Does $\mathbb{Q}(\sqrt{-1}, \sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7},\ldots)$ have countably many subfields? According to Example 3.10 of these notes, the field $L = \mathbb{Q}(\sqrt{-1}, \sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7},\ldots)$, where we adjoin $\sqrt{p}$ for every prime $p$ (and $p=-1$) has only countably many subfields. I think that this is only true if we require the subfields to have finite degree over $\mathbb{Q}$.
If we do not specify finite degree, then for any element $(a_{-1}, a_2,a_3,a_5,\ldots) \in \prod_p \{0,1\}$, we get a subfield $E = \mathbb{Q}(a_{-1}\sqrt{-1}, a_2\sqrt{2}, a_3\sqrt{3},\ldots)$. This defines an injection from $\prod_p \{0,1\}$ to $\operatorname{Gal}(L/\mathbb{Q})$, so the latter is uncountable.
Given that Keith Conrad and I disagree, the conditional probability that I am wrong is high. Can anyone spot any errors in my reasoning?
 A: Let $L = \mathbb Q(\sqrt{-1},\sqrt{2},\sqrt{3},\ldots)$
and let $G=\textrm{Gal}(L/{\mathbb Q})$.
Facts.

* $|L|=\aleph_0$.


* The set of subfields of $L$ that are
finite extensions of $\mathbb Q$ has cardinality $\aleph_0$.


* The set of all subfields of $L$ 
has cardinality $2^{\aleph_0}$.


* $|G|=2^{\aleph_0}$.


* The set of $2$-element subgroups
of $G$ has cardinality $2^{\aleph_0}$.


* The set of index-$2$ subgroups
of $G$ has cardinality $2^{2^{\aleph_0}}$.


* The set of all subgroups of $G$ has cardinality $2^{2^{\aleph_0}}$.


Let me briefly explain the 2nd, 3rd, and 6th of these.
The set of subfields of $L$ that are
finite extensions of $\mathbb Q$ has cardinality $\aleph_0$.

Let $\mathcal F$ be the set of intermediate extensions
$\mathbb Q\leq F\leq L$
of finite degree over $\mathbb Q$.
Map $L$ to $\mathcal F$ by $\alpha\mapsto \mathbb Q[\alpha]$.
This is a surjective map from a countable set $L$ onto $\mathcal F$,
so $\mathcal F$ is countable.
Since $\mathcal F$ contains infinitely many
distinct members, e.g.
$\mathbb Q[\sqrt{-1}], \mathbb Q[\sqrt{2}], \mathbb Q[\sqrt{3}], \ldots$,
we must have $|\mathcal F|= \aleph_0$.

The set of all subfields of $L$
has cardinality $2^{\aleph_0}$.
Since $|L|=\aleph_0$, the set $L$ has $2^{\aleph_0}$
subsets. The number of subfields must be $\leq 2^{\aleph_0}$.
However, the argument in the second
paragraph of the question statement shows that
the number of subfields is at least $2^{\aleph_0}$,
so we have equality.

The set of index-$2$ subgroups
of $G$ has cardinality $2^{2^{\aleph_0}}$.
$G\cong \mathbb Z_2^{\aleph_0}$ (see Example 3.10 of Conrad's notes).
This means $G$ can be viewed as an $\mathbb F_2$-vector space
of cardinality $2^{\aleph_0}$. When the cardinality of a
vector space is infinite and strictly exceeds the cardinality of the field,
then the cardinality equals the dimension, so
$G$ must have an $\mathbb F_2$-basis $\mathcal B$
of size $2^{\aleph_0}$. Different surjective
functions $f\colon {\mathcal B}\to \mathbb F_2$
extend to different surjective homomorphisms $\overline{f}\colon G\to \mathbb F_2$,
and these different surjective homomorphisms necessarily have different kernels.
Each kernel is a subgroup of $G$ of index $2$.
Since there are
$2^{2^{\aleph_0}}$-many surjective functions $f\colon {\mathcal B}\to \mathbb F_2$,
there are at least this many subgroups of $G$ of index $2$.
There can't be more since $G$ only has
$2^{2^{\aleph_0}}$-many subsets.


