groups-rings-fields

[tardy hedge]

Yeah that was a bit illuminating

Honestly not many examples. The Z to Q one aaaand …

Because it shows more clearly that Q(D) is really just extending D to include inverses but if it already has inverses then you havent done anything differently

Ill just try more questions ig

[delicate orchid]

I don't think that would help I think at this point you just need more examples lol

like what do you think the field of fractions of Q[x] looks like

[tardy hedge]

Isnt it just rational functions

When u ask a question like that im not totally sure what to say yk

It looks like fractions of polynomials idk

[tribal moss]

Yes, that's exactly the field of rational functions.

[tardy hedge]

Ok

Tbh im gonna try a few more exercises and then move on to next chapter

[naive whale]

@tardy hedge If you know universal properties, the fact that Q(D) is the "smallest" field that admits an embedding of D can be phrased as, any injective homomorphism from D to a field has to extend the injection D -> Q(D), so generally just take any extension field of Q(D) and you obtain an injection of D into a field that is bigger than Q(D)

[tardy hedge]

Thx

[white oxide]

from now on i'll be referring to polynomials as endomorphisms

[delicate orchid]

I used this idea a few weeks ago to help someone out with a finite fields question lol

well, the endomorphism part at least

[thick zealot]

if the SOn is a group how come when i move my mouse around in a game and come come back i am not in the same place i were

[rocky cloak]

Why?

[tribal moss]

Because it is not an abelian group.

[white oxide]

oh nah i was joking

[rocky cloak]

Also, I'm not sure that mouse movement really corresponds directly to multiplication by elements in SO(3). Then you'd probably have weird holonomy in every game.

[white oxide]

wait what game are you referring to

[thick zealot]

any first person or even 3rd person games

csgo or flight simulation

[naive whale]

Not sure if it's related since I'm not sure what mouse motions you mean, but gimbal lock is also an issue in game dev sometimes, because people who don't really care to understand the math can choose inferior systems for representing rotations eg Euler angles instead of quaternions or matrices

[chilly ocean]

My finals are over. Goodbye abstract algebra

[delicate orchid]

get back here at once

[white oxide]

bro got traumatized

[languid trellis]

no algebra is cool

i learnt the other day that pZ is a maximal subgroup of Z

that was quite cool

and I've made 0 progress on the homework you set me wew, so I am now officially giving up

[white oxide]

when we say that k[x] is a k-algebra we're just viewing it as a ring with scalar multiplication from k right

so in a way i guess a k-algebra A is just a k-module but instead of A being an abelian group we have A is a ring

a k-ring

!1

🤯

[delicate orchid]

what homework did I set you lol

[rocky cloak]

That's exactly what a k-algebra is yeah

[languid trellis]

if p divides |G| then there is a subgroup of order p in G. It was a follow up the case p=2 that you helped me with

[delicate orchid]

oh the whole element of order p thing

[languid trellis]

that bs combinatorial proof

[delicate orchid]

yeah so instead of pairing them up you form p-tuples

[rocky cloak]

A monoid in the category of vector spaces 😉

[languid trellis]

i had this idea but I didn't find any canonical way of doing it

[tardy hedge]

Oh man like how do i even interpret Dp\M

What does it look like?

Hard to think about

Q13

[delicate orchid]

for any g in G take (g, g^2, ..., g^p)

I think that's how it goes

[languid trellis]

i can't believe I didn't think of that

bollocks

[delicate orchid]

I'm improvising the details here. You still have to show that this is actually a partition of G into p-tuples still

[languid trellis]

this sounds like something I'll think about tomorrow

appreciate the hint wew much love my fellow northener

[delicate orchid]

I no longer believe that's the correct choice

[white oxide]

is this true because t if alpha is a root of a polynomial with coefficients in K then we can't deduce that all of its coefficients are zero

[delicate orchid]

the canonical proof is you take the set X of p-tuples such that the product of elements in those p-tuples is the identity, which is of size |G|^{p-1}. Then use the fact that this set is closed under the action of (12... p) and then fixed-point theorem your way to concluding that the number of elements with g^p = 1 (the number of diagonal elements in this set, which are the fixed points of the action) has to be a multiple of p, and can't be 0 because 1^p = 1 lol

[white oxide]

i didn't read allat only the first sentence but is that Cauchy's Theorem

[delicate orchid]

yur

[white oxide]

damn i don't know how somebody could prove that if they haven't seen it before

it involves hella insight doesn' tit

one of my favorite proofs tho

[delicate orchid]

the hard part is coming up with the idea for the tuples

but it is a generalisation of the case for p = 2 where the pairs are (g, g^-1)

[languid trellis]

Could you explain the sentence from 'the action' and also what is the fixed point theorem

[delicate orchid]

right so you let a cyclic group of order p act on the set X,

Burnside's fixed point lemma states that the number of orbits is a multiple of p, specifically it's 1/|C_p|\sum_{g \in C_p} Fix(g). The fixed point set for each element is either all of X (in the case of g = 1) or just the tuples (x, ..., x), implying x^p = 1. Define Y = {x in G : x^p = 1}. Then the number of orbits is 1/p(|X| + (p-1)|Y|) = 1/p(|G|^{p-1}+(p-1)|Y|), this is an integer so |Y| must be a multiple of p.

[rocky cloak]

I think instead of trying to interpret Dp/M you might want to construct a ring homomorphism Dp -> Q(D/p) with kernel M. Then first iso

[delicate orchid]

sorry you didn't actually need orbit stabiliser (I'm kind of reproving this as I go along)

it's also MUCH harder than I remember it being

many such cases!

[languid trellis]

Okay. I think this is getting put in the 'revisit much later on' box as I don't know what an orbit is, what a group action is, what a Burneside fixed point lemma is

But I appreciate yoy regardless wew

[tardy hedge]

Yeah i was thinking to do the question would be to use 1st iso thm somehow

But before trying to do that i wanted to interpret what that quotient ring looked like

Its hard to do such a thing tho , i suppose.

[rocky cloak]

I mean it will look like Q(D/P)

[tardy hedge]

Yes lol i guess it says it there

But like what if u didnt know that already

Just trying to think of Dp\M

Would u even bother with that?

[rocky cloak]

Well elements look like
a/b + M
Which in turn looks something like
(a + P)/b
Though that's a little handwavy

Once you define the homomorphism it gets more precise

[tardy hedge]

Thanks. Ill try actually solving the question by finding the map and maybe that will give me the insight im looking for

[glass zealot]

whats a good way to think about K[X1,...,Xn] / (M) for some finite M subset K[X1,...,Xn]

[delicate orchid]

you're defining a ring generated by x1... xn subject to the relations m = 0 for each m in M

basically the universal property

[south patrol]

the vanishing set of the polynomials defines some subspace V of K^n and this can be identified with the set of "polynomial functions" on V

[delicate orchid]

oh yeah you have the alg-geo interpretation of a coordinate ring as well

good quip

[glass zealot]

whats the vanishing set? M?

[south patrol]

It's the set of points of k^n on which all the elements of M vanish

(equivalently, on which all the elements of (M) vanish)

Also note that any ideal of K[X_1,...,X_n] is finitely generated so we can just ask about any subset of K[X_1,...,X_n] or any ideal etc wlog

[glass zealot]

ah cus K is a field

and noetherian

is this set of points on which (m) vanishes the coordinate ring? nvm

[dire siren]

it's basically reducing modulo elements of M; at least this is how I like thinking about it

[glass zealot]

yeah i kinda lack intuition for this "reduction" when the ideal is not generated by a single element

[tribal moss]

I wouldn't expect there to be a simple intuition that's much good. A large part of algebraic geometry could be described as laboriously building out the tools for understanding such quotient rings.

[delicate orchid]

more generally a good way of thinking of a quotient is as setting whatever is inside the thing you're quotietning by to 0. Which leads to interpreting a quotient as taking your original object and adding more relations between the elements of that object

the most clear cut example of this is the universal property of free groups and group presentation. When you write G = <X|R> for some set X and some relators R you're actually writing G has a quotient of the free group on X by the smallest normal subgroup containing everything in R

[delicate orchid]

hence why I gave the answer I did. I just find that this is the easiest way to work with quotients intuitively.

