groups-rings-fields

[coral shale]

practice

but it shouldnt take you 5 mins

[balmy belfry]

Isn’t it sufficient to just say the order of 5 is 12

[coral shale]

yes

[balmy belfry]

So obv <5> has 12 elements

[coral shale]

but what i was worried about is you were blindly doing this

[balmy belfry]

I’m not I know the proof of that

[coral shale]

cus to me, this is a bit of intuition thats important

to see why its 12

[balmy belfry]

Wait a second

So 5 is a generator of Zn

[coral shale]

n = 12

[balmy belfry]

Yes

n is 12 so far

[coral shale]

ok let me ask you this

since 5 in <2, 3>

is 1 in <2, 3> ?

[balmy belfry]

Yuh

[coral shale]

why

[balmy belfry]

2^-1 * 3

= 1

[coral shale]

ok yes thats one way

[coral shale]

but

[balmy belfry]

See that’s what confuse me

[coral shale]

using this fact

[balmy belfry]

I was confused by the multiply notation

[coral shale]

well dont use it

no one does

[balmy belfry]

I forgot it was really adding

[coral shale]

use additive notation

[balmy belfry]

Ok

[coral shale]

its 3 - 2

anyways, tho

1 . 2 + 1 . 3 = 5 yes?

[balmy belfry]

Oh shit

Wait

[coral shale]

how can we use this to write 1 in terms of 2 and 3

[balmy belfry]

So

If we can write 1 in terms of it

Wait but

Ok

[coral shale]

look at this

[balmy belfry]

Umm

Wait we just did

[coral shale]

5, 5+5, 5+5+5, ...

[balmy belfry]

5,10,3

[coral shale]

wait ur not like doing

30 mod 12

35 mod 12
...

are you?

[balmy belfry]

I’m confused

Wait lemme try to write it

[coral shale]

you can just add 5 to each of the last ones

and mod 12 the result

[balmy belfry]

That’s what it signaling here right

[balmy belfry]

Ik

[coral shale]

yh ok

[balmy belfry]

5,10,3,8,1,6,11,4,9,2,7,0

Ummm

Fuck wait

[coral shale]

8 + 5

[balmy belfry]

13

[coral shale]

is not 0

ok

[balmy belfry]

I hit a repeat

[coral shale]

9+5

[balmy belfry]

= 14

[coral shale]

its ite bro, arithmetic sucks

[balmy belfry]

Lol

I’m on my phone

I think that’s 12

[coral shale]

2+5

and then ur done

[balmy belfry]

Ok

[coral shale]

oh and 7+5

u didnt write it anywhere

anyways

[coral shale]

so uve written

[balmy belfry]

0

[coral shale]

0.5, 1.5, 2.5, ..., n.5, ...

right

[balmy belfry]

Yes

Basically

[coral shale]

1 . 2 + 1 . 3 = 5

use this

to now write 1 in terms of 2 and 3

[balmy belfry]

3-2

[coral shale]

no

[balmy belfry]

Wdym write it in terms of 3 n 2

[coral shale]

using the work uve done b4

[coral shale]

1 = ? * 5

[balmy belfry]

5

[coral shale]

1 = 5.5

yes

[balmy belfry]

(1.3+1.2).5

[coral shale]

yes

1 = 5.(1.3+1.2) = 5.3+5.2

thats what i was getting at

[balmy belfry]

Ok

[coral shale]

do you see how this generalises

[balmy belfry]

Lemme see

[coral shale]

ive had you show that once youve found a generator, then what

[balmy belfry]

<a,b>, 1=(a+b)a+(a+b)b?

[coral shale]

(this, by the way is unrelated to the original problem, but is an important intuition)

[coral shale]

no no

lemme write

[balmy belfry]

Ok

[coral shale]

1. Find a (single) generator g of G. Then <g> = G
2. Show <1> is also a generator

[balmy belfry]

Ofc 1 is a generator

1^g = g

[coral shale]

i wrote that bad

ahhh how do i say this

its like

<g> = {0g, 1g, 2g, 3g, ...}

then finding g as a linear combination

g = ax + by

gives u

[balmy belfry]

Lol

I was gonna say

Uhhh that’s not what I remember a Lin comb being

[coral shale]

<g> = {0(ax + by), 1(ax + by), 2(ax + by), 3(ax + by), ...} = {0x + 0y, ax + by, 2ax + 2by, 3ax + 3by, ...}

thats the general idea i wanted to show

[balmy belfry]

Then does it end up making x and y generators

Nvm obviously 😭

[coral shale]

ok heres better phrasing

once youve generated an element

everything that element generates

is also generated

that better?

[balmy belfry]

Is also generated?

[coral shale]

You wanna compute <S>. You've found a in <S>. Then <a> subset <S>

or

[balmy belfry]

Oh

[coral shale]

<S> = <{a} U S>

so back to our example

[balmy belfry]

Yeah I tried to use this but it didn’t work

[coral shale]

as soon as you found 5

and knew 5 generates Z_12

you knew <S> = Z_12

[balmy belfry]

Yes

[coral shale]

ite

[balmy belfry]

Ye

[coral shale]

ok, next S to choose

S = {4, 6}

[balmy belfry]

Wait

So if I have a generator, it’s factors are also generators right

Baisicsllt what you said

[coral shale]

i wouldnt call them factors

but if a^k is a generator, then so is a

[balmy belfry]

<7> = <5,2> right

[coral shale]

<7> subset <5, 2>

[balmy belfry]

Oh

[coral shale]

the other inclusion can be shown by saying 7 is a generator, sure.

[coral shale]

try this

compute what <4, 6> is

[balmy belfry]

Hm

0, 10, uh

[coral shale]

whats another way to write 10

[balmy belfry]

4+6

[coral shale]

-2

10 = -2

[balmy belfry]

Huh

Ok

Cuz equiv clas

[coral shale]

<2> = <-2> subset <4, 6>

[balmy belfry]

Why does this help tho

[delicate orchid]

I feel like we're prancing around the flowers here

[coral shale]

i think amukh needs to get his hands dirty with examples

to get intuition for this shit

[balmy belfry]

[delicate orchid]

amukh do you know bezout's lemma

[coral shale]

but yes, we're not doing the original problem, i agree

[balmy belfry]

No

Definitely not if it’s a lemma im@joke

Idk it tho

Is it number theory

[coral shale]

yes.

