I've been solving problems from my Galois Theory course, and I've had problems finishing this one. It says:
Being $a=(1+i)\sqrt[4] 5$, find the normal closures of these field
extensions:
- $\mathbb Q(a)/\mathbb Q(i\sqrt 5)$
- $\mathbb Q(i\sqrt 5)/\mathbb Q$
- $\mathbb Q(a)/\mathbb Q$
This is the work I've done so far:
The first I tried to find where the degrees of the extensions involved. It was easy to notice $X^2+5$ is the minimal polynomial of $i\sqrt 5$ over $\mathbb Q$, so then the extension $\mathbb Q(i\sqrt 5)/\mathbb Q$ has degree 2. Now, since it has degree two, then $\mathbb Q(i\sqrt5)$ is splitting field of $X^2+5$ over $\mathbb Q$, hence $\mathbb Q(i\sqrt5)/\mathbb Q$ is a normal extension and then it's normal closure is $\mathbb Q(i\sqrt5)$ itself.
Then, it was easy to see that $a$ is root of $X^4+20$, and it's irreducible for Eisenstein so $[\mathbb Q(a):\mathbb Q]=4$. Now, given that $a^2=2i\sqrt5\in\mathbb Q(a)$, it's clear that $\mathbb Q(i\sqrt5)\subset\mathbb Q(a)$, and the degree of $\mathbb Q(a)/\mathbb Q(i\sqrt 5)$ can only be 1 or 2 (because of transitivity), but it's clearly not $1$ because $a\notin\mathbb Q(i\sqrt5)$, so it's degree is 2, hence it's a normal extension so it's normal closure is $\mathbb Q(a)$ itself.
The last part, to be said, finding normal closure of $\mathbb Q(a)/\mathbb Q$ is where I'm not sure. What I did is first notice that, since $a$ is a complex numer, and $X^4+20$ is a biquadratic polynomial with rational coefficients, the it's 4 different roots are $a,-a,\bar a,-\bar a$. So the normal closure of $\mathbb Q(a)/\mathbb Q$ is $\mathbb Q(\pm a,\pm\bar a)$, but I want this expression to be reduced as much as possible. The first reduction is obviously removing the negative terms, to be said,
$$\mathbb Q(\pm a,\pm\bar a)=\mathbb Q(a,\bar a).$$
And here is where I don't know how to end. I'm not sure if $\mathbb Q(a,\bar a)=\mathbb Q(a)$ or not. I tried to find a way to make $\bar a$ out of combinations of elements from $\mathbb Q(a)$ but had no success, and something tells me $\bar a\notin\mathbb Q(a)$, but I don't know how to prove it. If this was true, then the normal closure of $\mathbb Q(a)/\mathbb Q$ would be $\mathbb Q(a,\bar a)$. If not, then it would be a normal extension and the normal closure would be $\mathbb Q(a)$ itself, but I don't think this is the case.
Is what I did correct? How can I finish the last part? If it's true, how can I prove $\bar a\notin\mathbb Q(a)$? If not, how can I prove the opposite? Any help will be appreciated, thanks in advance.
Best Answer
As you guessed, $\bar{a}$ is not in $\mathbb{Q}(a)$. If this is true, then $$ i = -\frac{\bar{a}}{a} \in \mathbb{Q}(a), \sqrt[4]{5} = \frac{a}{1+i}\in \mathbb{Q}(a) $$ so $\mathbb{Q}(a)\supseteq \mathbb{Q}(i, \sqrt[4]{5}) = K$. However, the field $K$ has degree 8 over $\mathbb{Q}$, so this is impossible.