Is $x^4+2$ irreducible over $\Bbb{Q}(i)$? Let $f(x)=x^4+2$. Using the Eisenstein test to $f(x+2)$, one can show that $f$ is irreducible over ${\Bbb Q}$. 
Let $\beta$ be a complex root of $f$. Then the question in the title is equivalent to 

Is $[{\Bbb Q}(\beta,i):{\Bbb{Q}(i)}]=4$?

I tried another somehow "equivalent" problem also:

Is ${\Bbb Q}(\sqrt[4]{2},i)={\Bbb Q}(\sqrt[4]{-2},i)$?

But I don't see how to go on. 
 A: The polynomial $x^4+2$ is irreducible over $\Bbb{Q}(i)$, because it is irreducible in the ring
$\Bbb{Z}_5[x]$. Normally this only implies irreducibility in $\Bbb{Z}[x]$, but here special circumstances apply. Namely $2$ and $3$ are both square roots of $-1$ modulo five, and thus we have $\Bbb{Z}_5[i]=\Bbb{Z}_5$.
Another way of saying the same thing is that if $f(x)$ were reducible over $\Bbb{Z}[i]$, then it would be reducible over the quotient ring $\Bbb{Z}[i]/\langle 2+i\rangle\cong\Bbb{Z}_5$.

Edit: In the comments I show how basic properties of finite fields imply that $f(x)$ is irreducible over $\Bbb{Z}_5$. It is very much possible that those properties have not been covered yet in a course, where this exercise came from. Hence I include a direct proof of that key fact. We easily see that $f(x)$ has no zeros in $\Bbb{Z}_5$. Therefore it cannot have any linear factors. The remaining possibility is that 
$$
f(x)=(x^2+ax+b)(x^2+cx+d)
$$
is a product of two irreducible quadratic factors in $\Bbb{Z}_5[x]$. The coefficient
of the cubic term in that product is $a+c$, so we can conclude that $c=-a$. The coefficient
of the linear term is $ad+bc=ad-ba=a(b-d)$. Thus either $a=c=0$ or $b=d$. The former case
is impossible, because
$$
(x^2+b)(x^2+d)=x^4+(b+d)x^2+bd
$$
can be $f(x)$ only when $d=-b$ and thus $2=-b^2$, which is impossible. Therefore we only need to worry about the putative factorization
$$
x^4+2=(x^2-ax+b)(x^2+ax+b)=x^4+(2b-a^2)x^2+b^2.
$$
But this implies $2=b^2$, which is impossible for $b\in\Bbb{Z}_5$.
It may be worth noting that both my proofs of this fact in a way depend on the fact that $\pm2$ are not quadratic residues modulo five. 
