# Integral domains with non-trivial group of units that are not fields

I'm looking for examples of integral domains that are not fields but at the same time have more units than just the multiplicative identity 1.

It's clear to me that by Wedderburn's little theorem, there are no finite examples of this type.

If I understand things correctly, then the ring of holomorphic functions on a domain is such an example: It's an integral domain because zeros of holomorphic functions are isolated, and it has more units than the 1-function because every constant function is invertible.

However, I can't find more examples, and I would like to see more.

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$\mathbb Z$ is a simple example since $-1$ is another unit. – Karl Kronenfeld Apr 11 '13 at 1:58

Examples arise from the following generalization of Euclid's theorem on infinitely many primes.

Theorem $\$ An infinite ring $\rm R$ has infinitely many maximal ideals if it has fewer units $\rm U = U(R)$ than it has elements, i.e. $\rm\:|U| < |R|.$

Thus, contrapositively, any infinite ring with finitely many maximal ideals has unit group having same cardinality as the ring (hence infinite). Examples abound, e.g. via localization.

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Excellent; this is really helpful in finding nice examples. For example, if you localize the integers at 2, you get an infinite local ring: the rationals with odd denominators. It's not a field since every even integer in this ring has no inverse, but the group of units is infinite (e.g. every odd integer is a unit). Very nice. – Tom Jonathan Apr 11 '13 at 10:15

The ring of Gaussian integers: http://en.wikipedia.org/wiki/Gaussian_integer

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$k [ x ]$ for a field $k$.

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In fact, all rings of algebraic integers in algebraic number fields are such examples: their unit group is always non-trivial, as they contain $-1$. And certainly they are not fields for there are prime elements within.
Regards.

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