The common zeroes of two homogeneous polynomials $F$ and $G$ in $\mathbb P^2$ are finite when they have no common factors.

My thought on this which might (not) be helpful:

I know we may use Bezout's theorem, which requires one of two polynomials to have no singularities. I feel like we need to somehow use the condition that they have no common factors to make one of them not singular?

I wish the proof can be an algebraic one rather than a geometric one. However, any solution will be appreciated!

  • 2
    $\begingroup$ What does it mean for polynomials to be in $\mathbb{P}^2$? Do you mean $F, G \in k[X_0, X_1, X_2]$? Anyway, does this post or this post answer your question? $\endgroup$ – André 3000 Jul 7 '17 at 22:24
  • $\begingroup$ Bézout's theorem does not require either of the two polynomials to have no singularities. Perhaps you are working with a more restricted version of the theorem? In any case, if you are looking for an algebraic proof, follow the links in the comment above. $\endgroup$ – Zach Teitler Jul 8 '17 at 3:27

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