# Tangents at singularities

Given an implicit polynomial function $f$ with singularity at the origin. How do I find the tangents to the curve at the point?

Wikipedia says that ignore all the terms except the ones with lowest degree. Why is this true?

Take for example the curve $x^3+x^4+y^3=0$. What are the tangents at the origin?

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Dear Grobber, your description of the tangent cone according to Wikipedia is mathematically correct, but could you please provide a link to the article ? – Georges Elencwajg Jan 4 '14 at 12:27
Thanks for the link, Grobber. Actually I had checked that page but missed the recipe for the tangents at a singular point. Probably because I was fascinated by the wonderful animation of the moving (and colour changing!) tangent at the beginning of the article. – Georges Elencwajg Jan 4 '14 at 19:45

Consider the curve $x^3 + y^3 + x^4 = 0$. For $(x, y)$ close to zero, the fourth-order term is negligible compared to the third-order terms, so near the origin the graph will strongly resemble that of $x^3 + y^3 = 0$. This factors as $(x+y)(x^2 - xy + y^2)$, so there will be a tangent line $x = -y$ and a couple tangent lines with complex slope.

To make this a bit more formal, we can use Hensel lifting to actually factor the expression $x^3 + y^3 + x^4$ as

$$(x + y + \cdots)\left(x - \left[\frac{1}{2} + \frac{\sqrt{3}}{2}i\right]y + \cdots\right)\left(x - \left[\frac{1}{2} - \frac{\sqrt{3}}{2}i\right]y + \cdots\right),$$

where each term is a power series. We can therefore view this algebraic curve as the union of three distinct analytic curves, each of which has its own tangent line at the origin.

That's the idea, anyway. If you want to do things in a more algebraic setting, it becomes a commutative algebra problem. But you should understand this way first.

(Also, to get a feel, it probably helps to start by considering a curve whose tangents at the origin have real slope, so that you can actually look at a picture. Try $x^2 - y^2 - x^4 = 0$ instead, for instance.)

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Dear Daniel, I don't think your Hensel factorization is correct: a couple of $i$'s seem to be missing. – Georges Elencwajg Jan 4 '14 at 10:04
Thanks! I fixed it. – Daniel McLaury Jan 4 '14 at 10:41

Wikipedia is right: the tangent cone to your curve $C$ is given by $x^3+y^3=0$.
If the base field is algebraically closed of characteristic $\neq 3$ and if $j\neq 1$ is a primitive cubic root of$1$, it consists of the three lines $x+y=0, x+jy=0, x+j^2y=0$, since $x^3+y^3=(x+y)(x+jy)(x+j^2y)$.

But where does the Wikipedia recipe come from?

Here is an explanation: we are looking at the intersection at $(0,0)$ of our curve with the line $L_{ab}$ given parametrically by $x=at, y=bt$ (with $a,b$ not both zero).
The values of $t$ corresponding to an intersection point $L_{ab}\cap C$ are those satisfying the equation $(at)^3+(at)^4+(bt)^3=0$ gotten by substituting $x=at, y=bt$ into the equation of $C$.
The equation for $t$ is thus $$t^3(a^3+at+b^3)=0 \quad (MULT)$$ The result is that $t=0$ is a root of multiplicity $3$ for all values of $a,b$ except for those with $a^3+b^3=0$ for which the multiplicity of the zero root of $(MULT)$ is $4$.
So we decree that the lines $L_{ab}$ with $a^3+b^3=0$ are the tangents to $C$ at the origin because they cut $C$ with a higher multiplicity, namely $4$, than all the other lines which only cut $C$ with multiplicity $3$.
This calculation is easily generalized to arbitrary curves through the origin and justifies Wikipedia's recipe.

[Purists will notice that the above is purely algebraic: no limits are involved and we don't need a topology on the base field.
But assuming that the base field is $\mathbb C$ with its metric topology and using limits is fine with me if it helps in understanding the situation. In mathematics as in love and war all is fair...]

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Actually I haven't read Wikipedia's entry, because I couldn't find it! What is a link for it? – Georges Elencwajg Jan 4 '14 at 12:29

Look at $0=f(x,y)$ with a magnifying glass $0=f(ε X,ε Y)=ε^3(X^3+ε X^4+Y^3)$ with $X$ and $Y$ always in the range $[-1,1]$. There you see that the higher order term has a coefficient that becomes smaller the closer you look. So in the first order approximation, $Y=-q^kX$, with $q\ne 1$, $q^3=1$ the third root of unity.

Now one try to find the next order of the approximation, $Y=-q^kX+δ$,

$$0=X^3+ε X^4+(-q^kX+δ)^3=ε X^4+3q^{-k}X^2δ+O(δ^2)$$

which gives $δ=-\tfrac13εq^k X^2$, so that now the approximation up to terms in $ε^2$ is

$$Y=-q^k(X+\tfracε3X^2).$$

Of course, in this case with only one term involving $y$, the faster solution is by directly computing the cube root, then

$$y=-q^kx\,\sqrt[3]{1+x}=-q^kx\,\left(1+\tfrac13 x-\tfrac19x^2+\tfrac{5}{81}x^3+\dots+\tbinom{1/3}{k}x^k+\dots\right)$$

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The tangent of the curve $x^3+x^4+y^3=0$ at the origin is $$y=-x.$$

Differenciate the both sides with respect to $x$ will give you $$3x^2+4x^3+3y^2y^\prime=0.$$ So, if $y\not=0,$ we have $$y^\prime=\frac{-3x^2-4x^3}{3y^2}.$$

Then, letting $a$ be the slope of the tangent we want, using l'Hôpital's rule will give you \begin{align}a&=\lim_{(x,y)\to(0,0)}\frac{-3x^2+4x^3}{3y^2}\\&=\lim_{(x,y)\to(0,0)}\frac{-6x+12x^2}{6yy^\prime}\\&=\lim_{(x,y)\to(0,0)}\frac{-6+24x}{6y^\prime y^\prime+6yy^{\prime\prime}}\\&=\frac{-6}{6a^2}\end{align} So, we have $$a=\frac{-6}{6a^2}\Rightarrow 6a^3+6=0\Rightarrow 6(a+1)\left\{\left(a-\frac 12\right)^2+\frac 34\right\}=0\Rightarrow a=-1.$$

If you understand the idea of the above answer, you'll be able to solve similar questions.

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