path: root/controle-20240410.tex

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\begin{document}
\ifcorrige
\title{ACCQ205\\Final exam — answer key\\{\normalsize Algebraic curves}}
\else
\title{ACCQ205\\Final exam\\{\normalsize Algebraic curves}}
\fi
\author{}
\date{2024-04-10}
\maketitle

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\vskip1truein\relax

\noindent\textbf{Instructions.}

This exam consists of three completely independent exercises.  They
can be tackled in any order, but students must clearly and readably
indicate where each exercise starts and ends.

\medbreak

Answers can be written in English or French.

\medbreak

Use of written documents of any kind (such as handwritten or printed
notes, exercise sheets or books) is authorized.

Use of electronic devices of any kind is prohibited.

\medbreak

Duration: 2 hours

\ifcorrige
This answer key has 8 pages (this cover page included).
\else
This exam has 4 pages (this cover page included).
\fi

\vfill
{\noindent\tiny
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Git: \input{vcline.tex}
\immediate\write18{echo ' (stale)' >> vcline.tex}
\par}

\pagebreak


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\exercise

We say that a set of eight distinct points $p_0,\ldots,p_7$ in the
projective plane $\mathbb{P}^2$ over a field $k$ is a
\textbf{Möbius-Kantor configuration} when the points $p_0,p_1,p_3$ are
aligned, as well as $p_1,p_2,p_4$ and $p_2,p_3,p_5$ and so on
cyclically mod $8$, and no other set of three of the $p_i$ is aligned.
In other words, this means that $p_i,p_j,p_k$ are aligned if and only
if $\{i,j,k\} = \{\ell,\; \ell+1,\; \ell+3\}$ for some $\ell \in
\mathbb{Z}/8\mathbb{Z}$, where the subscripts are understood to be
mod $8$.

The following figure (which is meant as a \emph{symbolic
representation} of the configuration and not as an actual geometric
figure!) illustrates the setup and can help keep track of which points
are aligned with which:

\begin{center}
\vskip-7ex\leavevmode
\begin{tikzpicture}
\coordinate (P0) at (2cm,0);
\coordinate (P1) at (1.414cm,1.414cm);
\coordinate (P2) at (0,2cm);
\coordinate (P3) at (-1.414cm,1.414cm);
\coordinate (P4) at (-2cm,0);
\coordinate (P5) at (-1.414cm,-1.414cm);
\coordinate (P6) at (0,-2cm);
\coordinate (P7) at (1.414cm,-1.414cm);
\draw[shorten <=-0.5cm, shorten >=-0.3cm] (P0) -- (P1) .. controls ($2.5*(P1)-1.5*(P0)$) and ($2.5*(P2)-1.5*(P1)$) .. (P3);
\draw[shorten <=-0.5cm, shorten >=-0.3cm] (P1) -- (P2) .. controls ($2.5*(P2)-1.5*(P1)$) and ($2.5*(P3)-1.5*(P2)$) .. (P4);
\draw[shorten <=-0.5cm, shorten >=-0.3cm] (P2) -- (P3) .. controls ($2.5*(P3)-1.5*(P2)$) and ($2.5*(P4)-1.5*(P3)$) .. (P5);
\draw[shorten <=-0.5cm, shorten >=-0.3cm] (P3) -- (P4) .. controls ($2.5*(P4)-1.5*(P3)$) and ($2.5*(P5)-1.5*(P4)$) .. (P6);
\draw[shorten <=-0.5cm, shorten >=-0.3cm] (P4) -- (P5) .. controls ($2.5*(P5)-1.5*(P4)$) and ($2.5*(P6)-1.5*(P5)$) .. (P7);
\draw[shorten <=-0.5cm, shorten >=-0.3cm] (P5) -- (P6) .. controls ($2.5*(P6)-1.5*(P5)$) and ($2.5*(P7)-1.5*(P6)$) .. (P0);
\draw[shorten <=-0.5cm, shorten >=-0.3cm] (P6) -- (P7) .. controls ($2.5*(P7)-1.5*(P6)$) and ($2.5*(P0)-1.5*(P7)$) .. (P1);
\draw[shorten <=-0.5cm, shorten >=-0.3cm] (P7) -- (P0) .. controls ($2.5*(P0)-1.5*(P7)$) and ($2.5*(P1)-1.5*(P0)$) .. (P2);
\fill[black] (P0) circle (2.5pt);
\fill[black] (P1) circle (2.5pt);
\fill[black] (P2) circle (2.5pt);
\fill[black] (P3) circle (2.5pt);
\fill[black] (P4) circle (2.5pt);
\fill[black] (P5) circle (2.5pt);
\fill[black] (P6) circle (2.5pt);
\fill[black] (P7) circle (2.5pt);
\node[anchor=west] at (P0) {$p_0$};
\node[anchor=south west] at (P1) {$p_1$};
\node[anchor=south] at (P2) {$p_2$};
\node[anchor=south east] at (P3) {$p_3$};
\node[anchor=east] at (P4) {$p_4$};
\node[anchor=north east] at (P5) {$p_5$};
\node[anchor=north] at (P6) {$p_6$};
\node[anchor=north west] at (P7) {$p_7$};
\end{tikzpicture}
\vskip-7ex\leavevmode
\end{center}

The goal of this exercise is to determine over which fields $k$ a
Möbius-Kantor configuration exists, and compute the coordinates of its
points.

