\section{Algebraic Geometry}
\subsection{Basics}
{\bf Theorem:}
Every conic in the affine space over $R$ is equivalent under an affine transformation to
one of the following:
(1) $X^2 + Y^2 + P = 0$ (ellipse, point, empty set),
(2) $X^2 - Y^2 + P = 0$ (hyperbola, intersecting lines),
(3) $X^2 + Y + P = 0$ (parabola),
(4) $X^2 + P = 0$ (parallel lines, point empty).  In projective space (1), (2), (3) are
equivalent. In the projective space over ${\mathbb C}$, they are all projectively equivalent.
\\
\\
{\bf Definitions:}
Let $R=k[x_1, \ldots, x_n]$, $W= k^n= {\mathbb A}^n$ for the \emph{affine} setup and
$R=k[x_0, x_1, \ldots, x_n]$ (where $f \in R$ is a \emph{homogeneous} polynomial),
$W= k^{n+1} \setminus \{ {\vec 0}\}= {\mathbb P}^n$ under the usual equivalence
for the \emph{projective} setup.
If $S \subseteq R$, $V(S)= \{ x \in W: f(x)=0, \forall f \in S \}$.  $V(S)$ is an
\emph{affine} (resp \emph{projective}) \emph{algebraic set}.  An algebraic set
is \emph{irreducible} if is cannot be expressed as the union of two non-trivial
algebraic sets.  An affine irreducible algebraic set is an \emph{algebraic variety}.
If $S= \{f\}$, $V(S)$ is called a \emph{hypersurface}.
If $V \subseteq W$, $I(V)= \{ f \in R: f(x)=0, \forall x \in V \}$.
If $\langle S \rangle= I$ as an ideal in R, $V(S)=V(I)$.
$rad(I)={\sqrt I}= \{a: a^n \in I\}$.  $I$ is a \emph{radical} ideal if $I={\sqrt I}$.
Roughly, radical ideals $\leftrightarrow$ varieties,
prime ideals $\leftrightarrow$ subvarieties,
maximal ideals $\leftrightarrow$ points.
If $V$ is an algebraic variety, $I(V)$ is a prime ideal (proof below)
and $\Gamma[V]= R/I(V)$ is
called a \emph{coordinate ring}.
${\overline R}= k(x_1, \ldots, x_n)$ is the quotient field of $R$; members of ${\overline R}$
induce \emph{rational} maps.
${\mathcal O}_P(V)$ denotes the rational functions on $V$ defined on $P$.
The \emph{Zariski} topology on $W$ is defined by identifying the $V(S)$ with
\emph{closed sets}.
\\
\\
{\bf Theorem:}
Let $k$ be algebraically closed.
There is a one to one correspondence between polynomial maps $\varphi: V \rightarrow W$
and the homomorphisms ${\tilde {\varphi}}: \Gamma[W] \rightarrow \Gamma[V]$.
Let ${\mathcal T}(V,k)= \{f: f:V \rightarrow W \}$.  If $\varphi: V \rightarrow W$,
$\tilde {\varphi}: {\mathcal T}(W,k) \rightarrow {\mathcal T}(V,k)$.
\begin{quote}
\end{quote}
{\bf Theorem:}
$k \subseteq \Gamma[V)] \subseteq {\mathcal O}_P(V) \subseteq k(V)$.
\begin{quote}
\end{quote}
{\bf Definitions:}
Two affine varieties
$V, W$ are \emph{isomorphic} if $\exists \phi, \psi:
\phi \circ \psi= id_W$. Suppose $F_i$ is a collection of functions.
The \emph{multiplicity of a root} at a point ${\vec a}$ is
$f(t)= gcd(F_{1}({\vec a} + L(t)), \ldots , F_{m}({\vec a} + L(t)))$ where $L(t)$ is a line
through $O={\vec 0}$.
By transferring $P$ to ${\vec 0}$ and express $F= F_m + \ldots + F_n$, $F_m$ is
a form of degree $m$.
$L$ touches $X$ at $O$ if its
intersection multiplicity is greater than 1.
Locus of points touching $X$ at $x$ is the \emph{tangent space}, $\Theta_{x,X}$.
Two varieties $V, W$ are \emph{birationally equivalent} if there are \emph{rational} maps
$\varphi_1: V \rightarrow W$ and $\varphi_2: W \rightarrow V$ between them.
$f(x,y)$ is \emph{rational} if $\exists \phi, \psi$: $f(\phi(t), \psi(t))=0$.
Any conic (2nd order equation) in two variables has either infinitely many
rational solutions or none.
\\
\\
{\bf Theorem:}
Two curves are birationally equivalent iff their fields of functions are isomorphic.
\begin{quote}
\end{quote}
{\bf Theorem:}
Every irreducible curve of degree 2 is rational.
$x^{n}+y^{n}=1$ is not rational for $n>2$.
\begin{quote}
\end{quote}
{\bf Definition:} Suppose $\phi: A^n(k) \rightarrow A^m(k)$, $f \in k[y_1, \ldots, y_m]$;
the \emph{pullback} is $\phi^*: \phi^* \circ f= f \circ \phi$.
A \emph{discrete valuation ring (``DVR'')} is a Noetherian,
local domain whose maximal ideal is principal.
If a form, F, does not vanish on an irreducible projective variety
X then $dim(X_{F})= dim(X)-1$.
\\
\\
{\bf Theorem:}
The following are equivalent: (1) The set of non-units in $R$
form an ideal; (2) $R$ has a unique maximal ideal.
The following are equivalent and define a DVR:
(1) $R$ is Noetherian and its maximal ideal is principal; (2)
$\exists t \in R:
\forall  0 \ne z \in R: z=ut^n$, where $u$ is a unit.
\begin{quote}
\emph{Proof:}
We show $1 \rightarrow 2$.  First uniqueness.  If $ut^m=vt^n$ for units $u,v$,
$ut^{m-n}=v$ but then $t^{m-n}$ is a unit.  For existance, since ${\mathfrak m}= (t)$,
for a non-unit, $z \in {\mathfrak m}$,
$z= z_1t$ and if $z_i$ is a non-unit, $z_{i+1}= z_i t$ so
$(z_1) \subset (z_2) \subset \ldots$ and since $R$ is Noetherian eventually
$(z_n)=(z_{n+1})$.  $z_{n+1}= vz_{n}, v \in R$ $z_n= t vz_n$ so $vt=1$ but $t$ is not a unit.
\end{quote}
{\bf Theorem:} The pole set is an algebraic variety.
$\Gamma[V]= \bigcap_{P \in V} {\mathcal O}_P(V)$.
\begin{quote}
\end{quote}
{\bf Theorem:}
If $S_1 \subseteq S_2$ then $I(S_1) \supseteq I(S_2)$.  Every algebraic set
is the intersection of hypersurfaces.
\begin{quote}
\emph{Proof:}
By the Hilbert basis theorem, $I(V)= (f_1, f_2, \ldots, f_r)$ and so
$V= V(f_1) \cup \ldots \cup V(f_r)$.
\end{quote}
{\bf Theorem:}
Every closed set is the union of finitely many irreducible ones.
Every irreducible closed set is birationally isomorphic to a hypersurface in $A^{n}$.
\begin{quote}
\end{quote}
{\bf Theorem:}
(1) An algebraic set, $V$, is irreducible iff $I(V)$ is prime.
(2) If $k$ is algebraically closed, $I(V(f))= (f)$.
(3) If $I$ is radical, $I(V(I))= I$.
\begin{quote}
\emph{Proof:}
(3) follows from Nullstellensatz.
\end{quote}
{\bf Theorem:}
Let $R=k[x,y]$. If $f, g \in R$ have no common factor, $V(f) \cap V(g)$ is finite.
\begin{quote}
\emph{Proof:}
$k(x)[y]$ is a PID, $Af+Bg=1$ over $k(x)$; now clear denominators.
\end{quote}
{\bf Theorem:}
If $I$ is prime, $V(I)$ is irreducible.  There is a $1-1$ correspondence between
prime ideals and irreducible algebraic sets; there is a $1-1$ correspondence between
maximal ideals and points.
\begin{quote}
\end{quote}
{\bf Theorem:}
Let $V$ be an algebraic set then $V= V_1 \cup \ldots V_r$, $V_i$, irreducible,
$V_i \nsubseteq V_j, i \ne j$.
\begin{quote}
\emph{Proof:}
We use the following (follows from axiom of choice).  Let
${\mathcal S}$ be a non-empty collection of ideals in a Noetherian ring, $R$,
then
${\mathcal S}$ has a maximal element.  From this we conclude and collection of
algebraic sets has a minimal element. \\
Now let
${\mathcal S}= \{ V:
\textnormal{algebraic, } V
\textnormal{ is not a finite union}
\textnormal{ of irreducible}
\textnormal{ algebraic sets} \}$ .
Let $V$ be a minimal element of
${\mathcal S}$.  $V= V_1 \cup V_2$ and this is a contradiction.  For the second condition,
throw out the $V_i$ violating it.
\end{quote}
{\bf Theorem:}
${\mathcal O}_P(V)$ is a Noetherian local domain.  Further, the maximal ideal
${\mathfrak M}_P(V)$ is generated elements of $f \in \Gamma[V]$ for which $f(P)= 0$.
\begin{quote}
\emph{Proof:}
STS $I \subseteq
{\mathcal O}(V)$ is finitely generated.
Since $\Gamma[V]$ is Noetherian, $(f_1, \ldots, f_r)= I \cap \Gamma[V]$.  Claim
$(f_1, \ldots, f_r)= I$ and an
${\mathcal O}(V)$ ideal, for if $f \in I$, $\exists b \in \Gamma[V]: b \ne 0$ and
$bf \in \Gamma[V]$ so $bf= \Gamma[V] \cap I$, $bf= \sum_i a_i f_i, a_i \in \Gamma[V]$ and
$f= \sum_i {\frac {a_i} {b_i}} f_i$.
\end{quote}
{\bf Definition:}
$I(P, F \cap G)= dim_k({\mathcal O}_P({\mathbb A}^2)/(F, G)$.
\\
\\
{\bf Theorem:}
$|V(I)| < \infty$ iff $R/I$ is finite dimensional over $k$.
\begin{quote}
\emph{Proof:}
If $P_1, \ldots, P_r \in V$, we can fine $f_i \in R: f_i(p_j)= \delta_{ij}$ so
$r \le dim_k(R/I)$.
On the other hand,
if $V(I)= \{ P_1 , P_2 , \ldots , P_r \}, P_i= (a_{i1}, a_{i2}, \ldots, a_{in})$,
define $f_j= \prod_i (x_i - a_{ij}), F_j \in I(V(I))$ so $F_j^N \in I, {\overline F}_j= 0$
and ${\overline X}_j^{rN}$ is a $k$-linear combination of
$ {\overline 1}, {\overline x}_j^{1}, \ldots, {\overline x}_j^{rN-1} $ and
$ {\overline x}_1^{m_1}, \ldots, {\overline x}_n^{m_n}; m_i < rN $ is a set of generators.
\end{quote}
{\bf Theorem:}
Let $I$ be an ideal of $R$ ($k$, algebraically closed), $V(I)= \{P_1 , \ldots , P_N
\}$ and $
{\mathcal O}_i= {\mathcal O}_{P_i}({\mathbb A}^n) $,
$\exists \varphi: R/I \rightarrow \prod_i
{\mathcal O}_i/(I{\mathcal O}_i)$ induced
from the natural homomorphisms $\varphi_i: R \rightarrow {\mathcal O}_i/(I{\mathcal O}_i)$.
\begin{quote}
\end{quote}
{\bf Theorem:} $P$ is a simple point of $f$ iff ${\mathcal O}_P(V(f))$ is a DVR.
If $L$ is a non-tangent line $l + {\mathcal O}_P(V)$ is a uniformizing parameter.
\begin{quote}
\emph{Proof:}
WLOG, $P=(0,0), L= ax$ and $y=0$ is the tangent.
${\mathfrak M}_P(V(f))= (x,y)$ whether $P$ is simple or not.
Suppose $f= y(1+g(x,y)) - x^2 h(x)$ and $yg= x^2h \in \Gamma(f)$ so
$y= x^2 h g^{-1}$ since $g(P) \ne 0$ and
${\mathfrak M}_P(V(f))= (x)$.
\end{quote}
{\bf Theorem:}
Suppose ${\mathfrak M}= {\mathfrak M}_P(F)$ denotes the maximal ideal of
${\mathcal O}= {\mathcal O}_P(F)$.
$ 0 \rightarrow {\mathfrak M}^n/{\mathfrak M}^{n+1}
\rightarrow {\mathcal O}/{\mathfrak M}^{n+1}
\rightarrow {\mathcal O}/{\mathfrak M}^{n} \rightarrow 0$.
$\chi(n)= dim({\mathcal O}/{\mathfrak M}^n)=$ Hilbert polynomial.
\\
\\
{\bf Theorem:}  Let $P$ be a point on an irreducible curve, $f$, for all sufficiently
large $n$, $m_P(f)= dim_k({\mathfrak M}_P(f)^n/{\mathfrak M}_P(f)^{n+1})$.
\begin{quote}
\end{quote}
{\bf Divisors:}
Suppose $E: y^2= x^3-x$ and $f(x,y)= {\frac x y}$.  On $E: {\frac x y}= {\frac y {x^2-1}}$.
If $u_P$ is a \emph{uniformizer}, $f= u_P^r g, r \in {\mathbb Z}, g(P) \ne 0, \infty$ and
we define $ord_P(f)=r$.  In this example, $u_{(0,0)}= y$ is a uniformizer at $(0,0)$ of
$E: y^2= x^3-x$.  Since $x= y^2({\frac 1 {x^2-1}})= y {\frac y {x^2-1}}= {\frac x y}y=x$,
$ord_{(0,0)}(x)= 2$.  In general, for $P=(x_0, y_0) \in E$, $u_P$ can be taken as any line
through $P$ not tangent to $E$, thus we can take $u_P= x-x_0$ when $y_0 \ne 0$ and
$u_p= y$ when $y_0=0$.  For example, if $E: y^2= x^3 + 72$ then $(-2, 8) \in E$.
Since $f(x,y)= x+y-6$ vanishes at $(-2,8)$ and
$f(x,y)= (x+2) + (y-6) = (x+2)( 1 + {\frac {(x+2)^2 - 6(x+2) + 12)} {y-8}})$,
$ord_P(f)=1$, $u_{\infty}= {\frac x y}$.
\\
\\
{\bf Weak Bezout:} If two curves of dimension $m$ and $n$ meet at more than $mn$
points (counting multiplicity) then they have a common component.
\begin{quote}
\emph{Proof:}
Strategy of Proof: (S-1) $\#(C_1 \cap C_2 \cap {\cal A}^2) \le
dim({\frac R {(f_1 , f_2)}}) \le n_1 n_2$, (S-2) first inequality is an equality,
(S-3) first inequality can be strengthened to
$I(C_1 \cap C_2,P) \le dim({\frac R {(f_1 , f_2)}})$,
(S-4) inequality in 4 is an equality,
(S-5) $I$ is invariant under projective transformations --- transform so the line at
infinity does not intersect $C_1 \cap C_2$.
Notation: Let $f_1 (x, y)$, and $f_2 (x, y)$, defining curves
$C_1$ and $C_2$, have dimension $m, n$ respectively.
$R= k[x, y]$, $(f_1 (x, y), f_2 (x, y))= R f_1 + R f_2$.
\\
\\
\emph{S-1:}
$C_1 \cap C_2 \leq dim_k ({\frac {R} {(f_1 , f_2 )}}) \leq mn$. [Argument:
If $P_1, P_2, \ldots P_r$ are distinct, $\exists h_i (x,y)$ with
$h_i (P_j ) = \delta_{ij}$, so if there are $r$ common roots of $f_1$ and
$f_2$,
$\sum_{i=1}^r c_i h_i (x,y) = r_1 f_1 (x,y) + r_2 f_2 (x,y)$ implies $c_i=
0$.] \\
Let $R_d$ be polynomials of degree $\leq d$ then
$dim_k (R_d )= \phi (d) = {\frac {(d+1)(d+2)} {2}}$.
Let $W_d = R_{d-m }f_1 + R_{d-n}f_2$, for $d \geq (m+n)$.
$R_{d-m}f_1 \cap R_{d-n} f_2 = R_{d-m-n} f_1 f_2$.
$dim_k ( R_d ) - dim_k (W_d )= mn$.
$g= \sum_i^l c_j g_j$ has a non-trivial dependency for $l>mn$ with
$g \in W_d$.
\\
\emph{S-2:}
Second inequality is equality if $C_1 \cap C_2$ don't meet at infinity.
Let $f^*$ be the homogeneous polynomial consisting of the highest degree terms
in $f$.  If $\infty \notin C_1 \cap C_2$ then $f_1^*$, $f_2^*$ have no common factor.
If $f_1^*$ and $f_2^*$ have no common factor then $(f_1 , f_2) \cap R_d= W_d$.
Under the conclusion of the previous sentence, if $d \ge n_1 + n_2$ then
$dim({\frac R {(f_1 , f_2)}}) \ge n_1 n_2$ which proves the result.
\\
Define ${\mathcal O}_P= \{ F \in K(x, y): F(P)$ exists $\}$,
${\mathfrak M}_P= \{ f \in {\mathcal O}_P: f(P)= 0 \}$.
${\mathfrak M}_P$ is a unique maximal ideal of $O_P$.  $(f_1, f_2)_P= f_1 O_P + f_2 O_P$.
Now define $I(C_1 \cap C_2; P)= dim( {\frac {{\mathcal O}_P} {(f_1 , f_2)_P}})$.
\\
\emph{S-3:}
${\frac {{\mathcal O}_P} {(f_1 , f_2)_P}} \le {\frac {R} {(f_1, f_2)}} < \infty$.
${\mathcal O}_P = (f_1 , f_2)_P +R$.
If $P \notin C_1 \cap C_2$ then $I(C_1 \cap C_2, P)=0$;
If $P \in C_1 \cap C_2$ then $(f_1 , f_2)_P \subset {\mathfrak M}_P$;
$I(C_1 \cap C_2; P) = 1 + dim({\frac R {(f_1 , f_2)_P}})$ iff
$(f_1 , f_2) = {\mathfrak M}_P$.
If $P \in C_1 \cap C_2$ and $r \ge dim({\frac {{\mathcal O}_P} {(f_1 , f_2)_P}})$ then
${\mathfrak M}_P^r \subset (f_1, f_2)_P$.  If
$P, Q \in C_1 \cap C_2 \cap {\cal A}^2, \psi \in {\mathcal O}_P$
then $\exists g \in R$:
$g= \psi \jmod{(f_1 , f_2)_P}$ and
$g= 0 \jmod{(f_1 , f_2)_Q}$ if $P \ne Q$.
\\
