There is an expository paper by Niven, I think, about formal power series, which covers these sorts of things

@AlexClark Then revise them :)

Formal power series by Nieven at MAA, I'm reading now. Thanks again.

If anything quotient groups are harder, so

Ah yup, did you just find it or you were reading it before? @DavidCardozo

I just find it, reading at the moment.

The quotient ring if you just look at addition is a group quotient

@AlexClark Right, and in the case of quotient rings, you look at cosets, but from the additive point-of-view.

Yep.

yeah

yup

the quotient thing is what you get when you set all of the things in the ideal = 0

It's essentially a quotient group if you forget about the multiplicative structure of $R$ and the multiplication-absorption property of $I$. (high-falutin' way to state that forgetful functor preserves quotient things)

since 0*anything=0, that means any multiple of a thing set equal to 0 is also 0. and since 0+0=0, the sum of two things set equal to 0 is also 0.

the multiples of x^2+1

are you familiar with the division algorithm?

$f(x)(x^2 + 1)$

@AlexClark yeah in R[x]

given a polynomial f(x) and another g(x), there exists a unique quotient q(x) and remainder r(x) such that f(x)=g(x)q(x)+r(x) with deg(r)<deg(g)

You should verify that $R/I$ is a ring, though.

That is what it looks like $r+\langle x^2 + 1\rangle$, if you do some polynomial division, you can show that everything can be "reduced" to something that is linear or constant.

That's where you need the multiplicative structure of $R$ and $I$.

So in a sense there are two coordinates, the "$x$' coordinate and the $c$ coordinate. so we get every polynomial $f(x) \equiv a+bx \mod \langle x^2 +1 \rangle $(sort of reminds you of complex right...)

Yes, since reals are the coefficients of the ring

$f(x)$ does not have to be degree $2$

every polynomial f(x) can be written as f(x)=q(x)(x^2+1)+ax+b for some quotient q(x) and remainder ax+b

but you can use the division algorithm (divide by $x^2+1$) to get that the non zero part is of the form $a+bx$

(the remainder)

Quadratic terms cancel out, yeah.

And now you see why the evaluation map $x \mapsto i$ is an isomorphism :)

Haha, is it all clicking

So this quotient ring, as you see, has an extra structure. It's $\Bbb C$, i.e., a field.

The upshot of this discussion, then, is that $\langle x^2 + 1 \rangle$ is maximal.

I did?

You did @BalarkaSen

That's weird, 'cause it's not.

I don't recall saying it.

whoa, I don't even recall.

what i said is false.

no idea what i had in mind.

It is true in general that ideal generated by any irreducible polynomial $f$ is maximal, 'cause $\Bbb R$ is a field and if you adjoin an algebraic to it, you still get a field.

I still can't figure out what I was thinking about, though. Probably I just messed up, ending up replacing $\Bbb R[x]$ by $\Bbb Z[x]$.

i cant get latex work properly in chat

Yes, you should prove the reverse implication.

@AbdouAbdou see "LaTeX in chat" on the starboard to the right --->

@AlexClark polynomials with coefficients in $k$, yeah

mhm

Recall what maximal means.

proper ideals, but yeah

@anon Has anyone noticed that the starboard is on the right?

And that starboard = right in nautical terms?

I think I grok the universals for quotients (the $A/\hskip -0.06in \sim$ stuff), products and coproducts at an okayish level now.

Alex: Check this out.

The pic on that page reminds me of what this chatroom feels like atm.

Try starring the picture.

Oh, nevermind.

anyone want to help me solve a probability problem?

@Newb Not if it's hard

it's about the sum of a uniform and a bernoulli random variable

Oh, too hard for me at least.

haha

@AlexClark this too

"Topology"

@robjohn later I'll show you in db my solution in the spirit of the art to that question I asked yesterday and you solved.

I wanna add it to my book, that's because it is simply nice (although some might argue that it's not hard enough - since harder often means nicer)

Eventually I also need in my book such questions, I don't wanna publish a collection of almost impossible-to-solve problems. :-)

That problem reminds me of another problem I proposed to myself some months ago and I didn't solve yet. Some more research is needed. (let me see where it is - a very hard job to find it through thousands of questions)

I've found it!

Well, I think I know how to solve it now though ... (BBL)

No, I can't yet.

@AlexClark, you on?

@Chris'ssis Are you writing a book? $$\text{Nearly-Undoable Integrals And Where To Find Them}$$ or something similar?

@SohamChowdhury I write a book on integrals, series and limits.

@Chris'ssis A collection of problems?

@SohamChowdhury yes, major part of them being created by me.

@Chris'ssis That's really cool. Writing books is hard, good luck!

@SohamChowdhury Especially (hard) for a self-educated ... ;)

@Chris'ssis An "artist" is an artist no matter what.

@SohamChowdhury Yes! :-)

Meanwhile, the rest of us are struggling with trivialities like $\int \sqrt{{\tan x}}$

@SohamChowdhury You can use a system of equations in integrals.

As in $I = \text{something} - I$?

And then solve for $I$?

That was just an example. I'm not really doing integrals now. :)

@anon cant make it work

@AbdouAbdou have you put the bookmarklet on your bookmarks bar?

@SohamChowdhury $I=\int \sqrt{{\tan x}} \ dx $ and $J= \int \sqrt{{\cot x}} \ dx$

do u mean favorite ?