We fix a field $k$.  The word “point”, in what follows, will refer
to an element of $\mathbb{P}^2(k)$, in other words, a point with
coordinates in $k$ (that is, a $k$-point).

We shall write as $(x{:}y{:}z)$ the coordinates of a point, and as
$[u{:}v{:}w]$ the line $\{ux+vy+wz = 0\}$.  Recall that the line
through $(x_1{:}y_1{:}z_1)$ and $(x_2{:}y_2{:}z_2)$ (assumed distinct)
is given by the formula $[(y_1 z_2 - y_2 z_1) : (z_1 x_2 - z_2 x_1) :
  (x_1 y_2 - x_2 y_1)]$, and that the same formula (exchanging
parentheses and square brackets) can also be used to compute the
intersection of two distinct lines.  (This may not always be the best
or simplest way\footnote{For example, one shouldn't need this formula
  to notice that the line through $(42{:}0{:}0)$ and $(0{:}1729{:}0)$
  is $[0{:}0{:}1]$.} to compute coordinates, however!)

\emph{We assume for questions (1)–(5) below that $p_0,\ldots,p_7$ is a
Möbius-Kantor configuration of points (over the given field $k$), and
the questions will serve to compute the coordinates of the points.}
We denote $\ell_{ijk}$ the line through $p_i,p_j,p_k$ when it exists.

\textbf{(1)} Explain why we can assume, without loss of generality,
that $p_0=(1{:}0{:}0)$ and $p_1=(0{:}1{:}0)$ and $p_2=(0{:}0{:}1)$ and
$p_5=(1{:}1{:}1)$.  \emph{We shall henceforth do so.}

\begin{answer}
No three of the four points $p_0,p_1,p_2,p_5$ are aligned, so they are
a projective basis of $\mathbb{P}^2$: thus, there is a unique
projective transformation of $\mathbb{P}^2$ mapping them to the
standard basis $(1{:}0{:}0), \penalty-100 (0{:}1{:}0), \penalty-100
(0{:}0{:}1), \penalty-100 (1{:}1{:}1)$.  Since projective
transformations preserve alignment, we can apply this projective
transformation and assume that $p_0=(1{:}0{:}0)$ and $p_1=(0{:}1{:}0)$
and $p_2=(0{:}0{:}1)$ and $p_5=(1{:}1{:}1)$.
\end{answer}

\textbf{(2)} Compute the coordinates of the lines $\ell_{013}$,
$\ell_{124}$, $\ell_{235}$, $\ell_{560}$ and $\ell_{702}$, and of the
point $p_3$.

\begin{answer}
Denoting $p\vee q$ the line through distinct points $p$ and $q$, we
get $\ell_{013} = p_0 \vee p_1 = [0{:}0{:}1]$ and $\ell_{124} =
p_1\vee p_2 = [1{:}0{:}0]$ and $\ell_{235} = p_5\vee p_2 =
[1{:}{-1}{:}0]$ and $\ell_{560} = p_5\vee p_0 = [0{:}1{:}{-1}]$ and
$\ell_{702} = p_2\vee p_0 = [0{:}1{:}0]$.  Denoting by $\ell\wedge m$
the point of intersection of distinct lines $\ell$ and $m$, we get
$p_3 = \ell_{013} \wedge \ell_{235} = (1{:}1{:}0)$.
\end{answer}

\textbf{(3)} Explain why we can write, without loss of generality, the
coordinates of $p_4$ in the form $(0{:}\xi{:}1)$ for some $\xi$
(in $k$).  (Note that two things need to be explained here: why the
first coordinate is $0$ and why the last can be taken to be $1$.)

\begin{answer}
The point $p_4$ is on $\ell_{124} = [1{:}0{:}0]$, so it is of the form
$(0{:}\tiret{:}\tiret)$ (its first coordinate is zero).  On the other
hand, it is \emph{not} on $\ell_{013} = [0{:}0{:}1]$, so it is
\emph{not} of the form $(\tiret{:}\tiret{:}0)$ (its last coordinate is
\emph{not} zero).  Since homogeneous coordinates are defined up to
multiplication by a common constant, we can divide them by this
nonzero last coordinate, and we get $p_4$ of the form
$(0{:}\tiret{:}1)$, as required.
\end{answer}

\textbf{(4)} Now compute the coordinates of the line $\ell_{346}$, of
the point $p_6$, and of the lines $\ell_{457}$ and $\ell_{671}$.

\begin{answer}
We have $\ell_{346} = p_3\vee p_4 = [1{:}{-1}{:}\xi]$.  Therefore $p_6
= \ell_{346} \wedge \ell_{560} = (1-\xi : 1 : 1)$.  Further,
$\ell_{457} = p_4\vee p_5 = [\xi-1 : 1 : -\xi]$ and $\ell_{671} =
p_1\wedge p_6 = [1{:}0{:}\xi-1]$.
\end{answer}

\textbf{(5)} Write the coordinates of the last remaining point $p_7$
and using the fact that we now have three lines on which it lies,
conclude that $\xi$ must satisfy $1-\xi+\xi^2 = 0$.

\begin{answer}
The point $p_7$ can be written as $\ell_{571} \wedge \ell_{702}$,
giving coordinates $(1-\xi:0:1)$, or as $\ell_{457} \wedge
\ell_{702}$, giving coordinates $(\xi:0:\xi-1)$.  That they are equal
gives the relation $\xi + (1-\xi)^2 = 0$ or $1-\xi+\xi^2 = 0$.
Alternatively, we can write $p_7$ as $\ell_{671} \wedge \ell_{457}$
with coordinates $(1-\xi : 1-\xi+\xi^2 : 1)$, and the fact that it
lies on $\ell_{702}$. we get $1-\xi+\xi^2 = 0$.
\end{answer}

\textbf{(6)} Deduce from questions (1)–(5) above that, if a
Möbius-Kantor configuration over $k$ exists, then there is $\xi\in k$
such that $1-\xi+\xi^2 = 0$.

\begin{answer}
As explained in (1), we can find a projective transformation of
$\mathbb{P}^2$ such giving $p_0,p_1,p_2,p_5$ the coordinates
$(1{:}0{:}0), \penalty-100 (0{:}1{:}0), \penalty-100 (0{:}0{:}1),
\penalty-100 (1{:}1{:}1)$, and as explained in (3) we then get $p_4$
of the form $(0{:}\xi{:}1)$, and as explained in (5) this $\xi$ must
satisfy $1-\xi+\xi^2 = 0$.  So if there is Möbius-Kantor configuration
over $k$ then there is such a $\xi$.
\end{answer}

\textbf{(7)} Conversely, using the coordinate computations performed
in questions (2)–(5), explain why, if there is $\xi\in k$ such that
$1-\xi+\xi^2 = 0$, then a Möbius-Kantor configuration over $k$ exists.
(A long explanation is not required, but at least explain what checks
need be done.)

\begin{answer}
Conversely, if a $\xi$ such that $1-\xi+\xi^2 = 0$ exists, then the
coordinates we have computed, namely
\[
\arraycolsep=1em
\begin{array}{cc}
p_0 = (1 : 0 : 0) & \ell_{013} = [0 : 0 : 1]\\
p_1 = (0 : 1 : 0) & \ell_{124} = [1 : 0 : 0]\\
p_2 = (0 : 0 : 1) & \ell_{235} = [1 : {-1} : 0]\\
p_3 = (1 : 1 : 0) & \ell_{346} = [1 : {-1} : \xi]\\
p_4 = (0 : \xi : 1) & \ell_{457} = [-1+\xi : 1 : -\xi]\\
p_5 = (1 : 1 : 1) & \ell_{560} = [0 : 1 : {-1}]\\
p_6 = (1-\xi : 1 : 1) & \ell_{671} = [1 : 0 : \xi-1]\\
p_7 = (1-\xi : 0 : 1) & \ell_{702} = [0 : 1 : 0]\\
\end{array}
\]
define a Möbius-Kantor configuration.  To check this, we need to check
that $p_i,p_j,p_k$ lie on $\ell_{ijk}$: most of these checks are
trivial, and the remaining few follow from $1-\xi+\xi^2=0$; but we
also need to check that no other $p_r$ lies on $\ell_{ijk}$: for
example, this requires checking that $\xi \neq 0$ (which follows from
the fact that $0$ certainly does not satisfy $1-\xi+\xi^2=0$) and $\xi
\neq 1$ (similarly).
\end{answer}

\textbf{(8)} Give examples of fields $k$, at least one infinite and
one finite, over which a Möbius-Kantor configuration exists, and
similarly examples over which it does not exist.

\begin{answer}
For fields of characteristic $\neq 2$, the usual formula for solving a
quadratic equation shows that a Möbius-Kantor configuration exists
precisely iff $-3$ is a square (since the discriminant of $1-t+t^2$
is $-3$).  This is obviously the case of fields of characteristic $3$
(with $\xi = -1$).

Some examples of fields with a Möbius-Kantor configuration are: any
algebraically closed field (e.g., $\mathbb{C}$), the field
$\mathbb{Q}(\sqrt{-3}) = \{u+v\sqrt{-3} : u,v\in\mathbb{Q}\}$, any
field of characteristic $3$ (e.g., $\mathbb{F}_3$), the field
$\mathbb{F}_4$ with $4$ elements (because it is
$\mathbb{F}_2[t]/(1+t+t^2)$), or the field $\mathbb{F}_7$ (because
$\xi = 3$ satisfies $1-\xi+\xi^2 = 0$).

Some examples of fields without a Möbius-Kantor configuration are: any
subfield of $\mathbb{R}$ (including $\mathbb{Q}$ or $\mathbb{R}$
itself), since $-3$ is not a square in $\mathbb{R}$, the field
$\mathbb{F}_2$ or the field $\mathbb{F}_5$ (checking for each element
that it does not satisfy $1-\xi+\xi^2 = 0$).