\emph{S-4:} Kernel of natural map $R \rightarrow \prod_{P \in (C_1 \cap C_2 \cap {\cal A}^2)}
{\frac {{\mathcal O}_P} {(f_1 , f_2)_P}}$ is just $(f_1 , f_2)$ where the natural map is:
$f \mapsto (\ldots, f \jmod{(f_1, f_2)}, \ldots)$.
$dim({\frac R J})
= \sum_P dim( {\frac  {{\mathcal O}_P} {(f_1 , f_2)}})
= \sum_P I(C_1 \cap C_2, P)$.  The last equality holds iff $J \subset (f_1 , f_2)$.
Define $L= \{ g \in R : gf \in (f_1 , f_2) \}$ and $1 \in L$.  $L$ is an ideal
$(f_1 , f_2) \subset L \subset R$.
$P \in {\cal A}^2$, $\exists g \in L: g(P) =0, P \in L$.
$\exists a \in k: 1 \notin L +R(x-a)$ and
$\exists b \in k: 1 \notin L +R(x-a) + R(y-b)$.
\\
\emph{S-5:}
Properties of intersection multiplicity.  $I((y-x^m), y; 0)= m$.  Show the definitions
make sense and that there is a line $L$ which does not contain any of the intersection
points.  The proof requires knowing there are only a finite number of points in the
intersection.
\end{quote}
{\bf Genus} for non-singular curve: $g_f = {\frac {(n-1)(n-2)} {2}} - d$.
$L(D)= \{ f: K(C)^*: div(f) \ge -D \}$.  $l(D)= dim (L(D))$.
\\
\\
{\bf Reimann Roch Theorem:} Let
$X$ be a non-singular projective plane curve.  $\exists g \ge 0: \forall D,
dim_k(L(D)) \ge deg(D)+1-g$.  The minimum such $g$ is called the genus.
\begin{quote}
\end{quote}
{\bf Resultants:}
The $r$ forms $f_1,f_2,...,f_r$ with indeterminate coefficients possess a resultant
system of integral polynomials $b_k$ such that for special values of the
coefficients in $K$ (algebraically closed).
The vanishing of all resultants is a necessary and
sufficient condition for
$f_1= f_2= ... =f_r= 0$ to have a solution $\ne 0$.  The $b_k$ are homogeneous
in the coefficients of every form $f_i$ and
satisfy $x_k^{s_r}b_k=0 \jmod {(f_1, f_2,...,f_r)}$.
\\
\\
{\bf Bezout's Theorem.}  If $f, g$ are two curves of degree $n,m$ respectively that
have no common component then they intersect in $mn$ points counting multiplicity.
Notes: A homogeneous system
$f_1= f_2= ... =f_r= 0$ has solutions
$(\xi_1^{(a)}, \xi_2^{(a)},...  \xi_n^{(a)}), a= 1,2,3,...,q$.
Set $l_x=u_1x_1+u_2x_2+...+u_nx_n$.
Form resultant system $b_1(u), ..., b_t(u)$.  The common zeros of $b_1, ...$ are
$\prod l_a$.  By Nullstellensatz,
$(\prod_a l_a)^{\tau}=0 (b_1(u),b_2(u),...,b_t(u))$
$\rightarrow D(u)= \prod {l_a}^{\rho_a}$ and
$(b_i(u))^{r_i}=0 (\prod l_a)$
$\rightarrow D(u)= (f_1,...,f_r,l)$.  $R(u)$ is the same as the u-resultant so
$\sum \rho_a$ is the degree of $R(u)= \prod deg(f_i)$.
\\
\\
\emph{Example:} $F_1(x,y,z)=x^2+y^2-10z^2=0$,
$F_2(x,y,z)=x^2+xy+2y^2-16z^2=0$, add $F_3(x,y,z)= u_0z+u_1x+u_2y$.
$Res_{1,2,2}(F_0,F_1,F_2)=(u_0+u_1-3u_2)(u_0+2{\sqrt 2}u_1+{\sqrt 2}u_2)
(u_0-2{\sqrt 2}u_1-{\sqrt 2}u_2)$.
Solutions are $(1,-3,1)$, $(-1,3,1)$,
$(2{\sqrt 2}, 2{\sqrt 2}, 1)$,
$(-2{\sqrt 2}, -2{\sqrt 2}, 1)$.
\\
\\
\emph{Example proof with generics:} $F_1(X,Y,Z)=X-Y^2$, $F_2(X,Y,Z)=XY-Z$,
$(X,Y,Z) \rightarrow (t^2,t,t^3)$ is generic because it is a solution for any
specialization of $t$ and any solution is obtainable this way.
\\
\\
Let $D$ be a domain and $\Omega=\Omega_D= {\overline {D(t_1,t_2,...)}}$ is called a universal
field.  Note that $\Omega \leftrightarrow$ {prime ideals over} $D[X_1,...]$.
\\
\\
{\bf Theorem:} $x_1, ... x_n \in \Omega$.  $I= \{ f: f(x_1,...,x_n)=0 \}$ is a prime ideal.
If $I$ is a prime ideal and $1 \notin I$ then $I$ has a generic 0.  Any extension
$K(\alpha_1 , ..., \alpha_m)$ can be embedded in $\Omega$.
\\
Hints: look at $E=D[X]/I$.  Under this homomorphism the image of $(X_1,...X_n)$ is generic.
\\
\\
{\bf Theorem:} If $\xi_1 ,..., \xi_n$ are elements of an arbitrary extension of $K$ then
If $\Re =K[X_1,...,X_n]$ and $\wp= \{ f: f(\xi_1,...,\xi_n)=0 \}$.
$1 \notin \Re$ and $\wp$ is a prime ideal.  Every prime ideal has a generic element.
\begin{quote}
\end{quote}
{\bf Theorem:}  Any ideal $g=(f_1, ..., f_n)$ which has no zeros in $\Omega$ is the unit ideal.
\emph{Proof:} Otherwise a maximal ideal would correspond to a non-zero generic point.
\begin{quote}
\end{quote}
{\bf Extension of Nullstellensatz}:  If $p_1, ..., p_s$ all vanish at the common zeros of
$(f_1,...,f_n)$, then $\exists q$ such that powers of the $p_i$'s of degree $q$
are in $(f_1,...,f_n)$.
\begin{quote}
\emph{Proof:}  For $s=1$, this is the simple Nullstellensatz.  Let the exponent for each $i$ be
$q_i$.  Set $q= q_1 + q_2 + ...+ q_n -n +1$.
Nullstellensatz bound: $\rho \leq 13 d^n$ where $d$ is the degree and $n$ is the number of
variables.
\end{quote}
{\bf Theorem:}
Let $N_q$ be the number of products $X_j$ of degree $q$.
Suppose $F_1, F_2,..., F_r$ are forms.  $(0,...,0)$ is the only common zero
iff all products $X_j$ can be expressed as linear combinations of the $X_{ki}F_i$
with coefficients in $K$.
Note:  This means they are linearly independent.  So there are other common zeros is
there are fewer than $N_q$.  Note that $X_1, ..., X_n$ satisfy the extension conditions.
If the $X_{ki}F_i= \sum a_{kij}X_j$
are not linearly independent, the determinant families, $R_i(a)$, form a resultant set.
\begin{quote}
\end{quote}
{\bf Multivariate resultants:}  If we fix degrees
$d_0,d_1, \ldots d_n$
then there is a unique
polynomial $Res \in {\mathbb Z}[u_i , \alpha]$ such that (a) if $F_0, F_1, \ldots , F_n$
are homogeneous polynomials of degrees
$d_0,d_1, \ldots d_n$ then $F_0= \ldots = F_n=0$ has a nontrivial solution over
${\mathbb C}$ iff
$Res(F_0, \ldots , F_n)=0$, (b)
$Res(x_0^{d_0}, \ldots , x_n^{d_n})=1$, (b)
, (c) $Res$ is irreducible
in ${\mathbb C}[u_i, \alpha]$.  If $PP=PP(x_1, x_2, \ldots, x_n)$ is a set of
power products in the $x_i$, there are $N_m= {{m+n-1} \choose {n-1}}$  $PP$'s of degree
$m$.  \emph{Example:}
$A_3=a_3x^2b_3y^2+c_3z^2$,
$A_2=a_2x+b_2y+c_2z$,
$A_1=a_1x+b_1y+c_1z$.  $S_i= {\frac {PP_i^d} {x_i^d}}$,
$S_1= \langle x^2, xy, xz \rangle, S_2=\langle y^2, yz \rangle, S_3= \langle z^2 \rangle$.
\\
\\
$\left(
\begin{array}{c|cccccc}
 & x^2 & xy & xz & y^2 & yz & z^2 \\
\hline
xA_1 & a_1 & b_1 & c_1 & 0 & 0 & 0 \\
yA_1 & 0 & a_1 & 0 & b_1 & c_1 & 0 \\
zA_1 & 0 & 0 & a_1&  0 & b_1 & c_1 \\
yA_2 & 0 & a_2 & 0 & b_2 & 0 & 0 \\
zA_2 & 0 & 0 & a_2 & 0 & b_2 & c_2 \\
A_3 & a_3 & 0 & 0 & b_3 & 0 & c_3 \\
\end{array}
\right)
$.
\subsection{Elliptic Curves}
{\bf Basic Definitions:}  An \emph{elliptic curve} is a smooth projective curve of
genus $1$ with a distinguished point $O$.  In the plane, (affine)
elliptic curves are described by an equation of the form
$E(F): y^2 +a_1 xy + a_3 = x^3 +a_2 x^2 + a_4 x + a_6$, $a_i \in F$
called the \emph{general affine Weierstrauss form (GWF)}.
An equation of the form
$E(F): y^2 = x^3 + a x + b$, $a, b \in F$, is said to be in
\emph{special affine Weierstrauss form (SWF)} and
sometimes we denote this curve by $E_{a,b}(F)$.
If $P=(x_P, y_P)$ is a solution of
the SWF, $P$ is said to be on the elliptic curve $E_{a,b}(F)$.
\\
\\
{\bf Definitions:}
\emph{Projective coordinates:}
$(X_1 , Y_1, Z_1) \sim (X_2 , Y_2, Z_2)$,
$X_1 = \lambda^c X_2$, $Y_1 = \lambda^d Y_2$, $Z_1 = \lambda Z_2$,
$c,d \in {\mathbb Z}^{>0}$ and usually $c=d=1$.
\emph{Jacobian projective coordinates:} $\infty= (1:1:0)$ and $-(X:Y:Z)=(X:-Y:Z)$.
\emph{Standard projective coordinates:} $\infty= (0:1:0)$ and $-(X:Y:Z)=(X:-Y:Z)$.
\\
\\
{\bf Addition Formula for SWF:}
$Y^{2}Z= X^{3}+aXZ+bZ^{3}$, $P_{i}= (x_{i},y_{i})$, $O=(0:1:0)$.
We want to calculate $R=P_1+P_2$.  If $P_1$ or $P_2$ is $O$, result is obvious.
If $x_1=x_2$ and $y_1= - y_2$, $R=O$.
If $x_1 \ne x_2$, set
$\lambda ={\frac {y_{2}-y_{1}} {x_{2}-x_{1}}}$.  If $x_1 = x_2$ and
$y_1 \ne - y_2$, set $\lambda= (3{x_1}^2+a)(y_1+y_2)^{-1}$.
In either case, $x_3= \lambda^2 - x_1 -x_2$,
$y_3= \lambda(x_1-x_3)-y_1$ and $R= (x_3 : y_3 : 1)$.
$|\epsilon_{p}| \leq 2{\sqrt p}$.  $E_{a,b}(F)$
defined by $f(x,y)= y^2-(x^3+ax+b)$ is \emph{non-singular}
if there is no point $(x,y)\in E_{a,b}(F)$ such that
${\frac {\partial f} {\partial x}} (x,y)=0$ and
${\frac {\partial f} {\partial y}} (x,y)=0$ or equivalently
provided
$x^3_ax+b$ does not multiple roots.
$f(x,y)$ has multiple roots iff $-(4a^3 + 27 b^2)=0$.
Usually we pick $Z$ axis tangent to
$O$ and then $(0,1,0)$ as the point at $\infty$.
\\
\\
{\bf Theorem:}
If $C_1$ and $C_2$ are two non cubic curves that meet
at eight points, they meet at nine points (counting multiplicity).
\begin{quote}
\emph{Proof:}
$C_1: a_1x^x+a_2y^3+ \ldots + a_{10}xyz$.
$C_2: b_1x^x+b_2y^3+ \ldots + b_{10}xyz$.  The set of cubics passing through eight
points form a two parameter family: $C= \alpha C_1 + \beta C_2$.
$C_1(P_i)=0=C_2(P_i), i \le 8 \rightarrow C(P_9)=0$.
\end{quote}
{\bf Observation:}  The ``eight point theorem'' gives a quick proof of associativity by
looking at the intersection of six lines (read down and across) in the following:
$$
\left(
\begin{array}{l | c c c}
 & l: & m: & n:\\
\hline
r: & P & X & Q+R\\
s: & Q & R & {\overline {QR}}\\
t: & {\overline {PQ}} & P+Q & O\\
\end{array}
\right)$$
Here, $X= {\overline {P(Q+R)}}= {\overline {R(P+Q)}}$ and the eight point theorem
is applied to
$C_1: l(X,Y,Z) m(X,Y,Z) n(X,Y,Z)= 0$ and $C_2: r(X,Y,Z) s(X,Y,Z) t(X,Y,Z)= 0$.
\\
\\
{\bf Definition:}  $Z=0$ is tangent to PSWF at $[0,1,0]$ but the tangent line intersects
the curve three times. This is called a \emph{flex point}.
\\
\\
{\bf Mordell's Theorem:}  If a non-singular cubic curve over $k$ has a rational point then
the rational points are finitely generated as a $k$-module.
Use $H({\frac m n})= max(|m|, |n|)$. Let $P=(x,y)$.  Define $H(P)= H(x)$ and
$h(P)= log(H(P))$.  From now on assume $C$ is given by
$y^2= x^3 + ax^2 + bx +c$.
\begin{quote}
\emph{Proof:} To prove it, need four lemmas:
\\
\\
\emph{Lemma 1:} There are a finite number of points $P$: $h(P) < M$.
\\
\\
\emph{Lemma 2:} Fix $P_0$ on $C$, $\exists K_0(P_0 , a, b, c):  h(P+P_0) \leq
2h(P)+ k_0$.
\\
\jt Show that if $P$ is on $C(Q)$, $P=({\frac m {e^2}}, {\frac n {e^3}})$.  Then
show $n \leq KH(P)^{\frac 3 2}$.  Use this to get
$k_0$.
\\
\\
\emph{Lemma 3:} Fix  $\exists K(a, b, c):  h(2P) \geq
4h(P) - K$.
\\
\\
\emph{Lemma 4:} $|\{C(Q):2C(Q)\}| < \infty$.
\\
\\
For lemma 4, assume $y^2= x^3+ax^2+bx$ (so the curve
always has a rational point), and use $\Gamma= C(Q)$ and
$\Delta=2\Gamma$.  Define the map
$\phi(x,y)= (x+a+{\frac b x}, y{\frac {x^2 - b} {x^2}})$.
Define $\psi$ similarly.  Note that $\psi ( \phi (P))= 2P$ and
$ker(\phi) = \{0, (0,0)\}$. ${Q^*}^2$ $\alpha(x,y)= x \jmod
{{Q^*}^2}$.  $im(\phi) \subseteq ker(\alpha)$.
Let $p_i | b$, $i= 1,2, \ldots t$ then $|\Gamma: \phi(\Gamma)| \leq
2^{t+1}$.
$|\Gamma:\phi(\Gamma)| \leq 2^{t+1}$.  Use the following lemma: If $A$ and
$B$
are abelian $A \rightarrow B \rightarrow A$ and
$|B:\phi(A)| < \infty$,
$|A:\phi(B)| < \infty$, then $|A:2A| \leq
|B:\phi(A)| |A:\phi(B)|$.
\\
\\
\emph{Proof given lemmas:} Let $Q_0 , \ldots, Q_{m-1}$ be the coset
representatives.  $P-Q_{i_1}= 2P_1$ is in the subgroup, $P_1-Q_{i_2}= 2P_2$,
repeatedly doing this yields:
$P= Q_{i_1} + 2 Q{i_2} + \ldots + 2^{m-1}Q_{i_m} +
2^m P_m$, $h(P_j) \leq {\frac 3 4} h(P_{j-1})$.
Since there are a finite
number of $Q_i$ there's a $k'$ so that $h(P-Q_i) \leq 2 h(P) + k'$ for all
$P$.  Using the inequalities $h(P_j) \leq {\frac {h(P_{j-1})} 2} +
{\frac {k+k'} 4}$.
So the group is generated by the $Q_i$ and the (finite number
of) points of $ht \leq {\frac {k+k'} 4}$.
\end{quote}
{\bf Theorem:}
Let $C$ be a non-singular cubic curve $C: y^2= x^3 + ax^2 + bx +c$.  Set
$D= -4a^3 c + a^2 b^2 + 18 abc -4b^3 -27c^2$.  Let $\Phi$ be the set of
points of finite order.  Let $\phi$ be the reduction map mod $p$.  If
$(p, 2D)= 1$ then $\phi$ is an injection into $C(F_p )$.
\\
\\
{\bf Nagel-Lutz Theorem:} Same as above with $P(x,y)$ as a rational point of finite order
$y=0$ or $y|d^2$.
\begin{quote}
\end{quote}
{\bf Functions on Elliptic Curves:}
$f_1(x,y) = {\frac {s(x,y)} {t(x,y)}}$ and
$f_2(x,y) = {\frac {u(x,y)} {v(x,y)}}$ are said to be \emph{equivalent} on the
elliptic curve $E$, denoted by $f_1 \sim_E f_2$ iff $(sv-ut)= 0 \jmod {E}$.
Polynomials can be uniquely $x$ or $y$ reduced but not rational functions.
\emph{Examples:}  Let $E: y^2-(x^3+x^2+x)$ then
$ {\frac {x}{y-x}} \sim_E {\frac {y+x}{x^2+1}}$.
Put
$F(X,Y,Z)= \{ {\frac x y} \}= [X, Y]$ then $F[0,1,0] =[0,1] \sim_E 0$ (a zero) while
if $G(X,Y,Z)= \{ {\frac y x} \}= [Y, X]$ then $G[0,1,0] =[0,1] \sim_E \infty$.
For $E: y^2= x^3-x$, $\{ {\frac y {x^2-1}} \} \sim_E \{ {\frac x y} \}$ and
$[YZ, X^2-Z^2] \sim_E [X,Y]$.
For $E: y^2= x^3-x$,
$\{x \}$ has a $0$ at $(0,0)$,
$\{y \}$ has a $0$ at $(-1,0), (0,0), (1,0)$,
$\{ {\frac x y} \}$ has a $0$ at $(-1,0), (1,0)$.
If $E(a,b)$ is non-singular, $E$ is irreducible and
we can embed $k[x,y]/(E)$ in the field of fractions $K(E)$
and we can define a map from $K(R) \rightarrow
{\overline K} \cup \{\infty\}$.
\\
\\
{\bf Definition:}
A \emph{morphism} between two elliptic curves $C_1$ and $C_2$ is a rational map
$\varphi: (C_1, I_1) \rightarrow (C_2 , I_2)$,
$\varphi(x,y)= (f(x,y), g(x,y)), g, g \in K(C_1 )$, such that
$\varphi(P+Q)= \varphi(P)+\varphi(Q)$.  $\varphi(P)= [n]P$ is a morphism.  A
morphism is an \emph{isomorphism} if $\exists \psi$, a morphism: $\psi \circ \varphi= [1]$.
