1 files changed, 126 insertions, 4 deletions
diff --git a/controle-20240410.tex b/controle-20240410.tex
index d3e54e3..edd2cd6 100644
--- a/controle-20240410.tex
+++ b/controle-20240410.tex
@@ -202,35 +202,157 @@ We denote $\ell_{ijk}$ the line through $p_i,p_j,p_k$ when it exists.
 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 $p_7$
-in two different ways (using two different pairs of lines) and
+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.)
 
-\textbf{(8)} Give two different examples of fields $k$, one infinite
-and one finite, over which a Möbius-Kantor configuration exists, and
-similarly two examples over which it does not exist.
+\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}
 
 
 %