[glass zealot]

yeah, im just kinda bamboozeled by geometric interpretations of this, in our course for example we showed this isomorphism

and i wondered how you see stuff like this if the ideal isn't as obvious as there

[south patrol]

What you wrote is the coordinate ring.

of V

[tardy hedge]

Can i do f: Dp -> Q(D/P) by f([a,b])=[a+P,b+P]

[tardy hedge]

Is this not well-defined?

[rocky cloak]

It is well defined, but that's something you should check.

[tardy hedge]

Ok i will check it, but then applying first iso to this shows that Dp\M = Q(D/P) right?

Its onto and kernel is M

[cobalt heath]

Doesn't it bring intuition from geometry?

[tribal moss]

Yeah, "simple" is the weasel word I'm hiding behind here.

[cobalt heath]

Ahh

Well tbf, I also don't see how geometric interpretation somehow works

Like how can we even differentiate using formal derivatives, and get "approximations" around "a point"

[cobalt heath]

Why does it work, it's like black magic

[tribal moss]

Even the word "around" there is just handwaving if your base field e.g. happens to be finite. Then there's only the formal similarity and "I can prove the analogous theorems" left.

[naive whale]

TIL an R-module M is actually just a ring homomorphism from R to M's endomorphism ring

mind = blown

[cobalt heath]

I guess it is similar to how group action can be characterized as such

[undone ledge]

the tensor ad homjunction

[winter shore]

What's the point of characteristic subgroups? I.e. why are they an important concept?

is there a useful/iluminating theorem about them? or something like that

[coral spindle]

They exist and have this strong property, so we're telling you about them. That's "the point" of them.

You can pull a few cute tricks with them. For example, it's not the case in general that a normal subgroup of a normal subgroup is going to be a normal subgroup of the bigger group. However, if the smallest subgroup is characteristic it works.

[rustic crown]

det hasn't seen any other trick than the one you mentioned

but det hasn't done much group theory either

[winter shore]

which strong property?

[coral spindle]

Being characteristic is a strong property

Read: typically there will not be many characteristic subgroups of a group

[winter shore]

oh, ok ty

[slim kayak]

Being invariant under some map is already a nice property, being invariant under all maps must be very nice

[chilly ocean]

if we take an arbitrary ring, at most how many distinct subrings can it contain?

[rocky cloak]

The power set of the cardinality is certainly an upper bound. And I think for example C achieves this.

Or Q for that matter

[delicate orchid]

wow that's way yuckier than my answer

[rocky cloak]

What was your answer gonna be?

[coral spindle]

"loads"

[south patrol]

lol

[delicate orchid]

it was way less precise, I was going to say that you can construct a ring with arbitrarily many subrings

[south patrol]

I don't think enumerating subrings is a particularly interesting question anyway, like ideals are more interesting

[delicate orchid]

like I noticed that if V has dimension n as a vector space, then End(V) has at least n subrings given by matrices of the form of "big block" followed by 1x1 blocks

[static glen]

As far as i know this is the best bound one can do in general

[delicate orchid]

kernels > subobjects indeed

[rocky cloak]

Hard to do better than a tight bound I guess, unless you have some bound that depends on more than the size of the ring.

[static glen]

I think i saw that the ring of bounded holomorphic functions on the unit disc even contains 2^c prime ideals ( c= cardinality of continuum)

[delicate orchid]

actually you can get way more than this following a similar idea - there's no reason the other blocks need to be 1x1 so you have at least \sum_{\lambda \in P(n)} #{the conjugacy class in S_n corrisponding to lambda, lol} where P(n) is the partitions of n.
swag.

[rocky cloak]

Now that's pretty crazy. Must be some crazy ideals.

[static glen]

The proof was non-constructive so yeah

[rocky cloak]

Also no reason you have to restrict yourself to a given basis, so you get a subring for any decomposition V = U(+)W

[chilly ocean]

hmm I see

but how do you know that?

[rocky cloak]

C has transcendence degree |C| over Q, so contains the polynomial ring in |C| many variables. Any subset of the invariants gives you a subring, so that's 2^|C| subrings.

I guess Q is an easier example because
Z[1/p : p in P] is a subring for any set of primes P

[delicate orchid]

I feel like it's easier to see for the Q example

yeah there we go

oops internet has died, sorry chat

and we're back

[static glen]

I guess it‘s an obvious estimate, a ring cannot contain more subrings than subsets

[chilly ocean]

oh right

it clicked

[naive whale]

Can someone give me a sanity check here? It's a pretty elementary exercise but I have become confuzzled by the different similar operations and having to remember left and rightedness

Also, I feel like there is a one line proof hiding, something something End l (M) = (End r (M)) op but it's not clicking

[rocky cloak]

Proof like good.

I don't know what your definition of End^r(M) is here, but yeah a map R -> S is exactly the same as a map R^op -> S^op, so that could be a one-liner.

[naive whale]

Thanks, and End^r(M) is just the ring of "right" endomorphisms aka endomorphisms where multiplication is composition left to right instead of right to left

Ok yeah then the original lambda is also just a map R^op -> End^r(M)

[tardy hedge]

Is the generator of any principle ideal some sort of “minimal” element in the set?

Like in F[x] its the polynomial with minimal degree

[tribal moss]

It's definitely (weakly) minimal wrt the divisibility ordering.

[tardy hedge]

Thanks

Minimal wrt to some divisibility type of way that makes sense

Tbh instead of going to the chapter about fields i shouldve done most problems in the ring theory section beforehand

[tardy hedge]

For proof of part a), im not sure why it concludes with “the image of phi must in fact be equal to K(u) …”

Oh i think i know why

It says before that the image must be contained in K(u). But the proof shows the image is in fact a field (that contains K and u) which implies its equal to K(u) (since there cant be a smaller field than K(u) that contains K and u, by defn)

Cool?

[delicate orchid]

yur

and it can't be a larger field because then it would have to contain elements in F other than those in K union {u}

[tardy hedge]

For b), why is theta an isomorphism

Why is the field of quotients of K[x] isomorphic to F

The conclusion of the theorem its referring to just says theta is a 1-1 ring homomorphism

[tardy hedge]

In this theorem why cant we immeditely say the image of phi_u is K(u)? The image consists of all a0+a1u+ … an u^n , which should be K(u) right?

But then it doesn’t necessarily contain inverses or smth?

[tardy hedge]

Yes, this is the problem right?

[tardy hedge]

I guess u can equivalently say for part a) K(u) is isomorphic to quotient field of K[x]… but in this case the quotient field of K[x] just turns out to be K[x]/<p(x)>

[blissful zinc]

What does wrt stand for? Please

[tribal moss]

"with respect to"

[icy bear]

artin's exercises are very cool

they really show the power of the theorems presented I think

[tardy hedge]

So isnt writing the elements as fractions kind of an abuse of notation kind of thint

Like … what does 1/(b0+b1u+…bmu^m) even mean here?

[delicate orchid]

Write it as [a, b] if you want but who cares

[tardy hedge]

Where [1,b] is just an element that acts as inverse of b

Right

Ik the answer is yes

I think my confusion with it is thinking it means something more than it is

[void knot]

How is that $x^2+3x+2=(x+1)(x+2)$ is irreducible in $\mathbb{Z}[[x]]$ (formal series)?

[cloud walrus]

RaD0N

[delicate orchid]

x+1 is a unit in Z[[x]]

it's inverse is uhhhh 1-x+x^2-x^3+...

[void knot]

Lmfao

Now I get it

But wait

[delicate orchid]

waiting

[void knot]

The given polynomial itself isn't a unit?

[celest furnace]

It is since x+2 is invertible too in Z[[x]]

Wait maybe not

[void knot]

It's not

2 isn't a unit of Z

Now I get it

[next obsidian]

Or so you think……

[south patrol]

Real

[tardy hedge]

Minimal polynomial of u over K is the irreducible polynomial that has u as a root right

[next obsidian]

Yeh

Probably monic

Which makes it unique

[tardy hedge]

Defn just said minimal degree but i wanted to make sure its irreducible too

Yea it said monic too

[next obsidian]

equivalent

[tardy hedge]

Yeah

Because … set of f so that f(u) = 0 is a prime ideal …

Is it: set of f(u)=0 is an ideal. Since K[x] is a pid then its a principle ideal. We can show this ideal is prime and then that implies its generated by an irreducible polynomial

Is all that necessary?

