SO that is A in terms of B?

Don't use notation that makes things difficult for you.

$\frac A B$ is $A$ divided by $B$..... (????)

If you use a fraction?

So Basically it means that 1 of B is equal to (A/B) of A?

$\frac{A}{B} = 1$

I don't understand what you're saying.

What do you mean by ratio?

Don't worry about it

I get it now.

But thanks

@PeterTamaroff happy now?

@Charlie Well, my screen doesn't taste that well, but whatevs.

@PeterTamaroff :(

@JacobBlack Mark this day!!!

@JacobBlack Did your campaign today end badly?

Hi @Charlie how are you?

@robjohn Duly marked >8(

@skullpatrol Hi, Skullito, I'm fine

@skullpatrol how are you?

@Charlie Fine, thanks for asking Chucky.

@skullpatrol :)

@Charlie :)

unmark this day

@skullpatrol HAHAHAHA

@Charlie HOHOHOHO

@Argon Yo

@skullpatrol HEHEHE

@skullpatrol What up?

@Argon Chillin' ... you?

@skullpatrol Same same

@JacobBlack I CANT WAIT UNTIL FRIDAY!

(here, that is)

some sooner than others

@JacobBlack Depressing. It's hard to cherish being forced to go to school.

maybe you should start exercising?

start slow and build up

Darn, so many answers to choose from

i mean reeeaally slow...

too many people jump in and try to do too much

@skullpatrol I think people try to do less and less

@Charlie True...

I'm always good pal

even when I'm bad

@skullpatrol yep

Is anyone familiar with the totient summatory algorithm?

bananas?

Hahahaa

Hohohoo

The quality of food depends on who cooks

No

@JacobBlack Do you do any cooking?

@user64283 mathoverflow.net/questions/3274/…

seen that already, thanks though

np

my post is math.stackexchange.com/questions/317323/…

@skullpatrol what about you?

No

@skull

@skull

@Charlie

@Charlie

Wazzup Chucky?

@skullpatrol ;DDD

@Charlie ;DDD

@skullpatrol I like Chucky

@Charlie Me too.

@skullpatrol nice :D

@Charlie Have you seen any of these Child's Play movies?

@skullpatrol of course!!!

I liked Chucky when I was a child

I watched all of them

@Charlie How 'bout this?

quick question: if i have a field extension $F/k$ and the tensor product $A\otimes_k F$ where $A$ is a $k$-algebra, am i incorrect in assuming that $a\otimes c=b\otimes c\implies a=b$?

@skullpatrol :D

@Charlie :D

rating please

@skullpatrol 8.63

@Charlie HAHAHA

:(

What did I do?

@Charlie $$\exp(8.63) = \exp(2.30325\cdot 2+ 1.0986\cdot 3+ 0.69314+ 0.03456) \approx 10^2\cdot 3^3 \cdot 2 \cdot 1.03456 = 5586.624$$

0.01 accuracy means you take your music very seriously

@argon ;)

@skullpatrol is that bad ?

@Charlie Haha I was impressing people today by computing exponentials of smaller values $0 \le x \le 7$ in my head or on paper :)

They don't know the tricks :D

@Argon good boy!

@Charlie No.

@Riem given $c\ne0$, $$a\otimes c=b\otimes c\iff (a-b)\otimes c=0\iff a-b=0\iff a=b$$

@skullpatrol does it mean I'm boring?

@Charlie Not at all :-D

@anon I think I have a guess for the behavior of the zeros of the Faulhaber polynomials. No clue how to prove it yet though.

@skullpatrol good :)

@anon hmm that makes sense, but ive also heard that tensor products of arbitrary fixed modules aren't injective, i.e. $\mathbb Z_2\otimes \mathbb{Q}=\mathbb Z_2\otimes\mathbb{R}$ but $\mathbb{Q}\neq\mathbb{R}$. What makes this different?

@AntonioVargas interesting. I was also curious as to what was causing the strangely arranged zeros inside the figure.

@anon Definitely numerical error. My computer is plotting the n=800 zeros now (taking a while)

@Riem since $F/k$ is an extension it has the same characteristic, and $A$ is a $k$-module. (${\bf F}_2$ isn't a $\bf Q$-module for instance)

@anon ahah! i understand now. thank you

Hi guys!

hi

@JacobBlack you are a book conoisseur right? could you recommend me an abstract algebra book which really helps me understand de subject?

