groups-rings-fields

[chilly ocean]

hmmm

i am not so sure...

[pastel cliff]

i have a proof in my notes

[coral shale]

Z3 is a good start

[chilly ocean]

show

[coral shale]

your field gotta be char 3

[chilly ocean]

Z[x] is an integral domain but..

🧐

[upper pivot]

R[x] is id if R is id

ok see in the proof they say R id

so clearly its a typo

[pastel cliff]

[chilly ocean]

"integral domain" is not the phrase you are looking for

[next obsidian]

That is super not true

[pastel cliff]

well

[next obsidian]

That should say Euclidean domain

[pastel cliff]

it did seem really odd that we could jump from an integral domain straight to a field

[chilly ocean]

please, always test your results against Z

please

[upper pivot]

oh yeah thats what they r doing

lol

[next obsidian]

Or even PID

[pastel cliff]

lesson learned

[next obsidian]

Being a PID is enough to guarantee R is a field

[upper pivot]

what is this book/notes

lets cancel the author

[pastel cliff]

prof lecture notes

[chilly ocean]

kill prof

[pastel cliff]

no he's nice :(

[chilly ocean]

speak to prof about error

[pastel cliff]

well the final is tomorrow

[chilly ocean]

still, do it

like right now

that's a big error and you don't want anyone going into the final with that

[pastel cliff]

,ti

[cloud walrus]

The current time for stμ₂dying is 10:18 PM (EST) on Wed, 07/12/2022.

[pastel cliff]

i'll email him but aint no way that man is awake rn

[next obsidian]

The onus is on you to read the proof and realize it makes no sense with the given assumptions IMO

[pastel cliff]

this is true

[chilly ocean]

reading proofs

[pastel cliff]

ive been skimming old lecture notes to make sure i have things mapped in my head correctly

by skimming i mean reading theorems and accepting them as law

but yeah lesson learned

[coral shale]

back to the original Q, is frobenius the way?

[pastel cliff]

idk what frobenius is but he sounds funny

[coral shale]

frobenius map

then maybe not

[pastel cliff]

you said Z/3 was a good start

[coral shale]

just cus it has char 3

i mean yeah, the answer probably has something to do with it

likely some quotient of Z3[x]

[pastel cliff]

quotient?

[coral shale]

quotient ring?

[pastel cliff]

i know what it means but why

[coral shale]

my cat ears say so

[pastel cliff]

[coral shale]

its just a guess

[pastel cliff]

[coral shale]

if no one else gives a hint, just try and see what u come up with

after all if any quotient of Z3[x] is a field

then it must be char 3

and therefore have order a power of 3

[next obsidian]

I’ll be honest

Have you seriously not done a classification of finite fields in your class?

[pastel cliff]

we didnt really do many things with fields

[next obsidian]

[pastel cliff]

a stray result here and there

[next obsidian]

Rip

[pastel cliff]

the class following this one is called fields and modules tho

[next obsidian]

Ah

[coral shale]

so whys this question now

[pastel cliff]

pain

[pastel cliff]

well one thm i do remember being true is R/I is a field iff I is maximal in R

that's probably useful here

so that quotient idea makes sense

goober question is Z/3[x] finite?

probably not but i also remember that in a finite ring prime => maximal

[chilly ocean]

x, x^2, x^3, ...

very not finite

[pastel cliff]

ok yeah thought so

a friend told me he used that prime ideals are maximal in Z/3[x]

but

i was p sure that's only true in finite R no?

[agile burrow]

any non-zero prime ideal is maximal in a PID, and Z/3Z[x] is a PID

[pastel cliff]

is this because primes are irreducible and (x) is maximal when x is irreducible

[agile burrow]

Sort of; you're hiding the fact that we're in a PID in the second statement because that certainly isn't true in general

otherwise every UFD would be a PID

[pastel cliff]

ahhh right

forgot the prime irreducibles thing is true in UFD's

[pastel cliff]

found this

it kinda makes sense

irreducible => prime => maximal => R/I is a field

and im assuming it being of degree n somehow gives us the desired order?

https://math.stackexchange.com/questions/1474378/constructing-finite-fields-of-order-8-and-27-or-any-non-prime

reading through this, though im not done yet

[ripe basalt]

sebbb that's right

not alking about Z_3[X] here though

[stoic rose]

I've heard that quiver representations do not come with a tensor product. However there is still a fairly natural (but maybe naive?) way to tensor two quiver representations, namely take pointwise tensor product of all vector spaces and tensor product of linear maps. Why is this not studied, is it uninteresting for some good reason?

[pastel cliff]

besides eisenstein, are there any other ways of showing that a polynomial is irreducible

also ive asked adjacent things before but are there tricks to showing that something is irreducible in a euclidean domain

8 hours remain

[burnt flower]

Rational root test

There’s also generalized Eisenstein if I remember right

[stoic rose]

That's not very helpful for irreducibility unless your polynomial has degree 2 or 3

[stoic rose]

Well given how hard it is to tell wheter a big integer is prime or not, I'd say it's an equally difficult problem in other euclidean domains

[south patrol]

Idk it is basically just hard rip

For degree <=3 suffices to show no roots and often stuff like taking it mod a prime helps to do that

Sometimes u can shift then eisenstein

Like with prime degree cyclotomics

[pastel cliff]

then what about the norm stuff that came up when i was trying specific examples before

ig that's not polynomials tho

[south patrol]

Like which

[pastel cliff]

.

[next obsidian]

That’s just a trick that works well here

[pastel cliff]

when else does it work well

Z[sqrt{-5}] aint even a euclidean domain

[next obsidian]

You have to exercise your creativity

[pastel cliff]

ok as an example, showing that x+1 is irreducible in Z[x]

is this just assuming x+1 = rs for non-units and finding contradiction

like can i just say that if it's the product of non-units it would also be a non-unit?

[next obsidian]

Just look at the quotient and it’s just Z

[pastel cliff]

saw that now ye

[formal ermine]

reduction modulo p

if R is a ufd and f in R[x] normed, then for a prime elemenet p in R if f is irreducible in R/p[x] then it is also irreducible in R[x]

[chilly ocean]

im a bit confused on what it means to say a polynomial is irreducible mod an ideal

[next obsidian]

Suppose I is an ideal of R

Let f in R[x]

By reducing all coefficients of f mod I, you get a polynomial f mod I living in R/I[x]

Is that polynomial irreducible?

[chilly ocean]

thank you! this is what i guessed but it did not appear anywhere in my notes

[coral shale]

its the same process when asking if a polynomial is reducible mod 3

the 'mod 3' at the end there indicates the ring is Z/3Z and the polynomial ring is (Z/3Z)[x]

[pastel cliff]

2 hours remain

[coral shale]

this one?

[pastel cliff]

oh nah that's fine

i mean until my exam

[coral shale]

oh right, maybe if u ask nicely u can take it next month

[white grotto]

what are all the groups that have a finite number of sub groups

ig all cyclic groups?

and are (Z,+) ,(Z^2,+) isomorphic?

