From 8319e0d50b4b52398d77bfd9173df25c87178ed8 Mon Sep 17 00:00:00 2001
From: "David A. Madore" <david+git@madore.org>
Date: Mon, 8 Apr 2024 18:33:04 +0200
Subject: Write answer key for third exercise.

---
 controle-20240410.tex | 100 ++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 100 insertions(+)

(limited to 'controle-20240410.tex')

diff --git a/controle-20240410.tex b/controle-20240410.tex
index 8935506..138d0fa 100644
--- a/controle-20240410.tex
+++ b/controle-20240410.tex
@@ -627,6 +627,27 @@ need to be checked, and do so.  Explain why $S$ is indeed a Zariski
 closed subset of $\mathbb{P}^n$: again, carefully state what needs to
 be checked before doing so.
 
+\begin{answer}
+For the point $(x_0 y_0 : \cdots : x_p y_q)$ to make sense, we need to
+check that not all its coordinates are zero.  But we know that at
+least one of the $x_i$ is nonzero and at least one of the $y_j$ is
+nonzero, so (as we are working over a field) the product $x_i y_j$ is
+nonzero.
+
+For the map $((x_0:\cdots:x_p),\, (y_0:\cdots:y_q)) \mapsto (x_0 y_0 :
+\cdots : x_p y_q)$ to make sense, we need to check that $(x_0 y_0 :
+\cdots : x_p y_q)$ does not change if we replace the $x_i$ and the
+$y_j$ by different coordinates for the same point, in other words, if
+we multiply all the $x_i$ by a common nonzero constant, and all the
+$y_j$ by a (possibly different) common nonzero constant.  This is
+indeed the case as $x_i y_j$ will be multiplied by the product of
+these two constants.
+
+Concerning $S$, we need to check that the equations $z_{i,j} z_{i',j'}
+= z_{i,j'} z_{i',j}$ are homogeneous: this is indeed the case (they
+are homogeneous of degree $2$).
+\end{answer}
+
 \textbf{(2)} Consider in this question the special case $p=q=1$ (so
 $n=3$).  Simplify the definition of $S$ in this case down to a single
 equation.  Taking $z_{0,0}=0$ as the plane at infinity in
@@ -635,10 +656,31 @@ $\mathbb{P}^3$, give the equation of the affine part $S \cap
 at infinity in $\mathbb{P}^1$, describe $\psi$ on $\mathbb{A}^1 \times
 \mathbb{A}^1$.
 
+\begin{answer}
+When $p=q=1$ the equations of $S$ are all trivial except $z_{0,0}
+z_{1,1} = z_{0,1} z_{1,0}$ (or equations trivially equivalent to
+this).  Taking $z_{0,0} = 0$ as plane at infinity, we get the equation
+of the affine part by dehomogenizing $z_{0,0} z_{1,1} = z_{0,1}
+z_{1,0}$, which gives $w_{1,1} = w_{0,1} w_{1,0}$ where $w_{i,j}$
+denotes the affine coordinate $z_{i,j}/z_{0,0}$ in $\mathbb{A}^3$.
+
+Concerning $\psi$, if we call $u = x_1/x_0$ the affine coordinate on
+the first $\mathbb{A}^1$ and $v = y_1/y_0$ that on the second, it is
+given by taking $(u,v)$, i.e. $((1:u),\, (1:v))$ to $(1:v:u:uv)$, that
+is $(v,u,uv)$.
+\end{answer}
+
 \textbf{(3)} Returning to the case of general $p$ and $q$, show that
 the image of $\psi$ is contained in $S$, that is, $\psi(\mathbb{P}^p
 \times \mathbb{P}^q) \subseteq S$.
 
+\begin{answer}
+If $(z_{0,0} : \cdots : z_{p,q})$ is given by $z_{i,j} = x_i y_j$, we
+just need to check that $z_{i,j} z_{i',j'} = z_{i,j'} z_{i',j}$: but
+this just says that $x_i y_j x_{i'} y_{j'} = x_i y_{j'} x_{i'} y_j$,
+which is obvious by commutativity.
+\end{answer}
+
 \textbf{(4)} Conversely, explain why for each point $(z_{0,0} : \cdots
 : z_{p,q})$ in $S$ there is a unique pair of points $((x_0 : \cdots :
 x_p), (y_0 : \cdots : y_q))$ in $\mathbb{P}^p \times \mathbb{P}^q$
@@ -646,6 +688,34 @@ which maps to the given point under $\psi$: in other words, show that
 $\psi$ is a bijection between $\mathbb{P}^p \times \mathbb{P}^q$
 and $S$.
 
