groups-rings-fields

[next obsidian]

When you hit 2048 I’d assume it blows up so hard it becomes virtually 100%

[lethal dune]

whaaat

[next obsidian]

MacHale (2020) shows that there are infinitely many values of n for which there are more groups than rings of that order

What the hell

[lethal dune]

some kind of AoC shit? feels like it defies common sense

[rustic crown]

is it because of some weird quirk of 2 or like is it like groups of order 2^n and 3^n are pretty same in number but 3^n just ends up taking far longer to appear if we look at group of order <= N

[next obsidian]

Apparently there’s at least 1.7quadrillion groups of order 2^n

Yeah so prime power ones get a lot

Because I think there’s a lot of freedom

You can see it like even with abelian groups

My shit-test for this is basically p-groups are simple

This means like the Cayley table isn’t as restricted

It’s “easier” to be a group

So a lot more binary operations end up giving you a group

Then as det said

2^n once n is liek… 10

Is just way smaller than 3^n

So you get that huge freedom from being p^n for a high n

And p^n is as small as it could be for that n

Since p = 2

[lethal dune]

huge I can see, but this huge really mind blowing

[rustic crown]

😵‍💫

[next obsidian]

I mean

There’s a lot

[rustic crown]

eevee fainted 🙈

[next obsidian]

Of operations

On a set of size 2048

Like…

[rustic crown]

yea

[next obsidian]

(2048^2)^2048

If I did that right

The number of operations on a set just grows so unbelievably fast

[rustic crown]

other way i think, but that would be number of functions

[lethal dune]

but like having more groups than rings?

[next obsidian]

Eh?

[lethal dune]

[next obsidian]

Didn’t what I wrote count the number of functions from

[lethal dune]

this

[next obsidian]

[2048] x [2048] to [2048]

Oh wait yeah

You’re right

Lmao

[rustic crown]

G x G --> G
should be like |G|^{|G x G|}

[next obsidian]

Yeah

[next obsidian]

Yeah this is fucked

I guess it means there’s a lot more groups compared to abelian groups for that

Along with some other fucked shit

[lethal dune]

yeah this makes more sense

[next obsidian]

Well anyway

💤

[rustic crown]

good night! 💤

[lethal dune]

[next obsidian]

Eevee boothy

[rustic crown]

[small karma]

@rustic crown test turned out to be easy lol 😂 but f i didnt do 1 half question. Read it afterwards

1st part was hard so i skipped that ques

And after submitting i noticed that i could have done the other half of that ques

But rest was easy

will do far better in mid sem

[rustic crown]

[lethal dune]

btw are left units i.e. a*u = e are also right units? i.e. u.a=e or maybe some other element v.a = e?

in a ring

[tribal moss]

Consider the ring of linear operators on real sequences. The shift-left operator is a left unit, but not a right unit.

[lethal dune]

nice

this is one misconception I had for ages

[hidden haven]

with non commutative unital rings there is a nice thing you can use for intuition and coming up with examples

every rings embeds into End(A) as a subring, where A is an abelian group

Similar to Cayley embedding for groups

Here you could use a free abelian group on N many generators, and take the endomorphism that sends each to the next

same idea as the shift example

And maps can be easier to think about

because you might already know that injective maps are left invertible and surjective maps are right invertible

[lethal dune]

[hidden haven]

so you just need an injective non surjective endomorphism of an abelian group to exist

[lethal dune]

nicee

[tribal moss]

Cool.

[lethal dune]

so for zero divisors, it's also false?

[rustic minnow]

How do I interpret the notation $\mathbb{Z}/12\mathbb{Z}$ ? Is this Z_mod12 with its corresponding cosets?

[cloud walrus]

Fredrikpiano
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[hidden haven]

yes

take a surjective endomorphism with non trivial kernel

in a free abelian group

[lethal dune]

ah yes right

[tribal moss]

Yes. 12Z is the set of all multiples of 12, which is an ideal you can quotient out.

(In general you'll sometimes see the notation aR for the principal ideal of R generated by a).

[hidden haven]

lol interesting that the same counterexample worked for both your questions

[rustic minnow]

@tribal moss what do you mean by quotient out?

[tribal moss]

It's an instance of the general concept of a quotient ring. If you haven't learned that yet, just take it on faith that it means modulo-12 arithmetic.

[rustic minnow]

ok, thanks. The def of an ideal makes sense when seeing this notation in the context of kernels as we know that ker(\phi) is closed.

[granite nymph]

The set of automorphisms of any object in any category forms a group, and it's easy to see that in fact all groups arise in this way. Similarly, the set of endomorphisms of an abelian group forms a ring. Do all rings arise in this way?

[hidden haven]

lol were you reading chat or is this just a coincidence

Any (possibly non commutative, but unital) ring R embeds into End(R^op, +)

I am not sure if you can make it an equality

and I write (R^op,+) because that might make the proof easier to see

[rustic crown]

maybe the correct equivalent is to look at End of preadditive categories

[hidden haven]

but (R^op, +) = (R, +) since we are forgetting the multiplication anyway

[hidden haven]

lol then it just becomes trivial

[tribal moss]

My gut feeling is that something like the free ring on countably many generators might be a counterexample to equality. A group with sufficiently many independent endomorphisms could end up having continuum many of them in total.

[hidden haven]

damn I guess embedding is the best we can probably do

[hidden haven]

yeah that seems like it would work

[rustic crown]

yea... i was comparing it to the group situation we just do the 1 object category with morphisms as G
similarly 1 object preadditive category with morphisms as R

[hidden haven]

Yeah

Are all finite cardinalities achieved by End(A)

they aren't for Aut(S)

but here it seems like they would be

Is the totient function surjective onto N?

[rustic crown]

isn't totient like always even

[hidden haven]

oh is it

[rustic crown]

except like 1, 2

[hidden haven]

idk I took a number theory course from a post doc 🤡

[rustic crown]

lol

[hidden haven]

Then maybe we can try and prove that like Z/3Z isn't End(A) or something

nvm it obviously is

for klein 4

I think

no its not

[rustic crown]

wait how do you see it so fast

[hidden haven]

i was thinking Aut(klein 4) lmao

and even then I did it wrong

very wrong

lmao

[rustic crown]

lol

also i think End(Z/nZ) = Z/nZ

so totient is aut again

[tribal moss]

Yes.

[hidden haven]

lol F

[granite nymph]

coincidence. I've been thinking about this for a while now

[granite nymph]

This is interesting! I'll get my hands dirty and try to prove this.

[tribal moss]

Actually, rather than a free ring it might be more promising to start with something like (the algebraic closure of?) Q with countably many transcendentals adjoined. Since it's commutative it ought to put pretty tight constraints on what the group and its endomorphisms can look like.

