Exercise 10.5

Question

\def\imp{\Rightarrow}
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}{\mathrm{N}}

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Let $f(x_0,x_1,x_2)$ be a homogeneous polynomial of degree $n$ in $F(x_0,x_1,x_2]$. Suppose that not every zero of $a_0x_0+a_1x_1+a_2x_2$ is a zero of $f$. Prove that there are at most $n$ common zeros of $f$ and $a_0x_0+a_1x_1+a_2x_2$ in $P_2(F)$. In more geometric language this says that a curve of degree $n$ and a line have at most $n$ points in common unless the line is contained in the curve.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow} \def\gcro{\mbox{[\hspace{-.15em}[}} \def\dcro{\mbox{]\hspace{-.15em}]}} ewcommand{\D}{\mathrm{d}} ewcommand{\Q}{\mathbb{Q}} ewcommand{\Z}{\mathbb{Z}} ewcommand{\N}{\mathbb{N}} ewcommand{\R}{\mathbb{R}} ewcommand{\C}{\mathbb{C}} ewcommand{\F}{\mathbb{F}} ewcommand{ e}{\,\mathrm{Re}\,} ewcommand{\ord}{\mathrm{ord}} ewcommand{ }{\mathrm{N}} ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}} \begin{proof} Let $\mathscr{C}$ be the curve with equation $f(x_0,x_1,x_2) = 0$. Since $a_0 x_0+a_1x_1+a_2x_2 =0$ is the equation of a line $l$, $(a_0,a_1,a_2) e 0$, so that we can suppose without loss of generality that $a_0 e 0$. Then \begin{align*} [u_0,u_1,u_2] \in &l \iff a_0 u_0+a_1u_1+a_2u_2 =0\ & \iff u_0 = -\frac{a_1}{a_0} u_1 - \frac{a_2}{a_0} u_2\ &\iff u_0 = \alpha u_1 + \beta u_2, \end{align*} where $\alpha = -\frac{a_1}{a_0},\ \beta = - \frac{a_2}{a_0}.$ Therefore \begin{align*} [u_0,u_1,u_2] \in \mathscr{C} \cap l &\iff \left\{ \begin{array}{ll} a_0 u_0+a_1u_1+a_2u_2 &=0,\ f(u_0,u_1,u_2) &= 0, \end{array} ight.\ &\iff \left\{ \begin{array}{ll} u_0 = \alpha u_1+\beta u_2,\ f\left(\alpha u_1+ \beta u_2,u_1,u_2 ight) &= 0. \end{array} ight. \end{align*} $\bullet$ Suppose first that every point $[u_0,u_1,u_2]$ of $\mathscr {C} \cap l$ is such that $u_1 e 0$. Since $f$ is homogeneous of degree $n$, $$0 = u_1^n f\left(\alpha + \beta \frac{u_2}{u_1},1,\frac{u_2}{u_1} ight),$$ and using $u_1 e 0$, $$0 = f\left(\alpha + \beta \frac{u_2}{u_1},1,\frac{u_2}{u_1} ight).$$ Consider the formal polynomial $P(x) = f\left(\alpha + \beta x,1,x ight) \in F[x].$ Then $\deg(P) \leq n$. If $P e 0$, then $P$ has at most $n$ roots $\lambda_1,\ldots,\lambda_k$, where $k\leq n$. In this case, $u_2 = \lambda_i u_1$ and $u_0 =\alpha u_1 + \beta u_2 = u_1\left(\alpha +\beta \lambda_i ight)$, therefore $$[u_0,u_1,u_2] =\left [\alpha +\beta \lambda_i, 1 , \lambda_i ight],\ 1\leq i \leq k,$$ so that $\mathscr{C}$ and $l$ have at most $n$ points in common. Therefore, if $|\mathscr {C} \cap l| >n$, then $P(x) = f\left(\alpha + \beta x,1,x ight) = 0$. \bigskip $\bullet$ Now suppose that some point $[u_0,u_1,u_2] \in \mathscr {C} \cap l$ satisfies $u_1 = 0$. Then $[u_0,u_1,u_2] = [\beta u_2, 0, u_2] = [\beta, 0, 1]$. There is exactly one point on $\mathscr {C} \cap l$ with $u_1 = 0$. By the previous computation, the other points $[u_0,u_1,u_2] \in \mathscr {C} \cap l$, such that $u_1 e 0$, satisfy $$0 = f\left(\alpha + \beta \frac{u_2}{u_1},1,\frac{u_2}{u_1} ight) = P\left(\frac{u_2}{u_1} ight),$$ Where $P(x) = f(\alpha + \beta x, 1 ,x)$ In this case, we prove that $\deg(P) \leq n-1$. Since $f$ is an homogeneous polynomial, with $\deg(f) = n$, we can write $$f(x,y,z) = \sum_{(i,j) \in \gcro 0, n\dcro^2, i + j \leq n} a_{i,j}\, x^iy^j z^{n-i-j}.$$ In the present case, $[\beta, 0,1]$ is on the curve $\mathscr{C}$, thus $$\sum_{i=0}^n a_{i,0} \beta^i = 0.$$ Moreover, \begin{align*} P(x) &= f(\alpha + \beta x, 1, x)\ &= \sum_{(i,j) \in \gcro 0, n\dcro^2, i + j \leq n} a_{i,j}\, (\alpha + \beta x)^i x^{n-i-j}\ &= \sum_{(i,k) \in \gcro 0, n\dcro^2} a_{i, k-i} (\alpha + \beta x)^i x^{n-k}\qquad \qquad (k = i+j). \end{align*} If $j>0$, then $\deg((\alpha + \beta x)^i x^{n-i-j}) < n$. Therefore the coefficient of $x^n$ in $P(x)$ is $\sum\limits_{i=0}^n a_{i,0} \beta^i = 0$, thus $\deg(P) \leq n-1$. If $P e 0$, then $P$ has at most $n-1$ roots, and so $\mathscr{C}$ has at most $n-1$ points $[u_0,u_2,u_2]$ such that $u_1 e 0$ on $l$. With the unique point $[\beta, 0 ,1]$ such that $u_1= 0$, we obtain at most $n$ points in $ \mathscr {C} \cap l$. \bigskip In both cases, if $| \mathscr {C} \cap l| > n$, then $P(x) = f(\alpha + \beta x, 1 ,x) = 0$. We show that this implies that $l \subset \mathscr{C}$. Let $[v_0,v_1,v_2]$ be any point on $l$. If $v_1 e 0$, $$f(v_0,v_1,v_2) = f(\alpha v_1 + \beta v_2, v_1,v_2) = v_1^n f\left(\alpha + \beta \frac{v_2}{v_1}, 1 ,\frac{v_2}{v_1} ight) = v_1^n P\left (\frac{v_2}{v_1} ight) = 0.$$ If $v_1 = 0$, then $[v_0,v_1,v_2] = [\beta, 0,1]$. Consider the reciprocal polynomial $Q(x) = x^n P\left(\frac{1}{x} ight)$ of $P(x)$. Then \begin{align*} Q(x) &= x^n \sum_{(i,k) \in \gcro 0, n\dcro^2}a_{i, k-i} \left(\alpha + \beta \frac{1}{x} ight)^i \frac{1}{x^{n-k}}\ &=\sum_{(i,k) \in \gcro 0, n\dcro^2}a_{i, k-i} \left(\alpha x+ \beta ight)^i x^{k-i}. \end{align*} If $P = 0$ ,then $Q = 0$, and $$f(\beta,0,1) = \sum_{i=0}^n a_{i,0} \beta^i = Q(0) =0.$$ This proves that $l \subset \mathscr{C}$. To conclude, if $l ot \subset \mathscr {C}$, then $|l \cap \mathscr{C}| \leq n$ : a curve of degree $n$ and a line have at most $n$ points in common unless the line is contained in the curve. \end{proof} {\bf Second proof} (with same beginning.) Let $\mathscr{C}$ be the curve with equation $f(x_0,x_1,x_2) = 0$. Since $a_0 x_0+a_1x_1+a_2x_2 =0$ is the equation of a line $l$, $(a_0,a_1,a_2) e 0$, so that we can suppose without loss of generality that $a_0 e 0$. Then \begin{align*} [u_0,u_1,u_2] \in &l \iff a_0 u_0+a_1u_1+a_2u_2 =0\ & \iff u_0 = -\frac{a_1}{a_0} u_1 - \frac{a_2}{a_0} u_2\ &\iff u_0 = \alpha u_1 + \beta u_2, \end{align*} where $\alpha = -\frac{a_1}{a_0},\ \beta = - \frac{a_2}{a_0}.