I am having some trouble in understanding where is used the fact that the algebra is taken to be commutative in the following
Theorem. Let $\mathcal A$ be a commutative Banach algebra (over $\mathbb C)$ with identity $e$ and let $x \in \mathcal A$. They are equivalent:
- $x$ is not invertible in $\mathcal A$;
- There existes a proper maximal ideal $\mathcal J \subset \mathcal A$, to which $x$ belongs.
The proof is as follows.
Proof.
NOT $1 \implies$ NOT $2$. Suppose $x$ is invertible. Then $\mathcal A x = \mathcal A$; hence, the only ideal containing $x$ is $\mathcal A$.
1 $\implies$ 2. If $x$ is not invertible, then $\mathcal A x$ is a proper ideal of $\mathcal A$, in fact, it does not contain $e$. Then Zorn's Lemma implies the existence of a proper maximal ideal which contains $\mathcal Ax$ and so $x$. $\square$
Now, how is made use of the fact that $\mathcal A$ is commutative, except for dropping qualifications right/left before the word "ideal"? Shouldn't "$x$ invertible" be enough to get $\mathcal A x = \mathcal A$?
Remark. The fact that $\mathcal A$ is Banach is obviously immaterial here; this theorem is taken from books on Banach algebras but it is known also in less specific contexts.
Yes, the theorem could be slimmed down a lot.
In any ring $R$ with identity, $x\in R$ is a unit iff $xR=Rx=R$.
The $\implies$ direction is obvious, and the other direction is about as obvious.
Negating, we find that $x$ is not a unit iff $xR\neq R$ or $Rx\neq R$. In case $xR$ is proper, we can find a maximal right ideal containing it, and if $Rx$ is proper, we can find a maximal left ideal containing it.
Now, it is not true that $x$ is a unit iff $(x)=R$ for noncommutative rings in general. You can have $xy=1\neq yx$, in which case $(x)=R$, but $Rx\neq R$. But the $\implies $ direction still holds, of course.
Assuming commutativity just smooths out these pathologies.