What relationship is there between repeated roots of discriminants and orders of roots of the original polynomials?

Question

Disclaimer:

I asked this problem several days ago on MSE, I'm cross-posting it here. The title sounds like a high school problem, but (as a grad student not in algebra) it feels subtle/deep.

Background/context:

If $S \subset \mathbb{CP}^1 \times \mathbb{CP}^1$ is a smooth curve with bidegree $(d_1, d_2)$, we know that its genus is $(d_1 - 1)(d_2 - 1)$ for example by the adjunction formula.

Alternatively, we can attempt to compute the genus via the Riemann-Hurwitz formula as follows:

1. Write $S = \{P(z,w) = 0\}$ where $z, w$ are in $\mathbb{CP}^1$.

2. The projection map $p$ onto the first factor is a $d_2$-sheeted branched cover. The branch points $z_i$ occur exactly when the polynomial $P(z_i, w)$ has repeated roots (when considered as a polynomial in $w$, treating $z_i$ as coefficients).

3. These repeated roots are detected by the discriminant of $P(z_i, w)$, which has degree $2d_2 - 2$ in the "coefficients". But the coefficients are homogeneous of degree $d_1$ in $z$, so the discriminant is a degree $d_1(2d_2-2)$ polynomial in $z$.

4. Computing $\chi(S) = 2d_2 - d_1(2d_2 - 2)$, we actually get exactly the correct genus for $S$! This means $d_1(2d_2 - 2)$ precisely counts the ramification $\sum (e_p - 1)$ of the branched cover.

Question:

Given the above context, it must be the case that the degree of the discriminant counts exactly the number of roots of $P(z,w)$ which are lost due to branching of $(z,w) \mapsto z$. Why is this the case? There are basically two opposing forces which must somehow cancel out:

• If the discriminant has a repeated root, then downstairs it corresponds to losing a branch point.
• If a branch point has ramification index greater than two, then upstairs it corresponds to a given root of the discriminant not accounting for all of the repeated roots which are lost at this branch point.

I have no idea why these two phenomena should cancel out, and would love some insight. Thank you!

Answer 1

$\def\P{{\mathbb{P}}} \def\A{{\mathbb{A}}} \newcommand{\O}{\mathcal{O}} \DeclareMathOperator{\Disc}{Disc}$I think the story goes like this. The multiplicity of a zero of the discriminant counts exactly how many fewer preimages we have than the typical number in the branched covering.

Let $Z\subset \P^1\times \P^1$ be a smooth curve of bidegree $(e,d)$. View your curve $Z\subset \P^1\times \P^1$ as being a family of divisors $\{Z_x:x\in \P^1\}$ of degree $d$ on $\P^1$ parameterized by $\P^1$. Any such divisor can be viewed as a point in $\P^d \cong \P H^0(\O_{\P^1}(d))$ (the space of homogeneous polynomials on $\P^1$ of degree $d$) just by associating it with its defining equation. Thus, $Z$ yields a curve $C$ in $\P^d$, of degree $e$.

Inside of $\P^d$ is the discriminant hypersurface $\Delta$ of polynomials with a repeated root. It is defined by the equation $\Disc(f) = 0$, a homogeneous polynomial of degree $2d-2$ in the coefficients of $f$. Given some $f \in \P^d$, you want to understand how any multiple roots of $f$ are connected with the multiplicity of $\Delta$ at $f$.

Claim: for a homogeneous polynomial $f$ of degree $d$, the multiplicity of $\Delta$ at $f$ is $$m = d- \#(\textrm{roots of $f$}).$$

To prove the claim, you need to see that if $g$ is a general homogeneous polynomial, then the polynomial $p(u,v) = \Disc(uf+vg)$ has a root at $v=0$ of order exactly $m$. For such a general choice of $g$, the corresponding line $\ell=\{uf+vg:[u:v]\in \P^1\}\subset \P^d$ will intersect $\Delta$ transversely except at $f$. Additionally, the curve $uf+vg=0$ in $\P^1\times \P^1$ will be smooth and the ramification of the projection will all be simple ramification except over the fiber $v=0$ (these types of things can be proved by a dimension count). Now $p(u,v)$ has degree $2d-2$, so the total number of intersections with multiplicity between $\ell$ and $\Delta$ is $2d-2$. Comparing with the Riemann-Hurwitz formula (as in the argument in the next paragraph), the only possibility is that the intersection at $f$ counts with multiplicity $m$.

Now consider how $C$ intersects $\Delta$. By Bezout's theorem, we must have $e(2d-2)$ intersections when counted with multiplicity. Whenever $x\in \P^1$ is a branch point of $Z$, there is a ramification point on the fiber $Z_x$, and the corresponding point $f\in C$ is a polynomial with a repeated root. The ramification contribution to Riemann-Hurwitz over $x$ is exactly the number $d-\#(\textrm{roots of $f$})$, and the intersection multiplicity between $C$ and $\Delta$ at $f$ is at least this big. But the sum of the all the ramification contributions to Riemann-Hurwitz is exactly $e(2d-2)$, and the sum of all the intersection multiplicities is at least this big, but no larger by Bezout. It follows that each intersection between $C$ and $\Delta$ counts with multiplicity exactly $d- \#(\textrm{roots of $f$}).$

Warning: Smoothness of the curve $Z$ is important. For example, working in $\A^1\times \A^1$, the polynomial $y(y-x)$, viewed as a family of quadratic polynomials in $y$, has discriminant $x^2$. But it gives a branched double cover where two sheets become one over $x=0$. This family of divisors defines a tangent line to the discriminant hypersurface $\Delta$, so the intersection multiplicity is larger than otherwise expected.

Answer 2

(1) There is the following indirect explanation:

For a generic curve neither of these phenomena happen - the discriminant has no repeated roots and the branch points all have ramification index two. The degree of the discriminant and the genus both agree for any particular smooth curve and in the generic case, so they are compatible for particular curves as well.

This gives for me a reasonably clear intuitive explanation, but I'm not sure how easy it is to check in practice that the discriminant has no repeated roots without doing the calculation I'm going to do next.

(2) The discriminant of a polynomial is the product of the squares of differences of the roots over the algebraic closure of the base field.

Here, the base field is a local field near a point where there is ramification. Each point of the ramified fiber corresponds to one or more roots over the local field. For two local field roots corresponding to different points of the ramified fiber, their difference is a unit in the local ring (or its integral closure) and this does not contribute to the order of vanishing of the discriminant. So the only pairs of roots that are relevant are two roots corresponding to the same point of the ramified fiber. This means the order of vanishing of the discriminant is a sum of local contributions at the points of the ramified fiber. Let's see that they agree with the ramification index.

Near a ramified point, for simplicity of notation $(0,0)$, the equation $P$ must mod $w$ have the form $ a z^e$ plus higher-order terms and since the curve is smooth the equation must have the form $a z^e + b w + $ higher-order terms (if the coefficient of $w$ was zero you would get a singularity). The roots of this equation over $\overline{k((w))}$ that reduce to $0$ at $w=0$ have the form $$z = (- b a^{-1} w)^{1/e}+ \dots $$

There are $e$ such roots, so $e (e-1)/2$ pairs of roots, and each pair has a difference of the form constant times $w^{1/e}$ plus higher-order terms, so a squared difference of the form constant times $w^{2/e}$ plus higher-order terms, for a product of the form constant times $$ w^{ \frac{e (e-1)}{2} \frac{2}{e}} = w^{e-1}$$ plus higher-order terms, i.e. an order of vanishing $e-1$.

So indeed the local contribution to the discriminant agrees with the local contribution $e-1$ to the Riemann-Hurwitz formula.