[lethal dune]

is there a way to identify Aut(U(n))?

U(n) has order phi(n), but the structure is not known in general I think

[tribal moss]

What is U(n) here? It doesn't look like the unitary group ...

[lethal dune]

multiplicative group of numbers relative prime to n

(Z/nZ)*

[cyan marten]

Its structure is known and follows from the Chinese remainder theorem applied to the ring Z/nZ, and then we can use the fact that Aut(A x B) = Aut(A) x Aut(B) when A, B have coprime orders.

[lethal dune]

ok so $$U(n)=(\mbb{Z}/n\mbb{Z})^{\times}=(\mbb{Z}/(p_1^{k_1}p_2^{k_2}\cdots p_n^{k_n})\mbb{Z})^{\times}$$

[lethal dune]

now I apply CRT to get

[lethal dune]

what's next?

for each factor, I know I have a group of order $p_i^{k_i-1}(p_i-1)$

[cyan marten]

Okay, and these are cyclic

[lethal dune]

they are?

[cyan marten]

Yes! Well, unless p = 2

Or something

[lethal dune]

it's not obvious to me really

[cyan marten]

It's not supposed to be obvious

It was proved by Gauss that (Z/nZ)* is cyclic iff n = 1, 2, 4, p^a or 2p^a for an odd prime p.

[lethal dune]

ok i'll look up the proof later,

I think it's related to primitive roots of Z/nZ

I'll have to check that chapter again

[cyan marten]

At any rate, my argument falls apart because the orders aren't prime

[past temple]

okay so im having some trouble understanding

the concept of a biproduct

[cyan marten]

Aren't coprime*
I don't even know how to compute something like Aut(C2 x C4)

[past temple]

in the screenshot, why is it true that if the identity on the last line holds, then the A1 oplus ... oplus An is both a product and coproduct?

[lethal dune]

[hidden haven]

Does first 2 conditions here refer to p_m i_n = δ_mn?

ok yes found the source of the screenshot

You can verify the universal property of the product or the coproduct

Suppose you have maps f_k: X → A_k

Then you can get a map (sum i_k f_k): X → direct sum

check using those equations that this is the unique map making everything commute

[hidden haven]

Think about vector spaces

[past temple]

yeah

i managed to figure it out though haha

i constructed the morphisms which would satisfy

the universal property of the product

and the universal property of the coproduct

and then the last line in the screenshot

guarantees that those morphisms are unique

which makes F(A1 oplus ... oplus An) the direct sum

[hidden haven]

right

[past temple]

it took me so long to figure out but its so satisfying

putting all the universal properties together like that

[hidden haven]

true 😌

[lethal cipher]

Also true

[barren sierra]

@sturdy marsh remember that problem you helped me with

so I showed the counterexample to my prof

and he said "The issue is that the statement of the problem assumes implicitly that under the isomorphism $A$ with $I \oplus J$ the ideal $I \oplus { 0 } \subset I \oplus J$. This is not satisfied in your 'counterexample'. "

[cloud walrus]

Spamakin🎷

[barren sierra]

I don't get how that is implicitly understood from this

[hidden haven]

If you read the direct sum as an internal direct sum then it is

But it's bad wording

They should then have said A is equal to that not just isomorphic

[hot lake]

it's not even true as additive groups

and I may not even have a unit

[wooden ember]

What’s meant by multiplication map here?

[next obsidian]

It’s an algebra

So it’s the ring multiplication

[wooden ember]

I’m dumb I was thinking of a vector space

[next obsidian]

The point is just that if you take the elements which are homogeneous (as in it looks like (0,…,n,…)

And multiply those

It goes up in degrees properly

[wooden ember]

Yeah I get it

I just misread lol

[next obsidian]

[barren sierra]

how though

I guess I'm not clear as to what internal vs external direct sums are

I was never taught a distinction, just "yea this is a direct sum"

[sturdy marsh]

that's precisely what liquid, kxrider and I asked yesterday

if it was an internal direct sum

"implicitly assumes that it was an internal direct sum" lol

the first part involved an external direct sum

[hidden haven]

Lol yeah and they say isomorphic not equal

Then you definitely assume external

[sturdy marsh]

as A/I is not a subthing of A in a natural way

so I + A/I cannot mean internal direct sum

[hidden haven]

Internal is like everything can be written uniquely as i+j where i is in I and j in J

And this + is the + of the ring

So this addition is happening internally

[sturdy marsh]

or in other words, the iso I + J ----> A is induced by the inclusion maps

did you figure out the M \simeq M + M thing @barren sierra

[barren sierra]

yea I used the ring you came up with yesterday

[sturdy marsh]

yup!

[barren sierra]

but yea ok

so it's internal

how does that imply that 0 is in J

[sturdy marsh]

?

[barren sierra]

for what my prof said

I + {0} subset I + J

[sturdy marsh]

now that the direct sum is internal, sigma^{-1}(I) = I

which is what we needed

[hidden haven]

J is an ideal so contains 0

[barren sierra]

yea

[hidden haven]

[barren sierra]

wait wait wait

how do we know J is an ideal?

J is an ideal?

that would make things easier yea

[sturdy marsh]

wasnt it assumed to be an ideal

[barren sierra]

no

[hidden haven]

[barren sierra]

J is just some set

[hidden haven]

First sentence of the problem

[barren sierra]

as is written in the problem

I is an ideal

[sturdy marsh]

then direct sum doesnt make sense...

[barren sierra]

J is

¯_(ツ)_/¯

[hidden haven]

Bruh

[sturdy marsh]

ask your prof what I + J means then lol

[hidden haven]

It says ideal

[sturdy marsh]

they said I is an ideal

[hidden haven]

oh

[sturdy marsh]

anyway, assuming J is a ring (possibly w/o identity), you can prove that it's an ideal

it eats all elements of the form (i,0)

[barren sierra]

hm ok

I guess I'll prove J is an ideal cause I've never seen that before

[sturdy marsh]

if J is a ring, (0,a)(0,b) is in J

[barren sierra]

yea (0, ab) in J

hm ok

but yea that solves it

poorly written question imo >_>

[sturdy marsh]

very poorly written

[hidden haven]

Wrongly written

[sturdy marsh]

the response is also

[hidden haven]

🙈

Brofib old pfp when

[sturdy marsh]

when I stop being

[barren sierra]

wdym the response is

[hidden haven]

I look at this one and think terra

[barren sierra]

lol

[sturdy marsh]

"implicitly assumes"

[barren sierra]

oh

yea >_>

I have mixed feelings about my prof lol

[sturdy marsh]

they should have just said that it's wrong as stated

and corrected it

[barren sierra]

I got points off on a midterm for not taking the most straightforward approach

which yes cost me time but it was correct >_>

[hidden haven]

[barren sierra]

hopefully next sem's algebra class goes better for me

[sturdy marsh]

what's the next sem on

[barren sierra]

the prof is a goat so I'm excited

uh

[sturdy marsh]

who

[barren sierra]

https://faculty.math.illinois.edu/~rezk/course-description-500-sp22.pdf

Rezk

[sturdy marsh]

ah

[barren sierra]

alot of people at my school like him

so I'm excited

[sturdy marsh]

yeah he seems pretty cool

[barren sierra]

class seems really small only like 10 people

[sturdy marsh]

his notes on quasicats are really good

[barren sierra]

also 50% HW

[sturdy marsh]

isnt this intro algebra?