An \emph{isogony} is a surjective morphism with finite kernel preserving the base point.
If $\psi: E_1 \rightarrow E_2$ induces a $\psi^*: K(E_2) \rightarrow K(E_1)$.
$\psi^*(f)= f \circ \psi$.  $deg(\psi)= ker(\psi)$.
\\
\\
\emph{Example of dual isogony:}
$E: y^2= x^3 + x^2 +x$,
$E': (y')^2= (x')^3 -2 (x')^2 -3 (x')$.
$\psi(x,y)=
( {\frac {y^2} {x^2}}, {\frac {y(1-x^2)} {x^2}})$,
$\varphi(x',y')= ( {\frac {(y')^2} {4(x')^2}}, {\frac {(y')(-3-(x')^2)} {8(x')^2}},)$
and $\varphi \circ \psi= [2]$.
\\
\\
{\bf Definition:}
Suppose $E= E_{a,b}(K), char(K) \ne 2,3$.  Let $x_1= \mu^2x$ and $y_1= \mu^3 y$ then
$(x_1 , y_1) \in E_{a', b'}(K)$ with $a'= \mu^4 a$ and $b'= \mu^6b$.  Two
elliptic curves related in this way are said to be \emph{isomorphic}.
\\
\\
{\bf Theorem:}
If $K= F_{p^m}, p \ne 2, 3$ and
$E_K^{(1)}(a, b) \cong E_K^{(2)}({\overline a}, {\overline b})$ iff
$\exists u \in K^*$ such that $u^4 {\overline a}=a$ and
$u^6 {\overline b}=b$ under the map $(x,y) \mapsto (u^2 x, u^3 y)$.
\begin{quote}
\end{quote}
{\bf Definition:}
$j(E)= 1728 {\frac {4a^3} {4a^3+27b^2}}$ is called the
\emph{$j$-invariant}.
\\
\\
{\bf Theorem:}
$j(E)$ is invariant under
the transformation above (i.e. - two isomorphic curves have the same
$j$-invariant) and, conversely, two curves with the same $j$ value are related in
this way (and are
thus isomorphic in the elliptic curve defined over the algebraic closure).
\begin{quote}
\end{quote}
{\bf Theorem:}
If $j(E_1)=j(E_2)$ then
$\exists \mu \in {\overline K}, \mu \ne 0: a_2= \mu^4 a_1, b_2= \mu^6 b_2$.
\begin{quote}
\end{quote}
{\bf Homogeneous forms:}
Let $G(u,v)$ be a homogeneous polynomial and $(u_0, v_0) \in {\mathbb P}^1_K$,
$\exists k \ge 0$ and $H(u, v)$ with $H(u_0, v_0) \ne 0: G(u,v)= (v_0 u -u_0 v)^k H(u,v)$.
Any line in ${\mathbb P}^2_k$ can be parameterized by
$(x,y,z)= (a_0 u + b_0 v, a_1 u + b_1 v, a_2 u + b_2 v)$.
$L$ intersects $C$ to order $n$ at $P=(x_0 : y_0 : z_0)$ if ${\overline C}(u,v) =
(v_0 u - u_0 v)^n H(u,v)$ in the foregoing theorem; denote this as $ord_{L, P}(C)=n$,
$ord_{L, P}(C)= \infty$ if ${\overline C}$ is identically 0.
If $L_1, L_2$ are lines, $ord_{L_1, P}(P)= 1$
or $\infty$.  If $C$ is a curve defined by $C(x,y,z)=0$, $C$ is non singular at $P$ if
$(C_x, C_y, C_z) \ne 0$ in which case the tangent line is $C_x X + C_y Y + C_z Z =0$.  If
$C$ is non-singular at $P$ there is a line in ${\mathbb P}^2_K$ that intersect $C$ to order at
least $2$.
\\
\\
{\bf Theorem:}
The number of equivalence classes of elliptic curves over $K$ is
$2q+6$, $2q+2$, $2q+4$, $2q$ according to $q = 1, 5, 7, 11 \jmod{12}$.
If $K= F_{2^m}$ and $E_K(a, b): y^2 + xy = x^3 + a x^2 +b$ then
$E_K^{(1)}(a, b) \cong E_K^{(2)}({\overline a}, {\overline b})$ iff
$b= {\overline b}, Tr(a)=Tr({\overline a})$ and if so $\exists s: {\overline a}= s^2+s+a$
under the map $(x,y) \mapsto (x, y+sx)$.
\begin{quote}
\end{quote}
{\bf Division Polynomials:}
Let $E(K): y^2+a_1xy+a_3y= x^3 +a_2 x^2 + a_4 x + a_6$ be an elliptic curve
$m \in {\mathbb Z}, P = (x,y)$,
$b_2 = a_1^2 + 4a_2$,
$b_4 = a_1 a_3 + 2a_4$,
$b_6 = a_3^2 + 4a_6$,
$b_8 =  a_1^2 a_6 + 2 a_4 a_6 - a_1 a_3 a_4 + a_2 a_3^2 -a_4^2$ then
$\exists \psi_m(x,y), \omega_m(x,y), \theta_m(x,y) \in K[x,y]$ such that
$[m]P=({\frac {\theta_m(x,y)} {(\psi_m(x,y))^2}}, {\frac {\omega_m(x,y)} {(\psi_m(x,y))^3}})$.
$\psi_1 = 0$, $\psi_1 = 2$,
$\psi_2 = 2y+a_1x+a_3$,
$\psi_3 = 3x^4+b_2x^2+3b_4x^2+3b_6+b_8$,
$\psi_4 = (2x^6+b_2x^5+5 b_4 x^4 + 10 b_8 x^2 +(b_2 b_8-b_4 b_6)x+ b_4 b_8 - b_6^2)
\psi_2$,
$\psi_{2m+1} = \psi_{m+2} \psi_m^3 - \psi_{m-1}\psi_{m+1}^3, m \ge 2$,
$\psi_{2m} =
{\frac {\psi_m} {\psi_2}}
(\psi_{m+2} \psi_{m-1}^2 - \psi_{m-2} \psi_{m+1}^2)$,
$\theta_m = x \psi_m^2- \psi_{m+1} \psi_{m-1}$,
$\omega_m =
{\frac {1} {4 y}}
(\psi_{m+2} \psi_{m-1}^2- \psi_{m-2} \psi_{m+1}^2), m>2$.
Note that $deg(\psi_m)= O(m^2)$.
\\
\\
{\bf Definition:}
An \emph{endomorphism} is a homomorphic map between and an elliptic
curve and itself that is expressible as a \emph{rational function}.
If $\alpha$ is an endomorphism and $P=(x,y),
\alpha(X+Y) = \alpha(X) + \alpha(Y), \alpha(x,y)= (r_1(x,y), r_2(x,y))$.  Because
$y^2= x^3+ax+b$, we may assume
$\alpha(x,y)= (r_1(x), y r_2(x))$; if $r_1(x)= {\frac {p(x)} {q(x)}}$, the degree of
endomorphism is $max(deg(p(x)), deg(q(x)))$.  This endomorphism
$\alpha$ is a \emph{separable endomorphism} if $r_1'(x) \ne 0$.
The \emph{Frobenius endomorphism} is
$\varphi(x,y)= (x^q , y^q)$.
\\
\\
{\bf Definition:}
$E[n]= \{P \in E({\overline K}): nP= \infty \}$.
\\
\\
{\bf Theorem:} (1) If $char(K) \ne 2$ $E[2]= {\mathbb Z}_2 \oplus {\mathbb Z}_2$;
if $char(K) = 2$ $E[2]= {\mathbb Z}_2$ or $0$.
(2) If $char(K) \nmid n$ or is $0$, $E[n]= {\mathbb Z}_n \oplus {\mathbb Z}_n$.  (3)
If $char(K)=p \mid n, n=p^r n'$ then
$E[n]= {\mathbb Z}_n \oplus {\mathbb Z}_{n'}$ or
$E[n]= {\mathbb Z}_{n'} \oplus {\mathbb Z}_{n'}$.  Proof
uses division polynomials.  Let
$E$ be an elliptic curve over $F_q$.  Then
$E(F_q)= {\mathbb Z}_n$ or
${\mathbb Z}_{n_1} \oplus {\mathbb Z}_{n_2}$ with $n_1 \mid n_2$.
\begin{quote}
\emph{Proof:}
Apply the above theorem.
$\phi_n$ induces a linear transformation on $E[n]$.
$E[n]= Z_{n_1} \oplus \ldots \oplus Z_{n_k}$ by ther FTOAG.  Let $l \mid n_1$ so
$l \mid n_j$. $E[l] \subseteq E[n]$, so it has order $l^k$.  But the
endomorphism induced by multiplication by $n$ has order $n^2$, so $k=2$.  Since
this map, annihilates $Z_{n_1} \oplus Z_{n_2}$, $n_1, n_2 \mid n$, and since
$n^2= \#E[n]$, $n_1=n_2=n$.
\end{quote}
{\bf Theorem:} If $\alpha$ be a separable endomorphism of $E_{A,B}(\overline{F_q})$,
$deg((\alpha)= |ker(\alpha)|$, otherwise $deg((\alpha)> |ker(\alpha)|$.
\begin{quote}
\emph{Proof:}
$\alpha(x,y)= (r_1(x), yr_2(x))$, $r_1(x)= {\frac {p(x)}{q(x)}}$; since $\alpha$
is separable, $p'q-q'p \neq 0$.  Let $S= \{x \in \overline{K}: (pq'-qp')q=0\}$.
There are $(a,b)$ on the curve: $a \neq 0 \neq b$, $(a,b) \neq O$.
$deg(p-aq)= deg(\alpha)$, and $a \notin r_1(S)$.  There are exactly $deg(\alpha)$
points: $\alpha(x,y)= (a,b)$.  STS there are no nultiple roots.  If $x_0$ is a multiple
root, $pq'-qp'(x_0) =0$.  If $\alpha$ is not separable, the same argument holds but
the equation has at least one solution with multiple roots so it has fewer roots than
$deg( \alpha )$.
\end{quote}
{\bf Theorem:} $\alpha: E_{A,B}(\overline{K}) \rightarrow E_{A,B}(\overline{K})$ is surjective.
\begin{quote}
\emph{Proof:}
If $p-aq$ is not constant and $x_0$ is a root, $q(x_0) \neq 0$.  Choose $y_0^2= x_0^3+Ax_0+B$.
$\alpha(x_0,y_0)= (a, b')$, $b'= \pm b$.  If $b'=b$, we're done.  If
$\alpha(x_0 , y_0 )= (a, -b')= (a, b)$.  Since $E(\overline{K})$ is infinite and
$ker( \alpha )$ is finite, only a finite number of points map onto a points
with a single $x$ coordinate.  Therefore either $p$ or $q$ is not constant.
There is a unique $a$ such that $p-aq$ is constant.  So there are at most two
points, namely, $(a, \pm b )$ not in the image.  Pick another
$P_1$: $\alpha(P_1 )=(a_1 , b_1 )$.  $(a_1, b_1 ) + (a,b) \neq (a, \pm b )$.
There is a unique $P_2$: $\alpha (P_2 ) =  ( a_1 , b_1 ) + (a, b)$ so
$\alpha (P_2 - P_1 )= (a,b)$ and
$\alpha (P_1 - P_2 )= (a,-b)$.
\end{quote}
{\bf Theorem:} $\phi_q^n-1$ is separable and $|ker(\phi_q^n-1)|= |\#E(F_q)|$.
\begin{quote}
\emph{Proof:}
$\phi$ is a homomorphism.  It has degree $q$ and $(x^q)'=0$.
\end{quote}
{\bf Theorem:}  Let $E_{A, B}$ be an elliptic curve.  Fix $(u,v) \in E$.  Define
$f(x,y), g(x,y)) = (x,y) + (u,v)$. Then ${\frac {\frac {df(x,y)} {dx}} {g(x,y)}}= {\frac 1 y}$.
\begin{quote}
Use the addition formula and the fact that $2y y'= 3x^2+A$ to get
$(x-u)^3u({\frac {df(x,y)}{dx}}-g(x,y))=
v(Au+u^3+v^3-Ax-x^3-y^2) + y(-Au-u^3-v^3+Ax+x^3-y^2)$.  Now use $B=y^2-x^3-Ax$ to get the result.
\end{quote}
{\bf Theorem:}  Let $\alpha_i, i= 1,2,3$ be endomorphisms.
Write $\alpha_i(x,y)= (R_{\alpha_i}, S_{\alpha_i}y)$.  Suppose $\exists c_{\alpha_1}, c_{\alpha_2}:
{\frac {R_{\alpha_i}'} {S_{\alpha_i}}}= c_{\alpha_i}, i=1,2$.  Then
${\frac {R_{\alpha_3}'} {S_{\alpha_3}}}= c_{\alpha_1} + c_{\alpha_2}$.
\begin{quote}
\emph{Proof:}
${\frac {\partial x_3} {\partial x_1}} = {\frac {y_3} {y_1}}$ and
${\frac {\partial x_3} {\partial x_2}} = {\frac {y_3} {y_2}}$.
${\frac {\partial x_j} {\partial x}} = c_{\alpha_j}{\frac {y_j} {y}}$.  Now apply the
chain rule.
\end{quote}
{\bf Theorem:} Let $E_{A,B}$ be an elliptic curve, $n \in {\mathbb Z}$ and
$n(x,y)= (R_n(x), yS_n(x))$.  Then
${\frac {R_{n}'} {S_{n}'}}= n$.
\begin{quote}
\emph{Proof:}
From the inverse formula if it holds for $n>0$, it holds for $n<0$.  It holds for $n=1$ and the
result above shows if it holds for $n$, it holds for $n+1$.
\end{quote}
{\bf Theorem:} $r, s \in \mathbb{Z}$, $r \neq 0 \neq s$ then $r\phi_n+s$ is separable iff $p \nmid s$.
$(x,y) \in E(F_q)$ iff $\phi_p(x,y)= (x,y)$.
\begin{quote}
\emph{Proof:}
$r(x,y)= (R_r(x), y S_r(x))$.
$(R_{r \phi_q(x)}(x), y_{r\phi_q}(x))= (R_r^q, y(x^3+Ax+B)^{\frac {q-1} 2} S_r^q(x))$.
Thus $c_{r \phi_q}= R_{r \phi_q}'/S_{r \phi_q}=0$.
$R_{r \phi_q +s} /S_{r \phi_1 +s} = c_{r \phi_q} + c_s = s \neq 0 \jmod{p}$.
\end{quote}
{\bf Theorem:} $deg(a \alpha + b \beta)= a^2 deg( \alpha ) + b^2 deg( \beta ) + ab (deg (\alpha + \beta )
- deg( \alpha ) -deg( \beta )$.
\begin{quote}
\emph{Proof:}
Consider $\alpha_n$ and $\beta_n$ given by the matricies when viewed as maps in $E[n]$.  The matrix calculation
is straigntforward.
\end{quote}
{\bf Theorem:} Let $E$ be an elliptic curve over $K$ and $n$ a positive integer, $char(K) \nmid n$.  There
is a pairing $e_n : E[n] \times E[n] \rightarrow \mu_n$ (the Weil pairing) with
(1) $e_n$ bilinear in each variable, (2) $e_n$, non-degenerate, (3) $e_n(T,T)=1$,
(4) $e_n(T,S)= e(S, T)^{-1}$, $e_n( \sigma T , \sigma S)$ for $\sigma \in Aut ( \overline{K} )$, and
(6) $e_n( \alpha (S) , \alpha (T))= e_n(s, T)^{deg( \alpha )}$, for all seperable automorphisms of $E$.
If the coefficients are in $F_q$, the statement holds for $\phi_p$.
\begin{quote}
\emph{Proof:}  The proof uses divisor theory.
\end{quote}
{\bf Theorem:} Let $\{T_1 , T_2 \}$ be a basis in $E[n]$.  Then $e_n(T_1 , T_2 )$ is a primitive $n$-th
root of unity.
\begin{quote}
\emph{Proof:}  Suppose $e_n(T_1 , T_2 )= \zeta$, $zeta^d=1$.  $e_n(T_1, d T_2)=1$ and
$e_n(T_2, dT_2)= e_n(T_2 , T_2 )^d= 1$.  Let $S= a T_1 +b T_2$.
$e_n(T_1, dT_2)= e_n(T_1 , dT_2)^a e_n(T_t, dT_2)^b=1$.  This is true for all $S$ so $dT_2= \infty$.  But
this can hold iff $n \mid d$, so $\zeta$ is  a primitive root.
\end{quote}
{\bf Theorem:} Let $\alpha$ be and endomorphism of $E$ over $K$ and $n$, a positive integer
not divisible by $char(K)$.  Then $deg( \alpha_n )= deg( \alpha ) \jmod{n}$.
\begin{quote}
\emph{Proof:} By the previous result, $\zeta= e_n(T_1 , T_2 ) $ is primitive. So
$\zeta^{deg( \alpha )}= e_n(T_1 , T_2 )= e_n(aT_1 + c T_2, bT_1 + d T_2)= \zeta^{ad-bc}$.
\end{quote}
{\bf Theorem:} $r, s \in \mathbb{Z}$, $(r,q) \neq 1$ then $deg(r\phi_n-s)= r^2q+s^2- r s a$, where
$a= q+1-\#E(F_q)$.
\begin{quote}
\emph{Proof:}
$deg(r \phi_q - s)= r^2 deg( \phi_q ) + s^2 deg(-1) + r( deg( \phi_q - 1) -
deg( \phi_q ) - deg(-1))$.
\end{quote}
{\bf Theorem:} $a= q+1-\#E(F_q)$.  $\phi_q^2-a\phi_q +q=0$ and $a=Tr((\phi)_q)_m \jmod{m}$.
\begin{quote}
\emph{Proof:}
Consider $\alpha= \phi_q^2 -a \phi_q +q$ as an endomophism in $E[m]$.
$|ker(\phi_q-1)|= deg(\phi_q-1)=det((\phi_q)_m-I)$.  By Cayley Hamilton,
$(\phi_q^2)_m -a \phi_q +q = 0 \jmod{m}$.  This is true for infinitely many
$m$ so $\alpha=0$.
\end{quote}
{\bf Hasse's Theorem:} Let $E_q$ be an elliptic curve then
$ q+1- 2 {\sqrt q} \le \#E_q \le q+1+ 2 {\sqrt q}$,
$\#E_q=q+2-t$, $t$ is the Frobenius trace.
\begin{quote}
\emph{Proof of Hasse:}
Let $\psi$ be the Frobenius map.
$\#E_p = |ker([1]-\psi)|$.  First note that $deg([1])=1$ (in fact, $deg([n]) =n^2$).
$deg(\psi)= p$.  Also note that $deg(a+b)-deg(a)-deg(b)=B(a,b)$ is bilinear.
$0 \le deg([t] + [2] \psi)=t^2 -4p -2tB[1, -\psi]= 4p-t^2$; so
$(deg([1]-\psi) - deg([1]) -deg(\psi))^2 \le 4p$ but the first term is $\#E(F_p)$.
\end{quote}
{\bf Theorem:}
If $\alpha \ne 0$ is a separable endomorphism of $E$,
$deg(\alpha)= \#ker(\alpha)$, otherwise
$deg(\alpha)> \#ker(\alpha)$.
The endomorphism $[n]P \mapsto Q$ has degree $n^2$; most endomorphisms are of this form.
If $char(K) \nmid n$ then $E[n]= {\mathbb Z}_n \oplus {\mathbb Z}_n$.
If $E[n] \subseteq E({\mathbb K})$ then $\mu_n \in K$.  Given $E_q(a,b), n \ge 1$, (1)
$ker(\phi_q^n -1)= \#E_{q^n}(a,b)$ and
$\phi_q^n -1$ is separable
$\#E_{q^n}(a,b)= deg(\phi_q^n -1)$.
If $\alpha$ is separable, then $deg(\alpha)= \#ker(\alpha)$.
$|E({\overline {F_p}})[m]|= m^2$ if $(m,p)=1$.
\begin{quote}
\end{quote}
{\bf Theorem:}
Let $E(F_q)$ be an elliptic curve $E(F_q) \approx {\mathbb Z}_n$ or
${\mathbb Z}_{n_1} \times {\mathbb Z}_{n_2}, n_1 \mid n_2 $.
The Frobenius endomorphism has degree $q$ and is not separable.
\begin{quote}
\end{quote}
{\bf Definitions:}
If $E_q$ is an elliptic curve over
a finite field of characteristic $p$, $E_q$ is said to be \emph{supersingular} if
$E_q[p]= \{ \infty \}$.
(1) $char(F) \ne 2,3$, $(x,y) \mapsto ({\frac {x-3a_1^2-12 a_2} {36}},
{\frac {y - 3 a_1 x} {216}} - {\frac {a_1^3 + 4 a_a a_2 -12 a_3} {24}})$,
