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\begin{document}
\centerline{\sc Bryden Cais. \\ Algebraic Geometry, HW 3.\\   \today}

\medskip\noindent
1.  Given a smooth curve $C$ and an open subset $U\subset C$ with a morphism $f:U\rightarrow \P_k^n$
we must show that $f$ extends uniquely to a morphism $C\rightarrow \P_k^n$.  Since $C$ has dimension
1 and $U$ is open, $U$ is the complement of finitely many points.  Thus, we can reduce to the case that
$C-U=P$ consists of a single point.  Working locally, we can replace $U$ by $\spec k(C)$
and $C$ by $\O_{C,P}$.  Now we have a morphism $f:\spec k(C)\rightarrow \P_k^n$ given by
$f=(f_0,f_1,\ldots,f_n)$ for some $f_i\in k(C)$ with $V(f_0,f_1,\ldots,f_n)=\emptyset$.  Since
$\O_{C,P}$ is a DVR, there is a valuation $v$ on $k(C)=\mathrm{frac}\,\O_{C,P}$ such that
$\O_{C,P}=\{f\in k(C):v(f)\geq 0\}.$
Now let $i$ be such that $v(f_i)\leq v(f_j)$ for all $j\ne i$.
Such a minimum exists since there are only finitely many $f_j$.  Then observe that
$v(f_j/f_i)=v(f_j)-v(f_i)\geq 0$ so that $f_j/f_i\in \O_{C,P}$ for each $j$.  Thus we define
$$\tilde{f}=(f_0/f_i,f_1/f_i,\ldots,f_n/f_i).$$  Clearly, since the $i$ th coordinate is 1, this gives
a well defined morphism $\O_{C,P}\rightarrow \P_k^n$.  Moreover, we have $f_i\tilde{f}=f$ so that $\tilde{f}$ extends
$f$.

\medskip\noindent
2. Let $X=\spec A$ and $\F=\tilde{M}$ for some finite $A$ module $M$.  We claim that $\supp \F=V(I)$ where
$I=\Ann M$.  Indeed, by definition $\supp \F=\{P\in X :\F_P\neq 0\}$.  Moreover, we have seen that $\F_P\simeq M_P$.
Thus, we are reduced to showing that $M_P=0$ if and only if $\Ann M \nsubseteq P$.  Indeed, suppose that $\Ann M\nsubseteq P$
and let $f\in \Ann M-P$.  Then for any $m/g\in M_P$ with $m\in M$ and $g\not\in P$ we have
$fm/g=0$ and $f\not\in P$ so that $m/g\sim 0$ in $M_P$.  Thus, $M_P=0$.

Conversely, suppose that $M_P=0$.  Then for all $m\in M$ we have $m\sim 0$ in $M_P$ (since, in particular, $M\subset M_P$).  That is,
there exists $f_m\not\in P$ such that $f_m(m-0)=f_m m=0$ for all $m\in M$.  Now since $M$ is a finitely generated $A$ module, we can find
an $A$ basis $m_1,\ldots,m_s$ for $M$.  Now let $f=f_{m_1}f_{m_2}\cdots f_{m_s}$.  By construction, $f\in \Ann M$.  Moreover, since
$f_{m_i}\not\in P$, we have $f\not\in P$.  It follows that $\Ann M \nsubseteq P$.

\medskip\noindent
3.  Let $X$ be a scheme and $\F$ a sheaf of $\O_X$ modules on $X$.  We show that $\F$ is coherent if and only if
there exists an open cover $\{U_i\}$ of $X$ such that, for each $i$, the sheaf $\F|_{U_i}$ is the cokernel of a morphism
of free $\O_X$ modules of finite rank.

One direction is trivial: if there exists an open cover $\{U_i\}$ of $X$ such that, for each $i$, the sheaf $\F|_{U_i}$ is the cokernel of a morphism
of free $\O_X$ modules of finite rank then certainly $\F|_{U_i}$ is coherent.  If $\{V_{i,j}\}$ is an open affine cover of $U_i$, for each $i$
then $\F|_{U_i}(V_{i,j})$ is a finite $A_{i,j}$ module, where $V_{i,j}=\spec A_{i,j}$.  Since the $V_{i,j}$ cover $X$, we see that $\F$ is coherent.