Good night @skull !

the*

@Charlie Good night Chucky ;*

@skullpatrol see you tomorrow ;)

@Charlie Yep ;)

@skullpatrol bye

@JorgeFernández what level of abstract algebra are you looking for?

@Charlie Good night

@Riem I have no intuition in it, I only no things like the definition of a group,ring,field,monoid,isomorphism,homomorphism,center,normalizer,subgroup.

But I am ot poised with it yet, I want a book that really makes you agile with it.

@JorgeFernández hmm i believe dummit and foote is the most used text (for good reasons), and fraleigh's isn't so bad either

I own Dummit and Foote's but I am finding it really boring

it depends on what you mean by boring then

if its the material, then im not sure what other algebra text would speak to you any differently :)

I also found D&F boring. It's a little more elementary, but I thought Pinter's A Book of Abstract Algebra was at least a good read.

for example in diestel's book (graph theory) he is allways saying like, oh a new proof of this was discovered 30 years ago, this is so cool. And in abstract algebra (the part I was on a month ago before I stopped reading it, there was a part of like 10 pages of. This is a centralizer, this is a center, this is a normalizer, and a ton of definitions along with some proofs

I had no clue why they where useful I was just spending hours thinking about the proofs without really getting to something juicy.

@riem mabye it's just that i'm too young to be reading these kind of books?

i have a question of my own: if i have a semiprimitive ring $R$, is the weyl algebra over this semiprimitive ring, $R[x,y]/\langle yx-xy-1\rangle$, also semiprimitive?

intuitively, i would assume that modding out by this ideal shouldn't ruin semiprimitiv(ity)

@AntonioVargas Do you know about Jacobson's BAI? I mean, what is your opinion?

@PeterTamaroff No, I haven't read it, sorry.

@AlexYoucis did you receive my last message?

@Sanchez are the explict formula for prime counting functions involving the roots of the zeta function just obtained by inverse mellin transforms?

@PeterTamaroff by whom?

@JayeshBadwaik Jacobson...

=)

@PeterTamaroff bleh, :P I mean who told you there were better places to learn from?

@Ethan, yes.

@Sanchez Why use the inverse mellin transform on say $$\frac{-\zeta'(s)}{s\zeta(s)}=\int_{1}^\infty\psi(x)x^{-s-1}\ dx$$, to get asymptotics for chebyshevs prime counting function, when it can also be expressed as a laplace transform $$\int_{1}^\infty \psi(x) e^{-sx} \ dx=\frac{1}{s}\sum_{n=1}^\infty\frac{\mu(n)\ln(n)}{e^{ns}-1}$$

I assume inverting one is easyier then the other?

I know you need analytic information about the proceeding functions to evaluate the inverse in terms of them, is it just easier to obtain this info using the zeta function or somthing?

@JayeshBadwaik Mariano

Is it appropriate in p-adic analysis to use $p$ when referring to a specific prime? or does that get too confusing?

@PeterTamaroff I see. Did he say why?

@Ethan, the goal of using Mellin transform was to relate the problem of estimating $\psi(x)$ to zeros of $\zeta$. I'm not familiar with Laplace transform. Does it transform the estimation problem to something potentially tractible?

@Ethan, your right hand side of Laplace transform seems sort of random to me. Perhaps if you can find a method to study its zero/poles then it would be useful.

If $M$ and $N$ are pure submodules of some larger module, does it make sense to ask when $M\cup N$ is a pure submodule? Not really, right? Or am I missing something...

@BenW., why would one talk about $M \cup N$ for modules?

@Sanchez I was thinking the same thing. I was reading an exercise which asked to prove that the intersection of pure submodules is pure, and the union of pure submodules is pure. The last case doesn't make much sense.

@BenW., where does the exercise come from?

@Sanchez Exercise 17 (d) page 161 of Steven Roman's Adv. Linear Algebra. I was rereading through it and it caught my eye as it didn't look right. I think his definition of a pure submodule in that exercise is misstated as well.

@BenW. The union of submodules is never a submodule. It probably meant direct sum?

@BenW. also, isn't a pure submodule come attached with a map?

