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Let $\mathbb Z$ be the ring of rational integers. Consider the power series ring $\mathbb Z[[x]]$. It is known that $\mathbb Z[[x]]$ is unique factorization domain. What are the primes in $\mathbb Z[[x]]$?

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up vote 10 down vote accepted

This is not a complete answer (but perhaps there is none). It is well-known that $\mathbb{Z}[[x]]$ is a $2$-dimensional regular noetherian domain. Let us look at the prime ideals. Considering the fibers of $\mathrm{Spec}(\mathbb{Z}[[x]]) \to \mathrm{Spec}(\mathbb{Z})$, there is a decomposition (as a set)

$$\mathrm{Spec}(\mathbb{Z}[[x]]) = \mathrm{Spec}(\mathbb{Z}[[x]] \otimes_{\mathbb{Z}} \mathbb{Q}) \sqcup \coprod\limits_p \mathrm{Spec}(\mathbb{Z}[[x]] \otimes_{\mathbb{Z}} \mathbb{F}_p).$$

In general one can show $R[[x]] \otimes_R R/(a) = R[[x]]/(a R[[x]])=R/(a) [[x]]$. In particular, $\mathbb{Z}[[x]] \otimes_{\mathbb{Z}} \mathbb{F}_p = \mathbb{F}_p[[x]]$ is a DVR, the only prime ideals are $(0)$ and $(x)$. The corresponding prime ideals in $\mathbb{Z}[[x]]$ are $(p)$ and $(p,x)$. Thus, the special fibers are easy to understand. But the generic fiber is more complicated: $\mathbb{Z}[[x]] \otimes_{\mathbb{Z}} \mathbb{Q} = (\mathbb{Z} \setminus \{0\})^{-1} (\mathbb{Z}[[x]])$ is a proper subring of $\mathbb{Q}[[x]]$; namely, it consists of those power series with rational coefficients whose denominators are bounded. For example, $\frac{2}{2-x} = \sum_{i=0}^{\infty} \left(\frac{x}{2}\right)^i$ is not contained. As far as I know, there is no easy description of this ring.

EDIT: Have a look at the paper Factoring formal power series over principal ideal domains by Jesse Elliott (arXiv). It contains an irreducibility criterion for elements of $\mathbb{Z}[[x]]$, as well as a factorization algorithm (actually with $\mathbb{Z}$ replaced by an arbitrary PID). This is rather involved and suggests that there is no "computable list" of all prime elements of $\mathbb{Z}[[x]]$.

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