Irreducibility in $\mathbb{Z}[\sqrt{-2}][X]$ I'm working on the following question:

(a) Prove that the ring $R = \mathbb{Z}[\sqrt{-2}]$ is Euclidean.
(b) Show that $R/(3+2\sqrt{-2})$ is a field. What is the characteristic of this field?
(c) Show that the polynomial $X^4+3$ is irreducible over the field $\mathbb{F}_{17}$ of $17$ elements; and deduce that $f(X) = X^4 − 170X^3 + 9 + 4\sqrt{−2} \in R[X]$ is irreducible.
(d) Is the polynomial $Y^4 −f(X)\in R[X,Y]$ irreducible? (Why?)

I have done part (a). For (b) I said that by (a), $R$ is a PID, hence prime and irreducible are equivalent. Thus $(3+2\sqrt{-2})$ is maximal if and only if $3+2\sqrt{-2}$ is prime. Now using the norm function defined by
$$\varphi\colon R\setminus\{0\}\to\mathbb{N}_{>0}$$ where $\varphi(a+b\sqrt{-2}) = a^2+2b^2,$ we have that $\varphi(3+2\sqrt{-2}) = 9 + 8 = 17,$ which is prime in $\mathbb{Z}$, hence $3+2\sqrt{-2}$ is prime in $R$. Therefore $R/(3+2\sqrt{-2})$ is a field. I'm not sure how to determine the characteristic. I had hoped to get some relation using $$3+2\sqrt{-2}=0,$$ but I wasn't getting anywhere.
For (c), I don't know the best way to do this. One way would be to check that for $n\in\{0,1,\dots,16\}$ we have that $n^4+3\not\equiv 0\bmod{17}$ to show that there is no linear factor. Then we could test that
$$X^4+3 = (X^2+aX+b)(X^2+cX+d)$$ has no solutions for $a,b,c,d\in\mathbb{F}_{17}$. Is there a better approach?
For (d), I'm thinking $Y^4-f(X)$ is irreducible by Eisenstein?
 A: For $(c)$ a quadratic factor $f$ cannot exist, else  in $\,\Bbb F_{17^{\large 2}} = \Bbb F_{17}[x]/f\,$ we get a contradiction
$$ \color{#c00}{x^{\large 4} = -3}\ \Rightarrow\  1 = x^{\large 17^{\Large 2}-1}\! = ((\color{#c00}{x^{\large 4}})^{\large 4\cdot 18}\! = (\color{#c00}{-3})^{\large 4\cdot 18}\! = (-4)^{\large 18}\! = (-4)^{\large 2}\! = -1$$
A: Trying the same as with the Gaussian Integers, and denoting $\;I:=\langle 3+2\sqrt{-2}\rangle\le\Bbb Z[\sqrt{-2}]\;$, define the following ring homomorphism (the canonical projection)
$$\phi:\Bbb Z\to\Bbb Z[\sqrt{-2}]/I\;,\;\;\phi(m):=m+I$$
From this, we clearly have that
$$m\in\ker\phi=\Bbb Z\cap I\iff m=(3+2\sqrt{-2})(a+b\sqrt{-2})=3a-4b +(3b+2a)\sqrt{-2}\iff$$
$$2a+3b=0\iff2a=-3b\implies \ker\phi=\left\{\,\left(3+2\sqrt{-2}\right)\left(-\frac32b+b\sqrt{-2}\right)\,\right\}=$$
$$\left\{\,-\frac92b-4b=-\frac{17}2b\,\right\}=17\Bbb Z\cong\Bbb F_{17}$$
since clearly it must be that $\;b\;$ is even (otherwise $\;2a+3b=0\;$ is impossible), and we get that characteristic of the field is $\;17\;$ .
To show $\;x^4+3\in\Bbb F_{17}[x]\;$ is irreducible: first, show it has not roots in $\;\Bbb F_{17}\;$. Now, if it factors as the product of two quadratics: as $\;f(x)=x^2-3\in\Bbb F_{17}[x]\;$ is irreducible (just checking what the squares modulo $\;17\;$ are...), then $\;\Bbb F_{17}[x]/\langle f(x)\rangle\cong\Bbb F_{17^2}\;$ . From here that both quadratics in the factoring of $\;x^4+3\;$ must split in the last field (remember: there is one unique field of any order a power of a prime), which means $\;\sqrt[4]{-3}\;$ exists there...There are some criteria for irreducibility over finite fields of cubics and quartics (By Skolem, Leonard and etc.), but I really don't know them.
A: Parts (a),(b),(c) have been handled by others. I do want to add the following point of view to Bill Dubuque's argument. 
Quadratic reciprocity shows that both $\pm3$ are quadratic non-residues modulo $17$. Because $17$ is a Fermat prime,
$17-1=2^4$, this already implies that both $\pm3$ are of order sixteen in $\Bbb{F}_{17}^*$ (by Euler the quadratic residues are the solutions of $x^8\equiv1$). As $16$ is even it follows that any zero of $x^4\pm3$ must be a root of unity of order exactly $64$. But $m=4$ is the smallest exponent such that
$$17^m\equiv1\pmod{64},$$
so the zeros of $x^4\pm3$ generate the field $\Bbb{F}_{17^4}$ and hence $x^4\pm3$ are both irreducible. 

This explains why Bill's argument will work, and shows that the same calculation will work with any quadratic non-residue in place of $3$. Also, I often use the same method to deduce the degrees of factors of binomials $m(x)=x^n-a$ modulo $p$. Find the order $\ell$ of $a$ modulo $p$, and then try and deduce the order $L$ of a zero $\alpha$ of $m(x)$ in some extension field of $\Bbb{F}_p$. The caveat is that we don't always have $L=n\ell$, because $\ell$ may have prime factors that not shared by $n$. Think about the rule familiar from cyclic groups:
$$
\ell=\operatorname{ord}(a)=\operatorname{ord}(\alpha^n)=\frac{\operatorname{ord}(\alpha)}{\gcd(n,\operatorname{ord}(\alpha))}=\frac{L}{\gcd(n,L)}
$$
that narrows down the possibilities for $L$. With possibilities for $L$ known from this, we can then identify the smallest extension fields of $\Bbb{F}_p$ containing the roots of unity of order $L$ as above.

Including both $\pm3$ above gives us the following way to settle (d). If
$$
Y^4-f(X)=g(X,Y)h(X,Y)
$$
then also $Y^4-f(0)=g(0,Y)h(0,Y)$ in $R[Y]$. Reducing this modulo $\mathfrak{p}=(3+2\sqrt{-2})$ and the fact that $f(0)\equiv3\pmod{\mathfrak{p}}$ leads to a factorization of $Y^4-3$ in $\Bbb{F}_{17}[Y]$. We just saw that such a factorization must be trivial, so $g(0,Y)$ (or $h(0,Y)$) must reduce to a constant modulo $\mathfrak{p}$. But the leading powers of $Y$ in $g(X,Y)$ must have coefficient $\pm1$, so we can conclude that $g(X,Y)\in R[X]$ which is absurd. 
You can, of course, use (c) and $f(X)=g(X,0)h(X,0)$ to conclude that either $g(X,0)$ or $h(X,0)$ must be a constant. I don't know which way is easier
A: for part (b) to find the characteristic of the field $J=R/(3+2\sqrt{-2})$ find the order of $1$ in the additive subgroup of $J$ . 
