16
$\begingroup$

I asked this question on Mathematics Stackexchange (link), but got no answer.

Let $K$ be a field, let $x_1,x_2,\dots$ be indeterminates, and form the $K$-algebra $A:=K[[x_1,x_2,\dots]]$.

Recall that $A$ can be defined as the set of expressions of the form $\sum_ua_uu$, where $u$ runs over the set monomials in $x_1,x_2,\dots$, and each $a_u$ is in $K$, the addition and multiplication being the obvious ones.

Then $A$ is a local domain, its maximal ideal $\mathfrak m$ is defined by the condition $a_1=0$, and it seems natural to ask

Is $K[[x_1,x_2,\dots]]$ an $\mathfrak m$-adically complete ring?

I suspect that the answer is No, and that the series $\sum_{n\ge1}x_n^n$, which is clearly Cauchy, does not converge $\mathfrak m$-adically.

$\endgroup$
  • 3
    $\begingroup$ I find it hard to visualise elements of $\mathfrak m^n$ (e.g. it is not true that $\mathfrak m = (x_1,x_2,\ldots)$). I think you might be right that $A$ is not $\mathfrak m$-adically complete, and I imagine this is somehow 'standard'. I'd be interested to see an example worked out of a Cauchy sequence that does not converge. $\endgroup$ – R. van Dobben de Bruyn Aug 14 '18 at 11:33
  • $\begingroup$ Why does $\sum_{n\geq 1} x_n^n$ not converge? It seems to be an element of $A$, by the way you described $A$. $\endgroup$ – Ehud Meir Aug 14 '18 at 12:44
  • 1
    $\begingroup$ I suspect that the answer is in fact yes, and that the phenomenon we have here is the other way around: there are sequences which are not Cauchy with respect to the $\mathfrak{m}$-adic topology, but which do converge (for example $\sum_{n\geq 1} x_n$. If $(y_n)$ is a Cauchy sequence with respect to the $\mathfrak{m}$-adic topology where $y_n = \sum_{n,u}a_{n,u}u$ one can argue that the $a_{n,u}$ stabilize for a big enough $n$. This is because all the elements in $\mathfrak{m}^N$ will be sum of monomials of degree at least $N$. $\endgroup$ – Ehud Meir Aug 14 '18 at 12:55
  • 1
    $\begingroup$ @EhudMeir - Thanks! You posted your second comment while I was writing the following answer to your first comment. In the definition of $A=K[[x_1,x_2,\dots]]$ that I gave (and which I found in Bourbaki), the notation $\sum_ua_uu$ is completely formal. In other words, this formal sum represents an element of $A$, but not the sum of a series. To say it in yet another way, $A$ is not defined as the $(x_1,x_2,\dots)$-adic completion of $K[x_1,x_2,\dots]$. $\endgroup$ – Pierre-Yves Gaillard Aug 14 '18 at 13:00
  • 1
    $\begingroup$ @EhudMeir: does the tail $\sum_{i \geq n} x_i^i$ lie in $\mathfrak m^n$? I do not know how to prove or disprove this. The problem is that elements of $\mathfrak m^n$ are by definition finite sums of products $f_1\cdots f_n$ of elements $f_1,\ldots,f_n \in \mathfrak m$. $\endgroup$ – R. van Dobben de Bruyn Aug 14 '18 at 16:16
8
$\begingroup$

[Edit: The lemma was revised and proved, changed the point of view from series to sequences.]

[2nd Edit: The proof of the lemma was improved, and now the argument can show that Cauchy series such as $x_1+(x_{1^3+1}^2+\dots x_{2^3}^2)+(x_{2^3+1}^{3}+\dots+x_{3^3}^3)+\dots$, diverge in the $\mathfrak{m}$-adic topology.]

Here is a construction of a Cauchy sequence which does not converge. It is based on the following lemma, a proof of which will is given at the end.

Lemma. Let $\mathfrak{m}_c$ denote the maximal ideal of $K[[x_1,\dots,x_c]]$. Then there exist sequences of natural numbers $(r_n)_{n\in\mathbb{N}}$, $(c_n)_{n\in\mathbb{N}}$ and a sequence of elements $(p_n\in(\mathfrak{m}_{c_n})^n)_{n\in\mathbb{N}}$ such that:

  • $\limsup r_n=\infty$ and
  • $p_n$ cannot be written as a sum of $r_n$ terms $\sum_{i=1}^{r_n} a_{i} b_i$ with $a_{i},b_i\in\mathfrak{m}_{c_n}$.

In fact, one can take $c_n=n^2$, $p_n=x_1^{n}+\dots+x_{n^2}^{n}$ and let $r_n=\lceil \frac{n^2}{2(n-1)}\rceil-1$ if $\mathrm{char}\,K\nmid n$ and $r_n=1$ when $\mathrm{char}\,K\mid n$.

It would be more convenient to replace $K[[x_1,x_2,\dots]]$ with the isomorphic ring $$S:=K[[y_{ij}\,|\,i,j\in \mathbb{N}]].$$ For every $n$, there is a ring homomorphism $\phi_n:S\to K[[x_1,\dots,x_{c_n}]]$ specializing $y_{n1},\dots,y_{n{c_n}}$ to $x_1,\dots,x_{c_n}$ and the rest of the variables to $0$.

