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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?

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Mike, I'm a little confused. First, unless I'm misunderstanding you, the 2nd sentence of your 2nd paragraph is wrong. A $k[x]$-module on which the action of $x$ is invertible is a $k[x,x^{-1}]$-module. A $k(x)$-module is one on which the action of every nonzero element of $k[x]$ is invertible. (The first sentence is also a bit funny.) More to the point, in your question 1, localization and completion are already general ring-theoretic constructions. Is what you don't like about them the fact that you have to name the multiplicative set at which you localize and the ideal at which you complete? – JBorger Jul 25 '11 at 11:27
Anyway, it seems unlikely that there will be a nice positive answer to question 1. From the geometric point of view, $k((x))$ is field of formal Laurent series at the point $x=0$ of the affine line. But this point is not special in any way. Indeed, the automorphism group of the line permutes the points transitively. The answer to the second question may very well be yes. You could ask that the $k((x))$-module be complete with respect to a topology defined by certain sub-$k[[x]]$-modules. You might want to look into Tate vector spaces, about which there is an MO question. – JBorger Jul 25 '11 at 11:55
Thanks for the correction; I clearly wasn't quite awake when I posted this. But why should the non-specialness of x=0 prevent there from being a positive answer to question 1? Shouldn't it just mean we'd have to make some choice in the construction? – Mike Shulman Jul 25 '11 at 18:49
I guess I just don't understand what you really want in question 1. As I said, localization and completion are already general ring-theoretic constructions. And if you don't mind making choices, then what don't you like about first completing at a chosen ideal and then inverting a chosen multiplicative subset? – JBorger Jul 25 '11 at 22:27
Here's a stab. Given a ring $R$ and an element $r\in R$, you can first complete $R$ with respect to the ideal $(r)$ and then invert the image of $r$ in the completion. Is that the kind of thing you want in question 1? – JBorger Jul 25 '11 at 22:30
  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.

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It seems to me that if I have a k((x))-module M, then x will act invertibly on M, and so the (x)-adic topology on M will be indiscrete. For the same reason, it seems that if I first localize M at the multiples of x and then complete it at (x), I should get 0. What am I missing? – Mike Shulman Jul 25 '11 at 18:46
What you said is true. Note that in the example I gave in (1), I am restricting the localization to open ideals only. These are ideals that contain the ideal of definition for the adic topology. – the L Jul 26 '11 at 11:36

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.

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What is the x-adic topology on k((x))? Since x is invertible, it doesn't generate a proper ideal in k((x)), does it? – Mike Shulman Jul 25 '11 at 18:47
You're right. Similarly people speak about the $p$-adic topology on $\mathbf{Q}_p$ even though $p$ is invertible. What you need to do is say that a subset $U$ of $k((x))$ is open if there exists an integer $n$ such that $x^nU$ is an open subset of $k[[x]]$. – JBorger Jul 25 '11 at 22:36

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