groups-rings-fields

[unkempt stream]

yeah

If a minimal poly’s leading coefficient is a unit, then you can just divide by it’s inverse to have an integral extension

which is always the case for a field

but yeah, this

i wonder if the evaluation map that sends X_n to \sigma_n(x) for some chosen x can be manipulated somehow

because applying an automorphism in G to that after you evalute it is the same as permuting the indeterminates

i have no idea

i know the proof of artin’s lemma is amongst those same lines

[white oxide]

why are we allowed to pull out the $r$? I thought by definition f was a R-module homomorphism from $L \to \text{Hom}\mathbb{Z}(R, J)$ and hence isn't in $\text{Hom}\mathbb{Z}(R, J)$

[cloud walrus]

okeyokay

[wraith cargo]

Say I have M = IM+N for fg R-modules M and N, and I is an ideal that is contained in the jacobson radical of R, does Nakayama's lemma then tell us that M=N?

[topaz solar]

What

[glossy crag]

I think so

[topaz solar]

It’s an R module hom, so you can pull out scalars

[wraith cargo]

thanks!
I looked elsewhere and found out that I was thinking of it wrong but now it makes sense

[white oxide]

oh yeah duh right thanks

[little root]

is there a trick for computing the 105th cyclotomic polynomial by hand, or is this problem not meant to be done manually?

[wraith cargo]

$x^n-1 = \prod_{d|n}\Phi_d$

oops lol

that ain't true

[formal ermine]

isn't 105 like the first without abs <= 1 coeff

[formal ermine]

make the lhs x^n - 1

or + 1?

probably + 1

[wraith cargo]

yea

x^n-1

I think if you put n=105 you could calculate it

[formal ermine]

right

now just gotta do loooooong division

[cloud walrus]

Irony Incarnate

[wraith cargo]

wtf 105 has a super nice factorization

[formal ermine]

,w 105

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by projective

[formal ermine]

bruh

[wraith cargo]

3*5*7

so the only divisors are 3,5,7,15,21,35 and 105

OH

YOU CAN USE

THE MOBIUS INVERSION FORMULA!!!

[wraith cargo]

you can calculate it as $\Phi_n = \prod_{d|n}(x^d-1)^{\mu(n/d)}$

[cloud walrus]

Irony Incarnate

[little root]

yeah but you would need a computer to actually compute that right

[chilly ocean]

bruh

[little root]

like theres not some trick that gives you the explicit answer easily?

im just confused since its an exercise in aluffi

[formal ermine]

what

[wraith cargo]

this a product with 7 terms so it's not that awful

[wraith cargo]

not really

[little root]

that's 7 too many hehe

ok cool lol

[little root]

wait what's the \mu here?

[wraith cargo]

the mobius function

[little root]

oh interesting

[little root]

wouldnt this formula have multiple factors of x - 1 though

[wraith cargo]

no?

[little root]

like x - 1 divides each (x^d - 1)^{\mu(n / d)}

[wraith cargo]

ahh
okay yes but I mean it'll be hidden in the computation

[little root]

wait wdym

[wraith cargo]

sry am very tired
I meant as in yes there are factors of x-1 but you can just multiply out the (x^d-1)

[void cosmos]

guys is 3 trivial

[wraith cargo]

no

[void cosmos]

wait how

[wraith cargo]

the a_i could generate some ideal that's only contained in M

[void cosmos]

i dont get it

tbh

[wraith cargo]

say the a_i generate some module M'

then when you map M' to R/M^2 it has the same image as M

[void cosmos]

you mean consider M'/M^2?

wdym by map

[wraith cargo]

R -> R/M^2

[void cosmos]

yea

the equivalence classes of the ai generate M/M^2

is that what ur saying?

just going mod M^2 or what

sorry ur just being weirdly minimalistic in ur explanation like

haha

[wraith cargo]

I'm was trying to come up with an example lol

but the problem is saying like

say you have generators of M/M^2 right
You have to show they generated M as well
Now this really isn't immediate since hypothetical what's stopping the a_i's from generating some subideal of M called let's say M' that just has the same image under the map R -> R/M^2 as M does?

I'm trying to come up with some tangible example but idk what rings are non noetherian and local lol

[void cosmos]

ig it would be

contained in M anyways

this subideal

that is

and i have to show that this + generating M/M^2 is enough

are*

[wraith cargo]

I don't think it need to contained lol

like modulo deletes a lot of info

[void cosmos]

isnt any ideal contained in a maximal ideal anywaays?

[wraith cargo]

yea

[void cosmos]

and M is our only maximal ideal right

[wraith cargo]

proper ideal

ah sure yeah in this case that's true
I got used to proving these kinds of results in group theory where it needn't be local at all

[void cosmos]

okay so now i have to show that M is a subset of this ideal generated by (a1,a2,...)

ugh idk hbow

how*

like

i dont know how to go from M/M^2 to R lmfao

maybe i can use nakayamas lemma or something idk

[wraith cargo]

that is really not a bad idea

[void cosmos]

idea*?

[wraith cargo]

yeah lol

[void cosmos]

yea maybe i can just

like

[wraith cargo]

Here's how I'd do it

[void cosmos]

the assumptoin that this ideal generated by a1a2

is not M

would just make me blow it up and make it = 0

[wraith cargo]

take M' to be the idal generated by the a_i

[void cosmos]

if i can show that MI = I

[wraith cargo]

and since it's local clearly M' subset M

[void cosmos]

yea

[wraith cargo]

now you need to show that M' = M
And the Way I'd do this is to take the quotient M/M' and use nakayama to show it has to be 0

You also very likely have to use the notherian condition too

[void cosmos]

probalby use like

third iso theorem to divide by M^2 up and down

and then i think any representative would be just a sum of the form ab where a is in M' and b is in M/M^2

lmao its far from trivial ig

and ofc

i cant find the solution

in any solution manual

[little root]

isn't this just nakayama's lemma?

or like specifically this corollary of nakayama's lemma

but this doesn't require R to be noetherian local so maybe im missing something

[void cosmos]

thats what i had in mind

when i thought it was trivial

but yea

thats enough

i think it being local noetherian is just for the nakayamas lemma hypothesis

[coral steeple]

Hi there, I'm looking for a group which has at least one subgroup of infinite index and at least two of finite index

[formal ermine]

Z x Z/2 x Z/2

[coral steeple]

My book hasn't gotten to product groups yet so I don't know how that works though

(Nor quotient groups)

[agile burrow]

Z itself should work, if you're familiar with computing the index of subgroups of Z

[coral steeple]

I tried the additive group of integers, and the multiplication groups of R and C

[delicate bloom]

multiplicative group of C should work

[coral steeple]

Wait but don't all subgroups of Z have finite index

[delicate bloom]

all of C itself is infinite, then {1,-1} and then {1,i,-1,-i}

[agile burrow]

There's exactly one subgroup of infinite index

[crystal turtle]

Or in other words, there's exactly one subgroup of finite order

[coral steeple]

Ah yes it's just 0 right

[delicate bloom]

oh I misread

[coral steeple]

@agile burrow

[agile burrow]

Yeah, that's right

[coral steeple]

Ok thanks!

[agile burrow]

Happy to help

[coral steeple]

This is not a commutative diagram, right?

Does this type of diagram have a name?

