groups-rings-fields

[pastel cliff]

https://tenor.com/view/steve-harvey-family-feud-kill-steve-harvey-kill-family-feud-kill-gif-21764693

is this better blitz

no i dont get that

[rustic crown]

so what's bugging you?

say we have a linear map
V --> W

[pastel cliff]

picking Z so that it works for every possible map

[rustic crown]

that's probably not going to work

:p

[pastel cliff]

i understand the definition of an epimorphism

i understand surjectivity

[rustic crown]

it would be easier to pick different Z for different V --> W

after all you know you can do it for every Z :3

[pastel cliff]

im overthinking

[rustic crown]

https://tenor.com/view/mala-mishra-jha-pat-head-cute-sparkle-penguin-gif-16770818

[pastel cliff]

ok let f be an epimorphism in Vect(R)

say f: A -> B

[rustic crown]

okie

[pastel cliff]

let g1, g2 be maps satisfying the "epic" part of this

so g1 \circ f = g2 \circ f implies g1 = g2

what then

[rustic crown]

hint: consider the vector space B/im(f)

and find two nice maps g1 and g2

[pastel cliff]

want to show that B/im(f) is empty right

wait not empty

just that B = im(f)

[rustic crown]

yep

so wanna show it's the trivial vector space

[pastel cliff]

oh so are we doing A -> B -> B/im(f)

[rustic crown]

what are two nice maps B --> B/im(f) which "agree on A"

[pastel cliff]

mmmmm

identity map and trivial map...?

when you say agree on A

[rustic crown]

yee, but that's not called identity lol

it's the natural projection B --> B/im(f)

[pastel cliff]

that's this part right

[rustic crown]

which sends b to the coset containing b

[rustic crown]

yep, i like being lazy :3

[pastel cliff]

based

the other map then is just the one sending everything to id

[rustic crown]

0 is the + identity

[pastel cliff]

but then is this the additive identity in B?

no right

it's the lone element of B/im(f) right

thank you for your patience btw

[rustic crown]

wait so what are the two maps again

[pastel cliff]

https://tenor.com/view/dora-themap-imthe-gif-22214617

this fucker can be one of them

[rustic crown]

aww lol

i once remember watching this show and i slept while this guy was telling me "I'm a map"

then i woke up after a while

and it was still saying "I'm a map" but possibly a different episode

i was so confused

[pastel cliff]

in any case

uhhhh

[rustic crown]

[pastel cliff]

natural projection B -> B/im(f)

[rustic crown]

[pastel cliff]

trivial map B -> B/im(f) sending everything to additive identity

[rustic crown]

do you see the agree when composed with A --> B?

[pastel cliff]

they kinda have to agree right

i wa about to say the map f : A -> B has no effect on either g

but g maps to B/im(f) so it's literally defined based on f

[rustic crown]

yep it's literally defined so that A --> B --> B/im(f) becomes 0

under the natural projection

[pastel cliff]

but we can say that since g1(f(x)) = g2(f(x)), it follows that g1(y) = g2(y)

[rustic crown]

so the natural projection and the trivial map are equal!

[pastel cliff]

x in A y in B

stumbled my way the wholee way through but i'll type it up

[next obsidian]

He was saying “I’m the map”

[rustic crown]

wait really

[pastel cliff]

yeah lol

[rustic crown]

woah

[pastel cliff]

this might be silly to ask, but then how does this give us that all epimorphisms are surjective

this isn't enough right

i think im missing pointing out that B/im(f) is just one elt but idk which part that follows from

[rustic crown]

because that means the projection map, which is surjective has the same image as the trivial map

so V2/im(f) = 0

V2 = im(f)

surjective

[rustic crown]

(typo at the end, should be V2)

[white oxide]

i did not know that

that makes sense

wait im confused

are you saying that i'm on the right track, or doing it manually is the better way to go?

oh wait

i just need to map the generators

or it's just the generators rather?

im still confused as to why the number of automorphisms is the number of generators

because the specification is that we must map a generator a generator

but then there can be different maps for different generators

so say we have {1, 2, 3, 4} as generators for Z5

[void cosmos]

its cuz ur just counting the number of maps as i said

[white oxide]

wait but there aren't 4 maps

[void cosmos]

u need to see how the maps being structure preserving

work

[white oxide]

wait but like 1 can map to 1

or 2

or 3

or 4

right

since those are generators

and a genereator must map to a generator

generator

[void cosmos]

try it out

let phi be an automorphism

if we let phi(1) =1

then the key here

is just to notice that

phi(l) is just l*phi(1)

[tender wharf]

so ||this will change the order|| which is ||bad|| and we might or might not want that

[white oxide]

ohhh

right

so it just like

yeah all an isomorphism does is essentially rename the elements

but the properties are the same?

like i'm thinking in terms of a table

[tender wharf]

automorphism?

automorphisms are isomorphisms and yes it must preserve all the group structure

[white oxide]

so like if we set phi(1) = 2

then we need to check whether 2 is a generator of Z5

but like

sorry im fucking lost lmao

oh wait

so every isomorphism is completely determined by phi(a) where a is a generator

and i literally proved that lol

so like

take 1 which is a genereator for Z5

[tender wharf]

oh you're working in Z_5

[white oxide]

then phi(1) has 4 cohices, 1, 2, 3, 4

[tender wharf]

i mean <2> generates that anyway so

[white oxide]

and then the rest is all decided by that i guess

or something

so that's why there are 4 automorphisms?

[white oxide]

yeah

[tender wharf]

its not a coincidence that

let T be eulers totient function

then T(5)=4

[white oxide]

i guess i'm just confused about the technical definition of completely determined then: the book states, that given $\phi: G \mapsto G'$ and $\psi: G \mapsto G'$ are two isomorphisms such that $\phi(a) = \psi(a)$ then $\phi(x) = \psi(x)$ for all $x \in G$

[cloud walrus]

okeyokay

[tender wharf]

this is not quite right for arbitrary groups

but idk

can you take a pic of the book

[white oxide]

yeah one sec

[tender wharf]

yeah well its for cyclic groups

[white oxide]

right sorry

[tender wharf]

and also a evidently has to be a generator

not some arbitrary element

[white oxide]

oh wait i thought i put that a is a generator

oops

apparently i didnt

but yeah

i'm still confused as to why

or how that is the technical meaning of completely determined

[tender wharf]

take a moment to think about it

hint: what is <phi(a)>?

