Dimension of splitting field of $(x^3+x+1)(x^2+1) \in \mathbb{Q}[x]$ over $\mathbb{Q}$ I am having trouble finding $[\Omega_f : \mathbb{Q}]$, where $\Omega_f$ is the splitting field of $f = gh = (x^3+x+1)(x^2+1) \in \mathbb{Q}[x]$. Since $g$ only has one real root we can show $[\Omega_g : \mathbb{Q}] = 6$, and similarly $[\Omega_h : \mathbb{Q}] = [\mathbb{Q}(i) : \mathbb{Q}] = 2$. It can also be shown that $g \in \mathbb{Q}(i)[x]$ is irreducible $\implies [\Omega_f : \mathbb{Q}] \in \{6, 12\}$.
However, how do I rule out that $[\Omega_f : \mathbb{Q}] = 6$ and not 12? Maybe some Galois theory could be useful but I am not sure.
 A: Let $\alpha$ be the real root of $g$. Then $g=(x-\alpha)(x^2+\alpha x-\alpha^{-1})$, the discriminant of the quadratic is $\alpha^2+4\alpha^{-1}=1+5\alpha^{-1}$, and the splitting field is $\Omega_g=\mathbb Q(\alpha,\beta)$ where $\beta^2=1+5\alpha^{-1}$.
Next, since $g(-1)=-1$ and $g(0)=1$ we know that $\alpha\in(-1,0)$, and so $1+5\alpha^{-1}<0$. Thus $\beta$ is purely imaginary. So, if $i\in\Omega_g$, then necessarily $i=u\beta$ for some $u\in\mathbb Q$. Squaring gives $-1=u^2\beta^2=u^2(1+5\alpha^{-1})$ inside $\mathbb Q(\alpha)$
We now take the field norm $N$ for $\mathbb Q(\alpha)/\mathbb Q$. For $v\in\mathbb Q(\alpha)$, multiplication by $v$ is $\mathbb Q$-linear, and $N(v)$ is the determinant of this linear map. Using the basis $1,\alpha,\alpha^2$ of $\mathbb Q(\alpha)$ we can quickly check that $N(\alpha)=-1$ (which is the negative of the constant term of $g$).
We compute $N(1+5\alpha^{-1})$. Since $\alpha^{-1}=-\alpha^2-1$, multiplication by $1+5\alpha^{-1}$ is given by the matrix
$$ \begin{pmatrix}-4&5&0\\0&1&5\\-5&0&1\end{pmatrix}, $$
and $N(1+5\alpha^{-1})=-129$. Clearly $N(-1)=-1$, so $u^2=-1/\beta^2$ yields $N(u)^2=1/129$. This is a contradiction, since $N(u)\in\mathbb Q$.
We have shown that $i\not\in\Omega_g$, and so $\Omega_f=\mathbb Q(\alpha,\beta,i)$ has dimension 12 over $\mathbb Q$.