[south patrol]

Z does not have a finite number of subgroups

every other cyclic group has a finite number of subgroups, since indeed any finite group does

[south patrol]

Is (Z^2,+) cyclic?

[chilly ocean]

only the finite groups I think

if such group weren't torsion, then it would contain Z which has infinite amount of subgroups

and otherwise, if it were infinite torsion group then consider all subgroups generated by one element

they're all finite and there is a finite amount of them, contradiction

yeah

[white grotto]

no, i think

[chilly ocean]

well this is really the theorem about finitely generated abelian groups

the isomorphism class depends on how many Z, Z/p^kZ you have

Z² has two disjoint cyclic subgroups but Z doesn't therefore they are not isomorphic

[white grotto]

wdym

[white grotto]

and this

[white grotto]

yeah seems good

[chilly ocean]

the theorem about finitely generated abelian groups tells you not only that all groups are isomorphic to a finite product of Z and Z/p^kZ for p a prime number, but also that such representation is unique (in the sense that it only depends on how many of those factors would you take)

[white grotto]

we can decompose Z^2 to Z+Z?, kind of?

[chilly ocean]

those two are isomorphic yes

more generally GxH is isomorphic to G+H for any abelian groups G and H

[white grotto]

waw

okay

and how does this help here

[chilly ocean]

Z^2 and Z are not isomorphic

[white grotto]

okay

[chilly ocean]

both Z and Z^2 are representations like in the theorem about finitely generated abelian groups

first group has one factor of Z while second has two

simple as that

[south patrol]

but also like if you take any element of Z^2 it clearly doesn't generate the group

that'd probably be how i'd do it

but ye many ways to do it

tensor with Q and then this would be an iso between Q and Q^2 ig

[chilly ocean]

then you'd still need to show Q not iso to Q²

[rotund aurora]

Given V vector space, is there a smallest algebra containing it?

[chilly ocean]

is V a vector subspace of some algebra then

[agile burrow]

you could always put the trivial lie bracket on V

unless you want associative algebra

[rotund aurora]

not necessarily. By containing it I mean like homomorphic injections

[chilly ocean]

well, smallest algebra implies we already are in some algebra, I think

[chilly ocean]

I think we can just inject V into its exterior algebra then

this is the "largest" unital such algebra from some canonical properties

I think

and then just take intersection of all subalgebras containing V

[agile burrow]

maybe tensor algebra instead of exterior?

[chilly ocean]

wait yeah. That's different

I was thinking of tensor algebra, sorry

[pastel cliff]

pop quiz, state 3rd iso without googling it

[agile burrow]

sebb is your final today

[pastel cliff]

*was

i blanked on 2nd and 3rd iso

[chilly ocean]

naming conventions differ from place to place

[agile burrow]

that's unfortunate

but i hope you did well overall

[rotund aurora]

(G/H)/(H/K)=G/K

I mixed up the H and the K haha

close

[pastel cliff]

yeah i took for granted that only first iso would appear and now i feel dumb

[rotund aurora]

but lol I think I never used that explicitly, I have used the correspondence theorem a lot, I think they are kind of the same, so I would just state that

[pastel cliff]

but it went better than i thought

[agile burrow]

glad to hear it

[rotund aurora]

The smallest algebra containing $V$ would be an algebra $A$ such that there exists an injective linear transfromation $f\c V\to A$ and such that if $B$ is another algebra then there exists an injective algebra homomorphism $g\colon A\to B$

I kind of meant this, but I guess this is too strong

[chilly ocean]

still got to answer walter's question

[rotund aurora]

What question?

[chilly ocean]

walter was asking if the algebras are required to be associative

[rotund aurora]

btw, looking at intersections of algebras in T(V) looks pretty good, so thanks

[chilly ocean]

or well, unital too

[rotund aurora]

okay I thought in general we considered algebras to be associative. But the tensor algebra T(V) is associative, then the answer would be very different no?

[chilly ocean]

Lie algebras

it's literally in the name

[rotund aurora]

yeah kind of stupid

what's the trivial lie bracket?

[chilly ocean]

sends everything to 0

[rotund aurora]

lmao

But if you wanted to give a vector space a non-associative algebra structure, is there a natural thing you could do?

[chilly ocean]

The natural Lie algebra structure on an algebra such as, the algebra of operators on a vector space, is given by the bracket [A, B] = AB - BA

this in general won't be associative

[rotund aurora]

I c I c

T(V) is associative right?

[chilly ocean]

yeah

I don't think we require much else than associativity

it's literally like a free construction

[rotund aurora]

yeah I know

but then Im confused about the universal property

I think thats why I assumed that algebras were associative

If there is a linear transformation from V to a non-associative algebra A, then this factors through T(V) by an algebra homomorphism, but this is an algebra homomorphism between an associative algebra and a non-associative algebra?

[chilly ocean]

concept of free non-associative algebra over a field F, exists I'm pretty sure

but I don't even want to imagine what it is

we know this from say, universal algebra

does a free non-associative algebra generated by some vector space exists? Hmm. Not sure

I mean it should be

[agile burrow]

you could take non associative monomials, right?

like similar to tensor algebra but just don't make it associative lol

[chilly ocean]

after adding all the vectors in V as nullary operations

oh, well, I was thinking of how to do this using the properties of universal algebras

[agile burrow]

fair enough, i don't know about universal algebras so I can't comment

[chilly ocean]

well this doesn't quite work since they could all "squish" together into a null subspace

non-associative monomials it is ig

[rotund aurora]

I was trying to ask what happens in this universal property when A is non-associative, because T(V) is associative, so many things will get killed?

ahhh

nvm

the universal property requires A to be associative

[chilly ocean]

yeah, it cannot be non-associative

[rotund aurora]

lololo that's why I was confused xD

so perhaps you would look at the analog of T(V) but for non-associative algebras

interesting

[chilly ocean]

unless we say that a non-associative algebra is just an algebra, then any algebra is non-associative
nomenclature issue

it's like, you have non-commutative rings but... those are just rings, so commutative rings are non-commutative ig?

[agile burrow]

i've always seen non-commutative rings reserved specifically for rings which are non-commutative

and just "rings" to encompass both commutative and non-commutative

[chilly ocean]

hmm... I guess wikipedia suggests the same

but with non-associative algebras it's a funny naming issue

associative algebras are non-associative

[rotund aurora]

maybe this is very open / dumb question, but given an algebraic structure (rings, vector spaces, etc.) then its easy to introduce relations (ideals, subspaces, etc.). But what about "deleting" relations. So for example, suppose your ring R is commutative, could you make it non-commutative in a natural manner, starting from the given R?

[chilly ocean]

well the biggest issue here is I think this
I don't think that a free (universal) algebra containing said object exists in general

I know free object exists in equational classes, but probably not the above

like what you are concerned with here is an extension problem

[solar glacier]

Can I verify a proof that if R is a PID every irreducible is prime. Let p be irreducible. We seek to show (p) is maximal as maximal ideals are prime in commutative rings with 1. Let I be an ideal in this PID then I = (m) and assume (p) \subset (m) then there exists and r \in R such that p = rm but as p is irreducible then either r or m is a unit. if m is a unit then I = R and if r is a unit then I = (p) thus (p) is maximal and p is prime.

