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\begin{document}
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\title{FMA-4AC05-TP / ACCQ205\\Final exam — answer key\\{\normalsize Algebraic curves}}
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\title{FMA-4AC05-TP / ACCQ205\\Final exam\\{\normalsize Algebraic curves}}
\fi
\author{}
\date{2026-04-15}
\maketitle

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\vskip1truein\relax

\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
indicate where each exercise starts and ends.

\medbreak

Answers can be written in English or French.

\medbreak

Use of written documents of any kind (such as handwritten or printed
notes, exercise sheets or books) is permitted.

Use of electronic devices of any kind is prohibited.

\medbreak

Duration: 2 hours

\ifcorrige
This answer key has \textcolor{red}{XXX} pages (this cover page included).
\else
This exam has \textcolor{red}{XXX} pages (this cover page included).
\fi

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\pagebreak

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\exercise

We say that a set of seven distinct points $p_1,\ldots,p_7$ in the
projective plane $\mathbb{P}^2$ over a field $k$ is a \textbf{Fano
  configuration} when the points satisfy the alignment conditions
depicted in the following figure:

\begin{center}
\vskip-2ex\leavevmode
\begin{tikzpicture}
\coordinate (P1) at (-2cm,0);
\coordinate (P2) at (2cm,0);
\coordinate (P3) at (0,0);
\coordinate (P4) at (0,3.464cm);
\coordinate (P5) at (-1cm,1.732cm);
\coordinate (P6) at (1cm,1.732cm);
\coordinate (P7) at (0cm,1.155cm);
\draw (P2)--(P4);
\draw (P1)--(P4);
\draw (P3)--(P4);
\draw (P1)--(P2);
\draw (P2)--(P5);
\draw (P1)--(P6);
\draw (P3) to[out=180,in=240] (P5) to[out=60,in=120] (P6) to[out=300,in=0] (P3);
\fill[black] (P1) circle (2.5pt);
\fill[black] (P2) circle (2.5pt);
\fill[black] (P3) circle (2.5pt);
\fill[black] (P4) circle (2.5pt);
\fill[black] (P5) circle (2.5pt);
\fill[black] (P6) circle (2.5pt);
\fill[black] (P7) circle (2.5pt);
\node[anchor=north east] at (P1) {$p_1$};
\node[anchor=north west] at (P2) {$p_2$};
\node[anchor=north] at (P3) {$p_3$};
\node[anchor=south] at (P4) {$p_4$};
\node[anchor=south east] at (P5) {$p_5$};
\node[anchor=south west] at (P6) {$p_6$};
\node[anchor=north west] at (P7) {$p_7$};
\end{tikzpicture}
\vskip-5ex\leavevmode
\end{center}

This means: the seven points are distinct; all the following sets of
points are aligned: $\{p_2, p_4, p_6\}$, $\{p_1, p_4, p_5\}$, $\{p_3,
p_4, p_7\}$, $\{p_1, p_2, p_3\}$, $\{p_2, p_5, p_7\}$, $\{p_1, p_6,
p_7\}$ and $\{p_3, p_5, p_6\}$; and no other set of three of the $p_i$
are aligned.

The goal of this exercise is to determine over which fields $k$ a Fano
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).

We shall denote by $(x{:}y{:}z)$ the (homogeneous) coordinates of a
point, and write $[u{:}v{:}w]$ for the line $\{ux+vy+wz = 0\}$.
%% Recall that the line through $(x_1{:}y_1{:}z_1)$ and
%% $(x_2{:}y_2{:}z_2)$ (assumed distinct) is given by the formula $[(y_1
%%   z_2 - y_2 z_1) : (z_1 x_2 - z_2 x_1) : (x_1 y_2 - x_2 y_1)]$, and
%% that the same formula (exchanging parentheses and square brackets) can
%% also be used to compute the intersection of two distinct lines.  (This
%% may not always be the best or simplest way to compute coordinates,
%% however!)