[wispy urchin]

resist the low hanging fruit challenge (impossible)

[delicate bloom]

think they call that a feynman diagram

[hidden haven]

It is commutative though no harm in calling it that

[ivory trail]

all DAG diagrams forests are actually commutative, true

[hidden haven]

No,
• → •
↓ ↓
• → •
is DAG but need not commute

[ivory trail]

fuck

there

[vague granite]

Does one need the axiom of choice to find $Y\subset X$ such that $X=\sqcup_{y\in Y}\mathrm{Orbit}(y)$ for a general group action?

[solar vessel]

yeah prolly

[cloud walrus]

Philka

[vague granite]

I suspect so to since you essentially need a representative for each orbit

If you have a.o.c. then you let $f:{\mathrm{Orbit}(x):x\in X}\rightarrow X$ be a choice function and you get $X=\sqcup_{y\in \mathrm{Im}f}\mathrm{Orbit}(y)$.

[cloud walrus]

Philka

[rocky cloak]

Let A_i be an indexed family of nonempty sets, let X be their disjoint union and let G be the product of symmetric groups S(A_i), acting on X in the obvious way. Then picking such a Y is exactly picking a choice function

[topaz solar]

If you can bound things in size (or maybe some structure?) you can get away with a weaker choice principle, but you’d have to have choice even if not a full power

[coral spindle]

My guess would be yes although I can't think of an immediate proof here. My reasoning is that under choice, given any partition of a set you could find a group action on that set whose orbits form that partition. This is obviously not sufficient to show equivalence, though.

[chilly ocean]

Hey I don't know how to prove the following :

On a G chart simple of order 2n,in which the degree of each edge is even,there is for each vertex another vertex connected to an even number of edges in common.

Thanks for help

[coral spindle]

This is not an abstract algebra question. Please move to #discrete-math for elementary graph theory questions.

[chilly ocean]

Ok ty

[topaz solar]

try doing it on partitoning UX by (disjoint) x in X?

or uh, UX u X kinda thing

[coral spindle]

I'm writing up my approach rn actually give me a mo!

@vague granite OK I got it – I think the following is a proof that the existence of representatives of a group action is equivalent to AoC. To prove it, I simply show that we may choose representatives of any partition assuming the group-theoretic version, since the latter is a convenient form of AoC.

Consider some partition of a set. For each part P of the partition, consider the group Sym(P) of bijections on P. Since we may construct a bijection that swaps any two elements of P, the action of Sym(P) on P is transitive. Now consider the group G defined to be the product of Sym(P) over all these parts P. We may explicitly construct elements of G that show that the orbits under the obvious action are the same as the parition, so we're done.

Obviously some details need checking but I'm confident

Stunningly simple tbh lol

Just had to choose the right groups

[vague granite]

I think jagr said something similar

[coral spindle]

Oh lmao that's what I get for not reading

[solemn dew]

herstein's book doesn't have subrings

but are subrings the same as ideals?

[solar vessel]

no

[coral spindle]

They are critically different

[solemn dew]

oh

[coral spindle]

A (unital) subring is almost never an ideal, and an ideal is almost never a unital subring.

[solemn dew]

unital subring

is it just {0}?

[coral spindle]

No

A unital subring is a subring, specifically that contains 1

Since we do sometimes care about subrings of a ring that don't contain 1, we need to make a distinction

An ideal is always a non-unital subring, but even non-unital subrings are only rarely ideals

[solemn dew]

wait what

an ideal of a field is the field itself or just 0

[coral spindle]

Yes

[solemn dew]

then the ideal can contain 1

[coral spindle]

Yes

That is the only ideal which is a unital subring

[solemn dew]

oh okay gotcha

thanks

any book recommendations on subrings?

[coral spindle]

None that I know

We don't tend to care very much about subrings

We care a lot more about ideals

[solemn dew]

alright thanks brother

[teal vessel]

so I'm having to prove that for some subgroup H of the additive group of rationals, the property that for all nonzero x, (x ∈ H) ⇒(1/x ∈ H) must mean that H is either {0} or H=Q. I know there are easier ways to do this, but (ignoring the H={0} case because it is incredibly easy to show) I showed the following:

assume 1 is a member of H, this directly implies all integers must be in H by closure.
this further implies that all unit fractions are in H by the given property
By closure, all multiples of unit fractions must be in H, i.e. all fractions, therefore, if 1 is in H, then H=Q

Further, assume some nonzero element x is in H, this implies that 1/x is in H as given
by closure, x/x must be in H (all multiples of 1/x must be in H, but this is the vital one), which is equal to 1, therefore if any nonzero x is in H, the above assumption starting at 1 applies.

thus, if there exists a nonzero x in H, then H must contain all rational numbers, and is therefore equal to Q.

(if there's only 0, then it's a trivial group where 0 is the identity, and it is exempt from the extra condition given because that's a universal, not existential quantifier).

there are easier more efficient ways to determine this, by starting from any arbitrary nonzero element instead of from 1, but I like this thought process.

[rocky cloak]

How are you concluding that x/x must be in H?

[teal vessel]

repeated addition of the unit fraction 1/x

[rocky cloak]

but x is rational?

[teal vessel]

fair. to be strictly rigorous, I suppose the proper statement is that it ends up being some integer by flipping the fraction and getting a multiple of the denominator, which then implies that a unit fraction in integer components exists, hence the rest follows.

[rocky cloak]

Right, yeah that works

[night onyx]

yeah it still works, just take p/q \in H, so that q * (p/q) = p \in H, so 1/p \in H, so p * (1/p) = 1 \in H, you can still get there

the proof does look good though, I agree

[teal vessel]

I think it's often better to show the process of breaking down the proof into easier parts, i.e. first show that 1 works, then any integer, then non-integers. The final proof given in texts is often quite dense and removes such lemmas

[crystal turtle]

yeah this is fine, not sure if there's a nice way to do it more efficiently

[teal vessel]

mostly just re-ordering for presentation.

[night onyx]

I wonder if you can prove that H has to be an ideal of Q, and Q being a field means H = 0 or Q (lol I don't know if it's true just a thought)

[teal vessel]

I would not know. I may return to that question when I learn of ideals.

isn't this literally just the notion that the union of two groups in a subgroup relationship is equal to the larger group, and therefore the union from 1 to n across chained subgroups H_i is just H_n, and in cases where an infinite chain exists, the "limit" of that unioning just converges onto G? No clue what the rigorous version of this will look like, but as a sketch that seems fairly obvious.

(also with the fact that any group is a subgroup of itself)

[ebon heron]

https://youtube.com/@4_laughs

[rocky cloak]

The limit doesn't need to converge to G, but you're right that the limit is the only thing you need to prove something about

It's also not very complicated

Just checking the definition of subgroup

[teal vessel]

oh, it could arbitrarily stop at any such subgroup and just be "unioning with itself" over and over right? damn axiom of choice.

[rocky cloak]

I'm not sure what you're saying, and the axiom of choice doesn't play a role. Consider for example

(1/2) < (1/4) < (1/8) < ... Q

[teal vessel]

if I didn't do all my work on a whiteboard and actually kept notes this would be much easier with a previous result from this section

[teal vessel]

I'm not sure what those groups are representing

(1/2) doesn't look like a group to me, it looks like a number

[rocky cloak]

The cyclic group generated by 1/2

I.e. numbers of the form n/2

[teal vessel]

ah, ty

so this ends up going to the group generated by all negative powers of 2, got it

OH given any subgroups H and K, H U K is a subgroup iff one is a subset of the other, and that's already established in the chain structure

[crystal turtle]

But that only could extend with induction to finite unions

Not infinite chains like this

[night onyx]

You can prove this is a subgroup pretty directly: obviously 1 is in each H_i so it's in the union, if x is in the union it's in some H_i, so it's inverse is also, so its inverse is in the union, and if you have x in H_i and y in H_j with i < j then because it's a chain of inclusions, both x and y are in H_j, so their product is. No fancy stuff required!

