the multiplicative identity of a cyclic ring must be a generator
Theorem.
Let R be a cyclic ring with multiplicative identity u. Then u generates (http://planetmath.org/Generator
) the additive group
of R.
Proof.
Let k be the behavior of R. Then there exists a generator (http://planetmath.org/Generator) r of the additive group of R such that r2=kr. Let a∈ℤ with u=ar. Then r=ur=(ar)r=ar2=a(kr)=(ak)r. If R is infinite, then ak=1, causing a=k=1 since k is a nonnegative integer. If R is finite, then ak≡1mod|R|. Thus, gcd(k,|R|)=1. Since k divides |R|, k=1. Therefore, a≡1mod|R|. In either case, u=r. ∎
Note that it was also proven that, if a cyclic ring has a multiplicative identity, then it has behavior one. Its converse is also true. See this theorem (http://planetmath.org/CyclicRingsOfBehaviorOne) for more details.
Title | the multiplicative identity of a cyclic ring must be a generator |
---|---|
Canonical name | TheMultiplicativeIdentityOfACyclicRingMustBeAGenerator |
Date of creation | 2013-03-22 15:56:59 |
Last modified on | 2013-03-22 15:56:59 |
Owner | Wkbj79 (1863) |
Last modified by | Wkbj79 (1863) |
Numerical id | 16 |
Author | Wkbj79 (1863) |
Entry type | Theorem |
Classification | msc 16U99 |
Classification | msc 13F10 |
Classification | msc 13A99 |
Related topic | CyclicRing3 |
Related topic | CriterionForCyclicRingsToBePrincipalIdealRings |
Related topic | CyclicRingsOfBehaviorOne |