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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?

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I think $(\sqrt{2} 5^{1/4} i)^4 + 20 = 40 \neq 0$. You should try to prove that $\Bbb Q(5^{1/4},i)$ is normal. –  mercio Nov 27 '12 at 9:22
    
Good catch, you're right! –  Kyle Schlitt Nov 27 '12 at 20:56

3 Answers 3

up vote 1 down vote accepted

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.

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Thanks very much! –  Kyle Schlitt Nov 27 '12 at 7:15
    
If you can use a little Galois Theory you can simplify considerably. Apply to the displayed equation for $\sqrt2$ an automorphism that takes $5^{1/4}$ to $5^{1/4}i$. By the same "examining the imaginary part" technique you get $\pm\sqrt2=a_1-a_3\sqrt5$. The calculations to show that this is impossible are not so bad. –  Gerry Myerson Nov 27 '12 at 11:37
    
@Gerry Alternatively, it simplifies nicely by applying simple results about linear independence of square-roots - see my answer. –  Bill Dubuque Nov 28 '12 at 0:19

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.

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we are doing Galois theory right now but we basically have only just started developing the ideas you're talking about, i'm actually trying to show that the extension is not normal over the rationals, and this just boiled down to very the last link of my argument (which is why i didn't bother to add the homework tag which I should have). I have a question though, if my extension is the splitting field of $x^{4} - 5$, then by definition wouldn't my extension be normal? –  Kyle Schlitt Nov 27 '12 at 6:23
    
Yes. ${}{}{}{}$ –  Gerry Myerson Nov 27 '12 at 6:27
    
I will add an edit with a more detailed argument but I think this causes a contradiction. –  Kyle Schlitt Nov 27 '12 at 6:43

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

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