The square-root law does not hold in the presence of zero divisors

Let R be a finite ring (with identity, not necessarily commutative) and define the paraboloid $P=\left\{\right(x_1,\dots ,x_d)\in R^d|x_d=x_1^2+\dots +x_{d-1}^2\}$. 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 $\left|\hat{P}\right(\psi \left)\right|\le C\left|R{|}^{-d}\right|P{|}^{\frac{1}{2}}$ for all nontrivial additive characters ψ, with C some fixed constant (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 $R^{d-1}\to R$ even when R is noncommutative), and set $V_f=\left\{\right(x_1,\dots ,x_d)\in R^d|x_d=f(x_1,\dots ,x_{d-1})\}$. 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 can establish that certain varieties do not satisfy a square root law even over finite fields.