# Prove via ultraproducts that the class of algebraic extensions is not axiomatisable in the language of rings

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.

• 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). Nov 26 '20 at 12:06
• 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. 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).