[teal vessel]

evidently I need to relearn induction conditions.

[crystal turtle]

I mean the general rule of thumb with something like unions or intersections is that, once you've proved it for the binary case, you can usually extend (at least under some circumstances) to any arbitrary finite case

But with (regular, not transfinite) induction, you won't every get the infinite case

[teal vessel]

my last exposure was entirely transfinite induction, I'm out of practice lol. that's what I get for doing studies in pure set theory before this 😅

[crystal turtle]

lmao

but yeah this specific case should just be an easy evercise in subgroup definition

[teal vessel]

yeah

[solemn dew]

any hints guys?

[rocky cloak]

If x is contained in an ideal then so is ix - xi

[south patrol]

Interesting problem btw

[rocky cloak]

Wedderburns theorem coming up maybe?

Not that the problem is really leading to it, just kinda hinting a bit

[solemn dew]

very nice

you show that 1 is in our ideal, hence it has to be the ring

[unkempt stream]

wedderburn’s theorem is neat

so is the fact that if for each x in a ring there is an n such that x^n - x is central, then the whole damn ring is commutative

doesn’t even need a multiplicative unit afaik

[unkempt stream]

quick question

by ideals is it referring to sided ideals or two sided ideals

[solemn dew]

two sided ideals

they call it left-ideal or right-ideal if it is not a two sided ideal

but i would argue that they should call it ideal-right and left-ideal

💀

[open sluice]

that doesn't sound ideal to me

[unkempt stream]

In my own stuff i usually call a two sided ideal a strong ideal

Otherwise left or right

Or a normal ideal

Given how it related to a normal subgroup almost

rH = Hr for normal subgroup H

rI = Ir (= I) for two sided ideals

[unkempt stream]

By the definition, 1 can be proven by showing that this ringc, call it Q_p, is a finitely generated module (with an associative algebra) over Z/pZ of degree 4

therefore it is module isomorphic to (Z/pZ)^4 which has cardinality p^4?

idk

WAIT

[delicate orchid]

it's p^4 because it's a dim 4 vector space over F_p

[unkempt stream]

Same proof different words given Z/pZ is a field

[unkempt stream]

Btw isn’t that the module algebra F_q[Q_8/Z(Q_8)]

or smth

idk

[delicate orchid]

are you asking if this doodad is iso to that?

[unkempt stream]

No it isn’t

i was going to but it isn’t

[coral shale]

we dont do politics here

[unkempt stream]

trying to think of a non-unit

[delicate orchid]

thankfully we're taking p odd here so this ring is actually nice (semisimple)

[unkempt stream]

assume x = a + bi + cj + dk is a unit

[delicate orchid]

stop pinging the dude mizalign

[unkempt stream]

i thought i had the reply pings off

srry

i had it off i must’ve turned in on again by accident

[delicate orchid]

Q_8/Z(Q_8) \cong C_2 x C_2, the group algebra F_q[C_2 x C_2] is commutative so no they're not iso

ok cool

[unkempt stream]

I’ll be back with a solution hopefully

i’m somewhat lazy so

x = a + bi + cj + dk
Then
x^-1 = (a - bi - cj - dk)/(a^2 + b^2 + c^2 + d^2) in F_q

so a^2 + b^2 + c^2 + d^2 ≠ 0 in F_p should be necessary and sufficient to prove an element unital

[unkempt stream]

thing is, idfk what examples to choose

I know for certain each element in a prime field is the sum of two squares

[delicate orchid]

wya wyd now boss

p = 11, a = 2 b = 3 lol

[unkempt stream]

so just chose any x, get x = a^2 + b^2

and -x = c^2 + d^2

So a + bi + bj + dk has F_p norm 0

So it’s a zero divisor

[delicate orchid]

seems like you knew exactly what examples to choose

in fact, you chose all of them at once

[unkempt stream]

Lol

[delicate orchid]

also I quite like that this shows that every prime can be written as the sum of 4 squares

[unkempt stream]

WAIT WHAT

[delicate orchid]

a very funny proof for that result

[unkempt stream]

dude someone gave me that sum of two squares as a challenge problem like 2 days ago, that’s the only reason why I thought of it

[unkempt stream]

the only other proof I’ve seen of that was with quadratic forms in Serre’s Course in Arithmetic

i have it printed out on my bedstand lol

[delicate orchid]

how would you show this without quadratic forms?

[glossy crag]

For the ideal bit you could argue like this: you know this is a 4-dimensional algebra (can be proved using the i,j,k relations) and any image of this algebra would again be an algebra satisfying the same assumptions (generated by F_p and i,j satisfying i^2=j^2=-1 and ij=-ji), so 4-dimensional => all its images are isomorphic to it => it has no ideals.

[unkempt stream]

consider the endomorphism f(x) = x^2

[solemn dew]

i just used @rocky cloak 's method and showed that 1 is in the ideal

[unkempt stream]

If f(x) = f(y), then x^2 - y^2 = (x+y)(x-y) = 0

[solemn dew]

but thanks for the advice

[unkempt stream]

follow from there

[solemn dew]

don't know what a algebra is yet though

[unkempt stream]

hint: no element is nilpotent of degree 2, sorry

[glossy crag]

Stuff akin to M_n(K) (a ring that is simultaneously a vector space so that ring multiplication and scalar multiplication commute). Prototypical examples include M_n(K), K[x] and the celebrated quaternions.

[unkempt stream]

WOrds mixed UP

[delicate orchid]

:uponthewitnessing: I cannot consider :tearsofthenile: a construction most mysterious :jubulantboltsofheaven:

[unkempt stream]

backslash colons

epic

[solemn dew]

i'll look into this as soon as i get to the vector space chapter

thanks 😄

Stuff akin to M n K a ring that is

[unkempt stream]

this is a great example of non-division simple rings

[glossy crag]

My bad, I spoke out of turn (didn't know you were just starting out).

[unkempt stream]

e.g having no nontrivial bi-ideals does not imply all elements are units

[delicate orchid]

doing rings before vector spaces is cracked

[glossy crag]

M_n(D) is better imo

It's very easy to see that there's plenty of non-invertible elements

[unkempt stream]

shut up about your endomorphism crap

[delicate orchid]

I could find an ideal in M_n(D)

[glossy crag]

What would be a witty comeback to this

Your mother?

[unkempt stream]

heteromorphism mapping you to my mother

[rocky cloak]

https://imgflip.com/i/7ubmzv

[glossy crag]

AKA Br(F_p)=1

Since I have you here @rocky cloak, mind if I run something by you?

[rocky cloak]

Shoot

[glossy crag]

I was trying to show an algebraic extension of a quasi-algebraically-closed field is QAC (QAC=homogenous polynomial polynomial of degree d in n variables with n>d has a non-trivial root). Clearly this reduces to the case of a finite extension, however I haven't made any headway there.

[rocky cloak]

Sounds like some algebraic geometry magic maybe

[glossy crag]

If f=\sum_j a_jx^j is in L[x], then write out the a_j wrt a K-basis b_i and get a b_i-linear combination of homogenous K-polynomials. However I don't know what to do afterwards lol.

[glossy crag]

Your mother's fundamental group will be Z*Z once I'm done with her.

[unkempt stream]

lol

[glossy crag]

I'll try there I guess.

[rocky cloak]

Yeah, I have no idea, but I'll give it a think

[glossy crag]

No need to strain yourself, I'll take it next door.

