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1:54 AM
How are morphisms in the category of probability spaces defined? Just measurable functions I guess?
2:32 AM
Is the zero ring is an integral domain?
Wiki says no, but {0} satisfies all properties for bein ring and an integral domain, so why "no" ?
By definition, being an integral domain requires 0 ≠ 1.
@DannyuNDos oh, so wait, I forgot 'bout it, thanks, but isn't that in the defn of the ring?
I mean when we define a ring, we assume, that 0 ≠ 1
AFAIK, most conventions of the definition of rings don't require that.
In Fraleigh, for example.
2:37 AM
So is that a point in the definition of integral domain as you mention, @DannyuNDos ?
Also, note that the definition of fields requires 0 ≠ 1.
@DannyuNDos The confusion arose becoz of this
This is an image from the book by Herstein
He didn't really make the point: 0 ≠ 1
In fact, he really doesn't assert that 1 should be in the ring. According to the definitions, the ring might not have a unit element. The ring just needs to be commutative and that's all!
Now, this is really confusing, imo
Oh, it seems that that defintion enables the zero ring to be an integral domain, when taken literally.
@DannyuNDos yes, so they consider that {0} is an integral domain. But there's a catch.
Different books may have different conventions, which can be very confusing. If you're in a class using that textbook, it's the best to follow the convention posed by that book.
2:43 AM
@DannyuNDos If we consider {0} be an integral domain then it's characteristic is not zero.
Neither is the characteristic a prime.
Yeah, the characteristic of the zero ring is 1.
The characteristic is finite and that's all
@DannyuNDos why?
Oh, wait ur right
In fact, it's the only ring (with unity) that has characteristic 1.
@DannyuNDos the thing is the claim made in the last line of the image above, that the characteristic of a ring is either 0 or a prime, is not valid then
For 1 is not a prime.
The book specifically says, that if the characteristic of an integral domain is finite, then it's gotta be a prime.
The books misses the fact that {0} is an integral domain which has a finite characteristic but it is not prime
Yeah, that book seems to have forgotten to specify 0≠1 for integral domains.
2:48 AM
@DannyuNDos That seems like the only possible explanation.
So, clearly this definition of an integral domain as mentioned in the book, leads us to a contradiction and hence it's precisely incorrect, right @DannyuNDos ?
If you chose the textbook, you'd better switch to Fraleigh.
@DannyuNDos So can you say, what's your definition of an integral domain? I mean a complete one.
Pretty strange, cause Herstein is such a good and a popular book.
@TedShifrin , @leslietownes Do you feel the same way, about it?
An integral domain is a commutative ring with unity, where 0≠1, admitting no zero divisors.
It's notable that every field is an integral domain.
@DannyuNDos ok, that's really good!
It makes sense...
3:10 AM
@DannyuNDos could also be an issue of the definition of a ring.
Many authors assume that a ring has an identity. Congrats with a "rng", which is a ring without Identity.
@DannyuNDos I know of a domain as a commutative ring with no zero divisors. Does the "integral" confer both the multiplicative identity and $0 \neq 1$?
@XanderHenderson in the book, the author does not assume that rings have an identity. They assume that all the rings are associative in the chapter.
3:26 AM
I have no idea what associativity has to do with whether or not a ring has an identity. The point I was making, which was not directed at you, was agreement with the idea that different books have different conventions.
The definition of integral domain might be up for grabs. But it could also be the definition of ring. Or any other term
@XanderHenderson The thing about associativity was just brought up by me without any reason and I said it just like that. I 'm sorry. But now, I got what you said.
And, unsurprisingly, this has come up before. math.stackexchange.com/q/1961652
Q: Are there domains that are not commutative?

user349449Let $A$ a domain, i.e. $ab\in A\implies a=0$ or $b=0$. It's written that all domains are commutative. Is it by definition, or can we prove that domains are commutative? I mean, do we only consider domaind for commutative rings, or is a ring that is a domain then commutative?

4:33 AM
How can I find all primitive roots of unity under a Galois field? $\mathbb{F}(2^{64})$ specifically?
By "all" I mean square roots, cubic roots, quartic roots, and so on.
