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Professor Penelope Maddy remarks without elaboration in her famous 'Believing the Axioms' essay that

'It should be mentioned that the Axiom of Inaccessibles also has a few extrinsic merits. It implies that ZFC has a standard model in the iterative hierarchy, and thus, that ZFC is consistent. This last is an arithmetic fact, and the Axiom of Inaccessibles, like other axioms of infinity, also implies the solvability of new Diophantine equations. '

I have never seen this result before (The Diophantine part). This is a truly remarkable result! I am wondering where I can read more about it, specifically to see a proof. However a preliminary search on google did not lead me to anything concrete. She also references Godel's famous 'What is the Continuum Problem?' essay where he states the exact same result without proof or elaboration.

Where can I find a proof for this remarkable claim?

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This is, sadly, not as cool as it sounds.

There are two facts here:

  • "Large cardinals imply consistency."

Large cardinal axioms have high consistency strength. For example, the existence of an inaccessible implies the consistency of ZFC, the existence of a weakly compact implies the consistency of ZFC + "there are a proper class of inaccessibles," etc. While at higher levels these arguments can be quite complicated, the lower-level instances are quite simple: e.g. to show that ZFC + "there is an inaccessible" proves Con(ZFC), you just have to argue that $V_\kappa$ satisfies ZFC whenever $\kappa$ is inaccessible.

  • "Consistency can be represented by a polynomial."

This is an improvement of Godel's result that "consistency can be represented by arithmetic;" while very cool, it should be understood as not too unexpected, given Godel's theorem. Precisely, if $T$ is a recursively axiomatizable theory then there is a (multivariable) Diophantine equation $e$ such that a very weak theory - say, PA - proves that $T$ is consistent iff $e$ has no solution. (This is closely related to the MRDP theorem, which says that arbitrary c.e. sets can be represented by polynomials - see e.g. these slides.)

So putting these together, we get for example (assuming that ZFC is consistent) that there is a Diophantine equation $e$ which ZFC + "There is an inaccessible cardinal" proves has no solution, but which ZFC alone does not prove has no solution.


This is really about the consistency strength rather than the actual large cardinal. The statement that a given Diophantine equation has a solution is quite simple, and in particular is absolute between $\omega$-models of pretty much the weakest set theory you can imagine. So we can never have an unsolvability statement imply the actual existence of a large cardinal.

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  • $\begingroup$ No not quite as cool :( . [Still pretty cool though :) ] Presumably such polynomials are impossible to actually exhibit/calculate? $\endgroup$ – Elie Bergman Aug 30 '18 at 21:44
  • $\begingroup$ @ElieBergman No, they can be constructed, although they're a bit on the big side; basically, the same situation as actually writing out the Godel sentence in full. I'll see if I can track down an explicit example. $\endgroup$ – Noah Schweber Aug 30 '18 at 21:45
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    $\begingroup$ @Elie: Stefan Geschke, I believe, had a student whose masters thesis was to write a polynomial which admits roots iff ZFC is inconsistent. $\endgroup$ – Asaf Karagila Aug 30 '18 at 21:45
  • $\begingroup$ Doesn't that pose problems for incompleteness? I mean, I'm pretty sure that for a single variate polynomial, there is an algorithm to determine whether or not it has roots in finite time. $\endgroup$ – Elie Bergman Aug 30 '18 at 21:49
  • $\begingroup$ @ElieBergman I believe the paper Asaf mentions is this one. $\endgroup$ – Noah Schweber Aug 30 '18 at 21:50

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