Because prime ideals arent necessarily generated by a single element so the pid part was needed

[delicate orchid]

I mean if you want to be stinky about it yeah that’s all needed

[tardy hedge]

I dont want to be stinky i need to be

I am learning!

[delicate orchid]

Take a BATH

[tardy hedge]

Maybe its time for one

[molten viper]

So, given a Noetherian module M with submodule M', I want to show M/M' is Noetherian. So, given a submodule of M/M', call it I, I need to show I is finitely generated. The elements of I are of the form x + M, so I can create a sort of "pre-image" ideal, which is all the x's. This ideal is finitely generated, let's say by (f1,...,fn), so I is finitely generated by (f1 + M',..., fn+M')

Does this argument make sense

replace "ideal" with submodule, sorry lol

[south patrol]

Ye, nice.

This is really a case of a more general phenomenon - if $N$ is a submodule of $M$, then submodules of $M/N$ correspond to submodules of $M$ containing $N$ (one direction is just taking preimage under the projection like you did, and the other is just image under the quotient map)

[cloud walrus]

potato

[south patrol]

This is super important! The correspondence theorem.

Using this you can also use the other definition of Noetherian (every ascending chain of submodules stabilises) since you can just pass to M and then stabilise there are pass back

@molten viper

[molten viper]

Interesting

I remember vaguely a correspondence theorem on rings

[south patrol]

In a sense I'm surprised you're doing stuff with Noetherian modules without having seen this but I'm not sure what context you're learning about modules in aha

[molten viper]

Well

[south patrol]

Hm it is probably the special case of this when you treat R as an R-module

[molten viper]

I'm probably in over my head

[south patrol]

There is a similar thing with groups, provided you take N to be normal

[molten viper]

I've got a pretty solid groups/rings foundation

[south patrol]

(oh and if R is non-commutative then take left modules or right modules; idk if it works for two sided ideals lol)

[molten viper]

I'm kind of learning about Noetherian rings and modules at the same time ig

[south patrol]

Yah i mean

A lot of the theory is the same since noetherian rings are noetherian modules over themselves lol

[molten viper]

I'm working through chapter 1 of Eisenbud, because reasons

[south patrol]

(ig i would recommend Atiyah Macdonald for a lot of this stuff if you've not looked at it)

ye noice

[molten viper]

I've looked more closely at Atiyah yeah

maybe I should spend more time familiarizing myself with the proofs in the chapter before doing exercises

[rocky cloak]

Unike groups, rings is something you've been learning about your whole life. So maybe just keep those familiar examples in mind, i.e. Z, C, polynomial rings, matrix rings, smooth functions.

[rocky cloak]

I'm not sure I have advice better than just to practice the problems you struggle with

[naive whale]

You can define a group action as a homomorphism from G to Sym S, similarly you can define an R-module as simply a homomorphism from R to the ring of endomorphisms of the module. So groups act on sets, rings act on modules

[cloud walrus]

The

[meager fractal]

need it for smt but my group theory is rusty'

[static glen]

How ist (Z/NZ)* a ring?

[meager fractal]

wdym

its a group

[static glen]

Nvm i just made up the word ring homomorphsim in my head sorry

[meager fractal]

real

hmmm

[meager fractal]

I think finding the case for p^1 is enough to work the rest out

this is what I have and so I want to find an n such that (Z/NZ)* has a subgroup isomorphic to Z/pZ

Ooohh

(Z/NZ)* is abelian of order φ(N)

so all I need is that p | φ(Ν)

by the fundemental theorem of abelian groups right?

[meager fractal]

and p^2 doesn't (?)

[wraith cargo]

See my post in #advanced-algebra

[coral shale]

i might be able to help. if i knew what a group ring was
lemme see

think the best way to view homomorphisms into automorphism groups is via actions. And to consider Cayley's theorem

R[G] = {f : G -> R | finite support}
group homomorphism into Aug(R[G]) is G acting on R[G]. Maybe its easier to view it as G x R[G] -> R[G] rather than G -> (R[G] -> R[G]) (currying).
They want this action to be constructed by conjugation. To me there's a natural action to try out and see if it works

idk if uve considered this already

(Also, group homomorphism into Sym(R[G]) would also be a group action, so its a special action that respects the group structure of R[G])

[delicate orchid]

the specific action we have here is $z = \sum_{g \in G} r_gg \in R[G]$ gets mapped to $xzx^{-1} = \sum_{g \in G} r_gxgx^{-1}$

[cloud walrus]

W;3w Lads Tbh

[delicate orchid]

you can translate any automorphism of G into an automorphism of the group algebra like this using the functoriality of taking group algebras (aka extending by linearlity lol)

[rocky cloak]

Conjugation by an element is an automorphism, so the map just sends g to "conjugation by g"

[delicate orchid]

so much rep theory recently

almost like people are revising for exams in the new year or something ;)

[coral shale]

i felt itchy thinking about this. This surely is the action used for the canonical module structure of the group ring i was thinking

but couldnt get there

[delicate orchid]

gimme a min I'm smoking my pen like an old timey pipe

[rocky cloak]

The cannonical module structure is left multiplication, no?

[coral shale]

wait did i mean something else

u have to define a multiplication of elements

for something

lemme look again

[delicate orchid]

the multiplication is standard "polynomial" multiplication

[delicate orchid]

@south patrol kunneth theorem

[coral shale]

oh im silly, i meant defining the multiplication in the group ring

[rocky cloak]

Like when you say canonical module structure, you mean viewing R[G] as a G-module?

[coral shale]

since a group ring is a ring

theres conjugation in that defn

or what looks like conjugation

[delicate orchid]

yur

[rocky cloak]

Not really

It's just g times h equal gh

[delicate orchid]

I think what shuri means is that there's a natural embedding of G into R[G]

[rocky cloak]

Sure, but I don't see how it "looks like conjugation"

[delicate orchid]

and we're implicitly passing through that embedding when we "move the conjugation" from G into R[G]

it's r_gxgx^-1 I don't see how that can be anything but conjugation

[coral shale]

like isnt this directly related to OP's question or am i off the mark

[delicate orchid]

wait you're viewing them as functions?

I mean I suppose you could do that

[coral shale]

ill play around with my thoughts on paper

til moment

[delicate orchid]

it's much clearer to view them as just formal linear combinations of group elements imo

but maybe that's just the perspective I am far more used to

[coral shale]

from rep theory?

ok

[delicate orchid]

like the function perspective makes the connection with cohomology easier to see ig

anyway I'm getting distracted

[south patrol]

Serre conjecture

[delicate orchid]

upon the

[tardy hedge]

im learning a bit about conjugacy in groups from some yt videos and it definitely helps clear up confusion about normal subgroups and the intuition behind them

in my textbook though it covers that stuff in chapter 7. im still on chapter 6

[delicate orchid]

I refuse to believe you've seen normal subgroups but not conjugation that's bizzare

[tardy hedge]

This guy talked about the parallel with similar matrices in lin alg and that was helpful

Yeah dude ikr

[rocky cloak]

I feel like the grass is always greener on the other side when it comes to which order you learn stuff in

[tardy hedge]

Yeah how this textbook laid out stuff worked for me

it likes to introduce concepts without introducing the concept

[rocky cloak]

Teaching through osmosis

[tardy hedge]

eg: it showed the isomorphism theorem before even really talking about quotient groups

[tardy hedge]

yeah

a lot of the meat and potatoes of group theory it doesnt cover until late in the book

[coral shale]

they just never called it that where i was (until we actually were doing conjugations as automorphisms)

iirc

maybe they call it undo do samwich in some places

[meager fractal]

opinions?

[spark veldt]

Hi can anyone explain to me whats going on here? It’s the subchapter in groups and they introduced Sn, but when dmonstrating S3 they somehow chose (123) and transposition withut explaining why, also why did they make that x^3=1 product etc.? Kinda lost, would be nice if someone can guide me through this.