[delicate orchid]

barely number theory

[coral shale]

this should help you compute what <4, 6> is

[balmy belfry]

<2>

?

[coral shale]

why

[balmy belfry]

Cuz it’s a subset

[coral shale]

[balmy belfry]

If a^k is a generator a is

U said so

[coral shale]

no, ur misthinking

[balmy belfry]

lol

[coral shale]

and its not i say so you say so, you have to be convinced

of a fact before you use it

[balmy belfry]

Ok

[delicate orchid]

can I just

[balmy belfry]

I convinced of this it pretty obvious

[delicate orchid]

show that if we take H = <a, b, c, ...> show that H = <gcd(a, b, c, ...)>

[coral shale]

[balmy belfry]

Oh fUCK

[delicate orchid]

I'm falling asleep here

[coral shale]

ok after ive finished this

,,\langle2\rangle\subseteq\langle4, 6\rangle

[balmy belfry]

Wait I’m confused

How is 5 gcd 3,2?

Or is it 1

Lol

Fuc

[balmy belfry]

If it is a factor is it a subset

[coral shale]

• why is this inclusion true (how did i conclude it above)
• show the other inclusion holds, hence deduce equality
• prove what wew just stated: show that if we take H = <a, b, c, ...> show that H = <gcd(a, b, c, ...)>

[balmy belfry]

Ok

Hm

Well this is obvious

2^2 = 4

2^3 = 6

So 2 generates waht 4,6 does

[coral shale]

but yeah, #elementary-number-theory is a helpful prerequisite for this stuff. You would have seen euclid's alg, gcd stuff, which is relevant here

[balmy belfry]

#1 finished?

[coral shale]

ok

uh

thats not how i did it above but it works

[balmy belfry]

I did I just forgot division algorithm cuz it’s a theorem not an algorithm

[coral shale]

I did

2 = -10 = -(4 + 6) = -1 . 4 + -1 . 6
<2> subset <4, 6>

[balmy belfry]

Yeah

I wanted to introduce my own too, cuz intuition

[coral shale]

ahhh ok i missed u

i think i misthought earlier

[coral shale]

which inclusion is this

[balmy belfry]

2 subset 4 and 6

Now to prove the other way

[coral shale]

uhhhhh

[balmy belfry]

(6-4)^n

2^n

Ok

[coral shale]

no amukh its the other way surely

[balmy belfry]

No

2 generates stuff that is in <4,6>

[coral shale]

i might have said sht the other way round earlier

[balmy belfry]

I’m showing stuff in <2> is in <4,6>

[balmy belfry]

XD

[coral shale]

<a^k> subset <a> ?

[balmy belfry]

Yeah

I knew what u meant tho

[chilly ocean]

Isn't (2)=(4,6)

[coral shale]

ok, but thats the reverse inclusion here

[coral shale]

yes, we're showing this

[balmy belfry]

If I tried to prove <a> subset <a^k> then I would have just proved the converse by accident

[balmy belfry]

I did both

[coral shale]

no what, im confused

how do u get this

,,\langle2\rangle\subseteq\langle4, 6\rangle

[delicate orchid]

up here shuri

[balmy belfry]

2^2 = 4, 2^3 = 6, thus all the elements of <4,6> is in <2> (part 2)

And part one

(6^1-4^1)^n = 2^n

So

[coral shale]

good ok

convinced 👍

[balmy belfry]

Yo@im actually getting this fr

[coral shale]

even if your notation is dodge

[balmy belfry]

Shuri actually best teacher motivator

[balmy belfry]

Lmfao

[coral shale]

ok to show what wew stated is harder than this obvs

take your time

or maybe it is easy, i havent done it

[balmy belfry]

Oh I forgot about that

Yeah I think it is

Ljust generalize it

Okay

<a_1,a_2,…,a_n> = <gcd(a’s)>

Soooo

Let’s say k = gcd

[coral shale]

d is convention but ok

d for divisor

[balmy belfry]

There is a b_j for every j s.t. d^(b_j) = a_j

That’s like

The definition of divisor

[coral shale]

write this additively

i mean like

normally

[balmy belfry]

d is better ur right

[coral shale]

oh actually nvm

[balmy belfry]

(b_j)d = a_j

[coral shale]

continue

[balmy belfry]

Ok

[coral shale]

use standard notation is fine this is a general proof

for all cyclic groups

[balmy belfry]

Ye

So now

I have written all a_j in terms of d

So now I’ll actually prove the first half

if a is in <a’s>, then a=(a_1)^(c_1)…(a_n)^(c_n) for some c’s

But then it equals

(d^(b_1))^(c_1)…(d^(b_n))^(c_n)

Right

[coral shale]

ok one second. You are using a fact here

that you havent stated

[balmy belfry]

That’s the definition

[coral shale]

why is a always in that form

not for a general group

[balmy belfry]

Oh

Cyclic groups is abelian

[coral shale]

kk

[balmy belfry]

So anyway

[coral shale]

ok

[balmy belfry]

Simply the exponents gives something like

d^(b_1c_1 + … + b_nc_n)

[coral shale]

yh

[balmy belfry]

So it’s in <d>

[coral shale]

yh

[balmy belfry]

First derection proved, <a’s> \subseteq <d>

[delicate orchid]

60000 words later...

