# 数学代写|现代代数代写Modern Algebra代考|Application: intersecting plane curves

#### Doug I. Jones

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## 数学代写|现代代数代写Modern Algebra代考|Application: intersecting plane curves

This and the following section discuss two applications of resultants which are not used later.

The historical purpose of the resultant was to solve geometric problems by elimination of variables. As an example, we want to determine the common roots of two polynomials in two variables, or, equivalently, the intersection of two plane curves. Suppose we are given $f, g \in F[x, y]$, where $F$ is a field, and want to intersect the two plane curves
$$X=\left{(a, b) \in F^2: f(a, b)=0\right}, \quad Y=\left{(a, b) \in F^2: g(a, b)=0\right} .$$
We eliminate the variable $y$ by considering the resultant $r=\operatorname{res}_y(f, g) \in F[x]$ with respect to $y$. We assume that $F$ is algebraically closed, so that every nonconstant univariate polynomial over $F$ has a root. This is often required to make general statements about geometric objects such as our curves $X$ and $Y$ true. The reader may imagine that $F=\mathbb{C}$. Now we let $Z$ be the projection of $X \cap Y$ onto the $x$-axis. If $a \in F$ and $\operatorname{lc}_y(f), \operatorname{lc}_y(g)$ do not both vanish at $x=a$, then Lemma 6.25 implies that
\begin{aligned} a \in Z & \Longleftrightarrow \exists b \in F(a, b) \in X \cap Y \Longleftrightarrow \exists b \in F f(a, b)=g(a, b)=0 \ & \Longleftrightarrow \operatorname{gcd}(f(a, y), g(a, y)) \neq 1 \ & \Longleftrightarrow r(a)=\operatorname{res}_y(f, g)(a)=0 . \end{aligned}

## 数学代写|现代代数代写Modern Algebra代考|Nonzero preservation and the gcd of several polynomials

In this section, we discuss the following problem: Given nonzero polynomials $f_1, \ldots, f_n \in F[x]$ over a field $F$, compute $h=\operatorname{gcd}\left(f_1, \ldots, f_n\right)$. Let $d \in \mathbb{N}$ be such that $\operatorname{deg} f_i \leq d$ for all $i$. We are particularly interested in the case where $d$ is close to $n$. A simple approach is to set $h_1=f_1$ and compute $h_i=\operatorname{gcd}\left(h_{i-1}, f_i\right)$ for $i=2, \ldots, n$. If $\operatorname{deg} h$ is fairly large, say $d / 10$, then this will take $n-1 \mathrm{gcd}$ calculations of polynomials of degree at least $d / 10$.

We now present a more efficient algorithm that uses only one ged calculation. The basic tool for this probabilistic algorithm is the following useful lemma. It says that a nonzero polynomial is likely to take a nonzero value at a random point. In other words, random evaluations probably preserve nonzeroness.

LEMMA 6.44. Let $R$ be an integral domain, $n \in \mathbb{N}, S \subseteq R$ finite with $s=# S$ elements, and $r \in R\left[x_1, \ldots, x_n\right]$ a polynomial of total degree at most $d \in \mathbb{N}$.
(i) If $r$ is not the zero polynomial, then $r$ has at most $d s^{n-1}$ zeroes in $S^n$.
(ii) If $s>d$ and $r$ vanishes on $S^n$, then $r=0$.
PROOF. (i) We prove the claim by induction on $n$. The case $n=1$ is clear, since a nonzero univariate polynomial of degree at most $d$ over an integral domain has at most $d$ zeroes (Lemma 25.4). For the induction step, we write $r$ as a polynomial in $x_n$ with coefficients in $x_1, \ldots, x_{n-1}: r=\sum_{0 \leq i \leq k} r_i x_n^i$ with $r_i \in R\left[x_1, \ldots, x_{n-1}\right]$ for $0 \leq i \leq k$ and $r_k \neq 0$. Then $\operatorname{deg} r_k \leq d-k$, and by the induction hypothesis, $r_k$ has at most $(d-k) s^{n-2}$ zeroes in $S^{n-1}$, so that there are at most $(d-k) s^{n-1}$ common zeroes of $r$ and $r_k$ in $S^n$. Furthermore, for each $a \in S^{n-1}$ with $r_k(a) \neq 0$, the univariate polynomial $r_a=\sum_{0 \leq i \leq k} r_i(a) x_n^i \in R\left[x_n\right]$ of degree $k$ has at most $k$ zeroes, so that the total number of zeroes of $r$ in $S^n$ is bounded by
$$(d-k) s^{n-1}+k s^{n-1}=d s^{n-1}$$

PRoOF. (i) We prove the claim by induction on $n$. The case $n=1$ is clear, since a nonzero univariate polynomial of degree at most $d$ over an integral domain has at most $d$ zeroes (Lemma 25.4). For the induction step, we write $r$ as a polynomial in $x_n$ with coefficients in $x_1, \ldots, x_{n-1}: r=\sum_{0 \leq i \leq k} r_i x_n^i$ with $r_i \in R\left[x_1, \ldots, x_{n-1}\right]$ for $0 \leq i \leq k$ and $r_k \neq 0$. Then $\operatorname{deg} r_k \leq d-k$, and by the induction hypothesis, $r_k$ has at most $(d-k) s^{n-2}$ zeroes in $S^{n-1}$, so that there are at most $(d-k) s^{n-1}$ common zeroes of $r$ and $r_k$ in $S^n$. Furthermore, for each $a \in S^{n-1}$ with $r_k(a) \neq 0$, the univariate polynomial $r_a=\sum_{0 \leq i \leq k} r_i(a) x_n^i \in R\left[x_n\right]$ of degree $k$ has at most $k$ zeroes, so that the total number of zeroes of $r$ in $S^n$ is bounded by
$$(d-k) s^{n-1}+k s^{n-1}=d s^{n-1}$$
(ii) follows immediately from (i).

# 现代代数代考

## 数学代写|现代代数代写Modern Algebra代考|Application: intersecting plane curves

$$X=\left{(a, b) \in F^2: f(a, b)=0\right}, \quad Y=\left{(a, b) \in F^2: g(a, b)=0\right} .$$

\begin{aligned} a \in Z & \Longleftrightarrow \exists b \in F(a, b) \in X \cap Y \Longleftrightarrow \exists b \in F f(a, b)=g(a, b)=0 \ & \Longleftrightarrow \operatorname{gcd}(f(a, y), g(a, y)) \neq 1 \ & \Longleftrightarrow r(a)=\operatorname{res}_y(f, g)(a)=0 . \end{aligned}

## 数学代写|现代代数代写Modern Algebra代考|Nonzero preservation and the gcd of several polynomials

(i)如果$r$不是零多项式，则$r$在$S^n$中最多有$d s^{n-1}$个零。
(ii)如果$s>d$和$r$在$S^n$上消失，那么$r=0$。

$$(d-k) s^{n-1}+k s^{n-1}=d s^{n-1}$$

$$(d-k) s^{n-1}+k s^{n-1}=d s^{n-1}$$
(ii)紧接(i)。

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