----- > [!proposition] Proposition. ([[the prime ideal principle]]) > Let $\mathcal{F}$ be a set of proper [[ideal|ideals]] of a ([[commutative ring|commutative]]) [[ring]] $R$ such that for every ideal $I$ of $R$ and $x \in R$,[^1] $( \ \ I + \langle x \rangle \notin \mathcal{F} \text{ and }(I : x) \notin \mathcal{F} \ \ ) \implies I \notin \mathcal{F}.$ Let $J$ be a maximal (by inclusion) element of $\mathcal{F}$. Then $J$ is a [[prime ideal]]. ^proposition **(b)** Let $I$ be an [[ideal]] of $R$, and let $x \in R$. Then if $I + \langle x \rangle=\langle a \rangle$ and $(I:x)=\langle b \rangle$, $a,b \in R$, then $I=\langle ab \rangle$. $\supset.$ $\langle ab \rangle= \langle a \rangle \langle b \rangle=(I+\langle x \rangle)\langle b \rangle=Ib + \langle x \rangle \underbrace{b}_{\in (I : x)} \subset I$. $\subset$. Let $i \in I$. Note that $I+\langle x \rangle=\langle a \rangle$ implies $i=ra$ for some $r \in R$. We are done if we can show $r=bc$ for some $c \in R$. Since $\langle b \rangle=(I : x)$, it suffices to show that $rx \in I$. By $I + \langle x \rangle = \langle a \rangle$, we deduce that $x=sa$ for some $s \in R$. Now $rx=ras=is \in I$. **(c)** If every [[prime ideal]] of an [[integral domain]] $R$ is [[principal ideal|principal]] then $R$ is a [[PID]]. Let $R$ be an [[integral domain]] that is not a [[PID]]. Let $\mathcal{F}$ denote the collection of proper non-principal ideals of $R$. By **(b)**, $\mathcal{F}$ satisfies the property that for every ideal $I$ of $R$ and $x \in R$, $( \ \ I + \langle x \rangle \notin \mathcal{F} \text{ and }(I : x) \notin \mathcal{F} \ \ ) \implies I \notin \mathcal{F}.$ If we can show existence of a maximal element $J$ of $\mathcal{F}$, then by the original proposition, $J$ is a [[prime ideal]] of $R$, and it is not principal, so we are done. To show $J$ exists, first note that any [[poset|inclusion-chain]] $\mathcal{C}=(I_{\alpha})_{\alpha}$ of elements in $\mathcal{F}$ is upper-bounded by the union $I=\bigcup_{\alpha}I_{\alpha}$ of the [[ideal|ideals]] in $\mathcal{C}$. Though the union of ideals is generally not an ideal (because the union of abelian groups need not be an abelian group) $I$ *is* an ideal. We claim $I \in \mathcal{F}$. Suppose not: write $I=\langle x \rangle$. Then $x \in I_{\alpha}$ for some $I_{\alpha}$ in $\mathcal{C}$, and so $\langle x \rangle \subset I_{\alpha} \subset I=\langle x \rangle$, i.e. $I_{\alpha} =\langle x \rangle$, and so $I_{\alpha} \notin \mathcal{F}$, a contradiction. Hence $I \in \mathcal{F}$, and [[Zorn's lemma]] applies to guarantee existence of a maximal element $J$ in $\mathcal{F}$. [^1]: Notation: $(I : x)=\{ r \in R: rx \in I \}$ is the [[ideal quotient]] $(I : \langle x \rangle)$. > [!proof]- Proof. ([[the prime ideal principle]]) > ~ Let $ab \in J$. Now, for every ideal $I$ of $R$ and $x \in R$, $I \in \mathcal{F} \implies ( \ \ I + \langle x \rangle \in \mathcal{F} \text{ or }(I : x) \in \mathcal{F} \ \ ) $ In particular, substituting $I=J \in \mathcal{F}$ and $x=a$: $J + \langle a \rangle \in \mathcal{F} \text{ or }(J : a) \in \mathcal{F} .$ If $J+\langle a \rangle \in \mathcal{F}$ then, by maximaxlity of $J$, $J+ \langle a \rangle=J$ and therefore $a \in J$, implying $J$ is [[prime ideal|prime]]. Otherwise $(J : a) \in \mathcal{F}$. Since $(J : a) \supset J$ and $J$ is maximal, this means $J=(J : a)=\{ r \in R : r a \in J \}.$ But the latter set contains $b$, and therefore $b \in J$, witnessing primality of $J$. ----- #### ---- #### 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 > ```