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group of units
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(Theorem)
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Proof. If $u$ and $v$ are two units, then there are the elements $r$ and $s$ of $R$ such that $ru = ur = 1$ and $sv = vs = 1$ . Then we get that$(sr)(uv) = s(r(uv)) = s((ru)v) = s(1v) = sv = 1$ , similarly $(uv)(sr) = 1$ . Thus also $uv$ is a unit, which means that $E$ is closed under multiplication. Because $1 \in E$ and along with $u$ also its inverse $r$ belongs to $E$ , the set $E$ is a group.
Corollary. In a commutative ring, a ring product is a unit iff all factors are units.
Examples
- When $R = \mathbb{Z}$ , then $E = \{1,\,-1\}$ .
- When $R = \mathbb{Z}[i]$ , the ring of Gaussian integers, then $E = \{1,\,i,\,-1,\,-i\}$ .
- When $R = \mathbb{Z}[\sqrt{3}]$ , then $E = \{\pm(2\!+\!\sqrt{3})^n\,\vdots\,\,\, n\in\mathbb{Z}\}$ .
- When $R = K[X]$ where $K$ is a field, then $E = K\!\smallsetminus\!\{0\}$ .
- When $R = \{0\!+\!\mathbb{Z},\,1\!+\!\mathbb{Z},\,\ldots,\, m\!-\!1\!+\!\mathbb{Z}\}$ is the residue class ring modulo $m$ , then $E$ consists of the prime classes modulo $m$ , i.e. the residue classes $l\!+\!\mathbb{Z}$ satisfying $\gcd(l, m) = 1$ .
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"group of units" is owned by pahio.
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Cross-references: residue classes, prime classes, residue class ring, field, Gaussian integers, iff, ring product, commutative ring, inverse, multiplication, closed under, proof, ring multiplication, group, ring, units
There are 22 references to this entry.
This is version 20 of group of units, born on 2004-10-05, modified 2008-03-30.
Object id is 6301, canonical name is GroupOfUnits.
Accessed 5889 times total.
Classification:
| AMS MSC: | 13A05 (Commutative rings and algebras :: General commutative ring theory :: Divisibility) | | | 16U60 (Associative rings and algebras :: Conditions on elements :: Units, groups of units) |
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Pending Errata and Addenda
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