groups-rings-fields

[somber bramble]

the statement that $G \cong G/H \oplus H$ is false, right?

[cloud walrus]

Sascha Baer:

[placid pond]

yes

[somber bramble]

it looks so true tho

[placid pond]

it doesn't

if it was true, every finite solvable group would be abelian

lol

[somber bramble]

I mean like, intuitively, G/H kinda means we group it into bits where the elements in h don’t do anything, and then we track them in the second component instead

it looks like the sorta thing that could be true as long as you don’t think about it too hard

[placid pond]

well, there is a whole subject studying exactly when this kind of statement is true

you have some category in which you have a short exact sequence of objects

when can you say that the short exact sequence splits, that is, the middle object is some kind of product or sum of the left and right objects?

in an abelian category there's an answer to that which is the splitting lemma

in the category of groups there's also an answer

but it's not quite identical

because there's the possibility of semidirect products

but generally, to conclude that a short exact sequence 0 -> A -> B -> C -> 0 splits, you need to give a morphism B -> A which is a left inverse to A -> B, or a morphism C -> B which is a right inverse to B -> C

if i recall correctly, in Grp you can conclude that a left inverse B -> A gives you that B is iso to the direct product of A and C, whereas a right inverse C -> B only gives you that B is a semidirect product of A and C

so that means if A is an injective object or C is a projective object, you always get splitting

and if you happen to be in an abelian category where every object is projective, say, like the category of vector spaces over a field, then this kind of statement is indeed true

but it's not true in Grp, or even in AbGrp

[somber bramble]

(I don’t know what a short exact sequence is)

I know very little category theory

[placid pond]

@somber bramble a short exact sequence is an exact sequence which only has three consecutive nonzero objects in it

it captures this correspondence between H, G and G/H

you have an injection H -> G

and a surjection G -> G/H

and the composition of these two maps is zero

H maps to the zero coset in G/H

and a short exact sequence 0 -> H -> G -> G/H -> 0 is just a way of writing all of that down nicely

in an exact sequence, the image of the map before is the kernel of the map after

so the image of 0 -> H is the kernel of H -> G, that is, H -> G is injective

the image of G -> G/H is the kernel of G/H -> 0, that is, G -> G/H is surjective

[somber bramble]

I see

seems like an utterly specific thing to have been given a name ^^

do they come up freqeuently?

[placid pond]

oh no, exact sequences are everywhere in algebra

surely you've heard of some big words related to it

like cohomology

or spectral sequence

[somber bramble]

cohomology I have heard of, aye. gonna take algtopo next semester

[placid pond]

yeah, (co)homology is a way of measuring how much a given sequence fails to be exact

if all of your (co)homology groups are zero, the sequence is exact

but sometimes you deal with sequences that aren't exact, so it's useful to have some way of measuring just how much they fail to be exact

in the topological context, there's some sequence called a chain complex associated to a topological space, and failure of it being exact roughly measures the difficulty of filling holes in the space

like, if you can embed a circle into your space, is there some disk which also embeds into it and has that circle as a boundary

but i shouldn't ramble on about topology in the algebra channel

[sharp sonnet]

So we know that if $a$, $b$ are coprime, the direct product of cyclic groups $C_a \times C_b = C_{ab}$. Is there a way to construct this isomorphism?

[cloud walrus]

Lochverstärker:

[mild laurel]

This is just Chinese remainder theorem

You would do it in the same way as you would find the unique solution mod 42 to the system x = 5( mod 6) and x = 2 (mod 7) for example

[sharp sonnet]

ah right

thanks

[smoky cypress]

@chilly ocean I finally understood the joke

Because

The ideals of a field is either {0} or the field itself...

[tame bear]

speaking of Ideals, i was struggling with a proof today but finally rederived it today$\$
Let $R$ be a commutative, unital ring, $I$ an ideal$\$
Then $I$ is maximal iff for all $a \not \in I$ there is a $b$ such that $ab - 1 \in I\$

[cloud walrus]

Sigma:

[tender mist]

Mm it seems like saying R/I is a field

[tame bear]

Assuming $I$ is maximal is the most involved direction$\$
So assume it is, then for any such $a$ form the set $J$ such that$\$
$J = { ar+i, r \in R, i \in I }\$
then $J$ is an ideal, and properly contains $I\$
so $J =R$ and $1 \in J\$
so there is some $b$ and $i$ such that $1 = ab + i\$
$ab - 1 = i \in I$

[cloud walrus]

Sigma:

[tender mist]

Very nice proof

[woven glade]

@chilly ocean this is what I was talking about, and confusion was on p(y)

[tender mist]

Maybe one can also argue that since R/I is a field
a not in I -> a+I has an inverse
So (a+I)(b+I)=ab+I=1+I
So ab-1 in I

[tame bear]

this is used to prove R/I is a field in the book im using

[tender mist]

Oh lol

Um I remember mine proved it showing R/I has only trivial ideals

[chilly ocean]

y is arbitrary element of X, just like x is

So, J = aR+I

why not write it like this 🤔

[woven glade]

Hmm can’t you create a set such that p(x) does not equal p(y) and the function F still exists?

[tame bear]

yes

[chilly ocean]

wdym

[woven glade]

Well you said that the sufficient and necessary condition would be that equivalence relation. But can’t their be a case that doesn’t follow what you said and the F function would still exist. Doesn’t that mean that that is not the condition?

[chilly ocean]

p(x) = p(y) implies f(x) = f(y) is the condition

do you have any counterexample?

If yes, then sure. People make mistakes sometimes (though I checked this myself)

[tame bear]

what if F just maps p(x) and p(y) onto the same output

[woven glade]

Yes that is what I was thinking. And f maps to different outputs

[chilly ocean]

Imagine that f = F*p and F(p(x)) = F(p(y)) but f(x) != f(y)

[sterile sable]

I need halp

I’m working with function and relations

[chilly ocean]

oh. Interesting

[sterile sable]

and ehhhh, one simple question

can 0 count in being

does it even exist

[chilly ocean]

wut

[tame bear]

be more specific

[sterile sable]

okay :,)) lemme try and word this right

can 0 make a relation when it’s listen in parentheses with another number

[chilly ocean]

yes

[sterile sable]

(x,y)

so it can go into a number?

not like that but

[chilly ocean]

0<1

Is this enough proof for you?

[sterile sable]

uhhhh

I’m not sure

I completely don’t understand it

[chilly ocean]

🤔

[random crag]

i would not say this is abstract algebra

[sterile sable]

Oop

[stone fulcrum]

An element like (-1, -9) can be thought of as an arrow that takes -1 to -9

[chilly ocean]

I guess it belongs to #proofs-and-logic more or #foundations

though relations are used so much in algebra that 🤷🏿

[solar vessel]

I would place it in facts-and-logic

[stone fulcrum]

Actually, first question - do you know what a domain and range are?