@Chris'ssis No hints!

@SohamChowdhury OK :-)

@AbdouAbdou Yes, I think.

dont know how to do this via chrome

ok i ll try more thx

Do you know how to drag and drop things?

yes

locally schlicht? wat

Go to this page: math.ucla.edu/~robjohn/math/mathjax.html

javascripts cant be bookmarked in chrome :(

Oh.

Are you sure?

Use Firefox then. ;)

i must use one of my tricky ways , maybe gonna fuse it somewhere in source code

ok ok i got it

thx

Good.

and aalll appeared by magic

lol

thank u so much

:)

well i have more mind-occupying question if u dont mind

about integer divisions

Interesting. I have one too.

about ?

integer divisions arnt submessive to any laws as rest of operations do

each operand has some calculation rules to follow , except integer divisions ! why ?

division is not an operation on the integers

the mathematical structure (ring) of the integers has defined addition and multiplication, and implicitly subtraction

yes i know , its either flooring or ceiling

im talking about integer-rounded divisions

like ... why $\frac{a}{b}^2$ doesnt equal $\frac{a^2}{b^2}$ in all cases ?

what are cases in which that doesn't hold?

math.stackexchange.com/questions/1293978/… hi guys can you put your thoughts to this.

what is rank in complex analysis in regard to mappings?

Woo. First chapter of Aluffi officially done! :)

i tried to create a relation between $\frac{a}{c}+\frac{b}{c}$ and $\frac{a+b}{c}$ , but failed nearly , are non successful attempts allowed in stackexchange ?

if we knock away downvote part

that was not a counter example

btw you will find that

(a/b)^2 = (a/b)(a/b) = (a^2/b^2)

ninja'd

faster than you ;)

wut about $\frac{7}{2}^2$ != $\frac{7^2}{2^2}$

how are those not equal?

(7/2)^2 = (7/2)(7/2) = 7^2/2^2

the second is 12

wait a moment

no its integer

yeah of course (7/2)^2 != (7^2)/(2^2)

im talking about rounded divisions

oh more of this stuff

bleh

ah , we v been misunderstood since first moment :D

@N3buchadnezzar depends on the map.

Let $\Phi$ be the map of the closed bicylinder into $\mathbb{C}^3$ defined as follows: $\Phi(z,w) = (z,zw,zw^2-w)$. Let $K$ denote the image of $B$ under $\Phi$. We assert:

the definition of "rank" I know comes from linear algebra. if $T$ is some linear transformation between vector spaces, then $\dim \im T$ is usually known to be the rank.

for where to where, @N3buchadnezzar?

$\Phi$ has rank 2 at each point of $B$.

\im nei

use \text

so it has been asked often here @Newb ?

@BalarkaSen The Bicylinder is defined as $B \ \colon \ |z|\leq 1$, $|w| \leq 1$.

it's a subspace of $\Bbb C^2$?

then the rank is probably what i have said above

my internet is fluctuating, sorry.

@SohamChowdhury you probably know more category theory than me now :)

@BalarkaSen Heya

how does |\text{Re} z < 1 | look ?

@BalarkaSen If nothing, go look at the universals of product and coproduct. They're really cool.

@N3buchadnezzar Open unit circle?

I fell asleep and forgot to put all the food back in the refrigerator. Damn.

@BalarkaSen And that will probably still take a while. :)

This place right now:

@SohamChowdhury really?

No.

It's an infinite rectangle

Bounded by $x = 1$. The left side of the plane, basically.

So, essentially: $x\in(-\infty,1), y\in(-\infty, \infty)$

@Hippalectryon

@Chris'ssis o/

@Hippalectryon I was thinking of something someone here said last days, ironically "@Chris'ssis BTW, I didn't know you did advanced analysis. I thought you just liked to do really complicated integrals " - Tobias

What about it ?

@Hippalectryon are you French by any chance?

@SohamChowdhury indeed

@Hippalectryon I was wondering how would those that starred the message (like 12 stars) would feel if they couldn't answer some elementary questions I might prepare anytime (they believing they know advanced analysis)

(Guessing from the fact that you put a space before the question mark)

:P

@Hippalectryon anyway, I don't wanna be bad :-)

@Chris'ssis You underestimate the natural human urge to belong or to feel part of a group. :P

Not all of them felt malicious toward you, I'm sure.

@Chris'ssis I don't think it was meant to be mean in any way (well, I don't know the context in which it was said). I've never seen a used named Tobias around as far as I know, so for someone coming casually to this chatroom, that's probably how you appear :-) someone who likes solving crazy and difficult sums and integrals ;-)

@Hippalectryon It was said ironically. Anyway, it.s OK. :-)

Back to my research.

@Chris'ssis Must've been just a joke, don't let it bother you!

@SohamChowdhury it would be a great joke! ;)

@Chris'ssis It's just like the one that's on the star board one now (Alex's). I starred it because it's kinda true :-) every time I come here you find a new awesome formula

@Hippalectryon That I like. ;)

@Chris'ssis however, in a sort of opposite way, not every film critic is a film director. It's entirely possible to have criticism of someone without being experienced in that field oneself.

But always take (almost) everything lightly. It helps.

:)

@SohamChowdhury I like you're shiny (in the sense of being happy). You look at things positively. That's good! :-)

Shinny?

What does that mean?

shiny*