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| -rw-r--r-- | controle-20260415.tex | 70 |
1 files changed, 39 insertions, 31 deletions
diff --git a/controle-20260415.tex b/controle-20260415.tex index 31847be..e196873 100644 --- a/controle-20260415.tex +++ b/controle-20260415.tex @@ -73,8 +73,8 @@ \noindent\textbf{Instructions.} -This exam consists of \textcolor{red}{XXX} completely independent exercises. They -can be tackled in any order, but students must clearly and readably +This exam consists of five completely independent exercises. They can +be tackled in any order, but students must clearly and readably indicate where each exercise starts and ends. \medbreak @@ -90,12 +90,20 @@ Use of electronic devices of any kind is prohibited. \medbreak -Duration: 2 hours +Duration: 2 hours. + +\medbreak + +Indicative and approximate grading scheme: 1 point per question, for a +final score out of 20 (so it will not be necessary to answer all +questions to get a perfect 20/20 score). + +\medbreak \ifcorrige -This answer key has \textcolor{red}{XXX} pages (this cover page included). +This answer key has 9 pages (this cover page included). \else -This exam has \textcolor{red}{XXX} pages (this cover page included). +This exam has 4 pages (this cover page included). \fi \vfill @@ -164,7 +172,7 @@ configuration exists, and compute the coordinates of its points. We fix an arbitrary 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). +with coordinates in $k$ (or “$k$-point”). We shall denote by $(x{:}y{:}z)$ the (homogeneous) coordinates of a point, and write $[u{:}v{:}w]$ for the line $\{ux+vy+wz = 0\}$. @@ -183,7 +191,7 @@ 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 $p_4=(1{:}0{:}0)$ and $p_2=(0{:}1{:}0)$ and $p_1=(0{:}0{:}1)$ and -$p_7=(1{:}1{:}1)$. \emph{We shall now do so.} +$p_7=(1{:}1{:}1)$. \emph{We shall now do so until question (4).} \begin{answer} No three of the four points $p_4,p_2,p_1,p_7$ are aligned, so they are @@ -248,13 +256,13 @@ $7$ lines, there are no other alignments than the prescribed ones. \medskip \textbf{(5)} \emph{Independently of all previous questions,} show that -the number of labeled projective bases in $\mathbb{P}^2(\mathbb{F}_q)$ +the number of “labeled” projective bases in $\mathbb{P}^2(\mathbb{F}_q)$ (in other words, $4$-uples $(a,b,c,d)$ of points such that no $3$ are aligned) is given by the formula: $q^3\, (q-1)^2\, (q+1)\, (q^2+q+1)$. \emph{Hint:} One possible approach is to count the number of possible choices for point $a$, then $b$, then $c$, then $d$; another possible -approach is to count matrices in $\mathit{GL}_3(\mathbb{F}_q)$ by +approach is to count elements of $\mathit{GL}_3(\mathbb{F}_q)$ by counting the possibilities for the first, then second, then third columns, and deduce the cardinality of $\mathit{PGL}_3(\mathbb{F}_q)$. Both approaches give the same formula (although in a slightly @@ -297,10 +305,10 @@ acts simply transitively on such. Let us now say that a \textbf{labeled Fano configuration}\footnote{In French: “configuration de Fano étiquetée”.