Finally, if we marry a scheme to an orbifold, the outcome is aI feel this way about most of algebraic geometry, but that's only because Penn has a high enough concentration of algebraic geometers that I get tired of hearing people walking around and talking about it all the time.stack. The study of stacks is strongly recommended to people who would have been flagellants in earlier times.

Also, I find myself screaming at the book quite often. But I do it in a good way; it's often some crucial insight that makes me think "why didn't anybody tell me that before?" (Of course, it's possible they did and I wasn't listening.) And sometimes it's "oh, damn, I thought I came up with that myself". For example, I just read the article on computational number theory, by Carl Pomerance, in which he explains a heuristic reason why Fermat's last theorem is true. I won't give it in full here, but it's basically the following. First, Euler showed it for n = 3. Second, consider all the positive integers which are nth powers for some n ≥ 4. The "probability" that a number

*m*is in this set is about

*m*

^{-3/4}. So replace the set of fourth-or-higher powers with a

*random*set S, which contains

*m*with probability

*m*

^{-3/4}. Then the probability that a givennumber

*n*can be written as a sum of

*two*such elements of this set S is proportional to n

^{-1/2};

*independently*, it has probability n

^{-3/4}of being in S. So the probability that n can be written as a sum of two elements of S and is also in S itself is proportional to n

^{-5/4}, and so we expect finitely many examples. This isn't quite true, because the set of fourth-or-higher powers has some nontrivial structure. But it also took a couple hundred words, instead of a couple hundred pages like the real proof.

But I figured out something like this when I was in college, and I was so proud of myself! So it saddens me to learn I'm not the only one who thought of it. (It also makes me happy, though, because the idea is due to Erdos and Ulam, and there are worse people to be imitating.)

## 3 comments:

But the same reasoning says you expect infinitely many counterexamples for n=3! Shouldn't that make you a little wary?

I've felt much that way about stacks from my little interaction with them...

Anyway, that's a nice heuristic for Fermat, except that finitely many doesn't mean zero...

The heuristic is really one that supports the earlier Mordell Conjecture / Faltings Theorem: it is suggestive, but it does not explain why there should be no solution at all (as the previous comment points out already), and it doesn't really address the dependency on n (except maybe that it suggests that the possibly finite number of solutions should decrease with n...), which is in a sense the most difficult challenge.

(Before Wiles's work, I've actually heard grouchy arithmetic geometers -- who apparently found distasteful the attraction that Fermat's problem held for the general public -- suggest that it would have been much better if someone had found a nice counterexample somewhere to show that, indeed, the Faltings-type result is best possible for those equations...)

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