Modules over Laurent series rings

Question

Let $k[x]$ be the ring of polynomials over a field k in one variable x. A $k[x]$-module is a k-vector space together with a linear endomorphism (the action of x).

The field $k(x)$ of rational functions is the maximal localization of $k[x]$, i.e. field of fractions. (Edit: yes, I was an idiot in what I wrote here first; thanks James.)

The ring $k[[x]]$ of formal power series is the completion of $k[x]$ at the ideal $(x)$. It is natural to consider only $k[[x]]$-modules which are likewise complete, meaning roughly that we can sum "infinite linear combinations" whose coefficients are increasing powers of x. Completeness of a $k[x]$-module automatically makes it a $k[[x]]$-module.

Finally, both $k(x)$ and $k[[x]]$ embed into the ring $k((x))$ of formal Laurent series. I have two questions, which I ask together because they seem related:

1. Is there a general ring-theoretic construction, akin to localization and completion, which produces $k((x))$ from $k[x]$?

2. Is there a natural condition to impose on $k((x))$-modules, akin to completeness for $k[[x]]$-modules, which would enable us to sum infinite linear combinations with coefficients increasing in powers of x?

Answer 1

1. In general, if $A$ is a commutative ring and $I$ is a finitely generated ideal in $A$, given an $I$-adically complete $A$-module $M$, and a multiplicative set $S \subseteq A$, when working in the adic category, one usually replaces the usual localization $S^{-1}M$ by exactly what you described - the complete localization $\Lambda_I (S^{-1}M)$ where $\Lambda_I$ is the completion functor. This behaves in many ways like the usual localization. For example, in ring theory a module $M$ is the zero module if and only if $M_p = 0$ for all $p \in Spec A$. If $A$ and $M$ are $I$-adically complete, one may replace this operation by the complete localization, and show that such a complete module $M$ is zero if and only if for all open primes $p$, one has that $\Lambda_I(M_p) = 0$. (Note that this is false if $M$ is not complete. For example, any injective module $J$ will have that $\Lambda_I(J_p)=0$).

2. Well, if you want to continue working with the ideal-theoretic stuff, I suppose you can cheat, view such a module $M$ as a $k[[x]]$-module (under the forgetful functor), and still demand your module to be $(x)$-adically complete.

Answer 2

Sorry, the first version of this answer was broken in a few ways.

For your first question, it seems that there is more than one construction that specializes to what you want. For example, you can take the completion $\hat{X}$ of a variety $X$ along a closed subvariety $Z$, and then take the tensor product $\mathscr{O}_{\hat{X}} \otimes_{\mathscr{O}_X} K_X$, where $K_X$ is the function field. Alternatively, if you have an effective divisor $D$ in a variety $X$, you can take the scheme whose underlying topological space is $D$, and whose sheaf of rings is given by $U \cap D \mapsto \varinjlim_m \varprojlim_n \Gamma(U, \mathscr{O}_X(mD)/\mathscr{O}_X(-nD))$. These two constructions are identical when we are presented with a codimension one subvariety, such as a point in a line. I do not know a succinct name for either construction.

For your second question, you can ask for the modules to have a $k$-linear topology, i.e., there is a basis of neighborhoods of zero formed by $k$-submodules. We can then demand that the action of $k((x))$ is continuous, where $k((x))$ is given the "usual" topology, with $k$ discrete and $ \{ x^n k[[t]] \}_{n \in \mathbb{Z} } $ forms a basis of neighborhoods of zero. A continuous $k((x))$-module is the same as a topological $k$-module equipped with a continuous invertible topologically nilpotent endomorphism.