On the splitting field over $\mathbb{Q}$ of $X^4 + 1$. My question can be stated very simply: why is the splitting field over $\mathbb{Q}$ of $X^4 + 1$ not the field $\mathbb{Q}[e^{i\pi/4}].$
After all, if one defines $\alpha = e^{i\pi/4}$, then
$$
X^4+1=(X - \alpha)(X- \alpha^3)(X-\alpha^{-1})(X-\alpha^{-3})
$$
I figure that the answer to my question will entail showing that what I'm calling "$\mathbb{Q}[e^{i\pi/4}]$" is essentially nonsense/invalid/ill-defined, etc., but I can't provide the argument myself.
BTW, I know (from my textbook) that the splitting field over $\mathbb{Q}$ of $X^4+1$ is $\mathbb{Q}[i,\sqrt{2}]$, but from this knowledge I'm not able to discern exactly where my reasoning above goes wrong.

Let me try to elaborate on my confusion here.
Suppose I treat $\alpha$ as an "opaque symbol".  All I know is that $\alpha^4 + 1 = 0$.  How do I determine that the degree of the extension $\mathbb{Q}[\alpha]$ is 4?
 A: In fact there isn't a confusion, as we have that $\mathbb{Q}\left(e^{\frac{i\pi}{4}}\right) = \mathbb{Q}(i,\sqrt{2})$
The inclusion $\mathbb{Q}\left(e^{\frac{i\pi}{4}}\right) \subseteq \mathbb{Q}(i,\sqrt{2})$ is trivial, as $e^{\frac{i\pi}{4}} = \frac 1{\sqrt{2}}(1+i)$. On the other side we have that $i = \left(e^{\frac{i\pi}{4}}\right)^2 \in \mathbb{Q}\left(e^{\frac{i\pi}{4}}\right)$. From here trivially $\sqrt{2} \in \mathbb{Q}\left(e^{\frac{i\pi}{4}}\right)$ and so we can conclude that $\mathbb{Q}(i,\sqrt{2}) \subseteq \mathbb{Q}\left(e^{\frac{i\pi}{4}}\right)$
A: Note that $\alpha^2=i$ so that $i$ is in the field you have defined.
Also $(\alpha+\alpha^{-1})^2=i+2+\frac 1i=2$ so that $\sqrt 2$ is in the field you have defined.
So your field contains $\mathbb Q (\sqrt 2, i)$, and you can just as easily show the inclusion the other way by writing $\alpha = (1+i)\frac {\sqrt 2}2$.
In fact $\alpha$ is a primitive element for the field $\mathbb Q(\sqrt 2, i)$, which is known to exist (by the primitive element theorem).
Fields differently specified can be the same without this being immediately obvious.
