# Are all units of a ring associates?

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..?

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Yes, your question is addressed here. en.wikipedia.org/wiki/Unit_(ring_theory) –  nullUser Aug 9 '12 at 14:55
Sorry, missed the word "finite." Yes, you are right, there are no finite UFDs that are not fields, and hence they are uninteresting. –  Thomas Andrews Aug 9 '12 at 15:19

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.

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Well, one obvious way to treat $-1$ would be to consider it an additional prime. Then $-7$ would be the product of the two primes $7$ and $-1$. However this way the prime factorization would no longer be unique because $(-1)^3=-1$ –  celtschk Aug 9 '12 at 16:00
@celtschk So, if you want to talk about "unique factorization domains," why would you want to treat $-1$ as a prime and then break unique factorization? The point is, "units" are uninteresting in rings when talking about factorizations. –  Thomas Andrews Aug 9 '12 at 16:04

"No" for your second question( "What about the units in a UFD, may they be factored into irreducibles?")

You'll notice in the definition of a UFD it says "all nonunit, nonzero elements factor into irreducibles." In a commutative domain the product of nonunits is a nonunit, so it's impossible for a unit to be a product of irreducibles (which are by definition nonunits).

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Generally, when studying factorization theory in a commutative cancellative monoid, it often proves convenenient to ignore units by working in the reduced monoid obtained by factoring out the unit group. This is the structural form of the congruences that you mention.

Many properties of domains are purely multiplicative so can be described in terms of monoid structure. Let R be a domain with fraction field K. Let R* and K* be the multiplicative groups of units of R and K respectively. Then G(R), the divisibility group of R, is the factor group K*/R*.

• R is a UFD $\iff$ G(R) is a sum of copies of $\rm\:\mathbb Z\:.$

• R is a gcd-domain $\iff$ G(R) is lattice-ordered (lub{x,y} exists)

• R is a valuation domain $\iff$ G(R) is linearly ordered

• R is a Riesz domain $\iff$ G(R) is a Riesz group, i.e. an ordered group satisfying the Riesz interpolation property: if $\rm\:a,b \le c,d\:$ then $\rm\:a,b \le x \le c,d\:$ for some $\rm\:x\:.\:$ A domain $\rm\:R\:$ is called Riesz if every element is primal, i.e. $\rm\:A\:|\:BC\ \Rightarrow\ A = bc,\ b|B,\ c|C\:,\:$ for some $\rm\:b,c\in R\:.$

For more on divisibility groups see the following surveys:

J.L. Mott, Groups of divisibility: A unifying concept for integral domains and partially ordered groups, Mathematics and its Applications, no. 48, 1989, pp. 80-104.

J.L. Mott, The group of divisibility and its applications, Conference on Commutative Algebra (Univ. Kansas, Lawrence, Kan., 1972), Springer, Berlin, 1973, pp. 194-208. Lecture Notes in Math., Vol. 311. MR 49 #2712

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