sends the general equation to
$E_q(a,b): y^2= x^3 + ax +b, \Delta = -16(4 a^3 + 27 b^2)$.
(2) $char(F) = 2, a_1 \ne 0$, $(x,y) \mapsto
(a_1^2x+{\frac {a_3} {a_1}}, y+{\frac {a_1^2 a_4 - a_3^2} {a_1^2}})$,
sends the general equation to
$E_q(a,b): y^2 + xy = x^3 + ax +b, \Delta = b$.  This is \emph{non-supersingular}.
(3) $char(F) = 2, a_1 = 0$, $(x,y) \mapsto
(x+a_2, y)$, sends the general equation to
$E_q(a,b): y^2 + cy = x^3 + ax +b, \Delta = c^4$.  This is supersingular.
(4) $char(F) = 3, a_1^2 \ne - a_2$, $(x,y) \mapsto
(x+{\frac {d_4} {d_2}}, y+ a_1 x + a_1 {\frac {d_4} {d_2}} +a_3)$,
$d_2= a_1^2+a_2, d_4= a_4-a_1 a_3$,
sends the general equation to
$E_q(a,b): y^2= x^3 + ax +b, \Delta = - a^3 b$.  This is non-supersingular.
(5) $char(F) = 3, a_1^2 = - a_2$, $(x,y) \mapsto
(x, y+ a_1 x + a_3)$, sends the general equation to
$E_q(a,b): y^2= x^3 + ax +b, \Delta = - a^3 $.  This is supersingular.
\\
\\
{\bf Theorem:}  $q=p^m$, $\exists E_q: \#E_q= q+1-t$ iff
(i) $t \ne 0 (p), t^2 \le 4q$; or, (ii) $m= 1(2)$ and either (a) $t=0$ or (b)
$t^2=2q, p=2$, or (c) $t^2=3q, p=3$; or, (iii) $m=0 (2)$ and either (a)
$t^2=4q$ or (b) $t^2=q, p \ne 1 (3)$ or (c) $t=0, p \ne 1 (4)$. $E_{p^m}$ is supersingular
iff $p \mid t$.  $E_q = {\mathbb Z}_{n_1} \oplus {\mathbb Z}_{n_2}$ and
$ n_2 \mid n_1 \mid (q-1)$.
\begin{quote}
\end{quote}
{\bf Divisors:}
Given $E(K)$, $P \in E({\overline K})$, define $D= \sum_j a_j [P_j], a_j \in {\mathbb Z}$ and
$deg(D)= \sum_j a_j$.  $Div^0(E)$ are the \emph{divisors} of degree 0.
If $f$ is a function on $E$,
$div(f) = \sum_P ord_P(f) [P] \in div(E)$.  Two divisors $D_1, D_2$ are said to be
\emph{equivalent} if $D_1-D_2 = (D)_E$.
\\
\\
{\bf Theorem:} If $P-Q$ is a divisor on an elliptic curve $E$, $P-Q= c$.
The theorem can be applied as follows.  Suppose
$div(\{ f \})= P_1 + P_2 + \ldots + P_n Q_1 - Q_2 - \ldots - Q_n$.  Let
$l_1$ be the line through $P_1, P_2$ and hence $-(P_1+P_2)$ while
$m_1$ is the line through $-(P_1+P_2)$ and $O$ and hence $P_1+P_2$ then
$div( \{f\} {\frac {m_1}{l_1}})= (P_1+P_2) + \ldots + P_n +O - Q_1 - \ldots - Q_n$ continuing
and doing the same for the $Q_i$, we get
$div( \{f\} \cdot
{\frac {m_1}{l_1}} \cdot
{\frac {m_2}{l_2}} \cdot \ldots
{\frac {m_{2n-2}}{l_{2n-2}}})=
(P_1+P_2 + \ldots + P_n) + (n-1) O - (Q_1 + \ldots + Q_n)
-(n-1)O =
(P_1+P_2 + \ldots + P_n) - (Q_1 + \ldots + Q_n)$ so
$f=c ( {\frac {l_1}{m_1}} \cdot {\frac {l_2}{m_2}} \cdot \ldots
{\frac {l_{2n-2}}{m_{2n-2}}})$ by the theorem.
\begin{quote}
\end{quote}
\emph{Example of divisor calculations:}
$div(\{x \})= 2(0,0)-2 \infty$,
$div(\{y \})= (-1,0)+(0,0)+(1,0) -3 \infty$,
$div(\{ {\frac x y} \})= (0,0)+ \infty -(-1,0)-(1,0)$.
Let $E: y^2= x^3 -2x -5$ and $D= (2,3)+(2,-3)+O-2(-2,1) -(29,156)$.  We
can confirm $(2,3) + (2, -3) +O= O$ and $2(-2,1)+(29, 156)= O$ are elliptic
curve divisors.  The line containing $(2,3), (2,-3),O$ is $x-2=0$.  The
line containing $2(-2,1)$ and $(25,156)$ is $y-5x-11=0$, so
$div({\frac {x-2} {y-5x-11}})= D$.
\\
\\
{\bf Theorem:}
$D= \sum_P n_P P$ is a divisor of an elliptic curve $E$ iff
$\sum_P n_P =0$ and $\sum_P [n_P]P =0$.
\begin{quote}
\end{quote}
{\bf Definition:}
$f \circ n (P)= f(nP)$.  If $T \in E[n], \exists T' \in E[n^2]: nT'=T$ and
$g=f \circ n, div(g)= \sum_{R \in E[n]} [T'+R] -[R]$.  $g(P+S)^n= g(P)^n$ so
$({\frac {g(P+S)} {g(P)}})^n=1$.
Define the \emph{Weil pairing} as
$e_n(S,T)= {\frac {g(P+S)} {g(P)}}$.  If $\sigma$ is an automorphism,
$e_n( \sigma S, \sigma T)= \sigma e_n(S,T)$.  If $\alpha$ is separable,
$e_n( \alpha S, \alpha T)= e_n(S,T)^{deg(\alpha)}$.
\\
\\
{\bf Theorem (Siegal):}  An elliptic curve over ${\mathbb Q}$ has only finitely many
integer points.
\begin{quote}
\end{quote}
{\bf Theorem (Mazur):}  The set of torsion points $T_E({\mathbb Q})$ of
$E/{\mathbb Q}$ is one of
${\mathbb Z}/n{\mathbb Z}, n= 1,2,\ldots,10,12$ or
${\mathbb Z}/n{\mathbb Z} \times
{\mathbb Z}/2n{\mathbb Z}$, $n= 1,2,3,4$.
\begin{quote}
\end{quote}
{\bf Theorem:}
For every isogony $\psi:E \rightarrow E'$ there is a \emph{dual} $\hat{\psi}:
\hat{\psi} \psi= [2]_E$.  $Frob_n:
{\overline {F_p}} \rightarrow
{\overline {F_p}}$ and $F_{p^n}= \{ \alpha \in {\overline {F_p}}: Frob_n( \alpha )= \alpha \}$.
\begin{quote}
\end{quote}
{\bf Theorem:}
$\#E(F_p)= p+1 + \sum_{x=0}^{p-1} ({\frac {x^3+ax+b} {p}})$.
\begin{quote}
\end{quote}
{\bf Point Counting:}
Counting points by
baby-step giant-step is ($O(q^{{\frac 1 4} + \epsilon})$). Set
$N= \#E_q$ then
$q+1 - 2 {\sqrt q} \le N \le q+1 + 2 {\sqrt q}$; if $[m]P= \infty$ then $N=m$, probably.
Put $m=q+1- 2{\sqrt q} +k, l= \lceil {\sqrt 4 {\sqrt q}} \rceil, k= al+b$, then
$[m]P= [c]P + [a]S + [b] P, c= q+1-2{\sqrt q}, S= [l]P$ or $[c]P+ [a]S = - [b] P$.  Baby
step computes LHS and stores it.  Giant step computes RHS and does a lookup.
\\
\\
{\bf Schoof point counting:}
Let $\varphi$ be the Frobenius automorphism $\varphi(x,y)= (x^q , y^q )$.
Schoof calculates
$t \jmod{l}$ for a set of primes $l \in {\cal P}$ with
$\prod_{l \in {\cal P}} l > {4 {\sqrt q}}$ and then
construct $t$ using CRT finally returning $q+1-t$.
Here's how:\\
\jt (1) For $l=2$, $t= 0 \jmod{l}$ iff
$(x^3+ax+b,x^q-x) \ne 1$.  \\
\jt (2) if $l$ is odd, set $q_l = q \jmod{l}, |q_l| < {\frac l 2}$;
find $(x', y')= \varphi(x,y)^2 + q_l (x,y) \jmod{\psi_l (x,y))}$;
for $j= 1,2, \ldots {\frac {l-1} 2}$:\\
\jt \jt
(i) Compute $(x_j , y_j)=j(x,y)$; \\
\jt \jt
(ii) if
$x'-x_j^q = 0 \jmod{\psi_l}$, go to iii, if not, try next $j$, if all such
$j$'s have been tried, go to (iv); \\
\jt \jt
(iii) Compute $y', y_j$, if ${\frac {y'-y_j} y} = 0 \jmod{\psi_l}$ then
$t= j \jmod{l}$ otherwise $t= -j \jmod{l}$; \\
\jt \jt
(iv) Let $w^2=q \jmod{l}$, if no such $w$ exists,
$t=0 \jmod{l}$;\\
\jt \jt
(v) if $(x^q - x_w, \psi_l )=1$ then
$t=0 \jmod{l}$, otherwise, set $g= numerator({\frac {y^q - y_w} y}, \psi_l)$, if
$g \ne 1$, $t= 2w \jmod{l}$ otherwise
$t= -2w \jmod{l}$.
\\
\\
{\bf Definition of Complex Multiplication:}
$E: y^2= x^3 - x$, $\mu: E \rightarrow E$, $\mu(x,y)= (-x, iy)$,
$\mu^2= [-1]$.
\\
\\
{\bf Definition:}
If $D= \sum_{P \in E} n_P [P]$, $deg(D)= \sum_P n_P$ and $sum(D)= \sum_P n_P P$.
$div(f)= \sum_{P \in E({\overline K})} ord_P(f) [P] \in Div(E)$.
\\
\\
{\bf Theorem:}  Let $E$ be an elliptic curve and $f$ be a function on $E$ not $0$ then
(1) $f$ has finitely many poles and zeros, (2) $deg(div(f))=0$ and (3) if there are no
poles or zeros, $f$ is a constant.
\begin{quote}
\end{quote}
{\bf Theorem:}  Let $E$ be an elliptic curve and $D$ a divisor on $E$ with
$deg(D)=0$ then $\exists f$ on $E$ with $div(f)=D, sum(D)= \infty$.
\begin{quote}
\end{quote}
{\bf Pairing:} $e_n: E[n] \times E[n] \rightarrow \mu_n$.  $\mu_n$ is the $n$th roots of
unity.
\\
\\
{\bf Definition:}
For simplified case of elliptic curves in Weierstrauss form ($E: y^2=x^3+Ax+B$),
all polynomials, $f(x,y)$, on $E$, denoted $f \in K[E]$,
can reduced to normal form,
$f(x,y)= v(x)+yw(x)$.  If
$f(x,y)= v(x)+yw(x)$, ${\overline f}= v(x)-y w(x)$ and $N(f)= f {\overline f}$.
$deg(f)= \textnormal{max}(2 deg_x(v), 3+2deg_x(w))$.
\\
\\
{\bf Theorem:}
If $f \in K[E]$ then $deg(f)= deg_x(N(f))$ and $deg(f)$ has the usual properties.
\begin{quote}
\end{quote}
{\bf Theorem:}
If $r \in K(E)$ and $P \in E$, $\exists u \in K(E)$ such that (1) $u(P)=0$ and
(2) if $r \in K(E)$, $\exists s \in K(E): r=u^ds$.  Further, $d$ does not
depend on the choice of $u$.
\begin{quote}
\emph{Proof:}
For the SWF curve, we can specify $u$ as follows:
(1) For $P=(a,b) \in E, b \ne 0$, $u(x,y)= (x-a)$;
(2) For $P=(w,0) \in E$, $u(x,y)= y$; and
(3) For $P= \infty$, $u(x,y)= {\frac x y}$.
\end{quote}
{\bf Definition:}
For the notation of the previous theorem, define $ord_P(r)= d$.
\\
\\
{\bf Theorem:}
Let $r \in K(E)$, $\sum_P ord_P(r) = 0$.  If $f \in K[E]$ then the sum of the
multiplicities of the zeros of $f$ equals the degree of $f$.
\begin{quote}
\end{quote}
{\bf Definition:}
$\Delta \in Div(E)$ is principal if $\Delta= div(r)$ for some $r \in K(E)$.  The
principal divisors are denoted $Prin(E)$ and $Pic(E)= Div(E)/Prin(E)$.
$|\Delta|= \sum_{P \in P- \infty} |m(P)|$.
\\
\\
{\bf Theorem:}
Let $\Delta \in Div(E), \exists \tilde{\Delta} \in Div(E)$ with
$\Delta \sim \tilde{\Delta}$ with
$deg(\Delta)=deg(\tilde{\Delta})$ and
$|\tilde{\Delta}| \le 1$.  \begin{quote}
\end{quote}
{\bf Theorem:}
$\forall \Delta \in Div^0(E), \exists ! P \in E$:
$\Delta \sim \langle P \rangle -\langle \infty \rangle$.
\begin{quote}
\end{quote}
{\bf Theorem:}
Let $[n]: E \rightarrow E$ be represented by the rational function
$[n](P)=(g_n(P), h_n(P))$ then if $(n,p)=1$,
${\frac {g_n} x} (O) \sim {\frac 1 {n^2}}$ and
${\frac {h_n} y} (O) \sim {\frac 1 {n^3}}$.
\begin{quote}
\end{quote}
{\bf Theorem:}
Let $P, Q \in E$ and suppose $u$ is a uniformizing parameter at $P$,
define $T_Q(u)](R)=u(R+Q)$ then $T_Q(u)$ is the uniformizing
parameter at $P-Q$.
\begin{quote}
\end{quote}
{\bf Theorem:}
Let $E$ be an elliptic curve with $deg(D)=0$.  $\exists f \in K(E): D= div(f)$ iff
$sum(D)= \infty$.
\begin{quote}
\end{quote}
{\bf Theorem:}
Suppose $m>n>0$ and $m, n, m-n, m+n$ are all prime to $p$, then
$div(g_m-g_n)=
\langle E[m+n] \rangle+
\langle E[m-n] \rangle-
\langle E[m] \rangle-
\langle E[n] \rangle$.
If $(n,p)=1$, $|E[n]|= n^2$.
\begin{quote}
\end{quote}
{\bf Theorem:}
If $(n,p)=1$, $div(\phi_n)= \langle E[n] \rangle - n^2 \langle \infty \rangle$.
\begin{quote}
\end{quote}
{\bf Theorem:}
Suppose $r \in K(E)$ is a non-constant function, then $r$ takes on all values
including $\infty$.
\begin{quote}
\end{quote}
{\bf Theorem:}
Suppose $f: E \rightarrow E, K(E)$ is a non-constant then $f$ is onto.
\begin{quote}
\end{quote}
{\bf Definition:} The \emph{ramification index} of $F$ at $P$ is defined by
$e_F(P)= ord_P(u \circ F)$.
\\
\\
{\bf Theorem:}
$ord_P(u \circ F) = ord_{F(P)}(r) \cdot e_F(P)$.
\begin{quote}
\end{quote}
{\bf Definition:}
$F^*: Div(E) \rightarrow Div(E)$ is
$F^*( \langle Q \rangle)= \sum_{F(P)=Q} e_F(P) \langle P \rangle$.
\\
\\
{\bf Theorem:}
$F^*$ is 1-1, $div(r \circ F)= F^*(div(r))$ and
$ e_{F_1 \circ F_2}(P)= e_{F_1}(F_2(P)) \cdot e_{F_2}(P)$.
\begin{quote}
\end{quote}
{\bf Theorem:}
Suppose $\alpha: E \rightarrow E$ is a non-zero endomorphism then
$e_{\alpha}(P)$ is independent of $P$.
\begin{quote}
\end{quote}
For $m \in {\mathbb Z}, r \in K(E)$ with $D$ the derivative then
$D(r \circ [m])= (m \circ D(r)) \circ [m]$.
\begin{quote}
\end{quote}
{\bf Theorem:}
If $\varphi$ is the Frobenius homomorphism over $GF(q)$ then $e_{\varphi}= q$.
\begin{quote}
\end{quote}
{\bf Definition:}
$e_{\alpha}=1$ means $\alpha$ is separable.
\\
\\
{\bf Theorem:}
If $(m,p)=1$ then $[m]$ is separable.
\begin{quote}
\end{quote}
{\bf Theorem:}
If $(m,p)=1$ and $(n,p)=1$ then $[m]+[n] \varphi$ is separable.
\begin{quote}
\end{quote}
{\bf Theorem:}
If $P \in E[m]$ then $[m]^*( \langle T \rangle - \langle \infty \rangle)$
is principal.
\begin{quote}
\end{quote}
{\bf Definition:}
A rational map $F: E \rightarrow E'$ that is a group homomorphism from
$E$ into $E'$ is a \emph{morphism}.
\\
\\
{\bf Theorem:}
Let $T_1$ and $T_2$ be a basis for $E[m]$ then $e+m(T_1 , T_2 )$ is a primitive
$m$-th root of unity.
\begin{quote}
\end{quote}
{\bf Definition:}
If $\alpha$ is an endomorphism then $\alpha(E[m]) \subseteq E[m]$
\\
\\
{\bf Definition:}
Let $M$ be an algebraic extension of $L$ and $A_r$ is the linear map from $M$
to $M$ represented by multiplication by $r$ with respect to the basis
$r_1 , \ldots , r_n$ and $f_r(x)= det(xI-A_r)$ with constant term
$c_n = (-1)^n det(A_r)$.  Define $N(r)= det(A_r)$.
\\
\\
{\bf Definition:}
Suppose $\alpha: E \rightarrow E'$ be an isogony $K'= \alpha^*(K(E')) \subseteq K(E)$.
Define $N= N_{K(E)/K'}$ then $N(r)](P)= \prod_{\alpha(Q)= \alpha(P)} r(Q)^{e_{\alpha}}$.
\\
\\
{\bf Definition:}
Let $r \in K(E)$ and $D \in Div(E)$.  Suppose $div(r)$ and $D$ have disjoint support
define $r(D)= \prod_{P \in E} f(P)^{n_p}$.  For $f, g \in K(E)$ further
define
$\langle f, g \rangle_P= (-1)^{mn} [{\frac {f^n}{g^m}}](P)$ where
$m= ord_P(f)$ and
$n= ord_P(g)$.  $\langle f, g \rangle_P$ is called the local symbol of $f$ and $g$.
\\
\\
{\bf Weil Reciprocity Theorem:} $\prod_{P \in E} \langle f, g \rangle_P =1$.
\\
\\
{\bf Construction of the Weil Pairing:}
For $T \in E[n]$,
$\exists f: div(f)= n[T]- n[\infty]$.  Choose $T' \in E[n^2]: nT'=T$ then
$\exists g: div(g) = \sum_{R \in E[n]} ([T'+R]-[R])$.  Put $f \circ n (P)= f(nP)$.
This gives $f \circ n(P)= g^n(P)$.  Put
$e_n(S,T)= {\frac {g(P+S)} {g(P)}} \in \mu_n,  S \in E[n], P \in E({\overline K})$ and
$e_n$ satisfies the pairing properties.
\\
\\
{\bf Example:}
If $E(F_7): y^2= x^3+2$ then $E(F_7)[3]= {\mathbb Z}_3 \oplus {\mathbb Z}_3$.  To
compute $e_3((0,3),(5,1))$, $D_{(0,3)}= [(0,3)]-[\infty]$ and
$D_{(5,1)}= [(3,6)]-[(6,1)]$, $div(y-3)= 3 D_{(0,3)}$ and
$div({\frac {4x-y+1} {5x-y-1}})= 3 D_{(5,1)}$.
$f_{(0,3)}(D_{(5,1)})= {\frac {f_{(0,3)}(3,6)} {f_{(0,3)}(6,1)}}=
{\frac {6-3} {1-3}}= 2 \jmod{7}$.
$f_{(5,1)}(D_{(0,3)})= 4$, $e_3((0,3),(5,1))= {\frac 4 2}= 2 \jmod{7}$.
\\
\\
{\bf Theorem:}
If $F= {\mathbb Q}, {\mathbb R}, {\mathbb C}$
$E(F)= {\mathbb Z}/m{\mathbb Z} \times {\mathbb Z}/m{\mathbb Z}$;