[unkempt stream]

ocean man, are you knowledgeable in galois theory

[glossy crag]

Depends on your definition of knowledgeable. Why?

[unkempt stream]

kinda wondering if you know if there is a way to generalize Artin’s Lemma

[glossy crag]

What exactly do you mean by Artin's lemma?

[unkempt stream]

if G is a finite subgroup of Aut(F)

and F^G is the fixed point subfield of F under G

then [F : F^G] = |G|

[glossy crag]

Also F/F^G is Galois with group G

I don't know anything about generalisations of this, I just know it as an auxiliary result in Galois theory used to prove the Galois correspondence in the finite case.

[unkempt stream]

yeah

[glossy crag]

But, maybe a potential avenue could be some kind of invariant-theoretic/group-action thing?

[unkempt stream]

i am wondering if there is some way to generalize galois theory to a larger class of rings via algebraic extensions (not just purely module extensions in the field case)

which finitely algebraicly generated <=> finitely module generated in the field case

[glossy crag]

Sorry, my man, I'm just a lowly undergrad far removed from such lofty ideas

[rocky cloak]

found a proof here https://www.jstor.org/stable/1969785

[unkempt stream]

i am not even an undergrad yet lol, just going off of D&F

[glossy crag]

That said I recall there was a chapter in Jacobson about Galois theory in the non-separable case or something like that. Not exactly what you're looking for, but it's something.

• someone more knowledgeable than me could tell you about Galois descent, afaik that's soooort of related (invariant stuff)

[rocky cloak]

Group actions on rings and the resulting extensions R^G -> R is something people study at least. Dont know if thats what your going for

guess it would fall under geometric invariant theroy

[delicate orchid]

or rep theory, lol

[rocky cloak]

There is also some generalization of the galois correspondence where you replace groups with hopf algebras

though I dont remember exactly how that works

from Hopf Algebras and Galois Theory - Chase--Sweedler

[south patrol]

If A is an R-coalgebra with unit (aka coaugmentation) A -> R, is there a name for the cokernel of the unit?

I just see it called JA but I wondered if e.g. "coaugmentation ideal" was common parlance lol

[delicate orchid]

coaugmentation is a funny name so I would use it

[unkempt stream]

ee major who does math as a hobby

Going into freshman year

never heard of a hopf algebra

[south patrol]

Thabks

[delicate orchid]

call it diminution for the music theory ref

[south patrol]

Oh that's funny I happened to ask a question motivated by Hopf algebras just under someone mentioning hopf algebras

Don't see them too often in this channel

[rocky cloak]

Is it supposed to be a section of the counit?

[south patrol]

the unit is yes

[glossy crag]

EE is pretty based, but math's baseder. Welcome to the light.

[unkempt stream]

fourier transforms go brrr

[glossy crag]

So the Brauer group of K describes finite-dimensional central division K-algebras. Do we have anything analogous for infinite-dimensional algebras?

[chilly radish]

what would even be the inverse in that case?

[dim bane]

please no, i was a music major and i already have suffered at musical isomorphisms

[pallid matrix]

(ii) states that $G_\mu$ is not compact. How does the fact that the kernel is a closed, normal subgroup of $G_\mu$ plus the fact that the quotient group is finite imply that $H$ is infinite? Also, how might I justify that the quotient group is finite?

[cloud walrus]

mdc

[pallid matrix]

I'm thinking the non-compactness is only used to extract the fact that G_\mu is not finite. But not sure this is correct

[topaz solar]

Well G mu is hard to be finite if it’s not compact

Any open cover would be finite there I think lol

[crystal turtle]

Could one have that G_mu is infinite, G_mu/H is finite, and H is finite?

[topaz solar]

If you want to justify why G_mu/H is finite too, it looks like a subgroup of permutation on r elements?

What text is this btw @pallid matrix

[pallid matrix]

" Products of Random Matrices with Applications to Schrödinger Operators" by Philippe Bougerol, Jean Lacroix

[pallid matrix]

Yes, that makes sense thanks

[topaz solar]

Interesting

[pallid matrix]

My algebra is extremely rusty so I'm having trouble parsing this

[topaz solar]

Yeah I mean it’s a little awkward looking to me with the typewriter setup lmao

And idk the action(s) etc

But if it’s not compact and finite, that’s an issue since P(G mu) is kinda finite

[pallid matrix]

G_\mu must be infinite, no?

Assuming it is not compact*

[topaz solar]

Right, and that’s the conclusion you draw from (ii)

If you’re reading what (ii) is correctly and all

I don’t feel like digging up and reading the whole text

[pallid matrix]

Yes, but how do I conclude that the H (the kernel) is infinite as well?

[topaz solar]

So, if H is finite, and G is infinite

How do you get G/H finite

Elements of G/H are equivalence classes right?

[pallid matrix]

yes

[topaz solar]

Union of them is G

And G is infinite, but you have finitely many classes

How big is each class?

Of course they’re all ||the size of H||, because of how they’re defined

This make sense?

[topaz solar]

This seems like an interesting topic, why dig up this clearly old paper?

[crystal turtle]

only from 1985, looks much older than that by the typewriting lmao

[pallid matrix]

Yes, I need to fill in some details however but I think I can continue by myself from here

[pallid matrix]

I am trying to understand Furstenberg's theorem on products of random matrices

[topaz solar]

Ah I see

[topaz solar]

I hate typewriter typesetting in math

[topaz solar]

Any particular holes I can help with?

[pallid matrix]

This is an alternative resource but it does not contain this proof https://www.mat.uc.cl/~jairo.bochi/docs/fur_revised.pdf

[topaz solar]

I see, and readable

Ok this is kinda interesting but above my pay grade on the measure shenanigans

[pallid matrix]

To argue that G/H is finite, I can use (a corollary of) the "First isomorphism theorem", which states that the G/H is isomorphic to the image of the homomorphism, which in this case is a subset of the set of all permutations of a finite set. No?

The corollary I'm referring to is this

[crystal turtle]

That looks correct

Since it's a map to a finite group with kernal H

[pallid matrix]

Thank you

[topaz solar]

You can indeed

I do not believe it is strictly necessary but ye

G = U(G/H) unioning the equiv classes, and each equiv class has cardinality |H|

And they’re disjoint

[crystal turtle]

This is for G/H to be finite

[topaz solar]

Ah right

Oop

I misread as G finite nvm me

Ye that’s exactly what I was suggesting earlier don’t mind my illiteracy

[pallid matrix]

Thanks to both of you

[rocky cloak]

One variant of Lagrange's theorem says that

|H| * |G/H| = |G|

Works for any cardinality, also infinite

[coral shale]

G/H. urgh

but ig thats the usual notation

[delicate orchid]

what

[coral spindle]

what

[coral shale]

well G/H is a set

[delicate orchid]

who gives one?

of course it is

[coral spindle]

ofc it's a set

everything's a set 😏

[coral shale]

ur not.

[coral spindle]

You don't know me

[delicate orchid]

would you rather jagr state "the size of a complete left transversal of H in G" rather than "|G/H|"

[coral spindle]

G/H is a G-set. It's very standard to write it as G/H

[coral shale]

idk what a G-set is

[delicate orchid]

basis of the burnside ring mmmmm yummyyy

[coral spindle]

Set with a G-action on it

U know group actions I'm sure

[coral shale]

probably

[coral spindle]

D:

[delicate orchid]

you don't get to have an opinion on notation then

[coral spindle]

As a side-note, I once mispronounced 'set' as 'sex' in a class while asking a question. It was the most embarassing thing in my life, I think.