[delicate orchid]

they do explain why, they pick those ones because you can write all of the other permutations by multiplying those two together in different ways

and x and y are defined to satisfy the same relations as (123) and (12) do

[spark veldt]

Ah i see, and how do they use cancellation law to show that those 6 elements are distinct?

[languid trellis]

because you can't cancel them down any further

if you set one equal to another and use cancellation law you get something contradicting group axioms, so they are distinct

e.g. sps x and x^2 weren't distinct

then x = 1, a contradiction

as x represents the permutation (123)

[meager fractal]

does this hold?

[tawny magnet]

I think there is a typo?? surely f is a map from B to A

[spark veldt]

Just curious, if we had chosen anything other than (123) and (12) would it not have been possible to express others in terms of these? And how did they know to choose these two permutations?

[blissful zinc]

Hi, I've seen different ideal's definition. But is it same?

Let A a ring, I an ideal
There is one which: for all a, b and for all x in I, we have axb in I.
And another version: for all a in A and for all x in I, we have ax in I.

[rocky cloak]

There's lots of different choices that would work, you just have to make a choice.

[meager fractal]

yep! thanks

[rocky cloak]

Definition 1 is a "two sided ideal", while definition 2 is a "left ideal". There is also a similar definition for "right ideal". These are equivalent if A is commutative, but not in general.

[blissful zinc]

Ah, I see. So my teacher wrote the "two sided ideal" definition.

[tawny magnet]

Is there much to be said on non commutative rings?

[rocky cloak]

Yes, a useful thing to note is that (two sided) ideals are the kernels of ring homomorphisms, while left ideals are the kernels of R-linear maps (aka homomorphisms of R-modules)

[rocky cloak]

Sure

[barren sierra]

Noncommutative rings can be useful

[tawny magnet]

are there properties of general non commutative rings

[rocky cloak]

Depends what you mean. When people talk about non-commutative rings, they usually mean not-necessarily-commutative rings.

There's a bunch of properties that hold for rings in general. But if you're asking for properties that only hold for non-commutative rings, I don't think you'll find very interesting stuff.

[blissful zinc]

If the ring isn't commutative, we can't even talk about euclidian / factorial / principal domains, right?

[barren sierra]

I mean

even if the ring is commutative you may not be able to talk about those things

although almost universally "domain" refers to commutative rings

[rocky cloak]

You can certainly extend the definition of UFD and PID to noncommutative rings, and there are examples. Wheter that's a useful thing to do... meh

I guess the easiest examples being that the noncommutative polynomial ring k<x1, ..., xn> is a UFD and that D[x] is a PID for any skew-field D.

[spark veldt]

Can you make the claim “Za is contained in S” when ur trying to prove S is a subset of Za here?

[delicate orchid]

"is this a legal move"
concerning

but 1 has proven that Za is a subset of S so it's fine

[spark veldt]

Oh i see okk thankss

[tardy hedge]

❌Illegal Move ❌

👮‍♂️

[long obsidian]

If I have two different polynomial rings R[x] and R[y] is it enough to define a map on x alone? So for instance is f:R[x]->R[y] well defined by f(x)=2y for example?

[chilly ocean]

Can we ask questions here?

Or only in help channels?

[round hull]

it would be well-defined as an R-algebra homomorphism

[round hull]

you can ask questions anywhere

[next obsidian]

If you specify the map is a map of R-algebras, yes

[round hull]

oh of course whoops

[delicate orchid]

equivalently, your map restricts to the identity on R

[chilly ocean]

I ended uo doing a #help-11, it's about group theory but it's so basic basic stuff that I don't think it belongs to "advanced mathematics" or even "early university"

[south patrol]

Well like

for any R-algebra A, an R-algebra map R[x] -> A is equivalently a choice of element of A

[next obsidian]

Module

Algebra map

Make it make sense

[long obsidian]

Does specifying f(x)=2y still make a well defined image for elements of the ring? For example how do I tell what is f(5)? If say R is the integers Z?

[next obsidian]

When you say it’s an algebra map, yes

You can set x to anything you want

Algebra map forces f(r) = r•f(1)=r•1 = r

[long obsidian]

Oh okay. I'm guessing using f is a map of algebras implies f(1)=1?

Do I get the same thing if I assume f is a map of rings?

[delicate orchid]

maps of rings also have to send 1 to 1

[south patrol]

a map of algebras is just a map of rings compatible with the maps from R

[long obsidian]

Is the map algebra Hom defined by f(x)=2y injective?

[rose prism]

depends on the ring?

[delicate orchid]

try and calculate the kernel

[woven panther]

How can i go about computing the automorphism group of S_3? I’m imagining i can simply use conjugation, but how do i know i’m “done”?

[delicate orchid]

automorphisms have to preserve the order of elements

[long obsidian]

Well I guess if my ring was the integers Z and I considered an algebra Hom f:Z[x]->Z[y] given by f(x)=2y and f(1)=1 then this basically induces f(p(x))=p(2y). I'm pretty sure if p(x) has any term with x then the image p(2y) has an indeterminate y as a term and so clearly it is not in the kernel. Therefore only elements of Z are possibly in the kernel yet the map f is the identity on Z and so I think only 0 is in the kernel.

[delicate orchid]

so you have your 6 conjugation maps, and then show that there's only 6 choices of where to send your generators (123) and (12)

[tardy hedge]

AGGH theres something with remainder theorem here isnt there

[delicate orchid]

let f(x) = 2y between R[x], R[y], let p be in R[x]
then f(p) = f(p_0+p_1x+...+p_nx^n) = p_0+p_1f(x)+...p_nf(x^n) = p_0+2p_1y+...+2^np_ny
so if f(p) = 0 then p_0+2p_1y+...+2^np_ny = 0, which is only possible if 2^kp_k = 0 for all k
so if your ring is an integral domain (or more generally 2 isn't a zero divisor), then your kernel will be trivial

[tardy hedge]

Theres always seems to be like 50 different ways of interpreting results

[long obsidian]

Thank you I see what you mean. Do you think if I considered an algebra Hom over several variables I could come up with an example of an algebra Hom between polynomial rings of several variables with a nontrivial kernel?

[delicate orchid]

well, yeah. Just set one of the variables to 0

R[x1, x2] -> R[y] given by x1 -> 0, x2 -> y has a pretty big kernel lol

[long obsidian]

All okay and I bet if I sent two of the same variables to an independent variable in the codomain poly ring that mights also work

[delicate orchid]

what we're basically doing here is matrix multiplication, if the matrix of coefficients you send stuff to isn't invertible then it's going to have a non-trivial kernel

[icy bear]

it's a simple proof but I just want to be sure if I didn't mess anything up
every group of even order has an order 2 element
well, every element has an inverse, and the inverse can only be the identity if it's the identity itself, so now, if we want no element to have order 2 (that is, to not be its own inverse), we need to pair up all of the 2n - 1 elements with its inverses, but since this is an odd number, you can't pair up every element with a different element, so one has to pair with itself, that is, has order 2

[delicate orchid]

yeah that's correct

[icy bear]

yey

it seemed so obvious after thinking about it

looked at the paper for 2 minutes and then it seemed obvious

2 or more idk

(by paper I mean sheet)

(thing you write and/or draw on)

[tardy hedge]

If u algebraic over K show u^-1 algebraic over K. Would it be just the polynomial for f(u)=0 being reversed like a0x^n+ … an to show g(u^-1)=0

[south patrol]

Yes nice

also given L/K and a in L, a is algebraic over K iff K(a)/K is finite

And clearly K(a) = K(a^-1)

[tardy hedge]

For q3, i have found a polynomial so that f(u+a)=0, and it is the same degree of the minimal polynomial for u. But how can i argue that the degree of u+a is the same?