@AlexYoucis Roman's particular definition there is that a submodule $N\leq M$ is pure if $v\in M\setminus N$ implies that $rv\notin N$ for all nonzero scalars $r$. This is specifically over a PID, not a more general scenario. I know that the union of submodules is not a submodule unless one submodule is contained in the other.

Maybe that's what he's getting at? If so, it seems odd. No matter, I just found it weird.

hello

Hello.

whats up

hi i have small question

i have 3 points given for a triangle

how to know clock wise rotating points

i know it has 3 combinations i just need one comnination

@JacobBlack Michael Artin's?

@JacobBlack I bought one few days ago.

@u2425, know what

yeah?

@user58512 i know i need to calculate the determinant

but what should i measure againsty?

@user58512 ???

determinant of 3x3 matrix?

$3\times3$

yeah 3*3 matrix

so you just need to find the matrix?

@JacobBlack Rings can be defined without $1$? Who does that?

why not 2x2?

@JayeshBadwaik Herstein I think does.

no i need to know the formula where i can input the three points and based on the result i should return either they are going in CW or CCW

oh!

you should use the cross-product

@peoplepower ahh, right, he calls them rings with unit elements.

you have 2D points A=(x1,y1), B=(x2,y2), C=(x3,y3)

yeah i use Point class in java to store points. Okay next?

you can form the 3D vectors X=A-B=(x1-x2,y1-y2,0), Y=C-B=(x3-x2,y3-y2,0)

aha

?

those are two sides of the triangle do you see?

now X x Y will either point into the screen or out of the screen depending on CW or CCW

@JayeshBadwaik It is not too outlandish, in my opinion, when one already knows how rings with 1 work.

i.e it will be (0,0,s) where s is positive or negative

@peoplepower yeah, I studied through it. It did not seem to make much difference to the theorems/results for me personally.

I anyway try to prove stuff without using more machinery than is necessary.

@user58512 i got the point what you are saying? so if s is positive then its clockwise? else counter clock wise right?

I don't know

any ways i ll try that then using your formula?

ok

@user58512 I was thinking about the problem of finding a well-founded relation of height $\alpha$. Ideally, one can do this inductively. For instance, if $(\mathbb N,<)$ is well-founded of height $\alpha$, attach an element $x$ to $\mathbb N$ and extend the relation such that $a<x$ for all $a\in\mathbb N$. Then use some bijection $\mathbb N\cong \mathbb N\cup\{x\}$ to get a relation of height $\alpha+1$. The problem is we cannot just take the union of these relations to reach limit ordinals.

that's a nice clean way to do it

there's probably a reason we can't find a uniform construction for a limit well ordering

I think I constructed $\omega^\omega$ well ordering though

but the construction is a bit weird so I am not sure about it

@user58512 I looked at the transcript, and I remember what you wrote; let me think about ti.

I didn't write about it in the chat yet

do you know haskell?

or similar programming language

I have not used it in a while, but yes.

pastebin.com/raw.php?i=UMnQgX3G

I tried sorting [1..100] by some of the well orderings to see what they look like

for $2 \omega$ its like this: 2,4,6,8,10,12,....,1,3,5,9,11,...

and for $\omega^2$ you get a string of infinitely many pentagonal-like numbers.., then shifted.. then shifted.. etc. so infinitely many sections

for higher ordinals the well orderings get a bit crazy, I think you would need to visualize them better than just printing out numbers

@peoplepower, so n_omega is the limit of adding n, omega_squared is the limit of multiplying by n, and omega_omega is the limit of omega^2, omega^3, omega^4, ... - just noticed now that actually the last two are very similar that seems odd

Yes, the cofinalities of all of these ordinals is $\omega$.

I don't understand that

does that mean I got it wrong?

> This definition of cofinality relies on the axiom of choice

oh that's why I don't understand it

I was trying to explain the pattern of them being limits of similar looking sequences.

ok

so as you described, whenever we have a natural well ordering for $\alpha$ we can get one for $\alpha+1$ too

If one has an $\omega$-sequence whose limit is $\alpha$, then the cofinality of $\alpha$ is $\omega=\aleph_0$.

so does that mean any well ordering of the natural numbers has it's rank cofinal $\omega$?

well doesn't mater anyway

It would be a good guess, I would say, since we should be able to break $\mathbb N$ under this ordering into pieces corresponding to a sequence of ordinals whose limit is $\alpha$. At most $\omega$ pieces are possible.

it seems by similar constructions we can take any well ordering of rank $\alpha$ and form well orderings with rank $\alpha+1$, $n \cdot \alpha$, $\alpha^2$, $\alpha^n$, $\alpha^\omega$ (not sure about the last one?)

