Why can we be sure that if $A$ is a Unique Factorization Domain and has at least one irreducible element, then $A$ is infinite?
I can't see how to prove this. Does it have any connection with primes?
Many thanks!
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user26857
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It would pretty much have to have something to do with primes, since being a UFD is about primes. – Thomas Andrews Apr 27 '15 at 22:48
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1Any domain with a nonzero nonunit is infinite, since finite domains are fields. More generally, If all but finitely many elements of a ring $\rm:R:$ are units or zero-divisors (incuding $0$), then all elements of $\rm:R:$ are units or zero-divisors, see here. – Bill Dubuque Apr 27 '15 at 23:00
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Hint. Consider all powers of that irreducible element.
user26857
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And how can I be sure that, for some $n$, the $n$-th power of that irreducible isn't already in the UFD? – Apr 27 '15 at 22:52
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Oh, I see. If that happened, then that irreducible would have to be a unit and that's a contradiction, right? I mean, is it true that $a^n$ unit implies $a$ unit? – Apr 27 '15 at 22:57
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