Room for Shine On You Crazy Diamond and Ultradark - 2019-12-10

last day (15 days later) »

Shine On You Crazy Diamond

7:33 PM

@Ultradark

Want to study now?

Or busy atm?

^_^

I could study now but

Shine On You Crazy Diamond

You're a butt, not me!

:D

can we spend a few minutes discussing something?

Shine On You Crazy Diamond

Sure

Shoot

waiting

Dude, you're not typing anything!

I'm a slow typer

Consider the matrix $$A=\begin{pmatrix} e^s&0\\ 0&e^{-s} \end{pmatrix}$$

Shine On You Crazy Diamond

7:36 PM

You need some electronic writing pad maybe lol

K where $s \in \Bbb{C}$?

$s\in \Bbb R$

Shine On You Crazy Diamond

K

What about it

You know that this is a rotation matrix?

Shine On You Crazy Diamond

It's a diagonal matrix

Oh, I see what you're askingt

it can be written in the form of hyperbolic cosines and such

Shine On You Crazy Diamond

7:37 PM

Oh, but not as regular cosines?

no

Shine On You Crazy Diamond

Do you have a reference for writing that in terms of cosh?

because this defines hyperbolic rotation of points in the plane

Shine On You Crazy Diamond

Link?

okay hold on

$$\begin{pmatrix}
\cosh(\phi) & \sinh(\phi) \\
\sinh(\phi) & \cosh(\phi) \end{pmatrix}$$

Shine On You Crazy Diamond

7:40 PM

where $\phi$ = ?

here's the hyperbolic rotation matrix

angle

Shine On You Crazy Diamond

What's the definition of $\cosh$?

Nvm

found it on goog

I see your formula

now what

I see now that it equals that diagonal matrix

okay great

Shine On You Crazy Diamond

When $\phi = 2\pi k$

Let's increase the power $$B=\begin{pmatrix} (e^{e^s})^{-1}&0\\ 0&(e^{e^{-s}})^{-1} \end{pmatrix}$$

$$C=\begin{pmatrix} (e^{e^s}) &0\\ 0&(e^{e^{-s}}) \end{pmatrix}$$

Shine On You Crazy Diamond

7:44 PM

Now what

I mean it's at $\phi = \pi k$

$k \in \Bbb{Z}$

Would you agree that $B$ and $C$

are images of a hyperbolic rotation?

Shine On You Crazy Diamond

I disagree

in other words a mapping of a hyperbolic rotation to some other space

Shine On You Crazy Diamond

Let $R(\theta) = $ your matrix

at $\theta$

I mean matrices compose by multiplying them but you took the output entries of one and plugged them into a paramete

let $e^s \to s$

Shine On You Crazy Diamond

7:47 PM

?

Ready to study CA?

by using this substitution

Shine On You Crazy Diamond

I don't think your conclusion is correct

You didn't multiply any matrices

Matrix(Rotation + Rotation') = Matrix(Rotation) * Matrix(Rotation')

Where Rotation = $\phi$.

hm

Shine On You Crazy Diamond

RotMatrix($\phi$) = a hom from $(\Bbb{R}, +)$ into $(GL_2(\Bbb{R}), \times)$

note the $\times$

That's just row by column dotting or matrix mul

?

Ready

I see where your coming from

yeah I'm just going to ask on the main site just to make sure and understand what's going on

because me and 2 others on the math chat came to the conclusion

Shine On You Crazy Diamond

7:54 PM

Oh cool

I could be wrong

Homework quiz

If $R$ is a commutative ring with $1$ and $I \leqslant R$ is an ideal, what can you tell me about the ideals of $R/I$?

they are also ideals in

R

I believe

so they form an additive group

and absorb into the ring

Shine On You Crazy Diamond

?

$S = R/I$ is a separate but related ring to $R$

SO the ideals are not equal

They're not even comparable

oh okay

Shine On You Crazy Diamond

An ideal in $R$ is a set of elements of $R$

While an ideal in $R/I$ is a set of cosets of $I$, a set of sets of elements of $R$ if you will

You can't naturally compare a set of sets of elements to a set of elements

See the text

what does it say on the issue

I'm working from the last position in the text

Give me one paragraph about ideals of $R/I$

Or a few sentences

:)

you're so fast

I'm trying to keep up

Shine On You Crazy Diamond

8:01 PM

Keep pen and paper handy

okay

Shine On You Crazy Diamond

To sketch proofs

then typing it will be easier

you sent me the new book right?

Shine On You Crazy Diamond

It's by Atiyah-McDonald

You got it ?

yeah

Shine On You Crazy Diamond

8:03 PM

Start the proof out

Let $\phi : A \to A/I$

be the natural surjection which is ?

It takes $x \in A$ to what?

I gotta go the bathroom

Shine On You Crazy Diamond

LOL

k

Log-a-rhythm

you're a regular mathematician

:D

There?

just got back

Shine On You Crazy Diamond

K

What is the natural surjection from $A \to A/I$?

given by?

okay so it looks like that mapping maps its coset to $x +a$

Shine On You Crazy Diamond

8:12 PM

Defined as...

Nope

that's not it

$x \mapsto $?