[white oxide]

G'?

oh

[tender wharf]

right, could you prove that?

[white oxide]

so it's saying that there's only a unique mapping

or there's one mapping

after a generator gets mapped

[tender wharf]

I mean you could send it to any other generator

[white oxide]

yeah but im too lazy lol

[white oxide]

oh yeah

right

[tender wharf]

but notice the end result is still the same

[white oxide]

right because the other generators are x

right?

[tender wharf]

they're not quite x

that's why I recommend you prove it

[white oxide]

okay

i'll do it in a moment

what are the number of automorphisms on Z?

[tender wharf]

on Z?

[white oxide]

would it just be how many prime num bers there are?

[tender wharf]

which is infinite

[white oxide]

well

i

yeah

[tender wharf]

I can tell you it'll be T(i) where T is euler's totient function (why?)

[white oxide]

idk what that is

[tender wharf]

counts the number of things coprime to i

and <= i

[sonic coral]

I think i nailed this one. I have shown H intersect K is a subgroup of G contained in H in a previous homework.

[white oxide]

atta boy

[sonic coral]

thank you. crazy this makes sense to me but i cant show a function is surjective to save my life

[white oxide]

how would i find a generator for a group under multiplication?

for instance C*, the group of nonzero complex numbers under multiplication

[void cosmos]

u cant always do that

not every group is cyclic

[upper pivot]

did someone say C*

oh

[white oxide]

what's wrong with that

[white oxide]

oh ok

[upper pivot]

just a joke, C* algebras is my field

(different then the C* u mentioned)

[white oxide]

cuz i'm trying to find the number of elements generated by the cyclic subgroup <i> of the group C* under multiplication

and i thought that you had to take some b in C* such that b = a^s where a is a generator of C*, and then take gcd(n, s) and then divide n/gcd(n, s) where n is the cardinality of C

so my reasoning was that i had to find the generator of C* first

[white oxide]

oh ok lol

[void cosmos]

<i> has |i| elements

cuz after |i| u will just go back

[white oxide]

wait im confused

|i| denotes the order of i right

oh wait

ur right

i can just count it

duh

thanks

[void cosmos]

yes

[white oxide]

wait no i can't count it

well ig you can make the argument that it is countable

but manually i can't

wait no

i'm actuallys tupid

ignore that

yeah ur rigth

so it has 4 elements

nope

it has 6 elements

yeah

[tender wharf]

truly abstract algebra

[rustic crown]

[somber sleet]

can somebody explain to me how groupoperations work and how to understand them

I'm kinda confused

Like in which context are they useful? What are the meanings of orbit, stabiliser, normaliser and so on

[rustic crown]

groups can be used to study symmetries.

[somber sleet]

If somebody had an easy example, it woulf be nice too

[rustic crown]

some people will argue this was the entire point of defining groups

say you had a counting problem

[tender wharf]

well theres the orb-stab theorem

[rustic crown]

and the naive way would do lots of overcounting

you would wanna count the number of distinct stuff

which owuld be number of orbits

okie a simple example would be to count number of rotations of a cube

the cube has 6 faces, and you know the rotation group acts transitively, so orbit of a single face would be all 6 faces

and the stablizer of a face would be the rotations that don't move that face at all.

[tender wharf]

ooh this actually has a really nice conclusion

[rustic crown]

if you're forced to fix a face, then the rotation but be around a vector perpendicular to that face

and from there it's easy to there are 4 such things

so orbit stablizer tells you the size of the rotation group is 4 * 6 = 24

[somber sleet]

may I ask how you "visualize" the orbit?

[tender wharf]

just consider the group of permutations not very visual

but you can kind of just imagine a cube rotating no?

[rustic crown]

i have a cube in my head, and it's rotating, then i see that any face can go to any other face by rotations

[somber sleet]

Okay, so in this case, we have 6 elements in the orbit, as you said, because we have 6 faces and by applying all the rotations on a face we get 6 faces back

[somber sleet]

actullay it is

cause you permute all the faces

[rustic crown]

yee

[tender wharf]

the orbit of a face would be a subgroup of the rotation group of the cube

so I thought it wouldn't be too visual to imagine elements of that subgroup since they are permutations but oh well

[somber sleet]

so my cube is the disjoint union of all the orbits right? this means that It's just equal to one orbit(?), because alle orbits are the same in this case

[rustic crown]

yep, there is a single orbit

[tender wharf]

in fact they'd actually all be disjoint!

[rustic crown]

[tender wharf]

because they form cosets

wait

oopsies

[rustic crown]

they're disjoint because that's the only set

[somber sleet]

what is then the stabiliser of a face? we get the identity for sure, but is it something else too?

I guess yes

[rustic crown]

the 4 rotaions around that face

[tender wharf]

imagine a line passing normally through the face

[rustic crown]

imagine the cube rotating around the vertical axis

[tender wharf]

and use that as the axis of rotation

[tender wharf]

thanks for the correction det

[rustic crown]

uwu

[somber sleet]

arent' there more rotations?

like the stabiliser is all the rotation who leave my face unchanged

[rustic crown]

it's basically the 4 rotations of square face we're fixing

[tender wharf]

now here's a really interesting fact

group of permutations acting on a cube is S_4

well isomorphic to

[somber sleet]

something stupid is gonna come

[rustic crown]

there's another really fun counting formula

[somber sleet]

so be ready

[rustic crown]

[somber sleet]

what is this then?

I can't categorize it

like this is just the composition of permutations right?