[barren sierra]

this is the grad version

[sturdy marsh]

interesting

[barren sierra]

undergrad is much larger

[sturdy marsh]

only 10 people

[barren sierra]

oh nah, 20

just checked

[sturdy marsh]

are there a lot of "algebraic" grad students at uiuc

[barren sierra]

¯_(ツ)_/¯

ngl

I know 0 math grad students

they are an enigma

[sturdy marsh]

if there arent a lot of them

[barren sierra]

I've met CS grad students through my courses

[sturdy marsh]

I might get in

[barren sierra]

👀

pull up

[sturdy marsh]

im applying to uiuc

for grad skool

[barren sierra]

nice nice

my only interaction with math grad students are they grade my HW

the CS classes seem much better about student TA interaction

but I've never met any of my math TAs

[sturdy marsh]

that's weird

[barren sierra]

yea

which sucks cause I'd like to talk to some math students about what I want to do for grad school 🤡

[sturdy marsh]

talk o your profs

[barren sierra]

yea

[sturdy marsh]

grad students probably dont know a lot about admissions

[barren sierra]

not admissions

just like

what to do

my algos TA has been a godsend in being able to try to figure out what I want to do

[sturdy marsh]

pick skool

fill out application

do the gre

[barren sierra]

skool depends on what field I want to do lol

[sturdy marsh]

oh youre trying to figure out what kind of math

[barren sierra]

I'm debating between theoretical CS and Math

kind of math I'm kind of set on something algebraic

which is hella broad

[sturdy marsh]

noice

[barren sierra]

but I do not know enough to say anything more

[sturdy marsh]

did you do analysis yet?

[barren sierra]

no

that's a next fall problem

from what I've self studied I greatly dislike it lol

it's very convoluted to me

[sturdy marsh]

you should probably do an analysis class before you decide lol

[barren sierra]

yea

I have time

I'm only a sophomore

so time is on my side

[sturdy marsh]

yeah

[barren sierra]

I could see my self doing CS stuff tho considering my research is algos rn

[sturdy marsh]

I had the opposite opinion right after I did some algebra and analysis

algebra seemed to have no substance

[barren sierra]

to be fair I am comparing taking a nice algebra course with self studying analysis

[sturdy marsh]

and then I read about the weil conjectures

[wooden ember]

what are the A_k's here?

[next obsidian]

It probably is defined above

But it looks like it should be wedge powers

[wooden ember]

alternating k-linear functions i forgot

[barren sierra]

just a quick sanity check

if I want to determine whether a group is abelian or not

if I have even just one example where x * y =/= y * x

it's not abelian right?

[sturdy marsh]

yes

[chilly ocean]

possibly

[barren sierra]

smh why did my prof take points off for that

he really takes off points for the dumbest shit

[chilly ocean]

what happened

[barren sierra]

4b

So I said my field is Z_3

showed everything has order 3 for the matricies

by BSing a diagonalization argument 🤡

but I found 2 specific matricies

that don't commute

but this was the answer key

so he showed it more generally, fine whatever

but like

_>

I hate this class sometimes

like

idk man

[frank lake]

why is G abelian?

pls help

😦

[tribal moss]

That's a very straightforward verification using the composition formula already provided in the image.

Start with stating what "G is abelian" means in terms of the T_a,b.

[subtle ivy]

a hopefully not too obvious hint is that addition is commutative.

[tribal moss]

<insert video clip of Tom Lehrer triumphantly declaring: "... because addition is commuative!">

[frank lake]

I got it resolved

tysm

😂

[frank lake]

would a working proof be something like: let a be the inverse of b, this implies that $ba = ab = e$, which implies that $a^{-1}e = a^{-1} \ne b$

[cloud walrus]

ChubbyMuffins

[frank lake]

therefore, there exists a group such that $a^{-1}ba \ne b$

[cloud walrus]

ChubbyMuffins

[frank lake]

idk 🤣

[chilly ocean]

why should a^{-1} not be equal to b?

i think they want you to give a specific example of a group here

[frank lake]

how would I construct the group?

Just get an inverse and an identity?

I don't get it 😂

what if ba is e

the identity element is ba, and the inverse of a is a^{-1}

so would that work?

[frank lake]

I mean according to the associative property, $a^{-1}ba \implies (a^{-1}a)b \implies eb \implies b$

[cloud walrus]

ChubbyMuffins

[frank lake]

I don't get how I'm supposed to construct a group that contradicts this

[chilly ocean]

are these \implies supposed to be ='s?

[frank lake]

?

yeah

[chilly ocean]

so in the first step you assumed that ba = ab, right?

that's not always true

[frank lake]

is it because not all groups are abelian?

[chilly ocean]

yes

so try looking at some non-abelian groups

you might find an example there

[frank lake]

hm

I have an idea

the dihedral group under composition?

would that work?

[chilly ocean]

probably

[frank lake]

I don't know much groups that aren't abelian except this one 😂

[chilly ocean]

GL(n, R) for n >= 2 works

that's a fun one

[frank lake]

GL(n, R)?

[chilly ocean]

invertible n by n matrices

[frank lake]

oh

I hate linear algebra 😰

[chilly ocean]

unfortunate

[frank lake]

😂

[waxen hedge]

S_3 is the smallest non abelian group !

[frank lake]

S_3?

[waxen hedge]

I have not read everything above but that might help

Permutations over {1, 2, 3}

[frank lake]

ahh I see

[waxen hedge]

It's the smallest with respect to cardinality (6 here)

[frank lake]

damn

man my braincells are dying

[frank lake]

I assumed ab = ba which isn't always true, so what do I change

[waxen hedge]

Well this doesn't work, since, if a is a inverse of b then a=b^-1 so a^-1=b

[waxen hedge]

a^-1 b a = b is equivalent to ba=ab so you just need to find two elements which doesn't commute

[frank lake]

yeah

that's what I'm struggling on

[waxen hedge]

Whats the context of this question ?