I doubt that the Schönhage-Strassen algorithm should work only modulo $2^n + 1$ so
4:47 AM
A: Finding the n-th root of unity in a finite field

Thomas PorninIn a finite field of size $q$, the multiplicative subgroup has order $q-1$ (i.e. all elements are invertible except $0$). If $n$ is relatively prime to $q-1$, then there is only one $n$-th root of unity, i.e. $1$ itself. If $n$ divides $q-1$, then there are $n$ roots of unity. In the remainder of...

2 hours later…
6:29 AM
@robjohn Are you aware of any other areas other than complex numbers which took a while to be accepted?
lord: non exhaustive list but non euclidean geometry (including not just outright denial of its utility, but lots of false proofs that euclid's fifth postulate followed from the others). maybe the proof of the four color theorem where a computer was used to do a lot of cases. maybe thomas hales proof of the kepler conjecture (at least at first).
the question borders on too vague because of the potential wide scope of 'took a while' and 'accepted'
maybe per enflo's banach space counterexample relating to the invariant subspace problem. maybe louis de branges proof of the bieberbach conjecture. (in these two examples "took a while" is on the order of "more than a couple of years, but definitely not decades" and "accepted" maybe also means "met with approval of the relatively small amount of people who worked in the area")
cardinality of reals greater than naturals too, I think
i'm reading into the inquiry a requirement that the math thing be the subject of active investigation by a whole lot of people, or at least known to be difficult by a group of people that is wider than the group that is working on it. maybe that isn't justified but if you go too small scale you can find this stuff everywhere.
fourier's results in fourier series maybe have this form, except with anything of that era (early 1800s) maybe he didn't state them precisely enough for it to be clear what "accepting" what he had done would have meant, exactly.
he had trouble getting as much praise for his work as he was expecting to, i think.
in modern times you see people adopting more of a 'wait and see' approach to things that seem revolutionary to the point of maybe being too good to be true. you don't have people getting into heated disagreements with warring camps where one side is like "this is fine" and the other side is like "we reject this." even historically i think that level of non-acceptance was pretty rare.
or at least rarer in math than it is in other fields. in economics, for example, a whole lot of stuff is contested at a level of one esteemed professor stating X and another esteemed professor stating something very close to not-X, where you might effectively have to pick a side just by working in certain areas.
maybe that's true in any field with a 'policy' dimension. i dunno.
6:50 AM
so I've read a decent chunk of this stats paper and I think I've only seen bounds on $f$ of the form $\lvert f(x) - f(y)\rvert \leq Cd(x, y)^{\alpha}$, which has to do with Hölder, not BV. But now I see an example where the authors describe a function as being Lipschitz. So I go look up Lipschitz and see that Lipschitz sometimes implies Hölder, but apparently there are examples of each that are not the other. So I dunno why you'd mention a third form of continuity at all.
Seems like it just confuses things
I realize I'm not a math guru, but it doesn't seem obvious that they are either
How is Robjohn's profile image a gravatar and not a picture?
If the profile image of current users of the room is dragged to the message box then it either shows a link of imgur or a link of gravatar
7:10 AM
novice: is there anything in the paper that is inconsistent with the author being completely unaware that "BV" has a special meaning in analysis? could they be using the term, on the fly, to mean something like: a class of F functions is "a class of bounded variation" if sup {|f(x) - f(y)| : f in F and x, y satisfy d(x,y) <= C} is finite for all C?
this is something ted was suspecting yesterday, and the more you say about the article, the more plausible it sounds to me.
I am not sure if this will answer your q, but "bounded variation" appears only 3 times in the paper, and I can post them here
ah, no need. but it might be a plausible explanation. my hypothetical definition would make lipschitz, holder, etc. all particular kinds of that kind of 'bounded variation'
okay, sure. I'm really only familiar with regular continuity and uniform continuity, so I might be missing something, but it seems like the authors are at minimum making it unclear exactly what assumptions they require
its not common for people to run into the real analysis notion of BV outside of pure mathematics [as far as i know], while i have seen both lipschitz and holder conditions in applications as particularly tractable ways of providing quantitative estimates
when I google "bounded variation statistics" I just seem to get pure math results, so if BV has some particular mean in stats that's escaping me, it's sure not obvious
anyway, thanks for your help
2 hours later…
9:38 AM
Let $A(t)$ be a matrix that depends on a real variable $t$. When we say $A(t)$ is differentiable, we mean differentiable with respect to all entries $a_{ij}$, right? In particular, we have $\frac{\mathrm{d}}{\mathrm{d}t}B=0$, where $B$ is any constant matrix, or? I guess we have that the multiplication by a scalar, sum rule and product rule hold, but not quotient rule since we can't divide by a matrix, right?