[chilly ocean]

seems good

[south patrol]

Yeah i think you can basically define irreducible elements a as those elements with (a) maximal among proper principal ideals right

And then for PID that just becomes equivalent to (a) being maximal

[rotund aurora]

Shouldnt these be called anti-symmetric, and alternating reserved for forms that are zero when two of the components are equal?

[chilly ocean]

if two of the components are equal then by transposing those two components you get f(that) = -f(that) so f(that)=0

well

unless you're in char 2

[rotund aurora]

but are they equivalent?

uhm idk are they?

[chilly ocean]

over char 2 no, not sure over other characteristics

i think they are equivalent over non-2 characteristic

[rotund aurora]

I think not, that's why I asked

ah no alternating maps are also antisymmetric

f(a+b,a+b,c,...)=f(a,b,c,...)+f(b,a,c,...)=0

[chilly ocean]

yeah

[chilly ocean]

anti-symmetric doesn't sound too good, does it

after all, an anti-symmetric relation is something like an order for example

[rotund aurora]

I have also heard skew-symmetric

[chilly ocean]

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

only heard of alternating tensors before

[formal ermine]

let $E/L$ and $L/K$ be two extension fields. \
if $E/K$ is algebraic, then any subfield of $E$ (such as $L$) must also be algebraic, as any element of $L$ is also an element of $E$ and can thus be expressed as the root of a polynomial in $K[x]$. \
if $E/L$ and $L/K$ are both algebraic, then any element of $E$ can be expressed as the root of a polynomial in $L[x]$, and any element of $L$ can be expressed as the root of a polynomial in $K[x]$. therefore any element of $E$ can be expressed as the root of a polynomial in $K[x]$, thus meaning that $E/K$ is algebraic.

did I make a mistake somewhere?

[chilly ocean]

How do you justify that any element of E can be expressed as a root of a polynomial in K[x]?

in both cases

It is true, but you didn't really give an explanation for why that is true

[white grotto]

why is this

i understand if the group H is a normal subgroup then G/H exists

but why is this equivalent to saying the set of left cosets forms a group

[formal ermine]

what part are you confused with

[white grotto]

what does it mean for left cosets of H to form a group

[chilly ocean]

G/H also exists if H is not normal, but it isn't a group then

[formal ermine]

to have a group structure

G/H is a group if and only if H is normal

[white grotto]

isnt each coset represented by a representative element?

[formal ermine]

like our definition of a normal subgroup directly follows from trying to define a group structure on G/H

[chilly ocean]

not by a unique one

[chilly radish]

How do you suggest we multiply cosets

[white grotto]

yeah... but how can we talk about a group of cosets

[chilly radish]

If using representatives, why is this well-defined

[chilly ocean]

a coset is a class of all equivalent elements

[chilly radish]

Rather, WHEN is this well-defined

if the multiplication doesn't depend on the representatives, we can form a group

[white grotto]

yeah

[formal ermine]

$G/H = \Set{g \circ H | g \in G}$

[white grotto]

doesnt it always not depend on the representative in cosets?

[chilly radish]

No

[formal ermine]

and we're trying to "make that set into a group" (define a group structure on it)

[chilly radish]

It is independent of the representatives IFF H is normal

[white grotto]

this isnt showing up

[chilly radish]

i.e., we can form a group out of them

Bot is dead

[formal ermine]

texit is down

[white grotto]

ouf

[white grotto]

mm.... i dont see why... i mean cosets are meant to represent the same structure of a subgroup

[chilly radish]

?

No they don't

[white grotto]

aH

[chilly radish]

Taking the quotient is 'setting H to 0'

All the elements of aH only differ by elements of H

It's exactly the complement of representing the subgroup's structure

[hot lake]

if the group operation of G induces a well-defined operation on G/H then that makes G/H into a group

and that happens if and only if H is normal in G

though it's a bit rude to say that G/H doesn't exist otherwise

[white grotto]

okay but i am confused about representatives...

[hot lake]

say you want to multiply aH and bH

[white grotto]

why when its not normal the representatives matter

[hot lake]

you pick a representative ah1 in aH, and a representative bh2 in bH

their multiplication is then a h1 b h2

[white grotto]

yeah

[hot lake]

and in the quotient that's in a h1 b H

I mean, its equivalence class

but if you had picked a and b as representatives instead

you would have ended in a b H

and it's not always the case that a h1 b H = a b H forall h1 in H

[white grotto]

i dont think i got it

[rotund aurora]

https://math.stackexchange.com/questions/14282/why-do-we-define-quotient-groups-for-normal-subgroups-only/14315#14315

You can read the answer of Arturo Magidin, I really liked it when I was learning this

[hot lake]

so in general, the equivalence class of the result of mulplying a representative of aH with a representative of bH

depends on the choice of representatives

so you can't define an operation on G/H with that

[white grotto]

why ah1b H != ab H

and why normality fixes that?

[chilly radish]

What is the definition of coset equality

Write out the definition

[white grotto]

well yeah turns out i am having a bit of a hard time i should restudy this

[chilly radish]

Look

You're almost there

Just spell out the definition

That's all you have to do

[south patrol]

yes but i meant as vector spaces those are clearly different by taking dimensions

but yeah it was a joke

[white grotto]

we have aH={ah; h in H}

wdym by coset equality exactly?

that for any ah in aH, ahH = aH?

[chilly radish]

No

I mean

When is aH=bH

[south patrol]

maybe a different way to see H must be normal if G/H is well-defined is that like

[chilly radish]

Well yes but more explicitly when does this happen

[south patrol]

the projection G -> G/H has kernel H assuming G/H is a sensible object

lol

[chilly radish]

And then apply that condition to ah1bH=abH

[white grotto]

aH= bH when b is in the same coset as a

[chilly radish]

Which happens when...

[white grotto]

i am not sure

[formal ermine]

E/K => E/L and L/K:
let a be in L, and therefore in E. because E/K is algebraic, there exists a nonzero polynomial f(x) in K[x] such that f(a) = 0; thus E/L is algebraic. similarly, since a is a root of f(x) in E, it is also a root of f(x) in K, since K subseteq E. this means that L/K is algebraic.

E/K <= E/L and L/K:
since E/L is algebraic, for an a in E there exists a nonzero polynomial g(x) in L[x] such that g(a) = 0. similarly with L/K; there exists a nonzero polynomial h(x) in K[x] such that h(b) = 0 for a b in L. we can combine the two polynomials into f(x) = g(h(x)). this polynomial lies in K[x] and has a root in E. therefore E/K is algebraic.

is that correct?

[white grotto]

what you mean

you want to say when H is normal?

