How many elements in a ring can be invertible?

Question

If $R$ is a finite ring (with identity) but not a field, let $U(R)$ be its group of units. Is $\frac{|U(R)|}{|R|}$ bounded away from $1$ over all such rings?

It's been a while since I cracked an algebra book (well, other than trying to solve this recently), so if someone can answer this, I'd prefer not to stray too far from first principles within reason.

Answer 1

$\mathbb{F}_p \times\mathbb{F}_q$ has $(p-1)(q-1)$ invertible elements, so no.

Since $\mathbb{F}_2^n$ has $1$ invertible element, the proportion is also not bounded away from $0$.

Answer 2

By the way, the number $\# U(R)/ \#R$ equals $\sum_{\mathfrak{m}} \left(1-\frac{1}{\# R/\mathfrak{m}}\right)$, where the sum ranges over all maximal ideals $\mathfrak{m} \subseteq R$.

Answer 3

The bound given by navigetor23 is tight for example in the case $R=\mathcal{O}/\langle p^2\rangle$, where $\mathcal{O}$ is the ring of integers of a finite unramified extension of the $p$-adic numbers of degree $n$: $|R|=p^{2n}$ and there are $p^n$ non-units consisting of the cosets in $p\mathcal{O}$.