TITLE: Why worry about the axiom of choice?
QUESTION [213 upvotes]: As I understand it, it has been proven that the axiom of choice is independent of the other axioms of set theory.  Yet I still see people fuss about whether or not theorem X depends on it, and I don't see the point.  Yes, one can prove some pretty disturbing things, but I just don't feel like losing any sleep over it if none of these disturbing things are in conflict with the rest of mathematics.  The discussion seems even more moot in light of the fact that virtually none of the weird phenomena can occur in the presence of even mild regularity assumptions, such as "measurable" or "finitely generated".
So let me turn to two specific questions:

If I am working on a problem which is not directly related to logic or set theory, can important mathematical insight be gained by understanding its dependence on the axiom of choice?
If I am working on a problem and I find a two page proof which uses the fact that every commutative ring has a maximal ideal but I can envision a ten page proof which circumvents the axiom of choice, is there any sense in which my two page proof is "worse" or less useful?

The only answer to these questions that I can think of is that an object whose existence genuinely depends on the axiom of choice do not admit an explicit construction, and this might be worth knowing.  But even this is largely unsatisfying, because often these results take the form "for every topological space there exists X..." and an X associated to a specific topological space is generally no more pathological than the topological space you started with.
Thanks in advance!

REPLY [3 votes]: Another useful example is the statement that  $M$ and $M^2$ have the same power for  any infinite set $M$. This is proved with the use of the axiom of choice.
This fact have some applications in supermathematics. If infinite set $M$ is the index set for generators  of a Grassmann-Banach algebra $G$ with an $l_1$-norm then the $l_1$-basis in $G$ has the same power as $M$. (See paper of mine in: 
Siegel W., Duplij S. and  Bagger J., eds. (2004), Concise Encyclopedia of Supersymmetry And Noncommutative Structures in Mathematics and Physics, Berlin, New York: Springer-Verlag; math-ph/0009006.) To circumvent the axiom of choice in this example one needs to consider $M$ to be a well-ordered set. But in general the construction of $G$ does not need a linear ordering in $M$. Of course, when physicists calculate Berezin's integral the power of $M$ and the axiom of choice are under the carpet of the calculations but the foundations (the flooring under the ``carpet'') need the axiom of choice.