\emph{We assume for questions (1)–(3) below that $p_1,\ldots,p_7$ is a
Fano configuration of points (over the given field $k$), and the
questions will serve to compute the coordinates of the points.}  We
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.}

\begin{answer}
No three of the four points $p_4,p_2,p_1,p_7$ 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 $p_4=(1{:}0{:}0)$ and $p_2=(0{:}1{:}0)$
and $p_1=(0{:}0{:}1)$ and $p_7=(1{:}1{:}1)$.
\end{answer}

\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 $p_6$.

\begin{answer}
Denoting $p\vee q$ the line through distinct points $p$ and $q$, we
get $\ell_{123} = p_1 \vee p_2 = [1{:}0{:}0]$ and $\ell_{347} =
p_4\vee p_7 = [0{:}-1{:}1]$.  Denoting by $\ell\wedge m$ the point of
intersection of distinct lines $\ell$ and $m$, we get $p_3 =
\ell_{123} \wedge \ell_{347} = (0{:}1{:}1)$.

Similar computations (or using the cyclic symmetry $p_1 \mapsto p_2
\mapsto p_4 \mapsto p_1$ which corresponds to a cyclic permutation of
coordinates) gives $p_5 = (1{:}0{:}1)$ and $p_6 = (1{:}1{:}0)$.
\end{answer}

\textbf{(3)} Using the last alignment condition that hasn't yet been
used, give a \emph{necessary} condition for a Fano configuration to
exist in $\mathbb{P}^2(k)$.

\begin{answer}
The remaining condition is the alignment of $p_3,p_5,p_6$.  This is
expressed by the vanishing of the determinant of their coordinates,
or, equivalently, by computing $p_3 \vee p_5 = [1{:}1{:}-1]$ and
expressing the fact that $p_6$ lies on it.  The necessary condition we
get is: $2=0$ in $k$, in other words, the field $k$ is of
characteristic $2$.

Thus, we have shown that a Fano configuration does not exist in a
field of characteristic $\neq 2$.
\end{answer}

\textbf{(4)} Conversely, use the previously computed coordinates to
explain why this necessary condition on $k$ is also sufficient for a
Fano configuration to exist.

\begin{answer}
If $k$ is any field, then setting $p_4=(1{:}0{:}0)$ and
$p_2=(0{:}1{:}0)$ and $p_1=(0{:}0{:}1)$ and $p_7=(1{:}1{:}1)$ and $p_3
= (0{:}1{:}1)$ and $p_5 = (1{:}0{:}1)$ and $p_6 = (1{:}1{:}0)$ ensures
six of the seven required alignments.  And if $k$ is of
characteristic $2$ then $p_3,p_5,p_6$ are also aligned for the reasons
explained in the previous question.  But furthermore, this gives an
identification of the $7$ points with the points of
$\mathbb{P}^2(\mathbb{F}_2)$ (where $\mathbb{F}_2$ is seen as a
subfield of $k$), and since $\mathbb{P}^2(\mathbb{F}_2)$ has
$7$ lines, there are no other alignments than the prescribed ones.
\end{answer}

\medskip

\textbf{(5)} \emph{Independently of all previous questions,} show that
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
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
different way).

\begin{answer}
First aproach: 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
to the line $ab$, which has $q+1$ points, we are left with $q^2$
possibilities.  Finally, for the last point $d$, there are three lines
to be ruled out ($ab$, $ac$ and $bc$), each one having $q+1$ points,
but as they meet pairwise in a single point, they have $3(q+1)-3 = 3q$
point together, and we are left with $(q^2+q+1)-3q = q^2-2q+1 =
(q-1)^2$ possibilities for $d$.  This means there are $(q^2+q+1)\,
(q^2+q)\, q^2\, (q-1)^2 = q^3\, (q-1)^2\, (q+1)\, (q^2+q+1)$ labeled
projective bases.