Matrix division is possible when the denominator is invertible. Left division and right division may differ tho.
ok, interesting
Also, I doubt the product rule for matrix multiplication would hold, for the matrix multiplication is better thought as a (linear) function composition.
@DannyuNDos I would say it does hold, since $$\begin{align}A(t + h)B(t + h) - A(t)B(t)&=A(t + h)B(t + h) - A(t + h)B(t) + A(t + h)B(t) - A(t)B(t)\\ &= A(t + h)(B(t + h) - B(t)) + (A(t + h) - A(t))B(t).\end{align}$$
Oh, that's cool.
9:52 AM
Is there a topological ring theory involving surreal numbers? $\mathbb{Q}(\omega)$ in order topology, for example?
2 hours later…
11:26 AM
@DannyuNDos what do you mean by that
$\mathbb{Q}(\omega)$ with order topology seems pretty far from surreal numbers
11:43 AM
I mean, the field is a subset.
12:32 PM
So are many others
But that doesn't explain what you would want
1 hour later…
1:43 PM
what is Ext(A,B), where A and B are Abelian groups? What is its proper concrete definition?
1:55 PM
I am reading Whitney Approximation Theorem. Suppose we have two smooth manifolds $M$ and $N$ and a continuous map $f: M \to N$. Roughly the theorem states that there exists a homotopy $F: M \times[0,1] \to N$ with $F_0=f$ and $g=F_1$ is a smooth map
Now let $f$ be injective, the theorem gives us no control over whether $g$ is injective or not
Suppose $f$ is smooth on a (possibly empty) closed subset $A$ and suppose that there
exists an open neighborhood of $\partial N$ that does not intersect $f(A)$
2:27 PM
Then the homotopy is a homotopy rel $A$
suppose that we are given a continuous map $\rho: M \to \Bbb{R}_{\geq 0}$ such that $\rho(x)>0$ for all $x \not\in A$
Then we can find a homotopy $F$ that also satisfes the condition: For every $(x,t) \in M\times [0,1]$, we have $d(F(x,t),f(x))<\rho(x)$ where $d$ is a metric on $N$ that induces the topology on $N$
So this is the approximation part
Now it also states that even if we let $f$ be injective, we cannot control whether $g$ will be injective or not
My question is what restrictions(if any) we should add to make $g$ injective
2:52 PM
Hoping someone can help me interpret the following:
In particular, I am lost at the definition of $X_n$. Two questions: (1) Why do they say "We write $X$..." but then write $X_n$, and not $X$? (2) Is the definition of $X_n$ simply that, if the subscript is 1, then we get the unity element (multiplicative identity) of $R[\![X ]\!]$ and otherwise we get the additive identity?
I don't think my understanding in (2) is correct, since that interpretation doesn't lead to (8.10). Under my interpretation, (8.10) would be independent of $m$ and just depend on if $n = 1$ or not.
3:22 PM
@EE18 $X_n$ is a sequence
they want to identify the $X$ in the definition of $R[[X]]$ and $X$ as the formal power series $(1, 0, 0, ...)$
basically, $X = (X_n)_{n=1}^\infty$, it's this sequence
@Koro read a text on homological algebra, I don't think any ad hoc definition helps
there's also a section in Hatcher explaining Ext
3:46 PM
OMG I see now. It confused me that they switched to using uppercase. But, to be sure, I am correct that $X_n$ is the unity element of $R[\![X ]\!]$ right? @Jakobian
Then I guess $^m_n$ really means $(X^m)_n$, i.e. the $n$th entry of $X^m$?
Oh no I see. $X := (0,1,0,0,...) \neq (1,0,0,...) := 1_{R[\![X ]\!]}$.
@Thorgott I want to know this in order to understand universal coefficient theorem.
@Thorgott I didn't find that gigantic section helpful :(.
4:24 PM
Let R be the ring of 2 × 2 matrices with rational entries. Prove that the only ideals of R are (0) and R.
Need some help with this.
@SoumikMukherjee Can you please take a look at this?
@ThomasFinley invertible or not necessarily?
