Factor $x^4-4$ into irreducible factors over $\mathbb{Q}$, over $\mathbb{R}$, and over $\mathbb{C}$ Factor $x^4-4$ into irreducible factors over $\mathbb{Q}$, over $\mathbb{R}$, and over $\mathbb{C}$
So for  $\mathbb{Q}$ and  $\mathbb{R}$, I can get some factors in the $\mathbb{C}$. 
but what I can do for $\mathbb{C}$?

Note that your polynomial is a difference of squares:
$$
  x^4 -4 = (x^2-2)(x^2+2).
$$
This gives you a factorization into irreducibles over $\mathbb{Q}$ since the two products, $x^2-2$ and $x^2+2$, have no roots in $\mathbb{Q}$. (This, plus the fact that they are degree 2, imply that they are irreducible over $\mathbb{Q}$.)
We can go further in $\mathbb{R}$ by splitting the first term:
$$
  x^4-4 = (x-\sqrt 2)(x+\sqrt 2)(x^2+2).
$$
The first two terms are clearly irreducible, and $x^2+2$ has roots $\pm i \sqrt 2$, which do not lie in $\mathbb{R}$. So, all three terms are irreducible over $\mathbb{R}$, so this is the factorization over $\mathbb{R}$.
Finally, over $\mathbb{C}$ each one splits, and so we have
$$
  x^4-4 = (x-\sqrt 2)(x+\sqrt 2)(x-i\sqrt 2)(x+i\sqrt 2)
$$
as irreducibles over $\mathbb{C}$.
