(v.1.1.5)
2.3: First Properties of Schemes
2.4: Separated and Proper Morphisms
Let $A$ be an abelian group, and define the constant presheaf associated to $A$ on the topological space $X$ to be the presheaf $U \mapsto A$ for all $U \neq \emptyset$, with restriction maps the identity. Show that the constant sheaf $\mathcal{A}$ defined in the text is the sheaf associated to this presheaf.
proof:
There exists $\phi: const_A\to \mathcal{A}$, by defining $a\in A\mapsto (f:U\to A, u\mapsto a)$ the constant map at $a$. We want to show that this is an isomorphism on the stalk for every $p\in X$.
We know that for every $x\in U$, there is a connected $x\in V\subseteq U$, therefore, the restriction of the colimit $\mathrm{colim}_{p\in U}F\cong \mathrm{colim}_{p\in V}F$ for all connected $V$. Apply this to $\phi$, we have that this is the identity on $A$, which is the stalk for both sheaf and presheaf.
Show that $\varphi: \mathcal{F}\to \mathcal{G}$ is surjective iff the following holds: for every open set $U \subseteq X$, and for every $s\in \mathcal{G}(U)$, there is a covering $ { U_i } $ of $U$, and there are elements $t_i \in \mathcal{F}(U_i)$, such that $\varphi(t_i)=s|_{U_i}$.
proof:
($\impliedby$) This shows that $\varphi$ is surjective on the stalk, so $\varphi$ is surjective.
($\implies$) we have that $\varphi$ is surjective on the stalk. Let $s\in \mathcal{G}(U)$, then for every $P \in U$, $s|_P\in \mathcal{G}_P$ has preimages $t|_P\in \mathcal{F}_P$. There is then an open set $U_i \subseteq U$ such that $\varphi(t|_{U_i})=s|_{U_i}$. Do this with all $P$.
Give an example of a surjective morphism of sheaves $\varphi: \mathcal{F}\to \mathcal{G}$ and an open set $U$ such that $\varphi|_U: \mathcal{F}(U)\to \mathcal{G}(U)$ is not surjective.
example:
Let $X=\mathbb{C}$, let $\mathcal{F=O}$ be the space of holomorphic functions and let $\mathcal{G=O}^{*}$ be the space of nonvanishing holomorphic functions, $\varphi(f)=\exp(f)$.
We see that the stalks $\mathcal{O}_P \to \mathcal{O}^{*}_P$ is surjective, as for every nonvanishing $g\in {\mathcal{O}^{*}}_P$, there is a small open disc $U$ such that $g(U)\subset Dom \ \ln$, the holomorphic domain of $\ln$. So this is surjective on stalk, hence surjective.
However, on $U=\mathbb{C}-{0}$, $g(z)=z$ is nonvanishing yet $\ln (z)$ is not holomorphic.
Let $\mathcal{F}$ and $\mathcal{G}$ be sheaves of abelian groups over $X$. For every $U\subseteq X$ show that the set $\Hom(\F|_U,\G|_U)$ of morphisms of the restricted sheaves has a natural structure of abelian group. Show that the presheaf $U\mapsto \Hom(\F|_U,\G|_U)$ is a sheaf. This is the Hom sheaf $\HomSh(\F, \G)$
proof:
For every $U \subseteq X$, define addition on $\psi, \phi\in\Hom(\F|_U,\G|_U)$ to be $(\psi+\phi)(V):=\psi(V)+\phi(V)$. Since the restriction maps are abelian group morphism, this is well defined.
Given $U$ and $V$ open, and $\phi|_U\in \Hom(\F|_U,\G|_U)$ and $\phi|_V\in \Hom(\F|_V,\G|_V)$, such that $\phi|_{U}|_V=\phi|_V|_U$. We define $\phi\in \Hom(\F|_{U\cup V},\G|_{U\cup V})$ to be for $W\subseteq U\cup V$ as following \[\phi(W): \F(W)\to \G(W)\ a\mapsto \phi|_{U\cap W}(a) \cup \phi|_{V\cap W}(a)\]
Here $\phi|_{U\cap W}(a) \cup \phi|_{V\cap W}(a)$ is the section glued from $\phi|_{U\cap W}(a)\in \G(U\cap W)$ and $\phi|_{V\cap W}(a)\in \G(V\cap W)$, which is doable because $\phi|_U|_V=\phi|_V|_U$.
Since restriction and gluing commutes, this defines us a morphism of sheaf $\phi: \F(U\cup V)\to \G(U\cup V)$. It is easy to see that the gluing of patches of morphism gives the original morphism.
Let $X$ be some variety over $k$ closed, $\Oc_X$ is the sheaf of regular functions
Let $Y$ be a closed subset of $X$. For each open set $U \subseteq X$, let $I_Y(U)$ be the ideal in the ring $\Oc_X(U)$ consisting of those regular functions which vanish on $Y$. Show $I_Y$ is a sheaf, the ideal sheaf
proof:
Suppose $f|_U\in I_Y(U)$ and $f|_V\in I_Y(V)$ that agrees on $U\cap V$, let $f\in \Oc_X(U\cup V)$ be the glued section, since $f(Y\cap U)=0$ and $f(Y\cap V)=0$, then $f(Y\cap (U\cup V))=0$, so $f\in I_Y(U\cup V)$. The uniqueness follows easily.
If $Y$ is a subvariety, then the quotient sheaf $\Oc_X/I_Y$ is isomorphic to $i_*\Oc_Y$, where $i: Y\to X$ is the inclusion.
proof:
For $U\subseteq X$, $i_{*}\Oc_Y(U)=\Oc_Y(i^{-1}(U))=\Oc_{Y}(Y\cap U)$.
There is map \[ \varphi: \Oc_X\to i_{*}\Oc_Y \] induced from precomposition of the inclusion, we see that $f\in \ker \varphi(U)$ iff $f$ vanishes on $Y\cap U$, so $f\in I_Y$. Since kernels are determined on the opensets, we have $\varphi$ determines the isomorphism we claimed.
Let $X=\Proj^1$ and let $Y= { P,Q } $ for $P\neq Q\in X$. There is then $\F=i_{*}\Oc_P\oplus i_{*}\Oc_Q$ and \[ 0\to I_Y\to \Oc_X\to \F\to 0 \] is exact. Show, however, the section map $\Gamma(X, \Oc_X)\to\Gamma(X, \F)$ is not surjective. (Leaves room for derived functors)
proof:
$\Gamma(X, \Oc_X)\cong k$, the constant functions. Yet $\Gamma(X, \F)\cong\Oc_P\oplus \Oc_Q$, clearly not surjective.
Let $X=\Proj^1$ and let $\G$ be the constant sheaf associated to the function field $K(X)$. Show that there is an injection $\Oc_X\to \G$. Show that the quotient sheaf $\G/ \Oc_X$ is isomorphic to the direct sum of sheaves $\bigoplus_{P\in X}i_P(I_P)$ where $I_P:=K(X)/\Oc_P$ and $i_P$ defines the skyscraper at $P$.
proof:
Define $\varphi: \Oc_X\to \G$ on $U$ to be \[ \varphi|_U: \Oc_X(U)\to \G(U) \ f\mapsto const_f \] The constant function at $f$. This is obviously a sheaf morphism and injective.
