Show that it is/is not a normal extension Let $a \in \mathbb{R}$ with $a^4=5$. Show that:


*

*$\mathbb{Q}(ia^2)$ is a normal extension of $\mathbb{Q}$.

*$\mathbb{Q}(a+ia)$ is a normal extension of $\mathbb{Q}(ia^2)$

*$\mathbb{Q}(a+ia)$ is not a normal extension of $\mathbb{Q}$


I have done the following:


*

*$Irr(ia^2, \mathbb{Q})=x^2+5$


The splitting field is $\mathbb{Q}(ia^2)$.
Therefore, $\mathbb{Q}(ia^2)$ is a normal extension of $\mathbb{Q}$.


*$Irr(a+ia, \mathbb{Q}(ia))=x^2-2ia^2$


The splitting field is $\mathbb{Q}(a+ia)$.
Therefore, $\mathbb{Q}(a+ia)$ is a normal extension of $\mathbb{Q}(ia^2)$.


*$Irr(a+ia, \mathbb{Q})=x^4+20$


Are the solutions $\pm \sqrt{2i}a, \pm \sqrt{-2i}a$ ??
These solutions are not in $\mathbb{Q}(a+ia)$, are they?? How can we prove it??
Is this correct??
 A: Hint. Prove this in three steps. 


*

*The four conjugates of $a + ia$ are $\pm a \pm ia$.

*The normal closure of $\mathbb{Q}(a + ia)$ is $\mathbb{Q}(a,i)$. (If you don't know about normal closure, prove this instead: If $\mathbb{Q}(a + ia)$ were normal, it would be equal to $\mathbb{Q}(a,i)$.)

*The degree of $\mathbb{Q}(a,i)$ over $\mathbb{Q}$ is greater than $4$.
A: To expand on user204305's answer, let $N$ be a normal extension
of ${\mathbb Q}$ containing ${\mathbb Q}(a+ia)$. We have to show that
$N\neq {\mathbb Q}(a+ia)$.
As indicated in the OP, the minimal polynomial of $a+ia$ 
is $M=X^4+20$. The roots of $M$ are $m_1=a+ia$,
$m_2=im_1=-a+ia,m_3=-m_1=-a-ia$ and $m_4=-im_1=a-ia$.
By definition, $N$ contains all the $m_i$, so $N$
contains both $\frac{m_1+m_4}{2}=a$ and $\frac{m_1-m_4}{m_1+m_4}=i$. So
$N$ contains ${\mathbb Q}(a)$, and the inclusion is strict
(because ${\mathbb Q}(a) \subseteq {\mathbb R}$ and $i\in N$). We deduce
that $d=[N:{\mathbb Q}(a)]$ is $\geq 2$, so 
$[N:{\mathbb Q}]=[N:{\mathbb Q}(a)][{\mathbb Q}(a):{\mathbb Q}] \geq 8$ and hence
$N\neq {\mathbb Q}(a+ia)$ as wished.