[chilly radish]

No

[formal ermine]

what property do the elements a and b need to fulfill such that aH = bH

[chilly radish]

When are two elements in the same coset

[white grotto]

b = ah

for some h in H

[chilly radish]

Another way to say this is that a^-1b is an element of H right

[white grotto]

yeah

[chilly radish]

Ok so ah1bH=abH implies that
b^-1a^-1ah1b= b^-1h1b is in H

But look, we just conjugated an element of H by some element of G and landed back in H

If this happens for all elements, this is exactly the definition of normality

[white grotto]

how did u use that a^-1 b is an element of H

[chilly radish]

That's the reason b^-1h1b is in H

I used the definition of coset equality

[chilly ocean]

Well, Z² also different from Z as free-Z-modules by taking dimensions

[white grotto]

why?

[chilly radish]

Because coset equality is the statement that the right thing inverted multiplied by the left thing is in H

Or equivalent to that statement

[white grotto]

i fail to understand what you mean

[tall surge]

if a group acts transitively on a set of $n$ objects, does it automatically act transitively on a set of $n-1$ objects?

[cloud walrus]

Winston_Smith

[south patrol]

ye

[cloud walrus]

ShiN

[chilly radish]

Why is it so SMOL

[white grotto]

Okaaaaay i see

thanks a lottt

[chilly radish]

[true ingot]

Hey Ive been reading something and come across this. Does anyone know why or in what sense does the fact that if G/H is isomorphic to Z this means "G and H give essentially the same information"? And what would the quotient have to be isomorphic to for that to not be the case?

[formal ermine]

.

[prisma ibex]

that's not really what this is saying, the two relevant facts here are that DivF/Div^0F=Z and PicF/Pic^0F=Z. Since PicF and Pic^0F are quotients of DivF/Div^0F by some equivalence relation, this is essentially saying that PicF is just like, Z-many copies of Pic^0F

[woven obsidian]

The second part is not correct. You want to prove that for any a in E, there is a polynomial in K[x] that has a as a root.

[formal ermine]

g(h(x)) is not necessarily in K[x] is it?

[woven obsidian]

You will probably have to write down a less direct proof

[white grotto]

how is this deduction immediate?

[formal ermine]

hH = H iff h is in H, try proving it

[woven obsidian]

You never define what b is, but nevertheless the composition need not be in K[x]

[formal ermine]

yeah ok

let me try something else then

[woven obsidian]

Do you want a hint?

[white grotto]

ah in H for all h in H, then ah1=h2, a=h2h1^-1 in H?

[formal ermine]

let me try something myself first, if I still can't figure it out - then sure c:

[white grotto]

wait what does this mean

just this

[woven obsidian]

If x=y then you can multiply both sides by y^(-1) to see that it is equivalent to xy^(-1) = e, where e is the identity

[white grotto]

yeah this i got... but the paragraph after it

normal subgroup is the set of elements equivalent to the identity?

[chilly ocean]

in the quotient group yes

taking the quotient G/N is basically saying: everything in N is now the identity

[white grotto]

under this logic isnt every subgroup containing the identity element a normal subgroup?

because being in the same coset corresponds to an equivalence relation no?

[chilly ocean]

every subgroup contains the identity element, so something's going wrong

it will be an equivalence relation but not a congruence

you want your relation to satisfy [a]+[b]=[a+b] for all a,b where [x] means the equivalence class of x

this is equivalent to being given by a normal subgroup

[white grotto]

proof

[white grotto]

wdym

[chilly ocean]

Look let's go step by step

Let's say you have a group G

you know what an equivalence relation is right?

Let's say we have an equivalence relation ~ on G

given an element a of G, call [a] it's equivalence class

now we'd like to somehow define multiplication of these equivalence classes

it would be natural to ask that [a][b] = [ab]

we need to check if this is possible

[white grotto]

im following

[chilly ocean]

because for some a~a' and b~b' it might happen that [ab] is different than [a'b']

then our definition wouldn't have made sense

because [a][b] must equal [a'][b'] since [a]=[a'] and [b]=[b']

[white grotto]

yeah okay we have to make sure it doesnt depend on my representative and we need it to be normal for that to happen

[chilly ocean]

yes

[white grotto]

Ah so the key here is that the equivalence relation is a subgroup already

so we know the relation is defined

so ofc the group equivalent to e is a normal subgroup

right?

[chilly ocean]

when you write G/H for H some subgroup of G the relation is given by a~b iff ab^-1 is in H yes

[white grotto]

continue with your explanation sry for interfering

yes

can i think of normal subgroups as a generalization for 2 elements are equal if xy^-1 is e?

but instead we think of them as being equivalent if they're in a shared subgroup?

[chilly ocean]

we think them as being equivalent if they're in the same coset

[white grotto]

or is it not normal, just cosets

[chilly ocean]

cosets are not subgroups except for the identity coset

[white grotto]

mean this

Ah

then no

[chilly ocean]

in G/N you have [a]=[b] iff ab^-1 is in N

[white grotto]

yeah so i am right?

[chilly ocean]

that is iff [ab^-1] = [e]

[white grotto]

Ah

so if ab^-1 is in the normal subgroup

[chilly ocean]

then a=b in the associated quotient group yes

[white grotto]

I think i see it much better now

thanks a lot

ill need to restudy everything with this point of view

[solar glacier]

question

how do I show 2,3 are irreducible in Z[i\sqrt{5}] using norms

so I assume 2 = ab where a and b are both non units

then do I take the norms to conclude N(a)N(b)=4

but the norm of no such element of Z[i \sqrt{5}] has norm of 2

[formal ermine]

you can also show it more explicitly

a = u + isqrt(5)v and b = x + isqrt(5)y, so under the norm it's 4 = (u^2 + 5v^2)(x^2 + 5y^2)

[solar glacier]

forcing either N(a) or N(b) equal 1 which implies a or b is a unit

does my method work

tho

and i can show no element of Z[i\sqrt{5}] has norm of 2 or 3

[solar glacier]

so eeitheer both factors need by 2 or one is 4 and one is 1? but both cannot eual 2

since the solution wont be in Z

[rotund aurora]

Is it usual to talk about "anticommutative polynomials"? By this I mean, for example, the polynomials R[x,y] where xy=-yx

[formal ermine]

yeah I'd need a hint please :v

[woven obsidian]

If E/K is an extension, then a in E is algebraic over K iff K[a] is contained in some finite extension of K

[chilly ocean]

And you can give them a non-discrete partial order!

[formal ermine]

do I do something with the minimal polynomial of a?

oh wait nvm

a needs to be algebraic in the first place for there to be a minimal polynomial

hmm

oh wait

I think I have an idea

[white grotto]

is there some intuitive understanding for why there is no homomorphisms other than the null homomorphism between (Q,+) and (Z,+)?

[rotund aurora]

What notation do you guys use for the exterior algebra in latex?

