More exercises.

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@@ -509,4 +509,200 @@ works for any two distincts points $A,B$.
 %
 %
 %
+
+\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}
+
+%
+%
+%
+
+\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}
+
+%
+%
+%
 \end{document}