[rustic crown]

here you're moving that shaded face right

[somber sleet]

yes exactly

like I was thinking about gx=x

[rustic crown]

image the third drawing rotating about the vertical axis

[somber sleet]

so x would be the first face

[rustic crown]

do a 90degree rotation

[somber sleet]

jaaa

[rustic crown]

that's a stablizer element because the top face remains at the top

[somber sleet]

do you mean the vertical rotation?

yes I understood this fact

So if we fix a face the stabilizers has 4 elements, the orbit has 6 so we get that our set G of rotations has 24 right?

sow we have 24 possible rotations of the cube?

damn

[rustic crown]

yee

rotations

[somber sleet]

yes sorry

[rustic crown]

say you wanna count the number of necklaces you can make with say 10 beads using beads of two colors. if you allow all positions of beads as different, then it would 2^10, but a lot of these are same under rotation.

so if you let Z/10Z act on the set of necklaces by rotation

then elements in the same orbit are to be considered the same

which means you want to find the number of distinct orbits

in general if G acts on the set X, then we want a nice formula for the number of orbits

[cloud walrus]

det

[rustic crown]

now if you swap the sums

[cloud walrus]

det

[rustic crown]

and the left most thing is calculating the number of orbits, because each element x contributes 1/|O_x| to the sum. so all elements in a single orbit would totally contribute 1 to the sum

this is called Burnside's counting formula

here X^g is the fixed point set, X^g = {x in X | gx = x}

[somber sleet]

I think it's this one here

like you get the formula you said before by using these counting formulas

[rustic crown]

this one is a little different

it's counting the size of X

[somber sleet]

This is nice though

[rustic crown]

the above formula counts the number of orbits

[somber sleet]

like you can easily see the connection between the things

[rustic crown]

btw you know you can prove fermat's little theorem using burnside's counting formula?

a^p = a mod p

say you wanna count number of necklaces with p beads of a colors. Z/pZ acts on this, so number of distinct such necklaces by the formula is 1/p(|X^0| + ...)
now X^0 is the fixed points under the identity, so X^0 = X and |X| = a^p. for any other i != 0, X^i would be fixed points under rotating by i, and since p is prime, this would only consist of necklaces where all the beads have the same color. so |X^i| = a

so we get the number of these are 1/p(a^p + (p-1)a) = a + (a^p - a)/p
this is an integer implies a^p = a mod p

[somber sleet]

okay wow that's really nice

If I have a proper subgroup H of G where G is finite. If I have to show that this is not the same as G, what is the best way to do so? I thought about defining a group actions, but somehow I got stuck

[rustic crown]

what's that notation?

[cloud walrus]

det

[somber sleet]

conjugacy classes of H

like gHg^-1

for g in G

[coral spindle]

${}^ab$ is fairly common notation for $aba^{-1}$

[cloud walrus]

Boytjie

[rustic crown]

oh okie, i wasn't aware >.<

[rustic crown]

so we wanna show that G is not the union of conjugates of H

since G is finite it suggests that you have to do some counting argument

and as soon as you say some conjugates, i have 3 actions in my mind
G acting on G by conjugation,
G acting of subgroups of G by conjugation
and G acting of G/H by left multiplication

the first one doesn't look that useful

only thing you know is that each conjugacy class is represented by some element of H, but i dont' see any nice thing from there

third one can be a bit useful

ig

because we wanna conclude something about the index

notice that the stablizer of aH would be all g such that g(aH) = aH, which is just aHa^-1

so we know every g fixes at least one element of G/H

okie any counting here isn't nice because it's a transitive action

what if you just make a very bad bound

like say number of conjugates of H is r, then |G| <= r * |H|

and we know r = [G:N(H)]

Oh

yea this does it

okie i won't say more >.<

this was weird

me got distracted by looking at actions

[cloud walrus]

Eso

[rustic crown]

it's probably (k, l) --> k - l

[pliant forge]

I feel like ive misinterpreted what direct sum means

[pliant forge]

ohhh

that seems pretty reasonable

[rustic crown]

yea that always creates some confusion at the start because of internal and external direct sums

i would just write K + L for the internal sum and K ⊕ L for external.

[pliant forge]

ig. but somehow distinguish that theyre separated

non-intersecting

[somber sleet]

At least you showed your passion haha

Thx for explaining though

[rustic crown]

[somber sleet]

I have another question though

Hahaha

[rustic crown]

[somber sleet]

Do you know how to 'calculate' the quantity of Homomorphisms of symmetric groups

Like S3->S4

[tender wharf]

every normal subgroup is the kernel of a homomorphism

now how do we count those?

[delicate orchid]

two homomorphisms may have the same kernel but be different maps

take any two embeddings of S_3 into S_4 for example

[somber sleet]

So I have to know how many normal subgroups are in S3 for example

Id say the last isomorphism thm?

Or something thede

Or Lagrange

[delicate orchid]

well for n > 4 the number of normal subgroups is well known, so just manually compute it for n <= 4

yes use the good words mr lads

[somber sleet]

To be honest

I have no idea how many are for n>4

[delicate orchid]

well yeah you haven't worked it out yet

[somber sleet]

You said that it's well known 😦

[rustic crown]

you know something about A_n for n >= 5?

[somber sleet]

[chilly ocean]

(123) should be mapped to one of the 3-cycles in S_4 for example

[rustic crown]

and in general to find maps out of something, it's usually going to be best to find a nice presentation of that group

since S3 = D3, it reduces to finding elements of order dividing 3 and 2 respectively such that their product has order dividing 2

[chilly ocean]

It's simple of index 2

[chilly ocean]

So a 3-cycle and product of disjoint translations

Here I think it's just as well good to work with cycles

[somber sleet]

Wait a sec, is it right that Sn/An always has order 2

[rustic crown]

yep

[somber sleet]

So it's normal I guess

[rustic crown]

sgn : Sn --> {1, -1} is surjective

[chilly ocean]

Or 1 for n = 1

[rustic crown]

A1 has 0.5 elements

[somber sleet]

Better

GUYS

[rustic crown]

uwu?

[chilly ocean]

Wlog we can multiply by 4 and assume (123) maps to (123). If a 2-cycle gets mapped to something with a 4 then conjugation leads to contradiction

So this is 4 times the self-homomorphisms of S_3

[somber sleet]

Is it for real that the only normal subgroups in Sn are {I'd} and An

[rustic crown]

and Sn

[chilly ocean]

Wait my bad. (123) can get mapped to 0

[somber sleet]

This looks so easy for you guys

I feel like I'm discovering the America's

[rustic crown]

my geography is terrible

[chilly ocean]

I'm rediscovering America

[somber sleet]

I'm refreshing all your minds haha

[rustic crown]

yea lmao

[somber sleet]

I feel hopeless

[rustic crown]

i thought about group actions after so long

no say that >.<

[somber sleet]

Do you want to write the exam for me?

[rustic crown]

[chilly ocean]

Okay so its either we go from (123) to (123) in S_3 multiplied by four, or (123) gets mapped to id in S_4, so we calculate Z/2Z to S_4

[rustic crown]

for the first one we can send a 2-cycle to any other 2-cycle because if you think in terms of D3, they're refelctions. can't send the 2-cycle to id as the product then would go to a 3-cycle.