(In fact, for finite fields, the law of quadratic reciprocity gives us
a complete answer of when a Möbius-Kantor configuration over
$\mathbb{F}_q$ exists: if $q \equiv 1 \pmod{4}$ we have
$\big(\frac{-3}{q}\big) = \big(\frac{-1}{q}\big) \,
\big(\frac{3}{q}\big) = \big(\frac{3}{q}\big) =
\big(\frac{q}{3}\big)$, while if $q \equiv 3 \pmod{4}$ we have
$\big(\frac{-3}{q}\big) = \big(\frac{-1}{q}\big) \,
\big(\frac{3}{q}\big) = -\big(\frac{3}{q}\big) =
\big(\frac{q}{3}\big)$; so if $q$ is neither a power of $2$ nor of $3$
this is $+1$ iff $q \equiv 1 \pmod{3}$.  For $q$ a power of $3$, a
Möbius-Kantor configuration always exists.  For $q$ a power of $2$, it
is not hard to check that it exists iff $q$ is an \emph{even} power
of $2$.  Putting all cases together, a Möbius-Kantor configuration
exists over $\mathbb{F}_q$ iff either $q$ is a power of $3$ or $q
\equiv 1 \pmod{3}$.)
\end{answer}


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\exercise

The focus of this exercise is \textbf{Klein's quartic}, namely the
projective algebraic variety $C$ defined by the equation
\[
x^3 y + y^3 z + z^3 x = 0
\]
in $\mathbb{P}^2$ with coordinates $(x{:}y{:}z)$.  Note the symmetry
of this equation under cyclic permutation of the
coordinates\footnote{To dispel any possible confusion, this means
simultaneously replacing $x$ by $y$, $y$ by $z$ and $z$ by $x$.},
which will come in handy to simplify some computations.  To refer to
it more easily, we shall denote $f := x^3 y + y^3 z + z^3 x$ the
polynomial defining the equation of $C$.

We shall work over a field $k$ having characteristic $\not\in\{2,7\}$.
For simplicity, we shall also assume $k$ to be algebraically closed
(even though this won't matter at all).

\textbf{(1)} The following relation holds (this is a straightforward
computation, and it is not required to check it):
\[
-27xyz\,\frac{\partial f}{\partial x}
+(28x^3-3y^2 z)\,\frac{\partial f}{\partial y}
-9yz^2\,\frac{\partial f}{\partial z}
= 28x^6
\tag{$*$}
\]
What does the relation ($*$), together with the other two obtained by
cyclically permuting coordinates, tell us about the ideal generated by
$\frac{\partial f}{\partial x}$, $\frac{\partial f}{\partial y}$ and
$\frac{\partial f}{\partial z}$ in $k[x,y,z]$?  What does this imply
on the set of points where $\frac{\partial f}{\partial x}$,
$\frac{\partial f}{\partial y}$ and $\frac{\partial f}{\partial z}$
all vanish?

\begin{answer}
The relation $*$ tells us that $28 x^6$, and consequently $x^6$ itself
(since $k$ is of characteristic $\not\in\{2,7\}$), belongs to the
ideal generated by $\frac{\partial f}{\partial x}$, $\frac{\partial
  f}{\partial y}$ and $\frac{\partial f}{\partial z}$.  By cyclic
permutation of coordinates, this is also the case for $y^6$ and $z^6$:
so this ideal is irrelevant: the set of points in $\mathbb{P}^2$ where
$\frac{\partial f}{\partial x}$, $\frac{\partial f}{\partial y}$ and
$\frac{\partial f}{\partial z}$ all vanish is empty (because
$x^6,y^6,z^6$ do not vanish simultaneously).  This implies that $C$ is
\emph{smooth}.
\end{answer}

\smallskip

\emph{The previous question implies that $C$ is a (plane) curve.  The
following picture is a rough sketch of an affine part of $C$ over the
real field.}

