Mathematics - 2018-10-18 (page 4 of 4)

Ted Shifrin

23:00

@dalbouvet No, $g$ is not defined at $0$. To have a removable discontinuity at $x=a$, the function must have a value at $x=a$. At least, that's the way most mathematicians use the language.

hi Lucas

dalbouvet

@TedShifrin ah yes, you're right

Ted Shifrin

I always have to remind calculus students that the function $f(x)=1/x$ is continuous, despite what their high school teachers may have said.

Kasmir Khaan

Hi Ted 1

@TedShifrin Ted! :D

Ted Shifrin

hi @Kasmir

Kasmir Khaan

yeey :)

I was waiting when you gonna show up

anyway I got a Q!

Ted Shifrin

23:07

I'm leaving soon, though.

Kasmir Khaan

damn it :D

Leaky Nun

hi @KasmirKhaan

Kasmir Khaan

@LeakyNun Hello leaky :D

Ok my Q is

we have a surjevtive ring hom from A to B

Leaky Nun

so, not just a ring epimorphism :P

Kasmir Khaan

the map I ---> f^-1 (I) is a bijection

damn it leaky we are over that now -.-

Leaky Nun

23:09

between what sets

Ted Shifrin

hush, Leaky

Kasmir Khaan

that is the problem

I think the proof of the teacher am following is wrong

Ted Shifrin

what does the letter $I$ signify?

Kasmir Khaan

since that map is not from A---> B

I is an ideal

Ted Shifrin

The teacher is right. YOu're not understanding.

Kasmir Khaan

23:10

I ideal in A, and f^-1( I) are the same thing

Ted Shifrin

No, $I$ is an ideal in $B$.

Kasmir Khaan

in that case it is clear

Ted Shifrin

This is a bijection from the set of ideals in $B$ to the set of ideals in $A$ containing $\ker f$.

Kasmir Khaan

elements of A mapping to elements of B

Ted Shifrin

Kasmir, you're saying garbage.

Kasmir Khaan

23:10

why is that

and too mean ????

Ted Shifrin

Well, you keep talking instead of reading what I've said. So, yes, mean.

Kasmir Khaan

I understand ur point TED

I did that before coming here

BUT , in writing it this way

Ted Shifrin

But you started out by writing $I$ an ideal in $A$ ... in which case $f^{-1}(I)$ makes zero sense.

Kasmir Khaan

the map f : A---> B , I mapped to the preimage f^-1 ( I)

exactly !!

that what I come here to ask , that does make no sense

Ted Shifrin

Reread everything I've typed.

And understand it.

Kasmir Khaan

23:13

I get it Ted i promise

Ted Shifrin

You clearly do NOT get it. Come on.

Kasmir Khaan

ideals in A that contain the kernel

are in bijection with ideals of B

this is no surprise

Ted Shifrin

Well, then why are you typing all this nonsense?

Kasmir Khaan

since ideals maps to ideal

Ted Shifrin

We're doing inverse image, not forward image.

Kasmir Khaan

23:14

am not, in proving this, the teacher messed up the roll of I being in A istead of being in B

Ted Shifrin

No, I is an ideal in B, dammit.

Kasmir Khaan

that what i said Ted

Ted Shifrin

If the teacher wrote $f^{-1}(I)$, the teacher knew that $I\subset B$.

Kasmir Khaan

the way he wrote it , it is in A , and that clearly make no sense AT ALL

Ted Shifrin

Maybe you messed up in your notes.

Where did the teacher say $I\subset A$?

Kasmir Khaan

23:16

nah nah, was online, but if you write it this way, the map f : A---> B , I --> f^-1( I)

Leaky Nun

show us

Kasmir Khaan

it make no sense ofc, since that map does not map A to B

Ted Shifrin

You're not understanding ... The teacher is right.

Kasmir Khaan

so we are allowed to use f going from A to B

and then take elements in B mapping tthem to A ?

Ted Shifrin

Precisely. That's what inverse image means.

Kasmir Khaan

23:17

preimage lives in A

I lives in B

Ted Shifrin

you're mapping sets to sets, not elements to elements.

Kasmir Khaan

why the rolls are reversed ?

