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Inspired by this question and its answer, I am curious whether or not Fermat's last theorem holds in the Grothendieck ring of the ordinals under Hessenberg (commutative) operations.

The excellent answer to the other question by Emil Jeřábek uses the fact that the sum of the coefficients of an ordinals Cantor normal form is $0$ iff the ordinal itself is $0$, which is no longer true in the Grothendieck ring. The 'Cantor normal forms' now have integer coefficients instead of natural number coefficients, so for example $\omega-1$ is nonzero with coefficients summing to $0$.

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    $\begingroup$ An idea: for each ordinal $\alpha$ let $X_\alpha=\omega^{\omega^\alpha}$. Isn't the ring you are considering isomorphic to $\mathbb Z[X_\alpha]$? If so, we can then apply Mason-Stothers theorem (a.k.a. abc for polynomials) to show that any counterexample to FLT has to have all polynomials constant in every variable, i.e. finite, so we can apply normal FLT. I can write details in an answer later. $\endgroup$
    – Wojowu
    Commented Jun 26, 2018 at 15:21
  • $\begingroup$ @Wojowu You are correct, and thank you. I asked Emil below and would like to ask you too, is it obvious whether FLT holds in the omnific integers? $\endgroup$
    – Alec Rhea
    Commented Jun 27, 2018 at 5:40

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It does hold. Assume for contradiction that $x^n+y^n=z^n$ is a solution with $x,y,z$ nonzero, and put $w=xyz$. It suffices to find a ring homomorphism $f$ to $\mathbb Z$ such that $f(w)$ is nonzero, as then $f(x)^n+f(y)^n=f(z)^n$ is a nontrivial solution in $\mathbb Z$ contradicting the Fermat–Wiles theorem.

Now, your ring is isomorphic to the ring of multivariate integer polynomials with the indeterminates being ordinals of the form $\omega^{\omega^\alpha}$. We thus get an evaluation homomorphism to $\mathbb Z$ whenever we fix all the indeterminates to some integers.

So, it suffices to show that a nonzero integer multivariate polynomial (the $w$ above) does not compute a constant zero function on $\mathbb Z$. This follows by induction on the number of variables, using the fact that a nonzero univariate polynomial can only have finitely many integer roots.

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  • $\begingroup$ Thank you! Not to create a moving target, but is it obvious to you wether FLT holds in the omnific integers? $\endgroup$
    – Alec Rhea
    Commented Jun 27, 2018 at 5:38
  • $\begingroup$ It fails. The omnific integers have real closed fraction field. $\endgroup$
    – Emil Jeřábek
    Commented Jun 27, 2018 at 6:56
  • $\begingroup$ Ah, I didn't realize that was a sufficient condition for failure, very nice. $\endgroup$
    – Alec Rhea
    Commented Jun 27, 2018 at 13:33
  • $\begingroup$ @AlecRhea More explicitly, since $1^3 + (\sqrt[3]{7})^3 = 2^3$, FLT would fail if $\sqrt[3]{7}$ were rational. But Theorem 32 of ONAG says that every number is the quotient of two omnific integers. In particular, $\sqrt[3]{7}$ is the quotient of the two omnific integers $\omega\sqrt[3]{7}$ and $\omega$. Thus $\omega^3 + (\omega\sqrt[3]{7})^3 = (\omega 2)^3$ is an omnific counterexample to FLT. $\endgroup$
    – Timothy Chow
    Commented Nov 16, 2021 at 4:04
  • $\begingroup$ @TimothyChow It's things like this that keep making me think the omnific integers are the 'wrong' discrete subring of the surreals to be looking at, at least for some purposes. $\endgroup$
    – Alec Rhea
    Commented Nov 16, 2021 at 12:33

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