I've been working through Alain Robert's A Course in $p$-adic Analysis and there's an exercise regarding $K((x))$ for a field $K$ in the first chapter that I'm hoping is more related to the topic than the obvious "It's another thing with an ultrametric."
The full context given by the exercise is: Define a metric $d$ on $K[x]$ for which $d(f,g)=0$ when $f=g$ and $d(f,g)=\theta^{\nu(f,g)}$ otherwise, where $\nu(f,g)$ is the smallest exponent of $x$ for which the coefficients of $f,g$ differ. Notice that this extends to formal series $K[[x]]$ and finally to Laurent series $K((x))$. Prove that $d$ is an ultrametric, that $K[[x]]$ is a completion of $K[x]$, and that $K((x))$ is a completion of $K[x,x^{-1}]$.
I've done all that, and the only thing that required $K$ to be a field was the assumption that $K[[x]][x^{-1}]=K((x))$.
What I'd really like to be able to do is, for any integer $n\ge 2$, construct the $n$-adic rationals $\mathbb{Q}_n$ by starting with $\mathbb{Z}[[x]][x^{-1}]$ and simply assigning $x=n$. (I originally feared that only the zero map could do this as a homomorphism, but Lubin reminded me that $\mathbb{Z}[[x]][x^{-1}]$ is not a field.) I know that $x-n$ is not an invertible element in $\mathbb{Z}[[x]][x^{-1}]$, so is it the case that $\mathbb{Z}[[x]][x^{-1}]/(x-n)\cong\mathbb{Q}_n$?
Here's a possible proof. Each equivalence class in $\mathbb{Z}[[x]][x^{-1}]/(x-n)$ contains a unique element with each coefficient $a_i$ such that $0\le a_i\le n-1$. To see this, let $a=\sum a_ix^i$ be an arbitrary representative and apply the division algorithm to the coefficient $a_i$ of the lowest-degree element $a_ix^i$ to find integers $q, r$ such that $a_i=qn + r$ and $0\le r\le n-1$. Then $a + q(x-n)x^i$ is another representative of the same class with coefficient $r$ of $x^i$ between $0$ and $n-1$ inclusive. Induct to find a representative for which all coefficients are between $0$ and $n-1$ inclusive. To see that this element is unique, notice that any nonzero element of $(x-n)$ has a lowest-order coefficient (say, of $x^i$) that is a nonzero multiple of $n$ and thus adding it to our representative produces an element with coefficient of $x^i$ no longer between $0$ and $n-1$ inclusive.
Create a bijection $\mathbb{Z}[[x]][x^{-1}]/(x-n)\to\mathbb{Q}_n$ using this unique representative, corresponding each equivalence class with the $n$-adic rational that has the same coefficients. This bijection is a homomorphism: It's clear that $1\mapsto 1$ since it's an element already in the proper form. The rules for addition in $\mathbb{Q}_n$ correspond to those in $\mathbb{Z}[[x]][x^{-1}]$ but with the extra carrying condition which is exactly the process described above. Same for multiplication.
Am I missing anything, or does this actually work the way I want? It even seems the ultrametric defined for $\mathbb{Z}[[x]][x^{-1}]$ corresponds exactly with that of $\mathbb{Q}_n$ as well, so they should be isomorphic as topological rings. It also seems that if all this is true, then $\mathbb{Z}[[x]]/(x-n)$ must be isomorphic to the $n$-adic integers $\mathbb{Z}_n$ as topological rings.
Too long for a comment:
I think you’re right that $K$ doesn’t need to be a field for that construction to work. And I find your “really like to do” interesting, but if by $\Bbb Z_n$ you mean the integers modulo $n$, you’re off the mark. But maybe you meant the $n$-adic integers?
An evaluation of the type you’re trying for has to be a ring homomorphism, and any such will have a kernel that’s nontrivial. Since I’m not precisely sure of your intent, I can’t say whether making the kernel an ideal containing $x-n$ or maybe generated by $x-n$: that would be just factoring out the $x-n$, as you fear. (Being an old fogy, I do not believe that “mod” can be used as a verb.)
Perhaps if you clarify your question, I can rewrite or cancel this answer. I don’t think I can have been very helpful at this stage.