The Square-Root Law Does Not Hold in the Presence of Zero Divisors

Let $R$ be a finite ring (with unit, not necessarily commutative) and define the paraboloid $P = \{(x_1, \dots, x_d)\in R^d|x_d = x_1^2 + \dots + x_{d-1}^2\}.$ Suppose that for a sequence of finite rings of size tending to infinity, the Fourier transform of $P$ satisfies a square-root law of the form $|\hat{P}(\chi)|\leq C|R|^{-d}|P|^\frac{1}{2}$ for some fixed constant $C$ (for instance, if $R$ is a finite field, this bound will be satisfied with $C = 1$). Then all but finitely many of the rings are fields. Most of our argument works in greater generality: let $f$ be a polynomial with integer coefficients in $d-1$ variables, with a fixed order of variable multiplications (so that it defines a function $R^{d-1}\rightarrow R$ even when $R$ is noncommutative), and set $V_f = \{(x_1, \dots, x_d)\in R^d|x_d = f(x_1, \dots, x_{d-1})\}$. If (for a sequence of finite rings of size tending to infinity) we have a square root law for the Fourier transform of $V_f$, then all but finitely many of the rings are fields or matrix rings of small dimension. We also describe how our techniques let us see that certain varieties do not satisfy a square root law even over finite fields.