[white grotto]

why is there no way for a function to preserve the structure of the group

[chilly ocean]

$\wedge$ or $\bigwedge$

[cloud walrus]

ТТерра

[white grotto]

(Q,+)

[chilly ocean]

small wedge if it's inline, big wedge i treat like a \sum or \prod symbol

[rotund aurora]

yeah, so I had this problem, when you are in equation mode bigwedge is too big

Im using \textstyle{\bigwedge}

[chilly ocean]

that probably belongs in #latex-help

[rotund aurora]

No, its algebra notation question. I can fix size, but if you are in equation mode you just write \bigwedge(V) for the exterior algebra of V?

the left seems too big to be standard, thats why I was asking, tho I may be wrong

especially because the size gets bigger in equation mode

[chilly ocean]

i would go with personal preference. i don't think there is a standard

the one on the right looks better inline

but it doesn't look as good if you're using the wedge symbol like you do \sum or \prod (i.e. writing something like a wedge over vector spaces V_i)

[white grotto]

how

[coral spindle]

Hint: for any a in Z_n, a + ... (n times) ... + a = 0.

[chilly ocean]

Can you please ping me next time

Anyway what do you mean by giving them a non-discrete partial order

[white grotto]

ok so n phi(1)=0 and phi(1)=0, but why did we need the embedding hypothesis?

[coral spindle]

Because that's the statement lmao

Fwiw

If the kernel is zero then it's injective, so there are only two possibilities

[white grotto]

i mean can we find any homomorphisms other than null, that are not injective, from Zn to Z?

[coral spindle]

(since Zn is cyclic)

[chilly ocean]

I wouldn't even know that this message exist if I didn't decide to browse in here

[white grotto]

can we prove it without the injective hypothesis?

can we prove that the only homomorphism from Zn to Z is the trivial one

[formal ermine]

let a in E. because a is algebraic over L, there exists an f(x) = sum_{i = 0}^n b_ix^i in L[x] such that f(a) = 0. it follows that a is algebraic over K(b_0, ..., b_n) and we therefore have [K(b_0, ..., b_n, a) : K(b_0, ...., b_n)] < infty. we have [K(b_0, ..., b_n) : K] < infty as b_0, ..., b_n in L are all algebraic over K. using the degree theorem (don't know the english translation), we get [K(b_0, ...., b_n, a) : K] = [K(b_0, ..., b_n) : K(a)] * [K(a) : K] < infty. therefore a must be algebraic over K.

is this correct? and if so, is there a more elegant proof to it?

[white grotto]

nvm i think i did that xd

[coral spindle]

U proved it without the injective hypothesis didn't you

[white grotto]

yep

[coral spindle]

Well there you go

You're in the clear

[woven obsidian]

Yeah that's essentially it, looks good

[formal ermine]

oke great

[woven obsidian]

By only considering one element at a time you can reduce to finite case

[chilly ocean]

Oh, sure, i just didn't want to bother because some people might be bothered by pings.

If G is a preordered group (as in G is a group that makes + monotone) and is finite then if a is a non-zero positive so will be -a because a has finite order. Therefore the preorder will not be antisymmetric because a>=0 and a<=0 but a is not 0. So there are no finite antisymmetric preordered groups aka finite partially ordered groups other than the discrete ones (x>=y iff x=y)

[chilly ocean]

No, I love being pinged, please always ping me

I see what you mean now. Thanks

[chilly ocean]

haha okay Blitzkrieg :)

Did you check the book ordered groups and topology?

[chilly ocean]

Just call me Blitz

I've never heard of it, but there's a topological functor from the category of preordered groups to the category of groups

so there's some relation with topology, for example given adequate morphisms of groups into/from ordered groups you can also put final and initial orders

It deals with some algebraic topology but first chapter or so is free of it and deals with some cool theory of ordered groups

You might be interested

Does it use ordered groups later in the book?

Ordered groups are the main point of the book

Topology refers to knot theory and orderability of fundamental groups of spaces, mostly

by Adam Clay right?

Yep

seems to be more about total orders but still interesting, thanks

It is

[white grotto]

an intuitive way of understanding this?

i mean i think i can see it clearly on a cayley graph

[chilly ocean]

let me make a little drawing for you fishe

[main needle]

If you kill all stuff that "aren't injective" you get bijection with the image

[white grotto]

ah. makes sense

[chilly ocean]

i wonder what's the motivation for left-orderable groups tho. For bi-orderable (even if not total) we can think of such groups as the automorphisms of some object in some preorder-enriched category, which makes them seem natural.

[coral spindle]

G/Ker(phi), as a set, is the equivalence classes of the relation a ~ b iff phi(a) = phi(b). There is then a bijection between these equivalence classes and the image of phi — this should be obvious. The isomorphism as groups basically just comes from the fact we're working with groups.

[white grotto]

yep. its clear now

thanks a lot

[chilly ocean]

When i was first learning group theory someone showed me a drawing like this which helped me a lot

Proof by picture:

[rotund aurora]

Its saying that the kernel determines the homomorphism completely. This is because phi(a)=phi(b) if and only if phi(ab^{-1})=1, that is, iff ab^{-1} is in the kernel. Thus if two homomorphisms have the same kernel, then the images are the same... (up to isomorphism I guess)

[white grotto]

yeaaaah i just had imagined this from the explanation of boytjie

thanks for the picture tho now its stuck in my head

[white grotto]

Ah very cool

we can construct the whole homomorphism from the kernel

[rotund aurora]

yes

[white grotto]

i didnt see it this way

[chilly ocean]

No!

[rotund aurora]

isomorphically yes? unless Im missing something

like thats first iso lol

ah he said function.. ok. But well we are doing algebra

But notice that this is only possible because elements in a group are invertible

[chilly ocean]

any Z->Z sending 1 to anything other than 0 have same kernel but are not same function

[white grotto]

i meant homomorphism ofc

the function, not "any" function

[white grotto]

right

[chilly ocean]

what the kernel determines is the image up to isomorphism as a group

[white grotto]

yeah

up to isomorphism

i see

makes sense ofc

[chilly ocean]

determining the image up to isomorphism is not the same as determining the whole homomorphism

[white grotto]

yeah i misspoke

thanks for the correction

[chilly ocean]

i m refering to croqueta's message, the first sentence is false

[rotund aurora]

I was speaking informally here

[formal ermine]

let $K$ be a field and $f = \frac{X^3}{X^2 + 1} \in K(X)$ we know that $X$ is algebraic over $K(f)$. how does that help us in showing that $K(X)/K(f)$ is an algebraic extension?

[cloud walrus]

i believe in mathemagic

[rotund aurora]

but I guess it would have been clear if I said image.

But you would still have to add "up to group isomorphism...", kind of annoying

[chilly ocean]

Who determines those things other than us mathematicians

[coral spindle]

Computers, sometimes 🤷

[chilly ocean]

Computers are humans too

[coral spindle]

I don't give them rights

I whip my laptop

[white grotto]

well pray they'll give them to you

[chilly ocean]

Wdym. My anime AI waifu has all the rights

[coral spindle]

She is an exception. I'll allow her rights.

