# Showing that $[\mathbb{Q}(\sqrt[\leftroot{-2}\uproot{2}4]{2}):\mathbb{Q}(\sqrt{2})]=2$

I came up with this question when considering the field extension $\mathbb{Q}(\sqrt[\leftroot{-2}\uproot{2}4]{2})|\mathbb{Q}$. I know that the minimal polynomial of $\sqrt[\leftroot{-2}\uproot{2}4]{2}$ over $\mathbb{Q}$ is $x^4-2$ due to Eisenstein, and thus the above extension has degree 4. Now I'm trying to look at this from another angle, considering the tower $\mathbb{Q}(\sqrt[\leftroot{-2}\uproot{2}4]{2})\supseteq\mathbb{Q}(\sqrt{2})\supseteq\mathbb{Q}$. It is obvious that the right extension has degree 2, but for the left extension I also what to show this. I think that the minimal polynomial of $\sqrt[\leftroot{-2}\uproot{2}4]{2}$ over $\mathbb{Q}(\sqrt{2})$ is $x^2-\sqrt{2}$. After all I know for this to be true I need to show that $\sqrt[\leftroot{-2}\uproot{2}4]{2}\notin\mathbb{Q}(\sqrt{2})$, but how do I do this?

• You already know that $\sqrt[4]{2}$ has degree 4 over ${\mathbb Q}$. So it can't be in ${\mathbb Q}(\sqrt 2)$, since that has degree 2 over ${\mathbb Q}$. – Magdiragdag Jan 17 '17 at 22:00
• Because the minimal polynomial of any algebraic element over $\mathbb{Q}$ in the extension field has to divide the degree of the minimal polynomial of the generating element of said extension? – azureai Jan 17 '17 at 22:13
• Well, yes, but I was more thinking of: the whole extension ${\mathbb Q}(\sqrt[4]{2}) : {\mathbb Q}$ has degree $4$; the intermediate extension ${\mathbb Q}(\sqrt{2}):{\mathbb Q}$ has degree $2$; so the other intermediate extension ${\mathbb Q}(\sqrt[4]{2}) : {\mathbb Q}(\sqrt{2})$ has degree $4/2=2$ (which is actually what you're after). After that, since that degree is not $1$, $\sqrt[4]{2} \not\in {\mathbb Q}(\sqrt{2})$ and $x^2 - \sqrt{2}$ must be the minimial polynomial of $\sqrt[4]{2}$ over ${\mathbb Q}(\sqrt 2)$. – Magdiragdag Jan 17 '17 at 22:32

Instead of first showing that $\sqrt[4]{2} \not\in {\mathbb Q}(\sqrt2)$ with the goal of establishing that $[{\mathbb Q}(\sqrt[4]{2}) : {\mathbb Q}(\sqrt2)] = 2$, it is probably easier to go the other way around.

In the tower of extensions ${\mathbb Q}(\sqrt[4]{2}) : {\mathbb Q}(\sqrt{2}) : {\mathbb Q}$, you have already established that $[{\mathbb Q}(\sqrt[4]{2}) : {\mathbb Q}] = 4$ and also that $[{\mathbb Q}(\sqrt{2}) : {\mathbb Q}] = 2$. Therefore $[{\mathbb Q}(\sqrt[4]{2}) : {\mathbb Q}(\sqrt2)] = 4/2 = 2$.

Now, if you're still interested in the minimum polynomial of $\sqrt[4]{2}$ over ${\mathbb Q}(\sqrt{2})$: you now know it must have degree $2$, so it must be $X^2 - \sqrt{2}$.

If $\sqrt[4]{2}\in\mathbb Q[\sqrt{2}]$, then there would be an element $k$ that has $k^2 = \sqrt{2}$.

Any $k$ can be written as $k:= a+b\sqrt{2}$. Consider the square of this: \begin{align*} k^2 & = (a+b\sqrt{2})^2 = a^2+2ab\sqrt{2}+2b^2 = (a^2+2b^2)+2ab\sqrt{2} \end{align*} So, for $k^2 = \sqrt{2}$, we need to have that $ab = \frac{1}{2}$, and $a^2+2b^2 = 0$. As the second equation is only satisfied when $a = b= 0$, it follows that no such $a,b$ exist.

The other answers already cover the problem very well, but I'd like to present another approach.

You want to prove that $x^2-\sqrt{2}$ is irreducible in $\mathbb{Q}(\sqrt{2})[x]$.

Note that $\mathbb{Q}(\sqrt{2})$ is the field of fractions of $\mathbb{Z}[\sqrt{2}]$, which is a Euclidean domain. Moreover, $\sqrt{2}$ is a prime element in $\mathbb{Z}[\sqrt{2}]$. Thus, you can apply Eisenstein's generalized criterion to $x^2-\sqrt{2}$ (with respect to the prime element $\sqrt{2}$) to show that it is irreducible in $\mathbb{Q}(\sqrt{2})[x]$.

This method does not generalize as well as the ones from the other answers, but I thought it worthwhile to point it out.