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It is sometimes the case that one can produce proofs of simple facts that are of disproportionate sophistication which, however, do not involve any circularity. For example, (I think) I gave an example in this M.SE answer (the title of this question comes from Pete's comment there) If I recall correctly, another example is proving Wedderburn's theorem on the commutativity of finite division rings by computing the Brauer group of their centers.

Do you know of other examples of nuking mosquitos like this?

33I once saw someone proving resolutions of singularities of curves by quoting Hironaka's theorem. – Richard Borcherds – 2010-10-17T15:23:37.673

18http://rjlipton.wordpress.com/2010/03/31/april-fool/ – Steve Huntsman – 2010-10-17T15:42:43.973

7Dear Mariano: The elementary proofs I've seen of Wedderburn's theorem are horrific in their complexity (hard to see the forest through the trees, so to speak), whereas the cohomological proof is simple and conceptual (and can be remembered!). Is there any "nice" elementary proof? – BCnrd – 2010-10-17T16:08:48.113

7The six color theorem as a corollary of the four color theorem. – muad – 2010-10-17T17:32:49.493

3I'm glad that Mariano was inspired rather than ticked off by my comment on our sister site. Regarding Wedderburn: I completely agree with BCnrd that the proof using Galois cohomology is the more natural one. Mariano's example is different and more interesting than, say, using Hironaka to resolve singularities of curves because he doesn't simply quote a more advanced / general result: rather, his argument proceeds "from scratch", albeit at a very high level of sophistication. – Pete L. Clark – 2010-10-17T17:49:37.607

3I do think that, for instance, Larry Washington's proof of the infinitude of primes (as came up here recently) is a good example. – Pete L. Clark – 2010-10-17T17:54:08.800

2Is it plausible that invoking Hironaka to deduce resolution of singularities for curves is non-circular? (Maybe it depends what one means by "resolution of singularities", but I am hard-pressed to imagine that one can get very far in algebraic geometry without the technique of normalization in general, let alone for curves.) – BCnrd – 2010-10-17T17:54:39.703

25Brauer groups and cohomology are certainly overkill for Wedderburn's theorem: if $D$ is a finite division algebra and $L$ is a maximal subfield, then the Noether-Skolem theorem shows that the multiplicative group of $D$ is a union of conjugates of that of $L$; hence $D$=$L$. – JS Milne – 2010-10-17T20:07:13.680

3Jim, great proof! – BCnrd – 2010-10-17T20:57:11.607

5A lot of textbook exercises in finite group theory can be killed by the classification of finite simple groups. For example every "Prove that a group of order such-and-such cannot be simple" can be answered that way. – Maxime Bourrigan – 2010-10-17T21:57:42.363

10@Maxime: I have trouble believing that such a proof is actually non-circular. Surely such proofs form a step, however easy, in the classification. – Qiaochu Yuan – 2010-10-17T21:59:55.613

2One can prove the parameterization of Pythagorean triples as a special case of Hilbert's Theorem 90 (as in Elkies' one page paper). – Ben Linowitz – 2010-10-18T00:59:37.647

3I'm not comfortable with the expression, "nuking mosquitos." It is commonly stated that the only survivors of World War Three will be the cockroaches, but I suspect they will have to share the smoking ruins with the skeeters. – Gerry Myerson – 2010-10-18T04:52:42.740

3

@BCnrd: the Wikipedia article http://en.wikipedia.org/wiki/Wedderburn's_little_theorem contains a very simple proof of Wedderburn's theorem that does not even use Noether-Skolem---it uses little more than the orbit-stabilizer theorem.

– Kevin Buzzard – 2010-10-18T19:15:33.0601Well, my wikipedia link doesn't work but you can guess what I mean. The proof on that page is apparently due to Witt. – Kevin Buzzard – 2010-10-18T19:16:13.823

@Kevin, that's the non-nuclear argument I had in mind (when you do it

ab nihilo, it seems to depend on a couple of magical observations; I explained it to my students the other day, and their faces surely made me think they thought that!); the one using Noether-Skolem is intermediate, in my eyes. – Mariano Suárez-Álvarez – 2010-10-18T20:21:17.6702Hironaka's proof is by induction on dimension. Therefore for curves it reduces considerably. You only need to define maximal contact and that a blowing-up solve the problem in dim zero. Or in other words, Applying Hironaka's theorem is circular reasoning since to get the full strength proof, which is by induction on dimension, you need to prove it first for curves. – Anna Taurogenireva – 2010-11-03T23:33:45.327

