diff options
| -rw-r--r-- | controle-20260415.tex | 95 |
1 files changed, 95 insertions, 0 deletions
diff --git a/controle-20260415.tex b/controle-20260415.tex index 8df5c9a..31847be 100644 --- a/controle-20260415.tex +++ b/controle-20260415.tex @@ -705,4 +705,99 @@ $P = (0{:}0{:}1)$, it is $\bar\psi((0{:}1))$. % % % + +\exercise + +In this exercise, we let $k$ be the field $\mathbb{R}$ or +$\mathbb{C}$, and we consider the curve $\bar C$ which is the +projective completion (in other words, the Zariski closure inside +$\mathbb{P}^2_k$) of $C := \{x^2 + y^2 + 1 = 0\} \subseteq +\mathbb{A}^2_k$. + +\textbf{(1)} Recall why $\bar C$ is a (smooth) curve and why the field +$K := k(\bar C)$ of rational functions on $\bar C$ is +$k(x)[y]/(x^2+y^2+1)$ (which can also be written, suggestively, as +$k(x)(\sqrt{-1-x^2})$). A short answer will suffice. + +\begin{answer} +The projective completion $\bar C$ of $C$ has homogeneous equation +$X^2 + Y^2 + Z^2 = 0$ given by homogeneizing that of $C$. It is +smooth because the partial differentials, $2X$, $2Y$, $2Z$ of this +equation never vanish simultaneously in $\mathbb{P}^2$ (they generate +the irrelevant ideal $(X,Y,Z)$). We have seen in the course that, +under such circumstances, $k(\bar C) = k(x)[y]/(x^2+y^2+1)$. +\end{answer} + +\textbf{(2)} Briefly recall why we can express elements of $K$ in the +form $f_0 + f_1 y$ with $f_0,f_1 \in k(x)$, and how sums and products +are computed. As an example of how to compute divisions, compute +$\frac{1}{y}$ and $\frac{1}{y-1}$ in this form. + +\begin{answer} +By performing Euclidean division by $x^2+y^2+1$ in $k(x)[y]$ and +keeping only the remainder, we express elements of $K = +k(x)[y]/(x^2+y^2+1)$ in as polynomials of degree $<2$ in $y$, which is +exactly of the form $f_0 + f_1 y$ with $f_0,f_1 \in k(x)$. Sums are +computed termwise ($(f_0 + f_1 y) + (g_0 + g_1 y) = (f_0+g_0) + +(f_1+g_1)y$), and products are computed by taking the Euclidean +division, which amounts to replacing the possible $y^2$ term by +$-1-x^2$: in other words, $(f_0 + f_1 y) \cdot (g_0 + g_1 y) = +(f_0 g_0 - (x^2+1) f_1 g_1) + (f_1 g_0 + f_0 g_1)y$. + +To compute $\frac{1}{y}$, we search for a Bézout relation $u\,y + +v\cdot (x^2+y^2+1) = 1$ in $k(x)[y]$, which is easy because $-y^2 + +(x^2+y^2+1) = x^2+1$ is an element of $k(x)$, so $-\frac{y}{x^2+1}\,y ++ \frac{1}{x^2+1}\,(x^2+y^2+1) = 1$, and this shows that +$-\frac{1}{x^2+1}\,y$ is the inverse of $y$ in $K$. + +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 +inverse of $y-1$ in $K$. +\end{answer} + +\textbf{(3)} Explain why $K$ is isomorphic to $k(t)$ (the field +$\operatorname{Frac}(k[t])$ of rational fractions in one +indeterminate $t$) when $k = \mathbb{C}$. Describe an explicit +isomorphism. (\textit{Hint:} What does $C$ become if we let $x' := +\sqrt{-1}\,x$ and $y' := \sqrt{-1}\,y$?) + +\begin{answer} +Letting $x' := \sqrt{-1}\,x$ and $y' := \sqrt{-1}\,y$ (which describes +an invertible linear transformation), the equation of $C$ becomes +$x^{\prime 2} + y^{\prime 2} = 1$. But we have seen in the course +that this is rational with an explicit isomorphism $k(t) \to +k(x')[y']/(x^{\prime 2} + y^{\prime 2} - 1)$ being given by $t \mapsto +\frac{y'}{x'+1}$ and conversely $x' \mapsto \frac{1-t^2}{1+t^2}$ and +$y' \mapsto \frac{2t}{1+t^2}$ (rational parametrization of the +circle). This shows that $K$ for $k=\mathbb{C}$ is isomorphic to +$k(t)$ by $t \mapsto \frac{y}{x-\sqrt{-1}}$ and conversely $x \mapsto +-\sqrt{-1}\,\frac{1-t^2}{1+t^2}$ and $y \mapsto +-\sqrt{-1}\,\frac{2t}{1+t^2}$. +\end{answer} + +\textbf{(4)} Show that $K$ is \emph{not} isomorphic to $k(t)$ when $k += \mathbb{R}$. (\textit{Hint:} The equation $u^2 + v^2 + 1 = 0$ has +no real solution, so none in $k(t)$, but it has an obvious one +in $K$.) + +\begin{answer} +In the field $k = \mathbb{R}$ the equation $u^2 + v^2 + 1 = 0$ has no +solutions because $u^2 + v^2 + 1 > 0$. Consequently, it also has no +solution in $k(t)$ because if $u^2 + v^2 + 1 = 0$ in $k(t)$, finding +some $\theta$ at which $u,v$ do not have a pole (which is possible as +a rational fraction has only finitely many poles), we can evaluate +$u,v$ at $\theta$, giving $u(\theta)^2 + v(\theta)^2 + 1 = 0$ in +$\mathbb{R}$, again a contradiction. + +On the other hand, $x^2 + y^2 + 1 = 0$ in $K$ by definition of $K$. +So the equation $u^2 + v^2 + 1 = 0$ has no solution in $k(t)$ and has +one in $K$, showing that $k(t)$ and $K$ cannot be isomorphic. +\end{answer} + +% +% +% \end{document} |