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hbghlyj 发表于 2024-3-16 22:54
对于几乎所有的$t_0 \in \mathbb{C}$，都有$\operatorname{deg}_y\left(g^{\star}(-i y+t_0, y)\right)=1$。
应用引理 1，并考虑到 $\mathcal{C}^{\star}$ 有无穷多实点故不可能是一对共轭虚直线，得出结论 $\mathcal{C}^{\star }$ 是一条直线或一个圆。
引理 1：设$f \in \mathbb{R}[x, y]$，$a\inC\setminus\Bbb R$，满足对于所有（除有限个外）复数$ t $ 有$$\deg_y(f(ay + t,y))= 1 $$则$ f(x,y)$定义直线或二次曲线。在后一种情况（即$\deg f=2$），$f$必定义椭圆或一对共轭虚直线。
若$f$无奇点，则$f$必定义椭圆；进一步，$f$是圆当且仅当$a=\pm i$.
例如$f=x^2+y^2-1$定义一个圆，$f(iy+t,y)=2ity+t^2-1$是$y$的一次式，即$\deg_y(f(ay+t,y))=1$。
证明$d\le2$将 $f$ 的各次项分别写出：$f=f_d+\cdots+f_0$，其中$f_j$为$j$次齐次多项式
将 $f_j$ 表示为$x-ay$ 的多项式$f_j(x)=\sum_{i=0}^{i=j} \frac{1}{i !} \frac{\partial f_j}{\partial x^i}(a y, y)(x-a y)^i$，
注意到$\frac{\partial f_j}{\partial x^i}(a y, y)$是关于$y$的$j-i$次齐次多项式，所以
\begin{aligned}
f(x, y) & =\sum_{j=0}^{j=d} f_j(x, y)\\&=\sum_{j=0}^{j=d} \sum_{i=0}^{i=j} \frac{1}{i !} \frac{\partial f_j}{\partial x^i}(a y, y)(x-a y)^i \\
& =\sum_{j=0}^{j=d} \sum_{i=0}^{i=j} \frac{1}{i !} \frac{\partial f_j}{\partial x^i}(a, 1)(x-a y)^i y^{j-i}
\end{aligned}把 $x$ 换为 $a y+t$ 得
\[
f(a y+t, y)=\sum_{j=0}^{j=d} \sum_{i=0}^{i=j} \frac{1}{i !} \frac{\partial f_j}{\partial x^i}(a, 1) t^i y^{j-i}
\]
改变求和指标$k=j-i$，
\[
f(a y+t, y)=\sum_{k=0}^{k=d} \sum_{i=0}^{i=d-k} \frac{1}{i !} \frac{\partial f_{k+i}}{\partial x^i}(a, 1) t^i y^k .
\]
Since for almost all $t \in \mathbb{C}$ the degree of $f(a y+t, y)$ w.r.t $y$ is one, it follows that for $k=2, \ldots, d$ the polynomials $\sum_{i=0}^{i=d-k} \frac{1}{i!} \frac{\partial f_{k+i}}{\partial x^i}(a, 1) t^i\in \mathbb{C}[t]$ vanish for infinitely many values of $t$. Hence, counting degrees, $\frac{\partial f_{k+i}}{\partial x^i}(a, 1)=0$ for $k=2, \ldots, d$ and $i=0, \ldots, d-k$. In particular, when $k+i=d$, one deduces that $\frac{\partial f_d}{\partial x^\ell}(a, 1)=0$ for $\ell=0, \ldots, d-2$. Therefore $(x-a)^{d-1}$ divides $f_d(x, 1)$. Now, since $a$ is a non-real complex number and $f_d(x, 1) \in \mathbb{R}[x]$, it follows that $(x-\bar{a})^{d-1}$ also divides $f_d(x, 1)$. Hence, $2(d-1) \leq d$, that is $d \leq 2$.
证明$d=2$时，$f$必定义椭圆或一对共轭虚直线We now proceed to analyze the conic defined by $f$, so we assume $d=2$. Then, taking into account that $(x-a y)(x-\bar{a} y)$ divides $f_d(x, y)$, one deduces that $f$ can be written (up to a multiplicative constant) as
\[
x^2-2 a_0 x y+\left(a_0^2+a_1^2\right) y^2+b_0 x+b_1 y+b_2
\]
for some $b_0, b_1, b_2 \in \mathbf{R}$ ($a_0$ and $a_1$ denote the real and the imaginary part of $a$, respectively), and that $(a: 1: 0),(\bar{a}: 1: 0)$ are the points at infinity of the conic. Hence, since $a \notin \mathbf{R}, f$ is neither a parabola nor a hyperbola. Therefore, $f$ has to be either a pair of conjugate complex lines (clearly, the lines are then $(x-a y+k),(x-\bar{a} y+\bar{k})$ for some $k \in \mathbb{C})$ or an ellipse; depending on the reducibility of $f$.
证明当$f$定义一个椭圆时，$f$定义一个圆$\Leftrightarrow a=\pm i$In order to distinguish between general ellipses and circles, we assume that $f$ is irreducible and we analyze the lengths of the axis of the conic, i.e. the eigenvalues of the matrix
\[
B=\left(\begin{array}{cc}
1 & -a_0 \\
-a_0 & a_0^2+a_1^2
\end{array}\right)
\]
Thus, since $(\operatorname{trace}(B))^2=4 \operatorname{det}(B)$ is the condition for coincident eigenvalues, it follows that $f$ is a circle if and only if $a_0^2+\left(a_1 \pm 1\right)^2=0$, i.e., if and only if $a= \pm i$.
很有用的引理 |
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