Now to get to the question: is there a mistake in
Example 3.10 of Conrad's notes?

I found no mathematical mistake in Example 3.10.
I did not read the rest of the notes.
I found it slightly misleading that Conrad would write
[[$\textrm{Gal}(L/{\mathbb Q})$ has uncountably many subgroups of order 2.
At the same time, $L$ has only countably many
subfields of each $2$-power degree over $\mathbb Q$.]]
Writing this way could lead the reader to believe that
that there is a reason to compare subgroups of finite order in
$\textrm{Gal}(L/{\mathbb Q})$ to intermediate extensions of $L/\mathbb Q$
of finite degree over $\mathbb Q$. Instead one should
compare subgroups of finite index
in $\textrm{Gal}(L/{\mathbb Q})$ to extensions
of finite degree over $\mathbb Q$.
I think it would have been clearer to write
[[$\textrm{Gal}(L/{\mathbb Q})$ has $2^{2^{\aleph_0}}$-many subgroups of index $2$.
At the same time, $L$ has only $\aleph_0$-many
subfields of degree $2$ over $\mathbb Q$.]]

Next, Conrad writes in the concluding two lines of Example 3.10
[[$L$ has only countably many subfields of each $2$-power degree over
$\mathbb Q$. Therefore the subfields of $L$ and the subgroups of
$\textrm{Gal}(L/{\mathbb Q})$ do not have the same cardinality.]]
The word Therefore connects mathematical claims.
These claims are correct.
But the final claim is not a consequence of former claims,
so the word Therefore is not the best choice.
In fact, Conrad does not determine the number
of subfields of $L$ nor the number of subgroups of
$\textrm{Gal}(L/{\mathbb Q})$ in this example, so there
is no place for Therefore in what he has written.
A: There is no error, but you misread the exercice. The finite degree subfields don't correspond to the finite index subgroups but to the open finite index subgroups. The point of this exercice is this subtlety which is essential when dealing with infinite Galois extensions.
A: I miswrote what I had intended, and will write a corrected version here.  Someday I will try to post an updated copy of the file where you found it. My aim there was to suggest a mismatch between subgroups and intermediate fields without aiming to make such a statement absolutely precise since, as you noted, the Krull topology was not yet introduced.
Let $L = \mathbf Q(\sqrt{-1},\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7},\dots)$, so
$G := {\rm Gal}(L/\mathbf Q) = \prod \{\pm 1\}$, which is a countable direct product of copies of
the group $\{\pm 1\}$ and $G$ is abelian.
Every quadratic extension of $\mathbf Q$ is in $L$, and there are countably many such extensions.
We expect by Galois theory that these extensions are in bijection with the subgroups of $G$ with index 2, but
we'll show there are uncountably many index-2 subgroups of $G$.
To each subgroup $H \subset G$ with index 2, $G/H \cong \mathbf F_2$ as groups in exactly one way, so
reduction mod $H$ gives us a nonzero group homomorphism $G \to \mathbf F_2$ with kernel $H$. The number of these $H$ is the number of nonzero group homomorphisms $G \to \mathbf F_2$.
Since all elements of $G$ square to the identity, we can think of $G$ as an $\mathbf F_2$-vector space
and think of nonzero homomorphisms $G \to \mathbf F_2$ as nonzero elements of the dual space
${\rm Hom}_{\mathbf F_2}(G,\mathbf F_2)$.  The group $G$ is uncountable (that is explained in other answers here), and this implies
the $\mathbf F_2$-dual space of $G$ is uncountable (also explained elsewhere on this page).
Therefore the number of index-2 subgroups of $G$ is uncountable.
