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\begin{document}
\ifcorrige
\title{ACCQ205\\Final exam — answer key\\{\normalsize Algebraic curves}}
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\title{ACCQ205\\Final exam\\{\normalsize Algebraic curves}}
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\author{}
\date{2023-04-12}
\maketitle

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\noindent\textbf{Instructions.}

This exam consists of a single lengthy problem.  Although the
questions depend on each other, they have been worded in such a way
that the necessary information for subsequent questions is given in
the text.  Thus, failure to answer one question should not make it
impossible to proceed to later questions.

\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 7 pages (cover page included).
\else
This exam has 4 pages (cover page included).
\fi

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\pagebreak


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\textit{The goal of this problem is to study a representation of lines
  in $\mathbb{P}^3$.}

\smallskip

We fix a field $k$.  Recall that \emph{points} in $\mathbb{P}^3(k)$
are given by quadruplets $(x_0{:}x_1{:}x_2{:}x_3)$ of “homogeneous
coordinates” in $k$, not all zero, defined up to a common
multiplicative constant, and that \emph{planes} in $\mathbb{P}^3(k)$
are of the form $\{(x_0{:}x_1{:}x_2{:}x_3) \in \mathbb{P}^3(k) : u_0
x_0 + \cdots + u_3 x_3 = 0\}$ (for some $u_0,\ldots,u_3$, not all
zero, defined up to a common multiplicative constant) which we can
denote as $[u_0{:}u_1{:}u_2{:}u_3]$ (a point of the
“dual” $\mathbb{P}^3$).  Our goal is to find a representation for
lines.

{\footnotesize It may be convenient, if so desired, to call $\langle
  w\rangle$ the point in projective space $\mathbb{P}^{m-1}(k)$
  defined by a vector $w\neq 0$ in $k^m$ (i.e., if $w =
  (w_0,\ldots,w_m)$ then $\langle w \rangle = (w_0{:}\cdots{:}w_m)$),
  that is, the class of $w$ under collinearity. \par}

\bigskip

\textbf{(1)} Given $x := (x_0,\ldots,x_3) \in k^4$ and $y :=
(y_0,\ldots,y_3) \in k^4$, let us define $x\wedge y := (w_{0,1},
w_{0,2}, w_{0,3}, w_{1,2}, w_{1,3}, w_{2,3}) \in k^6$ where $w_{i,j}
:= x_i y_j - x_j y_i$.  What is $(\lambda x)\wedge(\mu y)$ in relation
to $x\wedge y$?  Under what necessary and sufficient condition do we
have $x\wedge y = 0$?  What is $x\wedge(\lambda x+\mu y)$ in relation
to $x\wedge y$?

\begin{answer}
The $w_{i,j}$ are bilinear in $x,y$ (they are $2\times 2$
determinants) so $(\lambda x)\wedge(\mu y) = \lambda\mu(x\wedge y)$.
Vanishing of $w_{i,j}$ means $(x_i,x_j)$ is proportional to
$(y_i,y_j)$ so vanishing of all the $w_{i,j}$ means precisely that
$x$ or $y$ is zero or that $x$ and $y$ are collinear.  Again by
bilinearity, we have $x\wedge(\lambda x+\mu y) = \lambda(x\wedge x) +
\mu(x\wedge y)$ which is just $\mu(x\wedge y)$ since $x\wedge x$
is $0$.
\end{answer}

\textbf{(2)} Show that if $V \subseteq k^4$ is a $2$-dimensional
vector subspace, then the set of $x\wedge y$ for $x,y\in V$ is a
$1$-dimensional subspace of $k^6$.

\begin{answer}
Consider $u,v$ a basis of $V$: then $u\wedge v$ is nonzero, and any
element of $V$ can be written $\lambda u + \mu v$, and then $(\lambda
u + \mu v) \wedge (\lambda' u + \mu' v) = (\lambda \mu' - \lambda'
\mu)(u \wedge v)$ by (1) (or by composition of determinants).  In
other words, any $x\wedge y$ with $x,y\in V$ is collinear to $u\wedge
v$, and the latter is nonzero, and since $\lambda \mu' - \lambda' \mu$
can obviously take every value in $k$, we see that $\{x\wedge y :
x,y\in V\}$ is the line spanned by $u\wedge v$.
\end{answer}

