2

Proposition: In a ring $R\ni 1_R$ every proper ideal is contained in a maximal ideal.

The proof uses Zorn's Lemma.

Notation: $I\triangleleft R$ means that $I$ is and ideal of $R$

Proof:

Let $I\triangleleft R$, $I\subsetneq R$ be a proper ideal of $R$. We define $S:=\{J\triangleleft R\ :\ I\subseteq J\subsetneq R\}$ and consider the poset $(S,\subseteq)$. $S\ne\emptyset$ since $I\in S$. The conditions for Zorn's lemma are satisfied here because if $C$ is a chain of $S$, the union $\bigcup\limits_{J\in C}J$ is an ideal and an upper bound. (This is the part I'm confused about)

By the lemma $\exists M\in S$ a maximal element. Let's show that $M$ is a maximal ideal:

Let $N\subsetneq R$ be an ideal that contains $M$. Since it contains $M$ it also contains $I$ so we have $N\in S$.

But $M$ is a maximal element of $S$ so $N=M\ \square$

I get that the union $\bigcup\limits_{J\in C}J$ is an ideal because it the union of increasing nested sets, so the last one is an ideal (provided that it's a finite union). But what if it is an infinite or even uncountable union. Is it still true in that case that the union of nested ideals is an ideal? If it is not an ideal then it is not an upper bound of the chain and we cannot apply the lemma...

Community
  • 1
John Cataldo
  • 2,587
  • None of the elements of $S$ contain $1$. The union of the $J$'s still doesn't contain $1$ and therefore can't be $R$. The union of nested ideals is an ideal. If $a,b$ are in the union then they belong to some common $J$. Therefore, $a+b,ab$ belong to that same $J$ and therefore, to the union. –  Apr 04 '18 at 13:33
  • ok so 1 is not in the union. But that's not enough – John Cataldo Apr 04 '18 at 13:34
  • Added why it is an ideal. –  Apr 04 '18 at 13:35
  • if a,b are in the union they could belong to different J's – John Cataldo Apr 04 '18 at 13:39
  • 1
    It is nested, if they belong to different $J$'s they do belong also to the largest of the two $J$'s. –  Apr 04 '18 at 13:40

2 Answers2

2

Yes, the union of nested ideals is an ideal. If $a\in J$ and $b\in J'$ for some $J,J'\in C$, take $J''\in C$ such that $J,J'\subset J''$. Then $a+b\in J''\subset\bigcup_{I\in C}I$.

José Carlos Santos
  • 427,504
2

Just check all the axioms. If $x,y\in \bigcup_{J\in C}J$, then $x\in J_\alpha$, $y\in J_\beta$ for some $J_\alpha,J_\beta\in C$. Since $C$ is totally ordered, either $J_\alpha\subset J_\beta$ or $J_\beta\subset J_\alpha$. WLOG assume $J_\alpha\subset J_\beta$. Then $x,y\in J_\beta$, so $x-y\in J_\beta\subset \bigcup_{J\in C}J$, $xy\in J_\beta\subset \bigcup_{J\in C}J$ and if $r\in R$ then $rx\in J_\beta\subset \bigcup_{J\in C}J$, so $J$ is an ideal.

Aweygan
  • 23,232