[frank lake]

I gotta give an example of a group with elements a and b with the property a^(-1)ba is not equal to b

so I have to find a group that isn't abelian

and prove that it satisfies this property

[waxen hedge]

Ok, then I might have given you one above

[frank lake]

but like how do I use it to prove the property?

do I just put in the group and because it's non-abelian it satisfies the property and no further proof is needed?

[waxen hedge]

You can pick two elements a and b in this group
Compute ab and ba
And if you choose them well you will have ab≠ba

[frank lake]

I don't have to manipulate the elements or anything like that?

[waxen hedge]

Yes you need to

[frank lake]

I see

ohhhhhhh

I think I kind of get it

lemme try

[waxen hedge]

Because, even if the group is not abelian, you still can have ab=ba for some a and b
You just know that there exists a and b such that ab≠ba

Look at the neutral element e
Then for every x in the group, you have xe=ex=x

[frank lake]

yeah

[waxen hedge]

I have to go, I hope you will understand !

[frank lake]

ok

ty

[night hazel]

Hey, can anybody help me with this?
I want to find the largest number $c$ such that $\mathbb{R}^{3 \times 3}$ (the vector space of all $3 \times 3$ matrices with real entries) has subspaces $S_1, S_2, S_3$ with $S_1 \cap S_2 \cap S_3 = {0}$ and $dim(S_1) = dim(S_2) = dim(S_3) = c$.

How do I prove that (to maximise $c$) $S_1 \cap S_2 = {0}$ $S_2 \cap S_3 = {0}$?

[cloud walrus]

miku

[night hazel]

here is some more info https://math.stackexchange.com/q/4330525/1003675

[viscid pewter]

that's not necessarily true

that last statement

[night hazel]

I'm not really sure how to find the largest possible $c$ then

[cloud walrus]

miku

[night hazel]

My idea was to show that $S_1 + S_2 + S_3$ is a subspace of $\mathbb{R}^{3 \times 3}$ and thus has a dimension of at most $dim(3\times 3) = 9$

and then show $dim(S_1 + S_2 + S_3) = dim(S_1) + dim(S_2) + dim(S_3) = 3c$, which implies $c$ is at most $9$

However (my proof for) $dim(S_1 + S_2 + S_3) = dim(S_1) + dim(S_2) + dim(S_3)$ relies on that statement being true

[cloud walrus]

miku

[past temple]

so this guy says that its a well known result

that F(M) is isomorphic to M (x) F(A)

but why is that true

ok a more specific question

what does the author mean here by

"F is entirely determined by F(A)"

and how does it follow from the exact sequence?

[subtle ivy]

here's an idea: so since $\mathbb{R}^{3\times3}$ is isomorphic to $\mathbb{R}^9$, we may choose an orthogonal basis $e_1,\ \dots,\ e_9$ with respect to the standard inner product on $\mathbb{R}^9$. then, for any $c\leq 4$, we have that $9-(c+1)\geq c$, so we may set $V_1=$span$(e_1,\ \dots,\ e_c)$, $V_2=$span$(e_1,\ \dots,\ e_{c-1},\ e_{c+1})$, and $V_3=$span$(e_{c+2},\ \dots,\ e_{2c+1})$. since you can choose a basis of $V_3$ which is orthogonal to the bases of $V_1$ and $V_2$, their intersection is 0, and they clearly all have the same dimension. you can check that it fails for $c=5$, and so it fails for all values of $c\geq 5$

[cloud walrus]

shortcut

[subtle ivy]

something like this i think. i pinged the bot at first

[chilly ocean]

is $\text{Gal}(\bar{\mathbb{F}}_p/\mathbb{F}_p)$ uncountable

[cloud walrus]

wren

[chilly ocean]

yeah it is nvm

my galois theory instructor lied

not about Z_p being uncountable but about which elements in it don't represent ordinary integers

[delicate bloom]

what'd they say?

[chilly ocean]

she disguised the inverse of the frobenius automorphism to look like something that "can't be a power of the frobenius automorphism"

it was only considering the extension of F_p to contain all extensions of degree 2ⁿ

so basically she was like "take a map such that the restriction to each F_(p^(2ⁿ)) is frobenius^(1+2+...+2^(n-1))"

but that's just frobenius^(2ⁿ-1)

and frobenius^(-1) has all of those same restrictions

[delicate bloom]

well 1+2+...+2^(n-1) = -1 mod 2^n so I don't see the problem

in Z_2 you have that 1+2+2^2+... = 1/(1-2) = -1 still

[chilly ocean]

there is no problem

that's the thing

my instructor said there's a problem

[delicate bloom]

I guess I had to be there I didn't follow which part was the lie lol

[chilly ocean]

that the automorphism you get from the described restrictions is not a power of the frobenius map

[delicate bloom]

are we talking about F_(2^n) or the algebraic closure of F_2?

-1 is not a positive integer so there's no way to get it by repeatedly applying the frobenius map a finite number of times

[chilly ocean]

I'm talkimg about the union of degree 2ⁿ extensions of F_p

[delicate bloom]

whoops that's what I meant not alg closure

[chilly ocean]

also yeah it's not a positive integer but it's still the inverse of the frobenius map

when I say "power" I mean basically an element of <frob>

[delicate bloom]

oh I see, your professor should have motivated it with something more interesting

[chilly ocean]

so my instructor basically said frob^(-1) is not in <frob> 💔

[delicate bloom]

like -1/3 or something which is more obvious you can't get

[chilly ocean]

yes

I later figured that out by playing around

[delicate bloom]

ok I gotcha lol

[chilly ocean]

but -1 wasn't even a correct choice to pick was it?

[delicate bloom]

what do you mean

oh just any integer is a bad choice to pick

[chilly ocean]

like to find an automorphism that isn't in <frob> don't take frob^(-1) haha that's what I mean

[delicate bloom]

because the integers are generated by the cyclic group <frob>

[chilly ocean]

yes

[delicate bloom]

that's why I said -1/3 would be a good example

[chilly ocean]

besides non integers are way cooler to think about anyways 🥱

[delicate bloom]

cause -1/3 = 1/(1-4) = 1+4+4^2+4^3+...

[chilly ocean]

yeah

-1/3 is actually the first example I thought of hahaha

[delicate bloom]

or even $\sum_{n\ge 0} n!$ why not

[cloud walrus]

Merosity

[chilly ocean]

yeah that's works but it's harder to think about explicitly

at least for me

[delicate bloom]

I guess a simpler example might be sqrt(-7)

since that's clearly not rational

[chilly ocean]

I'm a fan of the rationals still

cause they are nice and have repeating patterns

[delicate bloom]

point is there is more to Z_2 than just rationals without 2 in the denominator

[chilly ocean]

yeah

I'm not that comfortable with it yet

[delicate bloom]

I guess as far as automorphisms are concerned, you can just randomly pick a sequence of 0s and 1s and put those on powers of 2 to get an arbitrary automorphism

[chilly ocean]

yessssss that's what I did

that's the exact thing I wrote down 😋

[delicate bloom]

so don't need to worry too much about it too much past that I suppose

[chilly ocean]

yeah

thanks

[delicate bloom]

yup

[lethal dune]

thanks everyone for helping me

I did well in my exams

@hidden haven you know what, the thing we discussed about free groups and not being injection but subjection was asked in the exam

[hidden haven]

[lethal dune]

[hidden haven]

what was the problem?

