proof that every subring of a cyclic ring is an ideal

The following is a proof that every subring of a cyclic ring is an ideal.

Proof.

Let $R$ be a cyclic ring and $S$ be a subring of $R$. Then $R$ and $S$ are both cyclic rings. Let $r$ be a generator (http://planetmath.org/Generator) of the additive group of $R$ and $s$ be a generator of the additive group of $S$. Then $s\in R$. Thus, there exists $z\in \mathbb{Z}$ with $s=zr$.

Let $t\in R$ and $u\in S$. Then $u\in R$. Since multiplication is commutative in a cyclic ring, $tu=ut$. Since $t\in R$, there exists $a\in \mathbb{Z}$ with $t=ar$. Since $u\in S$, there exists $b\in \mathbb{Z}$ with $u=bs$.

Since $R$ is a ring, $r^2\in R$. Thus, there exists $k\in \mathbb{Z}$ with $r^2=kr$. Since $tu=(ar)(bs)=(ar)[b(zr)]=(abz)r^2=(abz)(kr)=(abkz)r=(abk)(zr)=(abk)s\in S$, it follows that $S$ is an ideal of $R$. ∎

Title	proof that every subring of a cyclic ring is an ideal
Canonical name	ProofThatEverySubringOfACyclicRingIsAnIdeal
Date of creation	2013-03-22 13:30:52
Last modified on	2013-03-22 13:30:52
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	9
Author	Wkbj79 (1863)
Entry type	Proof
Classification	msc 13A99
Classification	msc 16U99
Related topic	ProofThatEverySubringOfACyclicRingIsACyclicRing