[chilly ocean]

humans are made by humans so they're artificial so humans are AI

[white grotto]

this is a very strong result

[chilly ocean]

They're not anime though

[chilly ocean]

There's an even stronger similar result for finitely generated abelian groups

[white grotto]

so freakin weird tho

[coral spindle]

mm

Idk if it's weird

But it's not easy to prove, by any means

[white grotto]

i just fail to see how abelian does anything

[chilly ocean]

Interestingly once you go into non-finitely generated abelian groups we can no longer give them any nice characterization

[rotund aurora]

Z is abelian

[white grotto]

abelian is a very weird property that i fail to grasp

the true meaning of

[coral spindle]

How is it weird sorry

[rotund aurora]

7 year olds do fine with commutativity

joke

[coral spindle]

This is an unusual take, that Abelianity is weird

[white grotto]

hold on people

lemme explain

like with cyclic groups, it represents some rotation

[coral spindle]

Abelianity is in fact a very nice property that makes things super neat (and this theorem demonstrates that to some extent)

[white grotto]

with Dihedral groups, some reflection

permutations, symmetric ....

[chilly ocean]

For non-abelian groups the order matters, but for abelian ones it doesn't. So if you have generators a, b for a group, if it is abelian every element is of the form a^n b^m. But for non-abelian ones is much more complicated: you can have weird sequences like abaabababababbb

[coral spindle]

So you're asking what symmetries do Abelian group represent?

Well this theorem answers that

[chilly ocean]

To describe it as weird I'd first have to say how natural numbers and their definition is weird

[coral spindle]

Products of "cyclic" symmetries

[chilly ocean]

Commutativity is nothing

[elder wave]

free product moment

that thing confused me for a while

[white grotto]

what does it mean for matrices for example that AB=BA, what does it imply about the nature of A and B and their relationship

[rotund aurora]

God gave the natural numbers to confuse humans

[coral spindle]

I means they have no relationship essentially.

This is the best interpretation

[white grotto]

yeah

its weird then

[coral spindle]

Not really

[white grotto]

it doesnt tell us much

[coral spindle]

No it tells you a LOT

It's a very strong property

[white grotto]

ok now i see why the theorem makes sense

[white grotto]

I know its strong

[chilly ocean]

I once had an idea. That makes abelian groups significant among groups. From a universal algebra perspective

[coral spindle]

Go on then blitz

Something to do with congruences?

In my head the nicest property of Abelian groups is that every subgroup is normal

[chilly ocean]

Nah. I'm not ready to reveal this great idea

[coral spindle]

Nooooooooo

[white grotto]

blitz is full of secrets

[chilly ocean]

also abelian groups = Z-modules

[chilly ocean]

That I am

so abelian groups are "vector spaces over Z"

[white grotto]

z-modules?

[white grotto]

what

[coral spindle]

The theory of modules generalises what is nice about Abelian groups

[white grotto]

ok i am not ready to go into it lol i have finals in 3 days

[coral spindle]

You know what a ring is, right, fishe?

[white grotto]

yeah

[coral spindle]

We can be brief

[white grotto]

like a group but with a multiplication

right?

[coral spindle]

Well a vector space is an abelian group that has multiplication from a field, right?

[white grotto]

yeah

[coral spindle]

And a field is in particular a ring

So a module over a ring R is just the same definition, but with a ring rather than a field

[white grotto]

yes every element has an inverse

[coral spindle]

So a Z-module is with multiplication from the integers Z

And you can work out that this is just every Abelian group

That's all Dishwasher means by this

[white grotto]

a module is an abelian group that has multiplication from a ring?

[coral spindle]

Yes that's right

there are some conditions on how this multiplication works ofc

[white grotto]

Ah

[coral spindle]

The theory of modules is very nice. Noether first proved the first isomorphism theorem for modules, in fact.

[chilly ocean]

I'll tell you since I don't do algebra anymore.
The idea is to look at the free algebra generated by n elements. This creates an increasing family of equational classes. For groups this sequence consists of two elements, abelian groups (n = 1) and groups (n>1)

[coral spindle]

Ha! I like that

[white grotto]

very cool boytjie thanks a lot

[coral spindle]

And Abelian groups are those that are quotients/subobjects of products of F(1)

a good point Blitz

I wonder if this is easily generalisable

[chilly ocean]

For rings it's the same. Lattices seem to form an infinite family though

[coral spindle]

For semigroups it's the same, right

[chilly ocean]

Yeah

[coral spindle]

And Lie algebras... yes those would be the Abelian Lie algebras

[chilly ocean]

free algebra as in Z[x]?

[coral spindle]

Ah no

This is in universal algebra

Blitz means "algebra" as in "algebraic structure"

in the language of universal algebra, e.g. a group would be an "algebra"

It's a horrible name imo.

[chilly ocean]

what does "free" mean in this context?

[coral spindle]

Satisfying the universal property of the free object

[chilly ocean]

Yeah

[coral spindle]

I know that's an annoying answer

But e.g. the free group is the free algebra in the variety of groups

[chilly ocean]

i know

[coral spindle]

The Kleene star is the free monoid

[white grotto]

to do basic arithmetic we need a field right?

[coral spindle]

Wym by basic arithmetic

[rotund aurora]

number theorists ded

[chilly ocean]

free algebra relative to what variety?

[formal ermine]

.

[white grotto]

gcd, smallest common multiple...

[chilly ocean]

the variety with just 1 constant?

[coral spindle]

I don't know what this means

[chilly ocean]

Fixed equational class

but which equational class?

[coral spindle]

WHOOPS WRONG MESSAGE

[chilly ocean]

Can be anything

[coral spindle]

No. gcd/lcm is trivial in any field.

gcd(a,b) = 1 for all nonzero a,b in any field.

[white grotto]

the euclidean algorithm then just requires a ring?

[chilly ocean]

The idea is that given variety $V$ we construct a countable sequence $$V_1\subseteq V_2\subseteq ...\subseteq V$$

[cloud walrus]

Blitz

[coral spindle]

No, it fails for general rings.

[chilly ocean]

it requires more. it requires an euclidean domain

[formal ermine]

it requires a euclidean ring

[coral spindle]

If you want to define the gcd/lcm in general (and have it possibly be nontrivial) the most general context I'm aware of it being in is a UFD.

[chilly ocean]

This gives us "distinguished" equational classes in V

[white grotto]

remind me whats a UFD

[coral spindle]

Unique factorisation domain.

[chilly ocean]

The ones that seem the most important

[white grotto]

a ring with something

[formal ermine]

ring where you can do prime factorization

[white grotto]

in a field the euclidean algorithm will always yield 1 right?

as the gcd

[coral spindle]

No, the euclidean algorithm will not be defined in general.

[white grotto]

ah yeah its not

[coral spindle]

Talking about the quotient and remainder is not meaningful

[white grotto]

god i get lost in the formalities and forget to make sense of what i am saying

[chilly ocean]

this is always possible, you can at least put every V_i=V. But i don't see what's the purpose of it?

[coral spindle]

Aiyah

[chilly ocean]

Uh

[coral spindle]

Blitz is defining V_i to be the variety defined by the identites of the free algebra on i elements.

If I'm not mistaken(!)

[chilly ocean]

identities?

