Let $A$ be a finite ring and $N$ the non-units of $A$. If $\vert N \vert \geq 2$, why $\vert N \vert \geq \sqrt{\vert A\vert }$ ?
I think that I must look for something to get a quotient that let's me the inequality, but I don't know which ideal must be the one I need. Can anyone help me?
EDIT: The ring is commutative with unit.
Hint: Let $n\in N$ be some nonzero element and use the ideal consisting of all $a\in A$ such that $an=0$.
A full solution is hidden below.
Let $n\in N$ be any nonzero element (this exists since $|N|\geq 2$) consider the $A$-module homomorphism $f:A\to A$ sending $a$ to $an$. Note that the image of $f$ is actually contained in $N$, since any factor of a unit is a unit. If $a\in\ker(f)$, then $an=0$, which means $a$ is not a unit since $n\neq 0$. Thus $\ker(f)\subseteq N$, which means that $|A/\ker(f)|=|A|/|\ker(f)|\geq|A|/|N|.$ But $A/\ker(f)\cong\operatorname{im}(f)\subseteq N$, so we get $|N|\geq|A|/|N|$, and $|N|\geq\sqrt{|A|}$ follows.
(This proof actually works without assuming $A$ is commutative. Indeed, the only step that might not work in a noncommutative ring is the assertion that the image of $f$ is contained in $N$, but that works in any finite ring. The reason is that if $an$ is a unit, then $n$ has a left inverse, which implies left-multiplication by $n$ is injective, which implies left-multiplication by $n$ is surjective since $A$ is finite, which implies $n$ also has a right inverse, which implies $n$ is a unit.)
Let $n$ be the number of zero divisors; then $|N| = n+1$ since $N$ also contains $0$.
A theorem of Ganesan proved here implies that $|A| \le (n+1)^2$ and therefore $$|N| = n + 1 \ge \sqrt{A}.$$
Here is the argument: For any $x \in A \backslash \{0\}$, the annihilator is contained in $N$ and therefore $|Ann(x)| \le n+1$. Also, the map $$A / \mathrm{Ann}(x) \rightarrow N, \; \; r + \mathrm{Ann}(x) \mapsto rx$$ is injective. Therefore $$|A| = |\mathrm{Ann}(x)| \cdot |A / \mathrm{Ann}(x)| \ge |N|^2.$$
