Factor $x^4-2$ into irreducible polynomials. Factor $x^4-2$ as a product of irreducible polynomials in $\mathbb{Z}_3[x]$, $\mathbb{Z}_{17}[x]$, $\mathbb{Q}[x]$, $\mathbb{R}[x]$, and $\mathbb{C}[x]$.
I think I figured out $\mathbb{R}[x]$ with $(x^2 + \sqrt{2})(x-\sqrt[4]{2})(x+\sqrt[4]{2})$, I don't think the first term can be factored any further.
I also got none for $\mathbb{Q}[x]$.
For $\mathbb{C}[x]$ I got $(x^2 - 2i)(x^2 - 2i)$.
$\mathbb{Z}_3[x]$ and $\mathbb{Z}_{17}[x]$, I got none from plugging in all their respective numbers and not getting 0 for any of them.
Could someone let me know if this is right? I am unsure if I am missing something for $\mathbb{Q}[x]$, $\mathbb{Z}_3[x]$ and $\mathbb{Z}_{17}[x]$ if it is reducible.
 A: You had a wrong sign in the factorization in $\mathbb{R}[x]$ (typo):
$$
(x^2+\sqrt{2}\,)(x-\sqrt[4]{2}\,)(x+\sqrt[4]{2}\,)
$$
is the factorization: the quadratic factor has no roots in $\mathbb{R}$, so it's irreducible.
The polynomial is indeed irreducible in $\mathbb{Q}[x]$: it has no rational root so it would be a product of two quadratic factors, but such a decomposition would also lead to a splitting in $\mathbb{R}[x]$ and the only way you can combine the previous factorization in the required way would be $(x^2+\sqrt{2}\,)(x^2-\sqrt{2}\,)$ and the factors are not in $\mathbb{Q}[x]$. Or you can use Eisenstein's criterion, which applies to $x^4-2$ for the prime $2$.
Sorry, but the factorization in $\mathbb{C}[x]$ is wrong: it should be a product of linear factors and you get them from the factorization in $\mathbb{R}[x]$.
I'll do the case $\mathbb{Z}_3[x]$. The polynomial has no roots in $\mathbb{Z}_3$, but it could be a product of two (monic) quadratic factors:
$$
x^4-2=(x^2+ax+b)(x^2+cx+d)
$$
Since $bd=-2=1$, you can have $b=1$ and $d=1$ or $b=2$ and $d=2$. In the first case we have
$$
(x^2+ax+1)(x^2+cx+1)=x^4+(a+c)x^3+(2+ac)x^2+(a+c)x+1
$$
and so you need $a+c=0$ and $ac+2=0$. Hence $c=-a$ and $a^2=2$, which is impossible.
In the second case we have
$$
(x^2+ax+2)(x^2+cx+2)=x^4+(a+c)x^3+(1+ac)x^2+(2a+2c)x+1
$$
Again $a+c=0$ and $1+ac=0$. Since $c=-a$, we obtain $a^2=1$. Well, this is indeed possible: we have $a=1$ or $a=2$. Thus the factorization is
$$
(x^2+x+2)(x^2+2x+2)
$$
A: The $\mathbb{Z}_{17}$ case remains.  The units of $\mathbb{Z}_{17}$ form a cyclic group of order $16$.  The order of $2$ is $8$, as $2^4=-1$.  Thus $\sqrt{2}\in \mathbb{Z}_{17}$, and your polynomial factorises as: $$x^4-2=(x^2+\sqrt{2})(x^2-\sqrt{2}).$$
It cannot factorise further as if there were a linear factor, then $\mathbb{Z}_{17}$ would contain an element $\alpha$ with $\alpha^4=2$, so $\alpha^{16}=-1$, which is impossible as for any element $\alpha \in \mathbb{Z}_{17}$ we have $\alpha^{16}=1$.
To complete the question you should work out $\sqrt{2}$ in $\mathbb{Z}_{17}$.  The easiest way to do this is check values of $2+17k$, for $k=0,1,2,3,\ldots$ till you get a perfect square.
