Topology of the ring of formal power series

Question

I'm interested in defining a topology on the ring $R[[X_i]]$ of formal power series in $(X_i)_{i\in I}$, where $R$ is a topological ring and $I$ is a (possibly infinite) index set. The wiki article discusses several options for this, and it seems that the most natural topology satisfies

$(x_n)_{n\in\infty}$ converges iff for every monomial $X^\alpha$ (i.e. $\alpha$ is a finite multiset of indexes in $I$), $([X^\alpha]x_n)_{n\in\infty}$ converges in the topology on $R$.

(The other main option discussed as a natural topology for $R[[X_i]]$ is equivalent to this one where $R$'s topology is ignored and replaced by the discrete topology.) My question is:

• How is this topology defined in terms of open sets? Can it be described as a special case of another topology, i.e. Krull topology or product topology? (The wiki answers this question in the univariate case but I'm more interested in the multivariate case.)

  • More generally, this can be seen as a question of how to translate a specification of a topology in terms of convergence to an explicit definition from open sets.

• Is this topology metrizable (assuming $R$ is)? Again, the wiki discusses this in the case when $R$ is discrete and there is only one variable via $d(x,y)=2^{-k}$ where $k$ is the smallest nonzero coefficient of $x-y$, but leaves the multivariate case for the reader.

Answer 1

The given topology is equivalent to the product topology, under the natural map $R[[X_i]]\to R^{{\cal M}(I)}$ (where ${\cal M}(I)$ is the set of all finite multisets of $I$, isomorphic to the set of monomials in $R[[X_i]]$) that maps a power series to its coefficient function, because a series in the product topology also converges iff all of its projections converge, which here is the same as saying that all of its coefficients converge.

If $R$ itself has a Krull ($I$-adic) topology with generating ideal $J$, there is a natural generating ideal for an $I$-adic topology given by $$G=(J,(X_i)_{i\in I})=\{x\in R[[X_i]]:[1]x\in J\},$$ the set of all power series with constant term in $J$, but only in certain cases does this generate the same topology as the first one. Note that in this case $G^n$ is the set of power series where each degree-$k$ coefficient is in $J^{n-k}$ for all $k\le n$, and by definition this defines a neighborhood base of $0$. An equivalent characterization of the product topology here takes as neighborhood base of $0$ the sets $$U_S=\{x\in R[[X_i]]:\forall\alpha\in S, [X^\alpha]x\in{\cal O}_R\}$$ for each finite $S$, where ${\cal O}_R$ is the topology on $R$. Since for each $S$ there is an $n$ such that $U_S\subseteq G^n$, we have that the $G$-adic topology is finer than the product topology, and if $I$ is finite each $G^n$ is a basic open set, so for discrete $R$ this topology matches the product topology.

When $I$ is infinite $G^n$ will not be a basic open set, and indeed is not open, so it generates a strictly finer topology. This shows in the convergence of series like $\sum_{i=1}^\infty X_i$ (supposing that $I=\Bbb N$). In the product topology this is a convergent series (since every monomial term converges), but in the $G$-adic topology we require that there is an $n$ such that $\sum_{i=n}^\infty X_i\in G^2$, i.e. all degree-$1$ coefficients must stabilize after some $n$, and this is not true.

Regarding the question of metrizability, note that ${\cal M}(I)$ is countable iff $I$ is countable. Then since the product topology $X^S$ is metrizable iff $X$ is metrizable and $S$ is countable (proof), it follows that $R[[X_i]]$ is metrizable under the given topology iff $I$ is countable. I don't think there is a "natural" metric, though, in the case when $I$ is countable, unless you already have well-ordering on $I$ on which to define a bijection $f:\Bbb N\to{\cal M}(I)$, from which you can define $$d(x,y)=\sum_{n\in\Bbb N}\frac{d([X^{f(n)}]x,[X^{f(n)}]y)}{2^{-n}}.$$