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Let $V(f)$ be a cubic projective curve and a $V(g)$ be a conic projective curve, with $f,g \in \mathbb{C}[X,Y,Z]$. If $V(f)\cap V(g)$ has more than six points, then $f$ and $g$ has a linear factor in common?

A friend of mine said that it's false, but I couldn't find any conterexample for this. Just need one conterexample, can you help me? Thanks.

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Your friend is absolutely correct - stated in that generality, the claim is false, and trivially so.

Take any irreducible conic $C$ and any line $L$, and let the cubic be $C \cup L$. Obviously the conic and the cubic have all of $C$ in common (infinitely many points!) but they don't have a line in common, since $C$ was chosen irreducible.

Bezout's theorem has some conditions under which it holds; obviously, my counter-example does not satisfy those conditions. For the claim to be true, it has to explicitly exclude extreme cases, such as my counter-example.

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Hint: what does Bézout's theorem say, and what are its assumptions?

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  • $\begingroup$ Bézout's Theorem says that $\#(V(f)\cap V(g)) = deg \, f. deg \, g$ for projective curves, witch in this case is 6 (because $deg \, f$ = 3 and $deg \, g$ = 2). That's what's confusing me, because how can I suppose that $\#(V(f)\cap V(g)) > 6$ without contradicting Bézout's Theorem? $\endgroup$ – Geaquinto May 8 at 18:12
  • $\begingroup$ Is $g= Z^2-XY$ a valid counterexample? Because it has no linear factor (is irreductible). $\endgroup$ – Geaquinto May 8 at 18:35
  • $\begingroup$ @Geaquinto (replying to your second comment) yes, you can produce a valid counterexample from that polynomial, as explained in the other answer. From Bézout's theorem you can deduce that $f$ and $g$ must share a common factor. If this is not a linear factor, then it would have to be an irreducible degree $2$ factor (otherwise they would share each of its linear factor). And then you only need to find a cubic polynomial having this quadratic polynomial as one of its factors. $\endgroup$ – Pedro May 9 at 8:15
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    $\begingroup$ @Geaquinto P.S. you should accept the other answer, since my hint was meant to suggest that the implication was true (I read the question too quickly and didn't pay attention to the word linear) $\endgroup$ – Pedro May 9 at 8:23
  • $\begingroup$ Ok, thanks for the help! $\endgroup$ – Geaquinto May 9 at 13:56

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