\begin{center}
\begin{tikzpicture}
\begin{scope}[thick]
\clip (-3,-3) -- (3,-3) -- (3,3) -- (-3,3) -- cycle;
\draw (-3.000,5.251) .. controls (-2.667,4.397) and (-2.333,3.618) .. (-2.000,2.946) ;
\draw (-2.000,2.946) .. controls (-1.833,2.610) and (-1.667,2.301) .. (-1.500,2.028);
\draw (-1.500,2.028) .. controls (-1.333,1.755) and (-1.167,1.519) .. (-1.000,1.325) ;
\draw (-1.000,1.325) .. controls (-0.833,1.130) and (-0.667,0.981) .. (-0.500,0.846) ;
\draw (-0.500,0.846) .. controls (-0.417,0.779) and (-0.333,0.716) .. (-0.250,0.638) ;
\draw (-0.250,0.638) .. controls (-0.208,0.600) and (-0.167,0.558) .. (-0.125,0.501) ;
\draw (-0.125,0.501) .. controls (-0.104,0.473) and (-0.083,0.441) .. (-0.062,0.397) ;
\draw (-0.062,0.397) .. controls (0,0.265) and (0,0.133) .. (0,0) ;
\draw (0,0) .. controls (0,-0.133) and (0,-0.265) .. (0.062,-0.397) ;
\draw (0.062,-0.397) .. controls (0.083,-0.441) and (0.104,-0.471) .. (0.125,-0.499) ;
\draw (0.125,-0.499) .. controls (0.167,-0.553) and (0.208,-0.590) .. (0.250,-0.622) ;
\draw (0.250,-0.622) .. controls (0.333,-0.684) and (0.417,-0.720) .. (0.500,-0.741) ;
\draw (0.500,-0.741) .. controls (0.667,-0.783) and (0.833,-0.755) .. (1.000,-0.682) ;
\draw (1.000,-0.682) .. controls (1.167,-0.610) and (1.333,-0.501) .. (1.500,-0.422) ;
\draw (1.500,-0.422) .. controls (1.667,-0.343) and (1.833,-0.288) .. (2.000,-0.248) ;
\draw (2.000,-0.248) .. controls (2.333,-0.168) and (2.667,-0.136) .. (3.000,-0.111) ;
\draw (-3.000,-5.140) .. controls (-2.667,-4.261) and (-2.333,-3.452) .. (-2.000,-2.694) ;
\draw (-2.000,-2.694) .. controls (-1.833,-2.315) and (-1.667,-1.962) .. (-1.500,-1.552) ;
\draw (-1.500,-1.552) .. controls (-1.458,-1.449) and (-1.417,-1.346) .. (-1.375,-1.209) ;
\draw (-1.375,-1.209) .. controls (-1.315,-1.013) and (-1.263,-0.817) .. (-1.375,-0.621) ;
\draw (-1.375,-0.621) .. controls (-1.417,-0.548) and (-1.458,-0.511) .. (-1.500,-0.477) ;
\draw (-1.500,-0.477) .. controls (-1.667,-0.339) and (-1.833,-0.295) .. (-2.000,-0.252) ;
\draw (-2.000,-0.252) .. controls (-2.333,-0.166) and (-2.667,-0.136) .. (-3.000,-0.111) ;
\end{scope}
\draw[->, shorten <=-0.1cm, shorten >=-0.1cm, thin] (-3,0) -- (3,0);
\draw[->, shorten <=-0.1cm, shorten >=-0.1cm, thin] (0,-3) -- (0,3);
\node[anchor=west] at (3,0) {$\scriptstyle x/z \,=:\, u$};
\node[anchor=south] at (0,3) {$\scriptstyle y/z \,=:\, v$};
\end{tikzpicture}
\end{center}

We now define the three points $a := (1{:}0{:}0)$, $b := (0{:}1{:}0)$
and $c := (0{:}0{:}1)$ (which obviously lie on $C$).

\textbf{(2)} List all points of $C$ where $x$ vanishes.  Do the same
for $y$ and $z$.

\begin{answer}
If $x$ vanishes on $C$ then $y^3 z = 0$, so $y=0$ or $z=0$.  So either
$x=y=0$ and we are at $c$, or $x=z=0$ and we are at $b$; so the set of
points where $x$ vanishes on $C$ is exactly $\{b,c\}$.  By cyclic
rotation of coordinates, the set of points of $C$ where $y$ vanishes
is $\{c,a\}$ and the set of points of $C$ where $z$ vanishes is
$\{a,b\}$.
\end{answer}

\textbf{(3)} Where do the points $a,b,c$ lie on the printed picture?
(If they do not lie on the picture, show the direction in which they
would be.)  What is the equation of the affine part of $C$ drawn on
the picture?  What is the tangent line at the point $c$?  What about
$a$ and $b$?

\begin{answer}
The point $c$ is at affine coordinates $(u,v) = (0,0)$ where $u =
\frac{x}{z}$ and $v = \frac{y}{z}$, that is, it is at the origin of
the printed picture.  The point $a$ is at infinity ($z=0$) on the axis
$y=0$ (or $v=0$ if we prefer), so it is at infinity in the horizontal
direction, whereas $b$ is at infinity on the axis $x=0$ (or $u=0$ if
we prefer), so at infinity in the vertical direction.

The equation of the affine part of $C$ is obtained by dehomogenizing
$x^3 y + y^3 z + z^3 x = 0$ with respect to $z$, i.e., by dividing by
$z^3$ and replacing $\frac{x}{z}$ by $u$ and $\frac{y}{z}$ by $v$,
giving $u^3 v + v^3 + u = 0$.

The tangent line at the origin $c$ of the affine part $\{z\neq 0\}$ is
given by $\frac{\partial g}{\partial u}|_{(0,0)}\cdot u +
\frac{\partial g}{\partial v}|_{(0,0)}\cdot v =0$ where $g := u^3 v +
v^3 + u$.  This simply gives $u=0$, so it is the vertical axis (as
could be guessed from the figure); as a projective line, this is
$x=0$.  By cyclic permutation of coordinates, we get $y=0$ as tangent
line at $a$ and $z=0$ as tangent line at $c$.  (Of course, one might
also compute these by taking affine charts around each one of the
points, but this would be more tedious.)
\end{answer}

