# Given an algebraic curve $F(x,y)=0$, why do the partial derivatives of $F(x,y)$ being zero at a point imply the plane curve has a singularity?

I'm looking at algebraic plane curves of the form $F(x,y)=0$ and trying to figure out why for points on the curve such that $\frac{\partial F}{\partial x} = \frac{\partial F}{\partial y}=0$, the plane curve has a singularity.

I've been looking at the surface $z=y^2-x^3$ and comparing it to $z=y-x^2$ and trying my darndest to understand what about having a flat tangent plane causes the corresponding curve to have a singularity right at the point where the surface 'dips' into the $z=0$ plane. So not being able to figure out geometrically/intuitively why one thing causes the other, I started looking for a proof of this fact hoping that it would show me the 'why' of it, but I can't find one online or in any of my books.

Can someone explain intuitively why this is, and maybe point me in the direction of a proof as well?

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There's a lot to unpack here. First, what is your definition of singularity? Second, are you looking at curves or surfaces? – Matt Jan 21 at 1:36
Maybe I wasn't clear enough in my wording. I'm talking about singularities of algebraic curves here, I'm just looking at surfaces because those are where you take your partial derivatives. As for a definition of singularity well I'm not sure I have a good one, beyond calling it any point which does not have a unique well defined tangent line. – lithium barbie doll Jan 21 at 1:44
I mean crossing singularities don't satisfy the implicit function theorem, but each line generally has a well defined tangent line, while cusp singularities it's the opposite. Both however satisfy the partial derivative criterion for being a singularity, so maybe I should take that as a definition instead of a theorem to be proven? Or is the best definition that no open set around the point is homeomorphic to $\mathbb{R}$ (for curves)? – lithium barbie doll Jan 21 at 1:46
@cat, I don’t understand what you mean by “looking at surfaces because those are where you take your partial derivatives”. – Lubin Jan 21 at 2:36
@Rahul, it certainly looks to me as if $x^3-y^3=0$ has a singularity at the origin, since the locus is three lines that intersect there. – Lubin Jan 21 at 2:38
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## 2 Answers

I’ve applauded someone else here at MSE for wanting to have an intuition different from others’, and I’ll say the same to you: differing intuitions lead to more interesting mathematics. But yet, I don’t think it’s at all productive to think of a curve $F(x,y)=0$ in the plane as the intersection of the surface $z=F(x,y)$ with the $(x,y)$-plane.

If you aren’t willing to accept the condition $\partial F/\partial x=\partial F/\partial y=0$ at a point $P$ as the definition of a singular point, then you need an independent definition of what it means for a point to be singular.

First, let’s look at cases where we agree that the origin is a singular or nonsingular point of some curve. If the origin is a nonsingular point, then any line passing very near to the origin in a direction not parallel to the curve’s tangent at the origin cuts the curve in only one point. If we move toward the origin with a line parallel to the tangent, though, the points of intersection are not comparably close (think of a horizontal line $y=\varepsilon$ and the parabola $y=x^2$) but when we hit, ¡plink!, the contact is multiple rather than single.

Compare this behavior with what happens at the origin with the curve $y^2=x^2(x+1)$, in other words $x^2+x^3-y^2=0$, which you and I agree has a clearly visible singularity at $(0,0)$, a node. Any line coming close to the origin, of no matter what slope, has two intersections with the curve comparably close to the distance $\varepsilon$ from the origin to the line. And in case the slope is $\pm1$, at the origin, ¡plink!, the intersection multiplicity there is three.

So here’s a proposed definition of singularity for point of a plane curve, independent of the standard one. A point $P$ on the locus of $F(x,y)=0$ is nonsingular if all but one of the straight lines passing through $P$ has only a single intersection with the locus in the immediate neighborhood of $P$. And $P$ is singular if all lines through $P$ have at least a double intersection with the locus, in the immediate neighborhood of $P$.

Now how does this work out in practice? Let’s again restrict our view to curves passing through the origin, so that the polynomial $F(x,y)$ has no constant term. And then we can write $$F(x,y)=a_{10}x + a_{01}y+a_{20}x^2+a_{11}xy+a_{02}y^2+\cdots\>,$$ where the ellipsis represents monomials of degree $3$ and greater. Now all is clear. The partial-derivative condition for singularity is exactly that the two linear coefficients $a_{10}$ and $a_{01}$ should vanish. And I hope that you see that this vanishing is exactly the condition that any line $\alpha x+\beta y=0$ through the origin should have at least a double intersection with the curve at the origin. I hope also that you’ll go through all this for the singular curve $y^2=x^3$, and see how the concepts fit together there.

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For an approach to your first problem, which was to understand why $\frac{\partial F}{\partial x}= \frac{\partial F}{\partial y}=0$ implies that there is a singularity, we begin by looking at the rank of the $1 \times n$ Jacobian matrix, since it would describe the partial derivatives with respect to another vector (in this case the vector is $\begin{bmatrix} x\\ y \end{bmatrix}$). Since the rank is the number of linearly independent components, a singularity is where there are no linearly independent components (hence $F(x,y)\neq 0$), and that is the "one point" w/ that. The orientation is preserved since $\frac{\partial F}{\partial x}= \frac{\partial F}{\partial y}$, and so we are still along the 2-dimensional vector. Setting the partial derivatives equal to zero, along that vector $F(x,y)=0$ still holds, and so continuity is also preserved.

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