Results with short, advanced proofs or long, elementary proofs

Recently I was preparing an undergrad-level proof of (a form of) the Jordan Curve Theorem, and I had forgotten just how much work is involved in it. The proof stored my head was just using Alexander duality plus some sanity-checks on the topology of the curve in question, which is a fine approach but does require a bit of algebraic topology machinery my audience didn't have access to. The more elementary proof (or at least the one I landed on) has a straightforward idea behind it, but turning that into a proper argument required slogging through quite a few details about polygons, regular neighborhoods, etc. Similarly, I couldn't help noticing just how much of a pain it is to prove the 2- and 3-dimensional versions of Stokes' theorem without some notion of manifolds, let alone the usual Stokes' theorem in some suitable setting.

Those are both elementary examples, but it got me thinking about the general topic of results that have very rough proofs from more elementary principles but have much clearer and smoother proofs with some more advanced background. Specifically, what are some examples of results from more advanced or narrow topics of mathematics can vastly simplify or explain in retrospect theorems that are encountered and proved laboriously in less specialized or more common areas of math? (If it helps clarify what I'm trying to get at, another example in my mind is May's "Concise Course in Algebraic Topology," which I think of as having the premise of, "So, now that you've gone through the standard intro algebraic topology course, here's what was secretly going on behind the scenes the whole time.")

• Szemerédi's theorem: the original proof is a very complicated, but elementary, induction; Furstenberg's ergodic theory based proof is much shorter but uses some existing theory. Nov 5 '21 at 15:09
• I think many topological results will fit this bill. For instance the fact that $\mathbb R^n$ are not homeomorphic for distinct $n$ has several "elementary" proofs which are rather technical, but can be also proven easily with help of homology. Nov 5 '21 at 15:16
• @LSpice, yes it can be done. You would never want to prove associativity of $Z/nZ$ by defining it to be $\{0,\ldots, n-1\}$ and the binary operation of addition followed by taking remainder. Instead, you should take the monoid of all words over X\cup X^{-1} and define an equivalence relation generated by deleting or inserting occurences xx^{-1} and x^{-1}x. Then you can prove that the equivalence classes form a group with the correct universal property in a few lines. Then you can observe that every word has a reduced form and prove by induction on the number of reductions uniqueness. Nov 5 '21 at 15:42
• I’m pretty sure you can find it formalized somewhere using the Diamond Lemma, aka Newman's Lemma, aka Church-Rosser condition, aka complete rewriting systems. I have a typed version I give students when I cover free groups Nov 5 '21 at 15:48
• Maybe Hindman's theorem? mathoverflow.net/questions/360924/… Nov 5 '21 at 20:54

The associativity of the group law on an elliptic curve can be proved in an elementary way by explicitly manipulating algebraic expressions, but this is not very enlightening. By using more advanced geometric ideas, one can prove associativity more conceptually.

• Thanks, that's a great example. If I remember correctly, the approach in Silverman derives it from Riemann-Roch (which I think is what the top answer in the linked question is referring to, though I don't have my copy of Silverman at hand at the moment). Definitely more motivated than just presenting a formula as a fait accompli, and I think elliptic curves are often encountered well before general algebraic geometry. Nov 5 '21 at 23:07

A famous example is the Abel-Ruffini theorem.

In this video, Fields medallist Richard Borcherds introduces Galois Theory and its background, mentioning that the proof by Ruffini was quite cumbersome and not quite clear, while Abel's was neat and short:

Ruffini had a sort of 500 page proof of it, except no one's really quite sure whether it was a proof of it or not and they sort of suspect it wasn't, and a bit later Abel came along and gave a very clear 6-page proof of it.

Instead of more cumbersome, geometry-based traditional methods (see e.g. this enjoyable YouTube video by Veritasium covering the Italian history of the cubic equation), knowledge on Galois theory makes much easier to see that there is no solution to quintic equations and upwards, by the analogy that their corresponding Galois group is not solvable, i.e. can't be split into Abelian groups.

• Did V. Arnold not give a nice elementary exposition to the insolubility of the 'general' quintic (i.e. without pinning a concrete quintuple of coefficients)? web.williams.edu/Mathematics/lg5/394/ArnoldQuintic.pdf Nov 7 '21 at 16:38
• Nice! feel free to edit that into the answer, and thanks a lot for sharing Nov 7 '21 at 17:55
• My comment is intended to argue that your answer does not give an example where the elementary proof is long or tedious. Nov 7 '21 at 21:24
• Fair enough. I was thinking chronollogically and not considering elementary proofs that came afterwards. My bad Nov 7 '21 at 21:42

Brouwer's theorem is immediately obtained as a 'by-product' of the development of integration on manifolds. The 'elementary' alternative is to prove and use Sperner's Lemma, which (to me at least) seems a more tedious road.

• While Sperner's Theorem certainly works, isn't the standard proof of Brouwer via noting that $\text{id}:H^{n-1}(S^{n-1}) \to H^{n-1}(S^{n-1})$ can't factor through $H^n(D^n) = 0$? I'm considering singular rather than de Rham cohomology above, but I think most students encounter de Rham cohomology and differential forms roughly around the same time. Nov 7 '21 at 18:11
• @anomaly That approach and the one via Stokes theorem are closely related. Nov 7 '21 at 18:33
• @anomaly Students going towards applied analysis will know Brouwer's theorem, as it leads them to the Schauder fix point theorem, which is used a lot to prove existence of solutions for all kinds of problems. But usually they know nothing about cohomology (at least not under the name) and may only have seen differential forms once from a distance. So to them the standard proof often is the one via Stokes theorem, though written explicitly, without differential forms.
– mlk
Nov 7 '21 at 18:59
• @ThibautDemaerel: Of course--- and maybe the equivalence of de Rham and singular cohomology is another example of the sort of thing I was talking about the original post, at least in terms of integration of forms to detect nonzero cohomology classes. Nov 7 '21 at 23:15