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.

score 0 · Accepted Answer · answered Aug 24 '21 at 15:58

0

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.

answered Aug 24 '21 at 15:58

Thank you for your clear answer. I just want to know if there is a theorem for the gcd part you wrote. I dig it up, but I can not find something like that. – ugurgozutok Aug 24 '21 at 18:32
@ugurgozutok What does it mean for two plane curves not to have a component in common? – Aug 24 '21 at 22:05
Okey, I got it. Thank you very much. – ugurgozutok Aug 25 '21 at 08:35

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

1 Answers1