if $F= {\mathbb F_p}, p \nmid m, \exists k:$
$E(F_{p^{jk}})= {\mathbb Z}/m{\mathbb Z} \times {\mathbb Z}/m{\mathbb Z}$.
\begin{quote}
\end{quote}
{\bf Note:}
If $m[P]=O$, $m[P]-m[O]$ is a divisor of $E(F)$.
\\
\\
{\bf Theorem:}  Let $P, Q \in E[m]$,
$div(f_P)= m[P]=m[O]$,
$div(f_Q)= m[Q]=m[O]$ then $e_m(P,Q)=
{\frac {f_P(Q+S)}{f_P(S)}}/
{\frac {f_Q(P-S)}{f_Q(-S)}}$ for $S \in E(F)$ is bilinear.
$E[m]= \langle aP_1+bP_2 \rangle$ and suppose.
$P= a_P P_1+b_P P_2 \rangle$, $e_m(P,Q)= \zeta^{det
\left(
\begin{array}{cc}
a_P &  a_Q \\
b_P &  b_Q \\
\end{array}
\right)
}$ where
$\zeta= e_m(P_1,P_2)$.
\\
\\
{\bf Theorem (for Miller's Algorithm):}  Let $E$ be an elliptic curve and
$P= (x_P, y_P)$ and
$Q= (x_Q, y_Q)$ are non-zero points on $E$.  Let $\lambda$ the the slope of
the line betwween $P$ and $Q$ and define
$g_{P, Q}= x-x_P$ if $\lambda= \infty$ and
$g_{P, Q}=
{\frac {y-y_p - \lambda(x-x_P)} {x+x_P+x_Q+\lambda^2}} $, otherwise.  Then (a)
$div(g_{P,Q})= [P]+[Q] -[P+Q]-[O]$ and
(b) For $m \ge 1$ with binary representation $[m_{n-1}, \ldots, m_0]$ satisfies
$div(f_P)= m[P]-[mP]-(m-1)[O]$ where
$g_{T,T}$ and
$g_{T,P}$ are defined by Miller's algorithm.\\
{\bf Miller's Algorithm:}\\
\jt 1. Set $T=P$ and $f=1$;\\
\jt 2. $\textnormal{for}(i=(n-2); i \ge 0; i--)$\\
\jt \jt 3. $f= f^2 \cdot g_{T,T}$;\\
\jt \jt 4. $T= 2T$;\\
\jt \jt 5. if($m_i=1$)\\
\jt \jt \jt 6. $f= f \cdot g_{T,P}$;\\
\jt \jt \jt 7. $T= T+P$;\\
\jt 8. $return(f)$.
\\
\\
{\bf Definition:}  Let
$P, Q \in E(F_q)[l]$, choose $f_P: div(f_P)= l[P]-l[Q]$ than the \emph{Tate pairing}
is $\tau(P,Q)= {\frac {f_P(Q+S)}{f_P(S)}}$ and the \emph{modified Tate pairing} is
$\hat{\tau}(P,Q)= \tau(P, Q)^{\frac {q-1}{l}}$.
\\
\\
{\bf Definition:}
Let $E$ be an elliptic curve over $F_p, m \ge 1, p \nmid m$, the \emph{embedding degree}
of $E$ with respect to $m$ is the smallest positive $k$ such that
$E(F_{p^k})[m]= {\mathbb Z}/m{\mathbb Z} \times {\mathbb Z}/m{\mathbb Z}$.
\\
\\
{\bf Theorem:}
Let $E$ be an elliptic curve over $F_p, p \nmid m$ and suppose $E$ has an element of
prime order $l \ne p$.  The embedding degree, $k$ is one of the following:
(1) $k=1$ ($l \le {\sqrt p}+1$);
(2) $k=l$ if $p=1 \jmod{l}$;
(3) if $p \ne 1 \jmod{l}$, $k$ is the smallest positive integer such that
$E(F_{p^k})= {\mathbb Z}/m{\mathbb Z} \times {\mathbb Z}/m{\mathbb Z}$.
\begin{quote}
\end{quote}
{\bf MOV Algorithm:}  Under the same conditions as the theorem:\\
\jt 1. Compute $N= \#E(F_{p^k})$, $l \mid N$;\\
\jt 2. Choose, at random, $T: T \in E(F_{p^k}), T \notin E(F_{p})$;\\
\jt 3. Compute $T'= (N/l)T$.  If $T'=O$, repeat step 2.\\
\jt 4. $lT'= O$, compute $\alpha= e_m(P, T')$ and $\beta= e_m(Q, T')$;\\
\jt 5. Solve $\beta= \alpha^n$ in $E(F_{p^k})$;\\
\jt 6. $Q=nP$.
\\
\\
{\bf Definition:}
Let $l \ge 3$, $P \in E[l]$, $lP=O$, $\psi: E \rightarrow E$.  $\psi$ is a
\emph{distortion map} for $P$ if (a) $\psi(nP)= n \psi(P)$ and (b)
$e_l(P, \psi(P))$ is a primitive $l$-th root of unity.
\\
\\
{\bf Theorem:}  If $E[l]=
{\mathbb Z}/l{\mathbb Z} \times {\mathbb Z}/l{\mathbb Z}$, TFAE:
(a) $P, Q$ is a basis for $E[l]$;
(b) $p \ne O$ and there is no $\alpha: Q= \alpha P$;
(c) $e_l(P,Q)$ is a primitive $l$-th root of unity.
\\
\\
Note: the \emph{decision ECDLP problem} is in $NP \cap co-NP$.
Attacks (1) Exhaustive Search -
to avoid, make sure $\#E_q=nh$, $n$ a large prime $>2^{160}$,
$h$, small; (2) Pohlig-Hellman/Pollard-$\rho$
use Pohlig to reduce from $n= p_1^{e_1} ... p_t^{e_t}$ to
$p$, since this step is easy, want $p$ large, Pollard costs $O({\sqrt p})$
[For Pollard, ``random'' function is $f(X)= X+ a_j P +b_j Q \jmod{p}$.];
(3) Isomorphism attack;
(4) MOV for anomalous curves - to avoid make sure $q=p^m$ and $p \nmid \#E_q$;
(5) Weil-Tate pairing - to avoid make sure $n \nmid (q^k-1), k \le C$ and that the
DLP problem for $F_{q^C}$ is intractable;
(6) Weil descent - to avoid, if $q=2^m$, make sure $m$ is prime.  Index calculus attack
is unlikely because the lifting required from $E_q(a,b)$ to
$E_{\mathbb Q} ({\overline a}, {\overline b})$ is unknown and the number of points of
small height in elliptic curves over ${\mathbb Q}$ is small.
\\
\\
{\bf Lenstra Elliptic Curve Factoring Method:}
\begin{enumerate}
\item $(n,6)= 1, n \ne m^r$.
\item Choose random $b, x_1, y_1$ between $1$ and $n$.
\item $c= {y_1}^2+{x_1}^3-bx_1 \jmod{n}$.
\item $(n, 4b^3+27c^2)= 1$.
\item $k= lcm(1,2, \ldots, K)$.
\item Compute $KP= ({\frac {a_k} {{d_k}^2}}, {\frac {b_k} {{{d_k}^3}}})$
\item $D= (d_k , n)$ If $D=1$, go to 5 and bump $K$ or go to 2 and select new curve.
\end{enumerate}
\subsection {Elliptic, Weierstrauss and Fermat}
{\bf Definition:}
Two curves $C, D$ are projectively equivalent if there is a projective
transformation $\phi$ with $\phi(C)=D$.
Every nonsingular cubic is equivalent to a curve which in affine coordinates is
$y^2= 4 x^3-g_2x-g_3= 0$.  This is the \emph{Weierstauss normal form.}
Note: To prove show that every non-singular curve has an inflexion point (triple tangent).
Map flex to $(0,0,1)$.
\\
\\
{\bf Elliptic Functions from Trigonometry:}
$S(x)= \int {\frac {dx} {\sqrt {1- x^2}}}$.
Let ${\frac {dx} {du}}= c(u)$,
$s(u)^2 + c(u)^2= 1$.  $s'(u)= c(u)$, $c'(u)=-S(u)$, $s(-u)=-s(u)$ and $c(-u)=c(u)$.
$s(x+y)= s(x)c(y)+s(y)c(x)$ and
$c(x+y)= c(x)c(y)-s(y)s(x)$. $\Omega: R \rightarrow S^1$ ($S^1$ is the 1-sphere - circle) by
$u \mapsto (c(u),s(u))$ is a morphism: $\Omega(x+y)= \Omega (x) \oplus \Omega (y)$.
$\Omega$ has a non-trivial kernel $K$ since $S^1$ is compact but $R$ isn't.
$K= 2 \pi {\mathbb Z}$.
These functions are periodic, satisfy the given derivatives, parameterize $S^1$ under the
indicated morphism and provide the integration property.
By analogy, set $F(k,v)= \int {\frac {dz} {\sqrt {(1-z^2 ) (1- k^2 z^2 )}}}$ and define
$sn$ by $F(k,sn(u))=u$.
$cn(u)= {\sqrt {1 - sn^2 (u)}}$,
$dn(u)= {\sqrt {1 - k^2 sn^2 (u)}}$.  $sn, cn, dn$ are doubly periodic with
periods $\omega_1 , \omega_2$.
\\
\\
{\bf Weierstrauss Parameterization:} $y(t)^2= x(t)^3+a x(t) + b$.
Let $\omega_1 , \omega_2 \in {\mathbb C}$ and
$\Lambda= \{ a \omega_1 + b \omega_2: a,b \in {\mathbb Z} \}$ which
is preserved under unimodular transformations,
$\Lambda'= \lambda - \{ {\vec 0} \}$.
${\mathbb C}/\Lambda$ is an equivalence class of complex numbers equivalent
to a torus.
\\
\\
{\bf Theorem:}
Any two basis of the same discrete group of an elliptic (doubly periodic)
function are related by \emph{unimodular} transformations.
The period module of a doubly periodic function is one of the following: (1) $0$;
(2) $n \omega, n \in {\mathbb Z}$;
(3) $n \omega_1+ m \omega_2, n,m \in {\mathbb Z}$.  In the latter case, there is a
canonical basis $(\omega_1 , \omega_2 )$ with $\tau= {\frac {\omega_2} {\omega_1}}$ such that
(i) $Im(\tau) >0$, (ii) $-{\frac 1 2} \leq Re(\tau) \leq {\frac 1 2}$, (iii) $|\tau| \geq 1$ and
$Re(\tau)>0$ if $|\tau|=1$.
\begin{quote}
\emph{Proof:}
Suppose neither (1) or (2) hold.  Let $\omega_1$ be the element of $M$ with smallest non zero modulus and
let $\omega_2$ be the element of $M \notin \{ m \omega_1 \}$ with the smallest non-zero modulus.  First,
if ${\frac {\omega_2} {\omega_1}} \in {\mathbb R}, \exists n:  n < {\frac {\omega_2}{\omega_1}} < n+1$ which
means $|n \omega_2 - \omega_1| < | \omega_1|$ which is a contradiction.  We may put
$\omega= \lambda_1 \omega_1 + \lambda_2 \omega_2, \lambda_1, \lambda_2 \in {\mathbb R} $
so $\exists m_1, m_2 \in {\mathbb Z}: | \lambda_1 - m_1 | \leq {\frac 1 2}, | \lambda_2 - m_2 | \leq {\frac 1 2}$.
But then $\omega'= \omega - m_1 \omega_1 - m_2 \omega_2 \in M$ and
$ |\omega'| \leq {\frac 1 2} |\omega_1|+ {\frac 1 2} |\omega_2| \leq |\omega_2| $.  To show (i, (ii), (iii) and
(iv), pick
$\omega_1$ and
$\omega_2$ as above.  We already have,
$ |\omega_1| \leq |\omega_2| $,
$ |\omega_2| \leq |\omega_1+\omega_2| $,
$ |\omega_2| \leq |\omega_1-\omega_2| $.
If $Im(\tau) <0$ replace $(\omega_1 , \omega_2)$ with $(-\omega_1 , \omega_2)$.
If $Re(\tau) = - {\frac 1 2}$ replace $(\omega_1 , \omega_2)$ with $(\omega_1 , \omega_1+\omega_2)$.
If $|\tau|=1$ and $Re(\tau) < 0$ replace $(\omega_1 , \omega_2)$ with $(-\omega_2 , \omega_1)$.
\end{quote}
{\bf Theorem:}  An elliptic function without poles is constant.
\begin{quote}
\end{quote}
{\bf Theorem:}  If $f$ is elliptic, $\sum_{a \in {\cal P}} Res(f,a) = 0$, ${\cal P}$ is the set of \emph{poles}.
\begin{quote}
\end{quote}
{\bf Theorem:}  If $f$ is elliptic,  $M$ is the module of periods,
$\langle a_1 , \ldots, a_n \rangle$ are the zeros,
$\langle b_1 , \ldots, b_n \rangle$ are the poles then
$\sum_i a_i= \sum_i b_i \jmod{M}$.
\begin{quote}
\end{quote}
{\bf Elliptic functions of order $2$:} $\wp(z)= z^{-2}+ a_2x^2+a_4 z^4 + \ldots $.
$\wp(z)= {\frac 1 {z^2}} + \sum_{\omega \ne 0} {\frac 1 {(z- \omega)^2)}}- {\frac 1 {\omega)^2}}$.  If
$|\omega| \geq 2|z|$,
$|{\frac 1 {(z- \omega)^2)}}- {\frac 1 {\omega)^2}}| \leq {\frac {10 |z|} {|\omega|^3}}$.
$\wp'(z)^2= 4 \wp(z)^3 - g_2 \wp - g_3= 4 (\wp(z)-e_1) (\wp(z)-e_2) (\wp(z)-e_3) $,
$e_1, e_2, e_3$ are distinct.  The equation can be solved by $z= \int^w {\frac {dw}
{\sqrt{4w^3 -  g_2 w - g_3}}}$.  Since $\wp(\omega_1 -z)= \wp(z)$,
$ \wp'({\frac {\omega_1} {2}})= 0 $; similarly,
$ \wp'({\frac {\omega_2} {2}})= 0 $ and
$ \wp'({\frac {\omega_1+\omega_2} {2}})= 0 $. So
$e_1=  {\frac {\omega_1} {2}}$,
$e_2= {\frac {\omega_2} {2}}$, and
$e_3= {\frac {\omega_1+\omega_2} {2}}$.  Put $\lambda( \tau )= {\frac {e_3 - e_2} {e_1- e_2}}$.
$\lambda$ is the quotient of two analytic functions in the upper half-plane.
\\
\\
{\bf Definitions:}
If $\lambda({\frac {a \tau + b} {c \tau + d}})= \lambda(\tau)$, the linear transformation is
an \emph{automorphism} of $\wp$.  The automorphisms
$\lambda({\frac {a \tau + b} {c \tau + d}})= \lambda(\tau)$ for
$
\left(
\begin{array}{cc}
a & b\\
c & d\\
\end{array}
\right)=
\left(
\begin{array}{cc}
1 & 0\\
0 & 1\\
\end{array}
\right) \jmod{2}
$ form the \emph{modular group}.
$\lambda(\tau + 1) = {\frac {\lambda(\tau)} {\lambda(\tau) -1}}$,
$\lambda({\frac 1 \tau}) = 1- \lambda(\tau)$.
$\wp(z, \Lambda)= {\frac 1 {z^2}}+ \sum_{\omega \in \Lambda'} ({\frac 1 {(z-\omega)^2}} -
{\frac 1 {\omega^2}})$.
\\
\\
{\bf Theorem:}
$\wp(z, \Lambda)$ converges uniformly for ${\mathbb C}/\Lambda$, $\wp(z)= \wp(-z)$ and
$\wp$ is doubly periodic.  As $z$ ranges over a fundamental region, $\wp$ takes on every
complex value twice.
$\wp(z, \Lambda)- {\frac 1 {z^2}}=
\sum_{\omega \in \Lambda'} ({\frac 1 {(z-\omega)^2}} - {\frac 1 {\omega^2}})=
\sum_{n=1}^{\infty} (n+1) z^n
(\sum_{\omega \in \Lambda'}
|{\frac 1 {\omega^{n+2}}}|)$. The final summation is an Eisenstein series of weight
$n+2$, denoted $G_{n+2}$.
$\wp(z)= {\frac 1 {z^2}} +3G_4 z^2+ 5G_6z^4 + \ldots$.
$P(z)= (\wp'(z), \wp(z))$ is a point on $y^2=4x^3-60 G_4 x -140 G_6$.
\begin{quote}
\end{quote}
{\bf Reimann surfaces:}
Glue two copies of $C$ to get ${\sqrt z}$.  For $N \in {\mathbb Z}, N>0$ define
$\Gamma_0 (N) \subseteq SL_2$ with $N|c$.
$\Gamma_0 (N)$ acts on $H$ and $H/
\Gamma_0 (N) \equiv X_0 (N) \backslash K$ where $K$ are the cusps.
$X_0 (N)$ is compact and
the members are the modular functions of level $N$.
\\
\\
{\bf Semi-Stable:} For all primes $l>3$, $l|Disc$ and only two of the roots are equal
$\jmod{l}$.
Frey curve: $C^F_{a,b}
{\buildrel\rm def\over =} \;
y^2= x (x-a^p) (x-b^p)$.
If $b$ is even and $a = -1 \jmod{4}$.  Frey curve is semi-stable.
\\
\\
{\bf Definitions:}
Denote $E_{A,B,C,D}({\mathbb Q}) {\buildrel\rm def\over =} \;
y^2= Ax^3 + B x^2 + CX +D, A, B, C, D \in {\mathbb Q}$.
Define $b_p$ to be the number of solutions to $E_{A,B,C,D}({\mathbb Q})= 0$. $E$
is \emph{modular} if
$\exists$ eigenfunction, $f(z)= \sum_n a_n e^{2 \pi i n z}$. $E/{\mathbb Q}$
is modular if $\exists f$ and eigenfunction with $a_p=p+1-b_p$ for all but finitely many
$p$.
\\
\\
{\bf Taniyama-Shimura Conjecture:}  Every elliptic curve is modular.
Alternate T-S: $E(A,B,C,D)$.  $\exists$ modular functions $f(z), g(z)$ such that
$g(z)^2= Af(z)^3+Bf(z)^2+C f(z) +D$.
\begin{quote}
\end{quote}
Define the \emph{conductor}
$Cond_{a,b,c}= \prod_{p|abc} p$.  Two elliptic curves are isomorphic
iff their \emph{$j$-invariants} are equal. The $j$-invariant of $C^F_{a,b} =
2^8 {\frac {(a^{2p} +b^{2p} +a^p b^p)^3} {a^{2p} b^{2p} c^{2p}}}$.
If $F({\frac {az+b} {cz+d}})= (cz+d)^2 F(z)$, $F$ is a modular form of weight $2$.
\\
\\
{\bf Proof of Fermat's Last Theorem:}  Suppose it's false and that $a^p + b^p= c^p$ is
a counterexample.  Let
$C^F_{a,b}$ be the Frey curve.  $Disc(C^F_{a,b})= a^{2p} b^{2p}c^{2p}$ so
$C^F_{a,b}$ is semi-stable.
Wiles proved every semi-stable elliptic curve is modular so
$C^F_{a,b}$ is modular and has a cusp form of weight $2$ and level $N$ where $N$
is the conductor.
If $l$ is an odd prime and $l|N$,
by Serre, we can obtain a new $F$ of
weight 2 of level $N/l$.  By induction, keep doing this until $N=2$.  The dimension
of the space of cusps is equal to the genus of compact Reimann surface $X_0(N)$.
But $Genus(X_0(2))= 0$, so there is no such cusp forms of weight $2$, level $2$.
This contradiction establishes the theorem.  Incidentially, the restriction of semi-stability
in Wiles Theorem has been removed.


