1. Something I realized yesterday when reading Wikipedia is that the fundamental groups of the complements of two homeomorphic topological spaces need not be isomorphic. Here’s an example:

The fundamental group of the complement of a circle in R^3 is isomorphic to Z.

The fundamental group of the complement of a trefoil knot in R^3 is given by the presentation <a, b | a^2 = b^3>. Clearly this is not isomorphic to Z.

2. When I was reading Algebraic Number Fields by Gerald Janusz yesterday, I came across the following rather mundane-looking result:

If A is a field and B a domain which is integral over A, then B is a field.

I was expecting an equally mundane proof based on showing that every nonzero element b in B has a multiplicative inverse. To show that would not be very difficult: we merely take the minimal polynomial of b, switch the order of the coefficients, multiply by some constant, and show that this polynomial has 1/b as a root. None of these things is difficult.

However, we had just shown the following result:

Let A subset B be integral domains with B integral over A and A integrally closed. If P is a nonzero prime ideal of B, then P intersect A is a nonzero prime ideal of A.

Therefore we prove the result like this:

If B were not a field, there would exist a prime ideal P which is nonzero (and not equal to B). By the lemma, P intersect A would be a nonzero prime ideal. Since A is a field, P intersect A = A, so 1 is in P, an impossibility.

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Hi. I'm Simon Rubinstein-Salzedo. I'm a mathematics postdoc at Dartmouth College. I'm also a musician; I play piano and cello, and I also sometimes compose music and study musicology. I also like to play chess and write calligraphy. This blog is a catalogue of some of my thoughts. I write them down so that I understand them better. But sometimes other people find them interesting as well, so I happily share them with my small corner of the world.

The fundamental group thing is cool.

For the knot case, replacing homeomorphism with isotopy seems to guarantee isomorphism of the fundamental groups. But then in R^2, the complements of (what I think are) isotopic and definitely homeomorphic spaces (0,1) and (0,\infty) have different fundamental groups. Is there another condition that would make more sense than isotopy?

And this might be nonsense since I don’t know these subjects well.

It would be very strange to except anything about the topology of the complement based on the topology of the space itself. To start with, if you can embed a space into R^n, then of course you can also embed it to *any* n dimensional manifold, or even any larger dimensional manifolds (which, needless to say, can have very complicated topology), using only a small patch of it (and most often you have embeddings which are not of this form). Also, the whole knot theory is about different (smooth) embeddings of the circle into R^3 (or S^3). A good notion of equivalence of embeddings is isotopy (homotopy between the embeddings which is an embedding at each level). IIRC, for two knots being isotopic is (almost) equivalent to having homeomorpic complements (almost = modulo orientation: there are for example two different trefoil knots, mirrors of each other).

Careful……

Careful…..

…..you’re correct, but using that logic will not work elsewhere.

Re: Careful……

What are you talking about?