Second approach: to construct a matrix in
$\mathit{GL}_3(\mathbb{F}_q)$, we first choose its first column, which
can be any nonzero vector, giving us $q^3-1$ possibilities; then we
choose its second, which can be any vector not collinear with the
first, giving us $q^3-q$ possibilities; then we choose the third,
which can be any vector not in the vector space spanned by the first
two, leaving us $q^3-q^2$ possibilities.  Thus, there are
$(q^3-1)\,(q^3-q)\,(q^3-q^2)$ elements in
$\mathit{GL}_3(\mathbb{F}_q)$.  But since
$\mathit{PGL}_3(\mathbb{F}_q)$ is its quotient by the subgroup of
order $q-1$ consisting of homotheties (multiplication by a nonzero
constant), there are $\frac{(q^3-1)\,(q^3-q)\,(q^3-q^2)}{q-1} =
(q^2+q+1)\,(q^3-q)\,(q^3-q^2) = q^3\, (q-1)^2\, (q+1)\, (q^2+q+1)$
elements of $\mathit{PGL}_3(\mathbb{F}_q)$.  This is also the number
of labeled projective bases because $\mathit{PGL}_3(\mathbb{F}_q)$
acts simply transitively on such.
\end{answer}

\medskip

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.)

\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
$d=2$ (that is, in $\mathbb{P}^2(\mathbb{F}_2)$ and
$\mathbb{P}^2(\mathbb{F}_4)$).

\emph{Note:} You can write numbers as products, there is no need to
fully compute the multiplications by hand.

\begin{answer}
We have seen in questions (1)–(4) that, over a field of
characteristic $2$, a labeled Fano configuration is constructed in a
unique way from a labeled projective basis (which serves as
$p_4,p_2,p_1,p_7$).  Thus, the number of Fano configurations in
$\mathbb{P}^2(\mathbb{F}_{2^d})$ equals the number of labeled
projective bases, which equals $2^{3d} (2^d-1)^2 (2^d+1)
(2^{2d}+2^d+1)$.  For $d=1$ this gives $8\times 3\times 7 = 168$; and
for $d=2$ this gives $64\times 9 \times 5 \times 21 = 60\,480$.
\end{answer}

\textbf{(7)} Deduce the number of \emph{unlabeled} Fano configurations
in $\mathbb{P}^2(\mathbb{F}_4)$.  (\emph{Hint:} The previous question
provides a way to count the number of labeled Fano configurations for
each unlabeled one!)

\begin{answer}
The number of ways to label a given Fano configuration, i.e., the
number of labeled Fano configurations for each unlabeled one, equals
the number of Fano configurations in $\mathbb{P}^2(\mathbb{F}_2)$,
which we have seen is $168$.  So we are left with $60\,480 \, / \, 168
= 360$ (unlabeled) Fano configurations in $\mathbb{P}^2(\mathbb{F}_4)$.
\end{answer}

%
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\exercise

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
work over the field $\mathbb{R}$ of real numbers, but we will also
consider some complex points (i.e., $\mathbb{C}$-points).

\textit{Definitions:} A \textbf{translation} of $\mathbb{A}^2$ is a
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) $u,v \in
\mathbb{R}^2$ 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).

\textbf{(1)} Describe the matrices, in $\mathit{PGL}_3(\mathbb{R})$ of
the extensions to $\mathbb{P}^2$ of the three kinds of transformations
we just defined (translations, vector homotheties, and vector
rotations).

\begin{answer}
The matrices in question are, with the same notations as in the
definition:
\[
\begin{pmatrix}1&0&0\\a&1&0\\b&0&1\\\end{pmatrix}\quad,\quad
\begin{pmatrix}1&0&0\\0&c&0\\0&0&c\\\end{pmatrix}\quad,\quad
\begin{pmatrix}1&0&0\\0&u&v\\0&-v&u\\\end{pmatrix}
\]
(each one being defined, of course, only up to multiplication by a
constant).
\end{answer}

\medskip

We define the complex points $I := (0 : 1 : \sqrt{-1})$ and $J := (0 :
1 : -\sqrt{-1})$ in $\mathbb{P}^2(\mathbb{C})$, also known as the
\textbf{cyclic points}.