[solar vessel]

I always hated the word range

[tame bear]

why would 0 being the in the ordered pair even be confusing in the first place

[solar vessel]

wat

[chilly ocean]

@woven glade if you want, you can try to attempt a proof, and I will guide you here (within reason)

[woven glade]

I appreciate I am new to algebra so having trouble with it, but I’m thankful for the help. I’ll work something up, when do you sleep haha

[chilly ocean]

oh. I probably should sleep

like rn?

well. whatever

so suppose that f = F*p and try to deduce that p(x) = p(y) implies f(x) = f(y) for any x, y in X

this is the easy direction

and explanation of the condition would be nice

it would be nice if F would always be the same when p(x) = p(y) i. e. the argument is the same

because then F is well defined

that's where the condition comes from

anything?

[woven glade]

Well for the first direction if you know that p(x) = p(y) then that must mean they all map to the same element in Q.

From there we know f= F*p so that means F is applied to only one element of of Q thus one element of Y. Then any x and y in f must be mapped to one element in Y and f(x) = f(y)

[chilly ocean]

Well I'd something more like

For any x, y in X, if p(x) = p(y), then f(x) = F(p(x)) = F(p(y)) = f(y)

But it's ok

Other direction, suppose that for any x, y in X, p(x) = p(y) implies f(x) = f(y) and try to construct F

[winter vigil]

For proving eisenstein's criterion, does this work ? However, I'm not sure how to justify that the final result is irreducible.

[somber bramble]

what is r₀? for eisenstein’s criterion, p has to divide a₀, but p² doesn’t

iirc the way I proved it was to assume it was reducible, then projecting it into 𝔽p and getting a contradiction

yea it follows pretty much immediately actually if you do that

I’m also not really sure about your conclusion here

would be nice if you could use a few more words

[somber bramble]

hmm I wonder what’s the hardest question on this algebra exam

I think “let G be of order 3^17 and let H be a subgroup of index 3. Is H normal?”

or, in a very similar vein “does A_17 have a subgroup of index 3?”

(the proofs to both are quite similar)

[placid pond]

easy algebra exam in that case

[somber bramble]

I’m not disagreing per se, but I also have to say that that’s a pretty condescending statement

I don’t think those are easy proofs to come up with if you’ve never seen the ideas before (which we hadn’t)

[placid pond]

@somber bramble that's odd

[somber bramble]

it was a fairly superficial course that kinda breezed through too many topics in too little time

[placid pond]

if p is the smallest prime s.t. p | |G| fora finite group G, then any subgroup of index p is normal; A_n is simple for n >= 5; any group with a subgroup of index n has a normal subgroup of index at most n!

all of these are the kinds of things you'd find in the exercises of a group theory textbook

[somber bramble]

p is the smallest prime s.t. p | |G| fora finite group G, then any subgroup of index p is normal
yes, that statement is not one we did in class

[placid pond]

i see

[somber bramble]

nor the last one, that one I haven’t even heard of

[placid pond]

it's straightforward: if H is a subgroup of G of index n, then G acts on the left cosets of H, that gives you a homomorphism G -> S_n whose kernel has index at most n!

and kernels are normal, so

by this proof you can also guarantee that said normal subgroup is contained in the original subgroup, in this case H

[somber bramble]

yea that’s essentially the proof I do here

either way, coming up with these ideas on your own is nontrivial

I will admit that most exercises on the list are fairly straightforward, and the fact that we have a list in the first place makes it even easier

[somber bramble]

ended up getting a followup question I couldn't answer on the spot, went well otherwise

(how many primitive elements are there in F_{2^n})

[placid pond]

@somber bramble i don't think that question has an easy answer, it's equal to n * (number of irreducible polynomials of degree n in F_2[x]), and the standard way to compute the thing in parantheses is mobius inversion

so you can give an explicit expression in terms of the mobius function but it's not really something that would show up in an algebra exam

[somber bramble]

he asked me to think about the subextensions and I said some true statements (each corresponding to a subgroup of Z/nZ and so on) but he decided to say stop. we never looked at möbius functions (except the presumably different ones in complex analysis ofc) so that can't be it

eventually he was like "well it wasn't on the question sheet"

and we moved on

[placid pond]

lol

[sweet moat]

guys what is the kernel of a homomorphism? not sure what im supposed to be looking for as ive seen different definitions of it

[chilly ocean]

homomorphism of groups?

[mild laurel]

If you take a homomorphism f : G -> H, the kernel of f is all elements g in G such that f(g) = e_H

where e_H is the identity of H

[sweet moat]

homomorphism of (M,+,x) to (R,+,x)

x is multiplication

[mild laurel]

Those are rings

The kernel is the exact same though, just all elements such that f(g) = 0_R

[sweet moat]

wait but

[chilly ocean]

A little difference that it maps to zero instead of identity

[sweet moat]

oh

cause i was wondering identity element is different for + and x

right?

[chilly ocean]

yes

[sweet moat]

so im looking for all elements that map to 0?

[chilly ocean]

yes. Same with vector spaces homomorphism etc

which can be thought of as the addition identity also

[sweet moat]

nice

thank you very much

[bleak finch]

What resource is "Group Theory for Dummies?"

[golden pasture]

more specific?

[bleak finch]

I need a resource that covers the important topics of group theory (groups, subgroups, normal groups, isomorphism theorems, group actions, automorphisms, and solvability) in a way that's so clear and simple that a patient 14-year-old can understand them.

[somber bramble]

does this 14-year old have any mathematical maturity or are they just a random high school student who’s interested in the topic but has never done any university-level math?

[bleak finch]

14-year-old is capable of doing proofs if they're presented in a straightforward and clear way.

[smoky cypress]

🤔

[wind parrot]

maybe look up the book Visual Group Theory

[smoky cypress]

The proofs in group theory are straightforward, that is if they’re comfortable using symbols and stuff

[bleak finch]

They often introduce definitions and it isn't obvious what motivates these definitions.

[mild laurel]

It's pretty hard to always motivate definitions, sometimes the motivation can only come after learning about it some

Like how do you motivate rings, they seem like a pretty weird combination of structure to consider

[smoky cypress]

I think if someone is there to teach the topic very slowly

With 1 definition per class/session

And a lot of examples

Then it might work

[bleak finch]

Whoever has officially lost his mind

[smoky cypress]

Why so

[bleak finch]

Only someone who had a psychotic breakdown could believe that a professor would go that slowly

let alone a tenured professor

[smoky cypress]

Lol

Did I ever say someone will do this?

Also

I think the ability to read proofs and understand definitions doesn’t come with age

It comes with exposure to higher math

[bleak finch]

Visual Group Theory is too visual and involves too much tinkering around.

You know what?

I'll just go to ProofWiki and perform a topological sort on the proofs and definitions there and work my way up.

[tame bear]

just read a goddamn book

almost any of them are fine to start with

[smoky cypress]

Are you the 14 years old we're talking about?

