How can we check that there exist such elements?

Question

We have that $f=x^4-2\in \mathbb{Q}[x]$ and $E$ is the splitting field of $f/\mathbb{Q}$. It holds that $E=\mathbb{Q}[\rho, i]$, where $\rho^4=2, i^2=-1$. We have that $[E:\mathbb{Q}]=8$. According to my notes, $E/\mathbb{Q}$ is Galois since it is the splitting field of $f$ and it is separable.

1) Why is the extension separable?

Then it says that we have to check that there exists a $\sigma \in G : \sigma (\rho)=i\rho$ and $\sigma (i)=i$ and that there extists a $\tau \in G: \tau (\rho)=\rho$ and $\tau (i)=-i$.

2) How can we check that?

After that it says that each element of $G$ is characterized by its values at $\rho$ and $i$.

3) What does this mean?

Answer 1

we have $f(x)=x^4-2=0$ then $x_1=+\sqrt[4]2,x_2=-\sqrt[4]2,x_3=+i\sqrt[4]2,x_4=-i\sqrt[4]2$ then roots of $f(x)$ is diffrent in $E$ and $E$ separable extension

$E$ splitting and separable field then $|Gal(E,\mathbb Q)|=[E,\mathbb Q]$

Galois group is $\{ id,\sigma,\sigma^2,\sigma^3,\tau, \sigma\tau,\sigma^2\tau,\sigma^3\tau\} ; \sigma: \sqrt[4] {2}\to i \sqrt[4]{2}; \tau: i\to -i $ and we have 8 subgroups $H_1=\{id,\sigma,\sigma^2,\sigma^3 \}, H_2=\{id,\tau,\sigma^2,\sigma^2 \tau \}, H_3=\{id, \sigma^2 \}, H_4=\{id,\sigma\tau,\sigma^2,\sigma^3\tau \} $ , $N_1=\{id,\sigma^3\tau \}, N_2=\{id,\sigma\tau\}, N_3=\{id,\sigma^2\tau \}, N_4=\{id, \tau\}$

Answer 2

Fields with characteristic $0$ (and finite fields) are perfect, i.e. every albebraic extension is separable. In other words: every algebraic element is a simple root of its minimal polynomial.

The existence of σ results from the chain of Galois extensions: $\;\mathbf Q\subset \mathbf Q(i)\subset E$. $\;\DeclareMathOperator{\Gal}{Gal}\Gal(E/\mathbf Q(i))$ is the subgroup of $\Gal(E/\mathbf Q)$ of automorphisms $\sigma$ of $E$ which fix $\mathbf Q(i))$. Thus in particular $\sigma(i)=i$. It is a group of order $4$, and it acts transitively on the root of the minimal polynomial of $\rho$ over $\mathbf Q(i)$, $x^4-2$.

Answer 3

1) A polynomial is separable if it has distinct roots. A field extension $E/F$ is separable if for each $\alpha \in E$, its minimal polynomial is separable. As DonAntonio says, for a field of characteristic zero, every irreducible polynomial $f$ is separable: if $f$ has degree $n$, then $f'$ has degree $n-1$ and since $f$ is irreducible this means that $f$ and $f'$ are relatively prime. (Recall that $f$ has a multiple root iff $\gcd(f,f') \neq 1$.) The reason this can fail in characteristic $p$ is because if $f(x) = x^p - a$ for some $a$, then $f'(x) = 0$. For more on separability, see here.

2) See Dummit and Foote, Theorem $8$, $\S 13.1$, p. $519$:

Theorem 8. Let $\varphi: F \overset{\sim}{\to} E$ be an isomorphism of fields. Let $p(x) \in F[x]$ be an irreducible polynomial and let $q(x) \in E[x]$ be the irreducible polynomial obtained by applying the map $\varphi$ to the coefficients of $p(x)$. Let $\alpha$ be a root of $p(x)$ (in some extension of $F$) and let $\beta$ be a root of $q(x)$ (in some extension of $E$). Then there is an isomorphism \begin{align*} \sigma: F(\alpha) &\overset{\sim}{\to} E(\beta)\\ \alpha &\mapsto \beta \end{align*} mapping $\alpha$ to $\beta$ and extending $\varphi$, i.e., such that $\sigma$ restricted to $F$ is the isomorphism $\varphi$.

In your example, to show the existence of $\sigma$ we can proceed as follows. Applying the theorem with $F = E = \mathbb{Q}(i)$, $\varphi$ the identity map $\operatorname{id}: \mathbb{Q}(i) \to \mathbb{Q}(i)$, $\alpha = \rho$ and $\beta = i \rho$, we get the desired isomorphism $\sigma: \mathbb{Q}(i)(\rho) \to \mathbb{Q}(i)(i \rho) = \mathbb{Q}(i)(\rho)$.

3) Every element of $\mathbb{Q}[\rho, i]$ is a polynomial in $\rho$ and $i$, i.e., given $\alpha \in E$ there exists a polynomial $g(x,y) \in \mathbb{Q}[x,y]$ such that $\alpha = g(\rho, i)$. An element $\sigma \in \operatorname{Gal}(E/\mathbb{Q}) =:G$ fixes $\mathbb{Q}$, so it is totally determined by its action on $\rho$ and $i$. In other words, given $\sigma, \tau \in G$, if $\sigma(\rho) = \tau(\rho)$ and $\sigma(i) = \tau(i)$, then $\sigma = \tau$. For instance, if $\alpha = 2 + 3i - 7 \rho^2$, then $$ \sigma(\alpha) = \sigma(2 + 3i - 7 \rho^2) = \sigma(2) + \sigma(3)\sigma(i) - \sigma(7) \sigma(\rho^2) = 2 + 3 \sigma(i) - 7 \sigma(\rho)^2 $$ since $\sigma$ is a ring homomorphism. Thus we can compute $\sigma(\alpha)$ just by knowing $\sigma(\rho)$ and $\sigma(i)$.