Ted Shifrin

You'll see that everywhere when you take topology, too. You talk about continuity in terms of preimages of sets.

Because preimages often behave way better than images.

The image of an ideal under a ring homomorphism is not necessarily an ideal. The preimage of an ideal always is.

Kasmir Khaan

i think you meant subgroup in the first one

but that is not my comfusion here ._.' the thing is , if one wrrite f :A--> B

Ted Shifrin

I meant what I said.

Kasmir Khaan

23:20

one would like to take elements in A mapping to elements in B

that what A---> B is

otherwise the map is not doing what it should

Ted Shifrin

And preimage maps subsets of $B$ to subsets of $A$.

Leaky Nun

show us your online notes.

5 mins ago, by Kasmir Khaan

nah nah, was online, but if you write it this way, the map f : A---> B , I --> f^-1( I)

show us this line

Kasmir Khaan

it is exactly liuke that leaky and it s a university course

cant share it

Ted Shifrin

I bet money it is not like that. There are some words.

Even when you first wrote it, you had words.

If $f\colon A\to B$ is a surjective ring homomorphism, then we get a bijection $I\rightsquigarrow f^{-1}(I)$ between ....

Kasmir Khaan

i would like if we had this , f^-1 : B---> A , I ---> f^-1 ( I)

Nebulae

23:23

Take a screenshot. Don't miss the money.

Ted Shifrin

That's garbage.

$f^{-1}$ is NOT a mapping.

Kasmir Khaan

i know

Ted Shifrin

Unless $f$ is an isomorphism to start with.

Kasmir Khaan

at least as sets

Ted Shifrin

You continue NOT to listen to what I've said multiple times.

Kasmir Khaan

23:23

TED ! I DO I SWEAR

MY COMFUSION IS NTO WHERE U THINK IT IS

Ted Shifrin

No, you have to talk about $f^{-1}:\mathscr P(B)\to\mathscr P(A)$.

Kasmir Khaan

OUPS CAPS

am not using this as fns

Nebulae

lmao!

Kasmir Khaan

as map of sets

f^-1 is just a premiage

Ted Shifrin

That does not give a mapping from set $B$ to set $A$.

Kasmir Khaan

23:24

why is that?

Ted Shifrin

I just told you it maps SUBSETS of $B$ to SUBSETS of $A$.

A function has to associate to each element of its domain a single element of the range.

This is basic, basic, basic stuff.

Kasmir Khaan

am not using that as a function

just wanted to know , the sourse and the target do agree

in that way

we take elements that lives in A to B

howver typing it this way, f : A--->B , then taking I in B mapping to it its preimage in A

seems not something that should be written that way, A---> B

Leaky Nun

then type verbatim what is said online

Kasmir Khaan

B--->A would be more appropriate

LEAKY ! THAT IS EXACTLY WHAT IS SAID GRRRRRRRRRR

Ted Shifrin

I give up, Kasmir. Seriously. You keep repeating the same stuff, totally not understanding what I've tried to say multiple times.

Kasmir Khaan

23:28

Ted !

Ted Shifrin

No, I mean it.

Kasmir Khaan

ask me anything about this

and ill answer to show i understood

ask me what u think i dont get

Leaky Nun

you don't get the fact that there is no way you quoted verbatim what the online course notes say

Nebulae

Quote the exact statement, in their own words?

Kasmir Khaan

am not saying that the notes are wrong just that particular detail

like Ted said

ideals in B are in 1:1 corres with ideals in A that contain ker

but that is not what I dont get

Leaky Nun

23:31

do you know what the word "verbatim" means

Kasmir Khaan

-.-

Leaky Nun

or the phrase "in their own words"

if yes, then do that and resolve this conversation

Kasmir Khaan

i litterly wrote what is in there 5 times now

that is the statement leaky

anyway ._. sorry about this'

Ill continue doing my stuff ._.'

Leaky Nun

quote the whole statement

Kasmir Khaan

explain to me this map

f : I ---> f' (I)

f' is the preimage

if you know that f : A-->B

A,B are rings

the full citation is this,

let f : A--> B be a surjective ring hom

the map f : I---> f'(I) is a bijection

Lucas Henrique

23:37

Back again

So I have a pretty simple question.

