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\newtheorem{problem}{Problem}
\title{Iterated Quadratic Extensions Over $\mathbb{Q}$\\ (Version 1.2)}
\author{Po-Shen Loh}
\date{7 May 2001}
\begin{document}
\maketitle

\begin{problem}

Let $\{c_1, c_2, \ldots, c_n\}$ be a set of distinct positive integers
such that no product of distinct elements is a square, and let $a_i =
\sqrt{c_i}$ for each $i$.  Prove that $[\mathbb{Q}(a_1, a_2, \ldots,
a_n):\mathbb{Q}] = 2^n$ and the extension is Galois over $\mathbb{Q}$
with Galois group isomorphic to $\mathbb{Z}/2\mathbb{Z} \times
\mathbb{Z}/2\mathbb{Z} \times \cdots \times \mathbb{Z}/2\mathbb{Z}$
($n$ times).

\end{problem}

\noindent \textbf{Solution:}

We use strong induction.  Our base case is $n=2$, which follows
immediately since it is a biquadratic extension.

Suppose that our problem is true for all $n \leq N$.  We show that it
is true for an $(N+1)$-set of $a_i$.  Let $\mathcal{S} = \{a_1, a_2,
\ldots, a_{N+1}\}$, let $K$ be the extension of $\mathbb{Q}$ by
$\mathcal{S}$ and consider a subset $\mathcal{T} \subset \mathcal{S}$
with $N$ elements.  By inductive hypothesis, we know that the degree
of the extension of $\mathbb{Q}$ with elements of $\mathcal{T}$ is
$2^N$.

Proceed by contradiction; suppose that $[\mathbb{Q}(a_1, a_2, \ldots,
a_{N+1}):\mathbb{Q}] \neq 2^{N+1}$.  It contains $\mathcal{T}$, so the
extension must have degree $2^N$, so it is exactly the extension field
generated by the elements of $\mathcal{T}$.  By inductive hypothesis,
it is Galois over $\mathbb{Q}$ and has elementary Abelian Galois group
$G$.

Since each subgroup $H \leq G$ is a subgroup of the elementary Abelian
group of order $2^N$, by the theorem of Finitely Generated Abelian
Groups, it is a direct product of cyclic groups, and since all
elements have order dividing 2, $H$ must also be elementary Abelian.

We can determine what the corresponding subfields look like.  $H$ is
of the form $\langle \sigma_1 \rangle \times \langle \sigma_2 \rangle
\times \cdots \times \langle \sigma_k \rangle$ where $\sigma_i$
conjugates its associated (quadratic) root $d_i$.  Then, its fixed
field will be an extension of $\mathbb{Q}$ by quadratics $d_i$ which
are products of the $a_i$.  Since $H$ is elementary Abelian, it is
generated by $N$ elements only if it is $G$.

Let $m$ be the number of intermediate subfields of the extension
generated over $\mathbb{Q}$ by $\mathcal{T}$.  Since we have a Galois
extension, there are exactly $m$ intermediate subfields under $K$.
Let $\{b\} = \mathcal{S} - \mathcal{T}$.  Clearly, $\mathbb{Q}(b)$ is
a subfield of the full extension $K$, so it must be one of the $m$
subfields, say $\mathbb{Q}(d_1, d_2, \ldots, d_k)$.  If $k = N$, then
from the previous paragraph, this corresponds the full group; this is
impossible because $[\mathbb{Q}(b):\mathbb{Q}] = 2 \neq 2^N =
[\mathbb{Q}(\mathcal{T}):\mathbb{Q}$].

So $k \neq N$.  If we could form a product of distinct generators
$d_i$ that was in $\mathbb{Q}$, then we could simply discard one of
them and have the same field.  Therefore, we can assume that $k < N$
and no product of distinct generators is rational.  Furthermore, by
the construction above for the subfields, no product of distinct
generators and $b$ can be rational, because each $b_i^2 \in
\mathbb{Q}$ so any product of distinct generators can be reduced to a
product of distinct $b_i$.  We can then apply the inductive hypothesis
on $\mathbb{Q}(d_1, \ldots, d_k, b)$, and conclude that
$\mathbb{Q}(d_i, \ldots, d_k) \neq \mathbb{Q}(b)$.  We have a
contradiction.

Thus, $[K:\mathbb{Q}] = 2^{N+1}$.  Since it is Galois, it must have
exactly $2^{N+1}$ automorphisms.  All automorphisms permute the roots
within each irreducible factor, and since there are exactly $N+1$ such
factors with two ways to make an map out of each, we find $2^{N+1}$
maps.  Since there are no other kinds of maps and the roots are
generators, each map must be an automorphism, and the Galois group is
indeed $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times
\cdots \times \mathbb{Z}/2\mathbb{Z}$ ($n$ times).

And we are done.



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