\textbf{(3)} Deduce from (2) that if $L \subseteq \mathbb{P}^3(k)$ is
a line, then $(w_{0,1}{:}w_{0,2}{:}w_{0,3} {:} \penalty0
w_{1,2}{:}w_{1,3}{:}w_{2,3})$, where $w_{i,j} := x_i y_j - x_j y_i$ as
above, and where $(x_0{:}x_1{:}x_2{:}x_3)$ and
$(y_0{:}y_1{:}y_2{:}y_3)$ are two distinct points on $L$, is a
well-defined point in $\mathbb{P}^5(k)$, not depending on the chosen
points on $L$ nor on the homogeneous coordinates representing them.

\begin{answer}
Calling $\langle w\rangle$ the point in projective space
$\mathbb{P}^{m-1}(k)$ defined by a vector $w\neq 0$ in $k^m$, if $L$
is a line in $\mathbb{P}^3(k)$ we can see it as $\{\langle v\rangle :
v\in V\}$ for a $2$-dimensional vector subspace $V \subseteq k^4$, and
we have seen that $\langle x\wedge y\rangle \in \mathbb{P}^4(k)$
exists when $x$ and $y$ are not collinear (so that $x\wedge y \neq
0$), i.e., when $\langle x\rangle \neq \langle y\rangle$, and does not
depend on the $x,y \in V$.
\end{answer}

\bigskip

The $w_{i,j}$ in question are known as the \textbf{Plücker
  coordinates} of $L$.

\textbf{(4)} Show that any point $(z_0{:}z_1{:}z_2{:}z_3)$ on the
line $L$ as above satisfies
\[
w_{0,1} z_2 - w_{0,2} z_1 + w_{1,2} z_0 = 0
\tag{$*$}
\]
— and analogously by replacing $0,1,2$ by three distinct coordinates
in $\{0,1,2,3\}$.

\begin{answer}
Expanding the $3\times 3$ determinant
\[
\left|
\begin{matrix}
x_0&y_0&z_0\\
x_1&y_1&z_1\\
x_2&y_2&z_2\\
\end{matrix}
\right|
\]
gives $w_{0,1} z_2 - w_{0,2} z_1 + w_{1,2} z_0$.  If $\langle
z\rangle$ is in the line through $\langle x\rangle$ and $\langle
y\rangle$, meaning the three vectors $x,y,z$ are linearly dependent,
then this determinant is zero.  The same holds, of course, for any
other choice of coordinates instead of $0,1,2$.
\end{answer}

\textbf{(5)} Deduce from (4) that the $w_{i,j}$ satisfy the following
relation:
\[
w_{0,1} w_{2,3} - w_{0,2} w_{1,3} + w_{0,3} w_{1,2} = 0
\tag{$\dagger$}
\]

\begin{answer}
By (4), we have $w_{0,1} x_2 - w_{0,2} x_1 + w_{1,2} x_0 = 0$ and
$w_{0,1} y_2 - w_{0,2} y_1 + w_{1,2} y_0 = 0$.  Adding $y_3$ times the
first to $-x_3$ times the second gives the stated
relation ($\dagger$).
\end{answer}

\bigskip

The projective algebraic variety defined by ($\dagger$) in
$\mathbb{P}^5$ is known as the \textbf{Plücker quadric}.  In other
words, we have shown above how to associate to any line $L$ in
$\mathbb{P}^3(k)$ a $k$-point $(w_{0,1}:\cdots:w_{2,3})$ on the
Plücker quadric.  We now consider the converse.

\textbf{(6)} Assuming $(w_{0,1}{:}w_{0,2}{:}w_{0,3} {:} \penalty0
w_{1,2}{:}w_{1,3}{:}w_{2,3})$ in $\mathbb{P}^5(k)$
satisfies ($\dagger$) (viꝫ. belongs to the Plücker quadric), and
assuming also that $w_{0,3} \neq 0$, show that the two points
$(w_{0,3}{:}w_{1,3}{:}w_{2,3}{:}0)$ and
$(0{:}w_{0,1}{:}w_{0,2}{:}w_{0,3})$ in $\mathbb{P}^3(k)$ are
meaningful and distinct, and that the line joining them has the
Plücker coordinates $(w_{0,1}:\cdots:w_{2,3})$ that were given.
(\emph{Hint:} \underline{first} compute $(w_{0,3},w_{1,3},w_{2,3},0)
\wedge (0,w_{0,1},w_{0,2},w_{0,3})$ and then use the result, with the
Plücker relation and the fact that $w_{0,3} \neq 0$ to conclude.)

