$\mathbb{Q}[x]/(x^4 + 1)$ is a field?
I understand, that in order to $\mathbb{Q}[x]/(x^4 + 1)$ be a field, polynomial $(x^4 + 1)$ must be irreducible. And I checked its roots, they are not in $\mathbb{Q}$: $\pm \frac{1\pm i}{\sqrt{2}}$. But how I can simply check that in $\mathbb{Q}[x]$ there are no divisors of $x^4+1$? There are infinitely many polynomials of degree less than $4$ in $\mathbb{Q}[x]$.
Since $\mathbb{Q}[x]$ is a PID, it is sufficient to prove that $x^4+1$ is irreducible as element of $\mathbb{Q}[x]$
(Observe that the same polynomial is not irreducible over $\mathbb{R}[x]$, it can be useful to find various counterexamples!)
Now $x^4+1$ is reducible if and only if $(x+1)^4+1$ is reducible, but the latter polynomial is irreducible by Heisenstein criterion with $p=2$.