\textbf{(4)} Considering $v := \frac{y}{z}$ as a rational function
on $C$, explain why it vanishes at order exactly $1$ at $c$, that
is\footnote{We write $\ord_p(h)$ for the order at a point $p \in C$ of
a rational function $h \in k(C)$.  By the way, please note that
$x,y,z$ themselves do not belong to $k(C)$ (they are not functions and
have no value by themselves), so we cannot speak of $\ord_p(x)$.},
$\ord_c(v) = 1$.  Explain why $\ord_c(u) = \ord_c(u^3 v + v^3)$ where
$u := \frac{x}{z}$ and deduce that $\ord_c(u) = 3$.  Deduce the order
at $c$ of $\frac{y}{x}$ (which is also $\frac{v}{u}$).

\begin{answer}
The coordinate $v$ vanishes with order exactly $1$ at the origin $c$
of the tangent line $u=0$ to $C$ at $c$; therefore it also has order
exactly $1$ at $c$ on $C$.  In other words, $\ord_c(v) = 1$.

Now $u^3 v + v^3 + u = 0$ on $C$, that is $u = -u^3 v - v^3$, so
$\ord_c(u) = \ord_c(u^3 v + v^3)$.  This shows that $\ord_c(u) =: k$,
which is $\geq 1$ because $u$ vanishes at $c$, satisfies $k \geq
\min(3k+1,3)$, so $k \geq 3$; but now $3k+1 \geq 10$, so $\ord_c(u^3
v) = 3k+1 \neq 3 = \ord_c(v^3)$, so in fact $k = \min(3k+1,3) = 3$, as
required.

Consequently, $\frac{y}{x} = \frac{v}{u}$ has order $\ord_c(v) -
\ord_c(u) = 1 - 3 = -2$ at $c$.
\end{answer}

\textbf{(5)} By using symmetry, compute the order at each one of the
three points $a,b,c$ of each one of the three functions $\frac{x}{z}$,
$\frac{y}{x}$ and $\frac{z}{y}$.  Explain why there are no points
(of $C$) other than $a,b,c$ where any of these functions (on $C$)
vanishes or has a pole.  Summarize this by writing the principal
divisors $\divis(\frac{x}{z})$, $\divis(\frac{y}{x})$ and
$\divis(\frac{z}{y})$ associated with these three functions.

\begin{answer}
We have seen that
\[
\arraycolsep=1em
\begin{array}{ccc}
\ord_c(\frac{x}{z}) = 3 &
\ord_c(\frac{y}{x}) = -2 &
\ord_c(\frac{z}{y}) = -1
\end{array}
\]
so by cyclic permutation we get
\[
\arraycolsep=1em
\begin{array}{ccc}
\ord_a(\frac{x}{z}) = -1 &
\ord_a(\frac{y}{x}) = 3 &
\ord_a(\frac{z}{y}) = -2
\\
\ord_b(\frac{x}{z}) = -2 &
\ord_b(\frac{y}{x}) = -1 &
\ord_b(\frac{z}{y}) = 3
\end{array}
\]
Now we have also pointed out earlier that none of $x,y,z$ vanishes on
$C$ outside possibly of $\{a,b,c\}$: so
$\frac{x}{z},\frac{y}{x},\frac{z}{y}$ have neither zero nor pole on
$C\setminus\{a,b,c\}$, i.e., their order is $0$ everywhere on this
open set.  This shows that
\[
\begin{aligned}
\divis(\frac{x}{z}) &= -[a] -2\,[b] + 3\,[c]\\
\divis(\frac{y}{x}) &= \hphantom{+}3\,[a] - [b] - 2\,[c]\\
\divis(\frac{z}{y}) &= -2\,[a] + 3\,[b] - [c]
\end{aligned}
\]
Two sanity checks can be performed: the degree of each of these
divisors (i.e., the sum of the coefficients) is zero, as befits a
principal divisor; and the sum of these three divisors is also zero,
as it should be because it is the divisor of the constant nonzero
function $1$.
\end{answer}


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\exercise

This exercise is about the \textbf{Segre embedding}\footnote{French:
  “plongement de Segre”}, which is a way to map the product
$\mathbb{P}^p \times \mathbb{P}^q$ of two projective spaces to a
larger projective space $\mathbb{P}^n$ (with, as we shall see, $n =
pq+p+q$).

Assume $k$ is a field.  To simplify presentation, assume $k$ is
algebraically closed (even though this won't matter at all).

Given $p,q\in\mathbb{N}$, the Segre embedding of $\mathbb{P}^p \times
\mathbb{P}^q$ is the map $\psi$ given by:
\[
\begin{aligned}
\psi\colon & \mathbb{P}^p \times \mathbb{P}^q \to \mathbb{P}^n\\
&((x_0:\cdots:x_p),\, (y_0:\cdots:y_q)) \mapsto (x_0 y_0 : x_0 y_1 : \cdots
: x_0 y_q : x_1 y_0 : \cdots : x_p y_q)\\
\end{aligned}
\]
where $n = (p+1)(q+1)-1$ and the coordinates of the endpoint consist
of every product $x_i y_j$ with $0\leq i\leq p$ and $0\leq j\leq q$
(in some order which doesn't really matter: here we have chosen the
lexicographic ordering).