[delicate orchid]

I would have continued pronouncing it as "sex"

[coral spindle]

Chad

[coral shale]

but what was the question

[delicate orchid]

"how is babey made"

[coral spindle]

Something like "can you explain the intuition behind these sets"

[coral shale]

so when one element and another element love each other very much...

[open sluice]

they create a product

[solemn dew]

guys is this a contradiction?

[delicate orchid]

ermm... what the deuce??

[coral shale]

suppose it isnt true. prove directly it is true. #. sotrue

[delicate orchid]

well it's obviously true if it's a division ring so assume that it isn't

now if ab \neq 0 for some a,b in R then (a) is a strictly larger ideal than (0) - could still be the whole got darn thang tho

but it's only the whole ring if there is NO non-zero b such that ab = 0 very curious

[solemn dew]

R=U says that 1 must be in U. So you know there must be a inverse element for say, b. But let us assume that ab = 0, then abb^-1=(ab)b^-1=0 = a(bb^-1)=a, hence contradiction, so it must be a division ring

[delicate orchid]

yeah zero-divisors cannot be units

[solemn dew]

How can it be a division ring or a ring with prime numbers if ab=0 is leads to a contradiction

[delicate orchid]

wait are your rings unital
must be since you're using inverses

[solemn dew]

i have not heard of a unital ring so far 😅

[delicate orchid]

cause if they are then a = 1a = 0 for all a in R lol

[solemn dew]

unital ring is a ring R s.t. R = {0}?

[delicate orchid]

no

[solemn dew]

Then the first argument holds, that U = {0}

oh okay

[delicate orchid]

it's a ring with a 1

although we saw yesterday that the quaternions mod p are a non-divison ring with no non-trivial ideals

so they're somehow in the latter category?

[solemn dew]

idk haha

[delicate orchid]

I just don't believe this result

[coral spindle]

I believe I mentioned this yesterday, but a unital ring is a ring that has a 1

We occasionally want to think about rings which do not contain unity (i.e., 1) but nowadays people are fond of calling them rngs or rungs.

[delicate orchid]

rungs

[coral spindle]

that's how they say rng anyway

[solemn dew]

hahah

[coral spindle]

might as well write it right, right

that's my right

[delicate orchid]

rite

[delicate orchid]

have you get any thoughts on an actual proof of this mofo though

[coral spindle]

Oh the thing posted

let me lookee

So yeah clearly talkin about rungs

[delicate orchid]

yeah, I thought so

in which case I do not care because they are fake

[coral spindle]

facts and also true

[delicate orchid]

I can believe it though, as I said before:

[coral spindle]

So ok it has to have a prime number of elements bc we'd have nontrivial additive subgroups

why the multiplication though hmmm

[delicate orchid]

if ab \neq 0 for some b in R then (a) is a strictly larger ideal than (0) - I'm not sure how you determine when the ideal is the whole ring as units don't exist? Left/right-cancellability I guess

[coral spindle]

OK let's think about that

[delicate orchid]

good spot

[coral spindle]

Actually, maybe there's a definition of 'division rung' that we're unaware of

where every nonzero element divides every other

[delicate orchid]

it quite clearly says "division ring" boss

[coral spindle]

Even so maybe we can argue why there's an identity, but I kinda doubt it

In the first place the zero ring kinda fits this

[delicate orchid]

also the mod p quaternions are a ring that do not satisfy the latter property lol - they're order p^4

[coral spindle]

uh-oh spaghetti-ohs

[delicate orchid]

and not a division ring

same goes for like idk M_n(F_p) n > 1

[solemn dew]

uhh

so the authors are wrong in this one?

[coral spindle]

Then there is either context not seen here or this is just wrong

great counterexample with the matrices btw

Or is that just two-sided ideals

I can't remember how the ideal structure works out

[delicate orchid]

yeah it's just two sided whoops

[coral spindle]

oh wait lmao

ok

[delicate orchid]

the one sided ones are in a weird bijection with column spaces or something I can't remember

[coral spindle]

Right ofc

Perhaps R is assumed to be finite?

[delicate orchid]

is a ring with a prime number of elements

does this not imply finiteness?

[coral spindle]

:(

[delicate orchid]

I'm convinced it's because of this though, like this is what they're getting at

[delicate orchid]

simmer down in the back, numerics boy

[elder wave]

Fuck offfff

[coral spindle]

yeah but I'm wondering if we should try and argue that a unit exists through finiteness or sth

I don't see how we can assume that there isn't a unit and conclude finiteness, that seems weird

[delicate orchid]

oh wait we're being stupid

[coral spindle]

I know

[delicate orchid]

if (a) is a strictly larger ideal than (0) it is a proper subgroup of the additive group

but there aren't any

[coral spindle]

Yeah?

[delicate orchid]

so it's the whole ring

[coral spindle]

Well yeah but how's that help

We can't guarantee we have a unit from that, can we?

[delicate orchid]

who cares about a unit

we just had to show the ring is simple

[coral spindle]

D:

but we need to show it's a division ring otherwise, right?

[delicate orchid]

I did do the implication backwards so there could be a more mysterious thingymajig lurking in da void

[delicate orchid]

yeah ok, so this is the part I no longer believe

there's just too many questions - does it mean two sided ideals only?

[coral spindle]

OK. If the ring has a unit then it's clear to see that it's a division ring from right simplicity – this is standard unital ring theory junk. If it doesn't, then let's say that for some a in R we have that the right ideal aR is nonempty, so it contains a. So there is some b in R such that ab = a. Does that get us somewhere?

Obviously it looks a lot like a unit.

[smoky ivy]

is x^6-2 irreducible over F7

[elder wave]

Try

[coral spindle]

Hint: every nonzero element of F7 is a member of a group of order 6

[smoky ivy]

and then?

[coral spindle]

As for non-linear factors... you're fucked

[smoky ivy]

wait, so it is irreducible?

[coral spindle]

Dunno man I just gave you a way to approach it

[smoky ivy]

oh

hm, idrk what to do with that hint tbh

[summer path]

You only have like 6 things you can do here...

[smoky ivy]

those are?

[prime sundial]

good advice

[prime sundial]

what is x^n for x in a group of order n

[smoky ivy]

0

so x^6 - 2 = -2 for any nonzero element in F7

but how does that help

[delicate orchid]

x = 1 is a counter example to that claim

[smoky ivy]

oh

[delicate orchid]

what we're trying to envoke is fermat's little theorem but that's alright

[coral spindle]

You're mixing up the additive group and multiplicative group

[delicate orchid]

again, that just takes care of the linear factors which is the easy part

what's that result again? Hensel's lifting lemma or something?

won't help us as we can't get all the way to Z

annoying

[smoky ivy]

hm, i’m confused

[delicate orchid]

lol

we should have just done difference of two squares hahhahaha

[summer path]

Wow you just spoiled the fun

[delicate orchid]

I got an answer which was the goal

this isn't and never will be "fun"

[delicate orchid]

so we got that 3^2 = 2 right
so x^6-2 = (x^3+3)(x^3-3) lol

[summer path]

Well it's easier to see if we just did a -7

[smoky ivy]

oh wow hahah

[crystal turtle]

Makes sense since each coset has the same cardinality, wasn't sure if that generalized to infinite groups

[unkempt stream]

Axiomatic sex theory

[unkempt stream]

OH NO NOT AGAIN

[teal vessel]

the axiom of depravity

[crystal turtle]

Axiom of rizz

[unkempt stream]

i would work off of the existence of an element NOT in any ideal YET is not a unit

[ember field]

Guys is hom(z/4,q/z) under addition, isomorphic to z/4

[solar vessel]

ye

generated by the map taking 1 to 1/4

[rocky cloak]

Here's a solution:

I assume R is meant to be a rng

||So if ab = 0 for every a and b, then an ideal is just a subgroup. And the only groups without proper nontrivial subgroups are those of prime order.