[coral shale]

ok

[balmy belfry]

What’s the latex for subsetEq

[coral shale]

its good to be

thorough when beginning out

[balmy belfry]

Lol

[coral shale]

\subseteq

[balmy belfry]

Oh damn

[coral shale]

anyways, id want all that written out in a first hwk yh

[balmy belfry]

I guess I still got latex in my blood after all these years

[balmy belfry]

What’s hwk

[coral shale]

homework

[balmy belfry]

Oh

[coral shale]

other inclusion then we done

[balmy belfry]

Ok Anywyas lemme prove the other way

Hmm I forgot what I did lemme remember

Ok

[coral shale]

uhhh so for this other conclusion

dont you need bezout @delicate orchid

[balmy belfry]

What’s Bezout say

[coral shale]

gcd is always a linear combination

[delicate orchid]

other direction?

is this <d> \subseteq <S>

[coral shale]

yh

[balmy belfry]

Yh

[delicate orchid]

yeah this is equivalent to bezouts

[balmy belfry]

Bezouts sounds hard

[coral shale]

gcd(S) in span(S)

ig

[balmy belfry]

Huh

[delicate orchid]

well, bezout's is usually stated for just two numbers but bleh just induct

[coral shale]

the gcd of some numbers is always a linear combination of them

[balmy belfry]

That’s stupid

Like what if I say

Do all 0s except the last number

The last coefficient can be the quotient ?

[coral shale]

forall x, y : exist a, b : gcd(x, y) = ax + by

[balmy belfry]

😭

[coral spindle]

an INTEGER linear combination of them

[balmy belfry]

Ah

[coral shale]

within an appropriate uh

ring

ig

[balmy belfry]

Now that makes sense

[coral spindle]

Believe it or not, mathematicians are not stupid(!)

[balmy belfry]

LOL

[coral shale]

euclidean domain ig

[balmy belfry]

But what about the one who renamed Pythagorean theorem to distance formula for HS geometry

[coral shale]

forall x, y in Z : exist a, b in Z : gcd(x, y) = ax + by

I hope 0 doesnt break this.

but this generalises to more rings

[balmy belfry]

Idk rings Yeye

[coral shale]

and induction can give you gcd of more numbers

[balmy belfry]

Yet*

I purged them from my mind

Like last week

[coral shale]

Z is a ring, thats all that matters

[balmy belfry]

Is the quotient of a kernel always a ring right

[coral shale]

no

what

kek

[balmy belfry]

Ok

Kernels are ideals right

[coral shale]

bro forget ring theory

and show this 2nd inclusion

[balmy belfry]

I never learned it right 😭😭

[balmy belfry]

Ummmmm

Isn’t it just that dudes theorem

That’s it

XD

[coral shale]

well yes i mean it is just it

[balmy belfry]

Yuh

I’ll formal it tho

Ok

[coral shale]

[balmy belfry]

If a is in <d>

Then a is of the from d^k for some k

Cuz bezouts theorem d=lin comb

[coral shale]

ok

[balmy belfry]

So

[coral shale]

yes then distribute the k

wooo we're done

[balmy belfry]

Yuh

I’m so happy

[coral shale]

so this proof

can be written bidirectionally

the last step was reversible with bezout

[balmy belfry]

What’s that mean

[coral shale]

So you do both inclusions at ince

[balmy belfry]

That’s sounds hard

Oh

Wait literally

[coral shale]

x in <a> <=> ... <=> x in <d>

[balmy belfry]

So bezout is an iff only iff?

[coral shale]

no... bezout

was the reverse direction

of one of your implications

[balmy belfry]

Ok

[coral shale]

in the first proof

the last implication

[balmy belfry]

Is reverse be out

Bezout

???!

[coral shale]

wat

[stark helm]

If there is an element a in an integral domain R, can we directly construct an ideal <a> in R?

[balmy belfry]

Idk

[coral shale]

[balmy belfry]

What’s an integral domain

Is it analysis

[coral shale]

bro this channel is algebra

[balmy belfry]

Yh cuz it had ring

[stark helm]

It's about ring theory

[balmy belfry]

Ok

So it has to do with integers

[stark helm]

isn't it group-ring channel?

[balmy belfry]

Not integrals

[coral shale]

i feel like i should be able to answer this

[balmy belfry]

Yh

[coral shale]

but uh

i cant get your question ngl

[balmy belfry]

Ur in the right spot

[coral shale]

You mean the smallest ideal containing a right

[balmy belfry]

Just wait for wew ig

[coral shale]

by <a>

[balmy belfry]

Math in the wild fr

[coral shale]

{ra : r in R} is it not this or am i high

[delicate orchid]

why does it need to be in integral domain?

[delicate orchid]

yeah, it is

exercise show that (a) is aR

[stark helm]

suppose there is an element a in integral domain R, i want to know if it is valid to construct an ideal <a> directly

[delicate orchid]

and I'm asking you why on earth it wouldn't be

or what's made you think it isn't?

[coral shale]

What do you mean by "construct"

[stark helm]

No, I think it is, but not very sure about it

[coral shale]

you mean state the object <a> <== this is an ideal?

[delicate orchid]

it's just this, like by definition

[coral shale]

prove <a> is always an ideal

[balmy belfry]

Ok gn guys I gtg to sleep

[stark helm]

suppose there must be <a> in R

[coral shale]

cya

i think if u show <a> is indeed an ideal, then youre good to go

with the definition
<a> = {ra : r in R}
presumably

there is another defn though

<a> = intersection of all ideals in R containing a

[delicate orchid]

yur

[stark helm]

it seems it must be ideal, since it absorbs all elements in R

[delicate orchid]

make sure to show it's an additive subgroup

[coral shale]

well i mean, show it is via the definition of an ideal directly

[delicate orchid]

and this is then an ideal because intersections of ideals are ideals

[coral shale]

So in a ring with no unity.
All ideals are subrings?
Hence it would work to define <a> as the intersection of all subrings containing a

(which you can prove is always an ideal)

which feels nicer idk

[delicate orchid]

I simply must have my quotient objects coincide with my subobjects or I will cry and stamp my feet

*kernel

not quotient

[coral shale]

then we can scrap "ideal"

and call them normal subring

...

[delicate orchid]

unfortunately sweaty rings have to have a unit because I said so

[coral shale]

wait is this actually true

<S> is always an ideal wtf

ugh "fun" exercise ig...

yh it should be

[delicate orchid]

the intersection of all subrings containing S contai- ok you got it

[coral shale]

intersection of all subrings = intersection of all ideals

but thats like whack

im thinking whack because it doesnt work for groups ig

but it does for rngs, vector spaces...