[lethal dune]

it was something of a T/F question

there is a surjective homomorphism from a ring to a ring which is not injective

[hidden haven]

interesting question for an exam lol

[lethal dune]

yeah, they do set some quality questions

specially linear algebra

[hidden haven]

Nice

[lethal dune]

another was about a group of order p³ which we discussed a day before that

😛

[hidden haven]

I think you need to be investigated

[lethal dune]

yeah those discussions helped me a lot

[lethal dune]

I wasn't able to answer this one, let K be the cantor set, then there is an injective homomorphism from C([0,1], R) to C(K, R)

T/F

[hidden haven]

There is a continuous surjection from K to [0,1] right

I hope I am not recalling this wrong

but that gives the injective homomorphism almost immediately if you know about hom functors

[lethal dune]

i don't know, is there?

these are category theory terminology?

continuous subjection from [0,1] to K?

[hidden haven]

Nah I am just saying if you are used to cats then there's motivation for this

[lethal dune]

left one is connected and right one is disconnected

oh it's from K to [0,1]

[hidden haven]

Let p be a surjection

Ye

Then an injective map as defined by

f ↦ fp

Precomposition by p

[hidden haven]

But if you don't have this fact then idk

[lethal dune]

ok then it's obvious

but is there a nice description of the surjective map p?

or is it like some ugly shit

like binary expansion?

to real number??

[cloud walrus]

Moldilocks ✓

[lethal dune]

I was close

[cloud walrus]

Moldilocks ✓

[lethal dune]

I thought first changing 2 to 1 in the expansion then converting the binary to decimal

oh it's the same one

[hidden haven]

Oh ye

Numerator should have e_i/2

[lethal dune]

now it looks like I could have answered that one as well

[hidden haven]

I think

Lol

[lethal dune]

nice

[upbeat juniper]

please send more of them lol. they're fun to think about

[paper flint]

What exam is it?

[cunning cargo]

Hey all! Here' s Wikipedia's definition of the ideal class group. I've seen this definition elsewhere but with an equivalence relation on regular ideals of R, not fractional ones. See https://math.stackexchange.com/a/296094 for example. Are these definitions equivalent, or is Wikipedia's definition wrong here?

[untold cloud]

Hi, guys, is there anything will go wrong if I have a set of left coset of N, {g_1 N, ..., g_n N}, but N is not a subset of {g_1, ..., g_n}

[barren swift]

if H is a normal subgroup of G and g in G has order m, are the only possibilities for |Hg| in G/H 1 and m?

[untold cloud]

Z/3Z, 1 has order infinity in Z, the order for 1+3Z will be 3?

[barren swift]

i meant if g has finite order m

[untold cloud]

maybe it is not true, because g^m = e, but perhaps g^k\in H, where 1<k<m

[hidden haven]

G = Z/4Z, H = <2>, g = 1 should work

[untold cloud]

for example, G={1, w, w^2, w^3}, then H={1, w^2} is a normal subgroup, but w has order 4 in G, but wHwH=w^2H

[hidden haven]

lol same example

[untold cloud]

lol

[hidden haven]

I don't get it, what are you trying to do with this set?

[untold cloud]

I actually want to show that if N is normal in G and G_1 is a maximal normal subgroup of G containing N, then G_1/N is maximal normal subgroup of G/N

[hidden haven]

This should follow from the correspondence theorem

normal subgroups of G containing N are in an order preserving bijection with normal subgroups of G/N

[untold cloud]

So i write it as this, but then i realise that is N is not a subset of K, then i cannot see any contradiction

[hidden haven]

definition of K is bad

it should be union of g_iN

[chilly ocean]

Sorry question 15b I'm a bit confused

[untold cloud]

So i should define K as the group of equivalent class of g_i?

[hidden haven]

as the union of equivalence classes

It is not a group as defined because you have a lot of freedom in choosing the g_i's

but I think you mean take all possible g_i's

but as written, it is wrong

[untold cloud]

sry, i see, i mean set....

[hidden haven]

ye but you want it to be a subgroup

not just a subset

[untold cloud]

yeah, I think i know it, thanks!

[tardy yacht]

Hey, I am studying automorphism groups, and I am struggling to come up with a way to compute them for integer rings like $\mathbb{Z}[\sqrt[4]{3}]$. Any tips?

[cloud walrus]

pastalover69

[chilly ocean]

You need to check what 3^(1/4) can map to, automorphism will have to be an identity on Z

[tardy yacht]

I don't need to consider the other roots (1/2, 3/4)?

[lethal dune]

tifr

[lethal dune]

after j wake up

[barren sierra]

So I've shown that no matter what the kernel of $\phi: K[x] \to K[x] / \langle q(x) \rangle \oplus K[x] / \langle d(x) \rangle$ is $\langle p(x) \rangle$. So this reduces down to showing $\phi$ is surjective if and only if $\gcd(q(x), d(x)) = 1$. I am struggling to show that if $\gcd(q(x), d(x)) \neq 1$ then the function is not surjective. Any tips?

[cloud walrus]

Spamakin🎷

[barren sierra]

In particular I want to show there is no $f \in K[x]$ such that $\phi(f) = (1, 1)$

[cloud walrus]

Spamakin🎷

[barren sierra]

At least that's my idea

I mean if some $f(x)$ existed like that then we would know that $q(x) \mid f(x) - 1$ and $d(x) \mid f(x) - 1$.

[cloud walrus]

Spamakin🎷

[barren sierra]

but I'm not sure where to go from here

or is there a better way to do this whole problem?

[lethal cipher]

No, because they are not part of your ring

[past temple]

given A = Z[sqrt(-5)] and a mapping pi: A x A -> (2, 1 + sqrt(-5)) given by pi(a,b) = 2a + (1 + sqrt (-5))b

how do i show that (2, 1 + sqrt (-5)) is isomorphic to ker pi ?

[tribal moss]

Isomorphic in which sense? Just as an additive group?

[past temple]

isomorphic as A-modules i think

@tribal moss

[tribal moss]

Okay, makes sense.

[past temple]

so..

im pretty lost on how to even start so

any help would be rly appreciated!

[tribal moss]

Sorry my algebraic number theory knowledge seems to be too rusty to be of help.

