I am trying to find the prime ideals of $\mathbb{Z}[[X]]$.
$\mathbb{Z}[[X]]$ is the ring of formal power series with coefficients in $\mathbb{Z}$. Addition and multiplication are defined analogously to $\mathbb{Z}[X]$
So far, I've got the following. Suppose $\mathfrak{p}$ is prime ideal of $\mathbb{Z}[[X]]$. Note that, $\mathfrak{p}^c = \mathfrak{p} \cap \mathbb{Z}$ is prime ideal of $\mathbb{Z}$. So we consider the following two cases -
Case 1: $\mathfrak{p} \cap \mathbb{Z} = (p)$
Then note that
$$\frac{\mathbb{Z}[[X]]}{\mathfrak{p}} \cong \frac{\mathbb{Z}[[X]]/(p)}{\mathfrak{p}/(p)} \cong \frac{\mathbb{Z_p}[[X]]}{\mathfrak{\overline{p}}}$$
But $\mathbb{Z}_p$ is a field and we precisely know all the ideals of $\mathbb{F}[[X]]$ when $\mathbb{F}$ is a field. So $\mathfrak{\overline{p}} = (X)$ or $(0)$. Hence $\mathfrak{p} = (p, X)$ or $(p)$ where $p$ is a prime number.
Case 2: $\mathfrak{p} \cap \mathbb{Z} = (0)$
I have not able to work-out this case completely but nonetheless, here is what I have uptil now (assume $\mathfrak{p} \neq (0)$) -
If $X \in \mathfrak{p}$ then $\mathfrak{p} = (X)$ because if $f \in \mathfrak{p}$ then let $$f(X) = \sum_{n \geq 0} a_n X^n = a_0 + Xg(X)$$ where $g(X) = \sum_{n \geq 1} a_n X^{n-1}$. Hence $$a_0 = f(X) - Xg(X) \in \mathfrak{p} \implies a_0 = 0 \implies f(X) = Xg(X) \implies \mathfrak{p} = (X)$$
If $X \not \in \mathfrak{p}$ then let $0 \neq f \in \mathfrak{p}$. Then $$f(X) = X^kg(X) = X^k\left(\sum_{n \geq 0} a_n X^n\right)$$ for some $k \geq 0$, such that $a_0 \neq 0$. This implies, $g(X) \in \mathfrak{p}$ and so $a_0 \neq \pm 1$. I don't know how to proceed further.
Besides, the only non-tirival map $\mathbb{Z}[[X]] \to \mathbb{Z}$, I can think of is $f(X) \mapsto f(0)$. Yet I don't see why $(X+2)$ is not a prime ideal of $\mathbb{Z}[[X]]$ (or is it?)