[rotund aurora]

Is this true? $K[X]\otimes_K L\cong (K\otimes_K L)[X]$ where $L/K$ is a field extension?

[cloud walrus]

Croqueta

[coral shale]

what kindof product is that

L as a vector space over K?

[rotund aurora]

tensor product

https://en.wikipedia.org/wiki/Tensor_product_of_algebras

[rotund aurora]

is a K-algebra

[coral shale]

i think i could try get it but yh sry 😅

[rotund aurora]

why sry?

[coral shale]

nothing

[rotund aurora]

🤔

[coral shale]

thought galois stuff id seen but no i havent seen it before

[rotund aurora]

What I was thinking was, $\left(\sum_{i=0}^n k_ix^i\right)\otimes_K\ell=\sum_{i=0}^n k_i(x^i\otimes\ell)$ so that you could send this to $\sum_{i=0}^n (k_i\otimes\ell) x^i$, maybe?

[cloud walrus]

Croqueta

[slim kayak]

isnt that just an extension by scalars? I've seen $A[X] \otimes_A B \cong B[X]$ before

[rotund aurora]

oh well yeah, it would collapse to that

because K otimes_K L=L

[coral shale]

im thinking here, is
K[X] = K tensor {0, 1}[X]

in some way or not at all

{0, 1} is uh F2 ig

[cloud walrus]

Kerr

[rotund aurora]

ig you can just send $x^i\otimes\ell\mapsto \ell x^i$

[cloud walrus]

Croqueta

[slim kayak]

or do it more stylishly by checking the universal property

[rotund aurora]

what is it lol. These are rings so its a bit confusing to me

[slim kayak]

any bilinear map factoring through it

so like the vector space case, just weakening it for rings

[south patrol]

if this is't true then nothing is good in the world

Anyway yes you can check that both have the same universal property

por ejemplo

I assume you want an iso of L-algebras

but yes an L-linear map out of K[x] (x)_k L into M is just a K-linear map K[x] -> M which is just a point of M

the same is true for L[x]

i am too category brained

[rotund aurora]

I have something slightly spicier. $\frac{K[X]}{P(X)}\otimes_K L\cong L[X]/P(X)$ where $L/K$ a field extension. This is true because $\frac{K[X]}{P(X)}\otimes_K L\cong \frac{K[X]\otimes_K L}{P(X)\otimes_K 1}\cong\frac{L[X]}{P(X)}$

[cloud walrus]

Croqueta

[rotund aurora]

this is correct, right

[south patrol]

Sure

[delicate orchid]

yes - although K (x)_K L should just be L lol

oh whoops I was scrolled up

[balmy belfry]

Prove there is phi(n) elements of order n in a cyclic group finite

Shuri is this possible for me

@shrui

@coral shale

[slim kayak]

is the group by any chance of order n

[balmy belfry]

Idk, ig not why

[rotund aurora]

maybe, maybe not

[balmy belfry]

Wait lemme change this

Group order = n, phi(k) is number of elements with order k

Ok

Soo

I assume k needs to divide n

For it to work or nah

[coral shale]

thats lit elementary number theory

💀

[balmy belfry]

Bro I ain’t ever take number theory!

[slim kayak]

once you cleared up the problem statement go through the definition

[coral shale]

wait i got pinged here

oops kek

[balmy belfry]

And IT HAS THE WORD GROUP IN IT?

[rotund aurora]

https://en.wikipedia.org/wiki/Carmichael's_totient_function_conjecture

[balmy belfry]

It’s call the toitent function?

[coral shale]

so isnt the actual statement. |Z_n^x| = phi(n)

[balmy belfry]

The definition is number of elements such that they are relatively prime to n

[slim kayak]

you are looking at this one https://en.wikipedia.org/wiki/Euler's_totient_function

[balmy belfry]

I don’t even know

No..?

[coral shale]

you can skip this regardless

[balmy belfry]

No it’s a theorem

My book say

[slim kayak]

isnt it in english anyways

[balmy belfry]

That the number of elements of order k in a cyclic group@is phi k

And I gotta prove it

How is it number theory tho

[coral shale]

oh i guess i misinterpreted

yeah that is true

i think

[balmy belfry]

Yeh

[coral shale]

anyways, you can skip this problem, trust me on this

[balmy belfry]

phi k is number of elements relatively prime

[coral shale]

and come back later

there are more important ones

[balmy belfry]

No my book might use it

I need to just get thru this one it’s the only one I’m stuck on

[coral shale]

you can come back when it uses it

[slim kayak]

have you heard of the bezout identity amukh

[coral shale]

but honestly, i dont think its important

[balmy belfry]

Nope

[slim kayak]

then maybe skip it for now

[coral shale]

we lit just talked about bezout

[balmy belfry]

I didn’t know it was the same one

[slim kayak]

with amukh?

[balmy belfry]

Bezout identity Bezout theorem

[coral shale]

its the same one right

[balmy belfry]

Ok

[coral shale]

the lemma

[balmy belfry]

Yuh

[slim kayak]

well?

[balmy belfry]

Ok then Ik it very well

[coral shale]

yes we just stated it

no he doesnt know how to prove it

[balmy belfry]

Ok whatever

It’s group theory not number theory!

