When studying UFDs I started to get confused…
If $u$,$v$ are units in $R$ then $u^{-1}$$v$ is a unit in $R$ and so $v$ = ($u$$u^{-1}$)$v$ = $u$($u^{-1}$$v$) hence u and v are associates..? Are really all units associates? So in every field all non-zero elements are associates?
What about the units in a UFD, may they be factored into irreducibles?
And also, since every UFD is an integral domain and every finite integral domain is a field, we can't really have any interesting finite UFDs? Since these are all fields..?
Best Answer
Yes, the units are all associates of $1$, so, since the relationship "$a$ is an associate of $b$" is an equivalence relationship on your ring, we can conclude that all units are associates of each other.
Units cannot be factored into irreducibles. Think of $1$ and $-1$ with the ring of integers, $\mathbb Z$. How would you "factor" $-1$? Do you treat $-7$ and $7$ as different primes, and, if so, what does it mean that $49 = (-7)(-7) = 7\cdot 7$?
With integers, we can easily solve this problem by only talking about positive integers, but in general rings, we don't have it so easy.
Yes, it is true that there are no interesting finite UFDs.