@Andy Putman: I am not sure I emphasized this enough, but my stress is on language and formal changes more than on results and discoveries. As I mentioned, my motivation is: will I still be able to read geometric articles being published in a neighborhood of algebraic topology without studying too much background? For example, if I got my PhD fifteen years ago, I guess I would have struggled a bit to learn infinity categories in the spare time (I am considering to leave academia). I found it nice to include big perspective shifts in the question, which have an interest oon their own.

@Deane Yang: to my surprise, tunless I am screwing something up, there is no tag "geometry". Free to edit if I am wrong!

I have been provoking and sloppy, but I just wanted to remember that involved definitions, despite appearing clear and obvious, are not, and I guess many philosophers have devoted quite a lot of effort to define and 'explain' perception, language, reality, meaning, thinking... All of them playing a role in the question "Which piece of mathematics contribute to the understanding of physics?". Despite I can understand the core of the question when thinking about Banch-Tarski paradox, I can't when thinking to trascendental numbers or even worse, topological quantum field theory.

I'd like to contribute with a minuscule bit of philosophy. Science is not about physical reality, which is inaccessible, but about our perception of such, then elaborated by our brain. I think that mathematics deals with, or at least live within, the laws of brain thoughts, and that's why it interacts with perception. For example, $S^1$ is how our brain interprets real circles (eve if that's a lie). Many doubts now arise, the main being: is the functioning of the brain, to which mathematics is subjected, less physical than a falling body? What does "outside reality" even mean?

I support this point of view. What I find nice of mathematics is that developing something beyond the (physically useful) boundaries actually yields something within the boundaries. See for example the Galois proof of non-existence for a général radical-based formula for the degree 5 equations. That needed complex numbers, which aren't that real. Some other times, as with the axiom of choice (see accepted answer), allowing for exotic objects actually make mathematics easier; the circle, for example, does not exist in nature, since it encompasses a transcendental number...

@VictorProtsak: when we will modernize elementary school and start mathematics right from the beginning (category theory), maybe that will happen

This is a nice variation of the inverse power method.

Of course, you are right. I am quite convinced that there is an error somewhere: the adjustment I made of my 'typo' is partial and should be revisited more thoroughly. The devil is in the details...

Yes, you are right: I forgot a -1. I was identifying $x^km$ with $m$, which corresponds of course to quotienting by $x^k-1$. Why do you claim that an extension of free modules by a free module is free? It seems like mine is a counterexample, but maybe you have a simple homological proof in mind, and that's why you are puzzled by my calculation.

When it comes down to teaching, I like to give several flavors of the determinant (at least abstract, computational, and geometric). Different perspectives are useful in different ways, and I think explaining the determinant is a perfect occasion to illustrate this idea.

not sure if 1+2 implies the permutation and minors formulae easily. The latter are far more easy in practice for calculations of matrices of size $\le 4$. I agree that the mathematician's definition would be via the exterior algebra (which is really easy from a conceptual point of view). I guess (4) is the way of explaining the exterior approach without mentioning it.

I like the emphasis on misdirection, that is indeed a key ingredient of magic tricks. However, in geometric tricks (e.g. Whitney), I have the impression that 'wonder' is the essence: the author considers a spectacularly crafted object that makes the solution evident.

(false proof that every module is quasi-free: consider $M_0 = M \otimes_R \mathbb{Z}_n$. Then $M_0 \otimes_{\mathbb{Z}_n} R \simeq (M \otimes_R \mathbb{Z}_n ) \otimes_{\mathbb{Z}_n} R \simeq M \otimes_R (\mathbb{Z}_n \otimes_{\mathbb{Z}_n} R ) \simeq M \otimes_R R \simeq M$ )

I think $M$ quasi-free and fgen can be characterized quite explicitly. Note that $M \otimes_R \mathbb{Z}_n \simeq M_0$ must be fgen too. By the classification theorem, $M_0 \simeq \bigoplus_{\lambda \in \Lambda} \mathbb{Z}_{q_{\lambda}}$ with $\Lambda$ finite and $q_{\lambda} \mid n$ prime powers. This implies $M \simeq \bigoplus R/q_{\lambda} R$, that is a finite sum of quotients for primes coming from $R \to \mathbb{Z}_n$.