Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 10 / 2

(A) Use the dimension of a bounded-total-degree polynomial space. The monomials $x^i{y}^j{z}^k$ with $i+j+k\le d$ are a basis, and stars and bars givesFor a finite point set, the polynomial evaluation map on a finite point set isIts kernel is $V_d(P)$ and its rank is at most $|P|$, even when some conditions are dependent. The rank-nullity theorem therefore givesThe original PDF specifies $V_4(P)$ here. With twelve points,No general-position assumption is required.

(B) Let $L=|\mathcal{L}|\ge 1$. We seek a polynomial vanishing on a finite set of spatial lines. For each line $\ell$, choose an affine parametrization ${\mathbf{a}}_{\ell }+t{\mathbf{v}}_{\ell }$ with ${\mathbf{v}}_{\ell }\ne 0$. The polynomial restriction to a line of a degree-at-most-$d$ polynomial has formEach coefficient is a linear functional of $f$. Setting all $d+1$ coefficients to zero is precisely the condition that $f$ vanish identically on $\ell$.

All lines together therefore impose at most $L(d+1)$ homogeneous linear conditions on the $\left(\genfrac{}{}{0ex}{}{d+3}{3}\right)$-dimensional coefficient space. A nonzero solution exists wheneverTake $d=\lceil \sqrt{6L}\rceil$. Then $d^2\ge 6L$, so this strict inequality holds. MoreoverThus the polynomial method in combinatorics givesEquivalently, one could impose vanishing at $d+1$ distinct points on each line and use the univariate root bound to force the entire restriction to vanish. For an empty line family, the constant polynomial one supplies vacuous vanishing; the strict degree comparison is understood for nonempty families.