# Is $\mathbb C$ "exponentially algebraically closed"?

Is it known whether there is any term (with parameters) $$t$$ with exactly one free variable $$x$$ in the language of exponential fields such that $$\forall x : t\neq 0$$ in $$\mathbb C$$ but $$\exists x : t=0$$ in some exponential field extension of $$\mathbb C$$?

(There seems not to be an exponential-fields tag so i put the exponential-function tag, not sure if adequate).

• The usual model-theoretic generalization of "algebraically closed" is "existentially closed". It is known that $\mathbb{C}$ is not existentially closed in the class of exponential fields. But your condition is weaker, since you only consider existential formulas of the special form $\exists x\,(t = 0)$ for a term $t$ in one free variable. (Also: I assume you allow parameters from $\mathbb{C}$?) I don't know whether $\mathbb{C}$ is exponentially algebraically closed in your sense (which is why I"m writing a comment, not an answer). Commented Nov 22, 2022 at 14:14
• @AlexKruckman When i thought about the question i didn't envision allowing parameters from $\mathbb C$. But an answer for that case would be interesting too. Commented Nov 22, 2022 at 15:03
• If you don't allow parameters from the field, your condition isn't a generalization of "algebraically closed" at all. Consider the condition on a field $K$: For any term $t$ in the language of fields with exactly one free variable $x$, if $F\models \exists x\,(t = 0)$ for some field extension $K\subseteq F$, then $K\models \exists x\,(t =0)$. This condition just says that $K$ contains the algebraic closure of its prime field (not that $K$ is algebraically closed), since terms without parameters only describe polynomials over the prime field. Commented Nov 22, 2022 at 15:26
• @AlexKruckman Actually forget that comment i was confusing things. Parameters should be allowed, i will update the question. Commented Nov 22, 2022 at 15:41