Mathematics

1:31 AM

Quick question, if we have an uncountable power set $2^S$, and we consider a subset $H\subseteq 2^S$ for which everything in $H$ is disjoint, is it possible for $H$ to be equinumerous with $2^S$?

Thorgott

if you assume choice, there is a choice function $H\rightarrow S$, which is injective by the disjointness condition, so $\# H\le\# S<\#2^S$

Sir Cumference

@Thorgott I see, thanks

Rithaniel

2:17 AM

So, here's a thought. If you have a ring $R$, does it make any sense to talk about $\lim\limits_{n\to\infty}R[x^2,x^n]$? Is that notion even well defined?

Thorgott

what's $R[x^2,x^n]$

Rithaniel

$R$ adjoin the indeterminates $x^2$ and $x^n$

Thorgott

a) $R[x,y]$ or b) the subring of $R[x]$ generated by $x^2$ and $x^n$?

Rithaniel

Ah, I see. It's the subring of $R[x]$

If $x^n$ were not referencing the same $x$ as $x^2$, this would be an easier thing to "answer?"

Thorgott

it would be boring then, cause all those rings would be the same

option a) was silly on my end

Rithaniel

2:31 AM

Ah, gotcha

Thorgott

anyhow, I don't see a natural map $R[x^2,x^n]\rightarrow R[x^2,x^{n+1}]$

Rithaniel

Okay, so the limit would be in terms of homomorphisms? I can see the logic in that.

I guess the larger thing is that I've never encountered the idea of a limit of a sequence of rings

Thorgott

the thing is that just a "sequence of rings" is not enough information to create a sensible notion of limit

you somehow need to "relate" the rings in your sequence

(if you look at a sequence in a topological space, the elements of the sequence are "related" by the given topology - I might be stretching analogy here)

Rithaniel

Ah, so we need to set up a metric space where the elements are rings

Or, maybe not even a metric space?

Thorgott

so what you want to do is look at a sequence of rings $(R_n)_n$ and homomorphisms $f_{ij}\colon R_i\rightarrow R_j$ for $i\le j$ such that $f_{ii}=\mathrm{id}_{R_n}$ and $f_{jk}\circ f_{ij}=f_{ik}$ for $i\le j\le k$

Rithaniel

2:39 AM

It's been a minute since I've thought about more abstract topological spaces

Thorgott

nah, this is something completely algebraic

if you think of it diagramatically, you just arrange the rings in a sequence and have an arrow from one to the next one

Rithaniel

Alright, so you can think about the sequence as a sequence of homomorphisms $(f_{1n})_n$

Then consider the images as $n\to\infty$

Thorgott

that is a great idea

for example, take $R_n=R[x_1,...,x_n]$ (where $R$ is some ring), then you have a natural inclusion $f_{ij}\colon R[x_1,...,x_i]\rightarrow R[x_1,...,x_j]$

this is essentially just a tower of rings where you just adjoin more and more indeterminates

the "limit" of this sequence should intuitively be $R[x_1,...,x_n,...]$, the polynomial ring in infinitely many variables

Rithaniel

Yeah, alright, I think I get the general idea

Thorgott

but that's not just because of the individual rings, but also because of how we know that they lie inside one another (which is given by the inclusion homomorphisms)

Rithaniel

2:44 AM

So, if I wanted to somehow talking about an indeterminant of infinitely high power, I would need to have these maps set in place beforehand

So, I could have the inclusion maps $$f_{ij}:R[x^3,x^{2^i}]\to R[x^3,x^{2^j}]$$

superscripts on superscripts

Thorgott

the idea behind constructing the limit object in the general case is to take the disjoint of all the $R_n$ and then identify all the elements that "become the same" under the given homomorphisms (this accounts for how the rings relate to one another under the given homomorphisms)

this works not only for sequences, but more generally for directed sets (or, even more generally, for filtered categories)

Rithaniel

Okay, so the limit of my $R[x^3,x^{2^n}]$ idea would be $R[x^3]\cong R[x]$?

Thorgott

I don't see what your inclusion maps are

it seems to me as if the arrows would have to go in the reverse direction

Rithaniel

Ah, right, yeah, they would, if they were inclusion

Thorgott

(in which case, you can rediscover the notion of an inverse limit)

Rithaniel

2:50 AM

I was thinking $f_{12}(x^{2})=x^4$, which wouldn't be inclusion

In fact, thinking about it, I am unsure if that is a well defined homomorphism

Thorgott

how would it act on a general polynomial?

Rithaniel

So, to start $x^{20}=f_{12}(x^2)^5=f_{12}(x^{10})=f_{12}(x^3)f_{12}(x^3)f_{12}(x^2)$, and so $f_{12}(x^3)=x^7$. Checked a few examples, and it seems to be consistent

Wait, I think I found a counterexample

$x^{21}=f_{12}(x^3)^3=f_{12}(x^9)=f_{12}(x^3)f_{12}(x^2)f_{12}(x^2)f_{12}(x^2)=x^7x^{12}=x^{19}$

So my hypothetical map $x^2\to x^4$ is not well defined

Faust

3:10 AM

@TedShifrin hows it going?

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$Mathematics$ Mathematics