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I was reading this answer, which says that:

In his Master's Thesis, Merlin Carl has computed a polynomial that is solvable in the integers iff ZFC is inconsistent. A joint paper with his advisor Boris Moroz on this subject can be found at http://www.math.uni-bonn.de/people/carl/preprint.pdf.

Note that the link is dead. Emil Jeřábek provided an alternate link here: A polynomial encoding provability in pure mathematics (outline of an explicit construction).

That phrase "solvable in the integers iff $\mathsf{ZFC}$ is inconsistent". If $\mathsf{ZFC}$ is inconsistent, then of course the polynomial is solvable in the integers - every statement in the model is true! So it seems to be just a fancy way of saying that the polynomial is not solvable in the integers.

I believe I'm missing some subtleties here, so I would like to have someone address this confusion of mine.

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  • $\begingroup$ Maybe ZFC is inconsistent but a weaker theory in which you can still do most math (i.e. Zermelo set theory ZC) is consistent? $\endgroup$
    – Monroe Eskew
    Commented Mar 22, 2022 at 9:48
  • $\begingroup$ Good link: projecteuclid.org/journals/… $\endgroup$
    – Sean Eberhard
    Commented Mar 22, 2022 at 9:55
  • $\begingroup$ It means that there is an explicit polynomial $P$ "encoding provability of inconsistency" in the sense that in principle you can check whether ZFC is consistent by just checking whether $P(z) = 0$ for all integers $z$ in turn. This is completely different from just saying "hey check it out $x^2 + 1 = 0$ doesn't have solutions in integers, assuming ZFC is consistent of course, because if not then anything goes". $\endgroup$
    – Sean Eberhard
    Commented Mar 22, 2022 at 10:03
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    $\begingroup$ You are mixing up theory and metatheory. It is perfectly possible to be in a model of ZFC in which the statement “ZFC is inconsistent” is true. (Moreover, I haven’t read the thesis, but the equivalence is probably proved in something much weaker than ZFC anyway, like PA.) The statement means what it literaly says: there is a polynomial $p\in\mathbb Z[\vec x]$ such that $(\exists\vec x\in\mathbb Z\,p(\vec x))\leftrightarrow\neg\mathrm{Con_{ZFC}}$. $\endgroup$
    – Emil Jeřábek
    Commented Mar 22, 2022 at 10:04
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    $\begingroup$ There is an explicitly-known Turing machine that halts iff ZFC is inconsistent. This is a problem of a similar nature. Imagine a Turing machine that will check values of the polynomial, and will halt when it finds a root... $\endgroup$
    – David Roberts ♦
    Commented Mar 22, 2022 at 10:09

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Edit: After I wrote this, the paper was posted in the commentary. What I wrote below is what is meant, however they use $\mathsf{GBC}$ instead of $\mathsf{ZFC}$. Since these two theories are equiconsistent and have the same first order consequences, this technical difference is immaterial.

However, when people say things like this there are a number of ways to interpret it, none of which are trivial. If I had to guess, probably what is meant is something like ``there is a polynomial $p(x)$ (often concretely computed) and $\mathsf{ZFC}$ proves ($\exists x \in \mathbb Z$ $p(x) = 0$ iff $\mathsf{ZFC}$ is inconsistent) ". By the second incompleteness theorem even if $\mathsf{ZFC}$ is consistent it cannot prove this fact so the equivalence is not trivial. Moreover, there are models on $\mathsf{ZFC}$ which think $\mathsf{ZFC}$ is consistent, and in such models there are no integer solutions to $p(x)$ and other models of $\mathsf{ZFC}$ which think that $\mathsf{ZFC}$ is inconsistent and in such models some (necessarily nonstandard) integer will satisfy $p(x)$.

The point is usually that the kinds of coding that lead to the incompleteness theorems can in fact be coded into surprisingly concrete polynomials. In the case of $\mathsf{PA}$ there are many examples of such coming from MRDP/Matiyasevich's theorem. In the wikipedia article on diophantine equations this is mentioned: https://en.wikipedia.org/wiki/Diophantine_set, see ``further applications".

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    $\begingroup$ I think it's worth noting that the theory that proves that the polynomial has a root iff $\mathsf{ZFC}$ is inconsistent can actually be much weaker than $\mathsf{ZFC}$ or even $\mathsf{PA}$. $\endgroup$
    – James E Hanson
    Commented Mar 22, 2022 at 19:19

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