\begin{answer}
We straightforwardly compute $(w_{0,3},w_{1,3},w_{2,3},0) \wedge
(0,w_{0,1},w_{0,2},w_{0,3}) = (w_{0,3} w_{0,1}, w_{0,3} w_{0,2},
w_{0,3}^2, \penalty0 w_{1,3} w_{0,2} - w_{2,3} w_{0,1}, w_{1,3}
w_{0,3}, w_{2,3} w_{0,3})$.  By Plücker's relation ($\dagger$),
$w_{1,3} w_{0,2} - w_{2,3} w_{0,1} = w_{1,2} w_{0,3}$, so we get
$w_{0,3}$ times $(w_{0,1}, w_{0,2}, w_{0,3}, \penalty0 w_{1,2},
w_{1,3}, w_{2,3})$.  Assuming $w_{0,3} \neq 0$, this is a nonzero
vector, which implies (by question (1)) that
$(w_{0,3},w_{1,3},w_{2,3},0)$ and $(0,w_{0,1},w_{0,2},w_{0,3})$ are
nonzero and non-collinear, so that the two points
$(w_{0,3}{:}w_{1,3}{:}w_{2,3}{:}0)$ and
$(0{:}w_{0,1}{:}w_{0,2}{:}w_{0,3})$ are meaningful and distinct, and
by the computation we have just done, the Plücker coordinates of the
line joining them is the set of coordinates $(w_{0,1}:\cdots:w_{2,3})$
that were given.
\end{answer}

\textbf{(7)} Deduce from (6) that any $(w_{0,1}{:}w_{0,2}{:}w_{0,3}{:}
\penalty0 w_{1,2}{:}w_{1,3}{:}w_{2,3})$ in $\mathbb{P}^5(k)$
satisfying ($\dagger$) (viꝫ. belonging to the Plücker quadric) is the
set of Plücker coordinates of a (clearly unique) line in
$\mathbb{P}^3(k)$.  (\emph{Hint:} what needs to be proved is that the
assumption $w_{0,3} \neq 0$ in (6) is harmless: explain how it can be
arranged by a judicious permutation of coordinates.)

\begin{answer}
We have seen in (6) that when $w_{0,3} \neq 0$ then
$(w_{0,1}{:}w_{0,2}{:}w_{0,3}{:} \penalty0
w_{1,2}{:}w_{1,3}{:}w_{2,3})$ are the Plücker coordinates of a line
in $\mathbb{P}^3(k)$.  But all $w_{i,j}$ cannot be zero (as they are
given in $\mathbb{P}^5$), and we can always permute coordinates in
such a way that any $w_{i,j} \neq 0$, which is sure to exist, becomes
$w_{0,3}$, and the formula $w_{0,1} w_{2,3} - w_{0,2} w_{1,3} +
w_{0,3} w_{1,2} = 0$ is invariant under any permutation of coordinates
(for this is is enough to check a cyclic permutation and a
transposition; keep in mind that $w_{j,i} = -w_{i,j}$ when rewriting
so as $i<j$), so we have confirmed the result in all cases.
\end{answer}

\bigskip

At this point, we have established a bijection between the set of
lines $L$ in $\mathbb{P}^3(k)$ and the set of $k$-points in the
Plücker quadric defined by ($\dagger$) in $\mathbb{P}^5$; we know how
to compute Plücker coordinates from two distinct points lying on $L$
(by definition).  We now wish to compute Plücker coordinates for a
line that is described as the the intersection of two planes.