[past temple]

@hidden haven

[barren swift]

what does this show, though? w^2H isn't equal to H, and so the order of wH isn't 2

consider G = Z12 x Z12. let H = <(2, 2)>.
how do i find the order of H + (5, 8) in G/H?

i know the order has to be divisible by the order of (5, 8) in G

but im not sure what to do

or rather, not divisible by

but has to divide that order

i know the order isn't 1, because (5, 8) is not in H

is there any way i can determine the actual order though?

am i just going to have to do computation until i reach H again

or is there a better way

[lofty trout]

In $R$ a commutative ring, when is $R_p$ a free $R$ module?

[cloud walrus]

glittersparkles

[tardy yacht]

If they weren't part of the ring, multiplication wouldn't be closed (3^1/4 squared is 3^1/2). But I see that they are determined by what 3^1/4 maps to

[thorn delta]

@barren sierra did you figure out your problem?

you're on the right track except (1,1) is not the point in the image you should care about. Rather, i think its || (1,0) and/or (0,1) ||

[barren sierra]

Yea I got it

But ye

Thanks @thorn delta

ohhhh 0 or 1 yea yea

Cause I want it to divide either or

[thorn delta]

@barren sierra Sorry for the ping, but just to offer an alternative perspective: if there exist elements $f,g$ such that $\phi(f) = (1,0)$ and $\phi(g) = (0,1)$, then given any elements $r,s \in K[x]$,
\begin{align*}
\phi(rf + sg) & = \phi(r)\phi(f) + \phi(s)\phi(g) \ &= (r + (q(x)), r + (d(x)))(1,0) + (s + (q(x)), s + (d(x)))(0,1) = (r + (q(x)), s + (d(x)))
\end{align*}
so the whole map has to be surjective.

[cloud walrus]

kxrider

[barren sierra]

Damn that's nice

[paper flint]

Ah, okay.

[barren swift]

let G be a group and H be a normal subgroup of G such that [G:H] = 20 and |H| = 7. consider an element x in G such that x^7 = e. how do i show that x is in H

my first thought is that every non-identity element of H has order 7

but that doesnt get me anywhere

[thorn delta]

do you know lagrange's theorem?

for paths' problem? That's ridiculously overkill

[barren swift]

i do know lagrange's theorem

can you give me a hint as to how that applies?

[thorn delta]

hmm my initial idea doesn't work

[barren swift]

https://math.stackexchange.com/questions/4140294/let-g-be-a-group-and-h-unlhd-g-such-that-gh-20-and-h-7-suppose-x

i can't seem to understand what the last answer at the bottom is trying to say

how could ord(xH) = 1? im assuming xH means the left coset of H by x

but all left cosets have order 7

[thorn delta]

oh yeah of course lol...

[thorn delta]

e is always a solution to x^7 = e.

[barren swift]

yeah

so if x = e then trivially its in H

[thorn delta]

no, the can have order 1, 4, 5 or 20

by lagrange ^

and since those are the only possibilities, x = e

[barren swift]

a coset of a subgroup can have a different order than the subgroup itself??

[thorn delta]

oh, by order, we mean the order of the coset as an element of the group G/H, i.e. the size of the subgroup <xH> of G/H

not the cardinality of the set xH

[barren swift]

oh

[thorn delta]

so the whole idea is that since H is a normal subgroup, you can apply lagrange's theorem to the group G/H

[barren swift]

ahhhhhh

i seeeee

thanks!

[thorn delta]

npnp

[barren swift]

for x in G such that x has finite order m, will every subgroup of G contain x raised to some power that divides m? if so, is there any intuition for why this is true?

[next obsidian]

This mf said id for the identity element of a group 🤣

[barren swift]

let G be a group and let H be a subgroup of index 2. im supposed to show that for every a in G, a^2 is in H. but i dont understand how this can possibly be true... if we square every element in G, we have |G| elements, but |H| = |G|/2

so why isnt that false?

[next obsidian]

G/H has order 2

Either a is in H so it’s identity in G/H anyway

Or it’s not, then a^2 is identity in G/H so it’s in H

[barren swift]

why is a^2 the identity in G/H?

oh, i think i see it

no, i dont get it

[barren swift]

why is this true?

[next obsidian]

Because if it wasn’t then G/H has more than 2 things

It can’t be a

If it isn’t identity

Then it’s a third thing

[barren swift]

ahhhhh

that makes sense

[next obsidian]

More generally

[barren swift]

but i still dont get how what i said earlier is wrong

like

[next obsidian]

If |G| = n

Then a^n = e

Always

Because it isn’t true that if we squares everything we have |G| things

[barren swift]

oh

[next obsidian]

Think about like

Z under addition

Squaring is actually multiply by 2

Wel okay wait

That’s bad

Since that’s the same cardinality cuz infinite lol

Take Z/4Z

Under addition

Squaring = double

That’s not everything

[barren swift]

im confused

Z/4Z is a group of sets

how is squaring doubling?

oh

do you mean squaring 4Z in Z/4Z? because that's still 4Z

[viscid pewter]

0Z + 0Z = 0Z
1Z + 1Z = 2Z
2Z + 2Z = 0Z
3Z + 3Z = 2Z

so it doesn't cover everything

[barren swift]

ahhh

yeah ok i see now

thanks

[past temple]

given A = Z[sqrt(-5)] and a mapping pi: A x A -> (2, 1 + sqrt(-5)) given by pi(a,b) = 2a + (1 + sqrt (-5))b

how do i show that (2, 1 + sqrt (-5)) is isomorphic to ker pi ?

as A-modules

[chilly ocean]

2a+(1+sqrt(-5))b=0

a,b that satisfies this are in kernel of pi

you can expand out a and b as such,
a = x + ysqrt(-5) and similarly for b

then you can multiply and combine like terms, and it may become easier to see for which values of a and b that pi(a,b)=0

[final pasture]

You first need to talk about why G/H is a group tho, to do that, no ?

[next obsidian]

I don’t care

[final pasture]

[next obsidian]

You can probably do it manually without passing to a quotient

And I don’t wanna justify it even tho it’s as simple as looking at gH and Hg

[final pasture]

Wait

Is G finite

If it is, we can do something simpler

(just considering the translation map)

It's not 😔

[past temple]

i did this and it rly didnt seem to get me any closer to reasoning about why ker pi is isomorphic to I

im now attempting to construct a mapping from ker pi to A whose image is I

[chilly ocean]

dont matter

what is C/R as a group under addition tho

[final pasture]

I'm saying

if G is finite

we can do a simpler proof

[chilly ocean]

does that make sense lol

wait it doesnt

[final pasture]

it does make sense

it's iso to R

[chilly ocean]

3+4i goes to what tho?