Take a non zero ideal $S$, try to prove $I \in S$
@EE18 $X_n^m = (X_n)^m$ it means literally, $X_n\in R$ taken to the power $m$
4:41 PM
I have reread the section closely and actually disagree, but perhaps that's because I didn't provide enough context. I'm now quite sure that $X^m_n$ means $(X^m)_n$ because of the reference to Example 5.12, wherein we refer to the inductive definition of $a^m$ for any $a \in Y$ with $Y$ a set with an associative operation defined thereon (here $Y = R[\![X]\!]$). $X_n\in R$ taken to the power $m$ would give $X^m_n = 1 \iff n = 1$. Nevertheless, I do understand now, and that's thanks to you!
It's my fault for not providing enough context before, so apologies for that.
@robjohn not necessarily
@SoumikMukherjee That's precisely where I am stuck. I don't know whether S is invertible . If it is so, then $I_2\in R$ and we are done
But if, $S$ is not invertible then that's a problem.
@robjohn The set may have non-zero matrices that are non-invertible and that's where the problem occurs. If the set has an invertible matrix S, then $I_2\in $ the ideal and we are done as then, the set R=the ideal.
But if, $S$ is non-invertible then we have the problem.
@SoumikMukherjee My comment :
4 mins ago, by Thomas Finley
@SoumikMukherjee That's precisely where I am stuck. I don't know whether S is invertible . If it is so, then $I_2\in R$ and we are done
has a typo
I wrote $I_2\in R$ instead of $I_2\in $ the ideal
So what to do?
4:58 PM
Think about multiplying by $E_{ij}$.
@TedShifrin Is it the 2×2 matrix A whose entry in the ith row and jth column is 1 and all other entries are zero?
Take an arbitrary element $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ in $S$ and try to show $\begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix}$ and $\begin{pmatrix} 0 & 0 \\ 0 & d \end{pmatrix}$ also belong to $S$
Yes @Thomas
5:20 PM
5:34 PM
Whelp... it is now 10:30am. At 7:30am, I started working through the deluge of email which arrived after 3:00pm yesterday. And now I am caught up. Time to do some, like, teaching related work (nearly all of those emails were related to curriculum, and not for any of the classes I teach).
5:47 PM
done: $\begin{pmatrix} 1 && 0 \\ 0
&& 0\end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 1 && 0 \\ 0 && 0\end{pmatrix}=\begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix}$ and $\begin{pmatrix} 0 && 0 \\ 0
&& 1\end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 0 && 0 \\ 0 && 1\end{pmatrix}=\begin{pmatrix} 0 & 0 \\ 0 & d \end{pmatrix}$ also belong to $S=$ set of all 2×2 rational matrices.
@SoumikMukherjee Now?
@robjohn Sorry, I didn't get how that was supposed to solve things ?
I mean I didn't get that reference:?)
Do you know what E_ij is that Ted mentioned?
@ThomasFinley Multiply something, add something, get something that is invertible
@Thorgott I figured it out now.
A paper defines $P^+$ as the set of symmetric, positive semi-definite $d\times d$ matrices, $\mathcal S$ as the set of symmetric $d\times d$ matrices, and then defines $\mathbb P^+$ as the set of $P^+$-valued Radon measures $\mu$ on $\mathbb R^d$ with finite $\text{tr}(d\mu(\mathbb R^d))$. They then say that "clearly" $\mathbb P^+ \subset (C_0(\mathbb R^d,\mathcal S))^*$ (where $C_0$ are the continuous functions decaying at infinity). Any ideas on how to see this?
Ignoring the Radon and finite trace parts, even. I don't even see how to type-check that they are the same type of objects. I thought $G\in\mathbb P^+$ would look like something that takes in subsets of $\mathbb R^d$ and spits out elements of $P^+$. That doesn't look like an element of $(C_0(\mathbb R^d,\mathcal S))^*$ to me?
6:01 PM
@ThomasFinley yes. So you are on the right track. Now, you just show that all E_ij are in the non zero ideal.
Unless it's like a Riesz thing, like $G$ being identified with $f\mapsto \int f\,dG$?
I don't quite understand where $\mathcal S$ comes into play in that case...
@Koro I dont understand it...
6:55 PM
@SoumikMukherjee how? That's the whole issue :?)
