44
open set intersects $U$. Thus every open set contains $(0)\in \spec A=U$,
45
so $X$ actually contains a generic point $\xi=(0)$. Furthermore,
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$\so_{\xi}\cong A_{(0)}$ is the quotient field of $A$.
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\begin{prob}[3.7]
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Let $f:X\rightarrow Y$ be a dominant, generically finite morphism of
50
finite type of integral schemes. Show that there is an open dense
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subset $U\subseteq Y$ such that the induced morphism
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$f^{-1}(U)\rightarrow U$ is finite.
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\end{prob}
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\proof
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Let $U=\spec B$ be an open affine subset of $Y$ which contains the
56
generic point of $Y$. Let $V=\spec A$ be an open affine subset of
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$f^{-1}(U)$. Since $f$ is of finite type $A$ is a finitely generated
58
$B$-algebra. The generic point of $X$ is in $V$ since every
59
open set contains the generic point. Let $\phi:B\rightarrow A$ be
60
the homomorphism corresponding to the induced morphism of affine schemes
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$f:V\rightarrow U$.
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Since $f$ is dominant, we know that $\phi$ is injective.
63
The induced map on stalks then gives an inclusion
64
of function fields $K(Y)=B_{(0)}\hookrightarrow A_{(0)}=K(X)$.
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Since $A$ is a finitely generated $B$-algebra, $K(X)$ is a finitely
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generated field extension of $K(Y)$. If this field extension is not of
67
finite degree then $K(X)$ must contain an element which is
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transcendental over $K(Y)$. Thus $A$ must contain an
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element $t$ which is transcendental over $B$. But then infinitely
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many primes of $A$ lie over $(0)$. Indeed, since $K(X)$ is
71
finitely generated over $K(Y)$, $K(X)$ is
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a finite algebraic extension of $k(t)$ for some field $k$.
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Then (since the algebraic closure of a field is
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infinite), there are infinitely many irreducible polynomials
75
in $k[t]$. Since $K(X)$ is finite algebraic over $k(t)$
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infinitely many of these must remain irreducible in $A$.
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Multiplying through denominators this gives infinitely
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many irreducible elements of $A$ which generate prime ideals
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which lie over $(0)$. This would contradict the fact that
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$f$ is generically finite. Thus $K(X)$ is finite over $K(Y)$.
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Let $x_1,\ldots,x_n$ generate $A$ as a $B$-algebra. Then,
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since $K(X)$ is finite over $K(Y)$ each $x_i$ satisfies
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some polynomial $f_i$ with coefficients in $B$. Let $b$ be the
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product of all of the leading coefficients of the polynomials
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$f_i$. If $b$ is a unit in $B$ then so are all of the leading
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coefficients of the $f_i$ so we can divide by them and hence assume
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the $f_i$ are monic polynomials. If not, replace $B$ by the localization
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$B_b$ and repeat the whole argument with $U=\spec B_b$.
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In either case we may assume the $f_i$ are monic from which
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we conclude that $A$ is a finitely generated integral extension
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of $B$, thus $A$ is a finite module over $B$.
93
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Now let $U=\spec B$ be an open affine subset of $Y$ which contains the
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generic point. Since $f$ is of finite type we may write
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$f^{-1}(U)=\cup_{i=1,\ldots,n}V_i$ where each $V_i=\spec A_i$ is
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finitely generated $B$-algebra. By the work above we may shrink
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$U$ so that we can assume each $A_i$ is actually a finitely
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generated $B$-module. To complete the proof we need to show that
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there is a distinguised open subset of $U$ (which necessarily
101
contains the generic point) whose inverse image under $f$ is an
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open affine which is the spectrum of a finitely generated $B$-module.
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Let $\phi_i:B\rightarrow A_i$ be the homomorphism which
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induces $f|_{V_i}$. Since $f$ is dominant, each $\phi_i$
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is an injection. Thus we may, for notational convenience,
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identify $B$ with it's images in the various $A_i$. The
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morphism $f$ is then induced by the inclusion map
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$B\hookrightarrow A_i$.