Let $f_n=p_n(y_{n1},\dots,y_{nc_n})$ and define in $S$ the (formal) partial sums $$ g_t:=\sum_{n=t}^\infty f_n. $$ Then, by construction, $(g_t)_{t\in\mathbb{N}}$ is a Cauchy sequence relative to the $\mathfrak{m}$-adic topology, but it does not converge.

Indeed, if $(g_t)_t$ converges in the $\mathfrak{m}$-adic topology, then it also converges relative to the (coarser) grading topology of $S$, and so must converge to $0$. However, if this is so, there exists $t\in\mathbb{N}$ such that $g_t=\sum_{n=t}^\infty f_n\in \mathfrak{m}^2$. In particular, we can write $\sum_{n=t}^\infty f_n=\sum_{i=1}^u a_{i}b_i$ with $a_{i},b_i\in \mathfrak{m}$. Choose $n\geq t$ sufficiently large to have $r_n\geq u$. Applying $\phi_n$ to both sides of the last equality gives $p_n=\sum_{i=1}^u (\phi a_{i})(\phi b_{i})$ with $\phi a_{i},\phi b_i\in\mathfrak{m}_{c_n}$, which is impossible by the way we chose $p_n$.

Back to the lemma: Given $f\in K[x_1,\dots,x_c]$, let $D_if$ denote its (formal) derivative relative to $x_i$. The lemma follows from the following general proposition.

Proposition. Let $f\in K[[x_1,\dots,x_c]]$ be a homogeneous polynomial of degree $n$, and let $r\in\mathbb{N}$ denote the minimal integer such that $f$ can written as $\sum_{i=1}^ra_ib_i$ with $a_i,b_i\in \mathfrak{m}_c$. Suppose that $D_1f,\dots,D_cf$ have no common zero beside the zero vector over the algebraic closure of $K$. Then $r\geq \frac{c}{2(n-1)}$.

Proof. Suppose otherwise, namely, that $f=\sum_{i=1}^r a_ib_i$ with $2r(n-1)<c$. Notice that $a_i,b_i$ are a priori not polynomials --- rather, they are power series. To fix that, we write each $a_i$ and $b_i$ as a sum of their homogenous components and rewrite the degree-$n$ homogeneous component of product $a_ib_i$ as a sum of the relevant components of $a_i$ and $b_i$. By doing this, we see that $f$ can written as $\sum_{i=1}^{r(n-1)}a'_ib'_i$ with $a'_i,b'_i$ being homogeneous polynomials in $\mathfrak{m}_c$.

Let $V$ denote the affine subvariety of $\mathbb{A}^c_{\overline{K}}$ determined by the $2r(n-1)$ equations $a'_1=b'_1=a'_2=b'_2=\dots=0$. It is well-known (see Harshorne's "Algebraic Geometry", p. 48) that every irreducible component of $V$ has dimension at least $c-2r(n-1)>0$. Furthermore, $V$ is nonempty because it contains the zero vector (because $a'_1,b'_1,a'_2,b'_2,\dots\in \mathfrak{m}_c$). Thus, there exists a nonzero $v\in \overline{K}^c$ annihilating $a'_1,b'_1,a'_2,b'_2,\dots$.

Now, by Leibniz's rule, we have $$D_jf=\sum_i D_j(a'_ib'_i)=\sum_i(D_ja'_i\cdot b'_i + a'_i\cdot D_j b'_i).$$ It follows that $v$ above annihilates all the derivatives $D_1f,\dots,D_cf$, a contradiction! $\square$

If it weren't for the passage from power series to polynomials, the proof would work for non-homogenous polynomials and also give the better bound $r\geq \frac{c}{2}$. [Edit: This is in fact possible, see the comments.] Such a bound would suffice to prove that the Cauchy series $x_1+x_2^2+x_3^3+\dots$ suggested in the question diverges in the $\mathfrak{m}$-adic topology.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks a lot!!! Sadly I don't know enough algebraic geometry to fully understand your answer. But I'll still ask you an elementary question. It seems to me you could set $r_n:=n$ throughout, because when you prove the lemma, you start with arbitrary $n$ and $r$. Am I right? $\endgroup$ – Pierre-Yves Gaillard Aug 14 '18 at 18:59
  • $\begingroup$ Yes, $r_n=n$ will do. The proof of the lemma could be summarized as "dimension count". $\endgroup$ – Uriya First Aug 14 '18 at 19:04
  • $\begingroup$ I managed to find an explicit construction for $c_n,r_n,p_n$ and updated the proof accordingly. I still feel there is room for improving the bounds, though. $\endgroup$ – Uriya First Aug 15 '18 at 7:51
  • 2
    $\begingroup$ Very nice! Also, is your passage from power series to polynomials really necessary? To my eye it looks as if you could run the same argument, but using dimension theory for power series rings instead. As you say, that would imply that the series in the question does not converge. $\endgroup$ – dhy Aug 15 '18 at 9:22
  • 1
    $\begingroup$ @dhy Thanks! One could indeed replace the condition that $D_1f,\dots,D_cf$ has a common zero with saying that these polynomials generate an ideal of finite codimension in $K[[x_1,\dots,x_c]]$, and this is of course impossible if this ideal is contained in the one generated by $a_1,b_1,\dots,a_r,b_r$ and $2r<c$ (notation as in the proof of the proposition). This means that the bound can really be improved to $r\geq \frac{c}{2}$! Hopefully I am not mistaken... $\endgroup$ – Uriya First Aug 15 '18 at 9:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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