[chilly ocean]

In the former case we need 2-cycle to map to 2-cycle

[rustic crown]

so 3*4 + (1 + 6 + 3)

[chilly ocean]

The latter is clear as we map 1 either to id, 2-cycle, or product of two 2-cycles

For the former, (123) = (13)(12) = (23)(13) = (12)(23)

[rustic crown]

wait a smol q, shouldn't you have a factor of 8 instead of 4?

[chilly ocean]

Are the only ways you can write (123) as product of two 2-cycles

[rustic crown]

(123) going to a (123) and (132)

[chilly ocean]

Yeah

Thx

there are only 4 3-cycles in S_3

[formal ermine]

you forgot the other 6

[rustic crown]

[chilly ocean]

This should tell us what (12) and (13) can be mapped to, only 3 choices.

And ig we could write this for (132) as well

Forcing what translations get mapped to what translations

We can go home now

[blazing viper]

is this equivalent to saying that in this ring, all elements have a right and left inverse?

it's not. But so what is it? Is it a generalization of that?

and why all elements that commute with an element of the ring verify this

[south patrol]

I think the important thing is how it behaves with flatness

i.e. every left module over this ring is flat

[blazing viper]

flat? never heard this word

[south patrol]

Well it's something in (i guess mostly commutative?) algebra which comes up a lot and is v important

[blazing viper]

yeah exactly

[south patrol]

Okay i misread

[oblique river]

What does this mean?

[south patrol]

lmao

sorry ignore me

Wait no I didn't misread it

[chilly ocean]

OP is asking if this property is true, then does it mean x is invertible?

[south patrol]

I think verify would conventionally be written satisfy here

[blazing viper]

no ik its not i want to know the significance of this property

[south patrol]

but i don't think this is true

Or like commutative algebra would be too nice

[blazing viper]

this ring

this ring verifies the property

and i am asked to prove that

[oblique river]

Are you assuming that A satisfies the property?

And you want to show that this subring of A also does?

[blazing viper]

ah

right

i didnt read the question properly. yes

ok nvm then. SO what's the significance of the property again? Is it an abstract generalization of invertibleness?

[oblique river]

Potato wrote earloer

But deleted the wikipedia link

https://en.m.wikipedia.org/wiki/Von_Neumann_regular_ring

You can read about them here

If you’re proving this fact in an introductory ring theory course though

[chilly ocean]

i wonder if it comes from operator algebra

[blazing viper]

i am

[oblique river]

You should probably just think of this as “practice working with rings”

These kinds of rings do have applications but they are quite advanced

And so i think this is more just like, practice proving something about the center of a ring (that’s the set Z)

[blazing viper]

No it's okay i am not looking for a deep understanding, just naming things and understanding what i ma attempting to prove

so commutativity preserves weak invertibility, or taking the ring of commutators preserves that

[oblique river]

“Commutator” isnt the right word

Z is called the center of the ring

x is not part of the definitoon of Z

Z is the set of all x which commute with everything in A

The definition does not depend on one single x

[blazing viper]

okaay

yeah yeah i saw

thanks a lot

[oblique river]

Np and gl

[somber sleet]

Guys

How do I directly show the existence of the inverse for the theorem m a maximal ideal iff R/m is a field

[blazing viper]

is any ring in which we can find for any 2 elements a gcd, a principal ring?

if not why not

[chilly ocean]

Suppose x+m non-zero is such that (x+m)(y+m) = xy+m is not a unit for all y. Prove that (m, x) is a proper ideal

If 1 is in (m, x), then ...

(m, x) means ideal generated by m and x

Then x is in m so x+m = 0+m, contradiction

[oblique river]

No

Such a ring (if it’s a domain) is called a “gcd domain”

The ring of algebraic integers is a counterexample

It satisfies the even stronger property that every finitely generated ideal is principal

But there are non-finitely generated ideals which are not principal

[south patrol]

Ig also all UFDs are GCD domains, and certainly not all UFDs are PIDs e.g. Z[x]

[rustic crown]

@lethal dune is this you ?

[lethal dune]

impostor

[rustic crown]

[lethal dune]

[south patrol]

Sus

[blazing viper]

whats lacking

i am going over the proof for why Z is principal

lets take an ideal I we take the smallest element call it a

[rotund aurora]

Z[x] is a gcd domain

[blazing viper]

ok let me ask another question then, is any ring that we can do the euclidean division in a principle ring?\

[south patrol]

Yes

Euclidean domains are PIDs

The proof goes through in basically the same way

[blazing viper]

okay

[south patrol]

Just we replace "| |" for Z with our Euclidean function

[blazing viper]

|| ?

[south patrol]

Well like

You pick the smallest element i.e. the element a minimising |a|

(One would probably say the smallest positive element a, but if we phrase it this way then it generalises)

But in a Euclidean domain, one replaces taking absolute values with applying the Euclidean function and the proof otherwise goes through in much the same way

To see this in action, compare the proof that Z is a PID with the proof that F[x] is a PID for F a field

[coral spindle]

Blitz' idea of N-embedded subgroups and their relation to semidirect products.

Starting a thread so as not to interrupt you :)

[blazing viper]

the euclidean function becomes the degree, okay got it thank you very much

[somber sleet]

thanks my dear, I got it

[south patrol]

No problemo

[rustic crown]

owo

[summer path]

uwu

[somber sleet]

is it true that if 1 is in x then there exists b in R s.t. 1=b(x)+m ? Then it would follow that 1-b(x) is contained in m which implies that 1-b(x)=0 (?)