5uniqueness of prime factorization of integers follows from uniqueness of the Jordan Holder decomposition of Z/n. Riemann Roch implies the dimension of the space of polynomials of degree ≤ n equals n+1. – roy smith – 2010-12-09T05:36:59.650

31I once convinced myself the Cantor set is non empty because it is a descending intersection of non empty closed subsets of a compact set, before noticing it contains 0. – roy smith – 2011-01-29T06:48:30.883

2I think a standard way to convince oneself of simple identities in Boolean algebra is by going through the case for algebras of sets first and then applying the Stone representation theorem. – Michael Greinecker – 2011-03-11T21:53:25.607

1@Qiaochu Yuan: some of those exercises might be part of the proof of the classification theorem, but as there are countably infinite such exercises and the proof is finite, some exercises aren't. – Zsbán Ambrus – 2012-03-04T12:35:45.223

The compactness theorem of first order logic states that if every finite subset of a set of first order statements is satisfiable, then the whole set is satisfiable. As discussed in the thread http://mathoverflow.net/questions/68788/completeness-vs-compactness-in-logic , there are at least two proofs for this: a simple using ultraproducts, and a more complicated one by proving the completeness theorem, which involves introducing a syntactic deduction system and several technicalities even after that.

– Zsbán Ambrus – 2012-03-04T14:25:10.1272You can make the cohomological computation of the Brauer group more difficult by expressing it in terms of the Brauer-Severi conic, and using the Weil conjectures to find a point. – Will Sawin – 2012-12-21T19:46:58.973

3$\pi \neq 3\frac{1}{7}$ because $\sin(3\frac{1}{7})$ is transcendental by Lindemann–Weierstrass theorem and $\sin(\pi)$ is not.... – Nick S – 2014-01-25T23:55:19.920

@MarianoSuárez-Alvarez A proof of FTA bases on stability of fredholm index with small perturbation: http://arxiv.org/abs/math/0509113

– Ali Taghavi – 2015-02-06T10:16:22.3801"it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet." Clearly none of these things is true, and five (or more) people are confused. Or attempting meta-humor by using the wrong tool for the job. – hobbs – 2016-08-21T23:02:40.070

2I am sadly unhappy about the closing of this question, since I've very much enjoyed this. – Simply Beautiful Art – 2016-12-25T02:48:03.453

1@ZsbánAmbrus: The ultraproduct proof of compactness is not simpler; you can't just use the properties of ultraproducts without first proving them, and moreover the main reason people often think the Henkin construction is cumbersome is because they choose to use a very cumbersome deductive system (usually Hilbert-style). Furthermore, ultraproducts require transfinite induction even if the language is countable, whereas the Henkin model can be constructed in much weaker systems. – user21820 – 2016-12-25T08:45:51.290

http://us.metamath.org/mpegif/2p2e4.html – PyRulez – 2017-06-27T11:33:14.537

Every group of order 5 or less is cyclic or dihedral. Proof: Let G be such a group. This is a finite subgroup of SO(3). Indeed, G is a subgroup of Sym(n), where n is the order of G: but if n is 4 or less, then G is a subgroup of Sym(4); otherwise, G is a subgroup of Sym(5), hence of Alt(5) as 5, the order of G, is odd. As Alt(4), Sym(4) and Alt(5) have order strictly larger than 5, G is cyclic or dihedral. – user56097 – 2017-08-17T07:20:47.167

As a corollary of my previous comment, every group of order 3 or 5 is cyclic, as dihedral groups have even order. Thus, 3 and 5 are square free. (Actually, being square free for n is equivalent to "every abelian group of order n is cyclic". One may also prove the particular case of abelian groups by using the non-commutativity of Alt(4), Sym(4) and Alt(5) instead of their cardinalities. But as we dealt with arbitrary groups, any time we build a non-cyclic group, we know that its order divides neither 3 nor 5, e.g. 6 does not divide 5.) One can also show that 4 is twice a square-free number. – user56097 – 2017-08-17T07:40:32.980