\textbf{(8)} Rephrase (4) to deduce that, if $L$ is a line with
Plücker coordinates $(w_{0,1}:\cdots:w_{2,3})$, then the planes
$[w_{1,2} : {-w_{0,2}} : w_{0,1} : 0]$ and $[0 : w_{2,3} : {-w_{1,3}}
  : w_{1,2}]$ both contain $L$.  Now consider these as points in the
dual $\mathbb{P}^3$ and show that the Plücker coordinates of the line
$L^*$ joining the two points in question are: $[w_{2,3} : {-w_{1,3}} :
  w_{1,2} : w_{0,3} : {-w_{0,2}} : w_{0,1}]$, provided $w_{1,2} \neq
0$.

\begin{answer}
The relation ($*$) of (4), namely $w_{1,2} z_0 - w_{0,2} z_1 + w_{0,1}
z_2 = 0$, means precisely that $(z_0{:}z_1{:}z_2{:}z_3)$ is on the
plane $[w_{1,2} : {-w_{0,2}} : w_{0,1} : 0]$.  Shifting coordinates
cyclically, it is also on the plane $[0 : w_{2,3} : {-w_{1,3}} :
  w_{1,2}]$.  Computing the Plücker coordinates as defined in
questions (1) to (3) for the line through these two (dual) points
gives $[w_{1,2} w_{2,3} : {-w_{1,2} w_{1,3}} : w_{1,2}^2 : {w_{0,2}
    w_{1,3} - w_{0,1} w_{2,3}} : {-w_{0,2} w_{1,2}} : w_{0,1}
  w_{1,2}]$ provided not all are zero.  By Plücker's
relation ($\dagger$), $w_{0,2} w_{1,3} - w_{0,1} w_{2,3} = w_{0,3}
w_{1,2}$, and now we can divide all coordinates by $w_{1,2}$ if it is
nonzero, giving $[w_{2,3} : {-w_{1,3}} : w_{1,2} : w_{0,3} :
  {-w_{0,2}} : w_{0,1}]$.
\end{answer}

\textbf{(9)} Deduce from (8) that if $L$ is a line in
$\mathbb{P}^3(k)$ with Plücker coordinates $(w_{0,1} : w_{0,2} :
w_{0,3} : \penalty0 w_{1,2} : w_{1,3} : w_{2,3})$, and if $L^*$
denotes the “dual” line in the dual $\mathbb{P}^3$, that is, the line
consisting of all points corresponding to planes containing $L$, then
$L^*$ has (dual) Plücker coordinates $[w_{2,3} : {-w_{1,3}} : w_{1,2}
  : w_{0,3} : {-w_{0,2}} : w_{0,1}]$.  (\emph{Hint:} the only thing
that needs to be proved is that the assumption $w_{1,2} \neq 0$ in (8)
is harmless: explain how it can be arranged by a judicious permutation
of coordinates.)

\begin{answer}
We have seen in (8) that when $w_{1,2} \neq 0$ then the line $L^*$
dual to $L$ is given by $[w_{2,3} : {-w_{1,3}} : w_{1,2} : w_{0,3} :
  {-w_{0,2}} : w_{0,1}]$.  But all $w_{i,j}$ cannot be zero
(question (3)), and we can always permute coordinates in such a way
that any $w_{i,j} \neq 0$, which is sure to exist, becomes $w_{1,2}$,
and the formula $(w_{0,1} : w_{0,2} : w_{0,3} : \penalty0 w_{1,2} :
w_{1,3} : w_{2,3}) \mapsto [w_{2,3} : {-w_{1,3}} : w_{1,2} : w_{0,3} :
  {-w_{0,2}} : w_{0,1}]$ is covariant under any permutation of
coordinates (for this is is enough to check a cyclic permutation and a
transposition), so we have confirmed the result in all cases.
\end{answer}

\textbf{(10)} If $[u_0{:}u_1{:}u_2{:}u_3]$ and
$[v_0{:}v_1{:}v_2{:}v_3]$ are distinct planes in $\mathbb{P}^3(k)$,
how can we compute the Plücker coordinates of their line of
intersection?  (\emph{Hint:} first describe the Plücker coordinates of
the line $L^*$ joining the corresponding points in the “dual”
$\mathbb{P}^3$, and then apply the result of (9), together with
projective duality.)