[coral spindle]

Then e.g. V_1 = Abelian groups when V is Groups

[chilly ocean]

The one generated by it but it's probably equivalent

[coral spindle]

It should be

By HSP

[chilly ocean]

Yeah

Didn't know you were into universal algebra

oh i think i get it now

[coral spindle]

I got into it a bit recently lmao

I work around a lot of semigroup theorists

and they're interested in varieties of semigroups

So I just kinda picked things up

[chilly ocean]

Quasivarieties I suppose

[coral spindle]

yeah I mean like

This guy I know is comparing the identities of a few semigroups

And he's cinched a particular one between two semigroups of tropical matrices

[rotund aurora]

Exterior algebras are not unital right? Like v*v=0

[chilly ocean]

Didn't even know there were ppl studying it

[coral spindle]

Yeah lol, in the UK there are a couple of places

[chilly ocean]

They are

[coral spindle]

The copy of the base field is the only case that doesn't happen, Croq

Remember it's like, k + V + V^2 + etc (where k is your field)

[chilly ocean]

I guess I saw a book from it once and a pretty experienced user on math.se from semigroup theory

[chilly ocean]

So its like Th(Z) except you only allow formulas that are universally quantified equations

[coral spindle]

Yes that's right. Universal algebra is weakened model theory

But it should be noted that it gives us a pretty nice theory (c.f. Birkhoff's HSP theorem)

[chilly ocean]

thats a really interesting theorem

[coral spindle]

I know right

Can you imagine something like that for model theory? Lmao

It's very strong

[chilly ocean]

You can replace products with subdirect products etc

[coral spindle]

How do subdirects work again

I've seen this once and promptly forgotten

[rotund aurora]

Im confused, I thought the main property of the exterior algebra was v*v=0 for all v in the algebra, so in particular, 1x1 wont ever equal 1. Why is the exterior algebra of a vector space unital then?

[chilly ocean]

Subalgebra of product with onto projections

Like coproduct of abelian groups, say

[coral spindle]

Huh ok how do we do that

[rustic crown]

the thing is constants have grading 0, the stuff from your vector space lives in grading 1, etc... so it's not really weird like you're thinking

[chilly ocean]

Only for v in V

isnt for example <(1,1)> as subgroup of Z² a subdirect product of Z and Z?

[coral spindle]

Try inspecting the definition of the multiplication again. It's totally true that v wedge v is zero, but it isn't always true that when we multiply we take the wedge!

[chilly ocean]

Where you asking me?

[coral spindle]

Yeah

[rotund aurora]

ok Ill check the definitions more rigorously

[coral spindle]

getting back to that

I mean like

for groups we have the trivial group

[chilly ocean]

What was your question?

[coral spindle]

So what's the precise setup, bc I can't see how we'd do this in general

do you need some extra conditions?

Like

[chilly ocean]

You get product of algebras

[coral spindle]

Do you need there to be a prime substructure, or whatever you'd call it

[chilly ocean]

Hmm?

[coral spindle]

So what I mean

[chilly ocean]

I think you're trying to think about irreducibility

[coral spindle]

For the direct sum of abelian groups it works because we have the identity in every group

No. That's not true.

[chilly ocean]

Well it doesn't need to be a direct sum but that's just a prime example that comes to mind

[coral spindle]

Let me see if I understand. This is the minimal substructure of Prod_i A_i such that the projections into each A_i are surjective; is this right?

[chilly ocean]

Doesn't have to be minimal in any way

[analog zephyr]

Finally

[coral spindle]

Oh, is this not unique?

[chilly ocean]

Nope

[coral spindle]

Ah well that clears things up

Interesting. I suppose I should look at Birkhoff's proof

[chilly ocean]

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

This is connected

I just now that Birkhoff theorem can be restated in few ways

[coral spindle]

This will likely take me some time to appreciate

[chilly ocean]

It's in Sankappanavar if you want to see

[coral spindle]

I'll try and get my hands on a copy. Whereabouts in it?

[chilly ocean]

It's free online

[coral spindle]

I just found out

very nice

[rotund aurora]

ok I think I got it. The identity is just the identity from the field right? Because the field also lives in the algebra, as vectors

[chilly ocean]

Idk where exactly but that's what I read from UA so must be there

[analog zephyr]

What u guys think about "A first course in abstract algebra by Fraleigh"??

For being a first introduction to abstract álgebra

[coral spindle]

It's fine.

It's not the best book out there if you're reading it cover-to-cover in my opinion

It was the first time I saw groups, and for that it's good, in my opinion.

However as always with these things: don't worry too much about the book, and just start reading!

[white grotto]

whats the motivation behind ideals

[rotund aurora]

For this same reason, the symmetric algebra is also unital right? In fact, if B is a basis for the vector space V, I think the symmetric algebra S(V) is just (isomorphic to) F[B], where the elements of B are taken as variables

[coral spindle]

The same as for normal subgroups. You can quotient by them.

[white grotto]

yey i am the first one to talk 9/12 on the abstract algebra channel. what a nerd

[coral spindle]

(Also they're submodules of the ring but shhhh)

[white grotto]

ts 12 am for me, forgot theres time zones

[coral spindle]

happy tomorrow

[chilly ocean]

9/12

[white grotto]

9 december

[chilly ocean]

What's on 9th december?

[white grotto]

yeah but they dont look like normal subgroups...

[white grotto]

it's a very important day

because its today

and today is usually the only day there is

[coral spindle]

Because rings don't look like non-abelian groups

[analog zephyr]

Today? Today is 8th

[white grotto]

lol

[coral spindle]

Let me tell you a secret about algebra

Normal subgroups are the things we quotient by, but they don't tell you the full story.

In reality we quotient by a certain kind of equivalence relation, called a congruence.

[white grotto]

Ah

[analog zephyr]

Yeah

[coral spindle]

The fact that congruences and normal subgroups are in correspondence is remarkable

and very nontrivial

[white grotto]

not any equivalence relation?

[rotund aurora]

👆

[chilly ocean]

Kinda trivial tbh

[coral spindle]

Not just any, because we need things to actually respect the group structure

[white grotto]

ok

so "congruence"

[coral spindle]

Lol I think it's nontrivial if you've never seen it before, but I agree in principal

Anyway

The point is

[chilly ocean]

this but nonironically, it's a good exercise to try proving it yourself

[coral spindle]

for rings

congruences correspond to ideals in rings

and that's why they're really the same thing

[white grotto]

idk what congruence exactly is tho

[coral spindle]

n.b. I mean two-sided ideals

[white grotto]

xy^-1 is in H?

[coral spindle]

A congruence on a "thing" A is an equivalence relation on A which is also a "sub-thing" in AxA

[analog zephyr]

a~b if ab^(-1) is in H

[chilly ocean]

Congruences are equivalence relations so that the natural operations on the quotient make sense

[analog zephyr]

If*

[coral spindle]

replace "thing" with the appropriate structure (group, ring, module)

[white grotto]

ah shit

Waw

i like it but its too abstract

[coral spindle]

Remember what I said about the first isomorphism theorem?

A congruence is the thing we quotient by.

