Intersecting the quadric surfaces with the K3 surface

$$Q : a_0 x^2 + a_1 x y + \cdots + a_9 w^2 = 0 \subset \mathbf{P}^3, \quad [L := \mathbf{Q}(\{a_i\}_i):\mathbf{Q}] = 168$$

Goal

Show that $Q \cap X$ decomposes into two quartic curves.

It suffices to show that the singular locus $S$ of $Q \cap X$ consists of 10 distinct reduced points.
Hopeless to do this directly! Operations in $L$ are seriously expensive!
Linear algebra. 😰 Gröbner basis. 😱
One needs to compute $S$ by hand, and clear denominators before that.
Working over $\mathbf{F}_p$ we find 10 distinct points.
Hence, $S$ is zero-dimensional and reduced, and $\deg S \leq 10$.
We conclude $\deg S = 10$ via Gotzmann regularity theorem,
by checking that $\dim L[x,y,z,w]_{\bullet}/I_{\bullet} = 10$ for $\bullet = 6,7$, where $I$ is saturated and $V(I) = S$.

SOURCE: L:L723-753

What it does. The candidates become curves. This is the moment the numerics turn into geometry.

PANEL FLAG: that Lambda has no degree-4 class of square 0 is asserted from the author's enumeration and was not independently certifiable from the supplied data. The exclusion of (2,2) + (2,2) rests on it.

Smoothness gives the two rulings only over the algebraic closure: a smooth quadric over L need not be split. The bidegree argument and the component degrees are geometric, so Lbar is where they belong; if the two rulings are wanted over L itself, splitness over L is an extra thing to establish.

The division of labour matters. The special fibre over F_p bounds the degree from above and shows S is reduced and zero-dimensional; it cannot on its own prove that the characteristic-zero intersection splits. The equality deg S = 10 is a characteristic-zero Hilbert-function claim, checked over L in degrees 6 and 7. The exclusion of other decompositions rests on the bidegree count, the absence of lines, conics and twisted cubics from slide 9, and separately on the absence of a degree-4 square-zero class in Lambda, which is what rules out (2,2) + (2,2); primitive norm 4 is below B, so the enumeration supplies it.

Gotzmann regularity, in one clause: it bounds the regularity of a saturated ideal in terms of its Hilbert polynomial, so the Hilbert polynomial is determined by finitely many explicitly known degrees. The saturation hypothesis is not optional: adding a degree-8 equation can leave the values in degrees 6 and 7 unchanged while shrinking V(I).

Spoken: $Q$ is smooth, so $Q_{\overline{L}} \simeq \mathbf{P}^1 \times \mathbf{P}^1$ and $Q \cap X$ has bidegree $(4,4)$ over $\overline{L}$; slide 9 excludes components of degree $1, 2, 3$, leaving $(1,3) + (3,1)$ and $(2,2) + (2,2)$.

Spoken: The second is excluded by the enumeration of $\Lambda$, which has no degree-4 class of square 0: the singularity count does not separate the two, since two nodal $(2,2)$ curves also give $8 + 1 + 1 = 10$ reduced singular points.

Spoken: That leaves $(1,3) + (3,1)$, whose components meet in $1 \cdot 1 + 3 \cdot 3 = 10$ points, so it suffices that the singular locus $S$ is 10 distinct reduced points.

TRANSCRIBED FROM: five-nomial-quartics/slides/mukai_leiden.tex:723-753