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Since $\cap_{i=1,\ldots,n}V_i$ is open we can,
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for each $i$, $1\leq i \leq n-1$, find $\alpha_i\in A_i$
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such that $\spec (A_i)_{\alpha_i}\subseteq \cap_{i=1,\ldots,n}V_i$.
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Since $A_i$ is a finite module over $B$ there is an
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integral equation
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$$\alpha_i^n+b_{n-1}\alpha_i^{n-1}+\cdots+b_0=0$$
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where each $b_j\in B$ and $b_0\not=0$.
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Let $b=\prod_{i=1,\ldots,n-1}b_i$. Then any
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prime of $A_i$ which contains $\alpha_i$ must
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also contain $b_i$ and hence $b$. Therefore
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$\spec (A_i)_{b}\subseteq \spec (A_i)_{\alpha}$.
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We then have that
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$g^{-1}(\spec B_b)
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= \cup_{i=1,\ldots,n-1}\spec (A_i)_b\cup\spec(A_n)_b
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= \spec(A_n)_b$.
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The latter equality follows since, for $1\leq i\leq n-1$,
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$\spec(A_i)_b\subseteq \cap_{i=1,\ldots,n} V_i \cap f^{-1}(U_b)
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\subseteq f^{-1}(U_b)\cap V_n = \spec (A_n)_b$.
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We thus see that $U_b$ is a dense open
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subset of $Y$ such that the morphism $f:f^{-1}(U_b)\rightarrow U_b$
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is finite (since $f^{-1}(U_b)$ is the affine scheme $\spec(A_n)_b$
131
which a finite $B_b$-module). This completes the proof.
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\begin{prob}[3.8]
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Normalization. Let $X$ be an integral scheme. For each open affine
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subset $U=\spec A$, let $\tilde A$ be the integral closure of $A$
136
in its quotient field, and let $\tilde U=\spec \tilde A$.
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Show that one can glue the schemes $\tilde U$ to obtain a normal integral
138
scheme $\tilde X$, called the normalization of $X$. Show also that there
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is a morphism $\tilde X\rightarrow X$, having the following universal
140
property: for every normal integral scheme $Z$, and for every
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dominant morphism $f:Z\rightarrow X$, $f$ factors uniquely through
142
$\tilde X$. If $X$ is of finite type over a field $k$, then the morphism
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$\tilde X\rightarrow X$ is a finite morphism.
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\end{prob}
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\proof
146
147
We first verify the universal property for affine schemes where
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it is clear what the normalization is.
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\begin{prop} Suppose $X=\spec A$, $\tilde X=\spec \tilde A$
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its normalization and $Z=\spec B$ is a normal integral scheme.
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Then every dominant morphism $f:Z\rightarrow X$ factors uniquely
152
through $\tilde X$.
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\end{prop}
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\proof
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Let $\phi:A\rightarrow B$ be the homomorphism corresponding to $f$.
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Then, since $f$ is dominant, $\phi$ is injective. Indeed,
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$f(Z)\subseteq V(\ker(\phi))$ so if $\ker(\phi)\not = 0$
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then $f(Z)$ doesn't meet the nonempty open set
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$X-V(\ker(\phi))$ (nonempty since $A$ is a domain
160
so $(0)$ is prime so $(0)\in X-V(\ker(\phi))$.
161
So there is a unique extension of $\phi$ to a
162
homomorphism from $\tilde A \rightarrow B$.
163
Indeed, define $\ol{\phi}(a/b)=\phi(a)/\phi(b)$. Then
164
since $\phi$ is injective this is well-defined since $b\not=0$
165
implies $\phi(b)\not=0$. Furthermore, if $\psi$ is another
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possible extension of $\phi$ to $\tilde A$, then
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$\psi(b)\psi(a/b)=\psi(a)=\phi(a)=\phi(b)\ol{\phi}(a/b)=\psi(b)\ol{\phi}(a/b)$
168
so cancelling shows that $\psi(a/b)=\ol{\phi}(a/b)$.
169
Thus there is a unique morphism $f':Z\rightarrow
170
\tilde X$ whose composition with the natural map $\tilde X\rightarrow X$
171
equals $f$. (The natural map is induced by the inclusion
172
$A\hookrightarrow \tilde A$.)
