Show that it is/is not a normal extension

Question

Let $a \in \mathbb{R}$ with $a^4=5$. Show that:

1. $\mathbb{Q}(ia^2)$ is a normal extension of $\mathbb{Q}$.
2. $\mathbb{Q}(a+ia)$ is a normal extension of $\mathbb{Q}(ia^2)$
3. $\mathbb{Q}(a+ia)$ is not a normal extension of $\mathbb{Q}$

I have done the following:

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

The splitting field is $\mathbb{Q}(ia^2)$.

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

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

The splitting field is $\mathbb{Q}(a+ia)$.

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

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

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??

Answer 1

Hint. Prove this in three steps.

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

2. 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 $\mathbb{Q}(a,i)$.)

3. The degree of $\mathbb{Q}(a,i)$ over $\mathbb{Q}$ is greater than $4$.

Answer 2

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.