Now assume there are some elements where the product is nonzero. Consider an arbitrary b, if ab = 0, then (a)b = 0, so if a is nonzero (a)=R, so Rb = 0. But then (b) = Zb and so the product of any two elements is 0, contradiction.

So for a and b nonzero ab is never 0, thus we have cancellation. Now consider the product of 3 nonzero elements abc. Since this is nonzero we have (abc) = R. In particular there exists r in R and n in Z such that (r+n)abc = c = (ra+na)bc.

Define 1 = (ra + na)b.

Note d1c = dc implies d*1 = d.

Assume 1*d does not equal d. Then 1*d - d is nonzero, so a(1*d - d) = (a*1 - a)d is nonzero. Cancelling d we get a*1 - a, which is 0, contradiction. So 1*d = d, and 1 is unity.

From there, using cancellation and 1 it's not hard to show that R is a division ring.||

[rocky cloak]

I swapped right ideal for left ideal. So I guess just reverse all multiplications

[static yew]

What is the use of cayleys theorem?

It didn't make sense to me at all until I realized that it said "a group of one-to-one" which just means that a group exists, so there are at least as many one-to-one functions as there are group elements, which seems kind of obvious

[formal ermine]

it's used in the construction of a galois extension for an arbitrary group

[static yew]

Wait what

Not a field, not even a ring, but a group?

Could you please direct me to an example?

[delicate orchid]

you can also get funny permutation representations!!!

[static yew]

Yeah the book is talking about permutations now

The things its saying don't make sense to me, which suggests that I misunderstood something

[delicate orchid]

it basically means you can think of any group as just permutations

[static yew]

Q(x) is all of the expressions of the form a/b where a,b in Q[x], b!=0, gcd(a,b)=1 right?

[delicate orchid]

don't see why you need the gcd bit

oh right you're implicitly dividing them

yeah ok

[coral spindle]

Deeply cursed, gg for working this out

[cloud walrus]

Dealersgrip

[delicate orchid]

holy smokes

[static yew]

Wait what's a plain old morphism

[delicate orchid]

the f(H) is correct at least

[delicate orchid]

for your purposes, a homomorphism

[winter thunder]

ok nice, that's at least 1/2

[static yew]

Imma troll some people by inventing the heteromorphism

[winter thunder]

Define it

[delicate orchid]

https://ncatlab.org/nlab/show/heteromorphism

[coral spindle]

classic

[winter thunder]

no fucking way

😂

[coral spindle]

Interesting, it's the funny adjunctions

[delicate orchid]

seems to be adjunction but for special snowflakes

[static yew]

Damn. I thought I was onto something

[coral spindle]

makes sense

ig

[delicate orchid]

no it doesn't just write out Hom(F(a), b) rather than Het(a, b)

[coral spindle]

As in the name makes sense

[delicate orchid]

I guess

[coral spindle]

I don't especially think it's terribly helpful but we

[winter thunder]

Is my proof for K correct?

[static yew]

"are you heteromorphic or himomorphic"

"Biomorphic, actually. 3.14159 on the Kinsey-Galois scale"

Are we saying that A<B means A is a strict subgroup of B?

Like subset but keeps operator and closure property?

[winter thunder]

yeah

[delicate orchid]

yeah it's right

[winter thunder]

perfect thanks a lot

[unkempt stream]

???

Oh

Gal(F(X1…X_N)/F) is iso to S_N

[static yew]

Yeah its talking about that

[night onyx]

You shouldn't write x = f^{-1}(x') or y = f^{-1}(y'), rather f(x) = x' and f(y) = y'. Seems like a nitpick but writing x = f^{-1}(x') implies f has an inverse f^{-1}. In general f^{-1}(x') = {x | f(x) = x'} so writing x = f^{-1}(x') doesn't quite make sense. Other than that your argument is fine, if f(x) = x' and f(y) = y' then f(xy) = f(x)f(y) \in K, so xy \in f^{-1}(K)

[winter thunder]

yes thank you very much, it's still an important detail

[unkempt stream]

Galois theory eludes me

[static yew]

Wait what are elementary symmetric functions

[night onyx]

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

[unkempt stream]

the polynomials invariant under permutation of their roots

*vars

[static yew]

Q(x,y,z) is basically all the polynomial fraction type deals where top and bottom are of the form a + bx + cy + dz

[night onyx]

elementary ones are specific ones which generate all symmetric polynomials

[static yew]

Wouldn't that be the constants then? Not sure I follow

No wait

These are transcendental

So it's all polynomials of all exponents and coefficients

This is a biiiig field

[formal ermine]

x + y

xy

x + y + z

xy + xz + yz

[night onyx]

xy isn't, xy + yx is though

[formal ermine]

f(y, x) = yx = xy = f(x, y)

[static yew]

When you say invariant

What does that mean exactly? That the counts of each exponent are the same?

E.g.

axy^2z^3
ax^3y^2z
ax^2y^3z

I don't have a good picture of this in my head

[formal ermine]

for all $\sigma \in S_n$ we have $f(x_1, \ldots, x_n) = f(x_{\sigma(1)}, \ldots, x_{\sigma(n)})$

[formal ermine]

however when we say elementary symmetric we mean a specific kind of symmetric polynomials

[crystal turtle]

or in words, if you swap the variables (i.e. taking f(y,x) instead of f(x,y)), the output is the same

[static yew]

Which means both indeterminates must have the same exponent in each of the components of the expression

[formal ermine]

and in fact those elementary symmetric polynomials generate every symmetric polynomial. in other words: if $e_1, \ldots, e_n$ are the elementary symmetric polynomials then $$\bC[x_1, \ldots, x_n]^{S_n} = \bC[e_1, \ldots, e_n]$$

[crystal turtle]

not necessarily in each component, since xy^2 + yx^2 is symmetric

[static yew]

Oh true

This seems difficult to make use of

[formal ermine]

symmetric polynomials are used in doing discriminant/resultant stuff and also algebraic geometry iirc

[static yew]

This book seems to be focusing on something related to geometry

A cyclic group is any one that has a generator, right?

[coral spindle]

Generated by one element, yes

So Z/nZ for some n

[static yew]

Is the additive group of a ring always cyclic? Trying to think of a way where the multiplicative identity

Wait

2x2 matrices are a ring, right? Even if the matrix elements are of a finite field, you cant generate the group from the 2x2 identity matrix

[coral spindle]

No

[static yew]

No to which

[coral spindle]

As in, your example is right

The additive group of a ring is almost never cyclic

In fact the only such rings are (gasp) Z/nZ

where n is possibly 0

[static yew]

I'm used to dealing with stuff isomorphic to Z/nZ

[static yew]

To be fair, there are a a lot of those. Aleph-null of them, even.

[coral spindle]

Yeah but there are a lot more rings that ain't

And I can't even put a size to that

bc it's so much bigger, I reckon

not that I have a proof

[static yew]

Aleph-one

[coral spindle]

Way bigger

like, not set-sized, I would bet

[static yew]

Aleph-three

[south patrol]

Hm can you never put a different multiplication on Z/nZ

I've never thought about it lol

Oh yeah sure it is determined by 1.1 = 1 lol

fair

[static yew]

Do you mean a different multiplicative identity?

