Bezout Theorem

Bezout in the plane

Suppose $\mathbb{F}$ is a field and $P,Q$ are polynomials in $\text{Poly}({\mathbb{F}}^2)$ with no common factor. Let $Z(P,Q):=\{(x,y)\in {\mathbb{F}}^2|P(x,y)=Q(x,y)=0\}$. Then the number of points in $Z(P,Q)$ is at most $(\text{Deg}P)(\text{Deg}Q)$.

Proof ${\text{Poly}}_D({\mathbb{F}}^n)$ is a vector space of polynomials of degree at most $D$. Suppose that $I\subset \text{Poly}({\mathbb{F}}^n)$ is an ideal, $I_D:=I\cap {\text{Poly}}_D({\mathbb{F}}^n)$, $R:=\text{Poly}({\mathbb{F}}^n)/I$ We define the vector space $R_D\subset R$ to be ${\text{Poly}}_D({\mathbb{F}}^n)/I_D$. That is, an element of $R$ lies in $R_D$ iff it can be represented by a polynomial $P\in {\text{Poly}}_D({\mathbb{F}}^n)$ and we get a chain $R_0\subset R_1\subset \cdots \subset R$.

Lemma

1. If $X\subset {\mathbb{F}}^n$ is any finite set, and $f:X\to \mathbb{F}$ is any function, then there is a polynomial $P$ of degree $\le |X|-1$ which agrees with $f$ on $X$.
2. Let $I\subset \text{Poly}({\mathbb{F}}^n)$ be an ideal, and let $R:=\text{Poly}({\mathbb{F}}^n)/I$. We write $\text{Dim}R$ for the dimension of $R$ as a vector space over $\mathbb{F}$, which maybe finite or infinite, then $|Z(I)|\le \text{Dim}R$.

Proof For any set $X\subset {\mathbb{F}}^n$, let $E_X:\text{Poly}({\mathbb{F}}^n)\to \text{Fcn}(X,\mathbb{F})$ be the map that restricts each polynomial $P\in \text{Poly}({\mathbb{F}}^n)$ to a function on $X$. If $X\subset Z(I)$, then $I\subset \text{Ker }{E}_X$, and so $E_X$ descends to a map $E_X:R\to \text{Fcn}(X,\mathbb{F})$. If $X\subset Z(I)$ is any finite set, then the first lemma says that $E_X$ is surjective. Therefore, $\text{Dim }R\ge |X|$ for any finite subset $X\subset Z(I)$.

Proof for theorem: Suppose that $P,Q\in \text{Poly}({\mathbb{F}}^2)$ have no common factor. Apply lemma to $I=(P,Q)$, $|Z(P,Q)|\le \text{Dim}R=\text{Dim}(\text{Poly}({\mathbb{F}}^2)/(P,Q)).$ So to prove the theorem is equivalent to prove for arbitrarily large degree $D$, $\text{Dim}({\text{Poly}}_D({\mathbb{F}}^2)/(P,Q{)}_D)\le (\text{Deg}P)(\text{Deg}Q),$ consider the following sequence of quotient maps: ${\text{Poly}}_D({\mathbb{F}}^2)\stackrel{\alpha }{\to }{\text{Poly}}_D({\mathbb{F}}^2)/(P{)}_D\stackrel{\beta }{\to }{\text{Poly}}_D({\mathbb{F}}^2)/(P,Q{)}_D.$

Lemma: If $P\in \text{Poly}({\mathbb{F}}^2)$ is a non-zero polynomial, then for all $D\ge \text{Deg}P$, $\text{Dim}(P{)}_D={\text{Dim Poly}}_{D-\text{Deg }P}({\mathbb{F}}^2)=(\genfrac{}{}{0ex}{}{D-\text{Deg }P+2}{2}).$ Define ${\mu }_P$ by ${\mu }_P(R)=PR$. The map ${\mu }_P$ is a linear map from ${\text{Poly}}_{D-\text{Deg }P}$ to $(P{)}_D$. We claim this linear map is an isomorphism. The kernel of the map is zero. Any element in $(P{)}_D$ can be written as $PR$ for some $R\in \text{Poly}({\mathbb{F}}^2)$. We have $D\ge \text{Deg}(PR)=\text{Deg }P+\text{Deg }R$, and so $\text{Deg }R\le D-\text{Deg }P$ and $R\in {\text{Poly}}_{D-\text{Deg }P}({\mathbb{F}}^2)$. This shows that ${\mu }_P$ is surjective. Therefore, ${\mu }_P$ is an isomorphism and the dimension of its domain equals the dimension of its range.

