Room for Shine On You Crazy Diamond and Ultradark - 2019-12-06

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4:17 AM

Say for all $m<1,$ and $a>1,$ you take all points $(a,b)$ and $(1/b,1/a)$ such that line $y=mx$ contains those points.

And for all $m>1,$ and $a>1,$ you take all points $(b,a)$ and $(1/a,1/b)$ such that line $y=mx$ contains those points.

For $m=1$ you take all points $c>0,$ $(c,c)$ and $(1/c,1/c)$ such that the line $y=mx$ contains that point.

Shine On You Crazy Diamond

Not sure I understand

which part?

oh

Shine On You Crazy Diamond

So let $X = $ I need set builder notation pls

okay

Shine On You Crazy Diamond

Please present in standard formats :D

4:20 AM

this is the best I can present it as tho

and all the points taken together trace out a curve

Shine On You Crazy Diamond

Okay

let me help there

thank you

Shine On You Crazy Diamond

So

When you say all points $(a,b),(1/b,1/a)$ such that $y = mx$ I'm thinking there's another way to say that

$y = mx$ contains those two points only if: b = ma and 1/a = m 1/b, but those two are equivalent

Thus you're stating a redundancy which is why intuitively it seemed wrong to me

Did you mean 1/b = m 1/a ?

Are you there?

Um

Shine On You Crazy Diamond

only if == then

4:25 AM

yeah I'm trying to process if that's equivalent to what i said

Shine On You Crazy Diamond

Under $m \lt 1, a \gt 1$ if $y = mx$ contains $(a,b)$ then $b = ma$ by definition.

okay I guess you're right

Shine On You Crazy Diamond

So you get $1/a = 1/b m$ or that $y=mx$ contains $(1/b, 1/a)$

automatically

okay

Shine On You Crazy Diamond

You shouldn't state redundancies in math, they'll be called red herrings

:)

4:27 AM

okay

Shine On You Crazy Diamond

It was an accident

Want to study on some algebraics?

Yeah it was a rough draft definition

Shine On You Crazy Diamond

What's a commutative ring?

Commutative algebra is where its at, bro

isn't it just a ring whose multiplication operation is commutative

Shine On You Crazy Diamond

Yep

Give an example

4:30 AM

$\Bbb Z$

Shine On You Crazy Diamond

Yep

sent you some books

Want to search around before we settle on a book or is Atiyah-McDonald okay?

For CA

it's pretty pop

yeah Atiyah-McDonald is good

but I really wanted to try to figure out that question I posed

I'm having trouble putting it down

Shine On You Crazy Diamond

That means you should give it a rest, and come back to it later

Write down everything that you need or ideas you had

otherwhise you'll just get mind loops

for days

okay

Shine On You Crazy Diamond

app.ziteboard.com/team/f1bc2375-c5f6-465a-a1f1-e16947e66fc5

That's the first page

How you say the red underlined part

is you say:

Let $R$ be a commutative ring with $1$.

All rings in this book will be such

Without looking at the book

what's the definition of ring homomorphism

4:38 AM

a mapping that preserves a ring structure

Shine On You Crazy Diamond

yep

gj

thanks now im looking at the specifics

Shine On You Crazy Diamond

K, so you understand that since $f(x + y) = f(x) + f(y)$ that it also preserves $x - y$ as well just basic proofs using group theory

$f(x - y) = f(x) - f(y)$

okay

Shine On You Crazy Diamond

Since rings are additive groups we have that $f(-x) = -f(x)$ prove that for me

whenever $f$ is a function that happens to also be a ring hom.

I guess it's possible to have group homs between rings that ignore the algebraic structure

like a forgetful functor in category theory

*ignore the mult. structure

$f(x - x) = f(0) = 0$ (by definition of ring hom $f(0) = 0, f(1) = 1$)

So $f(x + (-x)) = f(x - x) = 0 = f(x) + f(-x)$

$f(-x) = -f(x)$.

$\square$

That's one way to prove that

Are you there?

4:44 AM

yeah

Shine On You Crazy Diamond

What is an ideal of a commutative ring

And using $\Bbb{Z}$ and $(4) = 4\Bbb{Z}$ as an example

Enumerate all elements of $\Bbb{Z}/(4)$ = the ring of integer modulo $4$.

It's a subset of the ring whose elements absorb into the ring and whose elements themselves form a ring

Shine On You Crazy Diamond

Yep

It's enough that the elements form a group

but every Ideal is actually a subring, nice catch!

haha :)

Shine On You Crazy Diamond

Now

State the 1st isomorphism theorem for groups

It's actually known as "the first"

for structure X

monoid, group, ring, module = X

4:49 AM

every group is isomorphic to a permutation group

Shine On You Crazy Diamond

That's not it, but is probably true

I've heard that before

From AA

Say you have a sequence of groups $G \xrightarrow{f} H \xrightarrow{z} 0$

And group homomorphisms

We say that the sequence is exact at $H$ if $\ker z = \text{im} f$.