The nontrivial ones have cofinality equal to $\omega$.

what are trivial ones? like 7?

$\alpha+1$, $n\alpha$,$\alpha^n$.

im really not happy with this notion of cofinality

Assuming we can get it to work in general, they are outliers.

are you assuming that alpha is cofinal omega here too?

No. It is just that $\alpha^\omega$ is the limit of an $\omega$-sequence regardless.

ok

I think the construction of a well ordering with rank $\alpha^\omega$ works (recursively splitting natural numbers into pairs of natural numbers)

in that way you have arbitrarily long strings of natural numbers to compare using the alpha well ordering

in that way we can build a sequence of well orderings of ranks $\omega$, $\omega^\omega$, $(\omega^\omega)^\omega$, ..

that bracketing seems bad.. to get $\epsilon_0$ we actually want $\omega^\alpha$

Yes.

wait a sec

I think this builds a well ordering with rank $\omega^\alpha$ from well orderings for omega and alpha: pastebin.com/raw.php?i=qTjWLt3Q

in other notation $(n,a) \le_{\omega^\alpha} (n,b)$ iff $a \le_{\omega} b$ and $(n,-) \le_{\omega^\alpha} (m,-)$ iff $n \le_{\alpha} m$.

that's "alpha copies of omega"

so now we can build the sequence $\omega$, $\omega^{\omega}$, $\omega^{\omega^{\omega}}$, ...

oh wait.. that's just $\alpha \cdot \omega$ isn't it?

Yes. That's what I was thinking.

this is hard

What's $\omega$?

Ordinal number?

yes

0={},1={0},2={0,1},...,omega={0,1,2,3,4,5,...}

von Neumann's definition, well.

I think we can work in closer analogy to what you did for $\omega^{\omega}$ get it to work for $\omega^{\alpha}$ when $\operatorname{cf}(\alpha)=\omega$, but I am devoid of ideas for general $\alpha$.

Is that useful in major branches of mathematics?

Set theory is major right?

@peoplepower, I am still not sure my $\omega^\omega$ is right

Algebraic geometry, differential geometry, commutative algebra, harmonic analysis, functional analysis, etc.

@user58512 Well, rank $(n,0)=\omega^{n-1}+\omega^{n-2}+\dots$, but that will not be a problem.

the way it works is: $(\mathbb N^2,<_{\omega^\omega})$ we have $(n,a) <_{\omega^\omega} (m,b)$ iff $n <_{\omega} m$ or $n=m$ and $a <_{\omega^n} b$.

That's what I thought.

yeah we can put any ordinal strictly less than $\omega^\omega$ into this, so its good

ok thanks, feel more confident about that one now

how do you actually use the assumption that $\alpha$ has cofinality $\omega$ in the construction of $\omega^\alpha$?

(also I still don't understand that term)

Take a sequence $\alpha_n$ with limit $\alpha$.

ah

I see the problem, nevermind.

They in turn must have cofinality $\omega$. :)

just to check I follow: given $\alpha$ has cofinality omega you can produce a strictly increasing sequence of ordinals $\alpha_n$ ($n \in \mathbb N$) and the limit ordinal of that sequence is $\alpha$?

Yes. So it is weaker to say $\alpha$ has cofinality $\kappa$ than $\alpha$ has cofinality $\omega$.

ok!

it seems weird that there could be an ordinal that has cofinality bigger than $\omega$

so if $\beta < \alpha$, presumably there's some way to express $\beta$ in terms of a (finite?) combination of $\alpha_n$?

No.

is that lack of being able to express it an obstruction to building a well ordering?

I would certainly say so, though I do not have proofs backing me up.

What about the recursive definition of Borel sets?

@FrankScience Any infinite union implicitly uses (the existence of) $\omega$.

