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Show that a line $L$ is tangent to a projective plane curve $C$ at a smooth point $P$ if and only if $I_P(C \cap L) \geq 2$. If $P$ is a point of multiplicity $2$, and $L$ is tangent of multiplicity $2$ at $P$, show that $I_P(C \cap L) \geq 3$.

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  • $\begingroup$ I proved with using dimension. But question is seen to be proven with simply geometry. Geometric picture.. $\endgroup$
    – vudu vucu
    Feb 14, 2015 at 21:05
  • $\begingroup$ Can you give your proof using dimension? $\endgroup$
    – Yuyi Zhang
    Nov 6, 2018 at 3:36

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Let $P=(0,0)$ and tangent line is $L:x-y=0$. (We shall assume them with using natural transformations.)

Now write $C(x,y)=F_0(x,y)+F_1(x,y)+F_2(x,y)+...$ where $F_i$ are homegenous polynomial degree $i$.

If $F_0(x,y) \neq 0$, this means $P$ is not on the curve. If $F_1(x,y) \neq L$ then, $L$ is not a tangent. Hence we can put $x=y$ and eliminate $F_0$ and $F_1$. Then we get at least a term with $x^2$. This proves first statement.

In second, $F_0(x,y)=F_1(x,y)=0$ since multiplicity of $P$ is 2 and $F_2(x,y)$ need to equal $L^2$ since $L$ is a tangent of $P$ with multiplicity 2. Hence we can put $x=y$ and eliminate $F_0$, $F_1$ and $F_2$. Then we get at least a term with $x^3$. This proves second statement.

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  • $\begingroup$ I think that this is really geometric proof. $\endgroup$
    – vudu vucu
    Mar 10, 2015 at 9:58

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