4i?

yeah loo

under addition C is R^2 so this makes sense

[final pasture]

funkier: R^2 ~= C and R are isomorphic as groups, assuming choice

[chilly ocean]

lol I thought proof was that [G:H]=2 implies |G/H|=2 implies g^2 = e for g in G/H implies g^2= H in G/H implies g in H.

[chilly ocean]

thas odd to think about

[final pasture]

it is

[final pasture]

That works btw

I was just wondering if we could do a proof without talking about quotient group

[chilly ocean]

in terms of cosets only?

idk if index has meaning unless G/H is a group

you seem to know logic

what is term for that

where definitions dont apply to the object, is it just not well defined?

[final pasture]

it does make sense without G/H being a group

It's just the cardinality of the set of cosets

[chilly ocean]

o

ok

[final pasture]

And yeah ig ?

[chilly ocean]

which happens to have a group structure

but if we ignore that part

[final pasture]

Well, if the index of H is 2, indeed H is normal so G/H has a group structure, but yeah we ignore that

[chilly ocean]

lol what if we think in terms of group actions on cosets lol

lol xd :p

G x G/H -> G/H

group action is simply transitive and G/H has order 2

actually the group action doesnt need to be free

neither transitive lol

well nvm it does need to be transitive

if we define it like (a,gH) mapsto (ag)H

nah lol

doesnt work you need to use fact that G/H has group structure

[broken stirrup]

hi, that's an interesting fact but i can't come up with an example

[hidden haven]

Z/mZ x Z/nZ ≈ Z/mnZ iff m and n are coprime

Might be useful

[proud bear]

let H_1=Z/2ZxZ/2Z, H_2=Z/2ZxZ/2ZxZ/2Z, K_1=Z/2Z, K_2=(Z/2Z)^4

[hidden haven]

Lol that's a lot simpler

[proud bear]

kind of dumb but it works i think

[final pasture]

Yeah this

Any nontrivial finite group G works, with H1 = G, H2 = G^4, K1 = G^2 and K2 = G^3

[broken stirrup]

got it thanks

[next obsidian]

If A is an integrally closed domain, K its field of fractions, L a finite separable extension of K, and A’ the integral closure of A in L, why can we choose a basis of L over K consisting of elements in A’?

[hot lake]

Pick any basis and multiply its elements by all the denominators ?

[next obsidian]

Hmmmmmmmmmmm

Is Frac(A’) = L?

[sturdy marsh]

yes '

I thik so

[next obsidian]

I guess if you can choose the basis in A’

It implies that Frac(A’) = L

[waxen hedge]

Yes

[ancient rivet]

the p-addicts

[maiden ocean]

If R_p is free as an R-module then we have a basis E for R_p. now for any x1, ..., xn in E, r1 x1 + ... + rn xn = 0 implies the r_i all equal 0. But the x_i have the form a_i/s_i for s_i notin p so we can rewrite this sum as

$\frac{\sum r_i a_i \prod_{j \neq i} s_i}{\prod s_i} = 0$

[cloud walrus]

Kosher Nostra Chaya Moth

Kosher Nostra Chaya Moth

[maiden ocean]

so basically we can take $r_i = 0$ for $i \neq 1$ and $r_1 = 1$ and we see that all we need is some $t$ with $t a_1 = 0$. and since the $a_i$ can range across all the elements of $R$ (we only need to find some $r_1, \ldots, r_n$ not all 0 for our module not to be free) this means that if any element $R$ not in p is a zero divisor then $R_p$ cant be free

[cloud walrus]

Kosher Nostra Chaya Moth

[maiden ocean]

@lofty trout kind of long winded but basically the idea is that localizing introduces new relations between elements not contained in your prime ideal and being free necessitates having a generating set with no relations so unless you have some pretty specific criteria it wont work out

[lofty trout]

@maiden ocean excellent! Thank you for your help

[maiden ocean]

no problem

Note that an example that DOES work is when R is local, because then p is all the non-units

so elements of R not in p will not be zero divisors

[lofty trout]

Yes, I sent this to my prof and that’s the case where it’s true

[maiden ocean]

Which is good because if you localizing a local ring at its maximal ideal you recover the local ring

nice

[lethal dune]

is there a 'globalization' of rings?

[chilly ocean]

wtf a local ring

i forgot the condition

[rustic crown]

ring with a unique maximal ideal

[chilly ocean]

but fields are

one maximal ideal makes it a local ring

why the word local though?

whats the intuition behind the naming

[rustic crown]

so like when you do algebraic geometry, you can think of the set of prime ideas as a geometric object, so points here are prime ideals, and if you wanna understand "functions" which are elements of the rings near this point, you look at the stalk
the collection of these germs is a local ring

you can think similar to usual real numbers. if you look at germ of a function at x, then you basically know this function in a neighborhood of x, so like if this function is infinitely differentiable, then you would know all it's derivatives at x and stuff.
so the collection of these germs forms a ring and it has a unique maximal ideal corresponding to the germs which evaluate to 0 at x

i don't know a lot more about this tho

[hidden haven]

A more intuitive example might be manifolds

[rustic crown]

do non-commutative local rings have geometry in them?

[hidden haven]

If you want to think about local real valued functions around a point p in a manifold (or maybe just R^n) you'd define germs, which are functions defined on open neighbourhoods of p, and you identify 2 germs if they agree of the intersection of their domains (so germs are equivalence classes of pairs (U, f: U → R)). These form an R-algebra under pointwise addition and multiplication. There is also a unique maximal ideal of all the germs which are 0 at p, so this is a local ring

The AG example det gave is just a more 😵‍💫 version of the same kinda thing

[lethal dune]

germs

[barren swift]

whats the intuition for why there being an isomorphism between two groups means theyre "the same group" in a sense?

[waxen hedge]

You can "translate" everything through the isomorphism

[barren swift]

im having trouble seeing why, just because theres a bijection f such that f(ab) = f(a) * f(b) means theyre the same

[waxen hedge]

The inverse is also a morphism of group in this case

[rustic crown]

think of it like two languages, and f is the translation tool like Adrien says...
you'll say the two language are same iff f respects structure of both languages, right? i.e x and f(x) as words should mean the same thing, grammar should be respected and other stuff

[barren swift]

is there maybe an example you could show that seems intuitive?

[rustic crown]

you can consider all reals numbers under + and positive real numbers under *

[barren swift]

under what operations

ah

[rustic crown]

we have a nice map x --> e^x

it respects structure! x+y is sent to e^x * e^y

[barren swift]

what exactly do you mean by "structure"

[rustic crown]

so you can understand multiplicative things by taking logs and understanding additive things then take exponential again to finish

[rustic crown]

i'm using it kinda loosely, a set is just a nice collection of things, there isn't any "connections" between any two elements. it's rather arbitrary. but groups on the other hand have some extra "flavoring", you can take two elements and squish them together to get a whole new element

we want this "structure" which is an operation in this case to be reflected in the map

[barren swift]

can you show an example with {e, x, x^2, ..., x^n-1} and Zn ?

