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
$$\varphi\colon R\setminus\{0\}\to\mathbb{N}_{&gt;0}$$ 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?
For $(c)$ a quadratic factor $f$ cannot exist, else  in $\,\Bbb F_{17^{\large 2}} = \Bbb F_{17}[x]/f\,$ we get a contradiction

$$ \color{#c00}{x^{\large 4} = -3}\ \Rightarrow\  1 = x^{\large 17^{\Large 2}-1}\! = ((\color{#c00}{x^{\large 4}})^{\large 4\cdot 18}\! = (\color{#c00}{-3})^{\large 4\cdot 18}\! = (-4)^{\large 18}\! = (-4)^{\large 2}\! = -1$$
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 $J$ .