173
174
Now we prove the universal property holds when $Z$ is an arbitrary
175
normal integral scheme but $X$ is still affine.
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\begin{prop} Let $X=\spec A$, $\tilde X=\spec \tilde A$ its
177
normalization and $Z$ be any normal integral scheme. Then
178
every dominant morphism $f:Z\rightarrow X$ factors uniquely
179
through $\tilde X$.
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\end{prop}
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\proof
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Let $U_i$ be a cover of $Z$ by open affines. If $U=U_i$ is any
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$U_i$ then $U$ is a normal
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integral affine scheme and $f|_U$ is a dominant morphism. Indeed,
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$U$ is dense in $Z$ since $Z$ is irreducible (Proposition 3.1).
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Thus $f^{-1}(\ol{f(U)})\supseteq \ol{U}=Z$ so
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$f^{-1}(\ol{f(U)})=Z$ so $\ol{f(U)}\supseteq f(Z)$ so
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$\ol{f(U)}=\ol{f(Z)}=X$.
189
We can thus apply the above proposition to find a unique morphism
190
$g_i:U_i\rightarrow \tilde{X}$ such that $\psi\circ g_i=f|_U$ where
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$\psi:\tilde{X}\rightarrow X$. By uniqueness on a cover of $U_i\cap U_j$
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by open affines, $g_i|_{U_i\cap U_j}=g_j|_{U_i\cap U_j}$. We
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can thus glue the morphism $g_i$ to obtain a morphism
194
$g:Z\rightarrow \tilde{X}$ such that $\psi\circ g=f$. The
195
morphism $g$ is evidently unique.
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197
Now we can define the identification maps $\phi_{ij}$. Let
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$\{U_i=\spec A_i\}$ be the open affine subsets of $X$. Let
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$\{\tilde U_i=\spec \tilde A_i\}$ be the associated normalizations.
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Let $\psi_i:\tilde U_i \rightarrow U_i$ be the morphism induced
201
by the inclusion $A_i \hookrightarrow \tilde A_i$. Let
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$W_{ij}=\psi_i^{-1}(U_i\cap U_j)$. Then $W_{ij}$ is an open
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subset of a normal scheme hence normal. $\psi_i:W_{ij}\rightarrow
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U_i\cap U_j \subseteq U_j$ so there is a unique morphism
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which we call $\phi_{ij}:W_{ij}\rightarrow \tilde U_j$ such
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that $\psi_j|_{W_{ji}} \circ \phi_{ij} = \psi_i|_{W_{ij}}$.
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By uniqueness we see that $\phi_{ij}\circ\phi_{ji}=id$ so
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$\phi_{ij}=\phi_{ji}^{-1}$. Furthermore, for each $i,j,k$,
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$\phi_{ij}(W_{ij}\cap W_{ik})
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=\psi_j^{-1}(\psi_i(W_{ij}\cap W_{ik}))
211
=\psi_j^{-1}(\psi_i(W_{ij}))\cap \psi_j^{-1}(\psi_i(W_{ik}))
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=W_{ji}\cap W_{jk}$. By uniqueness,
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$\phi_{jk}\circ\phi_{ij}=\phi_{ik}$ on $W_{ij}\cap W_{ik}$.
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So by the glueing lemma (Exercise 2.12) we may glue to obtain
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a scheme $\tilde X$. We can also glue the morphisms $\psi_i$
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to obtain a morphism $\psi:\tilde X\rightarrow X$.
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Next, we must verify that the universal property holds in general.
219
Let $Z$ be an arbitrary normal integral scheme, and let $X$
220
and $\tilde{X}$ be as above and suppose $f:Z\rightarrow X$ is a morphism.
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Cover $X$ be open affines $U_i$. Then for each morphism $f|_{f^{-1}(U_i)}$
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we can apply the above proposition to find a morphism $g_i$ such
223
that $\psi\circ g_i=f|_{f^{-1}(U_i)}$. By uniqueness we can
224
glue these morphism to obtain the required morphism
225
$g:Z\rightarrow \tilde{X}$.
