# Bezout in $\mathbb C [x,y]$ [duplicate]

Let $f_1, f_2 \dots, f_r \in \mathbb C [x,y]$ and suppose $$\gcd(f_1, \dots , f_r)=1$$

Show that $V(f_1, \dots , f_r) \subset \mathbb A^2$ is finite.

Partial Solution: suppose $(x_0,y_0) \in V(p(x,y),q(x,y))$, then for $(x,y_0)$ we can use Bezout and say that exist $a(x),b(x),h(x) \in \mathbb C [x]$ such that $$a(x)p(x,y_0) + b(x)q(x,y_0)=h(x)$$

So there are finite $x_i \in \mathbb C$ such that $(x_i,y_0)\in V(p(x,y),q(x,y))$. The same holds for $(x_0,y)$.

You're on the right track in using Bézout's identity. Suppose $f, g \in \mathbb{C}[x,y]$ with $\gcd(f, g) = 1$. Now, $\mathbb{C}[x,y]$ is not a Euclidean domain, but $\mathbb{C}(x)[y]$ is. By Bézout's identity, then there exist $a,b \in \mathbb{C}(x)[y]$ such that $af + bg = 1$. Writing $$a(x,y) = \frac{a_1(x,y)}{a_2(x)} \quad \text{and} \quad b(x,y) = \frac{b_1(x,y)}{b_2(x)}$$ then \begin{align*} 1 &= af + bg = \frac{a_1(x,y)}{a_2(x)} f(x,y) + \frac{b_1(x,y)}{b_2(x)} g(x,y) \end{align*} and clearing denominators yields $$b_2(x)a_1(x,y) f(x,y) + a_2(x)b_1(x,y)g(x,y) = a_2(x) b_2(x) \, .$$ Since $a_2, b_2 \in \mathbb{C}[x]$ are polynomials in one variable, they have only finitely many zeroes. This shows that there are only finitely many possibilities for $x_0$ in a point $(x_0, y_0) \in \mathbb{V}(f,g)$. Exchanging the roles of $x$ and $y$, the same argument shows there are only finitely many possibilities for $y_0$.