[coral shale]

this study of cyclic groups may as well be

number theory

[slim kayak]

just call the elements in Z/nZ functions for now, should keep any number theorist away from you for a little bit

[balmy belfry]

Yeah so anyway

What does relatively prime have to do with other

Order

[coral shale]

just try it with examples and see

[balmy belfry]

Fuck wait

Okay

So if G = <a> has order n

|a^k| = n/(n,k)

Right

[coral shale]

thats a very funny 'example' u got there

[balmy belfry]

Is it true or not

[coral shale]

probably

[balmy belfry]

Ok

It is iirc

If

k and n is coprime

The order of the element is n

Right

[delicate orchid]

a^n = 1
(a^k)^(n/k) = a^n = 1

I win. You lose, bye bye

[slim kayak]

The evolution from drawing squares, to QED, to we win, to I win to "I win. You lose, bye bye"

[balmy belfry]

I’m confused

This was a theorem in my book

How can it be incorrect

[slim kayak]

Same here can you please just post a screenshot of it

[balmy belfry]

Of what

The theorem or the problem I’m tryna do

[slim kayak]

The problem

[balmy belfry]

I’m not hung up on it

Ok sir

Wait *

MY AUTOCORRCT

bro

I meant to say “ok wait”

It said ok sir

Ong

[slim kayak]

Okay

[balmy belfry]

Ye

Ok so G = <a> with order n

So an arbitrary element of it looks like a^k for some k

If k and n is coprime, the order of a^k is n

So the number of elements with order n is phi n

?

[white oxide]

is the first part of this proof literally 1_R x a = a = 1_K x a for a in K, and then you just use the cancellation law since R has no zero divisors and ig we're assuming that a is nonzero

[slim kayak]

basically. ab=b can hold if 1-a is a zero divisor

[white oxide]

oh yeah

i'm hoping that Z9 is a suitable counterexample for the second part

nvm

[slim kayak]

was wondering what you wanted your subring to be

Whats the standard example for zero-divisors you know?

except Z_n

[white oxide]

Z6 works

i think

was thinking {0, 3} or am i trippin

[slim kayak]

wont you get 2 in there

i was trippin

[white oxide]

lol u good

[slim kayak]

sure

[slim kayak]

ill publish my findings as to why 3^2=8 soon

[chilly radish]

Another example: take matrices of order >1 over a field, with the subring of diagonal matrices whose first entry on the diagonal is arbitrary and the rest of the entries are 0

diag(a,0,..,0) for a arbitrary

[slim kayak]

the easiest example is IMO a product of rings

R x S, then (1,0) acts like the identity on the canonical subring R

[balmy belfry]

International math Olympiad fr

[slim kayak]

what are you yappin about?

[tardy hedge]

[F : K] if its finite, the degree can be found by taking an element of F and finding the degree of its minimal polynomial over K?

every element of F would be algebraic over K because its finite extension right? And so do we also know that every element of F has the same degree over K? (same degree of its minimal polynomial?)

[rustic crown]

nope, not all finite extensions are simple

[tardy hedge]

Ok

[rustic crown]

simple = generated by one element

but you easily have finitely generated

[slim kayak]

it tells you the degree of a extension of the form F(a):F

[slim kayak]

really?

oh right, separability

[rustic crown]

yea lol

an algebraic extension F/k is simple iff there are finitely many intermediate fields

[slim kayak]

Do you have an example at hand?

[rustic crown]

yep

[slim kayak]

kinda forgot most of my galois skills :(

[rustic crown]

take F = F_p(x, y) and k = F_p(x^p, y^p)

so for any c in k, the pth power of the element x+cy is x^p + c^p y^p which lies in k. so degree of [k(x+cy):k] = p.

and none of the x+cy and x+c'y can generate the same intermediate field. else both x and y will be there and that would make the degree p^2

this shows there are infinitely many intermediate fields

and so this extension can't be simple.

[slim kayak]

huh

[rustic crown]

det occasionally has to remind himself too ><

[slim kayak]

hm, is this a standard example shown in course?

[rustic crown]

yup

[slim kayak]

it feels slightly familiar

[rustic crown]

you have to make it non-separable

[slim kayak]

makes sense then

[rustic crown]

so work over field of char p

and the degree can't be p, since prime degree are automatically simple

so need base field infinite, inseparable, of char p, and degree more than p for the example to hope to work

in that sense, the above example is "minimal" counterexample

[white oxide]

how did they get that n = 1 in the second to last line?

[maiden ocean]

galois theory is among the most forgettable parts of math i feel

[coral shale]

1st line

[white oxide]

sorry wdym

[dire siren]

n is a positive integer that divides 1

[white oxide]

oh

[coral shale]

oh wait

dang why did i think |z| = 1 for z unit

[balmy belfry]

Hi

[rustic crown]

because that's true

(but sadly exactly what we were trying to prove >.<)

z unit iff z | 1 iff N(z) | 1 iff N(z) = 1

[slim kayak]

its really interesting but somehow it all feels a bit disjointed from the usual stuff you keep re-using at leat for my

[coral shale]

did they just replace c and d with x and y halfway there

💀

[rustic crown]

yea x = c and y = -d

<

[lethal depot]

Clicking definition of unit, doesn't it mean there's a multiplicative inverse per the fourth line beginning there exist c and d?

[coral shale]

ignore me - what filip said, btw

[slim kayak]

that proof feels awkward

Dont you usually define Z[i] with its norm in mind already?

[rustic crown]

Z[i] is usually defined as a subring of C or Z[x]/(x^2+1)

N(-) comes later

[slim kayak]

I guess

[rustic crown]

but yea, the motivation for looking at Z[i] is to understand integers of the form a^2+b^2

[balmy belfry]

Shuri

[slim kayak]

feels like a situation where'd you prove the existence of N(-) first if you gone the polynomial ring route, or just cashed it in for free from C

then again I enjoy nuking my problems so

[rustic crown]

yea that's nicer ><

isn't the "cash it in" process same in either case tho? ig you meant one is more intuitive. the definition is N(a) = a * bar(a).

