Divisors and Associates of Commutative Rings

# Divisors and Associates of Commutative Rings

## Divisors

So far we have discussed the term "divisor" with regards to integers and polynomials over a field $F$. We will now extend the notion of a divisor to a general commutative ring.

 Definition: Let $(R, +, \cdot)$ be a commutative ring and let $a, b \in R$. Then $b$ is said to be a Divisor of $a$ denoted $b | a$ if there exists an element $q \in R$ such that $a = bq$.

Many of the proofs regarding divisors of integers and polynomials also applies to divisors of elements in a commutative ring.

 Theorem 1: Let $(R, +, \cdot)$ be a commutative ring. Then: a) If $a | b$ and $b | c$ then $a | c$. b) If $a | b$ then $a | bc$. c) If $a | b$ and $a | c$ then for all $x, y \in R$, $a | (bx + cy)$.
• Proof of a) Since $a | b$ and $b | c$ there exists $q_1, q_2 \in R$ such that $aq_1 = b$ and $bq_2 = c$. Plugging the second equation into the first yields:
(1)
\begin{align} \quad a(q_1q_2) = c \end{align}
• So $a | c$. $\blacksquare$
• Proof of b) Since $a | b$ there exists a $q \in R$ such that $aq = b$. Multiply both sides of this equation by $c$ to get $a(cq) = bc$. So $a | bc$. $\blacksquare$
• Proof of c) Since $a | b$ and $a | c$ there exists $q_1, q_2 \in R$ such that $aq_1 = b$ and $aq_2 = c$. So for any $x, y \in R$ we have that $a(q_1x) = bx$ and $a(q_2y) = cy$. Adding these equations yields:
(2)
\begin{align} \quad a(q_1x + q_2y) = bx + cy \end{align}
• So $a | (bx + cy)$. $\blacksquare$
 Theorem 2: Let $(R, +, \cdot)$ be a commutative ring and let $a, b \in R$. Then $aR \subseteq bR$ if and only if $b | a$.
• Proof: $\Rightarrow$ Suppose that $aR \subseteq bR$. Then since $a \in aR$ we have that $a \in bR$. Since $bR = \{ bq : q \in R \}$ we see that $a = bq$ for some $q \in R$. So $b | a$.
• $\Leftarrow$ Suppose that $b | a$. Then there exists an element $q \in R$ such that $a = bq$.
• Let $x \in aR$. Then $x = ay$ for some $y \in R$. Thus $x = b(qy)$ which shows that $x \in bR$. Therefore $aR \subseteq bR$. $\blacksquare$

## Associates

 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 3: 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:
(3)
\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 4: 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 Theorem 2 this means that $b | a$ and $a | b$. By Theorem 3 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 $aR = bR$. $\blacksquare$