From d811d610e9856f04df4d3dcab472d74dc437c13d Mon Sep 17 00:00:00 2001
From: "David A. Madore" <david+git@madore.org>
Date: Sun, 12 Apr 2026 11:25:35 +0200
Subject: Clarifications.

---
 controle-20260415.tex | 70 ++++++++++++++++++++++++++++-----------------------
 1 file changed, 39 insertions(+), 31 deletions(-)

(limited to 'controle-20260415.tex')

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
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