Technique for showing an element is not in a field?

Question

I have an extension $\mathbb{Q}(5^{1/4}, i)$, and I want to show that $4^{1/4}$ is not contained in it.

(I hope what I am trying to prove is true!)

Anyways, my natural starting point is to assume for a contradiction that there exist polynomials $p(x,y)$ and $q(x,y)$ both over $\mathbb{Q}$ such that $4^{1/4} = \frac{p(5^{1/4},i)}{q(5^{1/4},i)}$.

But I have no clue what to do from here, yet I suspect there is a standard technique for showing things like this? (assuming it's true)

Edit:

I have an extension, which I want to show is not normal, so I want to show that it contains $\mathbb{Q}(5^{1/4}, i)$, which is not normal over $\mathbb{Q}$, by the following argument:

First observe that $\mathbb{Q}(5^{1/4}, i)$ contains a root of the irreducible polynomial $p(x) = x^{4} + 20$ (namely $5^{1/4} + 5^{1/4}i$) $over $\mathbb{Q}$.

Another root of this polynomial is $\sqrt{2}(5)^{1/4}i$, which is not in $\mathbb{Q}(5^{1/4}, i)$, since $\sqrt{2}\notin\mathbb{Q}(5^{1/4}, i)$.

Therefore the extension is not normal over $\mathbb{Q}$.

Is my reasoning correct?

Answer 1

This is almost always a pain. I would argue as follows, first note that if $\sqrt{2} \in \mathbb Q(5^{1/4},i)$ then $\sqrt{2} \in \mathbb Q(5^{1/4})$. You can show this by examining the imaginary part.

So the problem reduces to a somewhat simpler problem. Here is where the painful computation comes in. Note that $1,5^{1/4},5^{1/2},5^{3/4}$ form a basis for $\newcommand{\Q}{\mathbb Q}$ $\Q(5^{1/4})$. In particular if $\sqrt{2} \in \Q(5^{1/4})$ for some $a_i \in \mathbb Q$ we can write

$$\sqrt{2}=a_1+a_25^{1/4}+a_35^{1/2}+a_45^{3/4}.$$ Squaring both sides and collecting coefficients since we know the representation of $2$, we get a system of equations: $$\begin{align*} 2&=a_1^2+5a_3^2+10a_2a_4\\ 0&=2a_1a_2+10a_3a_4\\ 0&=a_2^2+5a_4^2+2a_1a_3 \\ 0&=2a_2a_3+2a_1a_4. \end{align*} $$ Then all you have to do is show this set of equations doesn't have a solution over the rationals. Which breaks into several cases.

Answer 2

It might help to note that $4^{1/4}=\sqrt2$.

It also helps to know some Galois Theory (although maybe there's a nice way to do this problem without). Your extension is the splitting field of $x^4-5$, and its Galois group is the dihedral group of order $8$. If $\sqrt2$ is in there, then ${\bf Q}(\sqrt2)$ is a subfield. Quadratic subfields are fixed fields of subgroups of index $2$. The dihedral group has three such subgroups. You can work out their fixed fields by standard techniques, and either ${\bf Q}(\sqrt2)$ shows up, or it doesn't --- either way, you have the answer to your question.

Answer 3

It is quite easy. $\ $ Taking real parts, $\ \sqrt{2}\in \Bbb Q(5^{1/4},{\it i\,}) \:\Rightarrow\:\sqrt{2}\in \Bbb Q(5^{1/4}),\ $ therefore

$$\rm\begin{eqnarray} \sqrt{2} &=&\rm a + b\sqrt{5} + \rm c\, 5^{1/4} + d\, 5^{3/4}\quad for\ \ a,b,c,d\in\Bbb Q \\ \rm \sqrt{2} - a - b\sqrt{5} &=&\,\rm c\, 5^{1/4} + d\, 5^{3/4}\quad which\ squared\ yields\\ \rm -2\color{#C00}a\sqrt{2} - 2\color{#0A0}b\sqrt{10} + 2ab\sqrt{5} + e &=&\,\rm (c^2\! + 5d^2)\sqrt{5} + 10 cd,\ \ \ for\ some\ \ e\in \Bbb Q \end{eqnarray}$$

By this simple Lemma we know $\rm\:\sqrt{2},\,\sqrt{5},\,\sqrt{10} = \sqrt{2}\sqrt{5}\:$ are linearly independent over $\Bbb Q,$ therefore $\rm\:\color{#C00}a = 0 = \color{#0A0}b\:$ $\Rightarrow$ $\rm\:c^2\!+\!5d^2\! = 0\:$ $\Rightarrow$ $\rm\:c,d=0\:$ $\Rightarrow$ $\:\sqrt{2} = 0,\:$ contradiction. $\ \ $ QED