Finite commutative rings with number of units is greater than number of zero-divisors

Question

Let $R$ be a finite commutative ring with unity and $Z(R), U(R)$ denotes the set of zero-divisors and units in $R$ respectively. If $R$ is a finite local ring with unity, then $Z(R)$ is the maximal ideal of $R$ and the quotient ring $R/Z(R)$ is a field. So, every element in $u+Z(R)$ is a unit for every unit $u$ in $R$. Hence, number of units is greater than or equal to the number of zero-divisors.

If $R$ is a product of finite fields with more than $3$ elements also satisfies this. But generally, what are the finite commutative rings with unity satisfying $|U(R)|>|Z(R)|$?

Thank you in advance for your attention.

Answer 1

To expand on @TorstenSchoeneberg's comment, $R$ is a direct product of local rings $R_i$, $i = 1, \ldots, I$ (this is not true if $R$ is not commutative). We also have $|R_i| = p_i^{k_i}$ and $|Z(R_i)| = p_i^{m_i}$ (here $0$ is also a zero divisor) for some integers $k_i > m_i$.

The units in $R$ are the elements which project to units in all $R_i$, so $|U(R)| = \prod\limits_i (|R_i| - |Z(R_i)|)$, and the ratio $\dfrac{|U(R)|}{|R|}$ is $\prod\limits_i (1 - |Z(R_i)|/|R_i|)$. The ratio may very well become less than $1/2$ if we increase $I$ (= add new factors to the direct product).

However, keeping in mind the Ganesan paper condition $|Z(R_i)|^2 \ge |R_i|$, the quotient $|Z(R_i)|/|R_i|$ can get arbitrarily small, as in $\Bbb F_{p^d}[x]/(x^2)$ for $d \to +\infty$. So for every $I$ there exists a finite commutative ring $R$ with $\text{char}\, R = p$ that is a product of at least $I$ local rings and still satisfies $|U(R)|>|Z(R)|$.