\textbf{(2)} Show that $I$ and $J$ are fixed under every affine
similitude (extended to $\mathbb{P}^2$).

\begin{answer}
We compute the product of the matrices found in (1) with the column
vectors giving the coordinates of $I$ and $J$.  Translations and
homotheties fix $I$ and $J$ simply because they are on the line at
infinity ($T=0$).  For a rotation acting on $I$ we find, with the same
notations as previously, $(0 : u + v \sqrt{-1} : -v + u \sqrt{-1})$,
and we observe that these are the coordinates of $I$ multiplied by $u
+ v \sqrt{-1}$, so it is the same point; the same argument works
for $J$.
\end{answer}

\textbf{(3)} Conversely, show that every real projective
transformation of $\mathbb{P}^2$ (i.e., every element of
$\mathit{PGL}_3(\mathbb{R})$) which fixes $I$ and $J$ is an affine
similitude.  (\emph{Hint:} You may want to first observe that it
stabilizes the line $\ell_\infty := IJ$ and conclude that it is an
affine transformation.)

\begin{answer}
The line $IJ$ is the line at infinity $\{T=0\}$.  A real projective
transformation fixing $I$ and $J$ must stabilize the line through them
(because projective transformations preserve alignment).  This means
that it is, in fact, an affine transformation, or equivalently, given
by a matrix of the form:
\[
\begin{pmatrix}1&0&0\\a&m&n\\b&p&q\\\end{pmatrix}
\]
(with $m,n,p,q\in\mathbb{R}$).  Now fixing $I$ imposes the condition
that $(0 : m + n \sqrt{-1} : p + q \sqrt{-1})$ is $(0 : 1 :
\sqrt{-1})$, which means that $p + q \sqrt{-1} = \sqrt{-1}\,(m + n
\sqrt{-1}) = -n + m \sqrt{-1}$, and by identifying real and imaginary
parts we conclude $p = -n$ and $q = m$.  So our matrix is now of the
form:
\[
\begin{pmatrix}1&0&0\\a&m&n\\b&-n&m\\\end{pmatrix}
\]
and if we let $c = \sqrt{m^2+n^2}$ (a real number) and $u := m/c$ and
$v = n/c$ (which satisfy $u^2 + v^2 = 1$), the above matrix is the
product (i.e., the composition)
\[
\begin{pmatrix}1&0&0\\a&1&0\\b&0&1\\\end{pmatrix}\,
\begin{pmatrix}1&0&0\\0&c&0\\0&0&c\\\end{pmatrix}\,
\begin{pmatrix}1&0&0\\0&u&v\\0&-v&u\\\end{pmatrix}
\]
as required.
\end{answer}

\textbf{(4)} In this question, let $A := (0,0)$ and $B := (1,0)$ in
$\mathbb{A}^2$.  Compute the equation of the line $(AI\wedge BJ) \vee
(AJ\wedge BI)$.  (Here, $\ell\wedge m$ denotes the intersection point
of the lines $\ell,m$, and $P\vee Q$ or just $PQ$ denotes the line
through $P,Q$.)

\begin{answer}
Let us write as usual $[\lambda{:}\mu{:}\nu]$ for the line $\{\lambda
x+\mu y+\nu z = 0\}$.  We find $AI = [0 : 1 : \sqrt{-1}]$ and $BJ = [1
  : -1 : \sqrt{-1}]$, so $AI\wedge BJ = (2 : 1 : \sqrt{-1})$.  The
formulas for $AJ$, $BI$ and $AJ\wedge BI$ are obtained by exchanging
$\sqrt{-1}$ with $-\sqrt{-1}$ so it is not necesary to recompute them.
Finally, the sought-after line $(AI\wedge BJ) \vee (AJ\wedge BI)$ is
obtained by joining $(2 : 1 : \sqrt{-1})$ with $(2 : 1 : -\sqrt{-1})$,
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
$\mathbb{A}^2(\mathbb{R})$ 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
question, independently of all previous questions.  Alternatively, you
can use the previous questions and a fact from projective geometry
seen in the course.)

\begin{answer}
By Euclidean geometry arguments: using a translation we can place any
point of $\mathbb{A}^2(\mathbb{R})$ in any given place, so we can
assume without loss of generality that $A = (0,0)$; using a homothety
we multiply distances by a constant $c\neq 0$, so we can assume
without loss of generality that the distance $AB$ is $1$, and $B$ is
now a point on the unit circle; finally, we can rotate around the
origin to get $B$ in $(1,0)$.  This provides the required affine
similitude.

