MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $k$ be a field. For $R=k[x_1,\ldots]$ with countably infinite number of variables, [due to the discussion in the comments] we have to make the following distinction between $k[[x_1,\ldots]]$ and the completion $\hat{R}$ of $R$ at the ideal $(x_1,\ldots)$: note that the former admits elements which can have infinitely many monomials of the same degree whereas the latter can not (e.g. $\sum_i x_i\in k[[x_1,\ldots]]$ but $\notin\hat{R}$). There are two questions

1) Is the morphism $R\to\hat{R}$ flat? If $R$ were any Noetherian ring, the map $R\to \hat{R}$ to its completion is always flat (see e.g. Atiyah-MacDonald 10.14) but my intuition breaks down for non-Noetherian rings.

2) Is the morphism $R\to k[[x_1,\ldots]]$ flat? This is answered positively by ayanta below.

share|cite|improve this question
Just a comment: the situation you are considering is very bad. Not only the ring is not noetherian, but the ideal $m$ that you complete with respect to is not finitely generated. In this case, assuming $k$ is a field, the $m$-adic completion of the left hand side will not be $m$-adically complete! – the L Jan 2 '13 at 19:08
The question should be written inside the message as well, not only in the title (and use quantifiers. what is $k$? finitely many indeterminates?). See – YCor Jan 2 '13 at 20:18
I'm a bit confused about your second statement if you wouldn't mind elaborating. I am taking the completion of $k[x_1,\ldots]$ along the ideal $(x_1,\ldots)$ which is nothing other than $k[[x_1,\ldots]]$. – Frank Jan 2 '13 at 22:11
Right, I was not aware of this difference between the definition of the formal power series in infinitely many variables (i.e. which contains $\sum x_i$) as opposed to the completion of $k[x_1,\ldots]$ at $(x_1,\ldots)$ (which does not). Now that this is clear, I suppose it's not clear whether either of the two morphisms are flat! – Frank Jan 2 '13 at 22:28
@unknown (google): I'm not sure what "coherent regular ring" means (it cannot be "coherent ring that is regular", since regularity of a commutative ring includes a noetherian condition in my experience), but valuation rings of complete rank-1 valued fields are coherent as rings, so for example the valuation ring of $\mathbf{C}_p$ is of the type you indicate. But such examples feel a bit removed from the nature of the question that is posed. – user30180 Jan 3 '13 at 6:42

Let's suppose by $k[[x]]$ we mean the formal power series ring in variables $x_1, x_2, \dots$ which is literally the space of sequences of monomials that individually involve only finitely many variables per monomial (so set-theoretically a direct product of copies of $k$ indexed by such monomials, with a "cofinite" topology). This is of course different from the $(x)$-adic completion of $k[x] := k[x_1,x_2,\dots]$ as noted by Francois, since the latter has as a cofinal system of discrete quotients the rings $k[x]/(x)^m$ of infinite $k$-dimension whereas the former has as a cofinal system of discrete quotients the artinian $k[x_1,\dots,x_r]/(x_1,\dots,x_r)^m$ of finite $k$-dimension.

I claim that $k[[x]]$ in the sense I have specified is flat over $k[x]$ (though I also think this is probably completely useless and so I don't claim this is interesting -- maybe just amusing). The key input is buried near the end of volume 1 of SGA3. These methods have no relevance to the $(x)$-adic completion of $k[x]$ (which is an entirely different beast than $k[[x]]$ as defined above).

First, some preliminary reductions. We have to show that if $I$ is a finitely generated ideal of $k[x]$ then the injection $I \rightarrow k[x]$ remains injective after tensoring against $k[[x]]$ over $k[x]$. By finite generation, $I$ "comes from" an ideal $I' \subset k[x_1,\dots,x_r]$ for some $r$, and more specifically the natural map $$I' \otimes_{k[x_1,\dots,x_r]} k[x] \rightarrow k[x]$$ is injective since $k[x]$ is certainly flat (even free) over $k[x_1,\dots,x_r]$. So in fact $$I = I' \otimes_{k[x_1,\dots,x_r]} k[x],$$ and hence our problem is to show that the injection $I' \rightarrow k[x_1,\dots,x_r]$ remains injective after tensoring over $k[x_1,\dots,x_r]$ against $k[[x]]$. More specifically, we claim this latter ring map is flat.

This final scalar extension process decomposes as a composition of two scalar extensions: $$k[x_1,\dots,x_r] \rightarrow k[[x_1,\dots,x_r]] \rightarrow k[[x]].$$ Since the first step is known to be flat by usual commutative algebra with noetherian ring, we're reduced to proving flatness of the second map. But this is a special case of the Gabriel-Grothendieck theory of pseudo-compact rings in SGA3, in which they systematically develop a good theory of "pseudo-compact modules" and "topological flatness" for "pseudo-compact rings", which are topological rings that are arbitrary inverse limits of artinian rings. This theory includes as a key ingredient a relationship between topological flatness and ordinary flatness when the base ring is noetherian (analogous to completions in the noetherian setting, but logically requiring more work).

More specifically, since $A := k[[x_1,\dots,x_r]]$ is noetherian, so any finitely generated $A$-module is finitely presented (and is pseudo-compact for its max-adic topology), for any finitely generated $A$-module $M$ and pseudo-compact $A$-algebra $A'$ the natural map $$M \otimes_A A' \rightarrow M \widehat{\otimes}_A A'$$ is bijective (ultimately because the left side is a cokernel of a map between finite free $A'$-modules and any such map automatically has closed image by a variant of Artin-Rees proved in SGA3). Thus, the preservation of injectivity of the left as a functor in finitely generated $M$ (which is equivalent to $A$-flatness of $A'$) is reduced to topological flatness of $A'$ over $A$.

Note that one can "distribute" formal power series over other formal power series when extracting out a finite set of variables into the coefficients over infinitely many variables (think for a minute, using our running definition of "formal power series" for a possibly infinite set of variables). Thus, in our case of interest $A' = k[[x]]$ is a formal power series ring over $A = k[[x_1,\dots,x_r]]$ in infinitely many variables. Thus, we're finally reduced to the question: if $A$ is a pseudo-compact ring (such as $k[[x_1,\dots,x_r]]$) then is $A[[y_1,\dots]]$ topologically flat over $A$? The answer is "yes" because such formal power series rings (in the sense of our running definition) are "topologically free", and a basic fact in the theory is that topological freeness (suitably defined...) implies topological flatness.


share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.