I am not so sure about the definition of (m,x). Is it (m,x)={am+b(x) | a,b in R}

[rustic crown]

nu

m was your maximal right

[chilly ocean]

x is an element of R

[rustic crown]

ig m + (x) is a more appropriate notation

[chilly ocean]

Wdym my notation is pretty standard I think

[somber sleet]

don't fight over it

[chilly ocean]

(m, x) = {a+bx : a in m, b in R}

[rustic crown]

me only saw that notation when all thingies are elements

[chilly ocean]

Not sure why brackets around x here

[somber sleet]

I meant it as a principle ideal

[chilly ocean]

I saw it for (J_1, J_2, ..., x_1, x_2, ...) where J_i are ideals and x_i are elements

[rustic crown]

souka

[chilly ocean]

Idk it feels like a good notation to me so I use it

[somber sleet]

it's this one right?

the strange symbols on the lhs are ideals

[chilly ocean]

Yeah ig

But it's different notation

[somber sleet]

I didn't understand how to interpret this notation

gotta be honest

[chilly ocean]

(I, x) for me is ideal generated by I and x

And in your notation its I+(x)

[somber sleet]

I also have this one

It's more appropriate I think

[somber sleet]

you just wrote a in m because x*m for x in R is always in m right? (m is an ideal)

[chilly ocean]

Yeah

[somber sleet]

I ask so many simple question, sorry for disturbing guys haha

it's just to make sure

[coral spindle]

To be clear, the sum on the left is different to the one inside the set

if a,b are ideals then we can define a + b as the ideal generated by (n.b.!) sums of elements in a and b

[chilly ocean]

If 1 is in (x) then it'd mean R = (x)

Or in other words x itself is a unit

[rustic crown]

((n.b.!)?)

[coral spindle]

n.b.!

[chilly ocean]

Is this Latin

[coral spindle]

yeah, just means nota bene, i.e. note well

[chilly ocean]

Ah makes sense

[rustic crown]

https://tenor.com/view/anime-fuuko-ibuki-salute-girl-cute-gif-21773862

[somber sleet]

do you have an eays example for two ideals which are not proper?

cause for proper ideals it makes sense i guess

[chilly ocean]

2Z in Z

And 3Z in Z

2Z+3Z = Z

2Z = (2) and 3Z = (3) and 2Z+3Z = (2, 3) = Z

Note how we also sometimes write (n, m) for gcd of n and m

[toxic zephyr]

for vi: H/G' is a subgroup of G/G', so because G/G' is abelian then H/G' is a normal subgroup. so by the third isomorphism theorem, H is a normal subgroup of G?

[somber sleet]

I did a proof yesterday about this and I guess it's because aZ+bZ=gcd(a, b)Z

The same goes when you have aZ intersected bZ=lcm(a, b) Z

Or something like this

Does this hold generally for faktorial rings?

[rotund aurora]

aR+bR needn't even be principal in general

[somber sleet]

Well aR is in R

A factorial ring is not pid

So I guess it holds anyway(?)

[rotund aurora]

hold what?

You are saying if aR+bR=gcd(a,b)R

but aR+bR need not be principal

for example, 2Z[x]+xZ[x] is not a principal ideal, and Z[x] is factorial

[chilly ocean]

Z is a special kind of ring

Idk exactly how this all works because it gets confusing at this point

But it should be true that (a, b) is gcd in some sense when your ring is good enough

[formal ermine]

when is R[x] a pid? I know that R field <=> R[x] euclidean => R[x] pid, but is there a stronger implication?

[rotund aurora]

can R[x] be a PID if R is not a field?

I feel like no, take a a non-unit (different from zero) and consider the ideal (a,x)

so R[x] is PID iff R is a field I think

not entirely sure

[formal ermine]

so we have R[x] euclidean <=> R[x] pid?

[rotund aurora]

if what I said is true, yeah

because you can do usual polynomial division

and the euclidean function is the degree of the polynomial

[formal ermine]

I don't quite understand your argument

[rotund aurora]

Suppose R is not a field, let n!=0 be a non-unit. Consider the ideal generated by n and x in R[x], (n,x). This is not principal

[formal ermine]

ah, so we're showing R not field => R[x] not pid

[rotund aurora]

uhh

if there are zero divsiors maybe its problematic EDIT: nah its fine I think

[formal ermine]

contrapositive of R[x] pid => R field

[rotund aurora]

all the same

[formal ermine]

yeah ok looking it up

it's true

epic

R field <=> R[x] euclidean <=> R[x] pid

interesting

[somber sleet]

as far as I know it holds that if R is a ring euclidean => pid=>factorial

[chilly ocean]

Yeah

[somber sleet]

but the other direction do not have to hold

[chilly ocean]

They're talking about R[x] though!

So pid and Euclidean should be the same for polynomial rings

[formal ermine]

yeah

something that helped me when I first learned about those things was KEHFIR

Körper
Euklidischer Ring
Hauptidealring
Faktorieller Ring
Integritätsring
Ring

[somber sleet]

YEAH I SWEAR SAME

[formal ermine]

LOL DID YOUR PROF TEACH YOU THAT TOO?

[chilly ocean]

Factorial ring isn't the same as unique factorization domain I think

[somber sleet]

NO I KNOW A YOGURT

loook

[rotund aurora]

I thought factorial was the same as UFD

[formal ermine]

LOL

[formal ermine]

faktorieller ring is the german translation of ufd

[chilly ocean]

I think it doesn't have to be unique in factorial ring

[rotund aurora]

In mathematics, a unique factorization domain (UFD) (also sometimes called a factorial ring following the terminology of Bourbaki) is a ring in which a statement analogous to the fundamental theorem of arithmetic holds

[chilly ocean]

I read a definition like this somewhere

[rotund aurora]

wikipedia

https://en.wikipedia.org/wiki/Unique_factorization_domain#:~:text=In mathematics%2C a unique factorization,fundamental theorem of arithmetic holds.

[chilly ocean]

Well

It doesn't mean people call factorial rings only UFD's

But good to know

[rotund aurora]

https://encyclopediaofmath.org/wiki/Factorial_ring

Here they say unique too

if you just want factorization into irreducibles, I guess thats kind of like noetherian

[chilly ocean]

I'm almost sure there was a book which differed between factorial rings and UFD's

[formal ermine]

this is giving me 0 is not in N vibes

[somber sleet]

my measure theory prof says no

my algebra prof says no

who should I believe

[chilly ocean]

measure theory?