Leaky Nun

are you sure they used --> not |-->

Lucas Henrique

How do I (formally) prove that $\forall n \in \Bbb Z_+$ the set $\{r\in \Bbb Z_{\geq 0}: n \geq 2^r\}$ is finite?

Ted Shifrin

@Lucas: Can you show $\{2^r\}$ is unbounded above?

Lucas Henrique

Informally: If it was infinite, then $n$ would also be infinite

@TedShifrin Yeah that's simple

Ted Shifrin

Then $\{r: 2^r\le n\}$ has a least upper bound.

Lucas Henrique

23:41

Why?

Ted Shifrin

Because you just admitted it was bounded above.

Leaky Nun

$\Bbb N := \Bbb Z_{\ge 0}$

2

Ted Shifrin

Not to me, @Leaky.

Leaky Nun

I was confused by the latter for so long

I was like, what? integers?

Lucas Henrique

@LeakyNun Yeah.

Ted Shifrin

23:42

We've fought over that before in here. In Europe, $\Bbb N$ apparently includes $0$. In the US, it doesn't.

Leaky Nun

not talking about that

Lucas Henrique

@TedShifrin where exactly?

Ted Shifrin

When you admitted $\{2^r\}$ is unbounded above :)

Lucas Henrique

OHHH, and $n$ would be an upper bound.

Ted Shifrin

No, we have to be careful about domain and range here.

Leaky Nun

23:44

it's a subset of a finite set {r | 0 <= r <= n}

Nebulae

I thought $\mathbb{Z}_{\ge 0}$ was the same as $\mathbb{N} \cup \left\{0\right\}$?

Kasmir Khaan

@LeakyNun ofc I |---> f'(I)

Ted Shifrin

It is to me, @Nebulae. See what I just typed about US and Europe.

Kasmir Khaan

@TedShifrin how would you state that map ?

Leaky Nun

oh so now it's "ofc"

that's interesting

Kasmir Khaan

23:45

f :A-->B
I|---> f^-1 (I)

Ted Shifrin

NO.

Lucas Henrique

@TedShifrin I'm not really sure of how things relate.

Kasmir Khaan

I know its a NO

but how else would one write it in a correct way

the teacher did say I is an ideal of B

but the way it is written is wrong

Leaky Nun

you see, you told us that what you wrote is an exact quote

Ted Shifrin

$f\colon A\to B$ induces a map from $\mathscr P(B)$ to $\mathscr P(A)$.

Leaky Nun

23:46

and then later you quoted a longer thing

9 mins ago, by Kasmir Khaan

the full citation is this,

9 mins ago, by Kasmir Khaan

let f : A--> B be a surjective ring hom

9 mins ago, by Kasmir Khaan

the map f : I---> f'(I) is a bijection

which I'm sure is still wrong, but at least it's more correct than your first "exact quote"

Kasmir Khaan

that make no sense , i was typing it very fast

f should not be in second statement

Leaky Nun

there you go

Nebulae

@TedShifrin Oh, I see. I thought leaky was implying it means something else entirely. Probably didn't see the $\ge 0$.

Leaky Nun

@TedShifrin so that happened

Lucas Henrique

@Ted is a famous guy

Kasmir Khaan

23:48

That part is not the problem ! and Ted! @TedShifrin if one wrote this

f : A--> B

f' ( I) |---> I

and I is ideal of B

wont this make more sense then the reverse?

f' (I) is in A, I is in B

Leaky Nun

I think nobody will help you if you keep on not giving us the exact quote.

Kasmir Khaan

i did'

Leaky Nun

Yeah, right.

Nebulae

Kasmir, you should be a politician! xD

Kasmir Khaan

let f : A--> B be a surjective ring hom
I|---> f'(I) is a bijection

Leaky Nun

23:50

now that's even less words

Kasmir Khaan

forget i got it from a lecture how else would it be written

in these terms

Leaky Nun

You said it's online.

Kasmir Khaan

okay lets forget abnout it

thanks for all the help

Leaky Nun

You mean none of the help.

Nebulae

The famous Paxman-Michael Howard interview - Newsnight archives (1997)

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