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This is a major edit of the original post after receiving helpful comments.

It is often the case when one adds additional structure to make a problem more tractable. When one attempts to forget this added structure, this leads to symmetries. One then needs to look at the original solution of the tractable case and modulo cases congruent under symmetry.

The simplest example is when we study free modules. We add a basis to make the problem tractable, reducing to linear algebra. However, when attempting to make basis-free statements of linear spaces and maps, one then must talk about matrices up to similarity, and this is the congruence relation. The change of base are symmetries.

Another example is the use of spectral sequences. The original grading of the graded ring may not be amenable to computation. So we add the structure of a filtration to introduce another grading. Again, one needds to forget this structure if one wants the original grades of the ring.

Do people have other examples of such a situation?

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To the people who have downvoted this post: Why not leave a comment as to how Colin can improve the question, rather than sniping from the sidelines? – Tom LaGatta Feb 16 '10 at 3:50
One comment is that the relationship between algebra and geometry was explored long before commutative algebra and algebraic geometry - it began with Descartes and was a major theme of Klein's work. So I think that the downvotes are because the question is not particularly clear. What aspect of the relationship between algebra and geometry is being explored? – Qiaochu Yuan Feb 16 '10 at 4:41
For one thing, I don't agree that to construct an algebraic closure of R, one must explicitly choose a square root of -1. Suppose, for instance, that I define C as the splitting field of the polynomial $t^4-2$: where have I made such a choice? Similarly, I don't see how a free module with a given basis is "more algebraic" than a free module without a basis. Parts of the question are obscure; the parts that I can understand, I don't agree with... – Pete L. Clark Feb 16 '10 at 4:58
How can one define a free module algebraically without specifying a basis? If one says that a free module is freely generated by a set $X$, then one already has chosen a basis. If one says that a free module splits into the direct sum of rank one modules, then again, that is a basis. That we have a notion of free module really comes from geometry, where each direction is not any special from another. – user2529 Feb 16 '10 at 5:35
"How can one define a free module algebraically without specifying a basis?" It happens all the time. For instance, consider the set of all integers (x,y,z) in Z^3 such that x + y + z = 0. This is a free Z-module of rank 2. In this case choosing a basis breaks the S_3-symmetry, so is arguably "unnatural". As for your second comment, I don't understand how it's a response to what I said. – Pete L. Clark Feb 16 '10 at 7:40

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