226
227
Now we check that $\tilde{X}$ is a normal integral scheme.
228
Note first that each $\tilde{U_i}$ is the spectrum of
229
an integrally closed domain and is hence a normal integral
230
scheme (since the localization of an integrally closed
231
domain is integrally closed).
232
Let $x\in\tilde{X}$. Then $x$ is contained in some $\tilde{U_i}$.
233
But the local ring of $x$ in $\tilde{X}$ is the same as the
234
local ring of $x$ in $\tilde{U_i}$ which is integrally closed.
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This shows that $\tilde{X}$ is normal.
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Since $X$ is irreducible,
238
every $U_i$ intersects every $U_j$. Thus every $\tilde{U_i}$
239
intersects every $\tilde{U_j}$ after glueing.
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Since each $\tilde{U_j}$ is irreducible and they all overlap
241
this implies $\tilde{X}$ is irreducible. Indeed, if $\tilde{X}=A\cup B$
242
with $A$ and $B$ closed, then every $\tilde{U_i}$ is either completely
243
contained in $A$ or in $B$. If they are not all contained in one of
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$A$ or $B$ then we can find an open subset $U$ contained in $A$
245
and an open subset $V$ contained in $B$ but not contained in $A$. Then
246
$V=(U\cap V)\cup(U^c\cap V)
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=(A\cap V)\cup(U^c\cap V)$
248
which expresses $V$ as a union of two proper closed subsets of $V$.
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$A\cap V$ is a proper subset of $V$ since $V$ is not contained in
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$A$ and $U^c\cap V$ is a proper subset of $V$ since
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$U\cap V\not=\emptyset$. This contradicts the fact that $V$ is
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irreducible. Thus $\tilde{X}=A$ or $\tilde{X}=B$ whence $\tilde{X}$
253
is irreducible.
254
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Now we check that the structure sheaf has no nilpotents.
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Let $U$ be an open subset of $\tilde{X}$ and suppose
257
$f\in\so_{\tilde{X}}(U)$ is nilpotent. Then since $f$ is
258
nonzero, there is some point $x\in \tilde{X}$ so that the
259
stalk $f_x$ of $f$ at $x$ in the local ring $\so_{\tilde{X}}$ is
260
nonzero and nilpotent (use sheaf axiom (iii) and the definition of
261
$\so_{\tilde{X}}$.) Let $\tilde{U_i}$ be some $\tilde{U_i}$
262
which contains $x$. Then the local ring at $x$ is a localization
263
of the integral domain $\tilde{A_i}$ so it can't contain any
264
nilpotents. Thus the scheme $\tilde{X}$ is reduced.
265
266
Now we check that if $X$ is of finite type over a field $k$, then
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the morphism $f:\tilde X\rightarrow X$ is a finite morphism.
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The $U_i$ form an open cover of $X$ and $f^{-1}(U_i)=\spec \tilde A_i$
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is affine for each $i$, so we just need to check that
270
$\tilde A_i$ is a finite module over $A_i$. This follows from
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Theorem 3.9A of chapter I.
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\begin{prob}[3.12]
275
Closed Subschemes of $\proj S$.
276
277
(a) Let $\phi:S\rightarrow T$ be a surjective homomorphism of graded
278
rings, preserving degrees. Show that the open set $U$ of (Ex. 2.14)
279
is equal to $\proj T$, and the morphism $f:\proj T\rightarrow\proj S$
280
is a closed immersion.
281
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(b) If $I\subseteq S$ is a homogeneous ideal, take $T=S/I$ and let $Y$
283
be the closed subscheme of $X=\proj S$ defined as the image of the
284
closed immersion $\proj S/I\rightarrow X$. Show that different homogeneous
285
ideals can give rise to the same closed subscheme. For example,
286
let $d_0$ be an integer, and let $I'=\bigoplus_{d\geq d_0} I_d$. Show
287
that $I$ and $I'$ determine the same closed subscheme.