\begin{answer}
Le line $L^*$ joining the points $[u_0{:}u_1{:}u_2{:}u_3]$ and
$[v_0{:}v_1{:}v_2{:}v_3]$ of the dual $\mathbb{P}^3$ is given by the
Plücker coordinates $w_{i,j} = u_i v_j - u_j v_i$.  We then obtain the
Plücker coordinates of $L$ as $(w_{2,3} : {-w_{1,3}} : w_{1,2} :
w_{0,3} : {-w_{0,2}} : w_{0,1})$ (the formula given in (9) for
computing the Plücker coordinates of $L^*$ from those of $L$: it is
involutive and projective duality ensures that it also computes the
Plücker coordinates of $L$ from those of $L^*$), in other words $(u_2
v_3 - u_3 v_2 : {-u_1 v_3 + u_3 v_1} : u_1 v_2 - u_2 v_1 : u_0 v_3 -
u_3 v_0 : {-u_0 v_2 + u_2 v_0} : u_0 v_1 - u_1 v_0)$
\end{answer}

\bigskip

\textbf{(11)} Explain how, by expanding a $4\times 4$ determinant
expressing the fact that four points $(x_0{:}x_1{:}x_2{:}x_3)$,
$(y_0{:}y_1{:}y_2{:}y_3)$, $(z_0{:}z_1{:}z_2{:}z_3)$ and
$(p_0{:}p_1{:}p_2{:}p_3)$ in $\mathbb{P}^3$ are coplanar, we can
obtain a formula for the plane through a line $L$ defined by its
Plücker coordinates and a point $(z_0{:}z_1{:}z_2{:}z_3)$ not situated
on $L$.  (It is not required to go through the full computations: just
explain how it would work.)

\begin{answer}
Consider the $4\times 4$ determinant
\[
\left|
\begin{matrix}
x_0&y_0&z_0&p_0\\
x_1&y_1&z_1&p_1\\
x_2&y_2&z_2&p_2\\
x_3&y_3&z_3&p_3\\
\end{matrix}
\right|
\]
whose vanishing expresses the fact that $x,y,z,p$ are in a common
hyperplane in $k^4$, i.e., that the points $\langle x\rangle$,
$\langle y\rangle$, $\langle z\rangle$ and $\langle p\rangle$ are
coplanar.  Expanding it with respect to the final column $p$ gives a
linear condition for $p$ to be in the plane $P$ spanned by $\langle
x\rangle$, $\langle y\rangle$, $\langle z\rangle$, whose coefficients
are $3\times 3$ determinants, which are the coordinates of the
plane $P$ in the dual $\mathbb{P}^3$.  Expanding those $3\times 3$
determinants with respect to the final column $z$ writes them in terms
of the Plücker coordinates of the line $L$ through $\langle x\rangle$,
and $\langle y\rangle$, and the point $\langle z\rangle$.

To be precise (although this was not asked), we get:
\[
\begin{aligned}
\relax [\;
& - w_{1,2} z_3 + w_{1,3} z_2 - w_{2,3} z_1 \\
:\;
& w_{2,3} z_0 + w_{0,2} z_3 - w_{0,3} z_2 \\
:\;
& w_{0,3} z_1 - w_{1,3} z_0 - w_{0,1} z_3 \\
:\;
& w_{0,1} z_2 - w_{0,2} z_1 + w_{1,2} z_0
\;]
\end{aligned}
\]
(the last coordinate being precisely given by ($*$)).
\end{answer}

For your additional information: if $L$ and $L'$ are two lines in
$\mathbb{P}^3(k)$ with Plücker coordinates $(w_{0,1}:\cdots:w_{2,3})$
and $(w'_{0,1}:\cdots:w'_{2,3})$ respectively, then $L$ and $L'$ meet
in a common point (or equivalently, belong to a common plane)
iff\footnote{This relation (the “polarization” of the quadratic
  relation ($\dagger$)) can be interpreted by saying that the line
  in $\mathbb{P}^5$ joining the two points $(w_{0,1}:\cdots:w_{2,3})$
  and $(w'_{0,1}:\cdots:w'_{2,3})$ on the Plücker quadric is entirely
  contained in said quadric.}
\[
w_{0,1} w'_{2,3} - w_{0,2} w'_{1,3} + w_{0,3} w'_{1,2}
+ w_{2,3} w'_{0,1} - w_{1,3} w'_{0,2} + w_{1,2} w'_{0,3} = 0
\tag{$\ddagger$}
\]
(it is not required to prove this).