So my question is i have a polynomial that has u+a as a root but how do i know its the smallest degree polynomial possible

[delicate orchid]

if you've chosen the right f it should be obvious

what f have you found

[tardy hedge]

Well i just found that its the same polynomial for u but you modify the constant term

[delicate orchid]

the f I would have chosen is f(x) := g(x-a) where g is the minimal polynomial for u

[tardy hedge]

Oh yeah. Cool

[delicate orchid]

I don't think this is true btw. Consider the minimal polynomials for i and i+1 over Q

x^2+1 and (x-1)^2+1 = x^2-2x+2

brain struggled to compute (x-1)^2 there

I need some sleep lmfao

[tardy hedge]

Ohhh yeah ok thanks i know where i messed up with that

[tardy hedge]

This type of thing is common isnt it

Some shifting function thing

I think ive seen that sort of tactic used a lot

[delicate orchid]

not telling....

[tardy hedge]

Lol

[tardy hedge]

I think im still not sure why this means u cant have any smaller defree

Degree

[south patrol]

write it out lol

[delicate orchid]

u+a is in K(u) and a is in K(u+a), so K(u) = K(u+a)

[tardy hedge]

Did u mean K(u)

[delicate orchid]

not telling.... (yes)

[tardy hedge]

So they must have the same minimal polynomial by this theorem?

[south patrol]

this is unnecessary anyway right

[tardy hedge]

lol

[south patrol]

i mean if f(x) is a poly killing u then f(x-a) kills u+a

and vice versa

so they have min polys of the same degree

[tardy hedge]

I dont see this part

[south patrol]

what are you unsure about w it?

The first part shows that min poly of u+a has degree <= that of u

but then by symmetry the same works the other way round

[blissful zinc]

Hi, I'm stuck on this:
Have to prove that x^{3} -9 is irreductible in Z/31Z[X]

I've seen that a polynom of degree 2 or 3 is irreductible if and only if he doesn't have roots.

So I guess there is another way to solve it. (no way I have to computate 31 times)

[south patrol]

well you can also use fermat's little theorem to show there are no roots

[tardy hedge]

Ok thanks

x <= y and y <= x makes x=y

lol

[south patrol]

Ye

[silent bronze]

Hi, can someone explain how you would go about proving this statement

Specifically using the union of orbits thing

[slim kayak]

Well, which direction do you get stuck on?

[silent bronze]

Firstly I’m struggling to understand the concept of like

What the orbits would even look like

[delicate orchid]

conjugacy classes

[slim kayak]

Let x be N (normal), then acting on conjugation just sends it to another element in N. So the whole orbit lies in N

[silent bronze]

Because I’m used to the concept of like groups acting on sets

But this is a little trippy for me

[delicate orchid]

the orbits are just the conjugacy classes

[delicate orchid]

why lol? G is just a set

[silent bronze]

Because we have a group that’s acting on something else

[slim kayak]

the orbit of x is just ${gxg^{-1} : g \in G}$ here

[cloud walrus]

Kerr

[silent bronze]

Here the domain and co domain are like intermingled

[slim kayak]

or $g.x$ if you wanna write it more abstractly

[cloud walrus]

Kerr

[silent bronze]

Right

Is it possible for the orbit to contain g

[slim kayak]

The group G is acting on a set. That set just happens to be G again

[coral shale]

a group acts on itself via left multiplication. And right.

[silent bronze]

Would the orbits partition the entire group then ?

[slim kayak]

Which g?

[slim kayak]

Have you heard of equivalence relations and equivalence classes?

[silent bronze]

Like, is it possible that $gxg^-1 = g$

[cloud walrus]

Ash

[silent bronze]

Yes

[coral shale]

let x = g?

[silent bronze]

Yea

Is that allowed ?

[slim kayak]

sure

[delicate orchid]

why wouldn't it be?

[slim kayak]

just try to be clear about what are the variables and what are particular elements

[silent bronze]

So as i said before, would the orbits partition the entire group ?

[slim kayak]

Okay.

1. convince yourself equivalence classes partition sets.
2. convince yourself orbits are an equivalence class

[silent bronze]

Or would they only partition the Normal subgroup

[silent bronze]

Okay, i understand this

[coral shale]

You can visualize an action on a group on its cayley graph if that helps

[slim kayak]

so, is this clear?

[silent bronze]

We suppose that x lies in a subgroup N

[slim kayak]

whoopsie, yes

[silent bronze]

Then conjugation, sends x to another element in N

So for each x, its orbit lies in N

Since orbits partition a set, the orbits partition N

[slim kayak]

eh, close enough I hope

[coral shale]

this needs to be quantified more precisely

A group acts on a set.

You choose a fixed element in that group. g.

[silent bronze]

So N is the union of the orbits

That’s fine

Why does that imply

That N is a normal subgroup

[coral shale]

The orbits with respect to this fixed element partition the set.

[slim kayak]

no that was the direction of N being normal -> is union of orbits

[silent bronze]

Oh

Right

I accidentally did the forward direction

[slim kayak]

this holds true for any g you conjugate by. Right?

[silent bronze]

Yes

[slim kayak]

So you can send x to any spot in its orbit and it will still be inside N

[silent bronze]

Yea

[slim kayak]

So for any element its entire orbit lies in N

[silent bronze]

Yea yea, I get that part

Where in that argument have we used the assumption that N is normal

[slim kayak]

In that x gets send to another element in N

gNg^-1 = N, so if n is from N then gng^-1 must be in N

[silent bronze]

Ohhh okay okay

Yea yea i see that

That was an accident

[slim kayak]

Alr, then you should have everything for the other direction. gl

Make clear what your other statement is and think about why being an union of orbits means that you are normal

[silent bronze]

Okay so for the opposite direction, its easy then, you assume that N is the union of orbits of G under the conjugacy action of G on itself

[slim kayak]

a union*

"the union" is just G

[coral shale]

a partition?

[slim kayak]

partition?

[silent bronze]

Oh right

[coral shale]

i dont understand 'a union' either

[silent bronze]

Then you say that by def of a normal subgroup, N= gNg^-1

[coral shale]

nvm im high ignore me

[silent bronze]

So if all the orbits of elements of N lie inside N

[slim kayak]

chilling on #groups-rings-fields while on that good shit

[silent bronze]

Then by def, N is normal…?

[silent bronze]

I wanna write this out formally, would you mind critiquing it once I’m done?

[slim kayak]

i am fried for today, so i am not doing anything rly - yeah

[silent bronze]

Thank youuuuu

[coral shale]

man, rereading that last line has me confused now lmao

everything else is ite

[slim kayak]

why?

okay it misses one assumption. Let N be a group, then N <= G is normal ....

or at least the right side needs it

[coral shale]

ah ok happy now

[slim kayak]

or perhaps...

[coral shale]

maybe

wait no

[slim kayak]

yeah no

[coral shale]

i dont get how N is the union

it shouldnt be

[slim kayak]

anywas~

[coral shale]

???

[slim kayak]

oh no, i made dc zoom in 🤢

phew

[slim kayak]

||If a subgroup is a union of orbits, then that group is invariant under group action?||

[coral shale]

i find 'union of orbits' vague

what orbits

[ivory trail]

of the conjugation action

[coral shale]

General orbit G . x = {g . x : g in G}

g . x = x^g here

G . x = {x^g : g in G}

[slim kayak]

x^g is conjugation by g?

[coral shale]

yes

[slim kayak]

aight

[coral shale]

union over all x in G?

U_x G.x ?

[slim kayak]

N is a group and it happens to be the union of some orbits of G

[topaz solar]

No, x in N

[coral shale]

ok, that makes way more sense

[slim kayak]

yeah

[coral shale]

the wording just threw me

"... iff N is the union of orbits for the action of N on G by conjugation" would surely make more sense

[slim kayak]

I understand it as: let U be a subset of G, let N be the union of G.x for x in U and assume N is a subgroup, then N is normal.

but U can be taken as N, doesnt matter here

[coral shale]

well u can't pick any U

?

{e}

[slim kayak]

why not?

is the trivial subgroup not normal?