\lab{Applications}{Leontief Input-Output Models}{Leontieff Input-Output Models}
\label{Leontief}

\objective{Explain the basics of input-output models, and apply them to analyze a state economy.}

One of the primary applications of linear algebra is modeling interactions between a large number of variables in a concise way.
We can use it to simplify the analysis of very complicated interactions in a variety of settings.

For example, in economics, one might be interested in how one the inputs of one sector of an economy are reliant upon the outputs of another economy.
To build a car, you need refined steel, plastic, and glass (among other materials).
These are all outputs of other sectors of the economy.
If there is an increase in car production there will necessarily be increases in production in other sectors of the economy like steel, plastic, and glass.
This type of interaction can be described using what is called a \emph{Leontief Input-Output model}, or just an Input-Output model.

Consider a simple economy divided up into sectors representing agriculture, textiles, and construction.
Given sufficient data, we can estimate how much of the output from one economy is used as an input to another.
An array representing the required inputs in this simple economy is shown in \ref{IOCoefTable}.
This is called a consumption matrix.
It can be especially useful when determining the short term impacts of a given change in the economy.
Directly, its columns represent the inputs required to produce one additional unit of output.
For example, to produce one extra unit of agriculture output we need to input about .4 units of agriculture, .04 units of textiles and .08 units of construction.

