# stuck on computing points of ramification of a map of curves via sheaf of relative differentials

Let $C$ be the curve in $\mathbb{C}^2$ given by $f(x,y)=x^3+y^3-y=0$ and let g:$C \to \mathbb{C}$ be the projection $g(x,y)=x$. I want to find where this map $g$ is ramified, and do this by computing the sheaf of relative differentials. I have $(3y^2-1)dy=-3x^2dx$. Now, by the definition of relative differentials I thought I set $dx=0$, and since ramified points are where this sheaf is nonzero, I get that dy is not killed when $3y^2-1=0$ but I don't think that gives the right answer. On the other hand, the derivative $dy/dx=\frac{-3x^2}{3y^2-1}$ vanishes when x=0, and that gives the right answer. Apparently I should be setting $dy=0$ but that doesn't make sense since the target of $g$ is $Spec\ \mathbb{C}[x]$. I'm confused why my sheaf of differentials approach isn't working - clearly I just have a basic misunderstand about it and I was hoping someone could clear if up for me.

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I think your calculation seems fine. You are trying to find those values of $y$ for which the equation $y^3 - y + x^3$ has a multiple root. This occurs when the derivative $3y^2 - 1$ vanishes, which is what you computed. Note that the point $x = 0$ is not a ramification point: $y^3 - y = 0$ has 3 distinct roots (namely $y = \pm 1$ and $y = 0$). – Matt E Sep 22 '10 at 3:51

As you have thought you should solve \left\{\begin{aligned}3y^2-1=0,\\x^3+y^3-y=0\end{aligned}\right. which gives you 6 ramification points $y=\pm 1/\sqrt{3}, x=\pm e^{2\pi i k/3}\sqrt[3]{2}/\sqrt{3}$ where $k=0,1,2$.
This should not under any circumstances be considered rigorous, but intuitively here is what is happening in my mind. If you project onto the "$x$-axis", then think of the tangent space of the curve $C$ at some point $(a,b)$. If the tangent direction is "vertical", then any vector in that space will project to 0.
So after projecting, there is no "$y$-component" to any of the tangent vectors, and hence the $dy$ always evaluates to $0$. So in some sense $g$ is transverse to the projection precisely when there is some "$x$-direction". This is precisely when $-3x^2dx\neq 0$.