[coral shale]

uh

[coral shale]

then i fail to see where N comes into your statement

oh ok i do

ok ok

yikes

[delicate orchid]

if N wasn't the union of orbits we could find some g that conjugates some n in N out of N

[slim kayak]

yup

[delicate orchid]

this "union/sums" of orbits <-> invariant objects is very common

[coral shale]

whys it obvious to everyone but me 'union of orbits' refers to orbits of things in N

[slim kayak]

yeah, doesnt matter what your action is

[coral shale]

i mean, yeah, we know the answer, but i dont see why apriori

[delicate orchid]

an adjacent example is considering all of Aut(G) acting on G, subgroups which are invariant under this are called characteristic. And indeed they are unions of orbits

[slim kayak]

hm, how does that appear?

[delicate orchid]

centre of a group algebra comes to mind

[slim kayak]

I cant exactly imagine a scenario where I start checking all elements if their orbits lies in the subgroup

ah

alr

[delicate orchid]

Maybe I'm biased because of how much that idea comes up in my own personal research but I don't care RARRRRGGHHH

[slim kayak]

based

[coral shale]

`N is the union of orbits for the action of $G$ on itself by conjugation'
$$N = \bigcup_{x\in N}G.x$$ vs $$N = \bigcup_{x\in G}G.x$$

[coral shale]

im like failing to see how the english reads as the 1st

[slim kayak]

"of orbits"

[coral shale]

what orbits.

[slim kayak]

so any collection of orbits works

[delicate orchid]

it's a union of conjugacy classes

[coral shale]

oh

[delicate orchid]

there is some ${g_1, ..., g_n} \subseteq G$ such that $N = \bigcup_{i=1}^n \text{cl}(g_i)$

[ivory trail]

it's like saying "an open set is a union of basis elements"

[cloud walrus]

W;3w Lads Tbh

[slim kayak]

Its a union of elements, which happen to be orbits

[coral shale]

ah ok, that had me for a big loop

[slim kayak]

The union doesnt need to go over all elements (whatever that means in general)

[coral shale]

N is not constant

`N is a union of orbits for the action of $G$ on itself by conjugation'
$$N = \bigcup_{x\in S}G.x$$

[delicate orchid]

well you can take you set to be N and it's fine

[slim kayak]

Wait, finitenes?

[delicate orchid]

the first thing you wrote there wasn't wrong

[languid trellis]

Sanest wew message

[delicate orchid]

all groups are finite

[coral shale]

im happy reading it as this

[slim kayak]

ah yes my bad

[round hull]

🤔

[slim kayak]

The free groups left math, they couldnt be forced to stay

the other infinite group followed suit

[delicate orchid]

ah yes the conjugacy classes of free groups

such rich structure

[slim kayak]

?

all groups are finite
isnt even about conjugacy classes 😭

[delicate orchid]

RAAAAARRRRRGHHHHHHHH

ooh look chat. We want to define a quotient? What should the corrispond subobject b- and it's invariant under some set of maps! WOAH!

[round hull]

chat?

[elder wave]

Chat is this real

[slim kayak]

never heard of a fourth person pronoun?

[round hull]

fourth person 💀

wait really

[slim kayak]

yes

[delicate orchid]

ideals? Unions under the r -> xr orbits WOAH!

[round hull]

i need to stop thinking about this

chat no longer became a real word

[slim kayak]

https://en.wiktionary.org/wiki/fourth_person

"chat" behaves like a fourth person pronoun, just in the category of english

[delicate orchid]

wait it's ACTUALLY a fourth person pronoun?

I was shitposting in discussion homedawg I didn't mean it!

[slim kayak]

Too late, you made a linguistics reference!

[slim kayak]

i am little curious. What maps and what are you doing with them?

[delicate orchid]

Morphisms in a category and I'm summing over orbits

[slim kayak]

basically - formalize pwease

as a treat

[round hull]

"One does not simply walk into Mordor"
c.f.
"Chat does not simply walk into Mordor"

[silent bronze]

Forward implication

[slim kayak]

"Chat, do I simply walk into mordor?"

[silent bronze]

Is this fine …

Ignore the lemma 3.1 that’s specifically from my own notes

Jus a reference point

[delicate orchid]

as long as you make r large enough to cover all of N, yeah

[slim kayak]

cmon give us both halfs :(

[silent bronze]

I’m lowkey like struggling to write a conclusion fir the second half

[slim kayak]

Well? Take one element, what can happen to it under conjugation?

[silent bronze]

Here’s where I’m at

Tryna word it nicely

[slim kayak]

ah, you are kinda walking yourself into a loop there i feel like

[silent bronze]

Exactly

[delicate orchid]

I think all you need is a "thus the claim is proven" at the end lol

[slim kayak]

What does it mean to be normal in terms of conjugation here?

[round hull]

no!

[coral shale]

ok just to clear up my confusion. this is what we are proving right

,,N\trianglelefteq G\iff \left[,\exists S \subseteq G : N = \bigcup_{x\in S} G.x \leq G,\right]

[delicate orchid]

we got a QEDcel in chat...

[silent bronze]

Uhh

[delicate orchid]

yur

[slim kayak]

I write the statement of the proof and then follow it with "thus the claim is proven", a hollow square, a filled square and QED right after

Some might complain that I didnt show anything, but clearly you cant miss that the claim was proven by any of the symbols

[silent bronze]

Here’s a nicer phrasing if u want

[silent bronze]

I’m tempted

[round hull]

if N is a disjoint union of orbits, then by the definition of a union, if x is in N, it's in some orbit contained in N

[silent bronze]

Right

[round hull]

and conjugacy of that element is in...?

[silent bronze]

Also in N

[slim kayak]

you dont know that yet

[silent bronze]

Don’t we ?

[slim kayak]

So normal means gNg^-1=N aye?

[silent bronze]

Yea

[slim kayak]

So we want to show that if x is in N, then gxg^-1 is still in N. Right?

[silent bronze]

Yes

[silent bronze]

Doesn’t this show that?

[slim kayak]

So what means x is in N here?

Or rather

can you use "N is a sum of orbits" here?

[silent bronze]

If x is in N then it’s conjugate takes us to the orbit of x which is in N

Because N is the union of orbits

[slim kayak]

Exactly! (still)

the vibes were right

[silent bronze]

Oof

[slim kayak]

well?

[silent bronze]

So HENCE, we have shown that if x is in N then gxg^-1 is in N so by def, N is a normal subgroup of G

[round hull]

ya

[silent bronze]

Hooray

[round hull]

btw the result is true without the finiteness condition

[silent bronze]

Yea, I figured

[slim kayak]

why did it pop up anyways

confusing

[delicate orchid]

it popped up because all groups are finite

[round hull]

because all groups are finite

beat me to it

[silent bronze]

Hehe

Finitely generated

In this case

[slim kayak]

ah, my mistake

[slim kayak]

ofc its finitely generated, its finite!

[silent bronze]

Can an infinite group be finitely generated ?

[delicate orchid]

Z

[silent bronze]

By 1

[slim kayak]

yeah

[silent bronze]

There ya go

[slim kayak]

Z^n

[delicate orchid]

most are, in fact

well most of the ones we care about

[slim kayak]

plus any relations you like

[delicate orchid]

here's a fun question for you Kerr

[slim kayak]

oh no

[delicate orchid]

If F(n) is the free group on n variables, is F(3) a subgroup of F(2)

[slim kayak]

yeah

[delicate orchid]

yeah ur right lol

[slim kayak]

i like to forgot about my combinatorial group theory class 🥲

[delicate orchid]

ah so you know the result

sad!

[slim kayak]

ill join your religion, all groups are finite.

[delicate orchid]

the countable free group is a subgroup of F(2)

[slim kayak]

Oh? We just did it for all n

I guess thats homework then, it holds for F^\infty

[delicate orchid]

yur

[slim kayak]

actually nvm..

[delicate orchid]

I don't remember the algorithm that gives you the generators for F(n)

[slim kayak]

just add the generators together

[delicate orchid]

so true...

[slim kayak]

me neither, but we had that as an exercise so I can like kinda pretend i know how to prove it

theoretically

i blame dementia

[delicate orchid]

based

the only proper thing I can remember from my combintorial group theory class two ( ) years ago is the presentation of a group extenstion, Nielsen-Schrier, and coset enumeration lol

oh and a couple of cayley graph nonsenseses

[slim kayak]

oh, it was a mix up with everything

we had an extra dose of cayley graph nonsense

bass-serre theory actually seems kinda cool. We also had a version of nielsen-schreier where you can just draw graphs

[delicate orchid]

I know very little bass-serre theory I must admit

[slim kayak]

the combinatorial version looks like torture method by the CIA

[silent bronze]

There ya go

Complete

[delicate orchid]

and I should REALLY learn it considering the connection with coexter groups

[slim kayak]

oh, that group

poped up a few times, didnt give us much reason to care for it

[delicate orchid]

the coexter groups?