[tardy hedge]

I was trying to understand why [F(a) : K(a)] = [F : K] in this case

[slim kayak]

well you usually have looked at the norm on C already in the context of an analysis course

[rustic crown]

you figured out the other 3 degrees in the diamond, so just the arithemtic of degrees calculates the last one

[F(a):K(a)] = [F(a):K]/[K(a):K] = [F(a):F]*[F:K]/[K(a):K]

[tardy hedge]

So its F : K if [F(a) :F] = [K(a):K

Oh wait that also doesnt even matter for the problem anyway

[rustic crown]

yep

[white oxide]

what

is there any reason why he set v(g) = deg g + 1

was it just to be instructive lmfao because normally it's not like that

at least if i'm remembering the division algorithim for polynomials again lol

my guy lorenz included all of the basic aa stuff in his linear algebra book apparently

that's hilarious

[barren sierra]

Yes it's just to be instructive

v(g) = deg(g) also works

[white oxide]

ye

[barren sierra]

Which also works for the case of 0

Since deg(0) is traditionally -∞

[white oxide]

ye

is there a canonical example of a non-factorization domain?

or whatever you call an integral domain that doesn't have a factorization property

[cobalt heath]

You can find non-unique factorization quite easily, I think

[white oxide]

oh i meant like

existence of factorization

[cobalt heath]

Yea, that is harder, I think

[white oxide]

like there's an element that can't be decomposed into irreducible factors

[cobalt heath]

Huh

Sorry I misread..

[coral spindle]

I think you can take a direct limit of k[x] → k[x, y]/(y^2-x) → k[x,y,z]/(y^2-x, z^2-y) → ... to get a ring in which x cannot be decomposed into irreducibles, but I would need to check details.

[cobalt heath]

Yea, and I think it should at least be non-noetherian, which is hard to satisfy already

[cobalt heath]

This is k[x1, x2, x3, ...] quotiented by relations xi ~ x(i+1)^2, right

[glossy crag]

Sir, this is #groups-rings-fields

[coral spindle]

Oh yeah ofc lol I feel so silly

[cobalt heath]

No way, you should not feel silly. Infinity business is always so difficult to grasp

[stark helm]

can we just say that a_r=brc0+...b0cr and say a_r should not be divisible by p because both br and c0 are not divisible by p, in order to make contradiction?

and also I feel pretty confused why it can conlude that there is a least integer t such that p does not divide bt?

[south patrol]

why would that mean a_r isn't divisible by p

There is a nicer way to write this if you know a little bit of modular arithmetic though

like a more insightful proof lol

[stark helm]

i mean it should be divisible by p, but p does not divide br and c0

[stark helm]

to make contradiction

[south patrol]

but there are other terms you've summed up

not just b_r c_0

[stark helm]

I know, so I can conclude that p does not divide br and c0.but don't know why p does not divide br means there is a least t that p does not divide bt?

[south patrol]

wdym "i know"

[stark helm]

I mean know about modular algorithm

[south patrol]

Oh okay i might show you the proof in a sec then

[south patrol]

Well the set { t such that p does not divide b_t} is finite and non-empty so it has a least element

The proof I had in mind was: suppose $f = gh$ where $\deg g, \deg h \ge 1$. Now reducing mod $p$, $gh \equiv f \equiv a_n x^n \mod p$. So by unique factorisation we can write $g \equiv b x^r$ and $h = c x^s$ for some integers with $bc \equiv a_n \mod p$ and $r + s = n$. Now if $r > 0$, say, then you can check constant coefficients of $g,h$ to get that $p^2 \mid a_0$, a contradiction. So $r = 0$ and $s = n$, also a contradiction

[cloud walrus]

potato

[south patrol]

it's a bit cleaner if you assume a_n = 1

[slim kayak]

now that you remind me I never tried it even after learning about the nice quotient map argument

[south patrol]

This was actually the way I was taught

i think it's just like, this proof I can visualise in my head

like very follow your nose aftre you go mod p

[slim kayak]

but yeah, if f was reducible then f(x)=g(x)h(x), then the zeroth coefficient of g and h must also be divisible by p for things to work out

or a_0 is divisible by p^2

[south patrol]

You can also prove Gauss' lemma by similar ideas

[slim kayak]

Given maps $A \to B$ and $A \to C$ I have a pushout $B \otimes_A C$, what are some ways to get some insight what their prime ideals look like? Rn I wanna know the case for A,B and C being fields specifically

[cloud walrus]

Kerr

[mortal quarry]

Hi

[rustic crown]

if one is simple extension of fields (e.g. finite + separable), then you have a good hold of the tensor product

[slim kayak]

Yeah no, I just have some residue fields from some eldritch stalks here

[rustic crown]

are you trying to prove that the (underlying space of) pullback of schemes surjects onto pullback as spaces?

i.e. the map |X ×_S Y| --> |X| ×_|S| |Y|

[slim kayak]

No, that proof just involved seeing that spec k(x) ×_spec k(s) spec k(y) is a cone over the diagram? nvm i misread u mb

well, s being the image of some random y, x being the element i plucked from the pre-image yada yada

I am trying to see how the underlying space of the of the above looks like. I assume it has several points and looks distinct from how you'd expect it from the set theoretic pullback

[rustic crown]

yea, set theoretic pullback for the case of fields is just a point

[slim kayak]

ig the more relevant question is how you would approach this?

what happened to the king 😨

[stark helm]

it seems that p does not divide both br and c_s and p does not divide one of b0 and c0, I am confused about why we can get p^2|a0 by checking constant coefficients of g and h as r>0?

[slim kayak]

If p^2 doesn't divide a0, then only one of b0 and c0 can be killed off

Det don't abandon me 😿

[stark helm]

You mean either b0 or c0 can disappear?

[rustic crown]

gomen, det eepy ><

idk how it becomes 5am these days so easily

[slim kayak]

I mean, neither can too

Oh timezone buddies

We have problems

[stark helm]

I mean if p^2 does not divide a0, then p divides only one of b0 and c0, and I am confused about what does the sentence only one of b0 and c0 can be killed off mean?

[slim kayak]

If p divides a number, that number mod p is 0

[stark helm]

and if p^2 divides a0, then b0*c0 mod p is 0. But why is that related to the question? and we just suppose f is reducible over Q, which means reducible over Z. so I don't know see the relationship between r>0 and p^2|a

[slim kayak]

Hm it does seem a bit more complicated. If say b0 isn't divisible by p, then for any c_k we have b_0 c_k x^k = 0, which forces c_k to be divisible by p

[stark helm]

You mean for a specific term x^k?, but the coefficient might be addtion of bo_ck+b1*c^k-1+...