[formal ermine]

mass theory

[somber sleet]

YEAH DAMN

[chilly ocean]

0 is a natural number

thats because natural numbers are the finite cardinal numbers

[formal ermine]

that's what the government wants you to believe

[chilly ocean]

and empty set has zero elements

and other definitions are dumb

[formal ermine]

proof that 0 isn't in N: we have that Z = N u -N. because 0 isn't in Z, it's not in N either

[rotund aurora]

you cannot justify 0 being a natural number with set theory

[chilly ocean]

I just did

[chilly ocean]

tbh now I'm wondering where it was from. If you want I can try to search for it

[rotund aurora]

No, I don't care

I know Eisenbud only uses factorial in his com alg book

[barren sierra]

0 is a natural number because it's prettier when the natural numbers are a monoid wrt to addition and multiplication, rather than just multiplication

QED

[rotund aurora]

and Neukirch also uses factorial I think

these are the only places where I have encountered that terminology

[coral shale]

0

[celest cairn]

Hi

[south patrol]

Hi

[celest cairn]

I have a question that may be stupid:
I head that for an extension to be Galois, the degree of the extension equals the order of the Galois Group. I’m trying to use this logic for an nth cyclotomic extension, but I can’t tell if it applies or not.
The degree of an nth cyclotomic extension is phi(n)=n-1, for all n being prime.
If my calculations are correct, the Galois Group of Q(zeta_n) over Q has order n. But n obv doesn’t equal n-1, so why is it Galois?

[formal ermine]

in general we have that the order of the galois group is equal to the degree of the extension iff the extension is separable (and of course finite)

[formal ermine]

why should the galois group of Q(zeta_n)/Q have order n?

[rotund aurora]

1 is not a primitive nth root of unity

[south patrol]

Smh don't call it the galois group in that case

jk conventions differ

[formal ermine]

lol if only

before we did galois in my algebruh class

we called that set the

uhh

"galois set"

[south patrol]

Yeah nah we called them galois groups even when not galois too

[formal ermine]

and used the notation $\on{Hom}_\bK(\bL, \bar{\bK})$

[south patrol]

lol

[chilly radish]

Why not just use Aut like normal people

For non-galois automorphism groups

[delicate orchid]

they really aut to start...

[formal ermine]

https://tenor.com/view/byuntear-sad-sad-cat-cat-meme-gif-25617057

[floral kernel]

If you have 2 matrices M1, M2 and perform polar decompositions M1 = O1P1, M2=O2P2 is there anything we can say about the polar decomposition of the matrix M=M1M2 in terms of O1,O2 and P1,P2? In other words if M=O3P3 is there any way to relate O3 or P3 to O1,O2,P1,P2? (O_i - orthogonal, P_i - positive semi-definite, real matrices).

[formal ermine]

#linear-algebra ?

[floral kernel]

Sure, I'll ask there, thanks.

[molten viper]

So in class we’ve just learned about cyclotomic polynomials and Eisenstein’s criterion, my question is, put crassly, “who cares?”. Or more properly “what applications do these have and where do we go from here?”

[prisma ibex]

cyclotomic polynomials generate cyclotomic extensions Q(\zeta_n) of Q and these have all sorts of nice properties

[molten viper]

Innnnnteresting

Right now about the only thing I know is the p-th ones are irreducible in Z[x]

We haven’t proved it for general n

[prisma ibex]

for example the Kronecker-Weber theorem classifies all Abelian extensions of Q in terms of these cyclotomic extensions

namely every Abelian extension of Q (if the extension is Galois this means one with Abelian Galois group) is contained in some cyclotomic field Q(\zeta_n)

[molten viper]

Weird outlier question, do we say 1 is a monic polynomial?

In whatever ring

[molten viper]

I’m sure I’ll understand this soon 😅

[sonic coral]

how did i do on these two?

[chilly ocean]

you don't need to say "let <g> be a cyclic subgroup of G" when g is already given to you

you just need to say (and justify?) that it is a subgroup of order k

the rest of that one is fine

the second one is similar

[sonic coral]

okay, thank you

[sonic coral]

is every finite group of order p, where p is a prime, abelian?

[next obsidian]

no

try to find a group of order 8 which is not abelian

[sonic coral]

8 isn’t a prime though

[next obsidian]

oh hurb

sorry for some reason I thought you said p-group

which is order a power of p

every prime order group is abelian, yes

[sonic coral]

i’ve proven that every prime order group is cyclic, would that help me show what i’m trying to show?

[next obsidian]

are cyclic groups abelian?

[sonic coral]

that would do it for me, let me think about it

i think i’ve proven that on my last test

[next obsidian]

well try to prove it right now

'it isn't very hard

what's your definition of cyclic?

[sonic coral]

,rotate

[next obsidian]

there you go

[sonic coral]

dumb mistake with my integers but i did it right for the most part

[next obsidian]

yeah the point is just that

g^ng^m = g^mg^n cuz

uh

[sonic coral]

so i’m done then, nice

[next obsidian]

associative

[sonic coral]

sometimes i just need to talk about things

[next obsidian]

rubber duck

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

[sonic coral]

that’s exactly how i debug code, didn’t know it had a name

[solar glacier]

when a problem says to show

($\prod_{k=1}^n(x-k)$) $- 1$ is irreducible over z for all positive integers n

[cloud walrus]

MyMathYourMath

[solar glacier]

are the x values pulled from z?

cause you could argue by contradiction that ($\prod_{k=1}^n(x-k)$) $- 1$=$ab$ where $a,b$ are non units.

[cloud walrus]

MyMathYourMath

[solar glacier]

cause for integers $k$ with $1 \leq k \leq n$ wed have $a(k)b(k)=-1$ forcing one to be 1 one to be -1 ?

[cloud walrus]

MyMathYourMath

[pastel cliff]

i learned recently that a group can be observed as a category with a single element

is this done by just taking a single object (call it x) and taking the group action of G on it as the morphisms

so is this actually a characterization of any group?

[next obsidian]

WDYM characterization?

[prisma ibex]

sure

but like

there's nothing really deep here, it's literally the same definition

[next obsidian]

the category of one object groupoids and groups are equivalent

so like in some formal sense they are the same

(this uses the convention that all categories are locally small kekw)

[prisma ibex]

this is a useful perspective since you can talk about functors to/from a "group" in this sense

[next obsidian]

this is uses to generalize rep theory right?

[pastel cliff]

ok taking a step back

[next obsidian]

you say "fuck it, functor from BG"

[pastel cliff]

how can there be more than one morphism in a category with one object?

[prisma ibex]

endomorphisms

you can have lots of them

[next obsidian]

you have a Hom set Hom(*,*)

it can be whatever you want

[pastel cliff]

[pastel cliff]

this isn't obvious to me just yet, do you have an example

[next obsidian]

the category group

[prisma ibex]

what

[next obsidian]

you can have lots of maps from G to G

[pastel cliff]

that endomorphisms can give us a distinct maps from an object to itself

[next obsidian]

??????

