I asked this question on Mathematics Stackexchange ([link](https://math.stackexchange.com/q/2863312/660)), 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.



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


