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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} + % % |