Showing $x \in R$ is invertible if and only if $x$ is not contained in any maximal Ideal

Question

Here $R$ is commutative.

$\Rightarrow$ is fairly easy to prove as every Ideal containing a unit must be the entire Ring so it cannot be maximal. But I'm having some trouble with $\Leftarrow$, this is what I got so far:

We want to get a contradiction, so assume $x$ is not a unit so $xR$ is a proper Ideal, let $A$ be the set of proper Ideals containing $x$, $A$ is not empty and partially ordered by Inclusion. I want to use Zorn's lemma here but I cant figure out what the supremum of a chain in $A$ is, I know that the union of the Ideals in the chain is an Ideal containing $x$ but I cant figure out why it must be a proper Ideal i.e. not $R$.

This standard proof is discussed here many times, e.g. see this answer in the dupe for full details. Note $(x)$ is a proper ideal when $x$ is a nonunit, so it is a special case of the dupe. Note also that this fails for rings without identity. Please search for answers before posing questions. — Bill Dubuque, Apr 17 '21 at 20:14

score 1 · Accepted Answer · answered Apr 17 '21 at 19:16

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Let $\mathscr{C}$ be the a chain of proper ideals containing $x$, and let $U=\bigcup\mathscr{C}$. If $1_R\in U$, then there is some $I\in\mathscr{C}$ such that $1_R\in I$, which is impossible.

answered Apr 17 '21 at 19:16

Brian M. Scott

616,228

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Please strive not to add more dupe answers to dupes of FAQs. – Bill Dubuque Apr 17 '21 at 20:16

Showing $x \in R$ is invertible if and only if $x$ is not contained in any maximal Ideal

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