[slim kayak]

Actually no.. uh.

Yeah you are right

You know the proof of why R[X] is integral for R integral?

[stark helm]

You mean if integral domain R-> R[x] is integral domain?

[slim kayak]

It might be a version of that but in reverse. Assume c_0 mod p = 0, let k be minimal s.t. c_k isn't zero mod p...

[slim kayak]

Yeah

[stark helm]

so are you intended to just suppose that f=g*h=0? to use non-zero divisor?

both g and h are integral domain

[slim kayak]

I am not, we want to see why a_n x_n mod p implies that any factoring f=gh must have one be constant, even more, a unit

[slim kayak]

Huh?

[stark helm]

then since p does not divide br and cs, it seems unlikely that one of them is constant?

I mean both br and cs are nonzero, so they are unlikely to be a constant polynomial

[slim kayak]

Anyways, if such k is minimal then the k-th coefficient of f mod p is exactly b_0 c_k x^k. If k<n it must be 0, but since b_0 isn't divisible by p it forces c_k to be divisible by p.

This means that all c_k are zero mod p for k<n

So h is just c_n x^n. And g must then have degree 0

So g is an unit (Z_p is a field), f is hence irreducible in (Z_p)[X]

[rustic crown]

det can't eep

[rustic crown]

so you have the two projections of A^2_k --> A^1_k, given by k[s] --> k[s, t] and k[t] --> k[s, t].
Spec k(s)⊗_k k(t) are the points in A^2_k which map to the generic points in both affine lines.
so lets identify the fiber of a generic point under A^2_k --> A^1_k, that's the image of Spec k(s)[t] --> Spec k[s, t]. since the left ring is a pid, a non-zero prime is generated by a single irreducible p in k(s)[t]. you can assume p is a primitive t-variate polynomial in k[s][t] by clearing the denominators. so the inverse image of (p) under k[s, t] --> k(s)[t] is the ideal of polynomials which are divisible by p in the ring k(s)[t], since we assumed p is primitive, these are exactly the ones divisible by p in the ring k[s,t], i.e (p) in k[s,t].

so points in A^2_k which map to generic points under both projections are exactly (0) and (p) where p in k[s,t] is irreducible and primitive as both s-variate and t-variate polynomial. in other words, it shouldn't lie in k[s] or k[t] and should contain both variables.

[slim kayak]

Oh right

The fiber identification is the exercise right after 😎

[rustic crown]

6 am in 20 seconds

[slim kayak]

Yeah I've been feeling "at least give me something to work with" after reading the chapter and seeing the exercises

[rustic crown]

det no understand

[slim kayak]

Now 👉 👈

[rustic crown]

are the exercises not counted in "something to work with"?

[slim kayak]

I mean lemmas and such

[rustic crown]

ah

finally det understand

can't even fully open eyes anymore

[slim kayak]

Det finally entering the eepy era?

[rustic crown]

i'm eepy for the last 3 hours

[hidden haven]

How is det so much more professional in private chat than in public chat

[rustic crown]

oh that's cause moldi no like casual chat

[slim kayak]

He has to maintain his brand in public

[hidden haven]

No I just don't like uwu chat

[slim kayak]

Uwumathecis

[rustic crown]

have to consciously stop myself from writing ">.<". but it still sometimes slips

[hidden haven]

Man I wish I bullied you more when I had the chance

[rustic crown]

🙈

how would you bully me

[hidden haven]

[slim kayak]

When you had the chance?

[hidden haven]

Ik him irl

[rustic crown]

(moldi was det's senior in undergrad)

[slim kayak]

Huh

[gritty sparrow]

Referring to yourself in the 3rd person? It has gone too far

[rustic crown]

oh

hi saketh

[gritty sparrow]

Hello

[slim kayak]

No hope left

[rustic crown]

3v1

[slim kayak]

He will forever

[rustic crown]

@summer path back me up ><

[slim kayak]

4v1?

[rustic crown]

[summer path]

det is uwu

[rustic crown]

3v2

[slim kayak]

Shit, we need to recruit more people

[rustic crown]

do i lose by default if i eep

[summer path]

[rustic crown]

🥺

[summer path]

det should go eep

,ti 671237809029251103

[cloud walrus]

The current time for det.uwu is 06:33 AM (CET) on Sat, 30/12/2023.
tubularcat is 6 hours behind, at 12:33 AM (EST) on Sat, 30/12/2023.

[summer path]

almost time for tubu to eep too

[slim kayak]

What does 12:33 AM mean in normal time units

[rustic crown]

00:33 :3

[summer path]

:3c

[slim kayak]

That makes no sense

Why does the clock go from 11 PM to 12 AM to 1 AM

[gritty sparrow]

They were so used to modular arithmetic, they never noticed the difference

(Possibly more likely: there was no roman numeral for 0)

[slim kayak]

It's not like there is a singularity at noon

Switch PM and AM there

[gritty sparrow]

I guess the only reason I can think of is so the 12:XX always gets the same am and pm tag no matter wether XX=0 or 60>XX>0

[rustic crown]

R/24Z --> {AM, PM} is locally constant in lower limit topology

[slim kayak]

So true

[gritty sparrow]

Also the cts bijection from the disjoint union of two half open interval to the circle necessarily can’t have continuous inverse

The points of discontinuity are the places you feel the “singularity” namely the transition from the am half open interval to the pm one

[rustic crown]

[rustic crown]

asymmetric uwu

.uwU

[summer path]

[dull marsh]

The uwu group

Gwoup*

[summer path]

[keen shuttle]

if G is a finite group and H is a subgroup, i dont understand why we say for g, g' in G that g is equivalent to g' iff g' in gH

as it directly follows, i don't understand why each coset of H in G represents an equivalence class

i think im not used to the idea of equivalence classes in an abstract context... modular arithmetic clearly induces equivalences classes simply because the value after the mod operation is the same

e.g. the odd numbers are in the same equivalence class mod 2 (the value of any odd number mod 2 is 1)

[keen shuttle]

$g \sim g^\prime \Longleftrightarrow g^\prime \in gH$

[cloud walrus]

qing dynasty

[still ledge]

just verify the relationship defined is reflexivity, symmetry or transitivity.

