4
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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 $\overline{\mathbb{Q}}$.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $\sqrt[3]2$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.