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Let $L$ be the complex line ${ax+by+c=0}$ in $\mathbb{C}^{2}$,and let $C$ be the algebraic curve ${{f(x,y)=0}}$, where $f$ is an irreducible polynomial of degree $d$. Prove that $C\bigcap L$ contains at most $d$ points unless $C=L$.

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I guess you're not allowed to use B\'{e}zout's Theorem? – Derek Allums Nov 21 '12 at 2:54
No, not allowed. – Alex Nov 21 '12 at 3:41

Assume that $a\neq 0.$ For any point $(x,y)\in C\cap L$ we will have $x=-\dfrac{by+c}{a}$ and so $g(y)=f(-\frac{by+c}{a},y)=0.$ But $g(y)$ is a degree $d$ polynomial in one variable over $\Bbb C,$ so ...

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Oh,yeah,thanks a lot. – Alex Nov 21 '12 at 3:45

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