[bleak finch]

The books suck

They're like using a French textbook to learn French

How many people actually LEARNED french from those textbooks?

My point exactly

[smoky cypress]

wtf

A language is very different from math

[bleak finch]

it was an analogy about how useless math books are

[smoky cypress]

Lol

Good luck finding resources

[bleak finch]

ProofWiki is the best

I just need to reorganize everything for my own purposes

that's all

[tame bear]

#books-old

basically every abstract book starts with a review of logic and set theory

[wind steeple]

Il learnt french from France :kappa:

[tame bear]

i learnt Jap from anime

[wind steeple]

:cosmicbrain:

[smoky cypress]

Proofwiki doesn't even have calculus proofs

Oh when I searched somewhere else then the calculus stuff showed up

[tame bear]

best way to learn math is to do math

i always looked for the best book to read, but that kept me from actually reading

just pick one and run with it

[worthy kindle]

proofs in calculus

[smoky cypress]

Why read books when you can come up with definitions on your own

[tame bear]

ive done that before, is fun

[wind steeple]

Pick one which transcends you and learn

For me all books with "group" "geometry" "algebra" "category" and "galois" and without "applied" in the title transcend me xd

[smoky cypress]

Lol

Abstract algebra applied

Sounds interesting

[somber bramble]

Also, once you have a decent footing in a language, reading books in that language is one of the best things to do to get better

you can't start there, but you can do that fairly early

say at A2 level or sth (which is, for all intents and purposes, still "beginner")

[mild laurel]

@bleak finch you are not going to learn group theory like that

Good luck

[chilly ocean]

Learn Russian from Russian books 🤔 good luck with that

[somber bramble]

as said, not from 0

but once you have a good idea of the grammar, know a lot of the most common words... yea you can learn a lot by reading a book and looking up words you don't know

also, while I don't know if such a thing exists for Russian, one of the best resources for self-studying Latin is a book that's entirely in latin

it starts off by showing most basic sentence structures, explains words with images and little notes in the margins, and then increases the complexity gradually

(it's called Lingua Latina per se Illustrata)

[chilly ocean]

what's a good ring theory book with exercises? (not necessarily commutative or unital)

J.Lambek, Lectures on Rings and Modules - how is this

[wind steeple]

@smoky cypress you joke but I found a book of applied category theory

[smoky cypress]

[wind steeple]

With, say, automatization of bread baking

Lel

[smoky cypress]

Idk whether they existed

[wind steeple]

Applied algebra exists too

It's called

Cryptography

Lel

[smoky cypress]

But I’ve always thought that abstract algebra and category theory are too abstract to be applied

Loll

[mild laurel]

Algebra is like the most applied math field

Like it's used so, so much in chemistry and physics

[smoky cypress]

Oh ok

Like all those algebraic structures?

[mild laurel]

chem uses mostly group theory

[smoky cypress]

[mild laurel]

But like a lot of group theory

[wind steeple]

Symetric groups are use in molecule geometry

[smoky cypress]

Oh that’s interesting

[wind steeple]

Yeah

But algebra isn't the most applied maths field in engeneering

[smoky cypress]

I’ve never learned any science other than math (is it science?) 😅

[wind steeple]

Engineers use stupid maths

[smoky cypress]

[wind steeple]

Like calculus

:kappa:

[smoky cypress]

κ

[mild laurel]

Cat theory is used a decent amount in cs

[fluid iron]

Hello does anyone know how to do graphing quadratics????

[mild laurel]

Please read #❓how-to-get-help

Maybe read the rules of a server before you say anything

[wind steeple]

Cs ?

[smoky cypress]

Computer science

[wind steeple]

Oh yeah

[smoky cypress]

Hmm I might look into category theory, I’ve never seen anything

[wind steeple]

Cat theory is used is every fields of math

I'm also looking for good category theory references

The earlier I learn cat theory, the more confortable I'll be to understand more abstract mathematics

[smoky cypress]

🤔

I don’t know if that’s true

[mild laurel]

Yeah I don't really think that's true either

There are definitely plenty fields of math that don't use any category theory for the most part

[somber bramble]

Also, pretty much everyone suggests first studying at least one subject that motivates cat theory

studying abstract nonsense without motivation is just not that amazing

[chilly ocean]

Lie algebras are used in robotics if I recall

[wind steeple]

Studying abstract thing is fun

🤓

[smoky cypress]

Imagine learning congruence in a general setting 🤔

It's weird why the definition is like this

[chilly ocean]

Imagine learning congruence in a general setting 🤔

Congruence relation generated by a relation 🤔

[smoky cypress]

Congruence is a special kind of equivalent relationship on a monoid tho 🤔

[chilly ocean]

on semigroups

[smoky cypress]

Sure I guess

This

is kind of painful to see

Although I know Z is abelian

[mild laurel]

I've seen $\mathbb{Z}/(-d)\mathbb{Z}$ in a paper once

[cloud walrus]

Zopherus:

[mild laurel]

It made sense to have that though

[smoky cypress]

[chilly ocean]

why congruences?

Imagine you want to define [a]*[b] or [a]+[b] for some equivalence classes [a] and [b]

with congruences you can

the binary operation just comes from the normal binary operations on a given structure

[cloud walrus]

Blitzkrieg:

[chilly ocean]

being a congruence makes it so the above definitions are correct

[smoky cypress]

Well yeah

Also notation question

Do we usually denote the equivalent class of an element $a$ as $\bar{a}$ or $[a]$

[cloud walrus]

Whoever:

[tame bear]

yes

[chilly ocean]

I never seen \bar{a}

[tame bear]

could also write it as the coset

[mild laurel]

\bar{a} comes up a lot when you discuss reducing mod p

And especially when you reduce functions mod p

[smoky cypress]

Hmm ok

[cloud walrus]

Blitzkrieg:

[smoky cypress]

Well I'm reading Jacobson and he's using \bar{a}

Sure

Jacobson's notation is just ehhhh

So weird to me

[tame bear]

specifying the relation

[chilly ocean]

what's so funny I don't understand

[tame bear]

no one ever does that

[smoky cypress]

Jacobson has done this

[chilly ocean]

because no one ever talks about more than one equivalence relation at a time I guess

[smoky cypress]

This theorem says that a congruence can define a normal subgroup

And a normal subgroup defines a congruence

Right?

