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Let $R$ be a ring (commutative with unit).Let $f_1,...,f_n\in R$ elements that generate the unit ideal. The map $R\to R_{f_1}\times ...\times R_{f_n}$ is faithfully flat, since this is just a Zariski cover. Now, my question is

Is the map $R[[t]]\to R_{f_1}[[t]]\times ...\times R_{f_n}[[t]]$ faithfully flat?

Note that in general $R_f [[t]]\ne R[[t]]_f$, so it is not a special case of the previous map.

I general, I am quite confused about how to work with rings like $R[[t]]$. Objects such as $R_f[[t]]$ seem much more natural then, say, $R[[t]]_f$. Yet abstractly, $R[[t]]_f$ is just a localization, where $R_f[[t]]$ is something more complicated. I guess $R_f[[t]]$ is a "completed" version of $R[[t]]_f$, but what is the formal description that allows one to derive properties of it?

Edit: As said in the comments, every maximal ideal of $A=R[[t]]$ is of the form $\mathfrak{m}+(t)$ and hence its extension in $B=R_{f_1}[[t]]\times ...\times R_{f_n}[[t]]$ is not the unit ideal (looking at the factor where $f_i \not\in \mathfrak{m}$). So the question is whether $A\to B$ is flat or just whether $R[[t]]\to R_f[[t]]$ is flat, since this is trivially equivalent.

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One remark is that it's enough to check that the right-hand side $B$ is flat over $A=R[[t]]$. Indeed, we know what maximal ideals $m\subset A$ are and obviously $mB\neq B$. –  Anton Fonarev Oct 15 '13 at 12:53
    
And yes, I forgot to mention that yes, $R_f[[t]]$ is obviously the completion of $R[[t]]_f$ with respect to the natural filtration by powers of $t$. –  Anton Fonarev Oct 15 '13 at 13:00
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2 Answers 2

Ok, I've been thinking for a while about a counterexample for $R$ not Noetherian, but lack experience in this kind of questions. For the sake of completeness I repeat my original comments.

Let $R$ be Noetherian. Denote $A=R[[t]]$ and $B=R_{f_1}[[t]]\times\ldots\times R_{f_n}[[t]]$. Remark that for any maximal ideal $m\subset A$ we obviously have $mB\neq B$. Thus, it's enough to check that $B$ is flat over $A$. Next, we immediately reduce to the case $B=R_f[[t]]$.

Remark that $B$ is the completion of $A_f$ with respect to the ideal $I=(t)$. $A_f$ is flat over $A$. Also remark that $R$ Noetherian implies $A$ Noetherian implies $A_f$ Noetherian. Now, the basic result from commutative algebra states that for any finitely generated $A_f$-module $M$ one has $\hat{A_f}\otimes_{A_f}M\simeq \hat{M}$ and $B=\hat{A_f}$ is flat over $A_f$, thus flat over $A$. (See e.g. Eisenbud, CA.) This proves the statement for $R$ Noetherian.

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Ok, you say that $B=\hat{A_f}$ and this is flat over $A_f$ (since $A_f$ is noetherian) which in turn is flat over $A$ and by transitivity of flatness, $B$ is flat over $A$, right? So where do you use the "basic result from commutative algebra"? I wonder what happens in the non-noetherian case... –  KotelKanim Oct 16 '13 at 13:50
    
@KotelKanim The "basic result" is that the completion of a Noetherian ring is flat. In our case $B$ is the completion of $A_f$ in $(t)$. This is the point where flatness may fail without noetherian assumptions. Once again, I'm no expert in constructing counterexamples in commutative algebra… –  Anton Fonarev Oct 16 '13 at 14:43
    
I see. Thanks for the explanation! –  KotelKanim Oct 16 '13 at 15:01
    
Why is $mB \neq B$? –  Konstantin Ardakov Dec 20 '13 at 23:09
    
@KonstantinArdakov any $m\subset A$ is of the form $n+(t)$ for some $n\subset R$ maximal. One of the $f_i$'s doesn't belong to the latter. –  Anton Fonarev Dec 20 '13 at 23:22
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No. An example is to take $R = k[z, y, x_1, x_2, ...]/(zyx_1, zyx_2, ...)$ and $f$ equal to the residue class of $z$. Then $R$ is a coherent ring, hence $R[[t]]$ is flat over $R$. Hence the kernel of $y : R[[t]] \to R[[t]]$ is generated by $zx_1, zx_2, ...$

But the element $\sum x_it^i$ of $R_f[[t]]$ is in the kernel of $y$ and not in the ideal generated by $zx_1, zx_2, ...$ of $R_f[[t]]$. One is not allowed to take infinite sums of generators! Thus $R[[t]] \to R_f[[t]]$ is not flat.

More generally, let $A$ be a ring, let $I$ be a finitely generated ideal, and assume that $A$ is complete with respect to $I$. Let $f_1, ..., f_r$ be finitely many elements of $A$ which generate the unit ideal. We have just seen that $$A \to\prod A_{\{f_i\}}$$ is not a flat covering and hence we cannot descent for it right? Wrong! It is possible to descent along any universally injective ring map, see for example Section Tag 08WE. As far as this bot knows (source code to be released in the year 2999) it is an open question as to whether the displayed map is universally injective.

Still, even if true this is of academic interest only. Duh!

Edit: Actually, $R$ is not coherent and $R[[t]]$ is not flat over $R$. But, unless I made another error, this still gives an example. Namely, an element $g$ in the kernel of $y : R[[t]] \to R[[t]]$ looks like $g = \sum t^{n_i}zx_ig_i$ with $n_i \to \infty$ and $g_i \in R[[t]]$. An element $h$ in the kernel of $y : R_f[[t]] \to R_f[[t]]$ looks like $\sum t^{e_i}x_i h_i$ with $e_i \to \infty $ and $h_i \in R_f[[t]]$. The idea is then to pick $e_i = i$ and $h_i = f^{-i} = z^{-i}$ and show that $h$ cannot be a finite linear combination of elements $g$ as above with coefficients in $R_f[[t]]$ by looking at the powers of $z$ in front of the terms $t^ix_i$.

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