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I have a homework question that asks

Let $f$ and $g$ be two quadratic polynomials (with coefficients in $\mathbb{C}$) that share no common linear factors, and let $C_f$ and $C_g$ be the zero set of $f$ and $g$, respectively. Let $p$ and $q$ be distinct points in $C_f \cap C_g$, and let $L$ be the line through $p$ and $q$. Show that there are constants $c_1$ and $c_2$ (not both zero) such that $c_1f + c_2g$ is identically zero on $L$ and is the product of linear polynomials.

This problem has been confusing me for some time. Here's where I'm currently at with it. Since $f$ and $g$ have degree $2$, I can say that $C_f = \{ p_1, p_2 \}$ and $C_g = \{ q_1, q_2 \}$ for some points $p_i$ and $q_i$. Since $p, q \in C_f \cap C_g$ are distinct, can I not then conclude that $C_f = C_g = \{ p, q \}$? But now $f$ and $g$ share a common linear factor (actually it looks like they are associate). What am I doing wrong?

tylerc0816
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    Are you talking of two variables quadratics? It is not given, but otherwise $;C_f,C_g, C_f\cap C_g;$ are finite sets, and according to the given data the intersection is empty... – DonAntonio Feb 23 '14 at 17:33
  • @DonAntonio The phrasing in the book says "quadratic polynomials", but I think your right. What is the definition of a two-variable quadratic? Is it $ax^2y^2 + bxy^2 + cx^2y + \dotsb?$. – tylerc0816 Feb 23 '14 at 17:40
  • $$ax^2+by^2+cxy +dx+ey+f=0$$ What book is that, btw? – DonAntonio Feb 23 '14 at 17:59
  • @DonAntonio This is a question from Artin chapter 11, exercise 9.11. – tylerc0816 Feb 23 '14 at 18:03
  • What "Artin", @tyler ? There are several books by at least two different Artin (one of them the well-known Emil) – DonAntonio Feb 23 '14 at 18:06
  • @DonAntonio Sorry, the book is 'Algebra' by Michael Artin (2nd ed.) – tylerc0816 Feb 23 '14 at 18:06
  • I have the book, I can't find the exercise...I've edition 1, though. – DonAntonio Feb 23 '14 at 18:10
  • The question comes from the section on algebraic geometry. I just skimmed the 1st edition and couldn't find the problem there, either. – tylerc0816 Feb 23 '14 at 18:19

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