[chilly ocean]

yes

I mean

It says there is a 1-1 correspondence of set of congruences upon the set of normal subgroups

[mild laurel]

Do they give a definition of congruence

Because this isn't true for any equivalence relations

[smoky cypress]

Yes

[cloud walrus]

Blitzkrieg:

[smoky cypress]

Yeah

That's the definition

[chilly ocean]

or equivalently, if it's a left and right congruence

[smoky cypress]

A equivalent relationship that you can multiply

[chilly ocean]

i. e. congruence is an equivalence relation where you can define [a][b] = [ab]

that theorem is cool because it tells you there's no real distinction between a normal subgroup and a congruence

[woven glade]

Given a function f : X -> Y what subsets of X x Y occur as the graph of some function. Wouldn't this be all subsets, except for the empty set? Since a single point can be graphed as a constant function

As well given a graph is there a general way to define the function? Like imagine if you had a line you can find the slope and then create the equation but how would i go about doing this more generally?

hmm is there a more elementary way to say it 😦 I need to study relations

[cloud walrus]

Blitzkrieg:

[chilly ocean]

if you read it, there's nothing non-elementary about this

[cloud walrus]

Whoever:

[smoky cypress]

@chilly ocean

[chilly ocean]

yes

[smoky cypress]

ok then it makes sense

[chilly ocean]

x maps to unique y

[smoky cypress]

Yeah

That's the definition of function using relation

[chilly ocean]

idk. There was this argument that for a function you have to specify domain and codomain, and I kind of agree tbh

[smoky cypress]

I agree too

Does anyone use the term monomorphism and epimorphism?

[chilly ocean]

I don't. Unless someone else does, that is

just say 1-1 or upon

[smoky cypress]

Yeah

Well

I use surjective and injective

Which is slightly more intuitive in my opinion

[chilly ocean]

more intuitive

[fringe nexus]

1-1 or onto is less memorization

[smoky cypress]

Lol

I mean

I've seen more people use injective, surjective, bijective

[chilly ocean]

[smoky cypress]

What

Maybe 1-1 and onto is more used

But

[chilly ocean]

all you need, is homomorphism and isomorphism

rest is useless

mono and epi> inj and surj > 1-1 and onto
where > means ”more intuitive than”

injective or surjective is more intuitive than 1-1 or upon?

yes

[smoky cypress]

To me

[chilly ocean]

1-1 is more suited to bijection rather an injection

[smoky cypress]

Tbh, 1-1 sounds more like bijective

Lol yeah

[chilly ocean]

maybe. But since I know it's not, then there's no confusion

:pepesigh:

[tame bear]

this is the first time ive ever heard upon
and i hate it

[chilly ocean]

:pepesigh:

wtf

nah it's great

[tame bear]

just use onto like a normal person

[chilly ocean]

nah

[tender mist]

Well at least epi, sur, onto mean the same thing semantically
It's just greek french english lol

[chilly ocean]

into for non necessarily upon

if I used onto, it would be confusing with into

so I use upon

epi, sur, onto

[tame bear]

why not just use epic at that point

[tender mist]

Lol

[chilly ocean]

it's one more letter

so one more unit of effort

[tame bear]

talking to blitz

[chilly ocean]

ah

upon is classy

epi classy
it's even used in medicine so it must be

you don't get my style ok

[smoky cypress]

@tame bear normal person uses surjective

[chilly ocean]

lol blitz

ugh I'm edgy teenager no one understands me here

[smoky cypress]

teenager hmmmmmm

[tame bear]

normal people use one to one and onto

[chilly ocean]

normies heh

[smoky cypress]

@chilly ocean how old are you

[chilly ocean]

20

[smoky cypress]

😮 that's not teenager smh

[chilly ocean]

ik

[tender mist]

Anything from greek and latin is more classy, there's no challenge

[chilly ocean]

i agree

upon is a word that feels you with mystery

it's classy

no one can deny

You don't have to use greek for everything

λολ

Greek looks awful. I only approve greek letters application in math

btw. Algebra is here

από την άλλη πλευρά.

[smoky cypress]

Lol

[chilly ocean]

So to not distract from the topic, I'll include a theorem which relates to congruences and groupoids (or magmas as you people call them)

Theorem is called 'Principle of the Maximal Homomorphic Image of Given Type'

[cloud walrus]

Blitzkrieg:

[chilly ocean]

tbh idk how to interpret this theorem

I guess that intersection of them is exactly what you would think it would be 🤔

[chilly ocean]

Prove that there exist semigroups not being groups such that $a1 = a$ and $a^{-1}a = 1$. I'm thinking about bi-cyclic semigroup i. e. generated by elements $p, q$ such that $pq = 1,\ qp\neq 1$. I won't check rn because 7 am though

[cloud walrus]

Blitzkrieg:

[smoky cypress]

So the only way it’s not a group if there isn’t an element e such that ea=a for all a?

@chilly ocean

[chilly ocean]

Ok I think I know.

[smoky cypress]

Was my statement correct

[cloud walrus]

Blitzkrieg:

Blitzkrieg:

[chilly ocean]

@smoky cypress A semigroup $G$ is a group if an only if there is an element $1$ such that $a1 = a$ and exists $a'$ such that $aa' = 1$

[cloud walrus]

Blitzkrieg:

[chilly ocean]

for all a

we don't impose uniqueness

[smoky cypress]

[chilly ocean]

i. e. we have a right identity and ability to take right inverses

[smoky cypress]

But in your question, all the conditions are satisfied?

[chilly ocean]

right identity but left inverses in my question

[smoky cypress]

Oh

Wait I realized I don’t know what a semigroup is 🤔 is that just a set with an associative binary op?

[chilly ocean]

$E = {a, b}$, $a^2 = ab = a$, $b^2 = ba = b$

[cloud walrus]

Blitzkrieg:

[chilly ocean]

ab = a and bb = b -> b is right identity

ba = b and bb = b -> every element has a left inverse

to show it's a semigroup is trivial

it's not a group and it's trivial

[brisk granite]

why does a polynomial, of degree n, in Z/pZ have at most n roots

[chilly ocean]

Because Lagrange theorem

https://en.wikipedia.org/wiki/Lagrange's_theorem_(number_theory)

[brisk granite]

cool, thanks

[brisk granite]

wait, so, why does the proof break down when p isn't prime

?

[mild laurel]

Well consider x^2 = 0 mod 8

@brisk granite

[faint elm]

hi

Why the young symmetrizer is defined as $a_Tb_T$ where $a_T = \sum_{g \in P_T} g$ and $b_T = \sum_{g \in Q_T} sgn(g)g$ where $T$ is a young tableaux and $P_T$ is the group that permutes the elements in the rows of $T$ while $Q_T$ is the group that permutes the elements of the columns of the tableaux

[cloud walrus]

emme:

[chilly ocean]

I have a structure (S, 1, /) with equalities $a/1 = a$, $a/a = 1$, $(a/c)/(b/c) = a/b$. Define $ab = a/(1/b)$. How to prove $(ab)c = a(bc)$? Upon expanding it looks horrible

[cloud walrus]

Blitzkrieg:

Blitzkrieg:

Blitzkrieg:

[chilly ocean]

Maybe like

[cloud walrus]

Blitzkrieg:

Blitzkrieg:

Blitzkrieg:

[chilly ocean]

uh huh

[cloud walrus]

Blitzkrieg:

[chilly ocean]

👏🏿

hence a(bc) = (ab)c

[cloud walrus]

Blitzkrieg:

[chilly ocean]

I want to prove that a binary operation on this Boolean algebra defined as

[cloud walrus]

Blitzkrieg:

[chilly ocean]

I need some help here, frightening things happen when trying to write it out

I'm trying to make this Boolean algebra a ring

but proving associativity is a challenge here

[cloud walrus]

Blitzkrieg:

[chilly ocean]

so this addition resembles a bit https://en.wikipedia.org/wiki/Exclusive_or
so I guess it should have similar properties

[cloud walrus]

Blitzkrieg:

[chilly ocean]

so I need something like distributive property of * and \lor

[bleak belfry]

if i dream of maths is that abstract algebra?

