Degree of Extension from $\mathbb{Q}$ to $\mathbb{Q}(i, \sqrt[4]2)$ I wanted to find the degree of extension from $\mathbb{Q}$ to $\mathbb{Q}(i, \sqrt[4]2)$. The minimal polynomial of $\sqrt[4]2$ over $\mathbb{Q}$ is $f(x)=x^4-2$. Now the roots of $f(x)$ are $\{\sqrt[4]2, -\sqrt[4]2, \sqrt[4]2i, -\sqrt[4]2i\}$. So we can observe $\frac{\sqrt[4]2i}{\sqrt[4]2}=i$ so $i \in \mathbb{Q}(\sqrt[4]2)$ so we can already see the degree of the extension from $\mathbb{Q}$ to $\mathbb{Q(i, \sqrt[4]2)}$ is 4. But also there is another approach to this problem:
Since we know $[\mathbb{Q}(\sqrt[4]2):\mathbb{Q}]$=4, then we can say $\mathbb{Q}(\sqrt[4]2)=\{a+b\sqrt[4]2+c\sqrt[4]2^2+d\sqrt[4]2^3 | a,b,c,d \in\mathbb{Q}\}$. Now I am not sure if my understanding is correct from here or not, but I think we should be able to express $i$ as an element of this linear combination. I have tried a few variations and I cannot seem to show this is possible. There is no negative under the radical sign. Is there something wrong with this second approach or can we express $i$ as a linear combination of the basis?
 A: I don't understand how did you get to $i\in\mathbb{Q}(\sqrt[4]{2})$. It's not true that all the roots of the minimal polynomial of $\sqrt[4]{2}$ must belong to this field. Actually, in this example it is clearly false. By definition, $\mathbb{Q}(\sqrt[4]{2})$ is the intersection of all subfields of $\mathbb{C}$ which contain the element $\sqrt[4]{2}$. In particular, $\mathbb{R}$ is such a field, so it is one of the fields in the intersection. This means $\mathbb{Q}(\sqrt[4]{2})$ is a subfield of $\mathbb{R}$, and so $i$ can't be an element of this field. This tells us that we must have $[\mathbb{Q}({i,\sqrt[4]{2})}:\mathbb{Q}(\sqrt[4]{2})]=2$, and so by the tower rule the extension degree you are looking for is $8$.
A: $i \in \Bbb Q(\sqrt[4]{2})$ is not correct. Indeed, $\Bbb Q(\sqrt[4]{2})$ is completely contained in $\Bbb R$.
Note that this extension is not a normal extension, it does not contain all the roots of the minimal polynomial. This is why your analysis fails.
Since $i \notin \Bbb Q(\sqrt[4]{2})$, you can conclude that $[\Bbb Q(\sqrt[4]{2})(i):\Bbb Q(\sqrt[4]{2})] \ge 2.$ On the other hand, the polynomial $x^2 + 1 \in \Bbb Q(\sqrt[4]{2})[x]$ shows that the degree is exactly $2$.
Since the degrees multiply, we see that the extension $\Bbb Q(\sqrt[4]{2})(i) = \Bbb Q(\sqrt[4]{2}, i) \supset \Bbb Q$ is of degree $\boxed{8}$.
