My idea was to take the ultraproduct of $Q_n = \mathbb{Q}(\sqrt[n]{2})$ over some non-principal ultrafilter and then let $T_n$ be the (first-order) sentence "There are at least $n$ linearly independent elements over $\mathbb{Q}$". The ultraproduct is on one hand an algebraic extension of $\mathbb{Q}$ and on the other hand $T_n$ holds for the ultraproduct for each $n$, so it is of infinite degree of extension. Thus the ultraproduct cannot be an algebraic extension. But I found out that there are algebraic extensions of infinite degree, so this doesn't work at all..

Another idea I could use is the fact that algebraic extension of $\mathbb{Q}$ are countable, and then somehow prove that the ultraproduct is uncountable. But I don't know how to do this, or if it is even true.

  • 2
    $\begingroup$ Here is an idea: add a constant symbol $c$ to your language. Then let $\Sigma$ be a set of first-order sentences saying that $c$ is transcendental (convince yourself that this can be done). Use an ultraproduct to prove that $\Sigma$ would have a model in the class of algebraic extensions (so we conclude that that class cannot be first-order axiomatisable). $\endgroup$ Nov 26 '20 at 12:06
  • $\begingroup$ I suppose this is cheating, but you can just argue that all algebraic extensions of rationals are countable. On the other hand, Lowenheim-Skolem/compactness/ultraproducts (however you want to frame it) show that any FO theory with infinite models has uncountable models. $\endgroup$
    – tomasz
    Nov 29 '20 at 9:51

I will provide a rough proof, but I will leave some of the details to you. I will indicate whenever I skip some details that you should check.

Suppose, for a contradiction, that there is a first-order theory $T$ in the language of rings axiomatising the algebraic extensions of $\mathbb{Q}$.

First we add a constant symbol $c$ to our language. We build a set $\Sigma$ as follows: for every polynomial $P(x)$ with rational coefficients we add a formula $P(c) \neq 0$. Convince yourself that we can indeed write $P(c) \neq 0$ as a first-order formula in the language of rings (with constant symbol $c$).

Now for any finite $\Sigma_0 \subseteq \Sigma$ there is an algebraic extension $F_{\Sigma_0}$ of $\mathbb{Q}$ with an element $a \in F_{\Sigma_0}$ such that $a$ satisfies all the equations in $\Sigma_0$, when we replace the constant symbol $c$ by $a$ (check this). That is, $F_{\Sigma_0}$ together with $c$ interpreted as $a$ is a model of $T \cup \Sigma_0$.

The idea is now to mimic the proof of the compactness theorem. In fact, if you would have the compactness theorem this would be a lot easier, but you asked for a proof using ultraproducts.

Just to show how this would go with compactness.

Theorem (Compactness theorem). A set of first-order sentences $\Gamma$ has a model if and only if every finite subset $\Gamma_0 \subseteq \Gamma$ has a model.

Now the claim would follow because by the above $T \cup \Sigma$ would have a model $M$. But then $M$ would be an algebraic extension of $\mathbb{Q}$, because $M$ is a model of $T$. At the same time we would have that the interpretation of $c$ in $M$ is transcendental, because $M$ also satisfies $\Sigma$. Contradiction.

So to do this using ultraproducts we do the following. Write $\mathcal{P}_\text{fin}(\Sigma)$ for the set of finite subsets of $\Sigma$. We define $A_{\Sigma_0} = \{ \Sigma_1 \in \mathcal{P}_\text{fin} : \Sigma_0 \subseteq \Sigma_1 \}$. Let $U$ be an ultrafilter on $\mathcal{P}_\text{fin}(\Sigma)$ containing all sets of the form $A_{\Sigma_0}$ (check that this can be done). Consider the ultraproduct $$ M := \prod_{\Sigma_0 \in \mathcal{P}_\text{fin}(\Sigma)} F_{\Sigma_0} / U $$ We claim that $M \models T \cup \Sigma$, and then we would be done as discussed before. Since every structure in the ultraproduct is a model of $T$, we clearly have $M \models T$ by Łoś's theorem. Now for $\varphi \in \Sigma$ we note that $A_{\{\varphi\}} \in U$ and for all $\Sigma_0 \in A_{\{\varphi\}}$ we have $F_{\Sigma_0} \models \varphi$. So, again by Łoś's theorem, we have $M \models \varphi$ (check these last two sentences in detail).


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.