[chilly ocean]

@bleak belfry go troll in hell

[bleak belfry]

mean

[chilly ocean]

righteous

[chilly ocean]

I don't need help anymore.

[cloud walrus]

Blitzkrieg:

Blitzkrieg:

[chilly ocean]

basically a relation which preserves all the identities of an equational structure

[shrewd halo]

I know this is wildly abstract and I have absolutely nothing but intuition but I was curious: does it bear fruit (in any context) to think of the integers as an infinite-dimensional module where the primes are your base?

[mild laurel]

There's a group isomorphism between the rational numbers under multiplication and the group of infinite integer sequences with only finitely many non-zero terms under addition

This isomorphism behaves in basically the way you want

(a_1,a_2, \dots ) gets sent to 2^(a_1) 3^(a_2) \dots

[worldly flower]

what do you mean primes as your base?
the definition of integers has nothing to do with primes

it's more like the set of answers you get when you subtract two natural numbers

[mild laurel]

Every integer can be written uniquely as $\pm 2^a3^b 5^c \dots$

[cloud walrus]

Zopherus:

[mild laurel]

every non-zero integer I guess

[worldly flower]

I guess you can express it like that, but what's the point?

[mild laurel]

This is like the whole point of number theory lmao

[worldly flower]

I thought we were tackling this from an abstract algebra perspective

[mild laurel]

Wow it's almost like mathematical fields are heavily related or something

[worldly flower]

different fields may have different perspectives on things

[mild laurel]

And you say the definition of integers like there's one definition

[worldly flower]

i didn't say there was only one definition

[mild laurel]

"the definition" usually implies 1

But the point is that a definition based on primes is definitely not useless

And even if the definition of integers that you use has nothing to do with primes, saying that there's no point to writing the integers in terms of primes is ridiculous

[chilly ocean]

damn this nick is annoying

[smoky cypress]

Same

[bleak finch]

am I banned?

Nevermind

[stone fulcrum]

Yes

Bean

[bleak finch]

Can I conclude from the fact that |D_4| = 8 and |S_4| = 24 that there are some permutations that are not symmetries?

[fading wagon]

you mean

there are some permutations of {1, 2, 3, 4} that do not correspond to symmetries of a square

@bleak finch ?

[bleak finch]

yes

[fading wagon]

yeah

[bleak finch]

I can conclude that BASED ON THE MAGNITUDES OF THE CARDINALITIES ONLY?

[stone fulcrum]

Imagine swapping two corners of a square

Oh, just based on the order of the groups?

[fading wagon]

Well, you must make sure that each element in D_4 can act on the set {1, 2, 3, 4} as permutations

[bleak finch]

Yes, just based on the order of the groups

[fading wagon]

then, yes, you can conclude that

Like, you are not allowed to swap two adjacent corners of a square

(without changing anything else)

but you are totally allowed to swap {1, 2, 3, 4} into {2, 1, 3, 4}

[stone fulcrum]

You'd really have to know the relationship between Sn and Dn

[bleak finch]

this is actually a question about proving not about groups

I concluded that there are some permutations that are not symmetries because 8 < 24, and I don't know if that is a sound argument.

Was I right by accident?

[fading wagon]

you need to phrase it better

Maybe you can say D_n is a subgroup of S_n

[stone fulcrum]

A proper subgroup at that

[bleak finch]

Now I think I have the answer

[fading wagon]

unless n=3

then they are the same group

[bleak finch]

Some permutations are not symmetries because D_4 is a subgroup of S_4 and |S_4| > |D_4| is a valid argument.

[stone fulcrum]

True, you've got me there

[fading wagon]

you wouldn't want to call it that way

"Some permutations do not correspond to symmetries of a square"

you need to be specific as to symmetries of [object]

[bleak finch]

obvious in context

[stone fulcrum]

What are you writing the proof for? What class?

[fading wagon]

well, but S_4 is isomorphic to the symmetry group of rotations of a cube

[fringe nexus]

Let G be a finite group with an automorphism $\phi$ s.t $\phi(g) = g$ iff $ g = 1$ and $ \phi^{2}$ is the identity map. Show that G is abelian

[cloud walrus]

Victoria:

[fringe nexus]

idk if my proof works

wait nvm

im dumb

[woven delta]

squared means function composition or group product?

[fringe nexus]

is it true that $\phi(x) = xa$ for some a?

[cloud walrus]

Victoria:

[fringe nexus]

uhh $ \phi(\phi(a)) = 1 \forall a \in G$

[cloud walrus]

Victoria:

[woven delta]

Okay got it

And no

[fringe nexus]

I have no idea how to use this hint Im given

[woven delta]

What's the hint?

[fringe nexus]

Show that every element of G can be written in the form $ X^{-1}\phi(X)$

[cloud walrus]

Victoria:

[woven delta]

🤔

[fringe nexus]

scuffed latex

[woven delta]

What you wrote earlier is wrong btw

[fringe nexus]

oof which

[woven delta]

You meant a

[fringe nexus]

oh

yea

oops

ok nvm

i am wrong

yes

what you said is right i think

from the hint we have

[woven delta]

Oh hmm

[fringe nexus]

$a = x^{-1}\phi(x) \ \phi(a) = \phi(x^{-1})x = [x^{-1}\phi(x)]^{-1}$

[cloud walrus]

Victoria:

[woven delta]

Yeah the reason I thought it was wrong was because of stuff with elements of order 2

[fringe nexus]

then showing that its an automorphism implies abelian seems easy

[woven delta]

But that just might not apply here

[fringe nexus]

but how do i get to the hint part?