[south patrol]

no

Though that could be part of it ig

[static yew]

Oh a different mapping of a×b ?

[coral spindle]

It's not immediate that 1 is the unit, but you can think about the characteristic of R whose additive group is Z/nZ. It's clear that char R = n, so it must contain a copy of Z/nZ as a ring

So even if we choose a different unit, it doesn't matter

I explained that sloppily but I hope it makes sense

[static yew]

I think I got it

Even if you permuted 0..n-1

If the resulting set is still a ring, then that ring is isomorphic to z/nZ

[coral spindle]

Yeah

that's the idea

And to be clear, you can absolutely permute it

[static yew]

Wait what about phi(x)=-x

[coral spindle]

what about it

[static yew]

I'd like to see them stop me

Negation doesn't preserve multiplicative identity

Is "minus Z" still a ring?

[coral spindle]

I don't know what that means

If you mean that we can permute it in that way and still have an isomorphic ring, yes

[static yew]

Replace each element x in Z with -x

[coral spindle]

So that's just Z

Be more specific

[static yew]

But it's not anautomorphism

Cuz -(xy) != (-x)(-y)

[coral spindle]

Yeah, and?

I'm not seeing your point

[static yew]

Not all transforms that produce isomorphisms are automorphisms

[coral spindle]

I don't know what that means

That's not an isomorphism

Automorphisms are, by definition, isomorphisms

[static yew]

Which one preserves f(a)f(b) = f(ab) then

[coral spindle]

Both

Let me make the permutation point better.

[static yew]

But f(15) = -15

f(3)f(5) = (-3)(-5) = 15

[coral spindle]

Again, so what

I'll use your idea, maybe this is the way you're trying to express it

We define a new ring

We'll call it R,

its underlying additive group will be Z.

[static yew]

Okay

[coral spindle]

We will define the multiplication a * b = phi(a)b

Let's double-check this forms a ring

It's pretty clear that -1 is going to be the multiplicative identity. You can see the multiplication is going to be distributive since (a+b)*c = phi(a+b)c = phi(a)c + phi(b)c

ah just realised I used the same symbol. I'll fix that

Hopefully you can see how the rest will work

This ring, R, is isomorphic to Z.

[static yew]

I thought at least one morphism had to have phi(ab) = phi(a)phi(b)

Maybe homomorphisms?

[coral spindle]

Yeah, phi : Z → R is now an isomorphism of rings

[coral spindle]

You're getting confused. Homomorphisms, isomorphisms, automorphisms of rings all have this property

But again, phi : Z → Z is not a ring homomorphism.

[static yew]

But it's not a transform on Z

It's a new ring with the same set and different operators

[coral spindle]

It is a function on Z.

Oh you're not talking about 'it' being phi

Yes, this is a new ring

This is the whole point I've been making

[static yew]

A function on Z

The book I'm reading mostly focuses on transforms that map one group/ring/field to another

[coral spindle]

So you're not using the word transform in the way I thought you were aaaaaaa

Please use the usual terminology such as 'function'

[static yew]

Sorry this book called them transformations

[coral spindle]

OK

[static yew]

Or "mappings"

[coral spindle]

Anyway this was the point

[static yew]

Would "mapping" work?

[coral spindle]

mapping is standard.

phi is not a ring automorphism Z → Z

but it is a group automorphism Z → Z

And using that automorphism, we produced a new ring structure on Z

[static yew]

For the additive group

[coral spindle]

When people refer to the group Z, they mean the additive group without fail.

[static yew]

Ahhh ok, thanks.

[coral spindle]

Just clarification on the terminology:
A homomorphism is a structure-preserving function.
An isomorphism is a homomorphism with an inverse homomorphism.
An automorphism is an isomorphism from something to itself.

[static yew]

Right, like "mod n" is a homomorphism on Z, right?

[coral spindle]

No

Mod n is a relation on Z

[static yew]

Oh damn

Okay lemme rephrasr

[coral spindle]

You can think of the map n |-> n + mZ as being 'mod m'

but I don't like this!

This is misleading!

But indeed, that is a homomorphism Z → Z/mZ

fixed typo

[static yew]

The mapping phi(x) -> the r from x = qn+r (euclidean thingy) is a homomorphism

[coral spindle]

Sure

Remember when we talked about choosing representatives?

You've composed the map x |-> x + nZ with the map that chooses representatives there.

[static yew]

Yeah. That r is my representative

[coral spindle]

It's way more natural and easy to think of mod as a relationship between integers rather than a function that chooses remainders

[static yew]

Oh I see what you mean

[coral spindle]

it makes things so much easier

Like

It is very easy to show that if a = b (mod n) and c = d (mod n) then ac = bd (mod n) using the idea of the relationship

proving that in terms of the remainder function is a pain

[static yew]

In software, the operator that computes that 'r' is called the modulus operator

[coral spindle]

I know

It is truly awful. That terminology is poor

So many undergrads think that's what modular arithmetic is...

[static yew]

Just wait till you see that it gives different answers for negative inputs

[coral spindle]

It's disgustingly bad

It's funny because I've seen languages use quot_rem for the function that produces the quotient and remainder

and that's great terminology, 10/10

[static yew]

Okay I gotta run

Thx for the knowledge

[night onyx]

In theory if f : R -> X is a bijection (just of sets), R a ring and X just a set, you can put a ring structure on X by defining xy = f(f^{-1}(x) * f^{-1}(y)) and x + y = f(f^{-1}(x) + f^{-1}(y)), and in this case f is a ring isomorphism. Lol it's not super exciting but yeah you can put plenty of different ring structures on the same set

[south patrol]

I mean of course but I mean in such a way that Z/nZ is the additive group

[night onyx]

ahhh like same additive group, different multiplication

[coral spindle]

Well yeah I just gave an example for n=0

but in general

[south patrol]

But ye

And qutomorphisms of Z/nZ

And stuff

[coral spindle]

you can obviously do it in the same way: define the multiplication as a*b = -ab for a single example

Now this doesn't work for n=2, but that's for a good reason :)

[south patrol]

Ye lol

Classifying finite rings is weird

Well just never thought about it or seen any applications

[coral spindle]

kinda sick tho

[delicate orchid]

Z/nZ^op

[solemn dew]

Ur clever! Nice proof

thx!

[solemn dew]

does this mean that there are inverses in a commutative ring now??

[topaz solar]

there are for some elements

consider -1 in Z

but not all are, because 2 has no inverse in Z

[solemn dew]

alright so unit elements are these elements as you described

yes i understand

very cool thnks

so a unit in Z can be -1

[coral shale]

The set *cough* of units is denoted R^x. What is this for Z

[solemn dew]

but how does this explain

gcd(a,b)=-1??

[coral shale]

gcd function is defined to output non-negative numbers

like how it is defined.

[solemn dew]

right but this is a contradiction to that definition then

[coral shale]

That doesn't break the definition you've been given.

[topaz solar]

gcd in a euclidean ring shuri

[coral shale]

No.

[coral shale]

uhhh i forgot that shit

[coral shale]

gcd function in Z at least

[topaz solar]

relatively prime <-> gcd is a unit

not every unit will show up

not necessarily anyway

[solemn dew]

oh okay i get it

yes sir

sorry if i am annoying

been studying for at least 8 hrs

[coral shale]

well its best if you consider examples and non-examples for given defns

[solemn dew]

alright

[solemn dew]

product of units are also units right?

[coral shale]

prove it.

