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\begin{document}

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\centerline{\sc Problem Set Seven: Due Thursday, March 21}

See the course website for policy on collaboration. 
All references to the textbook refer to the February 21, 2024 edition of \href{http://math.stanford.edu/~vakil/216blog/FOAGfeb2124public.pdf}{\emph{Foundations of Algebraic Geometry}}, by Ravi Vakil.


\begin{Problem}
This problem builds up some useful lemmas about finiteness for future use. In particular, we will eventually use it to show ``projective source plus finite fibers implies finite" (so don't use that in solving this problem).

 Let $A$ be a commutative ring. Let $x_1$, $x_2$, \dots, $x_n$ be coordinates on $\PP^{n-1}_A$. Let $U$ be the open set $U:=\bigcup_{j=1}^{n-1} D_+(x_j)$ in $\PP^{n-1}_A$ and let $Z$ be the complementary closed subscheme $Z:=V_+(x_1, x_1, \ldots, x_{n-1})$. Let $X$ be a closed subscheme of $\PP^{n-1}_A$ such that $X \cap Z = \emptyset$.
\begin{enumerate}
\item Let $I \subset A[x_1, x_2, \ldots, x_n]$ be the graded ideal corresponding to $X$ (see Exercise 15.7.J).
Show that, for some integer $M$, the ideal $I$ contains a homogeneous polynomial of the form 
\[ x_n^M + \sum_{m_1+\cdots + m_n=M,\ m_n<M} a_{m_1 m_2 \cdots m_n} x_1^{m_1} x_2^{m_2} \cdots x_n^{m_n} .\]
Hint: Consider Exercise 4.5.L and the fact that $X \cap Z  = \emptyset$.
\item Let $\lambda : U \to \PP^{n-1}_A$ be the map given in homogenous coordinates by the map $(x_1 : x_2 : \cdots : x_{n-1} : x_n) \mapsto (x_1 : x_2 : \cdots : x_{n-1})$. Show that $\lambda : X \to \PP^{n-1}_A$ is a finite map. 
\item More generally, let $U_d$ be the open set $U_d := \bigcup_{j=1}^{d} D_+(x_j)$ and let $Z_d$ be the complementary closed subscheme $Z_d := V_+(x_1, x_2, \ldots, x_d)$. Let $X$ be a closed subscheme of $\PP^{n-1}_A$ such that $X \cap Z_d = \emptyset$ and let $\lambda_d : \PP^{n-1}_A \to \PP^{d-1}_A$ be the map $(x_1 : x_2 : \cdots : x_{n-1} : x_n) \mapsto (x_1 : x_2 : \cdots : x_d)$. Show that $\lambda_d : X \to \PP^{d-1}_A$ is finite.
\end{enumerate}
We'll stop the problem here because it is long enough for now; expect a sequel.
\end{Problem}

\begin{Problem}
In this problem, we'll compute the Picard group of an elliptic curve over a field. Let $k$ be a field where $2 \neq 0$. Let $f(x) = x^3 + a x^2 + bx + c$ be a cubic with $\text{GCD}(f(x), f'(x))=1$. Let $E$ be the curve in $\PP_k^2$ with equation $Y^2 Z = X^3 + a X^2 Z + b X Z^2 + c Z^3$, and let $E_0$ be the affine curve $y^2 = x^3 + a x^2 + bx + c$ in $\AA_k^2$. Let $\infty$ denote the point $[0:1:0]$ of $E$, so $E_0 = E \setminus \{ \infty \}$. Finally, write $R$ for the ring $\cO(E_0) = k[x,y]/\langle y^2 = x^3 + a x^2 + bx + c \rangle$.

You may assume without checking that $E$ is smooth over $k$, and that $v_{\infty}(x) = -2$ and $v_{\infty}(y) = -3$.

\begin{enumerate}
\item Show that $\{ x^j : 0 \leq j \} \sqcup \{ y x^j : 0 \leq j \}$ is a $k$ basis for $R$ and that $\{ x^j : 0 \leq j \leq d/2 \} \sqcup \{ y x^j : 0 \leq j \leq (d-3)/2 \}$  is a $k$ basis of $\Gamma(E, \cO(d \infty))$.  Show that $\dim \Gamma(E, \cO(d \infty))  = d$ for $d \geq 1$.
\item Let $D$ be an effective divisor on $E$ with $\deg(D) = d$. Show that $\Gamma(E, \cO((d+1) [\infty] - D)) \neq 0$. Conclude that $D$ is rationally equivalent to $(d+1) [\infty] - [\fp]$ where $\fp$ is a $k$-valued point of $E$.
\item Show that every divisor on $D$ is rationally equivalent to a divisor of the form ${m [\infty] - [\fp_1] + [\fp_2]}$ for some $k$-valued points $\fp_1$ and $\fp_2$ of $E$. 
\end{enumerate}
%At this point, we have already shown that $\Pic(E)$ is quite small -- especially if $k$ is a finite field! Now it's time to clean up and got the nicest result we can:

Let $\fp$ be a $k$-valued point of $E$ where $x=u$ and $y=v$. Let $\overline{\fp}$ be the $k$-valued point where $x=u$, $y=-v$.
\begin{enumerate}[resume]
\item Verify that $[\fp] + [\overline{\fp}]$ is rationally equivalent to $2 [\infty]$ on $E$, and thus $-[\fp] \sim [\fp]-2 [\infty]$. 
\item Deduce that every divisor on $D$ is rationally equivalent to a divisor of the form $m [\infty] + [\fp_1] + [\fp_2]$ for some $k$-valued points $\fp_1$ and $\fp_2$ of $E$. Then use one final trick to show that every divisor on $D$ is rationally equivalent to a divisor of the form $m [\infty] + [\fq]$ for $\fq$ a single $k$-valued point of $E$.
\end{enumerate}
\end{Problem}
In fact,  the different $k$-valued points of $E$ are not rationally equivalent to each other, so each divisor is equivalent to one of the form $m [\infty] + [\fq]$ for a unique $(m, \fq)$, but we don't have the right tools to prove that yet.

\end{document}


