----- > [!proposition] Proposition. ([[nonzero prime implies irreducible]]) > Let $R$ be a [[commutative ring|commutative]] [[integral domain]], and let $a \neq 0$ be [[prime element of an integral domain|prime]]. Then $a$ is [[irreducible element of an integral domain|irreducible]]. ^proposition > [!proof]- Proof. ([[nonzero prime implies irreducible]]) > ~ > (Note that Aluffi pursues this a bit differently, but I am pretty sure this proof works...) > > $a$ is prime, nonzero, and it is not a [[unit]]; we have access to all the equivalences in [[irreducible element of an integral domain]]. > > Let $b \in R$ such that $\langle a \rangle \subset \langle b \rangle$. Then $a=bc$ for some $c$ in $R$. The [[Euclid's Lemma]]-esque characterization of prime elements implies $a | b$ or $a | c$. If $a | c$ then $c=ak$ for some $k \in R$, so $a=bak$, so $1=bk$, so $b$ is a [[unit]], so $1 \in \langle b \rangle$, so $\langle b \rangle=R$, so $\langle b \rangle=\langle 1 \rangle$. > > Else $a | b$; then $b=ad$ for some $d \in R$, implying $b=bcd$, implying $1=cd$ and so $c$ is a [[unit]]; hence $\langle c \rangle=R$. This means that if we let $r \in R$ be arbitrary and consider $rb$, we can factor $r$ as $\ell c$ for some $\ell$ and instead have $rb=\ell b c$. But then $rb = \ell (b c)=\ell a$ > which shows that $\langle b \rangle \subset \langle a \rangle$. ----- #### ---- #### 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 > ```