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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} % |