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I am trying to understand singular points on a complex projective -algebraic curve. I remember that singular points on a affine algebraic curve are determined by taking the partial derivatives and finding where they equal to 0.

But then I found this definition of singular point on a complex projective-algebraic curve that says that the multiplicity $v_p(C)$ of a curve C at p is the order of the lowest non-vanishing term in the taylor expansion of f at p. And thus a point is singular when $v_p(C) >1$. This process just confuses me and I wondered if I can use the process for affine curves.

I'm just confused on why these are different, for example for the curve $x^2-y^3$, why can I not set this equal to 0 , find the partial derivatives and tell me this has a singular point at $(0,0)$.

Lastly the textbook I'm using mentions that examples for singular points are at double points and simple cusps. And explain that the equation for a double point is $xy =0$ which I'm not sure how this would help if I saw a curve.

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  • $\begingroup$ In the second paragraph, is there a connection between $C$ and $f$? $\endgroup$ – Matthew Leingang Oct 28 '18 at 18:44
  • $\begingroup$ yes sorry C has the affine equation $f(x,y)=0$ $\endgroup$ – Sasha Oct 28 '18 at 18:47
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    $\begingroup$ You absolutely can use either method to detect singular points on affine curves. These two approaches are equivalent because the coefficients in the Taylor expansion are the various partial derivatives. In your example, looking at the Taylor expansion is easier: there are no linear terms in the Taylor expansion $f = x^2 - y^3$, so you can immediately see that $(0,0)$, the center of the expansion, is a singular point, and in fact a double point (i.e., point of multiplicity $2$). $\endgroup$ – André 3000 Oct 28 '18 at 19:08
  • $\begingroup$ Ok thanks, I understand that it is a singular point now why is it multiplicity 2 or a double point. I thought the two main types of singular points occurred at cusps and double points so how can a cusp be a double point? $\endgroup$ – Sasha Oct 28 '18 at 19:18
  • $\begingroup$ A cusp can be a type of double point. If $P$ is a double point of a curve $C$, then $C$ has two (not necessarily distinct) tangent lines at $P$. If these tangent lines are the same (a double tangent line), then $P$ is non-ordinary or a cusp; if the tangent lines are distinct, then $P$ is ordinary or a node. I think you might like the treatment in Fulton's Algebraic Curves. $\endgroup$ – André 3000 Oct 28 '18 at 19:27
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You absolutely can use either method to detect singular points on affine curves. These two approaches are equivalent because the coefficients in the Taylor expansion are the various partial derivatives. In your example, looking at the Taylor expansion is easier: there are no linear terms in the Taylor expansion $f = x^2 − y^3$, so you can immediately see that $(0,0)$, the center of the expansion, is a singular point and in fact a double point (i.e., point of multiplicity 2).

If $P$ is a double point of a curve $C$, then $C$ has two (not necessarily distinct) tangent lines at $P$. If these tangent lines are the same (a double tangent line), then $P$ is non-ordinary or a cusp; if the tangent lines are distinct, then $P$ is ordinary or a node. For instance, your example $C_1 : x^2 − y^3 = 0$ has a cusp because the lowest order homogeneous term factors as $x^2 = x\cdot x$, so $C_1$ has $x = 0$ as a double tangent line at the origin. By contrast, the curve $C_2 : y^2 = x^3 + x^2$ has a node at the origin because the tangent lines are $y=x$ and $y=-x$, as can be seen by factoring the lowest order homogeneous term $y^2−x^2=(y−x)(y+x)$. Try plotting them to get a picture of what this means visually.

For more information I recommend Fulton's Algebraic Curves.

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