\begin{table}
\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
& Agriculture & Textiles & Construction \\ \hline
Agriculture & .4167 & .5357 & .25 \\ \hline
Textiles & .0417 & .0893 & .1667 \\ \hline
Construction & .0833 & .0714 & .0417 \\ \hline
\end{tabular}
\caption{Input Output Coefficients.
The columns represent the amount of each product needed to produce one additional unit of output.}
\label{IOCoefTable}
\end{center}
\end{table}

This sort of matrix can easily be computed from a table of inputs and outputs like the one shown in Table \ref{IORawTable}.
A matrix of input and output values like the one shown in Table \ref{IORawTable} is called an exchange matrix.
If we divide each column by the output of the corresponding industry we can obtain the coefficient matrix (consumption matrix).
We can do this in Python as follows, and we obtain the coefficients in Table \ref{IOCoefTable}.

\begin{lstlisting}
>>> import numpy as np
>>> IO = np.array([[250., 150., 30., 600.], [25., 25., 20., 280.], [50., 20., 5., 120.]])
>>> IOCoeff = IO[:,:3] / IO[:,3]
\end{lstlisting}

\begin{table}
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
& Agriculture & Textiles & Construction & Total Output \\ \hline
Agriculture & 250 & 150 & 30 & 600 \\ \hline
Textiles & 25 & 25 & 20 & 280 \\ \hline
Construction & 50 & 20 & 5 & 120 \\ \hline
Total Input & 325 & 195 & 55 & \\ \hline
\end{tabular}
\caption{Raw Input Output Data.
Columns represent inputs to an industry, and rows represent outputs.
Notice that not all the total output is used as input for the economy itself.}
\label{IORawTable}
\end{center}
\end{table}

We will now consider the sorts of things these matrices tell us.
By way of example, we will consider the total output needed to produce one extra unit of agriculture.
This certainly costs one unit of agriculture without accounting for the inputs required, but in order to produce one unit of agriculture, we also need inputs of .4167 units of agriculture, .0417 units of textiles, and .0833 units of construction.
We also need the inputs necessary to produce those inputs, and the inputs necessary to produce the inputs to the inputs for the inputs, etc.
Letting $d$ be the $3\times 1$ array of needed goods and $C$ be the coefficient matrix, the total cost of producing the outputs in $d$ will be

\[ d + C d + C^2 d + C^3 d + \dots = \left( \sum_{n=0}^{\infty} C^n \right) d = \left( I - C \right)^{-1} d\]

This is just a geometric series in $C$ (recall that $C^0$ is just the identity matrix).
The last formula for evaluating the geometric series can be derived in the same way as the formula for a geometric series of complex numbers.
It can be shown that such a geometric matrix series converges if and only if the absolute value of each of the eigenvalues of the matrix is strictly less than one.
This is a constraint that will be built into any real system, and the non-convergence of such a series would indicate a problem with the data.

Since the overall effect of the increase in demand from inputs to inputs to inputs, etc. may not happen immediately, it may often be helpful to just evaluate the first two terms in this summation.

\begin{problem}
What is the cost of producing an additional 50\% more units of construction?
Calculate the answer in two different ways.
For the first way, you will approximate the answer by using the geometric series.
Write a function named \li{geomSeries} that takes an $n \times n$ array $C$ and an $n$-dimensional
vector $d$ as inputs, as well as an integer $m$, and returns $(\sum_{i=0}^m C^i)d$.
You can then use this function to approximate the solution for $m=5$.
Also calculate the exact answer by evaluating $(I-C)^{-1}d$.
Notice how a distortion in the market for construction can drastically affect the need for other goods as well.
\end{problem}

If $X$ represents the total output of the economy, we can represent the ``net'' output to consumers $D$ (demand) by the following equation:
\[ D = X - C X \]
In other words, the amount of output used for consumption is the total output minus the portion that is reused as input for the economy itself.

This agrees with our geometric series since the total output of the economy must be the total output required to produce everything that isn't reused, i.e.
\[ X = \left( \sum_{n=0}^{\infty} C^n \right) D = \left( I - C \right)^{-1} D \]
Which is equivalent to $ D = X - C X $.

\begin{problem}
What are the demand levels for the three product economy?
\end{problem}

\begin{problem}
A city is trying to determine how to allocate funds for the celebration of the centennial anniversary of its founding.
It is expecting to take in \$100000 in space (lodging, etc.) \$100000 in consumable goods (food, and other retail items), and \$40000 in services.
Given that the coefficient matrix for space, consumable goods, and services is
\[ \begin{bmatrix}
.2 & .3 & .3 \\
.1 & .2 & .3 \\
.2 & .2 & .2 \\
\end{bmatrix} \]
Estimate the amount the city should be prepared to produce of each good.
\end{problem}

\section*{Importing CSV Data with Missing Values}
Before we get to the next problem, we review some important skills concerning reading and writing data with CSV files.
Recall that a CSV file, or a `Comma Separated Values' file, contains tabular data with the rows on separate lines
of the file, and the columns separated by some delimiter character, such as a comma or a white space. This is a
ubiquitous format for storing and transferring data, and Python has good support for it. We will use the \li{csv}
module to examine the contents of the file \li{missing_data.txt}.

\begin{lstlisting}
>>> import csv
>>> with open("missing_data.txt", "r") as csvfile:
>>>     myReader = csv.reader(csvfile, delimiter=';')
>>>     for row in myReader:
>>>         print row
['1.0', '2.2', '-3.0', '???']
['0.0', '2.1', '???', '5.1']
['0.4', '2.3', '-2.8', '4.8']
['0.3', '???', '-3.1', '5.0']
\end{lstlisting}

Notice the \li{with ... as ...} block. This is the preferred way to open and read from files, as it takes care of
closing the file once you exit the code block. We next created a \li{csv.reader} object around
the file, and specified the delimiter character to be a semi-colon. Then we iterated through the rows of our reader
and printed the contents.

Notice the entries that consist of three question marks. This indicates missing data. Such situations arise frequently
in real life, and we need to know ways to cope with imperfect or incomplete datasets. NumPy provides a very convenient
way to import homogeneous data from CSV files, even with missing data. We will use the \li{np.genfromtxt} function to
extract the data and fill in the missing values. In this example, we will set the missing value in the second column to
2.2, the missing value in the third column to -3.0, and the missing value in the fourth column to 4.9.

\begin{lstlisting}
>>> data = np.genfromtxt("missing_data.txt", delimiter=';',
                         missing_values = ['???'],
                         filling_values = {1:2.2, 2:-3.0, 3:4.9})
>>> data
array([[ 1. ,  2.2, -3. ,  4.9],
       [ 0. ,  2.1, -3. ,  5.1],
       [ 0.4,  2.3, -2.8,  4.8],
       [ 0.3,  2.2, -3.1,  5. ]])
\end{lstlisting}

We now have a NumPy array containing our filled-in data, and we can continue with whatever computations we desire. Notice
that we still needed to specify the delimiter character. The \li{missing_values} argument is a list of strings that
designate missing values, and the \li{filling_values} argument is a dictionary that maps the index of a column to the
value used to replace all the missing values in that column.
% The value of a \li{None} key in this dictionary will be used as a default fill value, except where you specify.
% Except the above statement doesn't seem to be true!!!
If you want to replace all missing values with the same
filling value, the \li{filling_values} argument can be set to this single filling value (if the desired filling value
happens to be 0, you must enclose it in quotes, i.e. \li{filling_values = '0'}).

If for some reason we want to omit certain rows from the beginning or end of the CSV file, we can use the \li{skip_header} or \li{skip_footer}
arguments, which are integers representing the number of lines at the beginning or end of the file to skip, respectively.
If we want to extract only certain columns from the CSV file, we can use the \li{usecols} argument, which is a list of
the column indices that we want to extract. In this next example, we omit the first row and the first column.

\begin{lstlisting}
>>> data = np.genfromtxt("missing_data.txt", delimiter=';',
                         missing_values = ['???'],
                         filling_values = {1:2.2, 2:-3.0, 3:4.9},
                         skip_header = 1,
                         usecols = (1,2,3))
>>> data
array([[ 2.1, -3. ,  5.1],
       [ 2.3, -2.8,  4.8],
       [ 2.2, -3.1,  5. ]])

\end{lstlisting}

\begin{problem}
The file \li{io2002table.txt} contains the raw input output data for the state of Washington in the year 2002.
The last column is the output vector. This is a CSV file with a tab delimiter, so you will set to set your
delimiter to \li{'\\t'}. The top row contains the names of each of the columns, and thus should be omitted when
you import the data. Similarly, the first column contains labels for the rows, and should be omitted. Finally,
there are some missing values in this file, and we assume that they are missing because there is not interaction
between the two corresponding sectors of the economy. Thus, set all missing values to 0. Import the data
to a NumPy array, and calculate the IO coefficient matrix and the demand vector.

Next, find the output vector if the demand for construction increases by ten percent.
Note that column 8 of the original CSV file corresponds to construction.
This will require increasing the correct entry of the initial
demand vector by ten percent and using \li{linalg.solve()} to find the new output vector.
\end{problem}

This method can be used to model economies of almost any scale.
One of the primary limitations is that this model assumes that production varies linearly in its inputs, which may not be realistic.


%
% OK
%
% setting up at Aug. third, 2009
% finished at Aug 5th, 2009
% seems to be OK
%
%
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\chapter{Introduction to delta function}
\label{sec:delta_function}

As what we can see, in the discussion of quantum mechanics or even the
quantum chemistry, we always meet the situation that the delta
function has to be used (for example, the normalization of free
particle wave function, the discussion of position and momentum
operator in \ref{sec:PRAMR_in_position_representation} etc. For the
case of the continuous basis functions, the delta function is
frequently used). Hence in this part, we are going to give some
mathematical introduction for the delta function.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Definition of delta function}
\label{sec:definition_delta_function}
%
%
%
%
The delta function can be considered as some ``peculiar'' function
which has two features below:
\begin{equation}
\label{DELTAeq:1}
\delta (x) = \begin{cases}
    \infty & x = 0 \\
    0 & x \neq 0
  \end{cases}
\end{equation}

\begin{equation}
\label{DELTAeq:2}
 \int_{-\infty}^{+\infty}\delta (x)dx =
\int_{-\epsilon}^{+\epsilon}\delta (x)dx = 1
\end{equation}
Here $(-\epsilon, \epsilon)$ is some ``small enough'' open interval
around the $0$. The (\ref{DELTAeq:1}) and (\ref{DELTAeq:2}) are also
the definition for the delta function, which indicates that the
delta function is infinity at $0$, but quickly tends to $0$ as is
apart from the $0$ point.

Now let's give some examples. Firstly, consider the function below:
\begin{equation}\label{}
\Psi_{n}(x) = \frac{\sin nx}{\pi x} \quad \text{$n$ is some natural
number }
\end{equation}
Firstly, we can prove that:
\begin{align}\label{}
\lim_{x \rightarrow 0}\Psi_{n}(x) = \lim_{x \rightarrow 0}\frac{\sin
nx}{\pi x} = \lim_{x \rightarrow 0}\frac{\sin nx}{n x}\frac{n}{\pi}
= \frac{n}{\pi}
\end{align}
So we can think that $\Psi_{n}(0) = \frac{n}{\pi}$. Here as $n
\rightarrow +\infty$, $\Psi_{\infty}(0) \rightarrow +\infty$. Hence
the (\ref{DELTAeq:1}) is satisfied.

Secondly, from the improper integral theory, we can know that:
\begin{equation}\label{}
\int_{0}^{+\infty}\frac{\sin nx}{x}dx = \frac{\pi}{2}
\end{equation}
Hence we have the integral as:
\begin{align}\label{}
\int_{-\infty}^{+\infty}\frac{\sin nx}{\pi x}dx &=
2\frac{1}{\pi}\int_{0}^{+\infty}\frac{\sin nx}{x}dx \quad
 \text{here the $\Psi_{n}(x)$ is even function}\nonumber \\
&=2\frac{1}{\pi}\times\frac{\pi}{2} \nonumber \\
&=1
\end{align}
So the (\ref{DELTAeq:2}) is satisfied. Thus we can have
$\Psi_{\infty}(x) = \delta(x)$.

From the definition of $\Psi_{n}(x)$, we can even prove that the
function below is also the delta function:
\begin{equation}\label{}
\Phi(x) = \frac{1}{2\pi}\int_{-\infty}^{+\infty} e^{ikx}dk
\end{equation}

This function is very important in discussing the property related
to plane wave functions. Firstly, as $x = 0$, it's easy to see tat
the $\Phi(0) = \frac{1}{2\pi}\int_{-\infty}^{+\infty} dk \Rightarrow
\infty$, so the (\ref{DELTAeq:1}) is satisfied.

As for the (\ref{DELTAeq:2}) since we have (suggest that $a$ is some
natural number):
\begin{align}\label{}
\Phi_{a}(x) = \frac{1}{2\pi}\int_{-a}^{+a} e^{ikx}dk &=
\frac{1}{i2x\pi}e^{ikx}\Big|^{a}_{-a} \nonumber \\
&=\frac{1}{i2x\pi}2i\sin ax \quad \cos(a) = \cos(-a)\nonumber \\
&=\frac{\sin ax}{x\pi}
\end{align}
Hence the integral for the $\Phi(x)$ in the whole real space has
been resorted to the $\Psi_{a\rightarrow \infty}(x)$; so we have:
\begin{equation}\label{}
\Phi(x) = \delta(x)
\end{equation}

Finally, let's consider the Gauss distribution function:
\begin{equation}\label{}
\Omega_{\sigma}(x) =
\frac{1}{\sqrt{2\pi\sigma}}e^{-\frac{x^{2}}{2\sigma}}
\end{equation}

Firstly, we can see that:
\begin{align}\label{}
\Omega_{\sigma}(0) &= \frac{1}{\sqrt{2\pi\sigma}}\times 1 \quad
\underrightarrow{\sigma \rightarrow 0} \nonumber
\\
\Omega_{\sigma \rightarrow 0 }(0) &= \infty
\end{align}

On the other hand, since from the improper integral theory, we have:
\begin{equation}\label{DELTAeq:20}
\int_{0}^{+\infty}e^{-x^{2}}dx = \frac{\sqrt{\pi}}{2}
\end{equation}
Then we have the integral for the $\Omega_{\sigma}(x)$ as:
\begin{align}\label{}
\int_{-\infty}^{+\infty}
\frac{1}{\sqrt{2\pi\sigma}}e^{-\frac{x^{2}}{2\sigma}} dx &=
\frac{1}{\sqrt{2\pi\sigma}}\int_{-\infty}^{+\infty}
e^{-\frac{x^{2}}{2\sigma}} dx \nonumber \\
&= \frac{2}{\sqrt{2\pi\sigma}}\int_{0}^{+\infty}\sqrt{2\sigma}
e^{-(\frac{x}{\sqrt{2\sigma}})^{2}} d\frac{x}{\sqrt{2\sigma}} \quad
\text{$\Omega_{\sigma}(x)$ is even} \nonumber \\
&=\frac{2}{\sqrt{\pi}}\times\frac{\sqrt{\pi}}{2} \quad
\text{From the \ref{DELTAeq:20}}\nonumber \\
&= 1
\end{align}
Hence we have $\Omega_{0}(x) = \delta(x)$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Characters of delta function}
\label{sec:character_delta_function}
%
%
%
%
In this section let's state some characters of delta function.

