Write answer key for second exercise.

1 files changed, 104 insertions, 0 deletions

diff --git a/controle-20240410.tex b/controle-20240410.tex
index edd2cd6..8935506 100644
--- a/controle-20240410.tex
+++ b/controle-20240410.tex
@@ -396,6 +396,19 @@ on the set of points where $\frac{\partial f}{\partial x}$,
 $\frac{\partial f}{\partial y}$ and $\frac{\partial f}{\partial z}$
 all vanish?
 
+\begin{answer}
+The relation $*$ tells us that $28 x^6$, and consequently $x^6$ itself
+(since $k$ is of characteristic $\not\in\{2,7\}$), belongs to the
+ideal generated by $\frac{\partial f}{\partial x}$, $\frac{\partial
+  f}{\partial y}$ and $\frac{\partial f}{\partial z}$.  By cyclic
+permutation of coordinates, this is also the case for $y^6$ and $z^6$:
+so this ideal is irrelevant: the set of points in $\mathbb{P}^2$ where
+$\frac{\partial f}{\partial x}$, $\frac{\partial f}{\partial y}$ and
+$\frac{\partial f}{\partial z}$ all vanish is empty (because
+$x^6,y^6,z^6$ do not vanish simultaneously).  This implies that $C$ is
+\emph{smooth}.
+\end{answer}
+
 \smallskip
 
 \emph{The previous question implies that $C$ is a (plane) curve.  The
@@ -443,12 +456,45 @@ and $c := (0{:}0{:}1)$ (which obviously lie on $C$).
 \textbf{(2)} List all points of $C$ where $x$ vanishes.  Do the same
 for $y$ and $z$.
 
+\begin{answer}
+If $x$ vanishes on $C$ then $y^3 z = 0$, so $y=0$ or $z=0$.  So either
+$x=y=0$ and we are at $c$, or $x=z=0$ and we are at $b$; so the set of
+points where $x$ vanishes on $C$ is exactly $\{b,c\}$.  By cyclic
+rotation of coordinates, the set of points of $C$ where $y$ vanishes
+is $\{c,a\}$ and the set of points of $C$ where $z$ vanishes is
+$\{a,b\}$.
+\end{answer}
+
 \textbf{(3)} Where do the points $a,b,c$ lie on the printed picture?
 (If they do not lie on the picture, show the direction in which they
 would be.)  What is the equation of the affine part of $C$ drawn on
 the picture?  What is the tangent line at the point $c$?  What about
 $a$ and $b$?
 
+\begin{answer}
+The point $c$ is at affine coordinates $(u,v) = (0,0)$ where $u =
+\frac{x}{z}$ and $v = \frac{y}{z}$, that is, it is at the origin of
+the printed picture.  The point $a$ is at infinity ($z=0$) on the axis
+$y=0$ (or $v=0$ if we prefer), so it is at infinity in the horizontal
+direction, whereas $b$ is at infinity on the axis $x=0$ (or $u=0$ if
+we prefer), so at infinity in the vertical direction.
+
+The equation of the affine part of $C$ is obtained by dehomogenizing
+$x^3 y + y^3 z + z^3 x = 0$ with respect to $z$, i.e., by dividing by
+$z^3$ and replacing $\frac{x}{z}$ by $u$ and $\frac{y}{z}$ by $v$,
+giving $u^3 v + v^3 + u = 0$.
+
+The tangent line at the origin $c$ of the affine part $\{z\neq 0\}$ is
+given by $\frac{\partial g}{\partial u}|_{(0,0)}\cdot u +
+\frac{\partial g}{\partial v}|_{(0,0)}\cdot v =0$ where $g := u^3 v +
+v^3 + u$.  This simply gives $u=0$, so it is the vertical axis (as
+could be guessed from the figure); as a projective line, this is
+$x=0$.  By cyclic permutation of coordinates, we get $y=0$ as tangent
+line at $a$ and $z=0$ as tangent line at $c$.  (Of course, one might
+also compute these by taking affine charts around each one of the
+points, but this would be more tedious.)
+\end{answer}
+
 \textbf{(4)} Considering $v := \frac{y}{z}$ as a rational function
 on $C$, explain why it vanishes at order exactly $1$ at $c$, that
 is\footnote{We write $\ord_p(h)$ for the order at a point $p \in C$ of
@@ -459,6 +505,22 @@ $\ord_c(v) = 1$.  Explain why $\ord_c(u) = \ord_c(u^3 v + v^3)$ where
 $u := \frac{x}{z}$ and deduce that $\ord_c(u) = 3$.  Deduce the order
 at $c$ of $\frac{y}{x}$ (which is also $\frac{v}{u}$).
 
