From 7c13374d3db61d2357dce85ce4c51f8adb96e4fa Mon Sep 17 00:00:00 2001
From: "David A. Madore" <david+git@madore.org>
Date: Sun, 12 Apr 2026 11:49:36 +0200
Subject: Mostly minor changes (typos, clarifications) suggested by an LLM.

---
 controle-20260415.tex | 24 ++++++++++++------------
 1 file changed, 12 insertions(+), 12 deletions(-)

(limited to 'controle-20260415.tex')

diff --git a/controle-20260415.tex b/controle-20260415.tex
index e196873..49a7c5a 100644
--- a/controle-20260415.tex
+++ b/controle-20260415.tex
@@ -206,7 +206,7 @@ and $p_1=(0{:}0{:}1)$ and $p_7=(1{:}1{:}1)$.
 
 \textbf{(2)} Compute the coordinates (i.e., equations) of the lines
 $\ell_{123}$ and $\ell_{347}$, and deduce the coordinates of the point
-$p_3$.  Analogously compute the coordiantes of $p_5$ and $p_6$.
+$p_3$.  Analogously compute the coordinates of $p_5$ and $p_6$.
 
 \begin{answer}
 Denoting $p\vee q$ the line through distinct points $p$ and $q$, we
@@ -269,7 +269,7 @@ Both approaches give the same formula (although in a slightly
 different way).
 
 \begin{answer}
-First aproach: there are $q^2+q+1$ possibilities for the point $a$,
+First approach: there are $q^2+q+1$ possibilities for the point $a$,
 because that is the cardinality of $\mathbb{P}^2(\mathbb{F}_q)$.  For
 the point $b$, since it needs to be different from $a$, we are left
 with $q^2+q$ possibilities.  For the point $c$, since it cannot belong
@@ -313,7 +313,7 @@ information of which is $p_1$, which is $p_2$, etc.)
 \smallskip
 
 \textbf{(6)} How many labeled Fano configurations are there in
-$\mathbb{P}^2(\mathbb{F}_{2^d})$?  Compute this number of $d=1$ and
+$\mathbb{P}^2(\mathbb{F}_{2^d})$?  Compute this number for $d=1$ and
 $d=2$ (that is, in $\mathbb{P}^2(\mathbb{F}_2)$ and
 $\mathbb{P}^2(\mathbb{F}_4)$).
 
@@ -352,7 +352,7 @@ which we have seen is $168$.  So we are left with $60\,480 \, / \, 168
 
 In this exercise, we consider the affine plane $\mathbb{A}^2$ with
 coordinates $(x,y)$ as a subset of the projective plane $\mathbb{P}^2$
-with coordinates $(T{:}X{:}Y)$ by $(x,y) \mapsto (1{:}X{:}Y)$.  We
+with coordinates $(T{:}X{:}Y)$ by $(x,y) \mapsto (1{:}x{:}y)$.  We
 work over the field $\mathbb{R}$ of real numbers, but we will also
 consider some complex points (i.e., $\mathbb{C}$-points).
 
@@ -361,11 +361,11 @@ map $(x,y) \mapsto (x,y) + (a,b)$ for certain (fixed) $(a,b) \in
 \mathbb{R}^2$.  A \textbf{vector homothety} is a map $(x,y) \mapsto (c
 x, cy)$ for certain (fixed) $c \in \mathbb{R}^\times :=
 \mathbb{R}\setminus\{0\}$.  A \textbf{vector rotation} is a map $(x,y)
-\mapsto (ux + vy,\, -vx + uy)$ for certain (fixed) $u,v \in
+\mapsto (ux + vy,\, -vx + uy)$ for certain (fixed) $(u,v) \in
 \mathbb{R}^2$ satisfying $u^2+v^2 = 1$.  An \textbf{affine similitude}
 is an element of the group generated by translations, vector
-homotheties and vector rotations (this is a subgroup of the group all
-affine transformations).
+homotheties and vector rotations (this is a subgroup of the group of
+all affine transformations).
 
 \textbf{(1)} Describe the matrices, in $\mathit{PGL}_3(\mathbb{R})$ of
 the extensions to $\mathbb{P}^2$ of the three kinds of transformations
@@ -458,7 +458,7 @@ and this gives $[1 : -2 : 0]$.  So it is the line $\{2X=T\}$, or (the
 projective extension of) $\{x = \frac{1}{2}\}$.
 \end{answer}
 
-\textbf{(5)} Show that for any any two distinct points $A,B$ in
+\textbf{(5)} Show that for any two distinct points $A,B$ in
 $\mathbb{A}^2(\mathbb{R})$ there is an affine similitude taking
 $(0,0)$ to $A$ and $(1,0)$ to $B$.  (\emph{Hint:} You can use simple
 arguments of standard elementary Euclidean plane geometry for this
@@ -511,7 +511,7 @@ $\mathbb{A}^2(\mathbb{R})$ can be brought to this position by an
 affine similitude, and since affine similitudes preserve perpendicular
 bisectors (because each one of translations, vector homotheties and
 vector rotations preserve angles and midpoints), the construction
-works for any two distincts points $A,B$.
+works for any two distinct points $A,B$.
 \end{answer}
 
 %
@@ -635,7 +635,7 @@ the two points $-1$ and $1$.)
 
 We now see $\mathbb{A}^2_k$ as a subset of the projective plane
 $\mathbb{P}^2_k$ with coordinates $(W{:}X{:}Y)$ by $(x,y) \mapsto
-(1{:}X{:}Y)$.  We call $\bar C$ the projective completion of $C$, in
+(1{:}x{:}y)$.  We call $\bar C$ the projective completion of $C$, in
 other words, the Zariski closure of $C$ inside $\mathbb{P}^2$.
 
 \textbf{(3)} What is the equation of $\bar C$?  What are its points at
@@ -705,7 +705,7 @@ coordinates on $\mathbb{P}^1$; furthermore, these satisfy $W Y^2 - X^3
 
 It is surjective because we have already seen in (1) that every point
 other than the point $P$ at infinity is in the image of $\psi$ (for
-points $M$ other than $P$ and $O$, take the slope of the line $OP$;
+points $M$ other than $P$ and $O$, take the slope of the line $OM$;
 and for $O$ we have seen that it is attained twice); as for the point
 $P = (0{:}0{:}1)$, it is $\bar\psi((0{:}1))$.
 \end{answer}
@@ -762,7 +762,7 @@ Similarly, to compute $\frac{1}{y-1}$, we search for a Bézout relation
 $u\,(y-1) + v\cdot (x^2+y^2+1) = 1$ in $k(x)[y]$, which is again easy
 because $-(y+1)(y-1) + (x^2+y^2+1) = x^2+2$ is an element of $k(x)$,
 so $-\frac{y+1}{x^2+2}\,(y-1) + \frac{1}{x^2+2}\,(x^2+y^2+1) = 1$, and
-this shows that $-\frac{1}{x^2+2} + -\frac{1}{x^2+1}\,y$ is the
+this shows that $-\frac{1}{x^2+2} -\frac{1}{x^2+2}\,y$ is the
 inverse of $y-1$ in $K$.
 \end{answer}
 
-- 
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