By a projective geometry argument: for any two distinct $A,B$ in
$\mathbb{A}^2(\mathbb{R})$, the points $A,B,I,J$ form a projective
basis of $\mathbb{P}^2(\mathbb{C})$ (in detail: the points $A,B$ are
not aligned with $I,J$ because they are not on the line $\ell_\infty$
at infinity, and the points $I,J$ cannot be on the line $AB$ because
$AB$ is a \emph{real} line and therefore also $AB\wedge \ell_\infty$
is a real point).  So if we let $A_0 := (0,0)$ and $B_0 := (1,0)$,
there is a unique complex projective transformation taking
$A_0,B_0,I,J$ to $A,B,I,J$; now that complex transformation is real
because its complex conjugate takes $A_0,B_0,J,I$ to $A,B,J,I$, so it
is the same.  But we have seen in question (3) that a real projective
transformation of $\mathbb{P}^2$ which fixes $I$ and $J$ is an affine
similitude, so we have answered the question.

Alternatively, it is also possible to answer the question by a direct
computation of the coefficients of the matrix.
\end{answer}

\textbf{(6)} Conclude that, for any two distinct points $A,B$ in
$\mathbb{A}^2(\mathbb{R})$, the perpendicular bisector\footnote{In
French: “la médiatrice”.  The perpendicular bisector of $[AB]$ is the
line of points at equal distance from $A$ and $B$ in Euclidean
geometry.  In this context, it is also the perpendicular line to $AB$
through the midpoint of $[AB]$.} of $[AB]$ can be constructed as the
line $(AI\wedge BJ) \vee (AJ\wedge BI)$.

\begin{answer}
We have seen in question (4) that the construction $(AI\wedge BJ) \vee
(AJ\wedge BI)$ gives the perpendicular bisector $x = \frac{1}{2}$ of
the two points $A = (0,0)$ and $B = (1,0)$.  Since we have seen in
question (5) that any two distinct points in
$\mathbb{A}^2(\mathbb{R})$ 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 $A,B$.
\end{answer}

%
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\exercise

\textit{Reminder:} A morphism $\varphi \colon \mathbb{A}^2_k \to
\mathbb{A}^2_k$ (over a field $k$) is simply given by two polynomials
$P,Q \in k[x,y]$; the morphism takes a geometric point $(x,y) \in
\mathbb{A}^2(k^{\alg})$ to $(P(x,y), \, Q(x,y))$.

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 $U P + V Q = 1$.

\begin{answer}
The condition that $\varphi(x,y) \neq (0,0)$ for all geometric points
$(x,y)$ means, by definition, that the affine algebraic variety
$Z(P,Q)$ defined by the equations $P=0$ and $Q=0$ has no geometric
points.  By Hilbert's Nullstellensatz, this is equivalent to demanding
that $P,Q$ generate the unit ideal in $k^{\alg}[x,y]$.  But we have
seen in the course that this is also equivalent to demanding that they
generate the unit ideal in $k[x,y]$ (because the latter is the
intersection of the former with $k[x,y]$): that is, that there exist
$U,V \in k[x,y]$ such that $U P + V Q = 1$.
\end{answer}

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\exercise

In the affine plane $\mathbb{A}^2_k$ with coordinates $(x,y)$ over a
field $k$ of characteristic $\neq 2$, we consider the affine algebraic
variety $C$ with equation $y^2 = x^3 + x^2$.

