Calculate $[E: \mathbb{Q}]$ and show that $E$ is a Galois extension on $\mathbb{Q}$

Question

Problem: Let $f(x) = x^5 - 2x^4 - 2x^3 + 4x^2 - 3x + 6 \in \mathbb{Q}[x]$. Let $E$ be the spliting field of $f(x)$. Calculate $[E: \mathbb{Q}]$ and show that $E$ is a Galois extension on $\mathbb{Q}$.

My attempt: We have $f(x) = (x-2)(x^2 - 3)(x^2 + 1)$. Roots of $f(x)$ are $2, \sqrt{3}, - \sqrt{3}, i, -i$. So $E = \mathbb{Q} (1, \sqrt{3}, -\sqrt{3}, i, -i) = \mathbb{Q} (\sqrt{3}, -\sqrt{3}, i, -i) = \mathbb{Q} (\sqrt{3},i)$, implies $[E: \mathbb{Q}] = 4$.

How to show that $E$ be a Galois extension?

Answer 1

You have done almost everything you need. You only need to find the useful results to apply them and conclude what you want.

Let me recall what a Galois Extension is.

Definition 1. Let $K\subseteq F$ be an extension. Then, $K\subseteq F$ is a Galois extension if it is normal and separable.

Definition 2. Let $K\subseteq F$ be an extension. Then, $K\subseteq F$ is a normal extension if

i) $K\subseteq F$ is algebraic;

ii) if $p(x)\in K[x]$ is irreducible and contains one of its roots in $F$, it implies that all its roots are in $F$.

Definition 3. Let $K\subseteq F$ be an extension. Then, $K\subseteq F$ is separable if all $\alpha \in F$ is separable, i.e. $\alpha \in F$ is algebraic and its irreducible polynomial does not contain repeated roots in its splitting field.

Let me recall also 2 distinct results which will be useful.

Theorem 1. If $[F:K]<\infty$, then $K\subseteq F$ is normal if and only if $F$ is the splitting field of some $p(x)\in K[x]\setminus K$.

Proposition 1. If $ch(K)=0$, then $K\subseteq F$ is separable if and only if $K\subseteq F$ is algebraic.

Now, since we are in $K=\mathbb{Q}$ and $E=\mathbb{Q}(\sqrt{3},i)$, we have that $\mathbb{Q}\subseteq\mathbb{Q}(\sqrt{3},i)$ is finite and hence it is algebraic. Thus, since $ch(\mathbb{Q})=0$, it is separable. Moreover, since $E$ is the splitting field of $p(x)$, which is not a constant polynomial, $E$ is also normal.

Then, we can conclude that $\mathbb{Q}\subseteq\mathbb{Q}(\sqrt{3},i)$ is normal and separable, i.e. it is a Galois extension.