+\begin{answer}
+The coordinate $v$ vanishes with order exactly $1$ at the origin $c$
+of the tangent line $u=0$ to $C$ at $c$; therefore it also has order
+exactly $1$ at $c$ on $C$.  In other words, $\ord_c(v) = 1$.
+
+Now $u^3 v + v^3 + u = 0$ on $C$, that is $u = -u^3 v - v^3$, so
+$\ord_c(u) = \ord_c(u^3 v + v^3)$.  This shows that $\ord_c(u) =: k$,
+which is $\geq 1$ because $u$ vanishes at $c$, satisfies $k \geq
+\min(3k+1,3)$, so $k \geq 3$; but now $3k+1 \geq 10$, so $\ord_c(u^3
+v) = 3k+1 \neq 3 = \ord_c(v^3)$, so in fact $k = \min(3k+1,3) = 3$, as
+required.
+
+Consequently, $\frac{y}{x} = \frac{v}{u}$ has order $\ord_c(v) -
+\ord_c(u) = 1 - 3 = -2$ at $c$.
+\end{answer}
+
 \textbf{(5)} By using symmetry, compute the order at each one of the
 three points $a,b,c$ of each one of the three functions $\frac{x}{z}$,
 $\frac{y}{x}$ and $\frac{z}{y}$.  Explain why there are no points
@@ -467,6 +529,48 @@ vanishes or has a pole.  Summarize this by writing the principal
 divisors $\divis(\frac{x}{z})$, $\divis(\frac{y}{x})$ and
 $\divis(\frac{z}{y})$ associated with these three functions.
 
+\begin{answer}
+We have seen that
+\[
+\arraycolsep=1em
+\begin{array}{ccc}
+\ord_c(\frac{x}{z}) = 3 &
+\ord_c(\frac{y}{x}) = -2 &
+\ord_c(\frac{z}{y}) = -1
+\end{array}
+\]
+so by cyclic permutation we get
+\[
+\arraycolsep=1em
+\begin{array}{ccc}
+\ord_a(\frac{x}{z}) = -1 &
+\ord_a(\frac{y}{x}) = 3 &
+\ord_a(\frac{z}{y}) = -2
+\\
+\ord_b(\frac{x}{z}) = -2 &
+\ord_b(\frac{y}{x}) = -1 &
+\ord_b(\frac{z}{y}) = 3
+\end{array}
+\]
+Now we have also pointed out earlier that none of $x,y,z$ vanishes on
+$C$ outside possibly of $\{a,b,c\}$: so
+$\frac{x}{z},\frac{y}{x},\frac{z}{y}$ have neither zero nor pole on
+$C\setminus\{a,b,c\}$, i.e., their order is $0$ everywhere on this
+open set.  This shows that
+\[
+\begin{aligned}
+\divis(\frac{x}{z}) &= -[a] -2\,[b] + 3\,[c]\\
+\divis(\frac{y}{x}) &= \hphantom{+}3\,[a] - [b] - 2\,[c]\\
+\divis(\frac{z}{y}) &= -2\,[a] + 3\,[b] - [c]
+\end{aligned}
+\]
+Two sanity checks can be performed: the degree of each of these
+divisors (i.e., the sum of the coefficients) is zero, as befits a
+principal divisor; and the sum of these three divisors is also zero,
+as it should be because it is the divisor of the constant nonzero
+function $1$.
+\end{answer}
+
 
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