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.
 A: 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).
