From 701fd3ccdfa29f5f0a4084781a426a8dad6f2fdd Mon Sep 17 00:00:00 2001
From: "David A. Madore" <david+git@madore.org>
Date: Sun, 12 Apr 2026 11:02:23 +0200
Subject: Another exercise.

---
 controle-20260415.tex | 95 +++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 95 insertions(+)

(limited to 'controle-20260415.tex')

diff --git a/controle-20260415.tex b/controle-20260415.tex
index 8df5c9a..31847be 100644
--- a/controle-20260415.tex
+++ b/controle-20260415.tex
@@ -702,6 +702,101 @@ 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}
 
+%
+%
+%
+
+\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}
+
 %
 %
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