MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am looking for examples of theorems that may have originally had a clunky, or rather technical, or in some way non-illuminating proof, but that eventually came to have a proof that people consider to be particularly nice. In other words, I'm looking for examples of theorems for which have some early proof for which you'd say "ok that works but I'm sure this could be improved", and then some later proof for which you'd say "YES! That is exactly how you should do it!"

Thanks in advance.

share|cite|improve this question
6  
Seems related to the earlier MO question mathoverflow.net/questions/43820/extremely-messy-proofs. – Tom De Medts May 3 '12 at 10:13
3  
I don't know the original proof, but I heard that the trick of Rabinovich provided a drastic improvement of the proof of Hilbert's Nullstellensatz. – Peter Arndt May 3 '12 at 21:42
6  
It would be also interesting to hear of theorems where people didn't think that the proof could be much improved, but then were proven wrong. – David Corwin Jul 9 '12 at 6:02
2  
Many answers here so far seem to have the form "The original proof of X was very complicated, but now one can prove it as a simple consequence of Y." -- But if the proof of Y is not itself simpler than the original proof of X, should this qualify? – Louis Deaett Jun 28 '13 at 2:57
2  
@LouisDeaett: Good question. There is something, though, still nicer about having a proof follow as a simple consequence of something complicated. It shows there is some larger (unifying?) idea. – Manya Dec 3 '13 at 6:50

41 Answers 41

  1. Chirka's proof ("On the propagation of holomorphic motions", 2004) of Slodkowski's theorem ("Holomorphic motions and polynomial hulls", 1991) is much simpler. (Slodkowski's paper is not that long, but uses a lot more difficult mathematics. A number of people had previously attempted to give alternative proofs, but these turned out to contain gaps.)

  2. The original proof by Baker that repelling periodic points are dense in Julia sets of transcendental entire functions used the Ahlfors five islands theorem (a very deep result). The proof by Duval and Berteloot (Une démonstration directe de la densité des cycles répulsifs dans l’ensemble de Julia), building on work of Schwick, takes less than a page, and uses only very elementary results (notably, Zalcman's rescaling lemma for normal families). Even for rational functions, this is probably the simplest proof (simpler than the original ones of Fatou and Julia) currently in existence.

share|cite|improve this answer

Faltings' theorem (aka Mordell conjecture) can be taken as such an example. Different methods have been used so far with various difficulties.

share|cite|improve this answer

Manjul Bhargava's proof of the 15 theorem was dramatically simpler than Conway and Schneeberger's original proof.

share|cite|improve this answer
2  
JAS, I think you are confusing the 15 theorem and the 290 theorem. CS never published a full proof of the 15 theorem because it was so messy (a sketch appears in Schneeberger's Ph.D. thesis), and did not think that the 290 theorem was provable at all. MB's proof of the 15 theorem is short and the computer calculations are pretty modest by modern standards. Granted, the proof of the 290 theorem by MB and Hanke is indeed highly computational, but again, CS didn't have a proof of it at all. – Timothy Chow Nov 8 '13 at 19:24
1  
@Turbo: For the 15 theorem, see Manjul Bhargava, "On the Conway-Schneeberger fifteen theorem," in Quadratic forms and their applications, Contemp. Math. 272 (1999), 27–37. For the 290 theorem, try math.stanford.edu/~vakil/files/290-Theorem-preprint.pdf – Timothy Chow Jan 21 at 21:22

The Krylov–Bogolyubov theorem states that a continuous map on a compact metric space admits an invariant measure. The original article is 50 pages long, but nowadays this is a one-liner. This is because all the measure theory involved has been neatly repackaged in functional analytic terms.

share|cite|improve this answer

The original proof of Muller-Schupp theorem saying that finitely generated groups with context-free word problem are exactly the virtually free groups, is really involved (though nice) and uses accessibility and Stallings results on ends.

But there is a short and elementary way to prove Muller-Schupp theorem using rewriting systems, as was recently done by Volker Diekert.

share|cite|improve this answer

The Amitsur-Levitski Theorem (the standard non-commutative polynomial of order $2n$ vanishes identically on $M_n(k)$) qualifies. The original proof (1950) is messy, with no clear logical structure and takes 17 pages. The natural proof was given in 1976 by S. Rosset in a 2-pages article.

share|cite|improve this answer

Some results by Donaldson were simplified via the Seiberg-Witten invariants.

From Wikipedea: Seiberg–Witten invariants are similar to Donaldson invariants and can be used to prove similar (but sometimes slightly stronger) results about smooth 4-manifolds. They are technically much easier to work with than Donaldson invariants; for example, the moduli spaces of solutions of the Seiberg–Witten equations tend to be compact, so one avoids the hard problems involved in compactifying the moduli spaces in Donaldson theory.

See also this MO answer by Dylan Thurston.

(Added Jan 7, '16) An additional piece of information I learned today from the Zabrodsky's lecture delivered by Peter Ozsváth is about the existence of exotic smooth structure on $\mathbb R^4$. The original proof was based on Freedman theorem and on Donaldson theorem. Results based on new knot invariants such as the Knot Floer homology (and simpler combinatorial descriptions of these invariants) can replace the "Donaldson side" of the proof by a much simpler argument.

share|cite|improve this answer

DeMoivre's theorem. The pre-calc version of the proof relies on a lot of triangle geometry to establish trigonometric sum formulas and then uses induction. If you use Euler's Identity, it's a one line proof. (Of course, then there's a large amount of analysis implicit in the background.)

share|cite|improve this answer

The first proof of the Hopf invariant one theorem due to Adams is very technical. It involves decomposing $sq^{2^n}$ as a composite of secondary cohomology operations when $n\geq 4$. Then Atiyah and Adams came up with a proof that uses $K$-theory which both admired for its elegance and simplicity.

share|cite|improve this answer

One historical example that should probably be on this list is the Abel-Ruffini Theorem, which states that there is no general solution in radicals to polynomials of degree 5 and higher. Attempting to clarify the bigger picture of why the proof works may have been one of Galois's motivations for his development of what is now Galois Theory- and the proof we have now is quite insightful and illuminating.

share|cite|improve this answer

See Ostrowski's proof of Luroth theorem in Schinzel's book "Polynomials with special regard to reducibility"

share|cite|improve this answer

protected by quid Jan 13 at 13:41

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.