} is a $7$-tuple of points $(p_1,\ldots,p_7)$ satisfying the same conditions as previously. (In -other words, the difference is that the unlabeled Fano configuration -is just the set $\{p_1,\ldots,p_7\}$ of seven points, whereas the -labeled Fano configuration has the information of which is $p_1$, -which is $p_2$, etc.) +other words, the difference is that the \textbf{unlabeled} Fano +configuration is just the set $\{p_1,\ldots,p_7\}$ of seven points, +whereas the labeled Fano configuration is the tuple: it has the +information of which is $p_1$, which is $p_2$, etc.) \smallskip @@ -519,8 +527,8 @@ $P,Q \in k[x,y]$; the morphism takes a geometric point $(x,y) \in Explain why a $\varphi$ as above takes values in $\mathbb{A}^2 \setminus\{(0,0)\}$ (i.e., $\varphi(x,y) \neq (0,0)$ for every -geometric point $(x,y)$) if and only if there exist $U,V \in k[x,y]$ -such that $U P + V Q = 1$. +geometric point $(x,y)$) \emph{if and only if} there exist $U,V \in +k[x,y]$ such that $U P + V Q = 1$. \begin{answer} The condition that $\varphi(x,y) \neq (0,0)$ for all geometric points @@ -587,15 +595,15 @@ this defines a morphism $\psi \colon \mathbb{A}^1 \to C$. Both $\psi(-1)$ and $\psi(1)$ equal $O$. \end{answer} -\textbf{(2)} Remembering how $\psi$ was constructed, given $(x,y)$ a -(geometric) point of $C$ other than $O$, how can we compute $t$ such -that $(x,y) = \psi(t)$? Deduce that there is a morphism $\tau \colon -C\setminus\{O\} \to \mathbb{A}^1\setminus\{\pm 1\}$ such that $\tau -\circ (\psi|_{\mathbb{A}^1\setminus\{\pm 1\}})$ is the identity on -$\mathbb{A}^1\setminus\{\pm 1\}$ and $\psi \circ \tau$ is the identity -on $C\setminus\{O\}$. Conclude that $C\setminus\{O\}$ is isomorphic -to $\mathbb{A}^1\setminus\{\pm 1\}$. On the other hand, is $\psi$ -itself an isomorphism (why or why not)? +\textbf{(2)} Given $(x,y)$ a (geometric) point of $C$ different +from $O$, how can we compute $t$ such that $(x,y) = \psi(t)$ (remember +how $\psi$ was constructed!)? Deduce that there is a morphism $\tau +\colon C\setminus\{O\} \to \mathbb{A}^1\setminus\{\pm 1\}$ such that +$\tau \circ (\psi|_{\mathbb{A}^1\setminus\{\pm 1\}})$ is the identity +on $\mathbb{A}^1\setminus\{\pm 1\}$ and $\psi \circ \tau$ is the +identity on $C\setminus\{O\}$. Conclude that $C\setminus\{O\}$ is +isomorphic to $\mathbb{A}^1\setminus\{\pm 1\}$. On the other hand, is +$\psi$ itself an isomorphism (why or why not)? \begin{answer} The construction of $\psi$ was to take the point other than $O$ in the @@ -676,8 +684,8 @@ direction. \end{answer} \textbf{(5)} Show that $\psi$ extends to a morphism $\bar\psi \colon -\mathbb{P}^1 \to \bar C$. Why is it surjective (meaning: on the -geometric points)? +\mathbb{P}^1 \to \bar C$. Why is it surjective (that is, surjective +on the geometric points)? \begin{answer} We can extend $\psi$ either by explicitly describing an equation for @@ -758,11 +766,11 @@ this shows that $-\frac{1}{x^2+2} + -\frac{1}{x^2+1}\,y$ is the inverse of $y-1$ in $K$. \end{answer} -\textbf{(3)} Explain why $K$ is isomorphic to $k(t)$ (the field -$\operatorname{Frac}(k[t])$ of rational fractions in one -indeterminate $t$) when $k = \mathbb{C}$. Describe an explicit -isomorphism. (\textit{Hint:} What does $C$ become if we let $x' := -\sqrt{-1}\,x$ and $y' := \sqrt{-1}\,y$?) +\textbf{(3)} Explain why, when $k = \mathbb{C}$, then $K$ is +isomorphic to $k(t)$ (the field $\operatorname{Frac}(k[t])$ of +rational fractions in one indeterminate $t$). Describe an explicit +isomorphism. (\textit{Hint:} If we let $x' := \sqrt{-1}\,x$ and $y' +:= \sqrt{-1}\,y$, what does $C$ become ?) \begin{answer} Letting $x' := \sqrt{-1}\,x$ and $y' := \sqrt{-1}\,y$ (which describes |