[paper flint]

Sure. Switch e with 0, x with 1,..., x^(n-1) with n-1 and your group behaves exactly the same way as Z_n

As has been suggested before, isomorphism between two groups means they're essentially the same things, modulo the way we "label" or "name" the elements-the underlying properties of the group and how the elements interact with each other is the exact same

[barren swift]

i get that thats what an isomorphism is supposed to show

[waxen hedge]

You could try to work out the isomorphism between Z/nZ and n-roots of unity

[barren swift]

but i dont get how showing that e^(x+y) = e^x * e^y shows that the operation/elements are just being relabeled

[paper flint]

Addition of x and y in (R,+) corresponds to multiplication of e^x and e^y in (R*,×), where I suppose R* denotes the set of positive reals

[barren swift]

oh

i think im beginning to see it now

[next obsidian]

So there’s something more you can say here

A localization of R is faithfully flat iff it’s actually just R

This shows R_p can’t be free unless R is local and p is maximal

[paper flint]

The group operation on the two elements in the first structure is translated to the group operation on the images of the two elements in the second structure

This is essentially what homomorphism is

If you can further ensure your map is bijective, you get an isomorphism

@barren swift You may find this very useful: https://math.stackexchange.com/questions/242348/intuition-on-group-homomorphisms/242370

[chilly ocean]

ok am satisfied with germ explanation

on isomorphisms tho

how many equivalence classes of groups are there

that are useful atleast

because there can be a lot, but idk if its useful to categorize this way

thats one thing ive wondered

[rustic crown]

what do you mean by equivalence classes?

[chilly ocean]

how many useful relations are there on algebraic objects

[rustic crown]

isomorphism classes? classes with same size? or something else...

[broken stirrup]

helo

any tips?

[chilly ocean]

is there an equivalence class of groups under the relation that there needs to exist atleast one nonzero morphism between them

so i guess the something else

[chilly ocean]

use definition of free group on X

[frank fiber]

i:X->F is the inclusion ?

[barren swift]

what if its only onto? what if its an epimorphism? what does that say about the two groups?

[chilly ocean]

i(X) generates can be interpreted in two different ways

brb going to aluffi

then the other interpretation is that any element in F can be written as a formal sum of i(X) elements

[paper flint]

That still gives you some data, but if your map is known to be not one-one, then some information about the structure is "destroyed", since the only way you can compromise injectivity in homomorphisms is by having a non-trivial kernel, i.e., sending "extra" elements to the identity in the second group. All data about them is lost in the process.

I'm more or less rewording that MSE post actually :p

[chilly ocean]

like what are examples of groups with |Hom(A,B)| = 2,3

[rustic crown]

i don't think this is transitive
Z/4Z --> Z/6Z --> Z/9Z
there is no non-zero map between Z/4Z and Z/9Z as they coprime

[chilly ocean]

so the composition of the maps is the zeromap?

[barren swift]

@paper flint so in {e, x, x^2, ..., x^n-1} i can say that the x corresponds to the 1 of Zn. but in the case where we had e^(x+y) = e^x * e^y, what corresponds to what?

[rustic crown]

it has to be

the image has size dividing both 4 and 9

[chilly ocean]

yeah ok

[paper flint]

x->e^x

[chilly ocean]

wait why 4?

i see 9 because it would be a subgroup

[rustic crown]

cuz first iso theorem |im| = |G|/|ker|

[chilly ocean]

o

tbh my attempt for the relation is sort unmotivated

so idk why it would be useful

but ive always felt that the bigger the hom-set the more related two objects are

related by what? i dont know? potentially isomorphic subgroups?

or ig substructures

[hidden haven]

I don't have any context but the only ring endomorphism of ℝ is identity

🙈

It's supposed to be very related to itself

[barren swift]

so in the way that if i were to multiply x and x^2 im doing the same thing as adding 1 and 2, if i were to add x and y in R, im doing the same as if i were multiplying e^x and e^y?

[chilly ocean]

aq

what does it mean then lol

i cant just accept that number of morphisms is meaningless

R is very related to itself id say

maybe the most related?

[paper flint]

Yep!

[lethal dune]

ok are commutative ring with 1 and algebra same thing?

[frank fiber]

no

[barren swift]

this makes soooo much more sense now. thanks a ton!

[rustic crown]

nah, algebra is a more general term... algebras can be even non-associative

[paper flint]

No worries

[lethal dune]

even more general than rings?

[rustic crown]

yee

like consider a vector space V/k and a billinear bracket [-, -] : V x V --> V satisfying some different conditions than a ring... this is called a lie algebra

(specifically it's [a, a] = 0 and [a, [b, c]] + [b, [c, a]] + [c, [a, b]] = 0)

[hidden haven]

I interpreted your statement as |hom(A,B)| being big means A and B are more similar

[chilly ocean]

yeh this is wrong tho

[hidden haven]

And A = B = ℝ in Ring is a counterexample

[chilly ocean]

ur example

[hidden haven]

[chilly ocean]

yeah i am trying to figure out what it means

deepor meanings

[hidden haven]

There is only the identity homomorphism from R to R

Same for Q

Q is simpler

R is more amazing

[chilly ocean]

maybe size of hom is irrelevant

[rustic crown]

Q is awesome >.<

[hidden haven]

I am saying the fact about R is more amazing

[rustic crown]

oh right

[hidden haven]

Stop being so det det

[rustic crown]

🙈

[hidden haven]

🙈

[chilly ocean]

sort of off topic

but what are coinvariants supposed to represent

measure of how transitive a group action is?

because i did presentation on group cohomology

and the takeaway was that people care about it because it tells you how much coinvariant functor fails to be exact

i have no intuition for what that means besides definitions

[glossy grotto]

Just a quick question, let $\phi:\oplus_{n\in \mathbb{N}}\mathbb{Z}\to \mathbb{Z}$ a homomorphism, are there any other non-trivial homomorphism $\phi$ aside from the one determined by $(\eta_n 1_n){n\in \mathbb{N}}\mapsto \sum{n\in \mathbb{N}}\eta_n?$

[cloud walrus]

blackiris

[chilly ocean]

multiples of that homomorphism?

[glossy grotto]

oh yeah, but up to multiples?

are there more?

[chilly ocean]

yeah def

i think there are more

you just need to respect the multiplication ig

[rustic crown]

what morphisms are you looking at?

[chilly ocean]

my natural guess would be multiples only

im assuming ring

[glossy grotto]

Well, I'm trying to

[chilly ocean]

ring homomorphisms?

[rustic crown]

left side isn't a ring