[slim kayak]

i am open for the sales pitch

yes

[delicate orchid]

see uh: The entire representation theory of lie groups/algebras lol (over C)

[slim kayak]

hmm hm hm

[delicate orchid]

I'm very tired so I can't be bothered going into details but they're also just swag

they're simply swag

[slim kayak]

cant wait to learn that.... and have forgotten any useful combinatorail/geometric group theory by then

[delicate orchid]

ok here's one actually

S_n is the most basic coexter group

[silent bronze]

What level of math are y’all at, out of curiosity

[delicate orchid]

I'm 2nd year PhD

[silent bronze]

Damn , big brain

[delicate orchid]

I wish

[silent bronze]

What about Kerr

[slim kayak]

marinating in my undergrad

[silent bronze]

Omg same

[slim kayak]

Wait, does building theory ring a bell?

[delicate orchid]

they're related to buildings yeah

[slim kayak]

I vaguely remember us having a course on it and them hyping it up using coexter groups

[delicate orchid]

tits buildings oh oh oh he he he ha ha ha

another thing I don't know much about

[slim kayak]

I guess i can check that one out, gives points for my degree

seemingly pretty obscure topic tho

[delicate orchid]

they come up in rep theory

and maybe other places? like the more general study of algebraic groups

which is a stupid name btw

[slim kayak]

yeah we sadly dont have that

ah yes, my favourite "groups"

[delicate orchid]

as opposed to the non-algebraic groups

ok there are a couple of them but it's still stupid

[slim kayak]

mfw when algebraic groups arent even regular (read: the english word) groups

[slim kayak]

you left the finite number of orbits in?

[delicate orchid]

well yeah... there's only 8 planets....

[slim kayak]

ok I remember know. You used the language of nielsen transformations and nielsen reduced sets. So you play a bit around until you can think of ways to combine x and y into terms which follow N2 and N3

which is just checking a few inequalities

[delicate orchid]

that nielsen guy needs to touch grass...

[slim kayak]

i hate him

[delicate orchid]

idk he's kinda cool...

although if you really don't like combo group theory that much maybe don't do the buildings stuff lol

[slim kayak]

okay he is alright

I just didnt like having to do stuff with words for weeks on end

the geometric bits were well appreciated

Plus I still dont understand his proof for generating tuples of F(n) being nielsen equivalent to (x1, ..., x_n , 1 , ... , 1)

[delicate orchid]

I don't recall this proof so ur on ur own boss lol

[slim kayak]

oh no i dont want to understand it these days

mostly out of spite

[south patrol]

Nielsen

[clear fiber]

Hi all. I wrote this disproof of the statement. I used Z[x] as my counterexample ring. However, my proof takes advantage of the fact that Z[x] is an integral domain. Is my proof still valid?

[wraith cargo]

Ye

It doesn't matter if you take a nagata ring lol it's still a counterexample

[clear fiber]

ok great

[round hull]

2 isn't in (x + 2)

[clear fiber]

isn't it, since 0x + (2)1 = 2?

[slim kayak]

Isn't that supposed to be (x,2) the whole time?

[clear fiber]

oh crap yeah

I wrote it wrong, that's what I meant to write

ideal generated by x and 2

[round hull]

no (x + 2) and (x + 3) works better

[slim kayak]

?

[round hull]

their elementwise multiplications have degree 2 (or 0)

[round hull]

but you can get degree 1 through subtraction, as shown

[slim kayak]

Can't you say that x=fg implies either f =ax and g=a^-1 or vice versa directly?

Or is your grader very strict?

[clear fiber]

oh, you mean without invoking all this stuff about their degree and whatnot
yeah probably, this is very granular

not in a class right now, just studying by myself

[slim kayak]

I just find it a bit awkward when proofs lose their momentum

[clear fiber]

ok I think I did learn an important lesson, which is that it's okay that my counterexample has extra structure
It is nevertheless still an example of a ring

[slim kayak]

I think it's pretty standard example actually, at least I remember seeing sth similar before

[clear fiber]

Yeah it's funny much style matters in proofwriting. And deciding how much detail to include or exclude

I tend to like having a verbose level of detail, it helps me understand when the individual steps are small

[slim kayak]

Actually

That's just the definition of the product ideal up there, isn't it?

[clear fiber]

I think the point of the exercise was to show why the product of ideals has to be defined this way

Because the definition of K doesn't always end up being an ideal

[slim kayak]

Ah, yeah

I feel like you might be able to potentially do this with ideals of Z no?

[clear fiber]

I'm not sure, it might. I struggled a lot with this problem and spent a bunch of time trying to find ideals that wouldn't work

It seemed like if one of the ideals is principal, then the set K actually would be an ideal

[slim kayak]

By cleverly taking differences to get an element that isn't divisible by either

[round hull]

no

[slim kayak]

Oh? What's the relevant property there?

[round hull]

oh Rmoney said it already

if one of the ideals is principal

[slim kayak]

Oh nvm I see it now...

[celest furnace]

Channel got moved?

[tribal moss]

Hmm, right, looks like something got moved around.
#get-advanced-access had ended in the middle of the category, and I've put that back to the top, but I have no reliable memory of what the order of the rest of them should be.

[cobalt heath]

Yea, I was when puzzled I saw that

..wait, you reverted all the orders back

[grim hollow]

Hey folks, does there exist a surjective homomorphism $\phi: A_4 \to C_3$? And can you verify this without defining such $\phi$?

My prof seems to suggests that there does exist a surjective homomorphism, and it can be checked via the First Isomorphism Theorem of Groups.

So here's my incomplete attempt at it ($C_3$ is a cyclic group of order 3):

If there exists a surjective homomorphism $\phi: A_4 \to C_3$, then, by the First Isomorphism Theorem, $A_4/\ker \phi \cong C_3$. Since $|A_4| = 12$ and $|C_3| = 3$, $\ker \phi = 4$. Now, the Klein four-group $V_4$ is a (in fact, the only) subgroup of $A_4$ with order $4$...

[cloud walrus]

Andrew

[cobalt heath]

Seems the order of proof is inversed

[grim hollow]

Yea, it certainly feels inversed. I'm just not sure how to use my prof's suggestion of using the First Isomorphism Theorem here

On second thought, I think this is a no: If such a surjective homomorph were to exist, then the kernel must be the Klein 4 group (because it's the only normal subgroup of order 4). However, when we observe the elements of the quotient A4/V4, it is not cyclic (first coset is V4 itself, second coset consists of four 3-cycles, third coset also consists of four 3-cycles). Hence it can't be isomorphic to C3, and no surjective homomorph can exist.

[tribal moss]

I think what is intended is that you say something like: Because there is a normal subgroup of order 4, the isomorphism theorem says there must be a surjective homomorphism to a group of order 12/4=3. And the only group of order 3 is C3.

However, when we observe the elements of the quotient A4/V4, it is not cyclic (first coset is V4 itself, second coset consists of four 3-cycles, third coset also consists of four 3-cycles).
I don't understand what your argument is here, but the conclusion is false.

[grim hollow]

That makes a bit more sense than my previous interpretation.

However, I still do not fully understand. The First Isomorphism theorem in my textbook (Aluffi) reads: Let $\phi: G \to H$ be a surjective homomorphism, then $H \cong G / \ker \phi$. Are you suggesting that there's sort of a "reverse" (not converse) interpretation of this? Does it follow naturally from the definition I'm given?

[cloud walrus]

Andrew

[tribal moss]

TBQH, I can never remember a numbering or exact statement of the isomorphism theorems. (According to the Wikipedia article, different books number them differently anyway).
The property I need here is just that |G| = |G/N|·|N| for a normal subgroup N. In this case, then we have |G|=12 and |N|=4, so |G/N| must be 3.

