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More generally, let $(R,m)$ be a Noetherian local domain with fraction field $K$. The $m$-adic topology turns $R$ into a topological ring. When $R$ is a discrete valuation ring, this topology extends to a field topology on $K$, in such a way that $R$ is open in $K$. I wonder how much of this can be done for more general $R$. The first natural question is:

Is there a field topology on $K$ inducing the $m$-adic topology on $R$?

On the negative side, if $\dim(R)≥2$, there is not even a ring topology on $K$ such that $R$ is an open subring. Indeed, for any nonzero $f\in R$, multiplication by $f$ in $K$ must be a homeomorphism, so $fR$ would have to be open in $R$, which it is not unless $f\in R^\times$.

On the other hand, there is a ring topology on $K$ making $R$ closed: view $K$ as the union of all principal fractional ideals $f^{-1}R$ ($f\in R\smallsetminus\{0\}$), each endowed with its $m$-adic topology (as a free $R$-module). The transition maps are closed embeddings, and it is easy to see that the colimit topology on $K$ works. If we choose this topology on $K$, the question becomes:

Is this a field topology? In other words, is the inversion map on $K^\times$ continuous?

For instance, take $R=k[[x,y]]$ where $k$ is a field, and consider the sequence $n\mapsto \frac{x}{x+y^n}$. Does this sequence converge to $1$ for the above topology? Equivalently, does the sequence $n\mapsto \frac{y^n}{x+y^n}$ converge to $0$?

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    $\begingroup$ No $x/(x+y^n)$ does not seem to converge, since otherwise there would exist $f\in k[[x,y]]$ such that $1/(x+y^n)$ it belongs to $(1/f)k[[x,y]]$ for all $n$. Writing this as an equation, removing maximal factors that are powers of $x$, and evaluating at $x=0$ yields a contradiction. $\endgroup$
    – YCor
    Commented Dec 21, 2015 at 14:25
  • $\begingroup$ @YCor: for the topology just defined, $(1/f)k[[x,y]]$ is closed, not open. $\endgroup$
    – Laurent Moret-Bailly
    Commented Dec 21, 2015 at 16:48
  • $\begingroup$ Yes I know, but in a colimit topology of an increasing sequence $(M_n)$, if a sequence converges, then it belongs to a single $M_n$. $\endgroup$
    – YCor
    Commented Dec 21, 2015 at 20:30
  • $\begingroup$ Oh, right. Simple, now you say it ;-). So this answers the second question. Thanks! $\endgroup$
    – Laurent Moret-Bailly
    Commented Dec 21, 2015 at 20:52

1 Answer 1

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If I am correct, your Noetherian local domain $(R,m(R))$ admits a place $D$ (maximal local domain) such that $R\subset D\subset K$ and $m(D)\cap R=m(R)$ (for a local ring $L$, we note $m(L)$ its maximal ideal). The fraction field $K$ is common to $R$ and $D$. As $D$ is a place there is a valuation $$ v\ :\ K\to \Gamma_{\infty} $$
such that $$ D=\{x\in K\,|\, 0\leq v(x)<\infty\}\ ;\ m(D)=\{x\in K\,|\, 0< v(x)<\infty\} $$ this valuation defines the same quotient topology ($R$ is endowed with the $m$-adic topology and $R'=R\setminus \{0\}$ with the induced topology) $$ R\times R'\to K\mbox{ and } D\times D'\to K $$ which automatically is that of a topological field, see the construction here. This topology induces the given ($m$-adic) topology on $R$.

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  • $\begingroup$ [Sorry for the late comment, I had somehow overlooked your answer.] Is it clear that the valuation topology on $D$ induces the $m$-adic topology on $R$? $\endgroup$
    – Laurent Moret-Bailly
    Commented Mar 10, 2018 at 17:11
  • $\begingroup$ At least, it WAS when I wrote it. Let me a couple of days to elaborate. $\endgroup$
    – Duchamp Gérard H. E.
    Commented Mar 11, 2018 at 23:22
  • $\begingroup$ At least, it works in the special case $R=k[[x,y]]$ since you can do it explicitly: you can inject $R$ into $D:=k(t)[[x]]$ by sending $y$ to $tx$, and one checks immediately that the $m(R)$-adic topology is induced by the $x$-adic topology on $D$. $\endgroup$
    – Laurent Moret-Bailly
    Commented Mar 13, 2018 at 16:47
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    $\begingroup$ Yes, because $D$ is a DVR. In fact, the only ideals of $D$ are the powers of $m(D)$. $\endgroup$
    – Laurent Moret-Bailly
    Commented Mar 14, 2018 at 20:02
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    $\begingroup$ In the special case, yes, but in general I don't see why $m(D)^{k+s}\cap R\subset m(R)^k$ (for all $k$ and $s$ large enough depending on $k$). $\endgroup$
    – Laurent Moret-Bailly
    Commented Mar 15, 2018 at 8:06

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