Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $2^\mathfrak{c}$.

share|improve this question
add comment

1 Answer 1

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.