There is also $\psi:\G\to \bigoplus_{P\in X}i_P(I_P)$ defined on connected $U$ $\psi(U)$ as $f\in K(X)\mapsto \sum_{P\in U} f \mod{\Oc_P}$. This makes sense because $f\mod \Oc_P=0$ iff $f\in \Oc_P$ a regular function at $P$, so $f$ is nonzero on $I_P$ iff it has a pole at $P$. But there are only finitely many poles to a rational function on $\Proj^1$, hence this makes sense. On the stalk, this is $f\in K(X)\mapsto f\mod \Oc_P$. So it is surjective. There is sequence \[ 0\to \Oc_X\to \G\to \bigoplus_{P\in X}i_P(I_P)\to 0\]
The $\ker \psi_P$ is the just $\Oc_P$, so the sequence is exact on stalks. Hence the sequence is exact.
Let $A$ be a ring, let $X = \spec A$, let $f \in A$ and let$ D(f) \subseteq X$ be the open complement of $V( (f))$. Show that the locally ringed space$ (D(f), \Oc_X|_{D(f)}) $ is isomorphic to $\spec (A)_{(f)}$ .
proof:
First $D(f)$ is the set of prime ideals not containing $(f)$.
There is inclusion of algebra $i:A\to A_f$. We have for $p\in \spec (A_f)$, $i^{-1}(p)$ is prime, and it does not contain $f$, since $f$ is an unit in $A_f$. So we have a continuous map \[ i:\spec A_f\to D(f) \]
We can also define inverse map that takes $p\mapsto pA_f$, since $A_f/pA_f\cong A/p$ so this is prime. And continuous since if we have $g/f^n\in A_f$ and $V(g/f^n)\subseteq \spec (A_f)$ is closed, then the preimage would be $V(g)$.
Not hard to see that $\spec(A_f)\cong D(f)$ under the maps we defined. There is then morphism \[ \varphi:\Oc_{X}|_{D(f)}\to i_{*}\Oc_{\spec(A_f)} \] And it is isomorphic on the stalk.
Let $(X,\Oc_X)$ be a scheme, and let $U \subseteq X $be any open subset. Show that $(U, \Oc_X|_U)$ is a scheme. We call this the induced scheme structure on the open set $U$, and we refer to $(U, \Oc_X|_U)$ as an open subscheme of $X$.
proof:
Suppose we have $p \in U \subseteq X$, there is an affine open $\spec(A)$ containing $p$ in $X$. Choose $f\in A$ so that $x\in D(f)\subseteq U\cap \spec(A)$, which always exist since $U\cap \spec(A)$ is open in $\spec(A)$. Therefore, we have an affine neighbourhood of $p$ in $U$, that is $D(f)$. Do this with all $p\in U$.
$(X, \Oc_X)$ is reduced if for every $U\subseteq X$, $\Oc_X(U)$ has no nilpotents.
Show that this is equivalent to every stalk has no nilpotents
proof:
$\implies$ Let $a\in \Oc_{X,p}$, we know $a=[f, U]$ for some $U$ and some $f\in \Oc_X(U)$ and $(g,V)\in [f,U]$ iff $g|_{U\cap V}=f|_{U\cap V}$.
Suppose $a$ is nilpotent, there is $a^n=0$ for some $n$. So $f^n|_{U\cap V}=0$ for some neighbourhood $V$ of $x$. $U\cap V$ is open, so $f$ cant be nilpotent, a contradiction.
$\impliedby$ Let $f\in \Oc_X(U)$, we know that $f^n|_p=(f|_p)^n$, since $(f|_p)^n\neq0$, $f^n\neq 0$.
For scheme $(X,\Oc_X)$ define $X_{red}:=(X, \Oc_X^{red})$ to be the locally ring spaced where $\Oc_X^{red}(U)$ is the reduction of $\Oc_X(U)$ (quotient of ideal of nilpotnets). Show that $X_{red}$ is a scheme and there is a morphism $i:X_{red}\to X$ that is a homeomorphism on the topological space.
proof:
We show first $\spec(B)_{red}\cong\spec(B^{red})$.
There is a map $\psi:\spec(B^{red})\to \spec(B)$ from the quotient $q:B\to B^{red}$. Since $0\in p$ for every $p\in \spec(B)$, nilpotents $I\subseteq p$ for every $p$. So the map is a bijection. It is continuous because if we have closed set $V(f)\subseteq \spec(B^{red})$. We have then $\psi(V(f))$ is the set of $q^{-1}(p)$ for $f\in p$. As it turns out, $\psi(V(f))=V(f+I)$, since if $f+I\subseteq p’$, then $f\in q(p)$, and $q^{-1}q(p)=p$.
Since the localization alway from some set commute with quotient, we see that $\spec(B^{red})\cong \spec(B)_{red}$.
The rest is then trivial, finding a cover $\spec(A_i)$ of $X$, we can easily show that $\spec(A_i^{red})$ covers $X_{red}$
Shows that $f:X\to Y$ for $X$ reduced factors uniquely through $Y_{red}\to Y$
proof:
The underlying continuous map is easily determined uniquely.
The map $\varphi: \Oc_Y\to f_{*}\Oc_X$, is a sheaf map over $Y$. We know that ring maps takes nilpotents to nilpotents, so $\varphi$ factors through $\Oc_Y^{red}=i_{*}\Oc_{Y_{red}}$ uniquely.
For $f:X\to \spec(A)$, there is $f^{\sharp}: \Oc_{\spec(A)}\to f_{*}\Oc_X$, taking the top section we have \[ \alpha: \Hom_{\Sch}(X,\spec(A))\to \Hom_{\Ring}(A, \Gamma(X, \Oc_X)) \] Show that this is an isomorphism.
proof:
We start by construction a function $\beta$ of the other direction.
For a map $f^{\sharp}: A\to \Gamma(X, \Oc_X)$, for every $p\in X$, there is a map \[ f^{\sharp}_p: A\to \Gamma(X, \Oc_X)\to \Oc_{X,p}/m_{p}\],
here $m_{p}$ is the maximal ideal of $\Oc_{X,p}$
The kernel of this map is prime, since the codomain is a field. So we have a function $g:X\to \spec(A)$ that takes $p$ to $\ker f^{\sharp}_p$. Is it continuous?
For $r\in A$, $D(r)$ is the set of prime ideals not containing $r$, $g(p)\in D(r)$, if $\ker f^{\sharp}_p$ not contains $r$. So $f^{\sharp}_p(f)\notin m_p$. Since $X$ is a scheme, $p$ is contained in some affine open $\spec(A)$, $f^{\sharp}_p(r)\notin m_p$ is equivalent to $f^{\sharp}(r)\mod p\neq 0$, or $f^{\sharp} (r)\notin p$ over $\spec(A)$. Internal to $\spec(A)$, we have $g^{-1}(D(r))\cap\spec(A)=D_{\spec(A)}(f^{\sharp}(r))$. So we have that $g$ is continuous. We see that $g^{-1}(D(r))$ is the nonvanishing open set of $f^{\sharp}(r)$ in $X$ (make sense if we think $f^{\sharp}$ as precomposition.)