(Which, now that I look it up, is not generally considered to be one of the isomorphism theorems, but a variant of Lagrange's theorem).

[grim hollow]

Ah gotcha, thanks! Just to clarify, the only group of order 3 is C3, so, since we found a group G/N with |G/N|=3, then it necessarily means that they isomorphic?

[tribal moss]

Yes.

(Every group of prime order is cyclic).

[cobalt heath]

This is the first isomorphism theorem, also the only iso. thm that I can recall

[mighty kiln]

That's cuz it's the only iso. thm

[cloud solar]

Hello

I have a problem that says: If A is a ring and the set of zero divisors is finite then A is finite.

But I think this is false

For the case where the only zero divisor is 0

Because we only know A is an integral domain

If we have at least one zero divisor ≠0 then the problem is true

[deft pumice]

Yep if the only zero divisor is 0 you can look at the real numbers as a ring. its not finite but only has 1 zero divisor

If there is more than 0 as the zero divisor but at the same time its finite then A has to be finite.

cause:

[cloud solar]

I know

[deft pumice]

ah ok

[cloud solar]

I did it

[deft pumice]

I have a question could somebody help me understand the following proof:

I dont understand how they get the formula sum_c\d (psi(c)) = d

[cloud solar]

I supposed that A is not finite. Then I took a zero divisor ≠0 d1 so it must be d2≠0 zero divisor s.t. d1d2=0 (or d2d1=0 doesnt matter). And then I took the set M={a in A{0} s.t. ad1≠0}. This set is not finite. First suppose there is no a in A{0} s.t. ad1≠0 so ad1=0 for all a in A{0} and this is false because we have a finite set of zero divisors. So M has at least one element. Now if we suppose M is finite, M is in A,but because A is finite we need to have an infinite number of elements a in A\M s.t. ad1=0 and again this is false. So M is not finite and now 0=d1d2=(ad1)d2 with a in M. But because M is not finite we have again an infinite number of zero divisors which is false. So A must be finite.

It is ok?

[deft pumice]

yep basically you do: there can only be finite zero divisors, if there are finite elements in total we are done. So we assume there are infinite elements with only finite zero divisors. Now you take such a d*b=0 while d=/=0 and b=/=0. you multiply a element that is not a zero divisor (since we have infinite elements and only finite zero divisors we must be able to find such an element). and multiply it by d. Now you have a*d=/=0. but also b=/=0. Now we multiply simply a*d*b. this must be 0 so a*d is a zero divisor. So it follows that we have infinite zero divisors.

[cloud solar]

Thanks

[coral shale]

im confused

is the question missing information

A is not a field?

[cloud solar]

Is just a ring

I saw too that the problem has some missing information (the case when we have only one zero divisor)

I saw something cool that the number of elements of a finite ring R can be upper bounded by some value which depends on the number of zero divisors

Is it true?

[deft pumice]

Yep

Imagine you have 2 zero divisors

We already know for every non zero divisor we get another one. At max we have 4 elements. +1 due to the 0

cause let a and b be zero divisors, now your other elements fullfil that ac=b , and bc=a, since those are zerodivisors but we only have 2.

actually it sorta depends on if its abelian

atleast from what im trying to write

but id definetly reckon that there is a way to upper bound it

[cloud solar]

Yes foe abelian

Finite commutative ring R with n+1 zero divizors.

The ring R doesn't have more than (n+1)^2 elements

[cloud solar]

Yea

If we take x a nontrivial zero divisor his annulator Ann(x) has at most n+1 elements. Now take phi: R/Ann(x)-->D, where D is the set of zero divisors. phi(r+Ann(x))=rx. And phi is surjective so n+1>= the order of R/Ann(x) and from here we have that the order of R is <= (n+1)^2

Now an example where the equality holds is Z/4Z

But

Can we generalize these type of commutative rings with (n+1)^2 elements?

[dire siren]

I think it depends on terminology. Some sources don't consider 0 as a zero-divisor, so chances are that in your problem they meant that the set of non-zero zero divisors is nonempty and finite.

[silent bronze]

Can someone guide me through part (c)?

Homomorphisms and kernels trip me up

[rocky cloak]

x^d = 1, has d roots. And each root has order dividing d. So there are exactly d elements that have order dividing d.

[silent bronze]

The elements of Z(G) would be such that they commute with all other elements in G

And the elements of the Kernel are such that the image of g is the identity permutation

So where the image of x is x…?

[coral shale]

Don't ping specific users for help

[silent bronze]

Oh sorry

[coral shale]

This question isnt too roundabout; write down all the definitions

[silent bronze]

He helped me once before with group theory so

[coral shale]

if ur stuck, show where ur at

[silent bronze]

I’m legit just trying to understand what to do

[tribal moss]

Dp you have a moderation problem?

[silent bronze]

No …?

[tribal moss]

Then don't ping.

[silent bronze]

Sorry, I didn’t know 🥲

[tribal moss]

It shouldn't take any particular knowlegde to refrain from harassing someone for having helped you once.

[silent bronze]

I’m not harassing you

I apologised

I won’t do it again

[round hull]

if θ(g) is the identity action, what must g be doing via conjugation

[silent bronze]

Sending x back to x?

[round hull]

wdym?

[silent bronze]

So if theta (g) is the identify action then gxg-1 = x

Right ?

[round hull]

what do you mean by gxg-1 = x

[silent bronze]

So the action g * x is defined as the conjugacy action of G on itself

[round hull]

harsh

[silent bronze]

Hence so I mean $gxg^_1$

[cloud walrus]

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

[round hull]

admittedly i can't see the deleted messages

[round hull]

i was looking for something else

[silent bronze]

I wrote “can you take a look at this please “with his @

[round hull]

if θ(g) is the identity action then it must be that the identity gxg-1 = x holds for all x in G

[silent bronze]

Uhhhh

[silent bronze]

Yes that’s what I meant but in a badly mathematical written way I guess

[round hull]

being clear helps when you're thinking about ideas you've written

[silent bronze]

Oh

So like

If we rearrange that

[round hull]

and here it basically gives you the answer

[silent bronze]

Then gx = xg

For all x in G

By definition that means that all x commute with all other elements in the group so that means all x in G are elements of Z(G)

[round hull]

uh

[silent bronze]

So the kernel = the centre

[round hull]

no

[silent bronze]

Oh

Hold on let me write this formally

[round hull]

take your time

[silent bronze]

So here’s what I mean

[round hull]

you wrote more but not clearer

[silent bronze]

Which bit needs to be clearer..

[round hull]

actually maybe i shouldn't be so harsh

[silent bronze]

No pls b harsh

[round hull]

well when you say "Elements of Z(G) are such that xg = gx" what are you referring to

[silent bronze]

So I’m saying that for some x in Z(G), xg=gx for all g in G

[round hull]

that is true

or you mean "if x is in Z(G), then xg = gx for all g"

[coral shale]

yeah, fully quantify statements

[silent bronze]

Yes, put another way

[round hull]

now i'm being overly pedantic of course, but they're not the same

[silent bronze]

Oh

Okay so one is a conditional statement

[coral shale]

its the difference between

[silent bronze]

And we wanna show that x satisfies a certain condition so we use if ?

[coral shale]

exist x in Z(G) : forall g in G : xg = gx
forall x in Z(G) : forall g in G : xg = gx

Those are the 2 statements im seeing, which mean different things

[silent bronze]

The second one doesn’t make a whole lot of sense

I see that now

[coral shale]

that's what this is

[silent bronze]

Aside from that the argument is fine right ?

Like mathematically

[coral shale]

I dont know what u mean though, isnt that the definition of ZG

maybe im under a rock here

[round hull]

unfortunately still not quite

[silent bronze]

What’s missssinnnnngggggg

[coral shale]

Z(G) = {x in G : forall g in G : xg = gx}

right?

[silent bronze]

We’re tryha show that it’s equal to ker (theta)

Refer to the question again

Part c for reference

[delicate orchid]

Z(G) in #groups-rings-fields and Z[G] in #advanced-algebra very lol

[coral shale]

I do remember your overall goal, I was confused by what this meant