**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:

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 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!