Irreducible Elements in a Commutative Ring

Recall from The Greatest Common Divisor of Elements in a Commutative Ring page that if $(R, +, \cdot)$ is a commutative ring and $a_1, a_2, ..., a_n \in R$ then a greatest common divisor of these elements is an element $d \in R$ which satisfies the following properties:

• 1) $d | a_1, a_2, ..., a_n$.

• 2) If $c \in R$ is such that $c | a_1, a_2, ..., a_n$ then $c | d$.

We proved some very important results. First, if $d' \in R$ is an associate of $d$, that is, $d \sim d'$, then $d'$ is also a greatest common divisor of $a_1, a_2, ..., a_n$.

Furthermore, if $a, b, d \in R$ with $d \neq 0$ and $aR + bR = dR$ then $d$ is a greatest common divisor of $a$ and $b$.

Lastly, if $(R, +, \cdot)$ is a principal ideal domain and $a, b \in R$ with $a \neq 0$ and $b \neq 0$ then there exists a greatest common divisor of $a$ and $b$ of the form:

(1)

\begin{align} \quad d = as + bt \end{align}

We have established the concepts of divisors and greatest common divisors to commutative rings, and now we will extend the concept of irreducible elements to commutative rings.

Definition: Let $(R, +, \cdot)$ be a commutative ring. An element $p \in R$ is said to be Irreducible in $R$ if $p$ satisfies the following properties: 1) $p$ is not a unit. 2) If $p = ab$ then either $a$ or $b$ is a unit.

Theorem 1: Let $(R, +, \cdot)$ be a principal ideal domain and let $a, b, p \in R$. If $p$ is irreducible and $p | ab$ then either $p | a$ or $p | b$.

• Proof: Let $p$ be irreducible and let $p | ab$ and suppose that $p$ does not divide $a$. Then $p$ is not a common divisor of $p$ and $a$. The only divisors of $p$ are units and $1$. In particular, every unit is an associate with $1$, so $1$ is a greatest common divisor of $a$ and $p$. So there exists $s, t \in R$ such that:

(2)

\begin{align} \quad 1 = as + pt \end{align}

• Multiply both sides of the equation above by $b$ to get:

(3)

\begin{align} \quad b = abs + bpt \end{align}

• Since $p | abs$ and $p | bpt$ we have that $p | b$. $\blacksquare$

Theorem 2: Let $(R, +, \cdot)$ be a principal ideal domain and let $p \in R$ with $p \neq 0$. Then $p$ is irreducible if and only if $pR$ is a prime ideal.

Recall that an ideal $I$ in a ring $R$ is prime if whenever $ab \in R$ we have that either $a \in R$ or $b \in R$ and secondly, $pR \neq R$.

• Proof: $\Rightarrow$ Suppose that $p$ is irreducible and let $ab \in pR$. Then $p | ab$. By the previous theorem we have that either $p | a$ or $p | b$. So $pR$ is a prime ideal.

• $\Leftarrow$ Suppose that $pR$ is a prime ideal. Then if $ab \in pR$ we have that either $a \in pR$ or $b \in pR$. Suppose that $p$ is not irreducible. Then there exists elements $a, b \in R$ such that $a$ and $b$ are not units and:

(4)

\begin{align} \quad p = ab \end{align}

• Clearly $ab = p \in pR$ so either $p | a$ or $p | b$. If $p | a$ then there exists an element $q \in R$ such that $a = pq$. Substituting this into the equation above yields $p = pqb$. So $1 = qb$. So $b$ is a unit which is a contradiction.

• Similarly, if $p | b$ then there exists an element $q \in R$ such that $b = pq$. Substituting this into the equation above yields $p = paq$. So $1 = aq$. So $a$ is a unit which is a contradiction.

• So if $p = ab$ then either $a$ or $b$ is a unit so (2) in the definition holds. Furthermore, $p$ itself is not a unit since $pR \neq R$ so (1) in the definition holds. Hence $p$ is irreducible. $\blacksquare$