[woven delta]

Oh apply phi to x^-1 phi(x)

And you get phi(x)^-1 x

Wait no

x phi(x)^-1

[fringe nexus]

why isnt it phi(x)^-1x

[woven delta]

(ab)^-1 = b^-1 a^-1

Wait I'm stupid

Yeah I was right the first time

[tame bear]

yeah.just take a = x^-1 phi (x)

i made a mistake actually

[woven delta]

Then phi(a)^-1=a

And I guess every element can be written as x^-1 phi(x) for some x

Is the hope

[fringe nexus]

ok i got it

i just show that

the function f : G - > G

f(x) = x^(-1)phi(x)

is bijective

[woven delta]

Yeah

That would do it

But I don't know if that's obvious

[fringe nexus]

oh yea

showing that seems its gonna be kinda hard

[woven delta]

Maybe it's its own inverse

Or something

[fringe nexus]

oh

i got it

we show its one to one

let a,b $\in G$ where f(a) = f(b). Then, $ a^{-1}\phi(a) = b^{-1}\phi(b)$

just simplify to

[cloud walrus]

Victoria:

[fringe nexus]

$ba^{-1} = \phi(ba^{-1})$

[tame bear]

its finite so injectivity suffices

[cloud walrus]

Victoria:

[woven delta]

It's not finite

[fringe nexus]

it is finite

[woven delta]

Why would it be?

[fringe nexus]

G is finite

[tame bear]

finite group

[woven delta]

Oof

[tame bear]

yowch

[fringe nexus]

then we have that the equality holds iff ba^-1 is identity so b = a

[woven delta]

Do you need finite?

Probably

[fringe nexus]

first grad course and hw is so hard

[woven delta]

Yeah so you showed it's injective, so you're done

I would be curious to see the infinite counterexample

Maybe GL_2(\mathbb{R})

Is a good counterexample

[bleak finch]

How does the general proof go that three algebraic objects are not isomorphic?

[mild laurel]

What does this even mean

[bleak finch]

Given groups G, H, K, I need to prove that G, H, K are distinct groups.

[mild laurel]

Do it pairwise

[bleak finch]

There are (3 choose 2) = 6 pairs.

[mild laurel]

That's not 3 choose 2

[bleak finch]

No wait it it's not

But this way of proving doesn't scale

To prove n groups are distinct I need to do (n choose 2) proofs which leads to combinatorial explosion

[mild laurel]

I doubt you'd ever be asked to show that many groups are distinct from each other

In general, you can look at group properties too

If all your groups have different orders, then you're done obviously

Or if all your groups have different max element orders, or some are abelian and some aren't etc

[somber bramble]

yea you’ll be able to at least group them into small groups where you have to take a closer look

…those are three different meanings of “group” now

[woven delta]

Why would you care about such a thing?

@bleak finch

[bleak finch]

Because I like to press the easy button when it comes to proving theorems.

[woven delta]

🤔

[mild laurel]

lmao

[bleak finch]

Show that addition and multiplication mod n are well defined operations. That is, show that the operations do not depend on the choice of the representative from the equivalence classes mod n.

How do proofs like these work?

[mild laurel]

By understanding what it means for something to be well-defined

They even explain it for you

[bleak finch]

not this book

but I think somewhere on overleaf I had a similar problem for modules whose proof I might be able to adapt.

it's like reusing your piece of code for a new project

Okay this is funny

There's a wiki titled "groupprops" coined from "group properties"

and I searched for "group properties invariant isomorphism"

and no article exists

L-O-L

[mild laurel]

that's just uh

dumb

Things are well-defined

[tame bear]

ive never actually seen a definition for well defined

i just picked it up from context

[chilly ocean]

well defined means our definition doesn't contradict itself

[mild laurel]

That's not the best explanation lmao

[somber bramble]

something being well-defined pretty much just means the definition isn’t contradictory.
the particular case where where you have to prove well-definedness is pretty much always the same scenario though:
•you have some set on which you have a function f
•you also have a partition of that set into chunks
•you want to define a function on those chunks by using f
•that is, you want to define F([x]) = f(x), where [x] is the set of all elements in the same “chunk“ as x
•however, for this definition to make sense, you must have that if [x] = [y], then f(x) = f(y), or you’d run into a contradiction

[tame bear]

x = y implies f(x) = f(y)

[somber bramble]

I never said x=y tho

I said [x] = [y]

[tame bear]

same thing

[chilly ocean]

lol

[tame bear]

different name

[somber bramble]

uh, no?

[x] is some set containing, among other things, x

for example, for modular arithmetic thingies, [3] might be all the numbers that are equal to 3 mod n

[tame bear]

a = b implies f(a) = f(b)

[somber bramble]

yes, that is a true statement

[smoky cypress]

Well defined is basically when a function isn’t multi valued right?

[split niche]

Ye think so

Also the definition of well defined is too subjective

And the definition itself is arguably not well define

[chilly ocean]

well-defined depends on the context, but basically means that everything you wrote is fine and just works

[split niche]

But that’s subjective

And your definition of it isn’t well defined

[chilly ocean]

no it's not subjective

technically everything is subjective, actually

but the point is, that's not an argument

well-defined doesn't need a mathematical definition

[split niche]

If you say fine means logical than it’s fine

But it needs a well defined definition XD

[chilly ocean]

it depends on the context

and should be self-explanatory when arrives

[inner acorn]

👀

[chilly ocean]

that's not a logic or math problem, that's just language

[split niche]

That’s why it was better in the caveman times before language

When there was only math

[mild laurel]

Lmao that's an opinion for sure

[split niche]

An opinion is not well defined

Unless it is

[grizzled canyon]

Hello smart people I would appreciate help on very simple homework I’m just very stupid

[mild laurel]

Nah I meant what you just said was dumb

[split niche]

For sure it was dumb

[chilly ocean]

@grizzled canyon you shouldn't say you are stupid in a futile attempt to someone take pity on you

[split niche]

Smart things are only interspersed in the shitstorm of nonsense that regularly pours from my mouth

[chilly ocean]

just ask and be done

[grizzled canyon]

That’s not what I was doing but okay big man

[split niche]

They call him the toxic for a reason

Ok what’s the homework

[grizzled canyon]

😂😂

[chilly ocean]

I wouldn't know, but you shouldn't say that regardless

even if it's true

[grizzled canyon]

I just feel stupid because this is called abstract algebra and I can’t even do simple that is all

But okay

[split niche]

THEN WHY TF U ASKING IN ABSTRACT

Go to the related channel plz

We’re in all the channels

We’re helpers

[chilly ocean]

I'm not

[split niche]

Now which channel should this be I

in

[grizzled canyon]

@ me in one

[split niche]

What’s the topic and level

[grizzled canyon]

Geometry

[chilly ocean]

lol

[fading wagon]

#geometry-and-trigonometry

[grizzled canyon]

I’ll delete my old messages in this channel though sorry for the inconvenience

[split niche]

Nah

[chilly ocean]

np man

[split niche]

Don’t waste time deleting

[grizzled canyon]

Blitz why you saying lol

[chilly ocean]

idk

[somber bramble]

That’s why it was better in the caveman times before language
Language is older than "caveman times"

[clear obsidian]

ok so If someone can confirm what im thinking

because Im reading james munkres pdf on topology (very early in still) and he seemed to imply that Z is a field, but I dont think thats correct since we dont have multiplicative inverses other than for unity and -1 (and we cant have a field without also being a division ring, right?, Im pretty sure the definition of a field is a commutative division ring)

[mild laurel]

Can you take a picture of the part where he says that?

