Is $ \mathbb{Q} \subset L=\mathbb{Q}[X]/(X^4+1)$ a Galois extension? Let $L=\mathbb{Q}[X]/(X^4+1)$. Is $L/\mathbb{Q}$ a Galois extension?
$[L:\mathbb{Q}] = 4$. How do I show that there there are 4 elements in $\mathrm{Aut}(L/\mathbb{Q})$?
 A: Seeing this as a subfield of $\mathbb C$ by sending $X$ to $e^{i\pi/4}$, we can use the polar decomposition and write
$$
e^{i\pi/4} = \cos(\pi/4) + i \sin(\pi/4) = \frac{\sqrt 2 + \sqrt 2 i}2. 
$$
So $L = \mathbb Q(\sqrt 2, \sqrt{-2})$, thus our only options are $[L:\mathbb Q] \in \{2,4\}$. If you are only interested in your answer, you can see that $\mathbb Q(\sqrt 2)$ is a real subfield of $\mathbb C$, and therefore could not contain $\sqrt{-2}$. Therefore $[L: \mathbb Q] > 2$, so it must equal $4$. Showing $L/\mathbb Q$ is Galois is now easy since the other roots of $X^4+1$ are $\zeta = e^{i\pi/4}$, $\zeta^3, \zeta^5$ and $\zeta^7$, all elements of $L$, so $L$ is a splitting field for $X^4+1$. The other powers of $\zeta$ are $\pm 1$ and $\pm i$.
For a technique that's a bit more abstract, you can try factoring $X^4 + 1$:
$$
X^4 + 1 = (X^2)^2 + 1 = (X^2 - i)(X^2 + i)
$$
(where $i$ is some square root of $-1$ here). Since we are trying to find the square root of $i$ and $-i$ which are primitive fourth roots of unity, let us consider a primitive eighth root of unity, say $\zeta$. We get
$$
X^4+1 = (X-\zeta)(X+\zeta)(X-i\zeta)(X+i \zeta) = (X-\zeta)(X-\zeta^7)(X-\zeta^3)(X-\zeta^5). 
$$
The field generated by the roots of $X^4+1$ is the same as that generated by the roots of $X^8 - 1$ because $X^8-1 = (X^4-1)(X^4+1)$ and roots of $(X^4+1)$ squared are roots of $(X^4-1)$ since $\zeta^4 = -1$ means $(\zeta^2)^4 = \zeta^8 = 1$. If $\zeta$ is a primitive root of unity, then the roots of $X^8-1$ are just $\zeta^i$ for $0 \le i \le 7$. So $L$ is spanned as a vector space by $1,\zeta,\zeta^2,\zeta^3$, and the other powers are just these times a minus sign, i.e. $\zeta^4 = -1$, $\zeta^5 = -\zeta$, and so on. In particular, $[L: \mathbb Q]$ has degree at most $4$.
But it has degree at least $4$ since the irreducible polynomial $X^4 + 1$ has a root in it. It's a bit hard to show it's irreducible over $\mathbb Q$ without looking directly at its roots over $\mathbb C$; one way is to apply the $\mathbb Q$-algebra automorphism of $\mathbb Q[X]$ sending $X$ to $X+1$ which maps $X^4+1$ to $(X+1)^4+1 = X^4 + 4X^3 + 6X^2 + 4X + 2$ and notice that this is Eisenstein with the prime $p=2$. Automorphisms preserve factorizations in UFDs, and therefore also preserve irreducible polynomials in free polynomial algebras like $\mathbb Q[X]$.
Showing that $L/\mathbb Q$ is Galois is actually the easy part once you know that $X$ in $\mathbb Q[X]/(X^4+1)$ is a primitive $8^{\text{th}}$ root of unity (which is easily seen since $X^4 = -1$ and $X^8 = 1$; the roots of unity form a group, and Lagrange's theorem says that the order of an element of a group divides the order of that group, so this root has order $8$, and therefore is primitive). So you can see $L$ as the splitting field of $x^8-1$ or $x^4+1$, your choice; in any case, it is a Galois extension, and therefore has $[L:\mathbb Q] = 4$ automorphisms.
Hope that helps,