Note that with the definitions given in this course, we cannot state
that $\psi$ is a morphism of algebraic varieties (although it
certainly \emph{should} be one), because we did not define a “product
variety”\footnote{In fact, the Segre embedding is one way of doing
this.} $\mathbb{P}^p \times \mathbb{P}^q$.  But we can still consider
it as a function.

Let us label $(z_{0,0} : z_{0,1} : \cdots : z_{p,q})$ the homogeneous
coordinates in $\mathbb{P}^n$ (that is, $z_{i,j}$ with $0\leq i\leq p$
and $0\leq j\leq q$), so that $\psi$ is given simply by “$z_{i,j} =
x_i y_j$”.

We finally consider the Zariski closed subset $S$ of $\mathbb{P}^n$,
known as the \textbf{Segre variety}, defined in $\mathbb{P}^n$ by the
equations $z_{i,j} z_{i',j'} = z_{i,j'} z_{i',j}$ for all $0\leq
i,i'\leq p$ and $0\leq j,j'\leq q$.

\medskip

\textbf{(1)} Explain why the map $\psi$ is well-defined, i.e., the
definition given above makes sense: carefully list the properties that
need to be checked, and do so.  Explain why $S$ is indeed a Zariski
closed subset of $\mathbb{P}^n$: again, carefully state what needs to
be checked before doing so.

\begin{answer}
For the point $(x_0 y_0 : \cdots : x_p y_q)$ to make sense, we need to
check that not all its coordinates are zero.  But we know that at
least one of the $x_i$ is nonzero and at least one of the $y_j$ is
nonzero, so (as we are working over a field) the product $x_i y_j$ is
nonzero.

For the map $((x_0:\cdots:x_p),\, (y_0:\cdots:y_q)) \mapsto (x_0 y_0 :
\cdots : x_p y_q)$ to make sense, we need to check that $(x_0 y_0 :
\cdots : x_p y_q)$ does not change if we replace the $x_i$ and the
$y_j$ by different coordinates for the same point, in other words, if
we multiply all the $x_i$ by a common nonzero constant, and all the
$y_j$ by a (possibly different) common nonzero constant.  This is
indeed the case as $x_i y_j$ will be multiplied by the product of
these two constants.

Concerning $S$, we need to check that the equations $z_{i,j} z_{i',j'}
= z_{i,j'} z_{i',j}$ are homogeneous: this is indeed the case (they
are homogeneous of degree $2$).
\end{answer}

\textbf{(2)} Consider in this question the special case $p=q=1$ (so
$n=3$).  Simplify the definition of $S$ in this case down to a single
equation.  Taking $z_{0,0}=0$ as the plane at infinity in
$\mathbb{P}^3$, give the equation of the affine part $S \cap
\mathbb{A}^3$.  Similarly taking $x_0=0$ (resp. $y_0=0$) as the point
at infinity in $\mathbb{P}^1$, describe $\psi$ on $\mathbb{A}^1 \times
\mathbb{A}^1$.

\begin{answer}
When $p=q=1$ the equations of $S$ are all trivial except $z_{0,0}
z_{1,1} = z_{0,1} z_{1,0}$ (or equations trivially equivalent to
this).  Taking $z_{0,0} = 0$ as plane at infinity, we get the equation
of the affine part by dehomogenizing $z_{0,0} z_{1,1} = z_{0,1}
z_{1,0}$, which gives $w_{1,1} = w_{0,1} w_{1,0}$ where $w_{i,j}$
denotes the affine coordinate $z_{i,j}/z_{0,0}$ in $\mathbb{A}^3$.

Concerning $\psi$, if we call $u = x_1/x_0$ the affine coordinate on
the first $\mathbb{A}^1$ and $v = y_1/y_0$ that on the second, it is
given by taking $(u,v)$, i.e. $((1:u),\, (1:v))$ to $(1:v:u:uv)$, that
is $(v,u,uv)$.
\end{answer}

\textbf{(3)} Returning to the case of general $p$ and $q$, show that
the image of $\psi$ is contained in $S$, that is, $\psi(\mathbb{P}^p
\times \mathbb{P}^q) \subseteq S$.

\begin{answer}
If $(z_{0,0} : \cdots : z_{p,q})$ is given by $z_{i,j} = x_i y_j$, we
just need to check that $z_{i,j} z_{i',j'} = z_{i,j'} z_{i',j}$: but
this just says that $x_i y_j x_{i'} y_{j'} = x_i y_{j'} x_{i'} y_j$,
which is obvious by commutativity.
\end{answer}

\textbf{(4)} Conversely, explain why for each point $(z_{0,0} : \cdots
: z_{p,q})$ in $S$ there is a unique pair of points $((x_0 : \cdots :
x_p), (y_0 : \cdots : y_q))$ in $\mathbb{P}^p \times \mathbb{P}^q$
which maps to the given point under $\psi$: in other words, show that
$\psi$ is a bijection between $\mathbb{P}^p \times \mathbb{P}^q$
and $S$.