\begin{theorem}
$\delta(x)$ is some even function: $\delta(x) = \delta(-x)$
\end{theorem}

\begin{proof}
the proof is straightforward. For (\ref{DELTAeq:1}), the
$\delta(-x)$ is naturally satisfied, and for the integral in the
(\ref{DELTAeq:2}) we have:
\begin{equation}\label{}
\int_{-\infty}^{+\infty}\delta(-x)dx =
-\int_{-\infty}^{+\infty}\delta(-x)d(-x) =
\int_{-\infty}^{+\infty}\delta(t)dt = 1
\end{equation}
Hence  $\delta(x) = \delta(-x)$. \qedhere
\end{proof}

\begin{theorem}
$\delta(ax) = \frac{1}{|a|}\delta(x)$
\end{theorem}

\begin{proof}
Firstly suggest that $a>0$, then:
\begin{equation}\label{DELTAeq:3}
\begin{split}
  \int_{-\infty}^{+\infty}\delta(ax)dx &=
  \frac{1}{a}\int_{-\infty}^{+\infty}\delta(ax)d(ax) \\
    &= \frac{1}{a}\delta(x)
\end{split}
\end{equation}

For the $a < 0$ (in this case, if we simply repeat the progress in
\ref{DELTAeq:3}; the upper limit and the lower limit in integral
should change), since we have $\delta(ax) = \delta(-ax)$; then we
can replace the $a$ with $|a|$ to repeat the demonstration in
(\ref{DELTAeq:3}). So finally we can get the conclusion. \qedhere
\end{proof}

\begin{theorem}
\label{DELTA:4}
$\int_{-\infty}^{+\infty}f(x)\delta(x)dx = f(0)$.
\end{theorem}

\begin{proof}
\begin{equation}\label{}
\begin{split}
  \int_{-\infty}^{+\infty}f(x)\delta(x)dx &=
  \int_{-\epsilon}^{+\epsilon}f(x)\delta(x)dx \\
    &= f(0)\int_{-\epsilon}^{+\epsilon}\delta(x)dx \\
    &= f(0)
\end{split}
\end{equation} \qedhere
\end{proof}

The theorem (\ref{DELTA:4}) is some important expression for delta
function. Now consider the delta function $\delta (x - x^{'})$,
where $x^{'}$ is some arbitrary number; it can be easily understood
that the (\ref{DELTAeq:1}) and (\ref{DELTAeq:2}) are converted as:
\begin{equation}
\label{DELTAeq:5} \delta (x - x^{'}) = \begin{cases}
    \infty & x = x^{'} \\
    0 & x \neq x^{'}
  \end{cases}
\end{equation}

\begin{equation}
\label{DELTAeq:6}
 \int_{-\infty}^{+\infty}\delta (x - x^{'})dx =
\int_{-\epsilon}^{+\epsilon}\delta (x - x^{'})dx = 1
\end{equation}
The (\ref{DELTAeq:1}) and (\ref{DELTAeq:2}) can be considered as the
special case while $x^{'} = 0$. Through this expansion, by using the
theorem (\ref{DELTA:4}) we can prove that the relation below is
true:
\begin{theorem}
\begin{equation}\label{DELTAeq:7}
\int_{-\infty}^{+\infty}f(x)\delta(x - x^{'})dx =
\int_{-\infty}^{+\infty}f(x^{'})\delta(x - x^{'})dx^{'} =f(x)
\end{equation}
\end{theorem}

\begin{proof}
For the first equation in the (\ref{DELTAeq:7}), we can directly use
the even function character to prove it. For the second equation in
(\ref{DELTAeq:7}), we have:
\begin{equation}\label{}
\begin{split}
  \int_{-\infty}^{+\infty}f(x)\delta(x - x^{'})dx &=
   \int_{-\infty}^{+\infty}f(t+x^{'})\delta(t)dt
   \quad t = x - x^{'}\\
    &= f(t+x^{'})\Big|_{t=0} \quad \text{From theorem \ref{DELTA:4}}\\
    &= f(x^{'})
\end{split}
\end{equation}
 \qedhere
\end{proof}

\begin{theorem}
$\delta(x) = \delta^{*}(x)$
\end{theorem}

\begin{proof}
Suggest that we have some arbitrary real function of $f(x)$,
\begin{equation}\label{}
\begin{split}
  \left[\int_{-\infty}^{+\infty}f(x)\delta(x)dx\right]^{*} &=
   \int_{-\infty}^{+\infty}f(x)\delta^{*}(x)dx \\
    &=f^{*}(0) = f(0) \\
    &= \int_{-\infty}^{+\infty}f(x)\delta(x)dx \Rightarrow \\
    &= \int_{-\infty}^{+\infty}f(x)
    \Big(\delta(x)-\delta^{*}(x)\Big)dx = 0
\end{split}
\end{equation}
Hence we have $\delta(x) = \delta^{*}(x)$. \qedhere
\end{proof}

\begin{theorem}
$\int_{-\infty}^{+\infty}\delta(x - x^{'}) \delta(x - x^{''})dx =
\delta(x^{'} - x^{''})$. Here $x^{'}$ and $x^{''}$ are both of
arbitrary real number.
\end{theorem}

\begin{proof}
From  (\ref{DELTAeq:7}), we set the $f(x) = \delta(x)$ so according
to the (\ref{DELTAeq:7}):
\begin{equation}\label{}
\begin{split}
  \int_{-\infty}^{+\infty}f(x- x^{''})\delta(x - x^{'})dx
  &= f(x- x^{''})\Big|_{x =x^{'}} \\
    &= f(x^{'}- x^{''})
\end{split}
\end{equation}
\qedhere
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Example to use delta function:
the normalization of plane wave functions}
\label{sec:normalization_delta_function}
%
%
%
%
Now by using the delta function let's solve some important problem
in quantum mechanics: how to normalize the plane wave function?

According to the (\ref{BASICeq:14}), the plane wave function can be
expressed as:
\begin{equation}\label{}
\Psi_{p}(x) = e^{\frac{ipx}{\hbar}}
\end{equation}
Actually it's the eigen state for the $\hat{p}$:
\begin{equation}\label{}
\hat{p}\Psi_{p}(x) = p\Psi_{p}(x)
\end{equation}

Now let's go to see how to normalize this function:
\begin{equation}\label{}
\begin{split}
  \langle \Psi_{p}(x^{'})|\Psi_{p}(x)\rangle
  =\int_{-\infty}^{+\infty}\Psi^{*}(x')\Psi(x)dp
  &= \int_{-\infty}^{+\infty} e^{\frac{ip(x-x^{'})}{\hbar}}dp \\
  &= \hbar\int_{-\infty}^{+\infty} e^{\frac{ip(x-x^{'})}{\hbar}}
  d\left(\frac{p}{\hbar}\right) \\
  &= 2\pi\hbar\delta(x-x^{'})
\end{split}
\end{equation}

Hence if we integrate over all the $x$:
\begin{equation}\label{}
\begin{split}
  \int_{-\infty}^{+\infty}
  \langle \Psi_{p}(x^{'})|\Psi_{p}(x)\rangle dx
  &= \int_{-\infty}^{+\infty} 2\pi\hbar\delta(x-x^{'}) dx \\
    &= 2\pi\hbar
\end{split}
\end{equation}

Thus finally we can express the plane wave function as:
\begin{equation}\label{}
\Psi_{p}(x) = \frac{1}{\sqrt{2\pi\hbar}}e^{\frac{ipx}{\hbar}}
\end{equation}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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\chapter{Eigenfunctions of CT systems}
To summarize the course so far given an input signal $x(t)$ and a LTI system described (equivalently) by a linear, constant coefficient differential equation, impulse response, or a block diagram, we can determine the output using convolution. This is referred to as \emph{time-domain} analysis.

The advantages of this approach are that the analysis is straightforward (if cumbersome) and it applies to all LTI systems, stable or otherwise. Time-domain representations of signals are also intuitive given their direct application in physical systems.

There are also some disadvantages. First, time-domain analysis does not scale well to larger systems since analysis with block diagram decompositions requires convolution, and in the case of the feedback motif dealing with inverse systems or de-convolution. Second, it is difficult to design an impulse responses for a given purpose. Finally implementing a system directly from an impulse response is not intuitive.

We can borrow a technique from mathematics to overcome these disadvantages by transforming the \emph{domain} of the representations to one in which the operation of convolution becomes one of multiplication. This approach, called generally \emph{frequency domain} analysis has a number of advantages and will be our focus for the remainder of the course.

\section{The Response of LTI Systems to Complex Exponentials}

Recall convolution can be viewed as a decomposition of a signal into an infinite sum of $\delta$ functions plus the linearity property.
\[
x(t) = \int\limits_{-\infty}^{\infty} x(\tau)\delta(t-\tau) \; d\tau \;\longrightarrow\; y(t) = \int\limits_{-\infty}^{\infty} x(\tau)h(t-\tau) \; d\tau
\]

We now consider a different decomposition based on the complex exponential, $e^{st}$ for $s \in \mathbb{C}$, rather than $\delta$ functions. As we will see this decomposition simplifies convolution, turning it into multiplication.

\subsection{Eigenfunction $e^{st}$ and Transfer Function $H(s)$}

Let $x(t) = e^{st}$ for $s\in \mathbb{C}$, then $y(t) = h(t) * x(t) = x(t) * h(t)$ and by the definition of convolution
\begin{align*}
  y(t) & = \int\limits_{-\infty}^{\infty}h(\tau)x(t-\tau)\; d\tau\\
  &= \int\limits_{-\infty}^{\infty}h(\tau)e^{s(t-\tau)}\; d\tau = \int\limits_{-\infty}^{\infty}h(\tau)e^{st}e^{-s\tau}\; d\tau\\
  &= e^{st} \int\limits_{-\infty}^{\infty}h(\tau)e^{-s\tau}\; d\tau\\
  &= e^{st}H(s)
\end{align*}
where $H(s) = \int\limits_{-\infty}^{\infty}h(\tau)e^{-s\tau}\; d\tau$ is the \emph{Laplace Transform} of the impulse response, $h(t)$. $H(s)$ is called the \emph{transfer function} or \emph{Eigenvalue} of the system and $e^{st}$ is the \emph{Eigenfunction} for CT LTI systems.

Similar to the impulse function, the complex exponential is a special signal because it's response is easy to determine. It is just the same signal scaled by a multiplicative factor as illustrated below:

\begin{center}
  \includegraphics[scale=0.6]{graphics/lti-ct-complex-exp.pdf}
\end{center}

\begin{example}
Suppose $H(s) = \frac{1}{s+1}$ and $x(t) = e^{(-4+j2\pi)t}$. Then the output is
\begin{align*}
  y(t) &= H(-4+j2\pi)e^{(-4+j2\pi)t}\\
  &= \frac{1}{-4+j2\pi+1}e^{(-4+j2\pi)t}\\
  &= \frac{1}{-3+j2\pi}e^{(-4+j2\pi)t} \; ,
\end{align*}
another complex exponential.\\
$\blacksquare$
\end{example}

Given $H(s)$ and inputs that are sums of complex exponentials, the output is easy to determine.

\begin{center}
  \includegraphics[scale=0.6]{graphics/ct-linear-response-complex-exp.pdf}
\end{center}
In some cases the sums are countably infinite while in others the uncountably infinite so that the sums become integrals.

\begin{example} Consider the CT system with impulse response response
  \[
  h(t) = e^{-5t}u(t)
  \]
  Determine the Eigenvalues that corresponds to the input $x(t) = \cos(t)$ and the output $y(t)$.\\

  Solution: We note the cosine can be decomposed into two complex exponentials as
  \[
  \cos(t) = \frac{1}{2}e^{jt} + \frac{1}{2}e^{-jt}
  \]
  Thus in terms of the general decomposition there are two terms with complex constants $s_1 = 0+j$ and $s_2 = 0-j$ and real constants $a_1 = a_2 = \frac{1}{2}$.
  \[
   x(t) = \sum_i a_i e^{s_it} = a_1 e^{s_1t} + a_2 e^{s_2t} = \frac{1}{2}e^{jt} + \frac{1}{2}e^{-jt} = \cos(t)
   \]
   Then the output is given by
   \[
   y(t) = \sum_i H(s_i) a_i e^{s_it} = H(s_1) a_1 e^{s_1t} + H(s_2) a_2 e^{s_2t} = H(j) \frac{1}{2}e^{jt} + H(-j)\frac{1}{2}e^{-jt}
   \]
   which requires we find the Eigenvalues $H(j)$ and $H(-j)$. To do so we use the Laplace integral
   \[
   H(j) = \int\limits_{-\infty}^{\infty}h(\tau)e^{-j\tau}\; d\tau = \int\limits_{0}^{\infty} e^{-5\tau} \, e^{-j\tau}\; d\tau = \int\limits_{0}^{\infty} e^{-(j+5)\tau}\; d\tau = \frac{-1}{j+5} e^{-(j+5)\tau} \Big|_{0}^{\infty} = \frac{1}{j+5}
   \]
   Similarly
   \[
   H(-j) = \int\limits_{-\infty}^{\infty}h(\tau)e^{j\tau}\; d\tau = \int\limits_{0}^{\infty} e^{-5\tau} \, e^{j\tau}\; d\tau = \int\limits_{0}^{\infty} e^{-(-j+5)\tau}\; d\tau = \frac{-1}{-j+5} e^{-(j+5)\tau} \Big|_{0}^{\infty} = \frac{1}{-j+5}
   \]
   Substituting back into the output equation gives
   \begin{align*}
     y(t) &= H(j) \frac{1}{2}e^{jt} + H(-j)\frac{1}{2}e^{-jt}\\
     &=  \frac{1}{j+5} \frac{1}{2}e^{jt} + \frac{1}{-j+5} \frac{1}{2}e^{-jt}
   \end{align*}
   We can simplify this expression using the polar form of the Eigenvalues
   \begin{align*}
     y(t) &= \frac{1}{j+5} \frac{1}{2}e^{jt} + \frac{1}{-j+5} \frac{1}{2}e^{-jt}\\
     &= Re^{j\theta} \frac{1}{2}e^{jt} + Re^{-j\theta} \frac{1}{2}e^{-jt}\\
     &= R \frac{1}{2}e^{jt + j\theta} + R \frac{1}{2}e^{-jt -j\theta}\\
     &= R\cos(t + \theta)
   \end{align*}
   where
   \[
   R = \left|\frac{1}{j+5}\right| = \frac{1}{\sqrt{26}} \mbox{ and } \theta = \angle{\frac{1}{j+5}} = -\arctan \frac{1}{5}
   \]
   Note for this system, given a sinusoidal input, the output is a scaled and phase shifted sinusoid at the same frequency, where the scaling factor and phase shift is system dependent. It is illustrative to compare this analysis to the time-domain analysis of the same impulse response and input using convolution.
   $\blacksquare$
\end{example}

\section{Decomposition of signals using complex exponentials}

In this course we consider the cases of stable CT systems. Recall a stable system is one in which a bounded input leads to a bounded output, or equivalently the impulse response is absolutely integrable. We will consider two decompositions of the input:

  \begin{itemize}
  \item \emph{Fourier Series}: When $x(t)$ is periodic with fundamental frequency $\omega_0$, $\Re{(s)} = 0$ so that $s = jk\omega_0$, and the decomposition is a countably infinite sum. This gives the input-output relationship
    \[
    x(t) = \sum\limits_{k = -\infty}^{\infty} a_k \, e^{j k\omega_0 t} \; \longrightarrow\; y(t) = \sum\limits_{k = -\infty}^{\infty} H(j k\omega_0)\, a_k \, e^{j k\omega_0 t}
    \]
    where $H(j k\omega_0)$ are the Eigenvalues, also called the \emph{frequency response}.
  \item \emph{Inverse Fourier Transform}: When $x(t)$ is a-periodic, $\Re{(s)} = 0$ so that $s = j\omega$, and the decomposition is an uncountably infinite sum (real integral over $\omega$). This gives the input-output relationship
    \[
    x(t) = \frac{1}{2\pi}\int\limits_{-\infty}^{\infty} X(\omega) \, e^{j \omega t}\; d\omega \;\longrightarrow\; y(t) = \frac{1}{2\pi}\int\limits_{-\infty}^{\infty} H(\omega) X(\omega) \, e^{j \omega t}\; d\omega
    \]
    where $H(j \omega)$ are the Eigenvalues, again called the \emph{frequency response}.
  \end{itemize}

  Other courses (e.g. ECE 3704) look at the general case of unstable systems and $s \in \mathbb{C}$ with decompositions:

  \begin{itemize}
  \item \emph{One-Sided Laplace Transform}: $x(t)$ is causal and the decomposition is an uncountably infinite sum (complex integral)
  \item \emph{Two-Sided (Bilateral) Laplace Transform}: $x(t)$ is non-causal and the decomposition is an uncountably infinite sum (complex integral). This is the most general case for CT LTI systems.
  \end{itemize}

  While the Laplace decompositions require complex integration, they can be understood and computed using algebra and a table of forward transforms, which only require integration of a complex function of a real variable $t$ (this is the general approach taken in upper level courses). However, this is outside the scope of this course because of time limitations.

Instead, we will be spending the next few weeks going through the CT Fourier decompositions in some detail. You will also learn how to find the CT frequency response for a stable system, and see how to use both for analysis.