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Set $m=\lceil p/100\rceil$ and $d=m-1$, so that $d<p/100<p$. The dimension of a bounded-total-degree polynomial space in $n$ variables over ${\mathbb{F}}_p$ is $\left(\genfrac{}{}{0ex}{}{n+d}{n}\right)$. If $\#N$ were smaller than this dimension, evaluation at the points of $N$ would impose fewer homogeneous linear conditions than unknown coefficients. The rank-nullity theorem would give a nonzero multivariate polynomial $P$ of total degree at most $d$, vanishing on $N$.

For every $x$, select one of the promised rich affine lines in a vector space through $x$, and write it as $\ell =\{a+tv:t\in {\mathbb{F}}_p\}$ with $v\ne 0$. The polynomial restriction to a line $Q(t)=P(a+tv)$ has degree at most $d$, and at least $m=d+1$ distinct roots of a polynomial from $N\cap \ell$. By the root bound for a polynomial, $Q$ is identically zero, so $P(x)=0$. Since this works for every $x$, $P$ vanishes at all $p^n$ points of ${\mathbb{F}}_p^n$.

The Schwartz-Zippel lemma says that a nonzero multivariate polynomial of total degree $d$ has at most $d{p}^{n-1}$ zeros on this grid. Here $d<p$, so that count is strictly less than $p^n$, a contradiction. The distinction between a formal polynomial and its function on a finite field is crucial: our degree bound is what rules out a nonzero polynomial vanishing everywhere.

We have proved the stronger quantitative rich line covering bound over a finite fieldThus $\#N{\gtrsim }_n{p}^n$, as required. The implied constant may depend on the fixed dimension $n$. The very large numerical lower bound on $p$ is more than this proof needs.

For a finite collection $\mathcal{L}$ of distinct affine lines in a vector space in ${\mathbb{R}}^n$, $n\ge 2$, a joint is a point incident to $n$ lines whose direction vectors are linearly independent. The joints theorem assertsIn the customary three-dimensional formulation this is $\#J\lesssim L^{3/2}$, with three noncoplanar incident lines at every joint. We prove the general form, which includes that formulation.

Let $L=\#\mathcal{L}$ and $M=\#J$. The conclusion is immediate when $M=0$. Otherwise set $D=\lceil n{M}^{1/n}\rceil$. Suppose for contradiction that $M>LD$. Repeatedly delete any line incident to at most $D$ of the currently retained joints, deleting those joints at the same time. Each deleted line loses at most $D$ current joints, so even deleting all $L$ lines could lose at most $LD<M$ joints. Therefore the process must stop with a nonempty set $J^{\prime }$ and a line collection ${\mathcal{L}}^{\prime }$ such that each retained line contains more than $D$ retained joints. Every retained joint still has its original $n$ independent incident lines: if any line through it had been deleted, the joint would have been deleted too.

There is a nonzero multivariate polynomial of total degree at most $D$ vanishing on $J^{\prime }$, becauseChoose such a polynomial $P$ of smallest possible total degree $d\le D$. This is an application of the polynomial method in combinatorics. Every line of ${\mathcal{L}}^{\prime }$ contains more than $D\ge d$ roots of a polynomial of its polynomial restriction to a line, so $P$ vanishes identically on every such line.

At a retained joint $x$, differentiating along each of its independent line directions $v_1,\dots ,v_n$ gives $v_i\cdot \nabla P(x)=0$. Their linear independence therefore forces $\nabla P(x)=0$. Each partial derivative of $P$ vanishes on all of $J^{\prime }$ and has smaller total degree. Minimality of $d$ forces every partial derivative to be the zero polynomial. Over the real numbers, a polynomial with all partial derivatives zero is constant; a nonzero constant cannot vanish on the nonempty $J^{\prime }$. This is the required contradiction.

It follows that $M\le LD$. Since $D\le (n+1)M^{1/n}$ for $M\ge 1$,This proves the joints theorem by the pruning and minimal-degree polynomial argument.

The exponent is sharp. Take all axis-parallel lines passing through the grid $\{1,\dots ,m{\}}^n$. There are $n{m}^{n-1}$ distinct lines and $m^n$ joints; the coordinate directions span ${\mathbb{R}}^n$ at every grid point. Thus no smaller power of the number of lines can bound all joint configurations.

(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.

A family of $L\ge 1$ lines in ${\mathbb{R}}^3$ imposes at most $L(d+1)$ linear conditions on degree-at-most-$d$ polynomials, by polynomial restriction to a line. The coefficient space has dimension $\left(\genfrac{}{}{0ex}{}{d+3}{3}\right)$. Whenever $(d+2)(d+3)>6L$, a nonzero common vanishing polynomial exists. Taking $d=\lceil \sqrt{6L}\rceil$ gives degree less than $4\sqrt{L}$.