@peoplepower, sure I just mean intuitively, I wouldn't expect a proof of such a soft statement

maybe I should try to build $$\omega^{\omega^\omega}$$

I guess we need to build a well ordering that can hold $\omega$, $\omega^2$, $\omega^3$, ..., $\omega^{\omega}$, $\omega^{2 \omega}$, $\omega^{3 \omega}$, ... $\omega^{\omega^2}$, $\omega^{\omega^3}$, $\omega^{\omega^4}$, ...

so I should build $\omega^{\omega^2}$ first

if $(\mathbb N, <_{\omega^\omega})$ has rank $\omega^\omega$, does the lexicographic product order $(\mathbb N^n, <_{\omega^\omega,n})$ have rank $(\omega^\omega)^n \simeq \omega^{n \cdot \omega}$?

No, it would be $n\omega^{\omega}$.

aw :(

thanks

Can anybody help me understand this?

it is easy to show that complementation of sets maps Gm into itself for any limit ordinal m;

@peoplepower, i think i will ask on the site about thsi

@user58512 Sure, it is an interesting question.

@FrankScience I think you can use the de Morgan laws and what they said in the previous sentence.

@peoplepower It seems that no complementary process is made.

@FrankScience I cannot parse.

@peoplepower I mean, $A\in G^\alpha$ so that $A^c\in G^\beta$, etc.

Hello there!

Just got a bicycle accident.

It seems that I have to be slert with the surrounding environent more then...

And it will take about 2 to 4 weeks to recover, so that I can stay with you guys!! :-)

@awllower, get well soon!!

Thanks. :D

@JacobBlack Wow, you have read about 9000 algebra books?!

canon=canconical

@user58512 Have you seen this? This.

wow!

@awllower, actually it is not quite relevant - but I saw ths very interesting question mathoverflow.net/questions/123223/…

Interesting result!

I had considered this before, but didn'tget anywhere

so I hope there will be some good ansewr on MO

So this i srelated to Gauß problem, mh...

@JacobBlack Yo. Me not Ora any more.

the harvest is over

Yeah! the harvest season will be over here in a few days.

And what is the season now?

It is vasant right now. The harvest season.

:)

Harvest moon

I have a question about the site math.stackexchange.com/questions/317081/… he posted an answer and I am very grateful for that but it doesn't help me what should I do

Does it help you in some way?

no :(

Then you should do nothing and wait for more answers!

ok

:D

:)

I should study my number theory but I dont feel like it:S

haha

maybe I should go eat then come back and work

but I dont want to go out......

Hm...

@awllower, I am learning about infinite ordinals

About R?

R? the real numbers? no

well it starts simply

0={}, 1={0}, 2={0,1}, 3={0,1,2}, 4={0,1,2,3}, ..., $\omega$={0,1,2,3,4,...}

I see: It reminds me of the set-theory of Bourbaki.

$\omega+1$={0,1,2,3,4,...,{$\omega$}}, $\omega+2$={0,1,2,3,4,...,{$\omega$},{$\omega+1$}}, ..., $2 \omega$ = {0,1,2,3,4,...,{$\omega$},{$\omega+1$},{$\omega+2$},{$\omega+3$},...}$

Oh

yes, I think it is covered in Bourbaki!

Good!

But I guess I did not read that section carefully...

now we can keep extending $\omega^2$ = {0,1,2...,{$\omega+1$},{$\omega+2$},{$\omega+3$},...,{$2\omega+1$},{$2\omega+2$}‌​,{$2\omega+3$},...}

That is fine: you can continue if you want.

Could you please explain what a set within a set means?

well this way we build up $\omega^2, \omega^3,$... and $\omega^\omega$ & you can keep building bigger ordinals

That set contains this set as an element?

yes, in fact $\alpha < \beta$ iff $\alpha \in \beta$

@user58512 OK

And we call these ordinals the infinite ones?

I'm trying to understand the $\omega^{\omega^{\omega^{\omega^{\ddots}}}}$

yes they are all infinite starting from $\omega$

Hm, and I suppose you already have some ideas?

not very much :D

one idea is that $\in$ is a well founded relation (by nature of set theory). So I should study well founded relation in general.

Oh

I didn't find a well ordered relation for $\omega^{2 \omega}$ yet - so I have not climbed very high

infinities within infinities within infinities...