[white grotto]

yeah

[coral spindle]

If you have that, the first iso thm becomes really clear

[chilly ocean]

I have trouble with this statement, I don't think it's true

[white grotto]

isnt it tho the same as yours

[coral spindle]

Let me think for a moment and see if it works

[white grotto]

wdym

[coral spindle]

Univeral algebra does it

[chilly ocean]

it needs extra conditions that just being a subthing of AxA

[coral spindle]

You'll have to forgive me if I'm wrong but I think this is precisely what it means for the operations to respect the equivalence classes

[chilly ocean]

Yeah, I think so too. At least in general

[coral spindle]

For example saying that a ~ b and c ~ d implies ac ~ bd, well, I think it's clear

[rotund aurora]

if A and B are algebras, A is unital and B is not unital, then any algebra homomorphism between them is just trivial right? If so, Im confused about the universal property of the symmetric algebra, since it is unital, but we have linear transformations from V to an algebra, which is not required to be unital, that we have to factor through an algebra homomorphism

[coral spindle]

I admit I don't see what extra conditions it would need

[chilly ocean]

for example it needs to contain the diagonal subgroup {(a,a)}

to be reflexive

[coral spindle]

But that is what it means to be an equivalence relation

So yes, that is included

[white grotto]

those 2 tho seem like very different things

[chilly ocean]

then it's more than being a subthing of AxA

not all subthings of AxA are equivalence relations

[coral spindle]

maybe you should note that I said it is also an equivalence relation here, @chilly ocean

[analog zephyr]

There other equivalent property to be a normal grupo too

[chilly ocean]

the condition g+H = H+g for subgroups of rings is automatic

[white grotto]

how are they the same thing in the point of view of congruences

[chilly ocean]

so ideal just adds a new condition to be compatible with multiplication

[coral spindle]

If you want to work this out on your own, it's not hard to prove. But unless you want to go through these definitions I suggest you take the intuition and leave the details behind.

[white grotto]

i want the intuition

[coral spindle]

The intuition is they're what you can quotient by.

[chilly ocean]

Hmmm... actually yeah. I think you're right after all

[white grotto]

any example for what goes wrong when we quotient by some random ring

[coral spindle]

Try and see

[white grotto]

the fact that i cant come up with any examples is sad

[coral spindle]

Or more to-the-point, look at the definition of a quotient ring and see where we use that it's an ideal.

[white grotto]

lets take the ring Z

[coral spindle]

Mmm not a great example

[white grotto]

cause any subring is already an ideal?

[coral spindle]

Yes, any non-unital subring of Z is an ideal

[chilly ocean]

Z[x]?

[white grotto]

isnt it the same thing as Z

[coral spindle]

Noooooo

Nooooo it's nooooooot

[south patrol]

Well any subring of Z is also an ideal

[white grotto]

i mean that any subring of Z[x] is also an ideal

[coral spindle]

Z[x] is a good example but I also think that just Q or R are easier examples

[chilly ocean]

Any subring of Z is Z. Especially 0

[white grotto]

but it seems not

[coral spindle]

Z is a subring of Z[x]. Is it an ideal?

[chilly ocean]

They are talking about non-unital rings

[south patrol]

Lol

[white grotto]

because of 0 its not

[coral spindle]

I don't like this question mark

Maybe you should check the definition of an ideal again and try and find if it is or not

[white grotto]

its by mistake?

[coral spindle]

OK well, why would 0 mean it's not an ideal?

[white grotto]

ok so lets take Q

[coral spindle]

OK

Choose some subring of Q.

Any ideas?

[white grotto]

im not sure xd

[white grotto]

Z

[coral spindle]

OK this is a good chance for you to learn I think. Prove or disprove: "Z is an ideal of Z[x]."

[coral spindle]

Sure. Now you need to think about why quotienting by this (go check the definition of a quotient ring) doesn't work.

[white grotto]

for Z to be an ideal, az for any a in Z[x] should be also in Z

which is not true

it has nothing to do with zero

right?

[coral spindle]

Well remember the first part of the definition of an ideal.

[white grotto]

Q/Z, then (x+Z)(y+Z)= xy+Z

right?

[coral spindle]

That's how we define the multiplication, yes

I will let you know that the problems will occur with this multiplication

I will also give you a strong hint. Let x = 1/2.

[white grotto]

x*y^-1 is in Z is the equivalence relation?

[coral spindle]

y^-1 is not defined.

[white grotto]

right

how do we define then the eq relation

[coral spindle]

Multiplicative inverses do not exist in rings, generally.

OK forget about the equivalence relation.

The quotient is done. The equivalence classes, if you must know, are the cosets a + I.

Your task is now to figure out why Q/Z isn't a ring

and I've given you a hint: the multiplication has issues.

[white grotto]

(1/2+Z)(2+Z)=1+Z=Z while (1/2+Z)(Z)=1/2+Z

[coral spindle]

Eyyyy

Well spotted

And what's the problem here?

[white grotto]

i am not sure

i mean it should be that Z is not an ideal

[coral spindle]

No.

I hope we've already established that Z is not an ideal of Q. Maybe you should convince yourself of that.

Again, what we're here to show now is that problems arise when we then try to quotient Q by Z.

[white grotto]

wdym

[coral spindle]

It's a subring. In particular, all ideals must contain 0.

[white grotto]

yeah

so?

[coral spindle]

Now back to the content here. You decided to write these equations up here. So what's up with this?

[coral spindle]

So in fact 0 is important; every ideal must contain 0.

[white grotto]

yeah

got it

[white grotto]

i meant above that if Z was an ideal, then we wouldnt have had this problem

[coral spindle]

Well yes, but I'm asking: why is it a problem

[white grotto]

so the problem arizes from the fact that multiplying Q by Z gives something outside of Z

[coral spindle]

OK so you don't seem to be seeing what I'm seeing here so I'm just going to point it out

[white grotto]

yeah plz

[coral spindle]

Multiplication isn't well-defined anymore! It's not a function!

Z = 2 + Z

so (1/2 + Z)(0 + Z) = 1 + Z = Z as you mentioned

but then also Z = 1 + Z

So (1/2 + Z)(0 + Z) = 1/2 + Z

But 1/2 + Z is not equal to Z!

[white grotto]

yeah this i understood

i thought you were asking why multiplication isnt defined

[coral spindle]

This is what fails when you take something that isn't an ideal. Multiplication is no longer well-defined.

[white grotto]

yeah but where exactly it went wrong

how idealism would have fixed that

[coral spindle]

Exercise: show that multiplication is well-defined in a quotient ring iff we quotient by an ideal

[white grotto]

i guess because (1/2+Z)(1+Z) wouldnt have been possible

[white grotto]

okay i shall

[rotund aurora]

Shouldnt A be unital here too?

[formal ermine]

?

K(X) is the field of rational functions fwiw

[rotund aurora]

I wanna cry lmfao

[formal ermine]

wait isn't it like really trivial lmao

X is the root of a polynomial in K(f)

so K(X)/K(f) is algebraic

no?

[coral spindle]

I reckon so

[formal ermine]

I was thinking way more complex than I needed to

[solar glacier]

in showing I+J is an ideal

does this work

let x,y \in I+J then x=a1+b1, y=a2+b2