---- > [!definition] Definition. ([[irreducible element of an integral domain]]) > Let $R$ be a [[integral domain]]. An element $a \in R$ is **irreducible** if $a$ is not a [[unit]] and $a=bc \implies (b \text{ is a unit or } c \text{ is a unit}).$ > Else $a$ is called **reducible**. Note that $0$ is always *re*ducible. ^definition > [!equivalence] Equivalence *for nonzero elements*. > Let $a \in R$ *be nonzero* and not a [[unit]]. The following are equivalent: > 1. $a$ is irreducible; > 2. $a=bc$ implies $a$ is an [[divides|associate]] of $b$ or of $c$; > 3. $a = bc$ implies that $\langle a \rangle=\langle b \rangle$ or $\langle a \rangle=\langle c \rangle$; > 4. $\langle a \rangle \subset \langle b \rangle$ implies $\langle b \rangle=\langle a \rangle$ or $\langle b \rangle=\langle 1 \rangle$; > 5. $\langle a \rangle$ is [[maximal]] *among proper [[principal ideal|principal ideals]]* (rephrasing the previous point). ^equivalence > [!proof] Proof of Equivalence. > > > $(1) \implies (2)$. Let $a$ be irreducible; suppose $a=bc$. Then $b$ or $c$ is a unit. Suppose $b$ is a unit, then $a$ and $c$ are associates; indeed, $c | a$ by assumption and $b^{-1} a=c$ entails $a | c$. $c$ has similarly easy argument. > > $(2) \implies (3)$. Suppose $a=bc$. Recall the [[unit characterization of associates in commutative integral domains]]. Assume $a$ and $b$ are associates; write $a=ub$ for $u$ a unit. But then since [[cancellation characterization of zero division|cancellation]] holds in [[integral domain|integral domains]] and $R$ is commutative, $a=bc=ub$ implies $a=c$... hmm > > $(3) \implies (1)$. If $(a) \subset (b)$, then $a=bc$ for some $c \in R$. Then $(a)=(b)$ or $(a)=(c)$. If $(a)=(b)$ then $b=ya$ for some $y$ and substitution yields $b=ybc$ hence $1=yc$ so $c$ is a unit. Else suppose $(a)=(c)$. Then $c=sa$ for some $s \in R$, and substituting we find $c=sbc$, from which cancellation+commutativity provide $1=sb$, so $b$ is a [[unit]]. > > > $(4) \implies (5)$. This is just a restatement. > > $(5) \iff (1)$. > > $\to$. Suppose nonunit $a \neq 0$ is irreducible with $(a) \subset (b)$. Let $r \in R$. We have $ra=sb$ for some $s \in R$. In particular if $r=1$ then $a=sb$ and $s$ or $b$ is a unit. If $s$ is a unit then $b=s ^{-1} a$, witnessing $(b) \subset (a)$ (indeed, given any $x \in R$ we have $xb=(x s ^{-1})a \in (a)$) and thus $(b)=(a)$ ; else $b$ is a unit, so $1 \in (b)$, so $(b)=R$. > > $\leftarrow$. Suppose $(a)$ is maximal among proper principal ideals. Let $a=bc$. Then for any $r \in R$, $ra=(rc)b$; conclude $(a) \subset (b)$. So $(b)=(a)$ or $(b)=(1)$. If $(b)=(a)$ then for any $z \in R$, $zb=sa$ for some $s \in R$. That is, $zb=sbc$. So $z=sc$. In the special case $z=1$ we get that $c$ is a unit. Else $(b)=(1)$, and obviously $b$ is a unit. > ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```