# General Ring Theory/Ideals and multiplicatively closed sets

Proposition (ascending union of ideals is an ideal):

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

$J:=\bigcup _{\lambda \in \Lambda }I_{\lambda }$ is a left- right- or both-sided ideal of $R$ respectively.

Proof: Indeed, let first $a,b\in J$ . Then $a\in I_{\lambda }$ and $b\in I_{\mu }$ for certain $\lambda ,\mu \in \Lambda$ . Set $\eta :=\max\{\lambda ,\mu \}$ , then $a+b\in I_{\eta }$ since we are dealing with a chain, and consequently $a+b\in J$ . The remaining property, closedness under multiplication by $R$ , 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 $I_{\lambda }$ are right ideals of $R$ , and let $r\in R$ . Then we have $ra\in I_{\lambda }$ whenever $a\in I_{\lambda }$ , but for each $a\in J$ there exists some $\lambda$ such that this is the case, hence the claim. $\Box$ Proposition (existence of prime ideal not intersecting multiplicative set and containing ideal):

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

(On the condition of the axiom of choice.)

Proof: Define $\Omega$ 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 $S$ and contain $I$ . This is a nonempty set, since $I\in \Omega$ . Moreover, given any ascending chain $(I_{\lambda })_{\lambda \in \Lambda }$ in $\Omega$ with respect to inclusion, where $\Lambda$ is any totally ordered set, we obtain that

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

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