[prisma ibex]

take literally any object that has nontrivial automorphisms

[next obsidian]

an endomorphism is by definition a map from something to itself

[pastel cliff]

oh

[prisma ibex]

in your favorite category

[next obsidian]

you can have many of these

[pastel cliff]

i confus with epimorphism

too many morphisms

[next obsidian]

But I mean agian you can literally just say

[prisma ibex]

so under this definition every group is an automorphism group in some 1 object category

[next obsidian]

Let Dumb Cat be the category with an object *

and with Hom(*,*) = {, , , fuck you, 3}

btw 3 is the identity

and I mean you ahve to say how they compose I guess

but like, you can be as stupid and arbitrary as you want

[pastel cliff]

i hate you

but also thank you

[tender wharf]

can we pin that

[next obsidian]

A composition law is literally just a monoid structure on this and OEIS says there's 228 on a set with 5 elements so there's 228 ways to make Dumb Cat into a category

[pastel cliff]

fuck you ->

[tender wharf]

3 -> ->

[sonic coral]

is there any insight as to why the first simple group has order 60

i was learning about it on monday and he told me that the first one appeared at order 60 and the second at 168… the numbers seem so arbitrary

[prisma ibex]

so I mean finite simple groups have been classified, these are the smallest ones, they are very particular concrete groups (A_5 and PSL_2(F_7) respectively)

[sonic coral]

other than the fact that they’re highly composite numbers, but like 36 and other numbers are also highly composite

[prisma ibex]

there are some kinda tedious arguments to show that groups of smaller order can't be simple

but this is through a lot of case by case analysis

or by classifying groups of small order

idk if there is a very satisfying reason for why these groups in particular

[sonic coral]

he said i should go through all of them under 100 and show which can and can’t be simple

i think we already knocked out all the prime cases

but that’s not even a dent

[prisma ibex]

right yeah, you'll be able to knock out most of the cases with Sylow theorems but it's kinda like

idk a lot of the structure in this regime is coincidences with small numbers and just exhaustively enumerating all the possibilities

[sonic coral]

so there’s nothing special about 60, just happens to be the magic number

[prisma ibex]

well so here's a quick argument

other than the cyclic groups of prime order which are simple

[sonic coral]

he mentioned sylow theorems, so maybe i will learn about them tomorrow

[prisma ibex]

if a group has order pq with p and q prime then it can't be simple

so the next case to check is order pqr

smallest case is 2*3*5=30 but this can't be simple

[sonic coral]

is r prime

[prisma ibex]

yes

[sonic coral]

yeah he was like writing down all the possibilities to multiply 3 primes to be less than 100

[prisma ibex]

next case is 2*3*7=42 but Sylow says you have a normal subgroup of order 7 so this doesn't work

so you just keep checking cases like this until you get one that works

[sonic coral]

why can’t a group with the order of the product of 2 primes be simple

and does this generalize to n primes

[prisma ibex]

Sylow tells you it has a cyclic normal subgroup

and no this doesn't generalize

otherwise there wouldn't be many simple groups lol

[lethal dune]

https://www.quora.com/Is-there-a-proof-showing-that-any-group-with-an-order-less-than-60-is-solvable?share=1
here's the tedious argument

[prisma ibex]

but yeah all the arguments involve a lot of casework, it's not particularly enlightening

[sonic coral]

gotcha okay

seems tedious to show

like knocking out the primes is nice but there’s still a lot left

[prisma ibex]

yeah this is pretty typical of a lot of proofs in finite group theory

[sonic coral]

the pq argument would get a lot more

[prisma ibex]

like once you have the Sylow theorems you can classify groups of certain orders

and depending on the order this involves a lot of cases

e.g. order p^2 * q is a common exercise

long and tedious

[lethal dune]

Burnside might be helpful for ruling out non abelians

but it's not "basic group theory" statement

unlike those nG mentioned

[prisma ibex]

I mentioned the classification of finite simple groups, that is not a basic group theory statement either

why are the sporadic groups there, what do they mean

[next obsidian]

That god hates us

[white oxide]

ok kind of a stupid question but say a generates a cyclic subgroup of order 2. Then would a be included as one of those two elements?

[tender wharf]

yes

in fact its literally the only nonidentity element

[white oxide]

yeah that's why i think this proof in the book is kinda silly

or rather question

[tender wharf]

to see this just recall how generators are defined

[white oxide]

because they proved that there's only one group of two elements up to isomorphism

[tender wharf]

well yes

[white oxide]

but also e can't generate

[tender wharf]

theres only one group of p elements up to isomorphism

[white oxide]

any group other than the trivial group

[tender wharf]

uhm

yeah

[white oxide]

which is of order 1

yeah

[tender wharf]

what book is it

[white oxide]

a first course in abstract algebra

fraleigh

[tender wharf]

ah

I mean it is a very gentle introduction so

[white oxide]

yeah seems very very gentle

what would be something more rigorous if i wanted to self study over summer?

[tender wharf]

d&f is a usual rec

I don't use it though

[white oxide]

dunnitt and foote?

[tender wharf]

oh sorry

wrong one

[white oxide]

oh lmao

that's the supplement for this course that the professor suggested

[tender wharf]

rudin?

[white oxide]

oh nah

d&f

[tender wharf]

oh

yeah d&f is classic but dry

I use gallian

has a ton of examples

although if you use it beware of the typos

I'm thinking of collecting all the typos and sending an email to him after finishing the book

[white oxide]

damn

yeah i'm looking at gallian rn

looks pretty good

[tender wharf]

just be careful of typos

love the examples tho

[white oxide]

yeah examples are great

last textbook i worked through was axler ladr

had so many examples

was very helpful

how would i find the number of generators of Z_n for some reasonably large n? this question i feel like would border more on number theory, since it's essentially asking how many numbers 1...n-1 are coprime to n

if i'm understanding correctly

[tender wharf]

@white oxide have you heard of euler's totient function

[somber sleet]

why are these two isomorphic?

also this one

like do I have to dream these isomorphisms?

[agile burrow]

In general, (R/I)[x] is isomorphic to R[x]/IR[x]. This follows from the first isomorphism theorem. Can you think of the surjection R[x] -> R[x]/IR[x] that induces this isomorphism?

[somber sleet]

well IR[x] is an ideal right?

so these are just restclasse modulo IR[X]

It's something like this right? we only did this for groups, but I guess, as I already discussed it once, I can also use it for Z-rings

[somber sleet]

f -> f+If (?)