288
\end{prob}
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\proof
290
(a) Since $\phi$ is graded and surjective, $\phi(S_{+})=T_{+}$ from which
291
it is immediate that $U=\proj T$. By the first isomorphism theorem
292
$T\cong S/\ker(\phi)$ so $f(\proj T)=f(\proj S/\ker(\phi))
293
=V(\ker(\phi))$ which is a closed subset of $\proj S$. (This is
294
just the fact that there is a one to one correspondence between
295
homogeneous ideals of $S/\ker(\phi)$ and homogeneous ideals of $S$
296
which contain $\ker(\phi)$.) The map on the stalk corresponding
297
to a point $x\in\proj T$ is the map
298
$S_{(\phi^{-1}(x))}\rightarrow T_{(x)}$
299
induced by $\phi$. This map is surjective since $\phi$ is surjective.
300
Thus the induced map on sheaves is surjective.
301
302
(b) Let $\phi:S/I'\rightarrow S/I$ be the natural projection homomorphism.
303
(This makes sense because $S/I$ is a quotient of $S/I'$. Indeed,
304
$S/I=(S/I')/\bigoplus_{0\leq d<d_0}I_d$.) Then $\phi$ is a graded homomorphism
305
of graded rings such that $\phi_d$ is the identity for $d\geq d_0$.
306
So by (Exercise 2.14c) $\phi$ induces an isomorphism
307
$f:\proj S/I \rightarrow \proj S/I'$. Since this is a morphism
308
over $\proj S$ (the corresponding triangle of homomorphisms commutes)
309
it follows that $I$ and $I'$ give rise to the same closed subscheme.
310
311
\begin{prob}[3.14]
312
If $X$ is a scheme of finite type over a field, show that
313
the closed points of $X$ are dense. Give an example to show that
314
this is not true for arbitrary schemes.
315
\end{prob}
316
\proof
317
Since $X$ is of finite type over $k$ we can cover $X$ with
318
affine open sets $U_i=\spec A_i$ where each $A_i$ is a finitely
319
generated $k$-algebra. Let $U$ be an open subset of $X$. We must
320
show that $U$ contains a closed point. Since the $U_i$ cover
321
$X$, $U$ must intersect some $U_i$. Then $U\cap U_i$ contains
322
a distinguised open subset of $U_i$. So, to show that every open
323
set contains a closed point, it suffices to show that every
324
nonempty distinguised open subset of each $U_i$ contains
325
a closed points of $X$. Since a distinguished open subset $(U_i)_x$ of
326
a $U_i$ is also the spectrum of a finitely generated $k$-algebra
327
$\spec (A_i)_x$ we can just add it to our collection $\{U_i\}$.
328
The problem thus reduces to showing that each $U_i$ contains
329
a closed point.
330
331
\begin{prop}
332
With the notation as above, if $x\in U_i$ is closed in $U_i$ (here
333
$U_i$ has the subspace topology) then $x$ is closed in $X$.
334
\end{prop}
335
\proof
336
Suppose $x\in U_j$. There is a natural injection
337
$U_i\cap U_j\hookrightarrow U_j$. Let $\spec (B_i)_f$ be
338
a distinguished open subset of $U_i$ contained in $U_i\cap U_j$
339
which contains $x$. Then we have a morphism
340
$\spec(B_i)_f\hookrightarrow U_j=\spec B_j$. We
341
thus get a ring homomorphism $\phi:B_j\rightarrow (B_i)_f$
342
of Jacobson rings. Since it is induced by a restriction of
343
the identity map $X\rightarrow X$ which is a morphism over
344
$k$, $\phi$ is a $k$-algebra homomorphism.
345
Since $(B_i)_f$ is a finitely generated $k$-algebra, $(B_i)_f$ is
346
also a finitely generated $B_j$-algebra. Since $x$ is closed
347
in $\spec(B_i)_f$, $x$ is a maximal ideal of $(B_i)_f$.
348
Thus by page 132 of Eisenbud's {\em Commutative Algebra}
349
$\phi^{-1}(x)$ is a maximal ideal of $B_j$. Thus $x$ is also
350
a closed point of $U_j$ in the subspace topology on $U_j$.
351
Thus $X-x=\cup_{i}(U_i-x)$ is a union of open subsets of $X$,
352
hence open in $X$, so $x$ is closed.
353
354
To finish we just need to know that $U_i$ has a closed point.
