DaRT/DoME - 2024-12-03

Lukas Heger

11:20

@rschwieb I don't really understand how $\Bbb Q$ gets in the way, to be honest

rschwieb

14:10

@LukasHeger Maybe I've got the wrong impression about how the elements look. Usually one would say "You can't define products between elements of the form $\sum_{q\in \mathbb Q^{\geq 0}}a_qT^q$ because the order in $\mathbb Q^{\geq 0}$ isn't well-founded. There's infinitely many contributors to any given index $q$ in the product.

If you just take the sums with supports that are well-founded, then things work again. I must be thinking something wrong about what the elements of $k[[\{T^{1/n}]]$ look like.

rschwieb

14:27

I'll back up another step and make sure I understand what even $k[[\{X_i\}]]$ looks like. Am I remembering right that this is just a power series with coefficients in $k[\{X_i\}]$?

Lukas Heger

I was asuming that we have some kind of Hahn series ring here

Hahn series

In mathematics, Hahn series (sometimes also known as Hahn–Mal'cev–Neumann series) are a type of formal infinite series. They are a generalization of Puiseux series (themselves a generalization of formal power series) and were first introduced by Hans Hahn in 1907 (and then further generalized by Anatoly Maltsev and Bernhard Neumann to a non-commutative setting). They allow for arbitrary exponents of the indeterminate so long as the set supporting them forms a well-ordered subset of the value group (typically Q {\displaystyle \mathbb {Q...

so yes, we only allow sums with well-founded support

@rschwieb yes that would be a power series ring

rschwieb

Yeah, I am definitely not sure what is being talked about anymore. Perhaps part of my confusion is the fact that the powers of $T$ aren't independent of each other like they are for $\{X_i|\mid i\in\mathbb N\}$

Lukas Heger

as you said, you need some restriction on the support if you want well-defined multiplication

rschwieb

BUt $k[[\{T^{1/n}|\mid i\in\mathbb N\}]]$ should just be a quotient of $k[[\{X_i|\mid i\in\mathbb N\}]]$, right? $X_i\mapsto T^{1/i}$. ugh... I can't even say for sure that's enough to define a map :). I know so little about power series beyond $k[[x]]$.

Lukas Heger

yeah there should be a quotient map I agree

but the problem of well-defined multiplication arises also for $k[[\{X_i|\mid i\in\mathbb N\}]]$

rschwieb

14:40

yeah...

i'm googling around for some basic resources

rschwieb

15:50

Hmm, so I think i at least understand $k[\{T^{1/n}\}]$ as "polynomials with rational power exponents of $T$, and multiplication there is no problem. In this ring, $(\{T^{1/n}\})$ is a maximal ideal, and $k[[\{T^{1/n}\}]]$ is supposed to be the completion of $k[\{T^{1/n}\}]$ at $(\{T^{1/n}\})$ right?

But isn't $(\{T^{1/n}\})^m=(\{T^{1/n}\})$?

Lukas Heger

16:55

yeah indeed

but isn't completion wrt an idempotent ideal kinda boring?

rschwieb

17:16

Well, my meaning was that given that completion is one of the ways one defines power series, the fact that the maximal ideal is idempotent doesn't seem to tell us anything. It's an inverse limit of a constant sequence of fields? How different is that from the field itself?

If defining the ring is not done using completion, I'm not completely sure I understand how it's intended to be defined. Maybe the suggestion I read wasn't kosher

and yet, I think $k[[\{X_i|\mid i\in\mathbb N\}]]$ exists, because in that case you can probably argue the powers of the maximal ideal have zero intersection. And then you can make a quotient with all the $X_i-T^{1/i}$, so maybe it is something interesting

rschwieb

17:49

I'm beginning to think I should make a post asking about it

Lukas Heger

I would construct $k[[\{X_i|\mid i\in\mathbb N\}]]$ as an inverse limit

rschwieb

In that case, one can argue that there are no monomials with total degree $n$ in $I^{n+1}$ (where $I$ is generated by the unknowns) right? and that turns into an argument that the intersection of powers is trivial, I think

rschwieb

18:13

Actually the next comment after that example was offered says something like " Lampert's comment conveys it as well (in his example, one can replace [[ with [ and replace ]] with ] "