Here, what are $\ker z$ and $\text{im} f$ (their general definition?)

$\ker$ stands for kernel

but I'

not really sure what it means

Shine On You Crazy Diamond

Kernel of any algebraic structure is usually the pre-image of identity for that structure

okay so the kernel of a group is the pre-image of the identity for that group

Shine On You Crazy Diamond

In this case $\ker f = \{ x \in G : f(x) = 0 \}$ if $H, G$ are both treated in additive notation (which would make $0$ the identity).

What does the $\ker f$ kernel form?

?

Be quick about this

one

?

4:57 AM

not sure

Shine On You Crazy Diamond

I will show you what I mean if you don't know the answer

It forms a subgroup of $G$ but not just any type of subgroup, a normal subgroup.

What's the definition of a normal subgroup?

It's just $a + N = N + a \ \forall a \in G$.

And $N$ is a subgroup of $G$.

Without normality, it's not true in general that $(a + G) + (b + G)$ can be well-defined

as $(a + b) + G$.

What is a group with no proper, non-trivial, normal subgroup?

Why don't we have to worry about normality in the ring setting?

When we transfer language over from normal subgroup of $G$ to ideal $I \subset R$?

an irreducible group

Shine On You Crazy Diamond

No, it's just called a "simple group".

So, if I take any subgroup $I \subset R$ that is an additive subgroup, then isn't it already a normal subgroup?

Since $R$ is commutative, we have $a + I = \{ a + b : b \in I \} = \{ b + a : b \in I \} = I + a$.

Therefore all ideals of a ring are already normal subgroups by definition

and you can quotient by them to obtain a ring structure in the natural way

A lot of times this makes a finite ring out of an infinite one

okay

Shine On You Crazy Diamond

You write $R/I = \{ a + I : a \in R\}$

it's the set of cosets

for $\Bbb{Z}/(4)$ that's

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

That set forms a ring

with well-defined multiplication and addition given

by multiplying and adding respectively the residues

the residues modulo $(4)$ are just $1, 2, 3, 0$.

You could also choose $6$ instead of $2$

so $1, 6, 3, 0$ is a complete set of residudes

why is that?

Also, before we proceed, pick a quotient ring operation either $+$ or $\cdot$ and prove that it is well-defined

Why that proof is necessary, explain

5:08 AM

$+$

Shine On You Crazy Diamond

sure

To prove well-defined ness w.r.t. quotienting

You need to prove that if $a + I = a' + I$ that either one will result in the same coset $(a + b) + I = (a' + b) + I$ for all $b \in R$.

Prove it

It should only take up to three sentences

Show me the money

Still waiting

Start typing it out

O___O

$a+b$ is in the ring

Shine On You Crazy Diamond

Yes

if $a, b$ are

but $a + I = $ a set

= just what it reads: elementwise addition: = $\{ a + i : i \in I\}$.

What happens when you do $(a + I) + (b + I)$?

What then?

you're adding the set of cosets to another set of cosets

Shine On You Crazy Diamond

yes, so prove that it equals $a + b + I$.

Can you?

If $x \in I$, what is $x + I$?

5:18 AM

$I4

$I$

Shine On You Crazy Diamond

Why?

Why does $x + I = I$ whenever $x \in I$?

?

it doesn't matter how many times you have the same element in a set

you only count it once

Shine On You Crazy Diamond

It's because this:

$x \in I \implies b - x \in I$ for all $b \in I$.

Thus $I = \{ x + (b - x) : b \in I \} \subset x + I$.

and $x + I \subset I$ since $I$ is a group!

and is closed under $+$

Thus you have $I \subset x + I$ and $x + I \subset I$ or they have to be equal

Now given that $I + I = \bigcup_{x \in I} (x + I)$

What is $I + I$?

$I$

Shine On You Crazy Diamond

Correctomundo!

Thus $a + I + b + I = a + b + I$!

See that now?

5:24 AM

yeah

Shine On You Crazy Diamond

It is therefore well defined

since if $a' + I = a + I'$

then $a' - a + I = I$ or in other words $a' - a = a' - a + 0 \in I$

Conversely if $a' - a \in I$ then $a' + I = a + I$

okay that was a good hour of studying

Shine On You Crazy Diamond

Thus we can say that $a' + b + I = a + b + I \iff $ subtract $b$ from both sides

$a' + I = a + I$

Thus $+$ is well-defined

Cool

Yes, thanks for learning

We're going to get you up to speed so that we can be on the same page of CA book

okay

Shine On You Crazy Diamond

It's not much further before we get to some higher math stuff

5:28 AM

is tomorrow morning good?

Shine On You Crazy Diamond

Like $\text{Spec}(R)$

Surce

*sure

okay probably around noon, but ill ping you

Shine On You Crazy Diamond

sweet

tty then

okay bye

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