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Statement

Theorem 5.3 (p. 8). Let AA be a set of hyperplanes of AG⁡(n,F)\operatorname{AG}(n,\mathbb F), and for 1≤i≤n1\le i\le n let DiD_i be a nonempty proper subset of a finite set Si⊆FS_i\subseteq\mathbb F. Suppose every point (s1,…,sn)(s_1,\ldots,s_n) with si∈Sis_i\in S_i lies on at least tt hyperplanes of AA, except at least one point of D1×⋯×DnD_1\times\cdots\times D_n, which lies on no hyperplane of AA. Then

∣A∣≥(t−1)max⁡j(∣Sj∣−∣Dj∣)+∑i=1n(∣Si∣−∣Di∣).|A|\ge(t-1)\max_j\bigl(|S_j|-|D_j|\bigr)+\sum_{i=1}^n\bigl(|S_i|-|D_i|\bigr).

The hypothesis is read as for Theorem 4.1: points outside D1×⋯×DnD_1\times\cdots\times D_n lie on at least tt hyperplanes of AA, and at least one point of D1×⋯×DnD_1\times\cdots\times D_n lies on none. The paper calls the theorem almost the dual of Theorem 5.1.

The remark after the theorem (p. 8). With Si=FqS_i=\mathbb F_q and Di={0}D_i=\{0\} the paper states that a set of hyperplanes covering every point of AG⁡(n,q)\operatorname{AG}(n,q) other than the origin at least tt times has at least (n+t−1)(q−1)(n+t-1)(q-1) members, and calls this the dual of Theorem 5.2. The theorem gives this only when the origin lies on no hyperplane of the set, a condition the remark leaves out and cannot drop: for n=2n=2, t=1t=1 and q>2q>2, the qq lines X1=aX_1=a cover the whole plane and q<2(q−1)q<2(q-1).

Proof pointer

P. 8. The product ff of the affine linear forms defining the hyperplanes of AA has degree ∣A∣|A|, a zero of multiplicity at least tt at the covered points, and a nonzero value at the uncovered point, so Theorem 4.1 applies.

Read depth

Claims checked: the statement and the remark after it were read clause by clause against p. 8 of the print, and the proof was followed.

Dependencies

Theorem 4.1.

Source. Simeon Ball and Oriol Serra, Punctured combinatorial Nullstellensätze, Combinatorica 29 (2009), 511–522, doi:10.1007/s00493-009-2509-z. Labels and page numbers are those of the corrected author manuscript dated 14 June 2011, the edition named on the source card.

Bears on

None recorded. The paper names no Erdős problem.