Let's say, for concreteness, that $R = k[[t_1, \dots, t_m]]$. Suppose $P$ is a prime of $R[[x]]$, satisfying $P\neq 0$ and $P\cap R = 0$, and write $\mathrm{Frac}(R) = F$ and $\mathrm{Frac}(R[[x]]/P) = E$. What's the structure of $E/F$?
I had initially suspected that $E/F$ was algebraic. My proof went as follows.
- Let $S = R\setminus \{0\}$; then we can localise $R[[x]]$ to get $S^{-1}R[[x]]$. We've assumed $P\cap S = \varnothing$, so extending $P$ gives us a prime ideal $S^{-1}P$, and it satisfies $S^{-1}P\cap R[[x]] = P$, by standard localisation properties.
- We can further complete $(x)$-adically to get $F[[x]]$. Extending $S^{-1}P$ gives us a nonzero ideal $Q$, and it satisfies $Q\cap S^{-1}R[[x]] = S^{-1}P$ by faithful flatness.
- All of the above shows that we have a natural inclusion $R[[x]]/P \hookrightarrow F[[x]]/Q \cong F[x]/(x^n)$ for some $n$.
Everything on the right is certainly algebraic, but that's not what worries me here: what worries me is that the image of $x+P$ is nilpotent, which implies $x\in P$. This obviously isn't the case generically. So I've messed up somewhere, but I can't put my finger on where.
What's gone wrong? And what's known about the structure of $E/F$? Or are there any similar results on the structure of $R[[x]]/P$, or is my suspicion way off?