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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.

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What do you mean by "bounded away from 1"? – user26857 Jul 12 '12 at 12:40
@user26857: I don't remember any longer, but it probably had to do with probability in choosing a random element. (I just now realised you likely aren't the user26857 I intended anymore.) – Vandermonde Nov 20 '14 at 6:18
up vote 19 down vote accepted

$\mathbb{F}_p \times\mathbb{F}_q$ 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 $0$.

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Wow, you guys are good -- I might have hit upon that, but not this fast! – Vandermonde Jul 10 '12 at 1:41

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 $p\mathcal{O}$.

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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 $\mathfrak{m} \subseteq R$.

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