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:
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Write $S = \{P(z,w) = 0\}$ where $z, w$ are in $\mathbb{CP}^1$.
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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).
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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 $z$.
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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!
Best Answer
$\def\P{{\mathbb{P}}} \def\A{{\mathbb{A}}} \newcommand{\O}{\mathcal{O}} \DeclareMathOperator{\Disc}{Disc}$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.