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单词 CriterionForCyclicRingsToBePrincipalIdealRings
释义

criterion for cyclic rings to be principal ideal rings


Theorem.

A cyclic ring is a principal ideal ring if and only if it has a multiplicative identityPlanetmathPlanetmath.

Proof.

Let R be a cyclic ring. If R has a multiplicative identity u, then u generates (http://planetmath.org/GeneratorPlanetmathPlanetmathPlanetmath) the additive groupMathworldPlanetmath of R. Let I be an ideal of R. Since {0R} is principal, it may be assumed that I contains a nonzero element. Let n be the smallest natural numberMathworldPlanetmath such that nuI. The inclusion nuI is trivial. Let tI. Since tR, there exists a with t=au. By the division algorithmPlanetmathPlanetmath, there exists q,r with 0r<n such that a=qn+r. Thus, t=au=(qn+r)u=(qn)u+ru=q(nu)+ru. Since ru=t-q(nu)I, by choice of n, it must be the case that r=0. Thus, t=q(nu). Hence, nu=I, and R is a principal ideal ring.

Conversely, if R is a principal ideal ring, then R is a principal idealMathworldPlanetmathPlanetmath. Let k be the behavior of R and r be a generator (http://planetmath.org/Generator) of the additive group of R such that r2=kr. Since R is principal, there exists sR such that s=R. Let a such that s=ar. Since rR=s, there exists tR with st=r. Let b such that t=br. Then r=st=(ar)(br)=(ab)r2=(ab)(kr)=(abk)r. If R is infiniteMathworldPlanetmathPlanetmath, then abk=1, in which case k=1 since k is nonnegative. If R is finite, then abk1mod|R|, in which case k=1 since k is a positive divisorMathworldPlanetmathPlanetmath of |R|. In either case, R has behavior one, and it follows that R has a multiplicative identity.∎

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