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Do $\mathbb{C}$ and $\mathbb{A}$ (algebraic complex numbers) satisfy the same sentences in the language of fields? This was an optional question on an assignment for a first course in logic, model theory and set theory.

From my research on Google, I've found a notion of "elementary equivalence" of language structures which then guided me towards "Tarski-Vaught Principle", which says the following:

Tarski-Vaught Principle: let $M$ be an $L$-structure and $N$ a substructure of $M$. Then $N$ is an elementary substructure of $M$ iff $M \vDash \exists x \phi(x, b_1, \dots, b_n)$ then there is an element $a$ in $N$ such that $M \vDash \phi(a, b_1, \dots, b_n)$, where $\phi(x, y_1, \dots, y_n)$ is a first order formula over $L$ and $b_1, \dots, b_n$ are in $N$.

Intuitively, to me this means that $\mathbb{A}$ should satisfy any sentence that $\mathbb{C}$ does: since $\mathbb{A}$ is defined to be closed under algebraic operations, and the only sentences that can be made in this language are sentences about algebraic operations on elements. The only bad sort of sentence I can think of is something like $\exists x \ \text{s.t.} \ x \ \text{is not a root of any polynomial}$ but this would really require infinitely many sentences to write in the language of fields.

As requested in the comments, I will mention that my course does not cover anything in particular about algebraically closed fields or the theory of algebraically closed fields. I've seen in an exercise that the theory of algebraically closed fields is not finitely axiomatisable, but beyond this I have no special knowledge of the theory of algebraically closed fields.

Tom Misch
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    Your intuition is correct, but is exactly making this precise that is the challenge. What facts about the theory of (algebraically closed) fields have you learned so far? (edit those into your post, as they are relevant to the question) – Mark Kamsma Apr 04 '22 at 14:25
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    hi Tom. on top of Mark's question there's one other thing I'd like to point out; you are asked only to show that $\mathbb{A}$ and $\mathbb{C}$ satisfy the same first-order sentences, ie that they are elementarily equivalent. the Tarski-Vaught principle actually shows something stronger, that $\mathbb{A}$ is in fact an elementary substructure of $\mathbb{C}$. this is in general a stronger thing than their being merely elementarily equivalent. (for an example, the structures $B=(\mathbb{N},<)$ and $A=(\mathbb{N}\setminus{0},<)$ are isomorphic, and hence ... – Atticus Stonestrom Apr 04 '22 at 15:25
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    ... elementarily equivalent, but can you see that $A$ is not an elementary substructure of $B$?) anyway, in this case it is true that $\mathbb{A}$ is an elementary substructure of $\mathbb{C}$, but I wanted to point out that this is a priori a stronger fact than the one you're asked to prove in your exercise – Atticus Stonestrom Apr 04 '22 at 15:26
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    I'll chime in with a meta-observation: certainly some special knowledge of algebraically closed fields is needed. For an arbitrary structure $\mathcal{M}$, let $\mathcal{M}{alg}$ be the substructure (but see below) of $\mathcal{M}$ consisting of all $m\in\mathcal{M}$ such that there is some formula $\varphi(x)$ with $\vert\varphi^\mathcal{M}\vert<\infty$ and $\mathcal{M}\models\varphi(m)$. In general, $\mathcal{M}{alg}$ can be quite different from $\mathcal{M}$ - for instance, if $\mathcal{M}=(\mathbb{Z};<)$ then $\mathcal{M}_{alg}$ is empty! (cont'd) – Noah Schweber Apr 04 '22 at 15:51
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    So something about algebraically closed fields will be crucial here. Really the key point is that $\mathbb{A}=\mathbb{C}_{alg}$ (note that this isn't obvious since we only use polynomials in defining $\mathbb{A}$); this uses quantifier elimination, and the proof of that (if memory serves) will in fact give the full Tarski-Vaught criterion. – Noah Schweber Apr 04 '22 at 15:53
  • See https://math.stackexchange.com/questions/1563/example-of-non-isomorphic-structures-which-are-elementarily-equivalent one of the comments mentions a fact that applies here: "Any two algebraically closed fields of the same characteristic are elementarily equivalent". – Sam Apr 18 '22 at 21:15

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