I was thinking that too, that the quotient might not actually need the series

Lukas Heger

okay that simplifies things

In $k[\{T^{1/n}\}]/(T)$ every element is some power of $T$ times a unit, right? because unit+nilpotent=unit

rschwieb

Yes, that was already clear since it's a local ring with nilpotent radical

well, i guess your description is sharper :)

That should buy us that it's uniserial = local+bezout

Lukas Heger

okay but then every ideal is generated by powers of $T$ and then the whole Bezout argument I gave applies to show that every f.g. ideal is prinicipal

and uniserial as well, yes

rschwieb

All that I actually buy

Lukas Heger

okay is there still some question left about this ring?

rschwieb

18:26

Not anymore, I don't think! Looks like since $q<q'$ implies $T^{q'}\in (T^q)$, I think that convinces me the ideals are linearly ordered..

@LukasHeger Lol wait: we still don't know it's coherent. It's Bezout but not a domain :) That's the thing I really wish we had

is there some reason it'd be semihereditary?

There's no nondomain, semihereditary uniserial ring in the db at the moment

rschwieb

18:44

Compared to here though maybe we're in better shape, because maybe the annihilators aren't so big. I think you might have suggested earlier that $ann(T^{q})=(T^{1-q})$, considering only rationals $0<q<1$?

That seems right...

Lukas Heger

19:16

yes that's what I suggest

I don't know if its semihereditary though

but coherent should follow from the explicit computation of the annihilators

rschwieb

@LukasHeger While we're on the topic, I wonder if the fractional exponent polynomials have anything particularly interesting about them as a construction on their own

I should just include it as an interesting superring of Q[x]

(nonnegative fractional exponents, I intended. I'm not quite ready for fractional exponent laurent polynomials :))

Lukas Heger

Is $(T^q)$ even flat? I think not for some reason

but I can't make my gut feeling precise

rschwieb

Lucky you to have a feeling for flatness :). Are you still looking to show it isn't semihereditary using that criterion?

Y'know what, I'm gonna punch in this example and I can let you know if the DB has rationale for saying one way or the other

Lukas Heger

If $(T^q)$ is flat for any fixed $0<q<1$, then $(T^q) \otimes (T^{r})=(T^{q+r})$ for all $r$

but then taking $r=1-q$ gives $(T^q) \otimes (T^{1-q})=(0)$

Now consider the injection $(T^{1-q-\varepsilon}) \hookrightarrow (T^{1-q})$ doesn't stay injective after getting tensored with $(T^q)$

@rschwieb yeah

rschwieb

19:32

Here's another one: a f.p. flat module is projective, and if the ring is additonally local, it's free

Lukas Heger

oh good point

that's easier

rschwieb

:mosh:

Lukas Heger

so it's not semihereditary

does this show generally that semihereditary coherent uniserial rings are domains?

wait, we only need semihereditary + coherent + local

seems like an interesting implication

I mean we do we need coherent here

semihereditary + local => domain

rschwieb

yes

I think my answer is not to your last question

Actually maybe I don't need a new ring: It hink this has all the same traits:
https://ringtheory.herokuapp.com/rings/ring/47/

Lukas Heger

I mean yeah that seems really similar, but simpler to understand

Have you seen that I left a suggestion for the Krull dimension of the adeles?

I know that linking my own answer seems like bad practise...

rschwieb

20:02

On the "drop box" for contributions on the site?

That's ok, as long as its reasoning anyone can follow

Lukas Heger

@rschwieb yeah

rschwieb

I see it: to be honest i check that one very infrequently because the suggestions are few and far between.

Lukas Heger

actually I tried to compute the Krull dimension of the adeles years ago, to no avail, so I'm happy I came up with something now

rschwieb

But it's fine to alert me here when you do

neat!

Lukas Heger

@rschwieb okay

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$Mathematics$ DaRT/DoME