[clear obsidian]

lemme post the context first because he makes reference to numbered properties

so he says this ^^^ but notice property 4 (the second line particularly)

maybe he means some of but not all of properties 1 through 6 ? but I dont know why he would phrase it that way, thats confusing to me

[somber bramble]

he’s not saying that the integers are defined by having properties 1-6 (for one, because they don’t, as you pointed out)

he’s saying that you can define them making only reference to those properties

which, presumably, he does in the following paragraphs

[clear obsidian]

he doesent clarify this after though by the way ^^^

[somber bramble]

I’m downloading the book now just to see

[clear obsidian]

he immediately starts talking about Z^+

[somber bramble]

the sentence “we now define the integers” is not a definition, but a statement of intent

[mild laurel]

Yeah Sascha's interpretation is correct

[somber bramble]

he’s saying that the integers will be defined in the following

[clear obsidian]

like in a proper subset of 1-6 right?

using only these properties, but not strictly all of

[somber bramble]

is this munkres, second edition?

[clear obsidian]

yes

it should be everything except the second part of property 4 I guess bc we have a commutative ring with unity in the integers

[somber bramble]

yea, so that sentence there is not part of the definition

what it’s saying is “we now have the tools needed to define the integers”

and then he jumps to the next definition that’s needed, which are inductive sets

[clear obsidian]

yes

[somber bramble]

and the point is that only properties (1) through (6) are actually used in the definition

[clear obsidian]

right

[somber bramble]

meaning you could define the things more generally for any structure satisfying those properties

[clear obsidian]

side note I like how he defines ordered pairs

[somber bramble]

as functions from {1,2} into a set?

presumably?

[clear obsidian]

I think it was, (1,2) = { {1}, {1,2} }

[somber bramble]

okay that’s wacky

[clear obsidian]

I think he did it that way to make the ordering clear

[somber bramble]

usually the way you define tuples is by treating them as functions

[mild laurel]

I think that definition is pretty common too

[somber bramble]

I’ve never seen it before

[clear obsidian]

but yeah I guess his pdf doesnt expect abstract algebra either because he could've just said "ordered field with the supremum property" for R

[somber bramble]

guess I’ll come across it if I end up tutoring topology next spring semester (definitely gonna apply for it), cause in that case I’ll be reading munkres along the year

[clear obsidian]

its definitely a comprehensive pdf from what I can tell

ive been trying to get farther into it so that I can sound more intelligible in my statements of purpose for phd school

like I know some basic definitions, but im trying to get an intuitive sense in order to at least prove some basic theorems

[somber bramble]

wait what

is that normal?

as in, you’re applying for a phd

but these things are new to you?

the first chapter of munkres more or less sums up my first three weeks at university

[clear obsidian]

Well I didnt have a topology in undergrad, I had complex ana though at least

no the first chapter is not new to me though

stuff after is a little

[somber bramble]

and topology is a mandatory 4th semester subject here

[clear obsidian]

where?

[somber bramble]

ETH Zürich

[clear obsidian]

in undergrad?

[somber bramble]

leading up to the (electable) Algebraic Topology course in 3rd year

yes

[clear obsidian]

I came from a smaller school so even basic top wasent offered there

I took pretty much everything I could that was offered

[somber bramble]

I’m curious what kinda classes you took

just the actual math classes in case you took additional stuff

[clear obsidian]

other than complex ana, vector ana, numerical math ana, (ODE & linear of course), topics in geometry (axiomatic of course)

[somber bramble]

over the course of how many years?

[clear obsidian]

also calc 3 ta

those were in 2 years, I had to retake some basic stuff before though

[somber bramble]

that seems like extremely little

[clear obsidian]

I had basically a philosophy & math dual major

[somber bramble]

did you not even have a linalg class?

[clear obsidian]

linear I had

see above I said ode and linear

[somber bramble]

nor any abstract algebra? real analysis? probability/stats?

[clear obsidian]

real analysis, stats I had also

[somber bramble]

oh I thought that was numerical analysis

[clear obsidian]

nah I had numerical analysis & real analysis (intro theory and logic before that)

I understand that the lack of abstract algebra might cause me problems so I read 300 pages of pinters abstract algebra (need to do more excercises yet though bc I went through it a little to fast) and I'm working on an online galois theory course as of recently

I got mostly A+'s in my last semester also so hopefully that will help me a bit

[mild laurel]

Is it true that in a general Euclidean domain with a norm function N, that N(x) = 1 implies x is a unit?

Okay the answer is yes

[mild laurel]

@chilly ocean Hey

@fickle brook There's a guy here that I just pinged that speaks French and needs some help with basic abstract algebra

[chilly ocean]

hi

[mild laurel]

@worthy kindle are you free

[chilly ocean]

so I was asking how to check if a ring is distributive in order to chdeck if it is really a ring

[worthy kindle]

Hello

[mild laurel]

He speaks french

[chilly ocean]

oui

c'est vrai

[worthy kindle]

Pour la distributivité, c'est avec le context qu'on fait ça

[chilly ocean]

ok dans mon cas l'exemple c'est Zn sachant que Zn c'est l'ensemble des restes dans l'arithmetique modulo n

mais pour le moment je veux just apprendre les proprietes de certains elements avant de les appliquer

je suis sur les les anneaux en ce moment

et je veux savoir comment verifier qu'un anneau est distributif

[worthy kindle]

Avec cette example-là, il s'agit de montrer que pour tout a, b et c, on a (a+b)c congru à ac+bc modulo n

[chilly ocean]

a,b et c sont dans l'ensemble?

et + c'est l'operateur ou juste un +?

[worthy kindle]

ce sont des entiers qu'on utilise comme représentants

[chilly ocean]

comme representants?

[worthy kindle]

Z/nZ c'est un ensemble de classes d'équivalence

[chilly ocean]

ok je vois

moi je travaills juste sur un ensemble

une chose apres l'autre

tu as dit "Avec cette example-là, il s'agit de montrer que pour tout a, b et c, on a (a+b)c congru à ac+bc modulo n
" donc je prend un element au hasard dans l'ensemble et je fais ca?

[worthy kindle]

oui, montrer une propriété pour des éléments arbitraires, c'est montrer la propriété pour tous les éléments

[chilly ocean]

ok je vois

tu peux me donner un set et je t edis s'il remplit la propriete de distributivite de l'anneau?

juste pour voir mes calculs?