$ Therefore \begin{align*} [u_0,u_1,u_2] \in \mathscr{C} \cap l &\iff \left\{ \begin{array}{ll} a_0 u_0+a_1u_1+a_2u_2 &=0,\ f(u_0,u_1,u_2) &= 0, \end{array} ight.\ &\iff \left\{ \begin{array}{ll} u_0 = \alpha u_1+\beta u_2,\ f\left(\alpha u_1+ \beta u_2,u_1,u_2 ight) &= 0. \end{array} ight. \end{align*} Consider the polynomial $g(x,y) = f(\alpha x + \beta y,x, y)$. Then $g$ is homogeneous of degree $n$. A point $[u_0,u_1,u_2]$ of $l$ is on $\mathscr{C}$ if and only if $g(u_1,u_2) = 0$. Suppose that $|\mathscr{C} \cap l | = m$, and write $[u_0^{(1)},u_1^{(1)},u_2^{(1)}],\ldots, [u_0^{(m)},u_1^{(m)},u_2^{(m)}]$ the $m$ distinct points of $\mathscr{C} \cap l$. Then $1\leq i < j \leq m \Rightarrow (u_1^{(i)},u_2^{(i)}) ot \sim (u_1^{(j)},u_2^{(j)})$ : the $m$ pairs $[u_1^{(i)},u_2^{i}]$ are distinct, since $u_0^{(i)} = \alpha u_1^{(i)} + \beta u_2^{(i)}$. Then $g(u_1^{(1)}, u_2^{(1)}) = 0$, where $u_1, u_2$ cannot be $0$ simultaneously . We show first that $g(x,y) = (u_2^{(1)}x - u_1^{(1)}y) h_1(x,y)$, where $h_1\in F[x,y]$ is a homogeneous polynomial. If $ u_2^{(1)} e 0$, then $g\left(\frac{u_1^{(1)}}{ u_2^{(1)}},1 ight) = 0$. Therefore $\frac{u_1^{(1)}}{ u_2^{(1)}}$ is a root of the polynomial $g(t,1) \in F[t]$, thus $g(t,1) = \left(t - \frac{u_1^{(1)}}{ u_2^{(1)}} ight) h(t)$, where $h(t) \in F[t]$. Then \begin{align*} g(x,y) &= y^n g\left(\frac{x}{y},1 ight) =y^n \left(\frac{x}{y} - \frac{u_1^{(1)}}{ u_2^{(1)}} ight) h \left(\frac{x}{y} ight)\ &= (u_2^{(1)}x - u_1^{(1)}y) h_1(x,y), \end{align*} where $h_1(x,y) = \frac{1}{u_1^{(1)}} y^{n-1} h\left(\frac{x}{y} ight)$ is homogeneous of degree $n-1$. Same proof if $u_1^{(1)} e 0$. Since $u_2^{(1)}u_1^{(2)} - u_1^{(1)} u_2^{(2)} e 0$, then $h_1(u_1^{(2)}, u_2^{(2)}) = 0$, and we can continue with $h_1$ in place of $g$. By induction $$g(x,y) = (u_2^{(1)}x - u_1^{(1)}y)(u_2^{(2)}x - u_1^{(2)}y)\cdots (u_2^{(m)}x - u_1^{(m)}y) h_m(x,y),\qquad h_m(x,y) \in F[x,y].$$ This equality shows that either $g(x,y) = 0$, or $n = \deg(g) \geq m$. This proves that if $m = |\mathscr C \cap l |>n$, then $g(x,y) = 0$. In this case, every point $[u_0,u_1,u_2] \in l$ satisfies $f(\alpha u_1 + \beta u_2, u_1,u_2) = g(u_1,u_2)=0$, thus $l \subset \mathscr{C}$. To conclude, if $l ot \subset \mathscr {C}$, then $|l \cap \mathscr{C}| \leq n$ : a curve of degree $n$ and a line have at most $n$ points in common unless the line is contained in the curve.

Exercise 10.5

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Exercise 10.5

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