[topaz solar]

Consider the obvious choice of inverse Holmes

[solemn dew]

hint?

[coral shale]

In fact, prove $R^\times$, the set of units form a particular structure, have a guess

[coral shale]

nah. not til uve had a good go. falls out from the defn of unit

[delicate orchid]

the obvious one

[solemn dew]

oh got it

[coral shale]

casually written but ite.

[delicate orchid]

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

[solemn dew]

but b^-1 doesnt have to exist

or am i wrong

[delicate orchid]

you asked if the product of two units was a unit...

[solemn dew]

oh so a unit has a inverse then?

[coral shale]

$r\in R$ is a unit iff $(\exists s\in R)(rs = sr = 1)$

[night onyx]

if a and b are units you know ax = by = 1 for some x,y. To prove (ab)z = 1 for some z, you should use commutativity, along with x and y

[delicate orchid]

no offense but what the fuck did you think a unit was

[solemn dew]

a is a unit if ab = 1 for some b

[coral shale]

what is an (multiplicative) inverse then...

[solemn dew]

hah...

[delicate orchid]

ok so now it should be obvious that the set of all units is a group

[wraith cargo]

I kinda feel like u guys are being sorta mean to jonathan rn

[delicate orchid]

I am

I'm mean and grouchy

[coral shale]

spoilers !

[delicate orchid]

grumpy even

[solemn dew]

it is ok

i am used to it

🥺

(bad joke9

[coral shale]

hum. it usually has even order if finite right.

must usually be Z_2 x ??? ...

[delicate orchid]

Meh it is for Z/2nZ so yeah sure

And for a vague enough definition of “usually”

[coral shale]

my intuition on rings is so weak even tho its supposed to be more 'familiar' than groups...

[delicate orchid]

This. So much this.

[night onyx]

You should prove a few things about units along the way:

1. Prove inverses are unique (this justifies writing a^{-1} for the inverse)
2. Prove (ab)^{-1} = b^{-1}a^{-1}
3. Prove 1^{-1} = 1
4. Conclude that the set of units is a group

[delicate orchid]

Probably because the only intuitive examples for rings include 3 fields, Z, the basic finite fields, and then disgusting non-integral domains and not really much in between

Ok maybe k[x] and Z[x]

[solemn dew]

thanks brother

i'll try to prove 4

[wraith cargo]

Idk to me rings were easies to grasp than groups because groups use awful combinatoric arguments that I could never fucking come up with

[coral shale]

gyuh.

I think cayley makes groups so much easier to understand

This classifies all of them.

[wraith cargo]

I hate permutations with my life so you are WRONG

[coral shale]

[wraith cargo]

doing stuff in S_n is actually the most unfun thing ever for me

I hate it so much lol

[coral shale]

idk what stuff ur doing...

im guessing u hate group actions...

[wraith cargo]

no

[delicate orchid]

Combinatorial group theory is THE fun

[coral shale]

i cant even.

[wraith cargo]

I hate S_n

[coral shale]

but permutations are lit actions

[delicate orchid]

What do you think a group action is

[wraith cargo]

yeah but you don't have to think of them that way

I pretend they aren't

[coral shale]

sully

[wraith cargo]

ala my life is better because of it

[delicate orchid]

sully

[wraith cargo]

our algebra class had like 4 fucking classes where we talked abt nothing but S_n it was hell

it was like

[coral shale]

is this trauma...

[wraith cargo]

Yes

It was like

[delicate orchid]

S_n… le bad!!
The regular rep:

[wraith cargo]

"In the C part of the class we'll use these results to prove that the quintinc in unsolvable with radicals"

[delicate orchid]

Yeah

Galois theory is bad

[wraith cargo]

Man the solvability and nilpotency of groups

it changed me

[delicate orchid]

Nilpotency is easy

[coral shale]

i really barely give a sht about that ngl

[wraith cargo]

Upper/Lower central series are dreadful

[coral shale]

slept in the last part of galois class

[delicate orchid]

Is it a direct product of its sylow subgroups y/n

So easy

[topaz solar]

There’s also function rings by extension of those understandable ones

[wraith cargo]

idf care that class destroyed group theory for me

[delicate orchid]

Awwww poor baby

[wraith cargo]

I wanted Sylow to choke

[delicate orchid]

I nearly failed first year group theory

[wraith cargo]

does wew need cuddles to feel better 🥺

[coral shale]

i didnt gaf about sylow...

[summer path]

i hate sylow proofs

[coral shale]

i dont feel like i missed anything

[wraith cargo]

nah sylow is actually cool I was fucking with you lmao

get fucked
OK I wasn't but it would be funny if I was wouldn't it

[delicate orchid]

Not from YOU… if you could put me in contact with your mother however…

Sylow is not cool until you extend the theorems to general fusion systems

[wraith cargo]

what if I inform you I am someones mother?

[coral shale]

u werent having any of us on

[wraith cargo]

the anime pfp doesn't understand high brow humor (ala seinfeld)

[coral shale]

the anime pfp invalidates you as a person.

and your opinion on permutations

[wraith cargo]

that's okay

[topaz solar]

Hater of automorphisms

[unkempt stream]

muncher of ice

[molten viper]

So, if a ring has a unique maximal ideal, m, can I say that x \notin m implies x is a unit? It seems so, because every non unit is contained in a maximal ideal, in this case, m.

[dim widget]

Yes

[molten viper]

I'm sure I will use this fact to great affect

[dim widget]

perhaps also to great effect!

[delicate orchid]

don't be absurd.........

[summer path]

wow tteg is back

[unkempt stream]

localization of prime go brrr

[white oxide]

yo how important are hom and duality in the study of tensor products

my school's aboutta start and i'm tryna study the more important stuff

[delicate orchid]

they're pretty handy imo

[rocky cloak]

The fact that $Hom(X\otimes Y, Z) = Hom(X, Hom(Y, Z))$ comes up a lot

[cloud walrus]

jagr2808

[white oxide]

are they like necessary to study tensor products or can i just skip that section

oh well i guess hungerford gave a dependence chart

[rocky cloak]

Depends what you're doing with tensor products I suppose

[white oxide]

``On the other hand, it is also convenient to think of $A \otimes_R B$ as a sort of dual notion to $\text{Hom}_R(A, B)$."

[rocky cloak]

,tex ``quote''

[cloud walrus]

jagr2808

[south patrol]

They are basically the definition

So yes

[cloud walrus]

okeyokay

[white oxide]

oh ok i'll check out that section then

this shit so abstract, truly an #groups-rings-fields moment 😭

[south patrol]

So I mean

Basically the intuition for tensor products is that there's a natural notion of multilinear maps out of a product

Like multilinear A x B -> C

[rocky cloak]

I guess the question is, why are you so eager to learn about tensor products, but not about the hom functor?

The hom functor is the cool one

[south patrol]

And it'd be nice if we could just view those as maps out of another module

[white oxide]

idk i always hear things about tensor products and i want to learn what i hear about the most often

[south patrol]

A (x) B is then the one doing that

[white oxide]

i have no clue what a multilinear map is

[molten viper]

what exactly is the hom functor 👀

[white oxide]

plus i have like two more weeks until school starts and then i have to concentrate on my other classes

[topaz solar]

Hom(A, B)

As in, space of maps A->B lol

[delicate orchid]

ermmm actually it's Hom(A, -) sweaty

[wraith cargo]

It's a bifunctor bucko

[topaz solar]

it’s a bifunctor

Dang it I was sniped

[delicate orchid]

contravariant functors are not functors

[molten viper]

yeah Idk what this means