\textbf{(0)} Draw an approximate picture of $C$ for $k = \mathbb{R}$.
(You may wish to answer the next question first.)

\begin{answer}
Over the real numbers, the not-necessarily-smooth affine curve $C$
looks like this:
\begin{center}
\begin{tikzpicture}[scale=2]
%% \draw[step=.2cm,help lines] (-1.25,-1.25) grid (1.25,1.25);
\draw[->] (-1.15,0) -- (1.15,0); \draw[->] (0,-1.15) -- (0,1.15);
\draw (0.777778,-1.037037) .. controls (0.481481,-0.555556) and (0.222222,-0.222222) .. (0,0) ; % t from -4/3 to -1
\draw (0,0) .. controls (-0.666667,0.666667) and (-1,0.333333) .. (-1,0); % t from -1 to 0
\draw (-1,0) .. controls (-1,-0.333333) and (-0.666667,-0.666667) .. (0,0); % t from 0 to 1
\draw (0,0) .. controls (0.222222,0.222222) and (0.481481,0.555556) ..  (0.777778,1.037037); % t from 1 to 4/3
\end{tikzpicture}
\end{center}

It intersects the $x$ axis in two points, $(-1,0)$ and $(0,0)$.  At
the former, it has a vertical tangent as one checks by computing
partial differentials.  At the latter, the partial differentials
vanish (the point is not smooth): one can get an idea of what it looks
like by considering the lowest order terms $y^2 \approx x^2$,
suggesting the two lines $y = \pm x$; or by taking the derivative
around $-1$ and $1$ of the parametric equation $\psi$ found in the
next question.
\end{answer}

\textbf{(1)} For $t \neq 1,-1$ (in $k^{\alg}$), show that the line $y
= t\,x$ intersects $C$ in a unique point $\psi(t)$ different from $O
:= (0,0)$, compute its coordinates, and explain why this formula
defines a morphism $\psi \colon \mathbb{A}^1 \to C$.  What are
$\psi(-1)$ and $\psi(1)$?

\begin{answer}
For $t \neq 1,-1$, a point other than $(0,0)$ satisfying $y = t\, x$
and $y^2 = x^3 + x^2$ satisfies $t^2 x^2 = x^3 + x^2$ so $t^2 = x +
1$, so $x = t^2 - 1$ and $y = t^3 - t$, and conversely these
coordinates define a point on $C$ and on the line $y = tx$ for any $t$
(even $t = 1,-1$).  Since $t^2 - 1$ and $t^3 - t$ are polynomials,
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)?

\begin{answer}
The construction of $\psi$ was to take the point other than $O$ in the
intersection of $C$ and line $y = t\, x$: so we recover $t$ as $y/x$.
This defines a morphism $\tau \colon C\setminus\{O\} \to
\mathbb{A}^1\setminus\{\pm 1\}$, namely $(x,y) \mapsto y/x$: it makes
sense because $x\neq 0$ on $C\setminus\{O\}$ (clearly, $(0,0)$ is the
only point of $C$ with $x=0$, since the equation then implies $y=0$ as
well); and it is $\neq 1,-1$ because the only solution to $y^2 = x^3 +
x^2$ with $y = \pm x$ is $(0,0)$ (which we ruled out).  The fact that
$\tau \circ (\psi|_{\mathbb{A}^1\setminus\{\pm 1\}})$ is the identity
amounts to checking the obvious fact $(t^3-t)/(t^2-1) = t$ (for $t\neq
-1,1$); and the fact that $\psi \circ \tau$ is the identity follows
from our computations in question (1).  So we have morphisms $\tau
\colon C\setminus\{O\} \to \mathbb{A}^1\setminus\{\pm 1\}$ and
$\psi|_{\mathbb{A}^1\setminus\{\pm 1\}} \colon
\mathbb{A}^1\setminus\{\pm 1\} \to C\setminus\{O\}$ which are inverse
of one another, letting us conclude that $C\setminus\{O\}$ is
isomorphic to $\mathbb{A}^1\setminus\{\pm 1\}$.

On the other hand, $\psi$ itself is \emph{not} an isomorphism because
both $-1$ and $1$ get mapped to $O$.

(Intuitively, $C$ is obtained by taking $\mathbb{A}^1$ and identifying
the two points $-1$ and $1$.)
\end{answer}

\smallskip

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
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
infinity (in other words, the points of $\bar C$ on the line $W=0$)?
Where (in which direction) should we imagine these points to be in the
drawing of question (0)?

\begin{answer}
The equation of $\bar C$ is obtained by homogeneizing the equation
$y^2 - x^3 - x^2 = 0$ of $C$ (substituting $x = X/W$ and $y = Y/W$ and
multiplying by $W^3$ to get rid of denominators): this gives us $W Y^2
- X^3 - W X^2$.  Intersecting with the line at infinity $W=0$ gives
$X=0$, so we have a unique point $(0{:}0{:}1)$ (remember that
$(0{:}0{:}Y)$ defines a \emph{unique} point in $\mathbb{P}^2$).  This
point is the point at infinity in the vertical direction (because it
lies, say, on the line $X=0$ which is the $y$ axis): this is
intuitively in accordance with the fact that the curve seems to go to
infinity “more vertically”.
\end{answer}

\textbf{(4)} For each point at infinity $P$ found in question (3),
determine whether $C$ is smooth at $P$ and, if it is, compute the
tangent line $T_P C$ to $C$ at $P$.  (\emph{Hint:} You will need to
switch to a different $\mathbb{A}^2$ so that the point $P$ is no
longer at infinity.)  Where should we imagine this tangent line $T_P
C$ to be in the drawing of question (0)?

\begin{answer}
We have found only one point at infinity, $P = (0{:}0{:}1)$.  We
introduce new affine coordinates where it is not at infinity by taking
$Y = 0$ to be our line at infinity, in other words we put $w' = W/Y$
and $x' = X/Y$.  In the complement $\mathbb{A}^{2\prime} :=
\mathbb{P}^2 \setminus \{Y=0\}$ of this line at infinity, the equation
of the curve $\bar C \cap \mathbb{A}^{2\prime}$ is given by
dehomogeneizing the equation of $\bar C$: it is $w' - x^{\prime 3} -
w' x^{\prime 2} = 0$.  Our point $P$ is now given by $(w',x') =
(0,0)$.  Differentiating $g := w' - x^{\prime 3} - w' x^{\prime 2}$
with respect to $w'$ and $x'$ at $(0,0)$ gives $\left. \frac{\partial
  g}{\partial w'} \right|_{(0,0)} = 1$ and $\left. \frac{\partial
  g}{\partial x'} \right|_{(0,0)} = 0$, so the curve $\bar C$ is
smooth at $P$ and has tangent line $w' = 0$ in $\mathbb{A}^{2\prime}$,
that is $W = 0$ in $\mathbb{P}^2$.  This tangent line is the line at
infinity of our original affine chart $\mathbb{A}^2$: in the drawing
of question (0), one should imagine that it is at infinity in every
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)?

\begin{answer}
We can extend $\psi$ either by explicitly describing an equation for
it: if $(U{:}V)$ are coordinates on $\mathbb{P}^1$ with $t = V/U$
being the coordinate formerly used on $\mathbb{A}^1$, then we can
homogeneize the formulas for $\psi$ as follows:
\[
(U{:}V) \; \mapsto \; (U^3 : UV^2 - U^3 : V^3 - U^2V)
\]
— the coordinates being given in order $(W{:}X{:}Y)$.  Clearly, $U^3$,
$UV^2 - U^3$ and $V^3 - U^2V$ are homogeneous of the same degree; and
they never vanish simultaneously, because if $U^3$ and $V^3 - U^2 V$
vanish then $U=0$ and $V=0$ which is meaningless for homogeneous
coordinates on $\mathbb{P}^1$; furthermore, these satisfy $W Y^2 - X^3
- W X^2$, so we have our morphism $\bar\psi \colon \mathbb{P}^1 \to
\bar C$.

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 $OP$;
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}

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\end{document}