show there is $\sigma_i\in Gal(K/F)$ such that $\sigma_i\sqrt{d_j}=\sqrt{d_j}$ if $j\neq i$, but $\sigma_i\sqrt{d_i}=-\sqrt{d_i}$

Question

I have an exercise which I think is very confusing, is there anyone who can help me?

Let F be a field of characteristic $\neq 2$, and suppose that we have $k\in \mathbb{N}$, and elements $d_1,...d_k\in F^{\times}$ such that:

for any $\emptyset \neq S\subseteq\{1,...,k\}$ the element $\prod_{i\in S} d_i$ is not a square in $F^{\times}$

We put $K:=F\left(\sqrt{d_1},...,\sqrt{d_k}\right)$.

We need to show that K/F is Galois, and that for each $i=1,...,k$ there is $\sigma_i\in Gal(K/F)$ such that $\sigma_i\sqrt{d_j}=\sqrt{d_j}$ if $j\neq i$, but $\sigma_i\sqrt{d_i}=-\sqrt{d_i}$.

We have already shown $[K:F]=2^k$, and that a basis of K over F is $\prod_{i\in S} \sqrt{d_i}$ with S running over all subsets of $\{1,...,k\}$

Answer 1

K is the splitting field of $f(x) = (x^2-d_1)\ldots(x^2-d_k)$ over $F$, so it is a normal extension. The polynomial $f(x)$ is separable over $K$, since its irreducible factors $x^2 - d_i$ have distinct roots (precisely because $\text{char} K \neq 2$). Thus, $K/F$ is a separable extension (since it is the splitting field of a separable polynomial). $K/F$ is normal and separable, hence Galois.

Now, suppose that $\varphi:K \to K$ is an $F$-automorphism (i.e. $\varphi \in \text{Gal}(K/F)$). Then for each $i$, we have

$$\varphi(\sqrt{d_i})^2 = \varphi(\sqrt{d_i}^2) = \varphi(d_i) = d_i,$$

so $\varphi(\sqrt{d_i}) = \pm \sqrt{d_i}$. Since the $\sqrt{d_i}$ generate $K$ as an $F$-algebra, the homomorphism $\varphi$ is determined entirely by the images of the $\sqrt{d_i}$. Since each $\sqrt{d_i}$ has two possible images, there are $\textbf{at most}$ $2^k$ possible such $\varphi$. Since the order of the Galois group equals the degree of the field extension, there are actually $2^k$ such automorphisms, and hence each possibility must be realised (we must attain the upper bound of $2^k$ different $\varphi$).

In other words, for every choice of numbers $\epsilon_1\ldots, \epsilon_n \in \{\pm 1\}$, there is a homomorphism $\varphi \in \text{Gal}(K/F)$ with $\varphi(\sqrt{d_i}) = \epsilon_i\sqrt{d_i}$ for each $i$. Since the maps you want to construct are of this form (with exactly one $\epsilon_i = -1$), they must exist.