We define $g^{\sharp}: \Oc_{\spec(A)}\to g_{*}\Oc_X$ that takes $r’/r^n\in A_{r}$ to $f^{\sharp}(r’)/f^{\sharp}(r)^n\in\Oc_X(U)$, here $U$ is the nonvanishing open set of $f^{\sharp}(r)$ so it makes sense. We denote the construction $\beta(f^{\sharp})$.
Obviously $\alpha\circ \beta=id$. We show that this is true in the other direction.
Assume for now $X=\spec(B)$. A map $f: \spec(B) \to \spec(A)$, is induced by its top section $f^{\sharp}:A\to B$, from $(f^{\sharp})^{-1}$ on the primes ideals. This is equivalent to the kernel of the map \[ A\to B\to B_{(p)}\to B_{(p)}/m_p \] As preimage of $m_p$ in $B$ is $p$.
So for affine case, the continuous map we constructed agrees. Now for a general scheme $X$, there is an affine cover $\spec(A_i)$. Since stalk is determined locally, we easily deduce that the continuous map $\beta\circ \alpha(f)$ agrees with that of $f$.
The case of sheaf map can also be deduced from the affine case.
For $f:\spec(B)\to \spec(A)$ associated to ring map $\varphi:A\to B$, $f^{\sharp}: \Oc_\spec(A)\to f_{*}\Oc_{\spec(B)}$ on standard opens $D(r)$ of $\spec(A)$ is just
\[ A_r \to B_{\varphi(r)}\]
Where $r’/r^n$ are map to $\varphi(r’)/\varphi(r)^n$, the map we have constructed.
The general scheme case follows.
If $X$ is a topological space, and $Z $ an irreducible closed subset of $X$, a generic point for $Z$ is a point $\xi$ such that $Z=\overline{ { \xi } }$. If $X$ is a scheme, show that every (nonempty) irreducible closed subset has a unique generic point.
proof:
First of all, schemes are $T_0$. It suffices to show this for affines. For $p\neq q\in \spec(A)$, there is $f\in p$ and not in $q$. So $q\in D(f)$ and $p\notin D(f)$.
Let $\spec(A)$ be an affine open of $Z$. There is an ideal $I=\sqrt 0$ such that $V(I)=\spec(A)$. Since $\spec(A)$ is irreducible, $I$ is prime, since for $f, g\in A$ such that $fg\in I$, we have \[ V(I)\subseteq V(fg)=V(f)\cup V(g)\] So $f\in I$ or $g\in I$.
Therefore, $I$ is a generic point of $\spec(A)$. Since $Z$ is irreducible and $\spec(A)$ is open in $Z$, so $I$ is then a generic point of $Z$.
If there are two generic point $\xi$ and $\xi’$, we have $\xi\in \overline{ { \xi’ }}$ and $\xi’\in \overline{{\xi } }$. So any open neighbourhood of one contains the other. Since schemes are $T_0$, this implies $\xi=\xi’$.
Let$ f: X\to Y$ be a morphism of schemes, and suppose that $Y$ can be covered by open subsets $U_i$ such that for each $i$, the induced map $f^{-1}(U_i)\to U_i$ isomorphism. Then $f$ is an isomorphism.
proof:
We have that $f$ is a homeomorphism on the underlying space. Since $f_i^{\sharp}: \Oc_Y|_{U_i}\to f_{*}\Oc_X|_{f^{-1}U_i}$ is an isomorphism of sheaves and we have that any open sets $V$ in $Y$ are unions of opens of $U_i$ similarly $f^{-1}(V)$ would be the union of those opens of $f^{-1}(U_i)$. The gluing property tells us that the overall sheaves are isomorphic.
A scheme $X$ is affine if and only if there is a finite set of elements $f_i \in \Gamma(X, \Oc_X)$ such that the open subsets $X_{f_i}$ are affine and $f_i$ generate $1\in A$.
proof:
($\implies$) If $X$ is affine, since $X$ is quasicompact, then there are finite $f_i$ that generate $i$, and $X_{f_i}=D(f_i)$.
($\impliedby$) Denote $A=\Gamma(X,\Oc_X)$, we want to show $X\cong \spec(A)$. There is a map $\varphi: X\to \spec(A)$ that is induced from the identity in $\beta:\Hom_{\Ring}(A,\Gamma(X,\Oc_X))\to \Hom_{\Sch}(X, \spec(A))$.
We first show that $X_{f_i}\cong D(f_i)$, which is natural as $X_{f_i}$ are affine, so top section $\Gamma(X_{f_i}, \Oc_{X_{f_i}})\cong A_{f_i}$ determines $X_{f_i}$ entirely.
Since $f_i$ generate $1$, then $D(f_i)$ covers $A_{f_i}$, and $\varphi^{-1}(D(f_i))=X_{f_i}\to D(f_i)$ are isomorphisms. By the previous exercise, we have $X\cong \spec(A)$.
Show that a morphism $f:X\to Y$ is finite if and only if for every open affine subset $V = \spec B$ of $Y$, $f^{-1}(V)$ is affine, equal to $\spec(A)$, where $A$ is a finite $B$-module.
proof:
($\impliedby$) Ah Doiii ($\implies$) Let $U_i$ be an affine cover of $Y$ such that $f^{-1}(U_i)$ are covered by affine $V_{ij}$ that are finite.
For any affine open $U$ in $Y$, $U$ is covered by standard opens $D(r_{ik})$ for $r_{ik}\in \Gamma(U_i)$. Since $U$ is quasicompact, we can choose finitely many $r_{ik}$. So we have that $r_{ik}\in \Gamma(U)$ generate $1$, and hence $f^{\sharp}(r_{ik})$ generate $1$ in $\Gamma(f^{-1}(U))$. Ex.2.2.17 then tells us that $f^{-1}(U)$ is affine.
Since $V_{ij}$ covers $f^{-1}(U)$ and all of them are finite algebra, then so is $\Gamma(f^{-1}(U))$.
Let $X$ be an integral scheme. Show that the local ring $\Oc_{\xi}$ of the generic point $\xi$ of $X$ is a field. It is called the function field of $X$ and is denoted by $K(X)$. Show also that if $U = \spec (A)$ is any open affine subset of $X$, then $K(X) $is isomorphic to the quotient field of $A$.
proof:
If $X$ is an integral scheme, take an affine cover $\spec(A_i)$ of $X$. Since for every open $U$, $\Oc_X(U)$ is integral domain, $A_i$ are integral domain. The unique generic point of $X$ is the generic point $(0)$ of $\spec(A_i)$. Since stalk are locally determined, we have $\Oc_{X,(0)}=Q(A_i)$. Here $Q(A_i)$ is the quotient field of $A_i$ for some $i$.
Let $k$ be a field, $\A^1_k:=\spec(k[x])$, show that $\A^1_k\times_{\spec(k)}\A^1_k\cong \A^2_k$ and show that the underlying point set of the product is not the product of the underlying point sets of the factors (even if $k$ is algebraically closed).
proof:
First of all, $\A^1_k\times_{\spec(k)}\A^1_k\cong \spec(k[x]\otimes_kk[y])\cong \spec(k[x,y])=\A^2_k$.
