Splitting field of $x^4 + 4$ over $\mathbb{Q}$

Question

I am reading the example in Dummit and Foote section 13.4 where they calculate the splitting field of $x^4 + 4$. They do a clever factorization of $x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2)$ and conclude that the splitting field is $\mathbb{Q}(i)$ and so has degree 2 extension of $\mathbb{Q}$

However, I am struggling to see what is incorrect if you think about it like this: It is easily seen that the roots are $i^{1/2}$$\sqrt{2}$, $i^{3/2}$$\sqrt{2}$, $i^{5/2}$$\sqrt{2}$, $i^{7/2}$$\sqrt{2}$. Thus, the splitting field is $\mathbb{Q}(i^{1/2},\sqrt{2})$. It is not too hard to see that this is the same as $\mathbb{Q}(i,\sqrt{2})$. Now, $[\mathbb{Q}(\sqrt{2}): \mathbb{Q}]$ = 2 and $i$ is a root of $x^2 + 1$ which is irreducible over the real field $\mathbb{Q}(\sqrt{2})$. Thus, I conclude that $[\mathbb{Q}(i,\sqrt{2}):\mathbb{Q}] = 4$.

Where am I going wrong?

Answer 1

$x^4+4=(x-(1+i))(x-(1-i))(x+(1-i))(x+(1+i))$

(Note that $\sqrt2 i^{1/2}=\sqrt2(\frac1{\sqrt2}+\frac1{\sqrt2}i)=1+i.$)

Thus, the splitting field of $x^4+4$ over $\mathbb Q$ is a subfield of $\mathbb Q(i).$ Since this is an extension of degree $2,$ there is no field between the extension $\mathbb Q(i)/\mathbb Q.$ Hence, the splitting field must be $\mathbb Q(i).$

Answer 2

How do you separate out $i^{1/2}$ and $\sqrt2$ from the four roots ?