\bigskip

\textbf{(12)} Briefly summarize all of the above, emphasizing how we
have obtained formulæ allowing algorithmic computation of all possible
geometric constructions between points, lines and planes
in $\mathbb{P}^3$.

\begin{answer}
For projective subspaces in $\mathbb{P}^3$ we can represent:
\begin{itemize}
\item \textbf{points} by their homogeneous coordinates
  $(x_0{:}x_1{:}x_2{:}x_3)$ (which are arbitrary not all zero, defined
  up to a common multiplicative constant),
\item \textbf{lines} by their Plücker coordinates
  $(w_{0,1} : w_{0,2} : w_{0,3}  :  \penalty0
  w_{1,2} : w_{1,3} : w_{2,3})$ (which are not all zero, and subject
  to the sole condition ($*$) of belonging to the Plücker quadric,
  defined up to a common multiplicative constant), or equivalently by
  their dual Plücker coordinates which are the same up to a
  permutation and some changes of sign, $[w_{2,3} : {-w_{1,3}} :
    w_{1,2} : w_{0,3} : {-w_{0,2}} : w_{0,1}]$,
\item \textbf{planes} by their dual coordinates
  $[u_0{:}u_1{:}u_2{:}u_3]$ (which are arbitrary not all zero, defined
  up to a common multiplicative constant).
\end{itemize}

We can then compute:
\begin{itemize}
\item whether a point lies on a line by checking the relation ($*$)
  and all its permutations of coordinates (cyclic permutations
  suffice),
\item whether a point lies on a plane by the relation $u_0 x_0 +
  \cdots + u_3 x_3 = 0$,
\item whether a line lies in a plane by checking whether the dual
  point of the plane lies on the dual line (this gives $w_{2,3} u_2 +
  w_{1,3} u_1 + w_{0,3} u_0 = 0$ and cyclic permutations thereof),
\item the line joining two distinct points by computing the Plücker
  coordinates as $2\times 2$ determinants as defined in question (1),
\item the plane though a line and a point not lying on it by the
  formula found as explained in question (11),
\item the intersection line of two distinct planes by computing dual
  Plücker coordinates as explained in question (10),
\item the point of intersection of a line and a plane by taking the
  dual of the formula for the plane through a line and a point,
\item a sample point on a line as $(w_{0,3} : w_{1,3} : w_{2,3} : 0)$
  or some coordinate permutation thereof (as shown in question (6)),
\item a sample plane through a line dually to the previous point,
  namely $[w_{1,2} : {-w_{0,2}} : w_{0,1} : 0]$ (as shown in
  question (8)),
\item whether two lines meet, or equivalently, belong to a common
  plane, by using the criterion stated before (12), in which case
  their point of intersection can be computed by intersecting one of
  the lines with a sample plane through the other (as explained in the
  previous items), and their common plane can be computed dually.
\end{itemize}
\end{answer}

\bigskip

\centerline{\hbox to3truecm{\hrulefill}}

\medskip

\textbf{(13)} Independently of all of the above, compute the number of
lines in $\mathbb{P}^3(\mathbb{F}_q)$ (for example by counting the
number of pairs of distinct points in $\mathbb{P}^3(\mathbb{F}_q)$ and
the number of pairs of distinct points in a given line
$\mathbb{P}^1(\mathbb{F}_q)$), where $q$ is a prime power and
$\mathbb{F}_q$ denotes the finite field with $q$ elements.

\begin{answer}
There are $\frac{q^4-1}{q-1} = q^3 + q^2 + q + 1$ points in
$\mathbb{P}^3(\mathbb{F}_q)$.  There are thus $(q^3 + q^2 + q + 1)(q^3
+ q^2 + q) = q (q^3 + q^2 + q + 1) (q^2 + q + 1)$ pairs of distinct
points in $\mathbb{P}^3(\mathbb{F}_q)$, each defining a line.  There
are $\frac{q^2-1}{q-1} = q + 1$ points in
$\mathbb{P}^2(\mathbb{F}_q)$.  There are thus $q(q + 1)$ pairs of
distinct points defining any given line.  Thus, there are $(q (q^3 +
q^2 + q + 1) (q^2 + q + 1)) / (q (q+1)) = (q^2 + q + 1)(q^2 + 1)$
lines in $\mathbb{P}^3(\mathbb{F}_q)$ (that is, $q^4 + q^3 + 2q^2 + q
+ 1$).
\end{answer}