@worthy kindle

[worthy kindle]

hein

[chilly ocean]

mais il va falloir faire vite je dois partir dans 5 minutes

je demandais si je pouvais faire un mini exo pour voir

[worthy kindle]

ah bah, euh, Z, avec + et × usuels

[chilly ocean]

ok on va dire que Z vaut {0,1,2,3,4,5,6,7,8,9,10,... infini}
je prend 3 elements au hasard: 2,3,4
si 3X(4+5) = 3x4 + 3 + 5
alors l'anneau est distributif

du moins avec cette propriete

@worthy kindle

[worthy kindle]

alors, 3, 4 et 5 ils sont pas arbitraires xd

[chilly ocean]

@worthy kindle je dois y aller. ce que j'ai dit est surement faux. on peut se voir dans 1 h je dois partir

[worthy kindle]

Ok, à bientôt peut-être

[chilly ocean]

ok a+

[chilly ocean]

How can you tell using Light's Test whether a binary operation has an inverse or identity?

Oh wait, I figured out the identity part.

Okay, but what about inverse?

🤔 that's not what Light's test is used for

maybe you want to use some alternative definitions of a group

I think one of them was

[cloud walrus]

Blitzkrieg:

[chilly ocean]

I used Light's Test to prove associativity

But he wants us to justify whether this set along with this binary operation is a group.

The set is nonempty, it's associative, and one of the elements when starred with any other element in both directions "leaves it alone"

I have no idea how I would go about showing (or disproving) that the elements have an inverse.

for every x you can try finding an element y such that xy = yx = 1
or use the above definition of a group

Ah.

Let's see if I was given that information.

1 meaning the identity, right?

yes

Idk which method is faster, probably the one with inverses

Okay, it's not a group because I can eyeball this and clearly see that a few elements wouldn't have inverses. Thanks!

Abstract Algebra is the first and only class where I do the later problems first, then realize I was over thinking the earlier problems.

the other definition is useful if you want to show something is a group, but can't necessarily eyeball out the inverses or identity

in a more abstract case or when semigroup is infinite

Good to know!

Okay, doesn't seem like this class covers infinite groups for the most part

yeah those are usually neglected for the finite case

probably because with the finite case you can work with cardinalities and such better

Lagrange theorem

with infinite case the same theory isn't really that useful

oh, and the other definition of a group (works for finite or infinite semigroups)

there is element 1 such that x1 = x for any x
and exists y such that xy = 1

i. e.

we only care about right identity and right inverses (without assumption of uniqueness, it follows from definition)

similarly we can care just about left identity and left inverses

but for example left inverses and right identity doesn't work, so you must remember to keep them the same 'side'

its more in the spirit of 'inverses and identity' definition, but weaker assumptions

[chilly ocean]

@chilly ocean We did that proof in class

And we also showed why they have to be the same side for both.

We did a counterexample where they weren't the same side and it was clear that it wasn't a group.

[chilly ocean]

I was checking if a ring with the set {0,1,2,3,4,5,6,7,8,9,10,... infinitre} is distributive.
I took randomly 2,3,4
if 3X(4+5) = 3x4 + 3x5
then the ring is distributive
but somebody told me 2,3,4 are not arbitrary.
what does he mean?
note I actually told 3X(4+5) = 3x4 + 3 + 5 by typo. it might have been confusing

if you read that tomorrow, then can you let a message?

[fading wagon]

The problem is, an example does not prove the claim in general

Claim: All numbers are 19.
Clearly, 19 works.
Hence, all numbers are 19.

[chilly ocean]

interesting I understand the problem

@fading wagon so to ensure R is a ring I have check each elements of the set

for example S = {0,1,2,3,4,5} then I have to do:

if 0x(1,2,3,4,5) = 0x1+0x2+0x3+0x4+0x5 then the ring is distributive

[fading wagon]

not like that

what does distributive mean?

it means for all $a, b, c$ in the ring, $a\times(b+c)=a\times b+a\times c$.

[cloud walrus]

Element118:

[chilly ocean]

it smeans a × (b + c) = (a+b) x c

mistake

from me

[fading wagon]

"for all" is important

[chilly ocean]

but in the last example I tried to say 0x(1+2+3+4+5) = 0x1+0x2+0x3+0x4+0x5

I made a typo

[fading wagon]

yeah, but that's not sufficient, I think

You need to prove that statement directly: For all $a, b, c$ in the ring, $a\times(b+c)=a\times b+a\times c$.

[cloud walrus]

Element118:

[chilly ocean]

where is the difference? in 0x(1+2+3+4+5) = 0x1+0x2+0x3+0x4+0x5 I say a = 0, b = 1, c=2 etc...

@fading wagon

[fading wagon]

you don't say a=0, b=1, c=2

you need to show for all

So, for all possibilities of a, for all possibilities of b, for all possibilities of c...

[somber bramble]

it could be that it works specifically for this particular sum you wrote out

but not if you e.g. leave out a number

so just cause it works here doesn’t mean it always works

evne though your example involves all numbers

[chilly ocean]

Ok so concretely how to proove the distribution of the ring?

[fading wagon]

you already have the distributive property of multiplication

on integers

so, if your ring has only integers with multiplication and addition defined as known, then just quote that

[somber bramble]

what’s actually the specific problem you’re trying to solve

[fading wagon]

@chilly ocean example?

[somber bramble]

I don’t see it written down anywhere

[tame bear]

if hes struggling with distribution hes not going to follow that

[wind steeple]

is this a children algebra channel ?

[fading wagon]

@chilly ocean maybe ask your teacher

[chilly ocean]

@fading wagon "you already have the distributive property of multiplication on integers" I need a way to determine the distribution with other sets. for example how to know if {1,3,4,2} is distributive? if I can not know the ... ah ok I might understand. you mean check if a set is distributive is not related to ring!

[somber bramble]

anyone else failing to parse that comment?

also, still

what’s actually the specific problem you’re trying to solve

[fading wagon]

@chilly ocean You need to see the definition of the multiplication and addition operation given to you

It may not be the same as normal multiplication and addition

[somber bramble]

a set can’t be distributive, something can only be distributive if you also have operations

namely two of them

[chilly ocean]

ah ok so it is as easy as check to see the operator and determine if the ring is ditributive???

easxy

[somber bramble]

what’s actually the specific problem you’re trying to solve

[chilly ocean]

I do not have any specific problem I want to understand ring properties. I give to myself exercises selfmade

[somber bramble]

ah

alright, let me put out a concrete example then

[chilly ocean]

"a set can’t be distributive, something can only be distributive if you also have operations" => lol it is a ring

[somber bramble]

a set isn’t a ring

a ring is a set plus some properties. but {1,2,3} is not a ring

it’s just a set with three elements

anyway, consider this then: take the integers modulo 5. that is, take the integers but then only consider their remainder when doing division by 5. so in this ring, 2+3 = 0, and 4*3 = 2, to make two examples

show that this is indeed a ring

[chilly ocean]

ok lets take an example {{1,4,5}, +, X} : if 1x(4+5) = 1x4+1x5 then the ring is distributive. and both are 9 then the ring is distributive

[cloud walrus]

Blitzkrieg: