What is meant by "$k[[\{T^{1/n}\mid n\in\mathbb N\}]]$"?

Question

(This is not my normal convention but I'd like to start $\mathbb N$ at $1$ in this post, please.)

An example of a ring was mentioned in a comment in mathoverflow. What was actually mentioned was $k[[\{T^{1/n}\mid n\in\mathbb N\}]]/(T)$ for a field $k$, but we can forget the quotient part of that construction for the purposes of this discussion. I would like to know what $k[[\{T^{1/n}\mid n\in\mathbb N\}]]$ means. No one disputed the notion on the spot on mathoverflow, so it seems like it might be real.

If the double brackets were really intended, this ought to refer to a power-series construction in many variables.

I understand that power series in many variables $k[[\{X_n\mid n\in \mathbb N\}]]$ can be described in terms of elements. The description I know of first defines monomials (each of which has a total degree $n$ for some natural number $n$) and then uses these to justify a multiplication operation between infinite formal sums of these monomials.

I also understand they can be described as an $I$-adic completion of $k[\{X_n\mid n\in \mathbb N\}]$ (single brackets intended, that's the regular polynomial ring), where $I=(\{X_n\mid n\in \mathbb N\})$ is a maximal ideal. This makes sense because $\cap I^n=\{0\}$.

But when I try to apply this directly to $k[[\{T^{1/n}\mid n\in\mathbb N\}]]$, neither of these seems to hold up. It feels like formal sums amount to $\sum_{q\in\mathbb Q^{\geq 0}}\alpha_qT^q$. But without restrictions on the support of the formal sum, it does not seem they have a meaningful product. I think something about the total degree grading is lost along the way.

If instead we try the $(\{T^{1/n}\mid n\in\mathbb N\})$-adic completion of $k[\{T^{1/n}\mid n\in\mathbb N\}]$, I think we have $I=I^n$ for every $n>0$, so this would be the inverse limit of the same field over and over. That's seemingly not interesting...

Finally, I also thought that possibly one could take $k[[\{X_i\mid i\in\mathbb N\}]]$ and form the quotient with relations $T=X_1$ and $X_i^i=T$ for $i>1$. That seems completely well-defined.

To summarize, I guess these are the possibilities for $k[[\{T^{1/n}\mid n\in\mathbb N\}]]$:

1. it isn't a thing that people use: someone must have been mistaken when suggesting it
2. it is conventionally some subset of formal series indexed by $\mathbb Q^{\geq 0}$ with some conventionally given restrictions on support
3. it's shorthand for that quotient of $k[[\{X_i\mid i\in\mathbb N\}]]$
4. a directed union of $R_n=k[[T^{1/n}]]$
5. it's something I haven't thought of.

If it is indeed something that people discuss in certain fields, then it would be helpful to have pointers to examples in the literature.

Answer 1

tl;dr Based on experience, I've seen the rings $K[[t^{1/n}:n \geq 1]]$ appear in the wild as the quotient of $K[[X_n:n \geq 1]]$ at the ideal generated by the relations $X_1 = t$ and $X_n^n = t$ for all $n > 1$.

Below is a concrete example of how these rings can arise in nearby and vanishing cycles, a description of how to extend this off the algebraically closed field of scalars case, and a short description of what exactly the quotient $K[[t^{1/n}:n \geq 1]]/(t)$ encodes and enables us to do. I know this answer got long, and I apologize for that, but this is one of my SpEcIaL iNtErEsTs and I'd feel bad if I didn't give as good an exposition as possible.

For the sake of being concrete and starting off in a simple case, let $k = \mathbb{C}$. The ring of formal power series $A = \mathbb{C}[[t]]$ is a strictly Henselian DVR and has fraction field $K := \mathbb{C}((t))$ of formal Laurent series and residue field $k = \mathbb{C}$.

Because $\mathbb{C}[[t]]$ is strictly Henselian (as in because the residue field $k$ is closed), every field extension of $K$ arises as a ramified extension of $\mathbb{C}[[t]]$. Moreover, the finite field extensions of $\mathbb{C}((t))$ take the form $\mathbb{C}((t^{1/n}))$ for some $n \geq 1$. The integral closure of $\mathbb{C}[[t]]$ in $\mathbb{C}((t^{1/n}))$ is exactly the ring of formal power series $\mathbb{C}[[t^{1/n}]]$ of power series in $n$-th roots of $t$. As the algebraic closure $\mathbb{C}((t^{1/n}:n \geq 1))$ arises as a filtered colimit of fields (this is the usual ultimate union of the tower of field extensions), taking the filtered colimit of the corresponding rings of integers gives us that the integral closure of $\mathbb{C}[[t]]$ in the field $\mathbb{C}((t^{1/n}:n \geq 1)) = \overline{\mathbb{C}((t))}$ is calculated by $$ \overline{\mathbb{C}[[t]]]} \cong \operatorname*{colim}_{n \in \mathbb{N}} \mathbb{C}[[t^{1/n}]] \cong \frac{\mathbb{C}[[X_n:n \geq 1]]}{(X_n^n-X_1: n >1)} \cong \mathbb{C}[[t^{1/n}:n \geq 1]]. $$ This appears in a commuting diagram $$ \require{AMScd} \begin{CD} \mathbb{C}((t^{1/n}:n \geq 1)) @<<< \mathbb{C}[[t^{1/n}:n \geq 1]] @>>> \mathbb{C} \\ @AAA @AAA @| \\ \mathbb{C}((t)) @<<< \mathbb{C}[[t]] @>>> \mathbb{C} \end{CD} $$ where the left-handed square is a pushout. However, the right-handed square is not a pushout. We can detect this by noting that the map $\mathbb{C}[[t]] \to \mathbb{C}$ is defined by sending $t \mapsto 0$. Thus $t$ becomes $0$ upon tensoring $\mathbb{C}[[t^{1/n}:n \geq 1]]$ with $\mathbb{C}$ over $\mathbb{C}[[t]]$ and we find $$ \mathbb{C}[[t^{1/n}:n \geq 1]] \otimes_{\mathbb{C}[[t]]} \mathbb{C} \cong \frac{\mathbb{C}[[t^{1/n}:n \geq 1]]}{(t)}. $$ Thus the quotient ring $\mathbb{C}[[t^{1/n}:n \geq 1]]/(t)$ detects the degree to which the commuting square $$ \require{amscd} \begin{CD} \mathbb{C}[[t^{1/n}:n \geq 1]] @>>> \mathbb{C} \\ @AAA @AAA \\ \mathbb{C}[[t]] @>>> \mathbb{C} \end{CD} $$ fails to be a pushout. When translating to the case $K$ is a more general field, we instead simply have $\overline{K((t))} \cong \overline{K}((t^{1/n}:n \geq 1))$ and $\overline{K[[t]]} \cong K[[t^{1/n}:n \geq 1]]$ while the rest of what I've described carries over mutatis mutandis save for we look at the diagrams $$ \require{AMScd} \begin{CD} \overline{K}((t^{1/n}:n \geq 1)) @<<< \overline{K}[[t^{1/n}:n \geq 1]] @>>> \overline{K} \\ @AAA @AAA @| \\ K((t)) @<<< K[[t]] @>>> K \end{CD} $$ instead. In this case the right-handed square still fails to be a pullback, but instead we have $\overline{K}[[t^{1/n}:n \geq 1]] \otimes_{K[[t]]} K \cong \overline{K}[[t^{1/n}:n \geq 1]]/(t).$

As an aside, I've actually worked with these rings before in the context of nearby and vanishing cycles over traits a la SGA 7 Expose XIII (in which case you replace the ring $K[[t]]$ with any other Henselian DVR). They also appear in various works in the representations of $p$-adic groups (see, for instance, the reference [1] I've placed below --- the ring $\mathbb{C}[[t^{1/n}:n \geq 1]]$ here appears as the ring $\overline{R}$ described in Section 7.1). I'm in the process of writing a long but extremely complete and thorough article on nearby and vanishing cycles over traits, but I keep discovering new things that seem to be at least not written down in the literature and so I do not know when the article will be complete. That being said, I'm happy to at least share the current draft (with the caveat it's incomplete) so if anyone is interested either message me privately here or send me an email (my institutional email can be found in my profile).

[1]: Cunningham, Clifton L. R.; Fiori, Andrew; Moussaoui, Ahmed; Mracek, James; Xu, Bin, Arthur packets for $p$-adic groups by way of microlocal vanishing cycles of perverse sheaves, with examples, Memoirs of the American Mathematical Society 1353. Providence, RI: American Mathematical Society (AMS) (ISBN 978-1-4704-5117-2/pbk; 978-1-4704-7019-7/ebook). ix, 216 p. (2022). ZBL1490.11001.