\textbf{(14)} Deduce from (13) and (1)–(7) the number of
$\mathbb{F}_q$-points on the hypersurface of degree $2$ (“quadric”)
$\{X_0 X_3 + X_1 X_4 + X_2 X_5 = 0\}$ in $\mathbb{P}^5(\mathbb{F}_q)$
with coordinates $(X_0:\cdots:X_5)$.  Assuming $q \equiv 1 \pmod{4}$,
deduce the number of $\mathbb{F}_q$-points on the hypersurface of
degree $2$ (“quadric”) $\{Z_0^2 + \cdots + Z_5^2 = 0\}$ in
$\mathbb{P}^5(\mathbb{F}_q)$ with coordinates $(Z_0:\cdots:Z_5)$
(\emph{hint:} $q \equiv 1 \pmod{4}$ means $-1$ is a square in
$\mathbb{F}_q$, so we can factor $Z^2 + Z^{\prime2}$).

\begin{answer}
We have seen in (1)–(7) that $\mathbb{F}_q$-points on the Plücker
quadric are in bijection with lines in $\mathbb{P}^3(\mathbb{F}_q)$,
and in (13) that there are $(q^2 + q + 1)(q^2 + 1) = q^4 + q^3 + 2q^2
+ q + 1$ of them.  Thus, there are that many points on the Plücker
quadric.  The equation of the latter can be written in the form $X_0
X_3 + X_1 X_4 + X_2 X_5 = 0$ by a simple linear coordinate change,
i.e., projective transformation ($X_0 = w_{0,1}$, $X_1 = w_{0,2}$,
$X_2 = w_{0,3}$, $X_3 = w_{2,3}$, $X_4 = -w_{1,3}$ and $X_5 =
w_{1,2}$) which certainly does not change the number of points.

As for $Z_0^2 + \cdots + Z_5^2 = 0$, if we call $\sqrt{-1}$ some fixed
square root of $-1$ (which exists when $q \equiv 1 \pmod{4}$ as this
means that the Legendre symbol $(\frac{-1}{q})$ is $1$), we can write
$Z_0^2 + Z_1^2 = (Z_0+\sqrt{-1}\,Z_1) (Z_0-\sqrt{-1}\,Z_1)$ and
similarly for $Z_2^2 + Z_3^2$ and $Z_4^2 + Z_5^2$, and since the
linear transformation $X_0 = Z_0+\sqrt{-1}\,Z_1$, $X_1 =
Z_2+\sqrt{-1}\,Z_3$, $X_2 = Z_4+\sqrt{-1}\,Z_5$, $X_3 =
Z_0-\sqrt{-1}\,Z_1$, $X_4 = Z_2-\sqrt{-1}\,Z_3$, $X_5 =
Z_4-\sqrt{-1}\,Z_5$ is invertible, there are still the same number of
points.
\end{answer}

\bigskip

\centerline{\hbox to3truecm{\hrulefill}}

\medskip

(This question is more difficult; it is independent of (13)\&(14).)

\textbf{(15)} Let $h \in k[t_0,t_1,t_2,t_3]$ be a homogeneous
polynomial, so that it defines a Zariski closed set (hypersurface) $X
:= \{h(x_0,x_1,x_2,x_3) = 0\}$ in $\mathbb{P}^3$.  Show that the of
lines contained in $X$ defines a Zariski closed subset $Y$ of the
Plücker quadric in $\mathbb{P}^5$.  (To be completely clear, this
means\footnote{Here $k^{\alg}$ denotes the algebraic closure of $k$,
  but feel free to assume that $k$ is algebraically closed ($k =
  k^{\alg}$) in this question.}: there is a Zariski closed set $Y$ in
$\mathbb{P}^5$, defined over $k$ and contained in the Plücker quadric
$Q$ (defined by $\dagger$), such that, for $w \in Q(k^{\alg})$, we
have $w \in Y(k^{\alg})$ if and only if $L_w \subseteq X(k^{\alg})$,
where $L_w$ denotes the line in $\mathbb{P}^3(k^{\alg})$ having
Plücker coordinates $w$.)

