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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} + % |