From b12302a363df02f77d24cb8a5e1570da02e1da14 Mon Sep 17 00:00:00 2001
From: "David A. Madore" <david+git@madore.org>
Date: Thu, 6 Apr 2023 23:05:27 +0200
Subject: Some additional questions, various rewordings.

---
 controle-20230412.tex | 121 ++++++++++++++++++++++++++++++++++++++++----------
 1 file changed, 98 insertions(+), 23 deletions(-)

diff --git a/controle-20230412.tex b/controle-20230412.tex
index c6100bb..9332301 100644
--- a/controle-20230412.tex
+++ b/controle-20230412.tex
@@ -79,9 +79,11 @@
 
 \noindent\textbf{Instructions.}
 
-The different exercises below are completely independent.  They can be
-answered in any order, but candidates are asked to label very clearly
-on their papers where each exercise starts.
+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
 
@@ -99,9 +101,9 @@ Use of electronic devices of any kind is prohibited.
 Duration: 2 hours
 
 \ifcorrige
-This answer key has \textcolor{red}{XXX} pages (cover page included).
+This answer key has 6 pages (cover page included).
 \else
-This exam has \textcolor{red}{XXX} pages (cover page included).
+This exam has 3 pages (cover page included).
 \fi
 
 \vfill
@@ -118,10 +120,10 @@ Git: \input{vcline.tex}
 %
 %
 
-\exercise
+\textit{The goal of this problem is to study a representation of lines
+  in $\mathbb{P}^3$.}
 
-\textit{The goal of this exercise 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
@@ -140,8 +142,10 @@ $w\neq 0$ 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.
 
+\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 call $x\wedge y := (w_{0,1},
+(y_0,\ldots,y_3) \in k^4$, 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
@@ -193,6 +197,8 @@ exists when $x$ and $y$ are not collinear (so that $x\wedge y \neq
 depend on the $x,y \in V$.
 \end{answer}
 
+\bigskip
+
 The $w_{i,j}$ in question are known as the \textbf{Plücker
   coordinates} of $L$.
 
@@ -233,15 +239,17 @@ w_{0,1} w_{2,3} - w_{0,2} w_{1,3} + w_{0,3} w_{1,2} = 0
 \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 and $-x_3$ times the second gives the stated
+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 how to associate to any line $L$ in
-$\mathbb{P}^3(k)$ a $k$-point on the Plücker quadric.  We now consider
-the converse.
+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 $\mathbb{P}^5(k)$
@@ -252,7 +260,7 @@ $(0{:}w_{0,1}{:}w_{0,2}{:}w_{0,3})$ 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 use the result, with the Plücker
+(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}
@@ -295,18 +303,20 @@ 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 defined as the the intersection of two plines.
+line defined 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 $[0 : w_{2,3} : {-w_{1,3}}
   : w_{1,2}]$ contain $L$.  Now consider these as points in the
-dual $\mathbb{P}^3$: show that the Plücker coordinates of the line
+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 $w_{1,2} \neq
 0$.
@@ -320,13 +330,11 @@ cyclically, it is also on the plane $[0 : w_{2,3} : {-w_{1,3}} :
 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
+  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 $[w_{2,3} : {-w_{1,3}} : w_{1,2} : w_{0,3} :
-  {-w_{0,2}} : w_{0,1}]$ (we already know that these six coordinates
-cannot vanish simultaneously, so $w_{1,2} \neq 0$ is the only
-condition we need).
+  {-w_{0,2}} : w_{0,1}]$.
 \end{answer}
 
 \textbf{(9)} Deduce from (8) that if $L$ is a line in
@@ -369,16 +377,21 @@ 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 $L^*$).
+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.)
+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
@@ -437,6 +450,8 @@ w_{0,1} w'_{2,3} - w_{0,2} w'_{1,3} + w_{0,3} w'_{1,2}
 \]
 (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
@@ -470,7 +485,7 @@ We can then compute:
   \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),
+  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
@@ -492,6 +507,66 @@ We can then compute:
 \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}
+
+
 
 
 %
-- 
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