# Countable Fields with No Countable Extension

Let $\mathscr{S}$ be the set of all countable subfields of $\mathbb{C}$. Clearly, $\mathscr{S}$ is a partially ordered set under inclusion, and if $K_1\subseteq K_2 \subseteq \cdots$ is an ascending chain of countable subfields, then $\bigcup_{i=1}^{\infty}K_i$ is a countable union of countable fields, and is hence an upper bound for $K_1\subseteq K_2 \subseteq \cdots$ which is in $\mathscr{S}$. But then $\mathscr{S}$ satisfies the conditions of Zorn's lemma, so there is some maximal element $K$. It would then seem that $K$ is a countable field such that whenever $K\subsetneq L\subseteq \mathbb{C}$ is a pair of field extensions, we have that $L$ is uncountable. This seems quite unintuitive to me. Has anyone exhibited such a subfield of $\mathbb{C}$ and proved that it has the properties required? It would seem that for any countable subfield $K\subseteq \mathbb{C}$, there will be some complex number $\alpha\notin K$, in which case $K\subsetneq K(\alpha)\subset\mathbb{C}$ and $K(\alpha)$ is also countable.

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I think you've proved S doesn't satisfy the condition of Zorn's lemma, though. – Charles Matthews Feb 9 '11 at 12:53
To apply Zorn's lemma you need to use all chains, not just the countable ones. – Angelo Feb 9 '11 at 12:57

Every countably infinite field $F$ not only has a countable extension $K$, but has one satisfying exactly the same first-order truths in the language of fields. Indeed, one can add any further structure to the language, such as predicates for relative transcendency, and maintain this feature. This is an immediate consequence of the Lowenheim-Skolem theorem, which asserts that every infinite first order structure in a countable language, such as any field, ring or group, has elementary extensions in every larger cardinality. The same argument produces proper elementary extensions in the same cardinality. The downward form of the theorem asserts that for any infinite first order structure $M$ in a countable language and any countable set $F\subset M$, there is a countable elementary substructure $N\prec M$ with $F\subset N$. In particular, any countable subset of an uncountable structure can be properly extended to a countable elementary substructure of that structure.

As Greg pointed out, the error in your argument is the assumption that the chain itself is countable, which is unwarranted, and the union of an uncountable chain of countable structures can be uncountable. For example, the union of all countable ordinals is $\omega_1$, an uncountable set, even though this forms a chain of countable sets. This example is canonical for this situation: every union of a chain of countable sets has size at most $\omega_1$, since the cofinality of the chain (the size of the smallest unbounded subchain) cannot be larger than $\omega_1$, as this would lead to uncountable objects in the chain.

A structure is $\omega_1$-like if it has an order relation all of whose initial segments are countable, and that is the basic situation here, if one considers the order arising from the chain.

And the problem has nothing to do with fields or any kind of algebra. After all, consider the collection of countable subsets of a fixed uncountable set. This collection is closed under the union of countable chains, for the reason given by the OP, but of course we will never have a maximal countable subset of an uncountable set.

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Your collection $S$ doesn't satisfy the conditions of Zorn's lemma. Every countable chain has an upper bound, but not every chain. Your argument would have worked just as well for subsets of $\mathbb{C}$ as for subfields, and then it's easier to see what is going wrong: There is no maximal countable subset of $\mathbb{C}$. The situation is not very different for subfields. For instance you can replace $\mathbb{C}$ with an uncountable set $X$ of algebraically independent elements of $\mathbb{C}$, and then there is an order-preserving injection from subsets of $X$ to subfields of $\mathbb{C}$.

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