[keen shuttle]

yes

but that is very algebraic

i feel like im lacking the intuition

its very muddy to me how that property (g prime is in gH) denotes some sense of equivalence

i suppose symmetry is an interesting property

if g' in gH then g in g'H

[tawny dune]

honestly this doesnt seem like very good motivation for defining quotient groups. you might wanna check out benedict gross’s lectures on youtube, where he shows how the intuition for this came about

somewhere in the first 10 lectures he talks abt them

[keen shuttle]

do u know specifically whichlecture?

its completely different tot my schools syllabus ahah

nvm got it

i will check it out

ty

[keen shuttle]

ok thanks

mans goated

[chilly ocean]

ㅡㅡㅡㅡ

Is this correct

[rocky cloak]

If the c means preimage with respect to some map A -> B, then yes

[coral spindle]

Think of g ~ g' as saying that you can go from g to g' using elements of H. So, there exists some element h in H such that gh = g'. In my head I like to think of this as saying "if I only have the power of the subgroup H, where can I go from g?"

This should be more obviously an equivalence relation. Now it should equally be far more obvious that the equivalence class of g is gH.

If you want a visual example, imagine the group Z^2 with additive notation, and the subgroup H = <(0, 1)> which is the vertical axis. The 'power' that this subgroup gives us is moving up and down. It should be clear what the equivalence classes here are: they are vertical lines in this grid.

[tardy hedge]

Wow thanks! Gave me a new perspective too!

[grizzled crane]

I just completed group theory and started rings

I have compiled examples of rings

can you give a non example for a ring?

[slim kayak]

Any noncommutative group?

[grizzled crane]

but what are the operations?

Like is there a known non example?

[slim kayak]

The group operation

What is your definition of "non example"?

[grizzled crane]

anything fails the ring axioms

[summer path]

Think about what groups you know, and which ones are not commutative

[grizzled crane]

but under familar sets

[slim kayak]

Even Z fails to be a ring if you view it as just a group...

So like, you want a group which isn't naturally the additive group structure of a ring?

[tardy hedge]

Maybe hes thinking of some set that inherently cant be a ring or smth

Even tho for a given set we can have different algebraic structures defined on them

Like Z could be a group or a ring depending on how u view it

[south patrol]

Or a hopf algebra

[tardy hedge]

Fax so much fax

[rose prism]

is Z a hopf algebra lol

[formal ermine]

witt group in characteristic 2

can't be a ring because we can't meaningfully define the tensor product between two arbitrary quadratic forms

[tardy hedge]

Lol^

[coral spindle]

The naturals (as opposed to the integers) with addition and multiplication are a good non-example.
Invertible matrices with the usual addition and multiplication are too.

[formal ermine]

and also because the 1 element is no longer a not degenerated form

[grizzled crane]

Sorry i was away

If you ask me for a non example of a group, i can say Z with multiplication ( it fails to posses inverses)

I’m looking for examples of similar spirit

[formal ermine]

this is because of the way we naturally define the tensor product in the first place, namely basing it on the tensor product of the underlying bilinear forms. in characteristic 2 we no longer have a bijection between the space of bilinear forms and the space of quadratic forms. (in case sharp is curious)

[coral spindle]

Another non-example that I like a lot is the even integers, under the usual operations.

[rotund aurora]

why is that not a ring? You could give it a ring structure

[coral spindle]

No unit :)

[rotund aurora]

ah wait ig you can't lol

[coral spindle]

Perhaps another cute example is the following: if S is a set, then consider P(S) — the power set of S — under unions and intersections. Since these things distribute over each other, you could be forgiven for thinking it's a ring...

[grizzled crane]

Identity is only required for addition right? and 0 is even.

[coral spindle]

Multiplicative identity is required for a ring, usually.

People tend to call rings without identity "rngs" or "rungs"

[summer path]

Unless you like rngs

[grizzled crane]

The definition I have says associativity holds for multiplication

[coral spindle]

Well, you should at least be aware of this important difference in definitions.

The unit is a very important part of a ring, and rungs behave quite differently.

Hej hej sharp

[grizzled crane]

So this is a non example of a ring with unity?

[coral spindle]

Yes, it does not have a unit

[grizzled crane]

makes sense

[topaz solar]

Rings with distinguished unit vs unital rings

[coral spindle]

Woof

Since we've already spotted that units are important, here's a slightly more complex example.

Let R be any ring. I define the direct sum R^(N) as the set of functions f : N → R such that f(n) = 0 for all but finitely many values n.

[grizzled crane]

Isn’t invertable matrices abelian under addition and group under multiplication?

[coral spindle]

No!

[topaz solar]

What’s A - A

[grizzled crane]

Makes sense

[coral spindle]

This is a rung, but not a ring.

[grizzled crane]

Non invertable zero matrix

[topaz solar]

0 is rather non-invertible

[coral spindle]

So it's not in the set.

I.e., closure fails

[topaz solar]

One you might be interested in is let’s take functions N -> N under pointwise addition and (fg)(n) = f(g(n))

Is this a ring?

[coral spindle]

Ooh now I have to think about this one

Ah yes I see

Nice tho

[grizzled crane]

Is this rather composition of function?

idk the term pointwise addition

[topaz solar]

It’s literally composition for multiplication

[topaz solar]

(f+g)(n) = f(n)+g(n)

[coral spindle]

Pointwise addition means (f + g)(n) is defined to be f(n) + g(n)

hence added, point-by-point

[grizzled crane]

It should be a ring

I suppose

[topaz solar]

[coral spindle]

Have you checked all the axioms?

[topaz solar]

Have you checked any of the axioms?

[grizzled crane]

I had it as an example, iirc

[topaz solar]

No you didn’t

[coral spindle]

Mmm, I don't think so

[grizzled crane]

Mb