355
356
\begin{prop}
357
Let $X=\spec A$ be an affine scheme with $A$ a finitely generated
358
$k$-algebra. Then any nonempty distinguished open subset of $X$
359
contains a closed point.
360
\end{prop}
361
\proof
362
The key observation is that $A$ is a Jacobson algebra since it finitely
363
generated over a field, so by page 131 of Eisenbud's {\em Commutative
364
Algebra} the Jacobson radical of $A$ equals
365
the nilradical of $A$. Let $D(f)$ be a nonempty
366
distinguised open subset of $X$. Then some prime omits $f$
367
so $f$ is not in the nilradical of $A$. Thus $f$ is not
368
in the Jacobson radical of $A$ so there is some maximal
369
ideal $m$ so that $f\notin m$. Then $m \in D(f)$ and
370
$m$ is a closed point of $X$ since $m$ is maximal.
371
372
Strangely enough I never used the hypothesis that $X$ is of
373
finite type over $k$ but just the weeker hypothesis that
374
$X$ is {\em locally} of finite type over $k$. Did I miss something?
375
376
Finally, we present a counterexample in the more general situation. Let
377
$X=\spec Z_{(2)}$. Then $X$ contains precisely one closed
378
point, the ideal $(2)$. So the set of closed points in $X$ is
379
not dense in $X$. In fact, if $X$ is the spectrum of any
380
DVR we also get a counterexample.
381
382
\begin{prob}[4.2]
383
Let $S$ be a scheme, let $X$ be a reduced scheme over $S$, and let
384
$Y$ be a seperated scheme over $S$. Let $f_1$ and $f_2$ be two $S$-morphisms
385
of $X$ to $Y$ which agree on an open dense subset $U$ of $X$. Show that
386
$f_1=f_2$. Give examples to show that this result fails if either (a)
387
$X$ is nonreduced, or (b) $Y$ is nonseparated.
388
\end{prob}
389
\proof
390
Let $g=(f_1,f_2)_S:X\rightarrow Y\times_S Y$ be the product of
391
$f_1$ and $f_2$ over $S$. By hypothesis the diagonal $T=\Delta_Y(Y)$ is
392
a closed subscheme of $Y\times_S Y$. Thus $Z=g^{-1}(T)$ is a closed
393
subscheme of $X$. If $h:U\rightarrow Y$ is the common restriction
394
of $f_1$ and $f_2$ to $U$, then, since $g|_U$ makes
395
the correct diagram commute, the restriction of $g$ to $U$
396
is $g'=(h,h)_S$ and $g'=\Delta_Y \circ h$ since
397
$\Delta_Y\circ h$ makes the correct diagram commute.
398
\diagram
399
Thus $\Delta_Y^{-1}(T)=Y$ implies
400
$$g^{-1}(T)\supseteq (g')^{-1}(T)=h^{-1}(\Delta_Y^{-1}(T))=h^{-1}(Y)=U.$$
401
Thus $g^{-1}(T)$ is a closed set which contains the dense
402
set $U$. Thus $g^{-1}(T)=X$ so $g(X)\subseteq T$.
403
So, by the proposition below, since $X$ is reduced, $g$
404
factors as $g=\Delta_Y\circ f$ where $f:X\rightarrow Y$.
405
From the definition of $\Delta_Y$, we see that
406
$\pi_1\circ \Delta_Y=\id_Y=\pi_2\circ \Delta_Y$.
407
Thus $f_1=\pi_1\circ g=\pi_1\circ\Delta_Y\circ f=f$
408
and $f_2=\pi_2\circ g=\pi_2\circ \Delta_Y\circ f=f$ so
409
$f=f_1=f_2$, as desired.
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\begin{prop}
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Let $X$ be a reduced scheme, $f:X\rightarrow Y$ a morphism, $Z$
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a closed subscheme of $Y$, $j:Z\hookrightarrow Y$, such
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that $f(X)\subseteq j(Z)$. Then $f$ factors uniquely as
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$$X\stackrel{g}{\rightarrow}Z\stackrel{j}{\hookrightarrow}Y.$$
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\end{prop}
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\proof
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First assume $X$ and $Y$ are affine, $X=\spec A$, $Y=\spec B$,
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$Z=\spec B/I$.