[glossy crag]

Maybe it's the early hour, but I'm blanking at this: if R is contained in a field K, why should K contain a unique copy of the field of fractions of R? That it contains one is clear, it's just uniqueness I'm not getting.

[south patrol]

I imagine the point is that the map R -> K extends uniquely to a map out of the field of fractions right?

Or like, inverses are unique

[glossy crag]

Yeah, that gives a copy of Q(R) in K, but why should it be the only one?

[frank cosmos]

Lang says that if $H,K$ are subgroups of $G$ and $H$ is in the normalizer of $K$ then $H \cap K$ is normal in $H.$ This is easily verifiable since if $x=hk$ where $k \in H \cap K,$ then $\exists k_1 \in H \cap K$ such that $x=k_1h$ because you have $k_1=hkh^{-1} \in H \implies \in H \cap K.$

He then says that $f: H \to HK/K$ defined by $f(h)=hK$ is a homomorphism with kernel $H \cap K$ which implies that $H/(H \cap K) \cong HK/K.$ What was the point of writing that $H \cap K$ is normal in $H$ when it is the kernel of the homomorphism he described which automatically implies it is normal? Is this a triviality or is there some finer detail that I am missing...?

[cloud walrus]

Hello1

[frank cosmos]

sorry for formatting

[delicate orchid]

If there’s two copies then there would exist an r in R that has two different inverses in K (one in each copy of Frac(R)), I think

[sharp marsh]

someone know the approach to this?

I tried to start by trying to prove isomorphism from Z/pZ to Z_p and then later tell that Z_p is field only when p is prime then pZ will turn out to be maximal
but this is wrong, since both these sets do not have same number of elements
also it would be tough to proof the converse
now I am stuck

[rustic crown]

it's correct

and moreover all steps are if and only if

[sharp marsh]

but I cant say that Z/pZ and Zp is ring isomorphic isnt it?

the number of elements they contain are hugely different

[rustic crown]

how do you define Z_p?

[sharp marsh]

i define Zp as the remainder classes of the integers divided by p

[rustic crown]

(most people would define that by the quotient Z/pZ)

[rustic crown]

why do you say so

so ig you see that Z_p has p elements

[sharp marsh]

yes

[rustic crown]

what about Z/pZ

[sharp marsh]

Z/pZ will be {0,+-p, +- 2p, ..... , +- np}

where n tends to infinity

[rustic crown]

oh that's just pZ

it's not the quotient Z/pZ

element of the latter are {0+pZ, 1+pZ, ..., (p-1)+pZ}

[sharp marsh]

oh wait then Z/pZ is {a + pZ | a from Z}

[sharp marsh]

actually this is where its not clear to me

pZ already has infinite many elements

and we add infinite many more withe ach elements of pZ

[rustic crown]

so a+pZ is the set of elements, which give the remainder of a on division by p

[sharp marsh]

yes thats how i define the homomorphism

[rustic crown]

but it's a single element in the ring Z/pZ

you can use a different notation if that helps

say [a] = a+pZ

[sharp marsh]

okay

[rustic crown]

and Z/pZ = {[0], [1], ..., [p-1]}

even though each equivalence class is infinite, the number of such is only p right

[sharp marsh]

yes okay

[sharp marsh]

yes

in this way is it legal in ring theory to conclude that Z/pZ = Zp

?

[rustic crown]

you'll have to produce an isomorphism between the two to be able to say that

so far we only saw both have p elements

[sharp marsh]

ookayy

wait i know this question might be wrong but..

Zp doesnt really have only p elements in total right?

if we count all those numbers inside the equivalence classes Zpactually too has infinite elements?

sorry if i am asking a very bad question. I am pretty new to this

[hot lake]

what even is Zp

[sharp marsh]

Z_p my bad

[rustic crown]

wait, i'm a little confused... it feels like you're literally saying Z_p := Z/pZ

[hot lake]

yeah

[rustic crown]

earlier i thought you meant Z_p = {0, 1, 2, ..., p-1} and Z/pZ = {[0], [1], ..., [p-1]}

[sharp marsh]

Z_n has n elements alright

[hot lake]

do you have a definition for Z_n

[sharp marsh]

and we can present the elements in Z/pZ like with that [] notation to say there also exist p-1 elements

[sharp marsh]

set of remainders when divided by n

[hot lake]

is -17 a remainder when divided by 9

[sharp marsh]

no..

[hot lake]

so it looks like you're saying Z_n is {0; 1; 2; ... n-1} ?

[sharp marsh]

$\mathbb{Z}_n$

[cloud walrus]

TumAro

[sharp marsh]

it contains all the remainders of integer divided by n right...?

[hot lake]

what are the remainders of integer dividd by n

[sharp marsh]

0,1,2,...n-1

[hot lake]

then it's {0;1;2;...;n-1}

so yeah it only has n elements

[sharp marsh]

that where i am finding trouble to understand if it has p elements and Z/pZ = {[0], [1], ..., [p-1]} actually contains much more than p though we have represented it like it has p elements, can we show bijectivity?

[hot lake]

no, Z/pZ also has p elements

[0], [1], ..., [p-1]

say p=3

[frank cosmos]

bump, also why is subgroups relavent here?

[hot lake]

[0] is the set of all integers that are 0 mod 3

so [0] = {0;3;-3;6;-6;...}

that's ONE set containing many elements

but it's still ONE set

and Z/3Z = {[0];[1];[2]}

= {{0;-3;3;-6;6;...} ; {1;-2;4;-5;7;...}; {2;-1;5;-4;8;...}}

it is three sets

it's only three objects

even though each object can be opened up and has infinitely many items inside

but we don't necessarily care about that

[sharp marsh]

i didnt know that we actually divide the whole thing into distinct sets/objects when defining a quotient group/ring

[hot lake]

when we count the number of things in a set we don't open the things inside

and if you do set theory you define integers as sets of sets of sets of sets ....

[sharp marsh]

alright ill try and complete the proof now and see if this time I get it right

[frank cosmos]

quotient groups are the set of cosets of a group wrt the subgroup

so in the case Z/3Z the only subgroup is {0,+}

call this S

the elements of (Z/3Z)/S are {0S,1S,2S}

which are

{{0,1,2},{1,2,0},{2,0,1}}

but these are all the same

@shell agate what?