Conversely, suppose that $\F$ is coherent.  Let $\{U_i\}$ be an affine open cover of $X$ with $U_i=\spec A_i$ such that $\F|_{U_i}\simeq \tilde{M}_i$,
with $M_i$ a finite $A_i$ module.  Let $m_{1,i}\ldots,m_{s(i),i}$ be a basis of $M_i$ over $A_i$.  Define a homomorphism $\phi_i:A_i^{s(i)}\rightarrow M_i$
by $(0,0,\ldots,0,1,0,\ldots,0)\mapsto m_{j,i}$, where the $1$ is in the $j$ th position.  Now $\ker \phi_i$ is a submodule of $M$ and hence finitely
generated.  So let $n_{1,i},\ldots,n_{r(i),i}$ be an $A_i$ basis of $\phi_i$.  Define $\psi_i:A_i^{r(i)}\rightarrow \ker\phi_i$ as above.
Then $\psi_i(A_i^{r(i)})\simeq \ker \phi_i$ so that $\coker\psi_i=A_i^{s(i)}/\ker\phi_i\simeq M_i$.  It follows that the sheaf $\F|_{U_i}$ is the
cokernel of a morphism of free $\O_X$ modules of finite rank.


\medskip\noindent
4.  Let $X$ be a reduced connected scheme and $\F$ a coherent sheaf on $X$.  Define $\phi:X\rightarrow Z$ by
$$\phi(x)=\dim_{k(x)}\F_x\otimes_{\O_{X,x}}k(x)$$
where $k(x)=\O_{X,x}/\m_{X,x}$.

\medskip\noindent
(a) We show that the set $\{x\in X:\phi(x)< n\}$ is open for any integer $n$.  To do this, it suffices to produce an open neighborhood $U$
of any point $x$ such that for all $y\in U$ we have $\phi(y)\leq \phi(x)$.  Let $V=\spec A$ be any affine open set containing $x$
with $\F|_{V}=\tilde{M}$ and let $P\subset  A$ correspond to $x$.  Then we have $\F_x=M_{P}$ and $\O_{X,x}=A_P$ so that
$\phi(x)=\dim_{A_P/PA_P} M_P/PM_P=s$, say.  Let $\a_1,\ldots,\a_s\in M_P$ be elements whose reductions modulo $PM_P$ form
a basis for $M_P/PM_P$ over $A_P/PA_P$.  Without loss of generality, we may take $\a_i\in M$ since we can just clear denominators.
By Nakayama's Lemma (Atiya Macdonald, p. 22) $\a_i$ generate $M_P$ as an $A_P$ module.

Since $M$ is a finite $A$ module, we let $m_1,\ldots,m_r$ generate $M$ over $A$.  Since $m_i\in M_P$, we can write
$$m_i=\sum_j \frac{a_{i,j}}{b_{i,j}} \a_j$$ for $a_{i,j},b_{i,j}\in A$ and $b_{i,j}\not\in P$.  Let
$t_i=\prod_j b_{i,j}$.  Then $t_im_i$ is in the $A$ span of $\a_i$ and $t_i\not\in P$.  Now set $f=\prod_i t_i$.
Then $f\not\in P$ so that $P\in D(f)$ and $D(f)$ is open.  Moreover, if $Q\in D(f)$ then $t_i\not\in Q$ so that
$t_i$ is a unit in $A_Q$.  Since $c_im_i$ is in the $A$ span of $\a_i$, it follows that $m_i$ is in the $A_Q$ span
of $\a_i$.  Since $m_i$ span $M$ over $A$, we have $m_i$ span $M_Q$ over $A_Q$ so that the $\a_i$ span $M_Q$
over $A_Q$.  It follows that the images of $\a_i$ in $M_Q/QM_Q$ span $M_Q/QM_Q$ as a $A_Q/QA_Q$ vector space; that is,
$\phi(Q)\leq s=\phi(P)$.

\medskip\noindent
(b)  Suppose that $\F$ is locally free and let $U_i$ be an open cover of $X$ on which $\F|_{U_i}=\tilde{M}_i$
with $\F_{U_i}$ a free $\O_{U_i}$ module.  Suppose that for some $i$, $U_i=\spec A_i$ and $M_i$ has rank $n$
as an $A_i$ module.  Now let $W$ be the union of all $U_j$ such that $M_j$ if free or rank $n$ and
let $V$ be the union of all $U_j$ with $M_j$ free of rank $m\neq n$.  By construction, $W,V$ are open
and $W$ is nonempty.  But if $x\in W\cap V$ then $\F_x\simeq M_P$ has rank $n$ and rank $m\neq n$, and this is absurd.
Hence we conclude that $W\cap V=\emptyset.$  Then since we clearly have $X=W\cup V$ and since $X$ is connected by hypothesis,
we must have $V=\emptyset.$  It follows that $\phi(x)=n$ is constant on $X$.

Conversely, suppose that $\phi(x)=n$ is constant for all $x\in X$.  We show that the stalk $\F_x$ is free.
Let $U=\spec A$ be an open affine set containing $x$ with $\F_x=M_P$ where $P\subset A$ corresponds to $x$.
Let $\a_1,\ldots,\a_n\in M$ have images forming a basis of $M_P/PM_P$ over $A_P/PA_P$.  By Nakayama's lemma, as above,
the $\a_i$ generate $M_P$ as an $A_P$ module; we must show they are linearly independent.
So suppose we have a relation
$\sum_{i=1}^n \frac{a_i}{b_i} \a_i=0$ in $M_P$.  We can clear denominators to suppose that we have a relation
$$\sum_{i=1}^n a_i\a_i=0$$ in $M_P$ with $a_i\in A$; that is, there is some $c\not\in P$ with $c\sum a_i\a_i=0$ in $M$.
Certainly we then have a linear relation in $M_P/P_M$, and since the $\a_i$ are linearly independent over $A_P/PA_P$
we have, for each $i$, some $c_i\not\in P$ with $c_ia_i=0$ in $M_P/PM_P$; that is, $c_ia_i\in P$.  It follows that
$a_i\in P$ for each $i$.  By our arguments above, there is an open set $D(f)$ with $P\in D(f)$ and for all
$Q\in D(f)$ the $\a_i$ generate $M_Q$ over $A_Q$. Now if $Q\in D(cf)$ then since $c,g\not\in Q$ we have
$$\sum_{i=1}^n a_i\a_i=0$$ in $M_Q$. Since $\phi$ is constant, the images of $\a_i$ form a basis
of $M_Q/QM_Q$ over $A_Q/QA_Q$ from which it follows, as above, that $a_i\in Q$.  We have thus shown that for all
$Q\in D(cf)$, we have $a_i\in Q$.  By Atiyah Macdonald p. 5, this implies that $a_i$ lies in the nilradical of $A_{cf}$
(since the prime ideals of $A_{cf}$ are exactly the elements of $D(cf)$).  But since $X$ is reduced, we have $A$ reduced and
hence $A_{cf}$ reduced.  Thus, $a_i=0$ in $A_{cf}$.  Finally, since $cf\not\in P$ we have $A_{cf}\subset A_P$ whence
$a_i=0$ in $A_P$.  It follows that the $\a_i$ are linearly independent over $A_P$ and hence $\F_x$ is free for each $x$.
We conclude that $\F$ is locally free.

\medskip\noindent
5.  Let $X/k$ be a scheme and $\L$ a locally free sheaf of rank 1.  Construct a scheme $\pi:L\rightarrow X$ as follows:
let $\{U_i\}$ be an open affine cover of $X$ and $\phi_i:\L|_{U_i}\tilde{\rightarrow} \O_{U_i}$ a trivialization
of $\L|_{U_i}$.  Set $U_{ij}=U_i\cap U_j$ and let $g_{ij}\in \O_X^{\times}(U_{ij})$ be the transition functions.
Set $V_i=U_i\times A^1_{t_i}$ and glue the natural maps $V_i\rightarrow U_i$ and $V_j\rightarrow U_j$ to form $\pi:L\rightarrow X$ by
$$U_{ij}\times \A^1_{t_i}\rightarrow U_{ij}\times\A^1_{t_j},\quad (u,t_i)\mapsto (u,g_{ij}t_i).$$
We claim that $\Gamma(U,\L)$ is identified with sections $\sigma:U\rightarrow L|_U$ with $\pi\circ\sigma=\id$.
Indeed, since $U_i$ is affine, for each $i$ we have $U_i=\spec A_i$ and $V_i=\pi^{-1}(U_i)=\spec A_i[t_i]$.  Now any section
$\sigma:U_i\rightarrow V_i$ induces an $A$-algebra ring homomorphism $\theta_{i,\sigma}:A_i[t_i]\rightarrow A_i$.  We thus have the correspondence
$$\{\sigma\} \mapsto \{s\in \Gamma(U,\L):s|_{U_i}=\theta_{i,\sigma}(t_i)\in A_i\}.$$
In the reverse direction, we have
$$\{s\in \Gamma(U,\L):s|_{U_i}=a_i\in A_i\}\mapsto \{\sigma: \theta_{i,\sigma}|_{A_i}=\id,\ \theta_{i,\sigma}(t_i)=a_i\}.$$
It is not difficult to see that $\sigma$ defined in this way is a section (follows from the requirement that $\theta_{i,\sigma}|_{A_i}=\id$)
and that the local pieces defining $\sigma$ glue together compatibly (indeed, we have $g_{ij}a_j=a_i$).  Moreover, we see that these
are inverse correspondences.



\end{document}
