# Associates of Elements in Commutative Rings

Definition: Let $(R, +, \cdot)$ be a commutative ring and let $a, b \in R$. Then $a$ is said to be an Associate of $b$ denoted $a \sim b$ if there exists a unit $u \in R$ such that $a = bu$. |

Theorem 1: If $(R, +, \cdot)$ is an integral domain and $a \neq 0$ then if $a | b$ and $b | a$ then $a \sim b$. |

**Proof:**Since $a | b$ and $b | a$ there exists elements $q_1, q_2 \in R$ such that $aq_1 = b$ and $bq_2 = a$. Substituting the first equation into the second yields:

(1)

\begin{align} \quad aq_1q_2 &= a \\ \quad aq_1q_2 – a &= 0 \\ \quad a(q_1q_2 – 1) &= 0 \end{align}

- Since $a \neq 0$ and $(R, +, \cdot)$ is an integral domain we must have tat $(q_1q_2 – 1) = 0$ so $q_1q_2 = 1$. Hence $q_1$ and $q_2$ are units.

- Therefore $a \sim b$. $\blacksquare$

Theorem 2: If $(R, +, \cdot)$ is an integral domain and $a, b \in R$ then $aR = bR$ if and only if $a \sim b$. |

**Proof:**$\Rightarrow$ Suppose that $aR = bR$. Then $aR \subseteq bR$ and $aR \supseteq bR$. By a theorem on the Divisors of Elements in Commutative Rings page we have that $b | a$ and $a | b$. By Theorem 1 we must have that $a \sim b$.

- $\Leftarrow$ Suppose that $a \sim b$. Then $a = bu$ for some unit $u \in R$ which shows that $a \in bR$. Similarly, $b = au^{-1}$ where $u^{-1} \in R$ is also a unit which shows that $b \in aR$. So $aR \subseteq bR$ and $bR \subseteq aR$. Hence $aR = bR$. $\blacksquare$

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