Is $[F(x) : F] = 4$ if $x$ is a root of $X^8 + 2 X^4 + 1$? Let $K / F$ be a field extension and let $x \in K$ be a root of the polynomial $X^8 + 2 X^4 + 1$. It is not difficult to prove that $[F(x) : F] \leq 4$, but can we write $[F(x) : F] < 4$ for any case? Since $X^8 + 2 X^4 + 1 = {(X^4 + 1)}^2$, then $x$ is a root of $X^4 + 1$. I needed this argument to show that $[F(x) : F] \leq 4$. I have thought on supposing that $[F(x) : F] < 4$ and arrive to a contradiction, but, for instance, how can I prove that $x \notin F$? It is the same than showing $[F(x) : F] \neq 1$ and I do not find any contradiction here to end my proof. Thank you very much in advance.
 A: It depends on the field $F$. If $F$ contains a fourth root of $-1$, then $x \in F$, and $[F(x):F] = 1$. If $F$ contains a square root of $-1$, then $X^4+1 = (X^2+\sqrt{-1})(X^2-\sqrt{-1})$, so $[F(x):F] \leq 2$.
An example of the first phenomenon is, say, $F = \mathbb{Q}(\sqrt[4]{-1})$, or if you prefer, $\mathbb{Q}\Big(\frac{1+\sqrt{-1}}{2}\Big)$, a concrete subfield of $\mathbb{C}$. Then $x \in F$ and $[F(x):F]=1$.
An example of the second phenomenon is $F = \mathbb{Q}(\sqrt{-1}) = \mathbb{Q}(\sqrt{-1})$. Then $x^2 \in F$ (because $x^2 = \pm \sqrt{-1}$) and $[F(x):F] \leq 2$.
For $F = \mathbb{Q}$ we have $[F(x):F]=4$. To see this, we have to show that the minimal polynomial of $x$ is $X^4+1$, so we just have to show that $X^4+1$ is irreducible. We can use Eisenstein’s criterion, with a substitution of $X+1$ for $X$:
[
  (X+1)^4 + 1 = X^4 + 4 X^3 + 6 X^2 + 4 X + 2,
]
which is irreducible over $\mathbb{Z}$, hence so is $X^4+1$ (and irreducible over $\mathbb{Q}$ by Gauss’s Lemma).
A: If $F$ is the two-element field, then $X^8+2X^4+1=(X+1)^8$, so the only root of the polynomial is $1$ and, of course, $F(x)=F$.
The degree can also be $2$ or $4$. For instance, if the base field is $\mathbb{R}$, then $X^4+1$ is reducible:
$$
X^4+1=(X^2-\sqrt{2}\,X+1)(X^2+\sqrt{2}\,X+1)
$$
and the two factors are irreducible, as they have negative discriminant. Thus any root of the polynomial has degree $2$ over $\mathbb{R}$.
If $F$ is the rational field, then $X^4+1$ is irreducible and roots have degree $4$ over $\mathbb{Q}$.
