Why do we need Bezout's Theorem in this statement?

Question

I'm analyzing a paper concerning algebraic geometry. In a proof, author uses Bezout's theorem. But I don't get it much. Here is the summary: Let $K(t,s)$ and $F(t,s)$ be polynomials in $t,s$ with $F$ irreducible. The zero set of $K$ includes the zero set of $F$. Here, the author says "Since $F$ is irreducible, Bezout's theorem implies that $F$ divides $K$.

Here is my question: Why do we need Bezout's theorem here? How does Bezout's theorem imply this divisibility?

Thanks in advance. Sorry for the English.

Answer 1

You're in an algebraically closed field, therefore the two curves intersect in infinitely many points (namely, the points of $\mathcal V(F)$). By Bézout, $\operatorname{gcd}(F,K)\ne 1$, and therefore $F\mid K$ because $F$ is prime.