# General Ring Theory/Ideals and multiplicatively closed sets

Proposition (ascending union of ideals is an ideal):

Let ${\displaystyle \Lambda }$ be a totally ordered set, and let ${\displaystyle (I_{\lambda })_{\lambda \in \Lambda }}$ be a chain of left, right or both-sided ideals of a ring ${\displaystyle R}$. Then

${\displaystyle J:=\bigcup _{\lambda \in \Lambda }I_{\lambda }}$

is a left- right- or both-sided ideal of ${\displaystyle R}$ respectively.

Proof: Indeed, let first ${\displaystyle a,b\in J}$. Then ${\displaystyle a\in I_{\lambda }}$ and ${\displaystyle b\in I_{\mu }}$ for certain ${\displaystyle \lambda ,\mu \in \Lambda }$. Set ${\displaystyle \eta :=\max\{\lambda ,\mu \}}$, then ${\displaystyle a+b\in I_{\eta }}$ since we are dealing with a chain, and consequently ${\displaystyle a+b\in J}$. The remaining property, closedness under multiplication by ${\displaystyle R}$, we prove for left ideals, since the proof for right ideals is the same, and the claim for both-sided ideals then follows by combining the other two claims (or is proved directly in the same way). Thus, suppose that the ${\displaystyle I_{\lambda }}$ are right ideals of ${\displaystyle R}$, and let ${\displaystyle r\in R}$. Then we have ${\displaystyle ra\in I_{\lambda }}$ whenever ${\displaystyle a\in I_{\lambda }}$, but for each ${\displaystyle a\in J}$ there exists some ${\displaystyle \lambda }$ such that this is the case, hence the claim. ${\displaystyle \Box }$

Proposition (existence of prime ideal not intersecting multiplicative set and containing ideal):

Let ${\displaystyle R}$ be a ring, let ${\displaystyle I}$ be a left, right, or both-sided ideal of ${\displaystyle R}$, and let ${\displaystyle S}$ be a multiplicatively closed subset of ${\displaystyle R}$ so that ${\displaystyle I\cap S=\emptyset }$. Then there exists a left, right or both-sided ideal ${\displaystyle J}$ of ${\displaystyle R}$ respectively such that ${\displaystyle J}$ contains ${\displaystyle I}$, is prime, does not intersect ${\displaystyle S}$ and is maximal (with respect to inclusion) among all ideals that do not intersect ${\displaystyle S}$.

(On the condition of the axiom of choice.)

Proof: Define ${\displaystyle \Omega }$ to be the set of all (left, right or both-sided; we'll omit this distinction for brevity in the remainder of the proof) ideals that do not intersect ${\displaystyle S}$ and contain ${\displaystyle I}$. This is a nonempty set, since ${\displaystyle I\in \Omega }$. Moreover, given any ascending chain ${\displaystyle (I_{\lambda })_{\lambda \in \Lambda }}$ in ${\displaystyle \Omega }$ with respect to inclusion, where ${\displaystyle \Lambda }$ is any totally ordered set, we obtain that

${\displaystyle K:=\bigcup _{\lambda \in \Lambda }I_{\lambda }\in \Omega }$;

indeed, ${\displaystyle K}$ is an ideal, it certainly contains ${\displaystyle I}$, and finally it does not intersect ${\displaystyle S}$. Thus, ${\displaystyle \Omega }$ is inductive with respect to inclusion, and we may choose a maximal element by Zorn's lemma, which we denote by ${\displaystyle J}$. Claim that ${\displaystyle J}$ in fact does have all the required properties. All that is left to show is that ${\displaystyle J}$ is a prime ideal. ${\displaystyle \Box }$

TODO: Distinguish between completely prime ideals and prime ideals in noncommutative rings