Set of maximal subfields not containing particular elements.

Instead of extending a field, by adjoining a new element, consider what happens if we remove an element or elements.

This started as a question on math.SE Field reductions where Pete L. Clark explained that there isn't a unique subfield that could be called a "field reduction", rather there is a set of maximal subfields not containing the element.

Let $\mathbb{R}(\setminus a)$ be the set of maximal subfields of $\mathbb{R}$ that don't contain $a$.

Question: What is the cardinality of $\mathbb{R}(\setminus a)$ ?

Pete L. Clark suggested that this sounds "ultrafiltery" so I thought it would be an appropriate question for Mathoverflow.

Let $\mathbb{A}'$ be the set of non-rational algebraic numbers.

In Field reductions. part two Arturo Magidin showed that the cardinality of $\mathbb{R}(\setminus\mathbb{A}')$ is $2^\mathfrak{c}$.

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1 Answer

I might be missing something but I think Arturo´s idea can also be used to show that $\mathbb{R}(\setminus a)$ has cardinality $2^\mathfrak{c}$:

Let $T$ be any set of reals algebraically independent over $\mathbb{Q}(a)$ with $|T|=\mathfrak{c}$. For any $X \subseteq T$ the set $T_X:=aX \cup (T \setminus X)$ is still algebraically independent over $\mathbb{Q}(a)$ and by Zorn´s lemma we can find $F_X \in \mathbb{R}(\setminus a)$ such that $\mathbb{Q}(T_X) \subseteq F_X$. If $X \neq Y$ then $F_X \neq F_Y$ because otherwise there would be a $t \in T$ such that both $t$ and $at$ (and hence $a$) are elements of $F_X$, a contradiction.

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