7:29 PM
I have some inequality $C||\phi||_{L^p(\Bbb R^n,\mu)}\leq ||\phi||_{L^q(\Bbb R^n,\nu)}$ fo all $\phi\in C_c^{\infty}$. Then they are saying that as $\mu,\nu$ are regular measures so the inequality is also true for all Borel measurable functions.
Why is that true?
I think I can now visualize why choosing where a dot on a ball should point is not the same as choosing an element of SO(3)
for example, if I want to move the dot from the "north pole" to the "south pole", there are many rotations that can accomplish this
But, if I cared how that rotation occurred, then maybe using SO(3) is not crazy
...unless I'm misunderstanding again
@ThomasFinley why is that not an ideal?
@ThomasFinley if you show that all E_ij are in the non zero ideal that that ideal is basically the whole R because you can write any element of R as a linear combination of E_ij 's.
@SoumikMukherjee If the dimension of $N$ is more than twice than the dimension of $M$, any continuous map is homotopic to a smooth embedding. This is probably not quite what you're looking for, but the only thing I can think of.
@Koro glad to hear
1 hour later…
9:06 PM
obata obata
Hi :) Does anyone have any feedback on this?
A: I believe in another logic where there is no Gödel incompleteness theorem

ShaunI think Gödel's Incompleteness Theorems can still be applied, but with the statement, This sentence is either false or meaningless. If it is meaningless, then it is true, a contradiction. If it is true, then it is either false or meaningless; either way, a contradiction. If it is false, then it...

I think the question is inquisitive and important.
9:24 PM
@Shaun, Hi!
I'm gonna get more into complex analysis study. It's a big gaping hole in my know how
Hi @MathCrackExchange :)
This book is great for complex analysis: amzn.eu/d/2HF6Poq
I'm liking: Serge Lang's (digital copy) or: amazon.com/dp/3540257241
because of elliptic / modular forms
@Shaun how are studies going?
I'm at work now
Get to go home soon (2 hour drive), then study maths :D
It's just today that I had to drive down to Phx. Usually working from home though I did stay down here for 3 weeks last month
9:45 PM
Two hours!?
It's going really well. Thank you for asking :)
I had my 9 month review recently, @MathCrackExchange.
You can find my slides on Academia.edu.
I improved my code over the last few days too.
I actually managed to compute the values of a function $\Delta$ for a handful of groups I'm interested in.
I give tutorials in the department these days. I taught undergraduate engineers some introductory mathematics today.
10:05 PM
@MathCrackExchange Phoenix? Yuck.
My trick was to trade memory for time, by computing conjugacy classes before other calculations, instead of recalculating them during those calculations.
I really want brisket
10:28 PM
@robjohn Hey rob, I wondered about the correctness of the following argument:

Let $\sum \alpha_k$ be an infinite series of reals.
Suppose there exists a subsequence of $(n)_{n=1}^{\infty} denoted by $(n_k)_{k=1}^{\infty} such that $lim_{k \to \infty} n_{k+1} - n_{k} = \infty$. If \sum \alpha_k$ is bounded then it converges.
I know that if we assume the partial sum has a convergent subsequence, instead of assuming the partial sum itself is bounded, then it is true
I missed some latex and cannot edit, now. This is more clear:
Let $\sum \alpha_k$ be an infinite series of reals.
Suppose there exists a subsequence of $(n)_{n=1}^{\infty}$ denoted by $(n_k)_{k=1}^{\infty}$ such that $lim_{k \to \infty} n_{k+1} - n_{k} = \infty$. If $\sum \alpha_k$ is bounded then it converges.
What do the two pieces of that story have to do with one another?
I forgot to mention the assumption that $\lim_{k \to \infty} \sum_{i = n_k}^{n_{k+1}-1} \alpha_i= 0$
are the $\alpha_i\ge 0$?
oh I am so tired
$\lim_{k \to \infty} \sum_{i = n_k}^{n_{k+1}-1} |\alpha_i|= 0$
I apologize
Nah, it’s
10:40 PM
I know this is not necessarily true without the absolute value, since we can have the sum 1 - 1 + 1/2 + 1/2 - 1/2 - 1/2 + 1\3 + 1\3 + 1\3 - 1\3 - 1\3 - 1\3 + ...
I have a history class first thing in the morning
Perfect time to nap
@TedShifrin You mean that is false?
11:01 PM
Don’t know.
I wanted to make a really crude joke but I’ll resist

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