Since we claim that for even closed $k$, this is not the product set. So let $k=\mathbb{C}$. There are points $ { p \in \mathbb{C} } $ and generic point $\xi$ for the underlying set of $\A^1_k$. The product is $(p,q)\in \C^2$, $(\xi, q)$, $(p,\xi)$, and $(\xi,\xi)$. But $\A^2_k$ has $(p,q)$ and only one generic point $\eta$.
Describe $\spec(k(s))\times_{\spec(k)}\spec(k(t))$.
description:
$\spec(k(s))\times_{\spec(k)}\spec(k(t))=\spec(k(s)\otimes_kk(t))$. This is not a field, since $s\otimes 1-1\otimes t$ is not invertible. It is in general not a one point space. It is a integral scheme.
A closed immersion is a morphism of finite type.
proof:
Let $f:Y \to X$ be closed, choose affine cover $\spec(A_i)$ of $X$. Since closed immersion is stable under pullback, we have $f^{-1}(\spec(A_i))$ is a closed immersion of $\spec(A_i)$. Hence there is ideal $I$ of $A_i$ such that $f^{-1}(\spec(A_i))\cong \spec(A_i/I)$. $A_i/I$ is finitely generated $A_i$-algebra. Hence $f$ is of finite type.
A quasi-compact open immersion is of finite type.
proof:
We simply identify an open immersion with an open set $U$ of $X$ that is a scheme. Quasicompact implies for all open affine $\spec(A)$ of $X$, $U\cap \spec(A)$ are quasicompact.
$U\cap \spec(A)$ is quasicompact implies that we just need to find an affine cover of , $U\cap \spec(A)$ so it is induced from a finitely generated $A$-algebra. We can cover $U$ by something of the form $D(f_i)$ of $\spec(A)$, and $A\to A_{f_i}$ is finitely generated.
A composition of two morphisms of finite type is of finite type.
proof:
If $C$ is a finitely generated $B$-algebra, and $B$ is finitely generated $A$-algebra, then $C$ is $A$-finitely generated. Finiteness of cover is simple.
Morphisms of finite type are stable under base extension.
proof:
Let $f:Y\to X$ be finite, and $g:X’\to X$ be any map. For open affine $\spec(B)$ of $X’$, let $\spec(A_i)$ be some affine cover of $g(\spec(B))$. It suffices to assume that $\spec(B)$ is contained entirely in one of $A$. We have that $f^{-1}(\spec(A))$ are covered by finitely many fin.gen. $A$-algebra $R$. The preimage of $\spec(B)$ in $Y\times_XX’$ then is covered by $R\otimes_{A} B$, which is $B$ fin.gen.
If $X$ and $Y $ are schemes of finite type over $S$, then $ X \times_S Y$ is of finite type over $S$.
proof:
For affine $\spec(A)$ in $S$, there are $\spec(B_i)$ and $\spec(C_j)$ in the $X$ and $Y$ resp. and the preimage in $X\times_S Y$ is then $\spec(B_i\otimes_A C_j)$. All are fin.gen. $A$-alg and there are finitely many of them.
Let $X$ be an integral scheme of finite type over a field $k$ (not necessarily algebraically closed).
For any closed point $P\in X$, $\dim(X)=\dim(\Oc_{X,p})$ , where for rings, we always mean the Krull dimension.
proof:
Let $X\to k$ be an integral scheme of finite type over field $k$, there is then a finite affine cover $\spec(A_i)$ of $X$, such that each of $A_i$ is a finitely generated $k$-algebra, hence there is a surjective $k[X_j]\to A_i$. Moreover, since $X$ is integral, then $A_i$ are integral domain, hence $A_i\cong k[X_j]/I_i$ for some prime ideal $I_i$ of $k[X_j]$. Note that a closed point $p\in X$ is closed in the cover it belongs.
We first show the statement for affine $\spec(B)$.
First of all, by Example 2.3.2.7, \[\dim(\spec(B))=\dim (B)\]
Suppose, $p\in \spec(B)$ is a closed point, so it is a maximal ideal, then it is a point belonging to the affine variety of $I$ in $\A^{m}_k$, so by Theorem 1.3.2 c), we have that \[ \dim(B)=\dim(\Oc_{\spec(B), p)}\]
By Ex. 1.1.10(b), we have that $\dim(X)=\sup_i\dim (\spec(A_i))$. We will show that $\dim(\spec(A_i))=\dim (\spec(A_j))$. Since $X$ is integral, it is irreducible, therefore, it is connected. So any two cover is connected through a chain of overlapping open cover $\spec(A_i)$. Find a closed point $p$ on the overlap, the local ring of $p$ is determined completely by $p$, so we have \[ \dim(\spec(A_i))= \dim(\Oc_{\spec(A_i), p})\cong \dim(\Oc_{\spec(A_j), p})=\dim(\spec(A_j)) \]
Let $K(X)$ be the function field of $X$, then $\dim(X) =\mathrm{tr.d} K(X)/k$ (transcendental degree over $k$)
proof:
We know for an integral scheme, the unique generic point comes from any one of the affine cover, so we just need to show that the statement is true for affine cases. This is just Theorem 1.3.2d).
If $Y$ is a closed subset of $X$, then $\mathrm{codim}(Y,X) = \inf { \dim \Oc_{X,p}| \ p \in Y } $.
proof:
We show that this is true first for irreducible closed set $Z$ of affine $\spec(A)$.
Over the affine $\spec(A)$, we have that if $Z$ is a irreducible closed set of $\spec(A)$, then by Hilbert’s Nullstellensatz, $Z=\spec(A/I)$ for some prime ideal $I$. The codimension of $\spec(A/I)$ in $\spec(A)$ is then the supremum of the strictly increasing prime ideals containing $I$. Now for any $p\in \spec(A/I)$, $p$ contains $I$, $ \Oc_{X,p}=\Oc_{\spec(A),p}$, so for any prime ideals chain in $\Oc_{\spec(A),p}$, there is a corresponding prime ideals chain containing $I$.
The generic point of $\spec(A/I)$ is $I$, so we have that $\inf {\dim \Oc_{\spec(A),p}| \ p \in \spec(A/I) ight}=\dim (\Oc_{\spec(A),\eta_I})$, here $\eta_I$ is the inclusion of the generic point of $\spec(A/I)$ in $\spec(A)$, and the definition is precisely $\mathrm{codim}(\spec(A/I), \spec(A))$.
Suppose $Z$ is irreducible closed of $X$, then first of all $Z$ has a generic point $\eta$. Since $Z$ is connected, find a subfamily of cover of $X$ that covers $Z$, the locallity property that we have shown for codimension implies, that \[ \mathrm{codim}(Z, X)=\dim (\Oc_{X, \eta}) \]
The generalization is straight forward by looking at the generic point for all irreducible $Z\subseteq Y$ and their generic points.
If $Y$ is a closed subset of $X$, then $\dim (Y) + \mathrm{codim}(Y,X) = \dim (X)$
proof:
This once again comes from looking at irreducible closed set of affine. The first term is the chains of primes contained in $I$, and the second is the chain of primes containing $I$. Generalization is easy.