The important part of this question is: how can we compute equations
for $Y$ given the equation $h=0$ of $X$?

\begin{answer}
We have seen in question (6) that, so long as $w_{0,3} \neq 0$, the
line $L_w$ defined by $(w_{0,1}{:}w_{0,2}{:}w_{0,3} {:} \penalty0
w_{1,2}{:}w_{1,3}{:}w_{2,3})$ in $Q$ is the line through
$(w_{0,3}{:}w_{1,3}{:}w_{2,3}{:}0)$ and
$(0{:}w_{0,1}{:}w_{0,2}{:}w_{0,3})$.  In other words, $\lambda,\mu$ it
is the line consisting of points $(\lambda w_{0,3} : \lambda w_{1,3} +
\mu w_{0,1} : \lambda w_{2,3} + \mu w_{0,2} : \mu w_{0,3})$.  This
line is included in $X$ iff $h(\lambda w_{0,3}, \lambda w_{1,3} + \mu
w_{0,1}, \lambda w_{2,3} + \mu w_{0,2}, \mu w_{0,3}) = 0$ for all
$\lambda,\mu$ in $k^{\alg}$, which just means that, seen as a
polynomial in $\lambda,\mu$ now seen as two \emph{indeterminates},
this is identically zero.  But this is a homogeneous polynomial in
$\lambda,\mu$ whose coefficients are homogeneous polynomials in the
$w_{i,j}$, so equating all these coefficients to $0$ gives homogeneous
equations in the $w_{i,j}$ (coordinates in $\mathbb{P}^5$) for $L_w$
to be included in $X$.  This only holds so long as $w_{0,3} \neq 0$
(when it is zero, some of the resulting equation will be trivial);
however, at least one $w_{i,j}$ must be zero in any case, so writing
the corresponding equations for all permutations of coordinates,
together with the equation ($\dagger$) of the Plücker quadric itself
(to ensure that the $w_{i,j}$ do correspond to a line in
$\mathbb{P}^3$) gives us the equations of the desired $Y$.

To illustrate that this is actually algorithmic, the following Sage
code computes the equations for the set $Y$ of lines inside the
“diagonal cubic surface” $X := \{x_0^3 + x_1^3 + x_2^3 + x_3^3 = 0\}$:
{\fontsize{8}{10}\relax
\begin{verbatim}
sage: R.<x0,x1,x2,x3,w01,w02,w03,w12,w13,w23,u,v> = PolynomialRing(QQ,12)
sage: xvars = [x0,x1,x2,x3]
sage: wvars = [[0,w01,w02,w03],[-w01,0,w12,w13],[-w02,-w12,0,w23],[-w03,-w13,-w23,0]]
sage: # Plücker equation:
sage: plucker = w01*w23 - w02*w13 + w03*w12
sage: # Equation of the surface X:
sage: h = x0^3 + x1^3 + x2^3 + x3^3
sage: deg = h.degree()
sage: # All possible permutations of (0,1,2,3):
sage: perm4 = [(j0,j1,j2,j3) for j0 in range(4) for j1 in range(4) for j2 in range(4)
....:  for j3 in range(4) if len(set([j0,j1,j2,j3]))==4]
sage: # Generate the ideal I of the set Y of lines in X, as above:
sage: I = R.ideal([plucker] + [h.subs(dict([(xvars[j0],wvars[j0][j3]*u), (xvars[j1],w
....: vars[j1][j3]*u+wvars[j0][j1]*v), (xvars[j2],wvars[j2][j3]*u+wvars[j0][j2]*v), (
....: xvars[j3],wvars[j0][j3]*v)])).coefficient({u:deg-k,v:k}) for k in range(deg) fo
....: r (j0,j1,j2,j3) in perm4])
sage: # Compute its radical:
sage: I0 = I.radical()
sage: # This really means Y is 0-dimensional in projective space:
sage: I0.dimension()
7
sage: # This computes its number of geometric points (i.e., geometric lines on X):
sage: hp = I0.hilbert_polynomial() ; hp.leading_coefficient()*factorial(hp.degree()) 
27
\end{verbatim}
\par}\noindent (notation is as above except that $\lambda,\mu$ have
been called \texttt{u},\texttt{v}); the above code proves that thare
are $27$ geometric lines in the surface $\{x_0^3 + x_1^3 + x_2^3 +
x_3^3 = 0\} \subseteq \mathbb{P}^3$ (over $\mathbb{Q}$).
\end{answer}




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