I have been thinking the following question for several days:

Let $\mathcal{L}$ be the first-order language of theory of rings. Also let $\exists x \forall y \varphi(x,y)$ be a sentence of $\mathcal{L}$ with $\varphi(x,y)$ quantifier-free. Consider the infinite algebraic extension L of $\mathbb{Q}$ and finite subextensions $\mathbb{Q} \subseteq K_i \subset L$ for all i .

From field theory we know

L is algebraic extension of $\mathbb{Q}$ if and only if L is direct limit of its finite subextensions.

Then L=$\varinjlim K_i=\bigcup_iK_i$. Now let $K_i \models \exists x \forall y \varphi(x,y)$ for all i and $\mathbb{Q} \models \exists x \forall y \varphi(x,y)$.

Is it necessary that $L \models \exists x \forall y \varphi(x,y)$ ?

I try to give a counterexample but failed.

Any hints or answers are welcomed. Thank you!


I believe the following is a counterexample:

Take $L$ to be the algebraic closure of $\mathbb{Q}$, $\overline{\mathbb{Q}}$. We can find a sequence of fields $K_i$ ($i\in\mathbb{N}$) with $K_0=\mathbb{Q}, K_i\subseteq K_{i+1}$, each $K_i$ is a finite extension of $\mathbb{Q}$, and $\bigcup K_i=\overline{\mathbb{Q}}$.

Now in each $K_i$ there is some element which doesn't have a square root; this is a $\exists\forall$-sentence which fails in $\overline{\mathbb{Q}}$.

  • $\begingroup$ Thank you!! But what if adding $\bar{\mathbb{Q}} \models \exists x \forall y \varphi(x,y)$ as assumption? It's just a further question. $\endgroup$ – Max CYLin Feb 5 '18 at 18:04
  • $\begingroup$ @MaxCYLin Good question - I'm sure we can still get a counterexample, but off the top of my head I don't see it. $\endgroup$ – Noah Schweber Feb 5 '18 at 18:30
  • $\begingroup$ @MaxCYLin So just to clarify, your new question is whether there is an intermediate field $\mathbb{Q}\subseteq L\subseteq \overline{\mathbb{Q}}$ and a $\exists\forall$-sentence $\psi$ such that $\overline{\mathbb{Q}}\models \psi$ and $K\models \psi$ for all $\mathbb{Q}\subseteq K\subseteq L$ such that $K/\mathbb{Q}$ is finite, but $L\models \lnot\psi$? $\endgroup$ – Alex Kruckman Feb 5 '18 at 21:59
  • $\begingroup$ @AlexKruckman Correct! And Andreas Blass has provided another counterexample for the further question. $\endgroup$ – Max CYLin Feb 6 '18 at 4:07

Here's a counterexample for the version of the question asked in the comments on Noah Schweber's answer. Let $\exists x\forall y\,\phi(x,y)$ say "either there is an $x$ with no square root or $2$ has a cube root." (So $\phi(x,y)$ is $(y^2\neq x)\lor(x^3=2)$.) Let $L$ be the field obtained from $\mathbb Q$ by iteratively adjoining square roots for everything (sometimes called the Euclidean closure of $\mathbb Q$ because you've adjoined just what you need to make Euclid's theorems of plane geometry true in the "plane" $L^2$), and let the $K_i$ be an increasing sequence of intermediate finite-degree extensions of $\mathbb Q$ with union $L$. Then $L$ doesn't satisfy $\exists x\forall y\,\phi(x,y)$; everything in $L$ has a square root there, by construction, but $2$ has no cube root. (The latter fact is usually expressed as the ruler-and-compass unsolvability of the classical problem of duplication of a cube.) But $\mathbb Q$ and all the $K_i$'s satisfy $\exists x\forall y\,\phi(x,y)$ because they don't have square roots for all their elements. And $\overline Q$ satisfies $\exists x\forall y\,\phi(x,y)$ because it contains $\sqrt[3]2$.

  • $\begingroup$ Thanks! It looks like that $\exists \forall$-sentences are more complicated than $\forall \exists$-sentences since there are not so many preservation theorems about $\exists \forall$-sentences. $\endgroup$ – Max CYLin Feb 6 '18 at 7:18

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