If $U$ is a nonempty open subset of $X$, then $\dim (U) = \dim (X)$.
proof: See a).
If $k\subseteq k’$ is a field extension, then every irreducible component of $X’= X \times_k k’$ has dimension $= \dim X$.
proof:
We just need to show this on affine open, for which the pullback is $\spec(B\otimes_{k}k’)$. Since $k$ and $k’$ are both fields, this does not changes the dimension.
Show that finite morphism are proper.
proof:
Let $R\to k$ be a valuative ring and its quotient field. There are two point $m, (0)\in \spec(R)$, $m$ for the closed point and $(0)$ for the generic point. If we have $g:\spec(R)\to X$, any open $U$ containing $g(m)$ would must contain $g((0))$, since $g^{-1}(U)$ is open and the smallest open in $\spec(R)$ containing $m$ is itself.
Since $f$ is finite, Ex. 2.3.4 shows that $f^{-1}(U)$ for any affine $U$ is affine. So we can work completely in affine case.
We want to use Theorem 4.7 for properness, in affine case is equivalent to given $f:A\to B$, $i:R\to k$, $\varphi:A\to R$, and $\psi: B\to k$, there is an unique $\phi: B\to R$.
The uniqueness follows from the fact that any $\phi$ would agree with $\psi$ after composing with $i$, but $i$ is injective.
The existence follows from the fact that $B$ is finite over $A$, so for every $b\in B$, there is a monic polynomial $p(x)$ over $A$ such that $p(b)=0$. If $x:=\psi (b)\notin R$, then we divide $\varphi(p(x))$ by $x^{n-1}$, where $n$ is the degree of $p(x)$. Since $x^{-1}\in R$ (as for a valuative ring $R$, for any $x\neq 0\in k$, $x\in k $ or $x^{-1}\in k$), we have that
\[ \frac{p(x)}{x^{n-1}}=x+\varphi(a_{n-1})+\varphi(a_{n-1})x^{-1}+…+\varphi(a_0)x^{-n+1}=0 \] Since all of $\varphi(a_i)$ and $x^{-1}$ are in $R$, then $x\in R$ as well.
Therefore, there exists an unique lift $\phi:B\to R$.
Let $S$ be a scheme, let $X$ be a reduced scheme over $S$, and let $Y$ be a separated scheme over $S$. Let $f$ and $g$ be two $S$-morphisms of $X$ to $Y$ which agree on an open dense subset of $X$. Show that $f = g$. Give examples to show that this result fails if either (a) $X$ is nonreduced, or (b) $Y$ is nonseparated
proof:
Denote $U\subset X$ the open dense subset $f$ and $g$ agree on.
Define $(f,g): X\to Y\times_S Y$, and we have that $(f,g)|_U: U\to Y\times_S Y$ factors through $Y$ by assumption.
Since $Y\subset Y\times_SY$ is closed ($Y$ is separated), we have that $(f,g)^{-1}(Y)$ is a closed set, and contains $U$. Therefore, the preimage must be $X$, hence $f=g$ on the continuous map.
Of course, this does not implies that the sheaf map would agree. There is map $f^{\sharp}-g^{\sharp}:\Oc_{Y}\to f_{*}\Oc_{X}=g_{*}\Oc_{X}$. Choose $s\in \Oc_{Y}(V)$ for some open $V\subseteq Y$, we have that $(f^{\sharp}-g^{\sharp})(s)|_{f^{-1}(V)\cap U}=0$. Does this imply that $(f^{\sharp}-g^{\sharp})(s)=0\in \Oc_X(f^{-1}(V))$.
We will show that for $X$ is reduced and $U$ is dense in $X$, then if for $s\in \Oc_X(V)$ such that $s|_{U\cap V}=0$, then $s=0$. We show this for the affine case.
Let $V=\spec(A)$, it is reduced so $A$ is a reduced ring. Since $U\cap V$ is open in $V$, there exists some standard open $D(r)\subseteq U\cap V$, for $r\in A$, $s|_{D(r)}=0$, means that there is $n$ such that $r^ns=0\in A$. However, $A$ is reduced so this can only mean that $a=0$.
Let $X=k[x]/(x^2)$, let $U=X$, and let $Y=\A^1_k$. Then the maps are determined entirely by $k[x]\to k[x]/(x^2)$. Choose $f$ determined by $x\mapsto 0$, and $g$ determined by $x\mapsto x$. They induce the same continuous map, as there is only one point $(x)$ in $X$, and its preimage respect to both maps are $(x)$. Yet they are not the same scheme map
Let $X=\A^1_k$ and let $Y$ to be the line with two origin, so that is two $\A^1_k$ glued together along $U:=\A^1_k -0$. $U$ is dense in $X$ and in $Y$, we can define $f,g: X\to Y$ to be the inclusion of the two different line. They agree on $U$ but are not the same.
Let $X$ be a separated scheme over an affine scheme $S$. Let $U$ and $V$ be open affine subsets of $X$. Then $U \cap V$ is also affine. Give an example to show that this fails if $X$ is not separated.
proof:
Suppose $S=\spec(A)$, $U=\spec(B)$ and $V=\spec(C)$.
The diagonal $\Delta: X\to X\times_S X$ is closed, and we have map $U\times_S V\cong \spec(B\otimes_A C)\to X\times_S X$
The preimage $\Delta^{-1}(U\times_S V)$ on $X$ is precisely $U\cap V$. Recall that closed immersion are affine (Ex. 3.11b)) so $U\cap V$ is affine.
counterexample:
Let $X$ be the affine plane with two origin, then choose $U$ to be one of the affine plane and $V$ the other one. Their intersection is $\A^2_k-0$, which is not affine.
Let $X$ be an integral scheme of finite type over a field $k$, having function field $K$. We say that a valuation of $K/k$ has center $x$ on $X$ if its valuation ring $R$ ($v(r)\geq 0$), dominates the local ring $\Oc_{X,x}$
If $X$ is separated over $k$, then the center of any valuation of $K/k$ on $X$ (if it exists) is unique.
proof:
Given a valuation $v$ on $K$, there is then a valuation ring $R\to K$. There is then a commutative diagram containing $\spec(K)\to \spec(R)$, $\spec(K)\to X$ into the generic point, and over $\spec(k)$. Since $X$ is separated over $k$, there is at most one lifting $\spec(R)\to X$ by Theorem 4.3. Moreover, we have by Lemma 4.4, this is equivalent of finding a point $x\in X$ such that $R$ dominates $\Oc_{X,x}$, we are done.
If $X$ is proper over $k$, then every valuation of $K/k$ has a unique center on $X$.
proof:
See a).
Prove the converses of (a) and (b).
proof:
Let $P\to F$ be any valuative ring that forms a commutative diagram with $X$ over $k$. Let $x\in X$ be the point of $\spec(F)$
There is the $\kappa(x)\cong K(X)\to F$, and we have that $P\cap K(X)$ is then a valuative ring of $K(X)$.