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(Use exercise 3.11 to see that every closed subscheme $Z$ of
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$Y$ is of the form $\spec B/I$.)
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Let $\phi:B\rightarrow A$ be the homomorphism which induces $f$.
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Since $f(X)\subseteq Z$, the inverse image of any prime of $A$
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contains $I$. Since $A$ is reduced the intersection of all
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primes of $A$ equals $\{0\}$. Thus
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$$\ker(\phi)=\phi^{-1}(\{0\})=\phi^{-1}(\cap_{\mathrm{primes p}} p)\subseteq I$$
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so $\phi$ factors uniquely through $B/I$.
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$$B\stackrel{j^{\#}}{\rightarrow}B/I\stackrel{g^{\#}}{\hookrightarrow}A$$
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This proves the proposition when $X$ and $Y$ are affine.
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Now suppose $X$ is an arbitrary reduced scheme. Cover $X$ by open affines
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$X=\cup_i U_i$. For each $i$ let $g_i$ be the unique map which factors
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$f|_{U_i}$ through $Z$. By uniqueness $g_i|_{U_i\cap U_j}=g_j|_{U_i\cap U_j}$,
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so we can glue the $g_i$ to obtain a morphism $g:X\rightarrow Z$
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such that $j\circ g=f$. Now suppose both $X$ and $Y$ are arbitrary. Cover $Y$ by open affines,
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take their inverse images in $X$, perform the construction locally
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for each one, use uniqueness and glue.
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\noindent{\em Counterexamples.}
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(a) Let $A=k[x,y]/(x^2,xy)$, let $X=Y=\spec A$ and let $S=\spec k$.
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Then $Y$ is affine hence seperable over $S$, but $X$ is not reduced.
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Let $f:X\rightarrow Y$ be the morphism induced by the identity homomorphism
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$\id: A\rightarrow A$. Let $g:X\rightarrow Y$ be the morphism induced
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by the homomorphism $\phi: A\rightarrow A: x\mapsto 0, y\mapsto y$.
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Let $U=D(y)=\spec A_y$. Then since $A_y\cong \spec k[y,y^{-1}]$,
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the localized homomorphisms agree, $\id_y=\phi_y$. Thus
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$f|_U=g|_U$. Now $X$ is irreducible since $A$ has just one
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minimal prime, namely $(x)$, so $U$ is dense in $X$. But, $f\not=g$.
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since $f^{\#}=\id\not=\phi=g^{\#}$.
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(b) Let $X$ be the affine line and $Y$ the affine line
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with a doubled origin both thought of as schemes over $S=\spec k$.
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Let $f_1:X\rightarrow Y$ be one of the inclusions of the affine
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line in $Y$ and let $f_2:X\rightarrow Y$ be the other one.
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Then $f_1$ and $f_2$ agree on $X$ minus the origin but not on $X$.
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[Reference, EGA, I.8.2.2.1.]
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\begin{prob}[4.4]
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Let $f:X\rightarrow Y$ be a morphism of separated schemes of finite type
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over a noetherian scheme $S$. Let $Z$ be a closed subscheme of $X$ which
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is proper over $S$. Show that $f(Z)$ is closed in $Y$, and that $f(Z)$
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with its image subscheme structure is proper over $S$.
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\end{prob}
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\proof
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First we show that since $X$, $Y$ and $Z$ are of finite type
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over $S$ and $S$ is noetherian, $X$, $Y$ and $Z$ are noetherian.
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Suppose $g:X\rightarrow S$ is the map from $X$ to $S$. Cover
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$S$ by finitely many $\spec A_i$, $A_i$ noetherian. Then for
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each $f^{-1}(\spec A_i)=\cup_j \spec B_{ij}$, with $B_{ij}$
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a finitely generated $A_i$-algebra. Since $A_i$ is noetherian
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each $B_{ij}$ is noetherian (this is the Hilbert basis theorem).
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Since $X=\cup_{ij}\spec B_{ij}$, $X$ is noetherian. One shows
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that $Y$ and $Z$ are noetherian in exactly the same way.