(\textit{Hint:} you may wish to observe that if $(z_{0,0} : \cdots :
z_{p,q})$ is in $S$, the point $(z_{0,j_0} : \cdots : z_{p,j_0})$ in
$\mathbb{P}^p$ does not depend on $j_0 \in \{0,\ldots,q\}$ such that
$\exists i.(z_{i,j_0}\neq 0)$; and similarly for $(z_{i_0,0} : \cdots
: z_{i_0,q})$ in $\mathbb{P}^q$.)

\begin{answer}
Assume $(z_{0,0} : \cdots : z_{p,q})$ is in $S$.  By the definition of
$\mathbb{P}^n$, at least one coordinate $z_{i_0,j_0}$ is nonzero.
Define $x^*_i = z_{i,j_0}$ (note that $x^*_{i_0} \neq 0$) and $y^*_j =
z_{i_0,j}$ (note that $y^*_{j_0} \neq 0$): then $x^*_i y^*_j =
z_{i,j_0} z_{i_0,j}$, which, by the equations of $S$, is also
$z_{i_0,j_0} z_{i,j}$: this shows that $((x^*_0 : \cdots : x^*_p),
(y^*_0 : \cdots : y^*_q))$ maps to the given $(z_{0,0} : \cdots :
z_{p,q})$ under $\psi$ (by dividing all coordinates by the nonzero
value $z_{i_0,j_0}$).  So $\psi$ surjects to $S$.

But in fact, if $((x_0 : \cdots : x_p), (y_0 : \cdots : y_q))$ maps to
$(z_{0,0} : \cdots : z_{p,q})$ under $\psi$, then we have $z_{i,j_0} =
y_{j_0} x_i$ so that $(x_0 : \cdots : x_p) = (z_{0,j_0} : \cdots :
z_{p,j_0})$ provided $y_{j_0} \neq 0$, which is tantamount to saying
$z_{i_0,j_0}\neq 0$ for some $i_0$: so we had no other choice than to
take the $(x^*_0 : \cdots : x^*_p)$ of the previous paragraph, and the
same argument holds for $(y^*_0 : \cdots : y^*_q)$.  This shows
uniqueness of the points $((x_0 : \cdots : x_p), (y_0 : \cdots :
y_q))$ mapping to $(z_{0,0} : \cdots : z_{p,q})$ under $\psi$.
\end{answer}

\textbf{(5)} Call $\pi\colon S\to \mathbb{P}^p\times\mathbb{P}^q$ the
inverse bijection of $\psi$, and call $\pi',\pi''$ its two components.
(In other words, if $s = (z_{0,0} : \cdots : z_{p,q})$ is in $S$ then
$\pi'(s) = (x_0:\cdots:x_p) \in \mathbb{P}^p$ and $\pi''(s) =
(y_0:\cdots:y_p) \in \mathbb{P}^q$ are the unique points such that
$(\pi'(s),\pi''(s))$ maps to $s$ under $\psi$.)  Show that the maps
$\pi' \colon S \to \mathbb{P}^p$ and $\pi'' \colon S \to \mathbb{P}^q$
are morphisms of algebraic varieties.  (If this seems too difficult,
consider the special case $p=q=1$, and at least try to explain what
needs to be checked.)

\begin{answer}
Given $j_0 \in \{0,\ldots,q\}$, consider the map $(z_{0,0} : \cdots :
z_{p,q}) \mapsto (z_{0,j_0} : \cdots : z_{p,j_0})$ which selects only
the coordinates $z_{i,j_0}$.  This is a partially defined map from
$\mathbb{P}^n$ to $\mathbb{P}^p$, and the components are homogeneous
polynomials of the same degree (here, $1$): the only thing that can go
wrong is that all the $z_{i,j_0}$ are zero, so this is well-defined on
the open set $\{z_{0,j_0}\neq 0\} \cup \cdots \cup \{z_{p,j_0}\neq
0\}$.  Now restrict this map to $S$: this gives us a morphism
$\pi^{\prime(j_0)}$ from the open set $U^{(j_0)} := S \cap
(\{z_{0,j_0}\neq 0\} \cup \cdots \cup \{z_{p,j_0}\neq 0\})$ of $S$
to $\mathbb{P}^p$.

Note that the union the union of $U^{0)},\ldots,U^{(q)}$ is all of $S$
because there is always at least one coordinate nonzero.

Furthermore, we have seen in (4) that if $s = (z_{0,0} : \cdots :
z_{p,q})$ then $\pi'(s)$ is given by $\pi^{\prime(j_0)}(s) =
(z_{0,j_0} : \cdots : z_{p,j_0})$ where $j_0$ is any element of
$\{0,\ldots,q\}$ such that $z_{i_0,j_0} \neq 0$ for some $i_0$, i.e.,
$s \in U^{(j_0)}$.  This shows that $\pi'$ coincides with
$\pi^{\prime(j_0)}$ on the open set $U^{(j_0)}$ where the latter is
defined, so $\pi'$ is defined by “gluing” the various
$\pi^{\prime(j_0)}$.  So $\pi'$ is indeed a morphism (to be clear: it
is simply defined by selecting the coordinates of the form $z_{i,j_0}$
for any one $j_0$ such that not all of them vanish).

The same argument, \textit{mutatis mutandis}, works for $\pi''$.
\end{answer}



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