$x\in A$

Shine On You Crazy Diamond

Yes

$A$ is our ring, I changed to $A$ from $R$

$I \subset A$ is an ideal of $A$

$x \mapsto x + $?

a

Shine On You Crazy Diamond

Where's $a$ defined?

I didn't define $a$

I said $I$ is our ideal

it's in the book

$\mathfrak a$

Shine On You Crazy Diamond

8:15 PM

Then $\mathfrak{a} = I$ here

You have to be comfortable with switching notations

SO $x \mapsto x + I$ under $\phi$

oh yes

Shine On You Crazy Diamond

Prove that $\phi$ is a ring hom

okay so $\phi$ maps our ring $A$ to a quotient ring

and we are quotienting by the ideal $I$

Shine On You Crazy Diamond

Yep, but you haven't seen that it's actually a hom as well as a set map

Yep

homomorphism?

Shine On You Crazy Diamond

8:18 PM

To prove hom

hom?

Shine On You Crazy Diamond

you have to prove $\phi(x + y) = \phi(x) + \phi(y)$ and $\phi(xy) = \phi(x)\phi(y)$ for all $x, y \in A$.

So do it on paper, then present here

I'll do addition

then you do multiplication

$\phi(x + y) = x + y + I $ by defintion $= x + I + y + I$ since $I + I = I$, and that equals $\phi(x) + \phi(y)$. Done

Multiplication will be different but similar

so essentially i have to prove that the map $\phi$ sends a ring to another ring

Shine On You Crazy Diamond

No, that's already proven

You defined the set map already

and the sets happen to be rings

now you have to prove that $\phi$ respects their ring structures

We already know that they are rings

we defined them when we said let $A$ be a comm. ring with $1$ and then we talked about quotienting and saw that $A/I$ is also a ring

BUt we haven't yet proved that there is a hom between the two structures

This will imply a hom $\Bbb{Z} \to \Bbb{Z}/n \Bbb{Z}$ for any $n \in \Bbb{Z}$ and where we took $A = \Bbb{Z}$ here.

a surjective hom in particular

It's surjective because the set map is surjective by definition

$x + I$ covers all cosets

A surjective hom is just a hom that is surjective (as a set map)

In other words its surjective function in the usual way: $\forall y \in $ codomain, there exists $x \in $ domain such that $f(x) = y$.

So prove multiplication is preserved by $\phi$ similar to my proof above

?

okay

I ran into some connection issues

Shine On You Crazy Diamond

8:26 PM

$\phi(xy) = (xy + I) = $?

Some times its easier to work backwards in math to get ideas

$\phi(x) \phi(y) = (x + I) (y + I) = xy + xI + yI + I\cdot I$

Now why does the right hand side equal $xy + I$ actually?

because our ideal $I$

Shine On You Crazy Diamond

Actually, ignore this approach

read this:

8

A: What does "defining multiplication in quotient rings" actually mean?

The multiplication in the quotient ring is not defined by $$ (a+I)(b+I)= \{\,(a+i_{1})(b+i_{2}): (i_{1},i_{2})\in I \times I\,\} $$ but by $$ (a+I)(b+I)=ab+I. $$ This is a definition, nothing else. Why do we define it in this way? Because it does what we want, together with $$ (a+I)+(b+I)=(a+b)+I...

We define $(a + I) (b + I) = ab + I$

otherwise it doesn't work out

we get $xy + xI + yI + I\cdot I = xy + S$ where $S \subset I$

Are you done scanning that answer?

yeah

Shine On You Crazy Diamond

THe only thing you have to show is that $\cdot$ is well-defined

but what talked a little about that last time

it will keep coming up this well-definedness of certain maps

Okay, now take an ideal $\mathfrak{b} \subset A/I$.

Prove that $\phi^{-1}(\mathfrak{b})$ is an ideal of $A$

based solely on the fact that $\phi$ is a ring hom.

Might also require surjectivity

Actually, take $f: A \to B$ to be any ring hom of rings $A, B$, and if necessary take $f$ surjective

Prove that $f^{-1}(I)$ is an ideal of $A$ for any ideal $I \subset B$.

So start simply

You say "let $I \subset B$ be an ideal."

Are you there ?

Let $I\subset B$ be an ideal

Shine On You Crazy Diamond

8:34 PM

Okay

Now

What does $f^{-1}(I)$ equal?

by definition of taking inverse-image of a function

?

$f(I) = \{ x \in A : f(x) \in I\}$

That is the set we want to prove is an ideal

To prove that it is an ideal

You only need to show subgroup, and scalar absorption

To show subgroup, you don't need to show both $+$ inverses and closure under $+$. You combine those two and you only need to show that for all $x, y \in f^{-1}(I)$ we have $x -y \in f^{-1}(I)$.

Showing closure under $x-y$ implies closure under inverses (take $x = 0$)

and closure under $+$

Because $-(-y) = y$

Does that make sense?

So show that $f^{-1}(I) = J$ is closed under $-$.

Are you there?

yeah this is just very advanced for me

last day (15 days later) »

Transcript for

Dec '1910

$Mathematics$ Room for Shine On You Crazy Diamond a…

join this room
about this room

00:00

06:00

12:00

18:00

all times are UTC

$Mathematics$

site design / logo © 2024 Stack Exchange Inc; legal

mobile