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:
Write $S = \{P(z,w) = 0\}$ where $z, w$ are in $\mathbb{CP}^1$.
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).
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 $z$.
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 "the vanishing 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!
After getting no engagement here, I cross-posted the question to Math Overflow. I received some answers there, and will accordingly close this post.
See https://mathoverflow.net/questions/466758/what-relationship-is-there-between-repeated-roots-of-discriminants-and-orders-of