As our argument would be implied from showing the affine case, we just need to show that if there is $R\to K(R)\to F\xleftarrow P$ over $k$, and a lift $R\to P$. Then it must factors through $K(R)\cap P$. Then our statement would follow
If $X$ is proper over $k$, and if $k$ is algebraically closed, show that $\Gamma(X,\Oc_X) = k$.
proof:
Let $a\in\Gamma(X,\Oc_X)$ such that $a\notin k$. $K(X)$ is a field containing $k$, since $k$ is closed, so we have that $a$ is transcendental over $k$. Therefore, we can choose a $R$ that is a valuation ring of $K(X)$ with maximal ideal of $R$ containing $a^{-1}$, so $a\notin R$.
We see there is $\spec(R)\xleftarrow{} \spec(K(X))\to X$ over $k$. Properness tells us there is a lift $\spec(R)\to X$, hence a map $\Gamma(X,\Oc_X)\to R$. This implies $a\in R$, a contradiction.
Let $f: X\to Y$ be a proper morphism of affine varieties over $k$. Then $f$ is a finite morphism.
proof:
Assume that $X=\spec(B)$ and $Y=\spec(A)$. We have $K(X)=Q(B)$. The map are $B\to Q(B)$ and $R\to Q(B)$ for any valuative ring $R$ is injective. Let $P:=\mathrm{im}(A)$ the subring that is the image of $A$ in $Q(B)$. Since $f$ is proper, there exists an unique lift $B\to R$. This is injective, hence $B$ is a subring of $R$. We do this for all possible $R$ containing $P$, we then have by Theorem 3.4.11A, $B$ is integral over $P$, hence over $A$. Moreover, since $B$ is finitely generated over $A$ as $f$ is proper, then $B$ is a finite $A$-algebra.
Let $R$ be a discrete valuation ring with quotient field $K$, and let $X = \spec R$.
To give an $\Oc_X$-module is equivalent to giving an $R$-module $M$, a $K$-vector space $L$, and a homomorphism $p:M \otimes_RK\to L$.
proof:
Let $\F$ be a $\Oc_X$-module.
$X$ has three open set, $X$, $\emptyset$ and $(0)$. Of course, a $\F$ is a sheaf, so $\F(\emptyset)=0$ for any $\F$.
Denote $M=\F(X)$, by definition this is a $\Oc_X(X)=R$-module. Denote $L=\F((0))$, this is a $\Oc_X((0))=R_{(0)}=K$-module.
Of course, there is the restriction map, and by definition, it is a module map \[ p: \F(X)\otimes_{\Oc_X(X)}\Oc_X((0))=M\otimes_R K\to \F((0))= L \]
That $\Oc_X$-module is quasi-coherent if and only if $p$ is an isomorphism.
proof:
By definition, there exists an affine open cover, such that $\F|_{U_i}\cong \tilde{M}$. However, the only open cover of $X$ is just itself. So we have
\[ \F\cong \tilde{M} \] Where the restriction is then localization \[ p:M\otimes_{R}K\to M_{(0)} \] Which is an isomorphism.
Let $f: X \to Y$ be a morphism of schemes.
Show by example that if $\F$ is coherent on $X$, then $f_{*}\F$ need not be coherent on $Y$, even if $X$ and $Y$ are varieties over a field $k$.
example:
Let $X=\mathbb{G}_m$ and $Y=\A^1_k$, and let $f$ be the inclusion. Let $\F=\Oc_X$, which is coherent. Then $f_{*}\Oc_{X}=k[x,x^{-1}]$. This is not finitely generated as $k[x]$-module.
Show that a closed immersion is a finite morphism
proof:
Let $U=\spec(A)$ be an affine open of $Y$, then $f^{-1}(U)\cong \spec(A/I)$ for some ideal $I$ of $A$, which is obviously a finitely generated $A$-module.
If $f$ is a finite morphism of noetherian schemes, and if $\F$ is coherent on $X$, then $f_{*}\F$ is coherent over $Y$.
proof:
We show this for $Y=\spec(A)$ affine, the general case follows easily.
There is an affine cover of $X$, $\spec(B_i)$ such that each $B_i$ is a finite $A$-algebra.
By definition, for any open $U$ of $Y$, $f_{*}\F(U)=\F(f^{-1}(U))$ by definition. Over $U_i$, $f_{*}\F(U)$ is then a finitely generated $B_i$-module. Since $B_i$ is finite over $A$, then $f_{*}\F(U)$ is a finitely generated $A$-module.
Let $A$ be a ring, let $M$ be an $A$-module, let $X = \spec A$, and let $\F = \tilde{M}$. For any $m\in M = \Gamma(X, \F)$, show that $\mathrm{supp}(m)=V(\mathrm{Ann}_A(m))$, where $\mathrm{Ann}_A(m)$ is the annihilator of $m = {a \in A: am = 0}$.
proof:
We have that by definition $p\notin \mathrm{supp}(m)$ iff $m= 0\in \F_{p}$. Since $\F\cong \tilde{M}$, then $\F_{p}\cong M_{p}$, and $m=0\in M_{p}$. Therefore, for some $a\notin p$, there is $n$ such that $a^nm=0$, hence $a^n\in \mathrm{Ann}_A(m)$. Since $p$ is prime, if $a^n\in p$, then $a\in p$, which is not the case, hence $\mathrm{Ann}_A(m)\not \subseteq p$. Therefore, $\mathrm{supp}(m)^c\subseteq D(\mathrm{Ann}_A(m))$, or equivalently, $V(\mathrm{Ann}_A(m))\subseteq \mathrm{supp}(m)$.
If $p\in D(\mathrm{Ann}_A(m))$, so $\mathrm{Ann}_A(m)\subseteq p$, as there exists $a\in \mathrm{Ann}_A(m)$ that is not in $p$. Hence $a$ is invertible in $\F_{p}$. Yet $a$ annihilates $m$, hence $m=0 \in \F_{p}$. So $p \notin \mathrm{supp}(m)$.
So $\mathrm{supp}(m)=V(\mathrm{Ann}_A(m))$
Now suppose that $A$ is noetherian, and $M$ finitely generated. Show that $\mathrm{supp}(\F)=V(\mathrm{Ann}_A(M))$ (not too sure why we need noetherian assumption)
proof:
We know that $\mathrm{supp}(\F)=\cup_{m\in M}\mathrm{supp}(m)$. Since $M$ is finitely generated, with generator $m_i$, we claim that $\mathrm{supp}(\F)=\cup_{i=1}^n\mathrm{supp}(m_i)$. Of course, if $p \in \mathrm{supp}(m_i)$, then $m_i\neq 0 \in M_p$ so $M_p\neq 0$, and if $p\in \mathrm{supp}(\F)$, then there is $m=\sum a_i m_i$ such that $\sum a_i m_i\neq 0 \in M_p$, so $m_i\neq 0 for some $i$.
Therefore, we have \[ \mathrm{supp}(\F)=\cup_{i=1}^n\mathrm{supp}(m_i)=\cup_{i=1}^nV(\mathrm{Ann}_A(m_i))=V(\cap_{i=1}^n\mathrm{Ann}_A(m_i)) \]
Now, if $a\in\cap_{i=1}^n\mathrm{Ann}_A(m_i)$,then $am=a(\sum a_i m_i)=\sum a_i am_i = 0$. So $a\in \mathrm{Ann}_A(M)$. If $a\in \mathrm{Ann}_A(M)$, then for every $i$, $am_i=0$.
We are done.
The support of a coherent sheaf on a noetherian scheme is closed.
proof:
We now that every quasicoherent module $\F$ over a noetherian scheme $X$ can be covered by some finite affine cover $U_i=\spec(A_i)$, where $\F|_{U_i}\cong \tilde{M_i}$ with $M_i$ finitely generated.
We have that \[\mathrm{supp}(\F)=\cup_{i=1}^n\mathrm{supp}(\F|_{U_i})=\cup_{i=1}^n V(\mathrm{Ann}_{A_i}(M_i)) \] Which is a finite union of closed set.
For any ideal $a\subseteq A$, we define a submodule $\Gamma_a(M)$ of $M$ by \[ \Gamma_a(M)={ m\in M : a^nm=0 \text{ for some }n} \] Assume that $A$ is noetherian, and $M$ an $A$-module. Show that $\tilde{\Gamma_a(M)} \cong \mathcal{H}^0_Z(\F)$, where $Z=V(a)$ and $\F=\tilde{M}$. Here $\mathcal{H}^0_Z(\F)$ is the sheaf that takes open set $U$ to $\Gamma_{Z\cap U}(U, \F|_ U)$.
proof:
We see that the result would easily generalize, so we will just show the affine case. Morever, it suffices to show they are isomorphic on the standard opens $D(f)$.
For $f\in A$, $\mathcal{H}^0_Z(\F)(D(f))= \Gamma_{Z\cap D(f)}(D(f), \F|_ {D(f)})$. We know that $D(f)\cong \spec(A_f)$ and $ \F|_ {D(f)}\cong \tilde{M_f}$, and $Z\cap D(f)=a\subseteq A_f$. We claim that $ \Gamma_{Z\cap D(f)}(D(f), \F|_ {D(f)})\cong \Gamma_{a}(M_f)$. This is a special case of Ex.3.3.3 (take $i=0$, here the noetherian condition does matter).
We notice that every element in $[ \Gamma_a(M)]_f$ are of the form $m/f^n$ for $m\in \Gamma_a(M)$. Hence $\Gamma_{a}(M_f)\cong [\Gamma_a(M)]_f$.
Let $X$ be a noetherian scheme, and let $Z$ be a closed subset. If $\F$ is a quasi- coherent (respectively, coherent) $X$-module, then $\mathcal{H}^0_Z(\F)$ is also quasi- coherent (respectively, coherent).
proof: Over the affines, $Z$ is intersection of $V(a)$, so $\mathcal{H}^0_Z(\F)$ is intersection of $\tilde{\Gamma_a(M)}$ as well, this takes care of the quasicoherent. If $A$ is noetherian, every submodule of a finitely generated module is finitely generated. This takes care of coherent.
Again let $X$ be a noetherian scheme, and $\F$ a coherent sheaf on $X$. We will consider the function \[ \varphi(x) = \dim_{\kappa(x)}(\F_x \otimes_{\Oc_{X,x}} \kappa(x)) \] where $\kappa(x):=\Oc_{X,x}/m_x$ is the residue field at the point $x$. Use Nakayama’s lemma to prove the following results.
The function $\varphi$ is upper semi-continuous,i.e.,for any $n \in \mathbb{Z}$, the set $S(n):={x\in X: \varphi(x)\geq n}$ is closed.
proof:
We assume $n\geq 1$ (the other cases are trivial). We will show the complement $S(n)^c$, i.e.${x\in X: \varphi(x)< n}$ is open. Moreover, since union of opens are still open, we only need to show the statement is true for affines $X=\spec(A)$ and $\F=\tilde{M}$ for some noetherian $A$ and finitely generated $A$-module $M$.
Suppose we have $p\in S(n)^{c}$, so $\dim (M_p/pM_p)<n$. We need to find $f\in A$ such that $p\in D(f)$ and $D(f)\subseteq S(n)^{c}$.
Hence, we just need $r$ elements $m_i\in M_p/pM_p$ to generate the entire vector space. We furthermore suppose that $m_i$ exists in $M_p$, which is possible as $M_p\to M_p/pM_p$ is surjective. We have $M_p/(m_i)$ goes to zero after modulo $p$, therefore, by Nakayama’s lemma, $M_p/(m_i)=0$.
Since every element in $A_p$ is a fraction, so $m_i$ lifts to elements in $M$, and we have that $[M/(m_i)]_p=M_p/(m_i)=0$. Since $M$ is finitely generated, $M/(m_i)$ is also, with finitely many generators $n_j$. There is then $f_j\notin p$ such that $f_jn_j=0$ in $M/(m_i)$. We define $f:=\prod_j f_j$, and since $p$ is prime $f\notin p$ and thus $p \in D(f)$. We will show that $D(f)\subseteq S(n)^c$.
Let $x\in D(f)$ be another prime ideal. Now we have that $fz_j=0$, so $[M/(m_i)]_x=0$ as well. We don’t know if there is $g\notin x$ such that $gm_i=0$ for some $i$. Hence we can only conclude that $\varphi(x)<n$, but that is all we need.
If $\F$ is locally free, and $X$ is connected, then $\varphi$ is a constant function.
proof:
Once again, since $X$ is connected and $\varphi(x)$ depends only locally, so affine is enough. We just need to show that for any prime $p\subset A$, and finitely free $A$-module $M\cong A^k$, $M_p/pM_p$ has the same dimension. Freeness condition tells us that all we need to show is for $M=A$, and we just need to show that $A_p/pA_p=\kappa(p)$ does not vanish, and it doesn’t.
Conversely, if $X$ is reduced, and $\varphi$ is constant, then $\F$ is locally free.
proof:
Once again, affine suffices. Following notation in a), suppose $\varphi(p)=r$, there is then an exact sequence \[ A^{q}\xrightarrow{\psi} A^{r}\to M\to 0 \] Where the second map determines $m_i$.
We have that for any prime ideal $x$, $\varphi(x)=r$ from the constant assumption. So we have that the localized map $\psi_{x}\equiv 0\mod x$ (or else, $\varphi(x)\neq r$).
Since $\psi$ over any local field is a linear transformation between finite dimensional vector spaces, it has matrix representation $(a_{ij})\in A$, and we have that $a_{ij}\equiv 0 \mod x$, for every $x$. Hence $a_{ij}\in x$ for every prime ideal. Hence $a_{ij}$ is in the nilpotent ideal.
Now $A$ is reduced, so the nilpotent ideal is just $0$, so $\psi=0$ over $A$, hence $M$ is free.
A morphism of scheme $f:X\to Y$ is affine if there is an affine cover $V_i$ of $Y$ such that $f^{-1}(V_i)$ are affinte as well.
Show that $f: X \to Y$ is an affine morphism if and only if for every open affine $V$, $f^{-1}(V)$ is affine.
proof:
($\impliedby$) Ah Doiii ($\implies$) Let $V_i$ be the affine cover of $Y$ such that $f^{-1}(V_i)$ are affines. Suppose $V$ is an affine subscheme of $Y$, we want to show that $f^{-1}(V)$ is affine.
Suppose $U$ is an affine open of $Y$ such that $f^{-1}(U)$ is affine. We have that all opens of the form $D(r)$ in $U$ have preimage $f^{-1}(D(r))=D(f^{\sharp}(r))$.
Given affine open $V$ of $Y$, we look at its intersections $V_i\cap V$. Since $V$ is affine, it is quasicompact, and hence there exists finitely standard opens $D(r_{ij})$ for each $i$ that covers $V$.
Therefore, by Ex.2.2.17, $r_{ij}\in \Gamma(V)$ generates $1$, so $f^{\sharp}(r_ij)\in \Gamma(f^{-1}(V))$ generate $1$ as well, and $f^{-1}(D(r_{ij})$ are affines, so $f^{-1}(V)$ is affine.
An affine morphism is quasi-compact and separated. Any finite morphism is affine.
proof:
1) since affines are quasicompact, the quasicompactness follows. Affineness of $f$ implies that a morphism $\spec(k)\to \spec(R)$ of valuative ring in $f: X\to Y$ is reduced to a morphism of affines $\spec(A)\to \spec(B)$. So we are trying to find a lift $A\to R$, which exists iff $A\to k$ lies in $R$, and is unique.
2) Ex. 2.3.4
Let $Y$ be a scheme, and let $\F$ be a quasi-coherent sheaf of $\Oc_Y$-algebras (i.e., a sheaf of rings which is at the same time a quasi-coherent sheaf of $\Oc_Y$-modules). Show that there is a unique scheme $X$, and a morphism $f:X \to Y$, such that for every open affine $V\subseteq Y$, $f^{-1}(V)\cong \spec(\F(V))$, and for every inclusion $U\subseteq V$ of open affines of $Y$, the morphism $f^{-1}(U)\to f^{-1}(V)$ corresponds to the restriction homomorphism $\F(V)\to \F(U)$. The scheme $X$ is called $\spec(\F)$.
proof:
For $Y=\spec(A)$ affine and $\F\cong \tilde{B}$ for some $A$-algebra, we define $X=\spec(B)$. We see that the localization on standard open $D(r)$ for $r\in A$ of $\F$ is $B_r$, which is precisely $X_{r}$.
For a general $Y$, we choose an affine cover $U_i$ of $Y$ and $\F|_{U_i}\cong \tilde{B_i}$, and we construct $X_i:=\spec(B_i)$. Since restriction of $\F$ are well-defined and compatible with $X_i$ and their intersections, we have $X:=\cup_i X_i$.
If $\F$ is a quasi-coherent $Y$-algebra, then $f: X=\spec\F \to Y$ is an affine morphism, and $\F\cong f_* \Oc_X$· Conversely, if $f:X\to Y$ is an affine morphism, then $\F:=f_*\Oc_X$ is a quasi-coherent sheaf of $Y$-algebras, and $X=\spec(\F)$ .
proof:
($\implies$) Affineness is by our construction in c). Moreover, we have that $f_* \Oc_X(U)=\Oc_X(f^{-1}(U))=\F(U)$ for standard open $U$, hence all opens.
($\impliedby$)
Let $U_i=\spec(A_i)$ be an affine cover. Then $V_i=f^{-1}(U_i)=\spec(B_i)$ are also affine.
We have that $f_{*}\Oc_X|_{U_i}\cong \tilde{B_i}$ by comparing them on the standard opens, so $f_{*}\Oc_{X}$ is a quasicoherent $Y$-module that is also an algebra. Lastly, $X=\spec(f_{*}\Oc_X)$.
(*) $X$ is a noetherian integral separated scheme which is regular in codimension one.
Let $k$ be an algebraically closed field, and let $X$ be a closed subvariety of $\Proj_k^n$ which is nonsingular in codimension one (hence satisfies(*)). For any divisor $D = \sum n_i Y_i$ on X, we define the degree of $D$ to be $\sum n_i \mathrm{deg}(Y_i)$, where \mathrm{deg}(Y_i); is the degree of $Y_i$;, considered as a projective variety itself.
Let $V$ be an irreducible hypersurface in $\Proj_k^n$ which does not contain $X$, and let $Y_i$ be the irreducible components of $V \cap X$. They all have codimension $1$ in $X$. For each $i$, let $f_i$ be a local equation for $V$ on some open set $U_i$ of $\Proj_k^n$ for which $Y_i \cap U_i \neq \emptyset$, and let $n_i = v_{Y_i} (\overline{f}_i)$, where $\overline{f}_i$ is the restriction of $f_i$ to $U_i \cap X$. Then we define the divisor $V.X$ to be $\sum v_{Y_i}(\overline{f}_i)Y_i$. Extend by linearity, and show that this gives a well-defined homomorphism from the subgroup of $\mathbf{Div} \Proj_k^n$ consisting of divisors, none of whose components contain $X$, to $\mathbf{Div} X$.
proof:
Since $V$ and $X$ are closed, $V\cap X$ is closed. Since $X$ is integral, irreducible components $Y_i$ of $V\cap X$ is integral.
The most important part of the argument is that whether or not $n_i$ depends on the local equation. If is does not, then it is the group homomorphism is obvious.
Let $\eta_i$ be the generic point of $Y_i$, which determines the valuation $v_{Y_i}$ on $K(X)$. Suppose we have two open set $U$ and $V$ of $\Proj_k^n$, with local equation $f$ and $g$ resp. that determines $U\cap Y$ and $V\cap Y$. Since $Y$ is irreducible, we can assume that $U$ and $V$ intersects. We have that over the intersection $f$ and $g$ gives the same vanishing locus, they generate the same ideal that describes $Y_i$. Therefore, they differ by a factor of $u\notin \eta_i$. $v_{Y_i}(u)=0$, so $v_{Y_i}(f)=v_{Y_i}(g)$.
If $D$ is a principal divisor on $\Proj^n_k$, for which $D.X$ is defined as in (a), show that $D.X$ is principal on $X$. Thus we get a homomorphism $\mathrm{Cl}\Proj_k^n \to \mathrm{Cl} X$
proof:
Let $D$ be principal, hence there is rational $f\in K(\Proj^n_k)$, such that \[ D=(f)=\sum_Y v_Y(f) Y \]
Let $f_X$ be $f$ restricted to $X$, we will show that $D.X=(f_X)$.
Let $Y_i$ be one of the irreducible component of $Y\cap X$. Obviously $f(Y_i)=0$. On the other hand, if we have a prime divisor in $(f_X)$, say $V$. We first note that $f=F/G$ for $F$ and $G$ homogeneous and of the same degee, so if $V\subset X\subset \Proj^n_k$ are zeros of $f$, hence of $F$. It must be part of the prime divisor $W$ of $\Proj^n_k$ such that $F(W)=0$ and $V= X\cap W$.
Now move on to the coefficients. Since $Y$ and $Y\cap X$ are determined by the same generic point $\eta_Y$, just in different scheme, the valuation on $\Proj_k^n$ would be the same after restriction to $X$.
Suppose we have a prime divisor $V$ of $\Proj_k^n$ that contain all of $X$, we define its value on the divisor group to be $0$. Now if we have some other $V’$ such that $V~V’$, then there is $f$ such that $V’=V+(f)$, but both term in rhs are zero.
So we have a group homomorphism $\mathrm{Cl}\Proj_k^n \to \mathrm{Cl} X$