+(\textit{Hint:} you may wish to observe that if $(z_{0,0} : \cdots :
+z_{p,q})$ is in $S$, the point $(z_{0,j_0} : \cdots : z_{p,j_0})$ in
+$\mathbb{P}^p$ does not depend on $j_0 \in \{0,\ldots,q\}$ such that
+$\exists i.(z_{i,j_0}\neq 0)$; and similarly for $(z_{i_0,0} : \cdots
+: z_{i_0,q})$ in $\mathbb{P}^q$.)
+
+\begin{answer}
+Assume $(z_{0,0} : \cdots : z_{p,q})$ is in $S$.  By the definition of
+$\mathbb{P}^n$, at least one coordinate $z_{i_0,j_0}$ is nonzero.
+Define $x^*_i = z_{i,j_0}$ (note that $x^*_{i_0} \neq 0$) and $y^*_j =
+z_{i_0,j}$ (note that $y^*_{j_0} \neq 0$): then $x^*_i y^*_j =
+z_{i,j_0} z_{i_0,j}$, which, by the equations of $S$, is also
+$z_{i_0,j_0} z_{i,j}$: this shows that $((x^*_0 : \cdots : x^*_p),
+(y^*_0 : \cdots : y^*_q))$ maps to the given $(z_{0,0} : \cdots :
+z_{p,q})$ under $\psi$ (by dividing all coordinates by the nonzero
+value $z_{i_0,j_0}$).  So $\psi$ surjects to $S$.
+
+But in fact, if $((x_0 : \cdots : x_p), (y_0 : \cdots : y_q))$ maps to
+$(z_{0,0} : \cdots : z_{p,q})$ under $\psi$, then we have $z_{i,j_0} =
+y_{j_0} x_i$ so that $(x_0 : \cdots : x_p) = (z_{0,j_0} : \cdots :
+z_{p,j_0})$ provided $y_{j_0} \neq 0$, which is tantamount to saying
+$z_{i_0,j_0}\neq 0$ for some $i_0$: so we had no other choice than to
+take the $(x^*_0 : \cdots : x^*_p)$ of the previous paragraph, and the
+same argument holds for $(y^*_0 : \cdots : y^*_q)$.  This shows
+uniqueness of the points $((x_0 : \cdots : x_p), (y_0 : \cdots :
+y_q))$ mapping to $(z_{0,0} : \cdots : z_{p,q})$ under $\psi$.
+\end{answer}
+
 \textbf{(5)} Call $\pi\colon S\to \mathbb{P}^p\times\mathbb{P}^q$ the
 inverse bijection of $\psi$, and call $\pi',\pi''$ its two components.
 (In other words, if $s = (z_{0,0} : \cdots : z_{p,q})$ is in $S$ then
@@ -657,6 +727,36 @@ are morphisms of algebraic varieties.  (If this seems too difficult,
 consider the special case $p=q=1$, and at least try to explain what
 needs to be checked.)
 
+\begin{answer}
+Given $j_0 \in \{0,\ldots,q\}$, consider the map $(z_{0,0} : \cdots :
+z_{p,q}) \mapsto (z_{0,j_0} : \cdots : z_{p,j_0})$ which selects only
+the coordinates $z_{i,j_0}$.  This is a partially defined map from
+$\mathbb{P}^n$ to $\mathbb{P}^p$, and the components are homogeneous
+polynomials of the same degree (here, $1$): the only thing that can go
+wrong is that all the $z_{i,j_0}$ are zero, so this is well-defined on
+the open set $\{z_{0,j_0}\neq 0\} \cup \cdots \cup \{z_{p,j_0}\neq
+0\}$.  Now restrict this map to $S$: this gives us a morphism
+$\pi^{\prime(j_0)}$ from the open set $U^{(j_0)} := S \cap
+(\{z_{0,j_0}\neq 0\} \cup \cdots \cup \{z_{p,j_0}\neq 0\})$ of $S$
+to $\mathbb{P}^p$.
+
+Note that the union the union of $U^{0)},\ldots,U^{(q)}$ is all of $S$
+because there is always at least one coordinate nonzero.
+
+Furthermore, we have seen in (4) that if $s = (z_{0,0} : \cdots :
+z_{p,q})$ then $\pi'(s)$ is given by $\pi^{\prime(j_0)}(s) =
+(z_{0,j_0} : \cdots : z_{p,j_0})$ where $j_0$ is any element of
+$\{0,\ldots,q\}$ such that $z_{i_0,j_0} \neq 0$ for some $i_0$, i.e.,
+$s \in U^{(j_0)}$.  This shows that $\pi'$ coincides with
+$\pi^{\prime(j_0)}$ on the open set $U^{(j_0)}$ where the latter is
+defined, so $\pi'$ is defined by “gluing” the various
+$\pi^{\prime(j_0)}$.  So $\pi'$ is indeed a morphism (to be clear: it
+is simply defined by selecting the coordinates of the form $z_{i,j_0}$
+for any one $j_0$ such that not all of them vanish).
+
+The same argument, \textit{mutatis mutandis}, works for $\pi''$.
+\end{answer}
+
 
 
 %
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