0
$\begingroup$

Although the term dévissage has an specific meaning in algebraic geometry I have often heard of the expression "dévissage argument" as a proof that takes a quite general statement and, in a series of simple reductions, translates the original problem into a very easy one.

I always wanted to explain how nice this is to non mathematical audience and I was wondering if there is some sort of an accesible problem with an easy dévissage proof. Maybe some sort of classical geometry statement would be nice?

$\endgroup$
  • 2
    $\begingroup$ How about the fact that the medians in a triangle intersect in 1 point? Since an affinity preserves ratio of lengths and colinearity if it is true for 1 triangle it is true for any triangle obtained by applying an affinity to it. But the result is obviously true for an equilateral triangle by symetry and every triangle is obtained by applying an affinity to the unit equilateral triangle. It is of course a bit artificial. But i believe it really "captures" what devissage is: you show that something of interrest is preserved by some maps, and use this maps to reduce to an "obvious" case. $\endgroup$ – Ahr Jan 11 '18 at 14:12
  • $\begingroup$ Your definition of a dévissage argument seems to differ from the one given in this answer to the question what is a “dévissage” argument?. Quoting a part of this answer, "The simpler case could be very difficult and indeed the term is often used in a way that suggests the reduction is the easy step that pares the problem down to a smaller hard core", which is quite different from translating the problem into a very easy one. $\endgroup$ – J.-E. Pin Jan 11 '18 at 20:14
  • $\begingroup$ Well, the simpler case to which you reduce the problem to could be difficult or easy (I have in mind examples in both cases), but in any case it is a case where you can actually "compute" or "do" something. For instance in Riemann-Roch the devissage procedure (i.e deformation to the normal cone which to me is the "clever part") gives you two baby cases easy to compute and to deal with. On the other hand, for, say, the theorem of the cube, I agree that the "simple" case, is where the real difficulty lies. But to me both procedures are devissage. $\endgroup$ – Ahr Jan 12 '18 at 12:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.