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I'm not sure I understand, what you mean by "proofs-based upper-level course". I'm teaching an algebra course right now and it certainly contains plenty of proofs. I don't ask students to reproduce standard proofs at home, because I know that they will find them in the literature if they want to. I set my homework in such a way that it will be much more hassle for the students to find it in the literature (yes, even using google) than to try solving the questions themselves. Also, I rely more on exams than on homework to assess the students' knowledge.
As Franz explained, there are at least two reasons, why the original paper is not the right place to look for the letter 'T'. Firstly, why would Chebyshev (sic) call a polynomial after himself (assuming that the modern 'T' refers to his name) and secondly, in Russian the name is Чебышев and starts with a 'Ч'. It must have been a German or a French who introduced the letter 'T'.
I don't see how google is supposed to have changed anything. If you asked students standard proofs, they would have always been able to find them somewhere, if not on Google, then in text books.
Olivier, while I appreciate the cleverness of the approach avoiding to pick a lattice, I don't explicitly see exactly what information these conjectures provide about the Galois module structure of the units. That is to say, it is clear that the answer depends on the Galois module structure, but the dependence is - for my liking - not explicit enough. I am now looking at some of the articles of Bley and Burns and will report back, if a manage to extract the sort of thing I am looking for. Thanks for the suggestion!
I agree that it's not a bad place to start from and it's almost certainly the place they really do start from, but it's not a good place to stay at forever, especially for maths students. That's why I was saying that the fact, that their familiar identification with $R^n$ is arbitrary should be the main insight that they take from their first abstract linear algebra course. Matthew was proposing to short change them on exactly this point, which I think is a bad idea.
In your situation, the proof of the above result will construct all elements of GL_n(Z) of order k, whenever k is cube-free. If it's not cube-free, anything you can say would be interesting. I can put this up as an answer, if you would like me to.
The striking result you are looking for reads, in full generality, as follows: a finite group has finitely many isomorphism classes of indecomposable integral representations if and only if for any prime $p$, all its Sylow $p$-subgroups are cyclic of order at most $p^2$. This result is effective in the sense that when there are finitely many isomorphism classes, the proof constructs them for you. But as far as I know, there is nothing remotely resembling a classification or even any hope of obtaining one, if this condition is violated.
Since research moves at a much faster pace than the turnover of papers in prestigious journals, particularly in fashionable areas, it is not uncommon for a paper to have two or three generations of references (i.e. A references B, which references the paper in question) before it appears in print. You should of course be completely convinced that the result you are citing is true, if you don't want your paper to break down, should the reference turn out to be wrong.
The map that assigns to a matrix the discriminant of its characteristic polynomial is a polynomial function in the entries, hence continuous. Over an algebraically closed field, the set of non-diagonalisable matrices is the pre-image of 0 under this map, since a matrix is diagonalisable if and only if it has distinct eigenvalues. The assertion now easily follows. You have two options: either working over the complex numbers (a pretty cute little argument, also only 3 lines, reduces to this case) and then you can even use the Euclidean topology on C^{n^2}, or working with the Zariski topology.