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Since the following diagram commutes 4.8(e) implies
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$f|_Z$ is proper. \diagram
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Thus $f|_Z(Z)$ is closed in $Y$.
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(I'm assuming $Z$ is an $S$-subscheme of $X$ so that
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the diagram must commute.)
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We now have $f(Z)\hookrightarrow Y\rightarrow S$. We must
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show the composition $f(Z)\rightarrow S$ is proper. By Corollary 4.8a the
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closed immersion $f(Z)\hookrightarrow Y$ is proper.
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We are given that the map $Y\rightarrow S$ is seperated
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and of finite type. Since the composition of seperated morphisms
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is seperated and the composition of finite type morphisms is of
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finite type, the morphism $f(Z)\rightarrow S$ is seperated and
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of finite type. The hard part is to show that it is universally closed.
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Since the morphism $Z\rightarrow S$ is proper it is closed
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and from above the morphism $Z\rightarrow f(Z)$ is closed so
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the morphism $f(Z)\rightarrow S$ is closed. Let $W$ be an
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any scheme over $S$. We must show that the map
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$f(Z)\times_S W\rightarrow W$ is closed.
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We have the following diagram.
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\diagram \diagram
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If we can show that $g_1$ is surjective we will be done. For then
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if $A$ is closed subset of $f(Z)\times_S W$,
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$$g_2(A)=g_3(g_1^{-1}(A)),$$
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which, since $g_3$ is closed, is also closed. (We wouldn't have
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equality in the above expression if $g_1$ weren't surjective.)
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In order to establish the surjectivity of $g_1$ we prove that
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the property of being a surjective morphism is preserved under
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base extension. It will then follow, since $g_1$ is a base
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extension of the surjective morphism $Z\rightarrow f(Z)$,
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that $g_1$ is surjective.
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\begin{prop}
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Let $X$ and $Y$ be schemes over $S$. Suppose $x\in X$ and $y\in Y$
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both lie over the same point $s\in S$. Then there exists
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$\alpha\in X\times_S Y$ such that $p_X(\alpha)=x$ and
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$p_Y(\alpha)=y$.
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\end{prop}
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\proof
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Let $g_1:\spec(k(x))\rightarrow X$ and
519
$g_2:\spec(k(y))\rightarrow Y$ be the natural maps
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with $g_1((0))=x$ and $g_2((0))=y$.
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Let $Z=\spec(k(x))\times_{\spec(k(s))}\spec(k(y))
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=\spec(k(x)\otimes_{k(s)}k(y))$, and let $\pi_1$
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be the projection to $\spec(k(x))$, $\pi_2$
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the projection to $\spec(k(y))$.
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Let $g=g_1\times_S g_2$ be the product of $g_1\circ \pi_1$
526
with $g_2\circ \pi_2$. So $g:Z\rightarrow X\times_S Y$.
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See the following diagram.
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\diagram
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Since $Z$ is the spectrum of the tensor product of two
530
fields over a common base field, $Z\not=\emptyset$ so
531
there is some $z\in Z$. By the definition of $g$,
532
$g_1\circ\pi_1=p_x\circ g$ and
533
$g_2\circ\pi_2=p_y\circ g$ so
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$x=g_1\circ\pi_1(z)=p_x\circ g(z)$ and
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$y=g_2\circ\pi_2(z)=p_y\circ g(z)$ so
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we may take $\alpha=g(z)$.
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538
\begin{prop}
539
If $f:X\rightarrow Y$ is a surjective $S$-morphism then
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$f\times 1: X\times_S S' \rightarrow Y\times_S S'$ is surjective.
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\end{prop}
542
\proof
543
We have the following diagram.
544
\diagram
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Let $y'\in Y\times_S S'$. Then
546
$q(y')\in Y=f(X)$ so there is $x\in X$
547
such that $f(x)=q(y')$. Then by the above proposition
548
there is some $\alpha\in X\times_S S'$ such that
549
$p(\alpha)=x$ and $(f\times 1)(\alpha)=y'$. Thus
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$f\times 1$ is surjective.
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\end{document}
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