2
$\begingroup$

Let $F$ be an algebraic closure of $\mathbb{Q}$ and let $E\subset F$ be a splitting field over $\mathbb{Q}$ of the set $\{x^{2}+a|a\in\mathbb{Q}\}$ so that $E$ is algebraic and Galois over $\mathbb{Q}$. The goal is to show that $[E:\mathbb{Q}]<|\mathrm{Aut}_{\mathbb{Q}}E|$. I've shown that $\mathrm{Aut}_{\mathbb{Q}}E$ is uncountable ($|\mathrm{Aut}_{\mathbb{Q}}E|=|\mathcal{P}(X)|$). I've also shown that

(i) $\mathrm{Aut}_{\mathbb{Q}}E$ has uncountably many subgroups of index 2,

(ii) the set of extension fields of $\mathbb{Q}$ contained in $E$ of dimension 2 over $\mathbb{Q}$ is countable,

(iii) the set of closed subgroups of index 2 in $\mathrm{Aut}_{\mathbb{Q}}E$ is countable. (In Hungerford, a subgroup $G$ of $\mathrm{Aut}_{K}F$ is closed if $G''=G$, where $G'$ is the fixed field of $G$, and for an intermediate field $E$, $E'=\mathrm{Aut}_{E}F$.)

Finally I need to show that $[E:\mathbb{Q}]\leq\aleph_{0}$ but I can't figure this out from the above. Or does it follow from $|E|\leq\aleph_{0}|\mathbb{Q}|=\aleph_{0}$?

$\endgroup$

2 Answers 2

2
$\begingroup$

The answer to your last question is "yes" and this solves your problem.
More generally for any vector space $V$ of infinite dimension $\dim (V)$ over a field $k$ one has: $$ |V|=|k|\cdot \dim(V) $$ In particular if $k$ is finite or denumerable (as is your field $k=\mathbb Q$), one has the surprising result $$ |V|= \dim(V)$$ These remarkable displayed formulae don't seem to be as well known as they deserve and the only reference I know is Jacobson's Lectures in Abstract Algebra, vol.2, p.245.

$\endgroup$
0
1
$\begingroup$

Hint: Prove that $F$ is countable, therefore $E$ is countable. It is impossible that a countable field has an uncountable dimension over $\Bbb Q$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .