Understanding Ramification Points

Solution 1:

Let's look at your second example. Let $p(z, w) = w^2 - z^3 + z^2 + z$, and let $Y = \{ p(z, w) = 0 \}$. Then, $$p(z, i) = -z^3 + z^2 + z - 1 = -(z - 1)^2 (z+1)$$ so I claim $(1, i)$ has ramification index $2$ while $(-1, i)$ has ramification index $1$. Indeed, observe that \begin{align} \frac{\partial p}{\partial z} & = -3 z^2 + 2 z + 1 \\ \frac{\partial p}{\partial w} & = 2 w \end{align} so by an inverse function theorem argument, we find that $(z, w) \mapsto z$ is locally a chart near both $(-1, i)$ and $(1, i)$. In this chart, your function $g : Y \to \mathbb{C}$ is given by $z \mapsto \sqrt{z^3 - z^2 - z}$. Let us take Taylor expansions around $\pm 1$: \begin{align} g(z) - i & = -i (z-1)^2 + O((z-1)^3) \\ g(z) - i & = -2 i (z+1) + O((z+1)^2) \end{align} Hence, the ramification index at $(1, i)$ is indeed $2$ and at $(-1, i)$ it is $1$.

Morally, what is going on is that your curves are dense open subsets of projective curves. Indeed, your first curve is given in homogeneous coordinates by $$w^3 - z (z^2 - u^2) = 0$$ and your second curve is given by $$w^2 u - z^3 + z^2 u + z u^2 = 0$$ and one can check by hand that these curves are smooth "at infinity", so we have the desired embedding of the original affine algebraic curves into projective (hence compact) algebraic curves. Degree is well-defined on the latter, so is well-defined on the former by restriction; the only trouble is that there may be "missing" preimages and so the equation relating degrees and ramification indices becomes an inequality: $$\text{deg}(g) \ge \sum_{x \in g^{-1} \{w\}} \nu_x (g)$$ For example, take the affine hyperbola $z w - 1 = 0$ and the projection $(z, w) \mapsto w$; this function has degree $1$ (once we embed it in the projective closure), but obviously there are no preimages of $0$ in the affine hyperbola.

Let's develop a generic method of dealing with affine plane curves. Let $p : \mathbb{C}^2 \to \mathbb{C}$ be a polynomial function in two variables, and suppose $Y = \{ p(z, w) = 0 \}$ is a smooth algebraic curve. Let $f : Y \to \mathbb{C}$ be the projection $(z, w) \mapsto w$. For each fixed complex number $b$, we get a polynomial function $p(-, b)$, say of degree $d$. Now, because $\mathbb{C}$ is algebraically closed, we can write $$p(z, b) = c (z - a_1)^{e_1} \cdots (z - a_n)^{e_n}$$ for some distinct complex numbers $a_1, \ldots, a_n$, $c \ne 0$, and positive integers $e_1, \ldots, e_n$, such that $e_1 + \cdots + e_n = d$. Suppose also that $$\frac{\partial p}{\partial w}(a_i, b) \ne 0$$ for all $a_i$; then an inverse function theorem argument shows that $(z, w) \mapsto z$ is a chart near each $(a_i, b)$. I claim that the ramification index of $f$ at $(a_i, b)$ is $e_i$ under these hypotheses. Indeed, when $z$ is a local parameter, we have $$0 = \frac{\partial p}{\partial z} + \frac{\mathrm{d} w}{\mathrm{d} z} \frac{\partial p }{\partial w}$$ so if $e_i > 1$, we have $\frac{\partial p}{\partial z} (a_i, b) = 0$, so we must have $\frac{\mathrm{d} w}{\mathrm{d} z} (a_i) = 0$ because $\frac{\partial p}{\partial w} (a_i, b) \ne 0 $ by hypothesis – implying $f(z) - b = O((z - a_i)^2)$. Playing around with total derivatives more, we eventually find that the first non-zero coefficient of $f(z) - b$ around $a_i$ is the coefficient of $(z - a_i)^{e_i}$, as required.

On the other hand, when we have $\frac{\partial p}{\partial w} (a_i, b) = 0$, then by non-degeneracy we must have $\frac{\partial p}{\partial z} (a_i, b) \ne 0$, and we must have $e_i = 1$ and $(z, w) \mapsto w$ is a chart near $(a_i, b)$. But then obviously the ramification index of $f$ at $(a_i, b)$ must be $1$. So in either case the ramification index of $f$ at $(a_i, b)$ is equal to $e_i$. Convenient, no?