The map $\alpha$ is clearly surjective, and so we get $\text{Dim }({\text{Poly}}_D({\mathbb{F}}^2)/(P{)}_D)=\text{Dim Image }\alpha ={\text{Dim Poly}}_D({\mathbb{F}}^2)-\text{Dim Ker }\alpha .$ Plugging in our results for these dimensions, we get \begin{align*} \text{Dim } (\text{Poly}_D(\mathbb{F}^2)/(P)_D) &= \text{Dim Poly}_D(\mathbb{F}^2) - \text{Dim}(P)_D \\ &= \binom{D+2}{2} - \binom{D - \text{Deg } P + 2}{2}. \end{align*} For all $D\ge \text{Deg }P$, we can expand $\left(\genfrac{}{}{0ex}{}{D-\text{Deg }P+2}{2}\right)$ and $\left(\genfrac{}{}{0ex}{}{D}{2}\right)$ to get $\text{Dim }({\text{Poly}}_D({\mathbb{F}}^2)/(P{)}_D)=(\text{Deg }P)D+(1/2)(3\text{Deg }P-(\text{Deg }P{)}^2)$ The exact form of the constant term is not important for our argument, so we abbreviate it as $c(P)$: \begin{equation} \text{Dim } (\text{Poly}_D(\mathbb{F}^2)/(P)_D) = (\text{Deg } P)D + c(P). \end{equation} Now we study the map $\beta$, estimate the dimension of $\text{Ker }\beta$. Lemma: $\text{Dim Ker }\beta \ge \text{Dim }({\text{Poly}}_{D-\text{Deg }Q}({\mathbb{F}}^2)/(P{)}_{D-\text{Deg }Q}).$ Let ${\mu }_Q$ be the linear map ${\mu }_Q(R):=QR$. The map ${\mu }_Q$ is well-defined on various spaces. In particular, \begin{align*} \mu_Q &: \text{Poly}_{D-\text{Deg } Q}(\mathbb{F}^2) \to \text{Poly}_D(\mathbb{F}^2). \\ \mu_Q &: (P)_{D-\text{Deg } Q} \to (P)_D. \end{align*} Therefore, ${\mu }_Q$ descends to the quotient: ${\mu }_Q:{\text{Poly}}_{D-\text{Deg }Q}({\mathbb{F}}^2)/(P{)}_{D-\text{Deg }Q}\to {\text{Poly}}_D({\mathbb{F}}^2)/(P{)}_D.$ From now on, we fix this domain and range for ${\mu }_Q$. The image of ${\mu }_Q$ lies in the kernel of $\beta$. Using the fact that $P$ and $Q$ have no common factor, we will prove that ${\mu }_Q$ is injective. This will imply that $\text{Dim Ker }\beta \ge \text{Dim Image }{\mu }_Q=\text{Dim }({\text{Poly}}_{D-\text{Deg }Q}({\mathbb{F}}^2)/(P{)}_{D-\text{Deg }Q}).$ It just remains to prove ${\mu }_Q$ is injective. Suppose $r\in \text{Ker }{\mu }_Q\subset {\text{Poly}}_{D-\text{Deg }Q}({\mathbb{F}}^2)/(P{)}_{D-\text{Deg }Q}$. Let $R\in {\text{Poly}}_{D-\text{Deg }Q}({\mathbb{F}}^2)$ be a polynomial representing $r$. We know that $QR$ vanishes in ${\text{Poly}}_D({\mathbb{F}}^2)/(P{)}_D$, and so $QR$ lies in $(P)$. Since $P$ and $Q$ have no common factor, Proposition \fbox{6.2} implies that $R\in (P)$. But then $r=0$ in ${\text{Poly}}_{D-\text{Deg }Q}({\mathbb{F}}^2)/(P{)}_{D-\text{Deg }Q}$. This shows that ${\mu }_Q$ is injective, finishing the proof of lemma. The map $\beta$ is surjective, and so we see that $\text{Dim }({\text{Poly}}_D({\mathbb{F}}^2)/(P,Q{)}_D)=\text{Dim Image }\beta =\text{Dim }({\text{Poly}}_D({\mathbb{F}}^2)/(P{)}_D)-\text{Dim Ker }\beta .$ Plugging in above Lemma , we get \begin{align*} \text{Dim } (\text{Poly}_D(\mathbb{F}^2)/(P, Q)_D) &\leq \text{Dim } (\text{Poly}_D(\mathbb{F}^2)/(P)_D) \\ &- \text{Dim } (\text{Poly}_{D-\text{Deg } Q}(\mathbb{F}^2)/(P)_{D-\text{Deg } Q}). \end{align*} Now if $D$ is sufficiently large, we can plug in \begin{equation} \text{Dim } (\text{Poly}_D(\mathbb{F}^2)/(P)_D) = (\text{Deg } P)D + c(P) \end{equation} to the right-hand side, getting \begin{align*} \text{Dim } (\text{Poly}_D(\mathbb{F}^2)/(P, Q)_D) &\leq [(\text{Deg } P)D + c(P)] - [(\text{Deg } P)(D - \text{Deg } Q) + c(P)] \\ &= (\text{Deg } P)(\text{Deg } Q). \end{align*}