Mathematics - 2023-12-15

Thorgott

00:24

@Rithaniel sounds annoying :P

@TedShifrin fiber bundle? that's a closed inclusion!

Rithaniel

Yeah, I think I've got a proof

Thorgott

@oscarmetalbreak the same idea shows that a finite-dimensional algebra over a field is an integral domain if and only if it is a field

oscarmetal break

01:26

@Thorgott Can I prove this? That a finite commutative ring is a finite dimension vector space over a field? I tried on my own just now, seeing no clues :(.

Thorgott

it's not true that a finite commutative ring is a finite-dimensional vector space over a field. it's true for finite integral domains, however. but I didn't really intend to get at that.

in case you wanna know, though, the reason is that any integral domain is an algebra over its prime field and a finite integral domain is necessarily finite-dimensional over that prime field for cardinality reasons

LandonZeKepitelOfGreytBritn

Does anybody by any chance know what kind of graph this (its name) is and how you read it?

My guess is that the x-axis is just time, but don't have much info about the other axis. It is supposed to represent different samples done throughout time

Xander Henderson

@LandonZeKepitelOfGreytBritn Without context, it is impossible to know.

LandonZeKepitelOfGreytBritn

No wait, I have a better guess...

Xander Henderson

Presumably, this is the "graph" of a function $\mathbb{R}^2 \to \mathbb{R}$, where the horizontal and vertical axes correspond to some independent variables, and the color assigned to each pixel corresponds to a dependent variable.

LandonZeKepitelOfGreytBritn

01:40

the x-axis represents the value of the sample and the y-axis the number of samples with that specific value. Kinf of like a histogram. Just don't know yet what the colors mean

Xander Henderson

But without knowing where that graph comes from, it could be anything.

oscarmetal break

@Thorgott Thanks. I guess I was trying to say finite integral domain is a finite vector space over a field. But I didn't know it was a prime field. Where I should look up for the proof of this fact?

LandonZeKepitelOfGreytBritn

@XanderHenderson hmm, ok thanks... Could make sense. Will keep that in mind. Did not think about it this way

ow ow ow, I am starting to understand it!

little by little

Yhea I think it is a 2d representation of a memory area

and the color represents the measurements' value

Thanks @XanderHenderson !

Now just gotta find the name of such a graph, that way I can try to construct a similar one myself

Thorgott

02:13

for any ring $R$, there is a unique homomorphism $\mathbb{Z}\rightarrow R$ (uniquely determined by the fact it has to map $1\mapsto 1_R$). if $R$ is finite, this cannot be injective, so it has a kernel of the form $n\mathbb{Z}$ for some $n>0$ and factors through an injection $\mathbb{Z}/n\mathbb{Z}\hookrightarrow R$.
if $R$ is an integral domain, $\mathbb{Z}/n\mathbb{Z}$ has to be too, which can only happen if $n=p$ is prime, in which case you have obtained an injective ring homomorphism $\mathbb{F}_p\hookrightarrow R$, i.e. $R$ is an $\mathbb{F}_p$-algebra.

oscarmetal break

02:44

Ok, this things looks good to me. But I am not familiar with algebra though. How does it become a $\mathbb{F}_p$-algebra at the end? I got that this is the prime field.

no I mean I got the $\mathbb{F}_p$ is a prime field

CowperKettle

> Google DeepMind has used a large language model to crack a famous unsolved problem in pure mathematics. In a paper published in Nature today, the researchers say it is the first time a large language model has been used to discover a solution to a long-standing scientific puzzle—producing verifiable and valuable new information that did not previously exist. technologyreview.com/2023/12/14/1085318/…

oscarmetal break

Should I enforce myself to consider this $\mathbb{F}_p$-algebra as a vector space with a specific dimensions in mind? Or because it satisfies the axiom of the vector space and it is also finite so I can assign it as a finite dimensional vector space.

Thorgott

a) there are many equivalent ways of defining a $k$-algebra for a field $k$. my preferred definition is simply that a $k$-algebra is a ring $R$ together with a ring homomorphism $k\rightarrow R$. the multiplication on $R$ restricted along this homomorphism in the first component then makes $R$ into a $k$-vector space in a way so that its multiplication becomes $k$-bilinear.

b) the latter. it is finite as a set, so certainly finite-dimensional as a vector space. the dimension will be $\log_p|R|$, but of course that's only true post hoc.

oscarmetal break

03:28

I see. I was also interested in how you get $\log_p|R|$ for the dimension of algebra, I have checked that it is trivially true for $\mathbb{Z}/2\mathbb{Z}$. How is it true in general?

NVM I get it now

oscarmetal break

04:56

Given integer $a$ and any $a^2+1$ objects $x_0,\dots, x_{a^2}$ then there are always $a+1$ distinct indices $v$ for $0\leq v\leq a^2$, such that the corresponding $a+1$ objects $x_v$ are either all equal to each other or mutually distinct. I can't quite understand this assessment. I only got it partially if it is to put $a^2+1$ balls to $a$ container such that at least $a+1$ are in the same container. But how is the assessment at the beginning true without anymore input to the condition?

leslie townes

08:02

oscar some of the confusion may be in the question's poor choice of wording. e.g. if you "have" 10 "objects" but 4 of them "are all equal to each other," there is a very real sense in which you don't have 10 "objects" at all; you have some smaller number of objects. it might be clearer to think of "labeling" objects with 10 indices. it is probably clearer to visualize one thing potentially receiving multiple labels, than one thing potentially being multiple things that are "equal to each other."

the argument sketch you provided is addressing one of two alternatives. if you're labeling objects with a^2 + 1 labels. one thing that can happen is that a+1 different objects receive labels. in that case there is nothing to show. another thing that can happen is that no more than a objects receive labels, where the desired conclusion is that in this case, at least one object receives a+1 labels. that's where pigeonholing comes in.

edsger dijkstra wrote an informative essay against the "pigeons and holes" conception of the pigeonhole principle that you or others might enjoy. cs.utexas.edu/users/EWD/transcriptions/EWD09xx/EWD980.html

Unknown x

09:18

0

Q: $\frac{\partial}{\partial t}\frac{\partial }{\partial s}=\frac{\partial}{\partial t}\frac{\partial s}{\partial t}+\kappa^2\frac{\partial}{\partial s}$

Prove that $$\frac{\partial}{\partial t}\frac{\partial }{\partial s}=\frac{\partial}{\partial t}\frac{\partial s}{\partial t}+\kappa^2\frac{\partial}{\partial s}.$$ My attempt:- $$\frac{\partial}{\partial t}\frac{\partial }{\partial s}=\frac{\partial}{\partial t}\Big(\frac{1}{v}\frac{\partial }{\...

oscarmetal break

@leslietownes Thank you. I haven't read your link yet but it does feel more reasonable after you have explained it with labeling. So to make the assessment clearer. The objects $\{x_0,\dots, x_{a^2}\}$ does not exclude the possibility of repeated objects. So If there are $\leq a$ objects are not repeated then there are always $a+1$ of the labeled objects are equal. On the other hand if there are $>a$ non-repeated objects then there are always $a+1$ labeled objects are distinct.

Mad

https://math.stackexchange.com/questions/4602863/an-operator-is-compact-iff-it-maps-any-weakly-convergent-sequence-to-a-convergen

why in this proof, he uses the theorem that states that every sub sequencee has a subsequence that converges. wouldnt it not be sufficient to use the fact that if every subsequence converge to x then the sequence converge to x, IE why do we need the x_n_k_m instead of x_n_k

Jakobian