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Zilber’s field $\mathbb{B}$ is a field of the same size as the complex numbers $\mathbb{C}$, which satisfies the same first-order sentences about $+$ and $\cdot$. If $\mathbb{B}$ also satisfies the same first-order sentences about $+$, $\cdot$ and $^\wedge$ (exponentiation) as $\mathbb{C}$, then there must be an isomorphism $\mathbb{B}\simeq\mathbb{C}$.

What are the simplest sentences which might plausibly distinguish the two structures?

The answer could be:

  • a statement about exponentiation which is first-order but not settled by Schanuel’s conjecture; or

  • a statement about fields which is almost first-order but uses infinite conjunctions and disjunctions, and is not settled by the countable closure property.

In either case I’m not sure what that sentence would look like.

One key reference is Jonathan Kirby’s Note on the Axioms for Zilber’s Pseudo-Exponential Fields, but it’s hard for me to see if the answer is there.

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    $\begingroup$ I'm confused by the first bullet -- isn't Zilber's exponential field essentially axiomatized by Schanuel's conjecture? Moreover, the theory of Zilber's exponential field is complete. So any first-order statement which is true of $\mathbb B$ is implied by Schanuel's conjecture. $\endgroup$
    – Tim Campion
    Commented Dec 26, 2022 at 3:25
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    $\begingroup$ Well, a very simple kind of sentence that could distinguish the two structures would be a explicit counterexample to Schanuel's Conjecture. But maybe by "plausibly", you mean "assuming Schanuel's Conjecture is true"? If Schanuel's Conjecture is true but $\mathbb{B}\not\cong \mathbb{C}$, then $\mathbb{C}$ fails "strong existential-algebraic closedness", so you can distinguish the two fields with an explicit counterexample to this condition. $\endgroup$
    – Alex Kruckman
    Commented Dec 26, 2022 at 18:06
  • $\begingroup$ Also, regarding your second bullet point: No statement about fields, even in infinitary logic, can distinguish $\mathbb{B}$ and $\mathbb{C}$, since they are isomorphic as fields (being algebraically closed fields of characteristic $0$ and cardinality $2^{\aleph_0}$). $\endgroup$
    – Alex Kruckman
    Commented Dec 26, 2022 at 18:08
  • $\begingroup$ @AlexKruckman, will you either put the strong existential-algebraic closedness as an answer, or send me a link about it so that I can do it? I would accept that answer however it gets there. $\endgroup$
    – user44143
    Commented Dec 26, 2022 at 19:19
  • $\begingroup$ @MattF. I didn't think it was worth more than a comment, because it's right there in the paper by Kirby that you linked to. See Axiom 4 on p. 2, with more details in Section 2.3, pp. 3-4. $\endgroup$
    – Alex Kruckman
    Commented Dec 26, 2022 at 19:43

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Kruckman's comment is a good answer. A counterexample to Schanuel's conjecture would be one such sentence. For example, a non-zero polynomial $p(x,y)$ with integer coefficients such that $p(\pi,e) = 0$.

The other plausible simple counterexamples are counterexamples to strong exponential-algebraic closedness, or just to Exponential Algebraic Closedness.

With current knowledge, we need at least 3 variables. So maybe there is an algebraic variety $V$, given by polynomials over $\mathbb Z$, in variables $z_1,z_2,z_3,w_1,w_2,w_3$, of dimension 3, which satisfies the freeness and rotundity properties you can find in the paper you cite, but such that in $\mathbb C$ there are no $a_1,a_2,a_3$ such that $(a_1,a_2,a_3,e^{a_1},e^{a_2},e^{a_3}) \in V$.

The simplest cases where this is open are when the projection to the $z$-variables has dimension 2.

Maybe there is an example where the polynomials are defined over $\mathbb Z[t]$ for a variable $t$, and there are solutions if and only if $t$ is real. We don't know.

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