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!
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.
@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
@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 ?
@DannyuNDos I know of a domain as a commutative ring with no zero divisors. Does the "integral" confer both the multiplicative identity and $0 \neq 1$?
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.
Let $A$ 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?
In 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...
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")
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.
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
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.
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
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?
Also, I doubt the product rule for matrix multiplication would hold, for the matrix multiplication is better thought as a (linear) function composition.
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 $f(A)$
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
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.
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. $X := (0,1,0,0,...) \neq (1,0,0,...) := 1_{R[\![X ]\!]}$.
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.
@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.
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 $S$
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).
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 ?
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?
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.
@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.
I 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.
My trick was to trade memory for time, by computing conjugacy classes before other calculations, instead of recalculating them during those calculations.
@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.
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 + ...
Let $A$ 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?
I 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...
