I'm trying to find the simple/multiple points of the projective curve with the defining polynomial $F(X,Y,Z)=XZ-Y^2$. As far as I understand, I need to dehomogenize $F$ in each of the three standart charts $U_i$ and determine the simple/multiple of the resulting affine curve.

  • In the chart $U_0$ the dehomogenized polynomial is $G_0:=F(1,Y,Z)=Z-Y^2$, so all points on this affine curve have multiplicity one. This corresponds to the fact that all points $[1:Y:Z]\in V(XZ-Y^2)\subset \mathbb{P}^2$ are simple points of $F$.

  • The case of $U_2$ is similar: the dehomogenized polynomial is $G_2:=F(X,1,Z)=X-Y^2$, and so all points $[X:Y:1] \in V(XZ-Y^2)\subset \mathbb{P}^2$ are simple points of $F$.

  • In the case of $U_1$, the dehomogenized polynomial is $G_1:=F(X,1,Z)=XZ-1$. The point $(0,0)$ has multiplicity $0$, and the other points are simple. So all points $[X:1:Z] \in V(XZ-Y^2)\subset \mathbb{P}^2$ are simple points of $F$.

Is this correct? I'm unsure whether above is a full description of simple/multiple points.

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    $\begingroup$ In $U_1$, is your dehomogenization correct? Shouldn't it be $XZ-1$? $\endgroup$ – Mohan Apr 1 '17 at 15:06
  • $\begingroup$ I've fixed that. $\endgroup$ – user428554 Apr 1 '17 at 15:59
  • $\begingroup$ So, every point has multiplicity on the curve has multiplicity one. I wish you wouldn't use notations like $Z(XZ-Y^2)$, where the two $Z$s have different meanings. $\endgroup$ – Mohan Apr 1 '17 at 17:14
  • $\begingroup$ I've fixed that as well. Is it a comprehensive description now? In particular, is it sufficient to say that in all of the three charts, all points on the corresponding affine curves are simple, and then refer to the fact that $\mathbb{P}^2$ is covered by these 3 charts (and the fact that the points of each $U_i$ correspond bijectively to the points of $\mathbb{A}^2$ so that every point in every chart is simple for $F$) ? $\endgroup$ – user428554 Apr 1 '17 at 18:28
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    $\begingroup$ Yes, that is enough. $\endgroup$ – Mohan Apr 1 '17 at 19:23

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