TITLE: Structure of $\mathbb{Z}[[x]]/(x-n)$
QUESTION [12 upvotes]: Is $\mathbb{Z}[[x]]/(x-6) \cong \widehat{\mathbb{Z}}_2 \times \widehat{\mathbb{Z}}_3$?

It seems intuitive that $\mathbb{Z}[[x]]/(x-p)$ is the p-adic numbers, and I think this is not too hard to show formally.  Really, it doesn't seem too hard to show this is true for any number p, not just primes.  This gives a nice explicit description of $\mathbb{Z}[[x]]/(x-n)$, when combined with the well known (and hopefully well remembered) facts that $p^k$-adics is just a fancy name for p-adics, and that n-adics in general are the direct product of the p-adics for the distinct primes p dividing n.
Everything was fine and good until I tried to show this direct product decomposition directly in the ring $\mathbb{Z}[[x]]/(x-6)$.  I want to use Chinese remainder theorem on the ideals $(x-2)$, $(x-3)$, and $(x-6)$.  Clearly $(x-2)+(x-3) = (1)$, but I think it is also pretty clear that $(x-6) \neq (x-2) \cap (x-3)$.
Indeed, I don't think $x-6$ is even an element of $(x-2)$, since $$\frac{x-6}{x-2} = 3 + x + \tfrac12 x^2 + \ldots + \tfrac1{2^n} x^{n+1} + \ldots \notin \mathbb{Z}[[x]]$$
What has gone wrong?

REPLY [4 votes]: I left this as a comment above, but it is at the end of a long thread, so I will also note it here: the isomorphisms being discussed here are a special case of the
more general isomorphism discussed in this question, which states that for any Noetherian ring $R$, if $I = (a_1,\ldots,a_n)$ is an ideal,
then $R[[x_1,\ldots,x_n]]/(x_1-a_1,\ldots,x_n-a_n) =$ the $I$-adic completion of $R$.