groups-rings-fields

[cloud walrus]

okeyokay

[agile burrow]

The bilinearity relations might give you different relations in the tensor product over R than if you just consider them as abelian groups. For instance, Z/2 (x) Z/4 = Z/2 as Z-modules, but viewed as Z/2-modules, the tensor product is Z/4

Explicitly, as abelian groups we have the relation (1, 3) ~ (3, 1) = (1, 1) whereas when we view them as Z/2-modules we can't write (1, 3) ~ (3, 1) since we don't have as many scalars in Z/2

My example might not be correct bc I'm sleepy but it seems like you've got the right idea anyway

[rocky cloak]

I think you'll struggle to make Z/4 a Z/2 module

[crystal turtle]

Is Z/4 a Z/2 module?

Okay

I was doubting myself for a sec

[rocky cloak]

But for example if R = Z[x], then R tensor R over R is just R, but R tensor R over Z is Z[x, y]

[summer path]

Wow new wew pfp

[delicate orchid]

yeah I lost no nostalgia novemeber

[white oxide]

i mean

[elder wave]

R tensor R over R is just R, but R tensor R

[rocky cloak]

And if Z -> R is an epimorphism of rings, then the tensor products should coincide

[white oxide]

could i take $(a6, b) - (a, 6b) = 0 \in A \otimes_{\mathbb{Z}} B$ and just take any ring $R$ not equal to $\mathbb{Z}$ and say that $(a6, b) - (a, 6b) \neq 0 \in A \otimes_R B$ or na

[cloud walrus]

okeyokay

[white oxide]

because while every R-module is a Z-module we don't have $R \subseteq \mathbb{Z}$ for every ring $R$

[cloud walrus]

okeyokay

[sturdy mirage]

Question, is there always a homomorphism from N, a free A-module, to N' , a submodule of N, such that on N' the homomorphism acts as identity? (here A is commutative and has 1, if it matters)

[white oxide]

because i'm a bit confused. if we fix R-modules A and B, don't we write A tensor_R B to signify that we're quotienting by the (ar, b) - (a, rb) relation?

[crystal turtle]

Consider A=Z, N=Z, and N'=2Z

[agile burrow]

Oops yes, my example does not make sense

[crystal turtle]

For A a field, I think this should be true?

[white oxide]

like i'm not sure understand. if A is a right R-module and B a left R-module, don't we by definition also quotient by (ar, b) - (a, rb) in the tensor product A tensor_Z B?

[white oxide]

just considering the first paragraph of this definition

like what the modules are over determines the relations we quotient by

so if they're both over R then for both of them we're gonna quotient by all elements of the form (ar, b) - (a, rb)

hence they have to be equal

[delicate orchid]

for A semisimple it should be true

[sturdy mirage]

Have I misunderstood something then, or is this exercise plain wrong? Use A=N=Z, and N'=2Z like you said.

[delicate orchid]

take the projection map from N to N' setting the appropriate simple modules to 0

[crystal turtle]

Yeah well I don't know what that word means so it doesn't exist

[rocky cloak]

No, only if r is in Z. The R in tensor_R is indicating which ring we pick the rs in (ar, b) = (a, rb) from

[white oxide]

i see, i guess that makes sense

[white oxide]

so this would work then

[crystal turtle]

Do they give any limitations on A? Otherwise yeah I question that exercise since I'm pretty sure my example worka

[sturdy mirage]

no, only that A is commutative and with identity

[delicate orchid]

can you not form a SES 0 -> N' -> N and then use universal property of N to get a map going backwards?

hmm

[chilly ocean]

Let $M$ and $N$ be normal subgroups of a group $G$ and let $N \le M$. Then $M/N \lhd G/N$ and $(G/N)/(M/N) \cong G/M$.

[cloud walrus]

Gev

[chilly ocean]

sorry for being late

[delicate orchid]

ah no, I'm assuming submodule of free is free I think

[white oxide]

why is it that the typical element of $A \otimes_R B$ is not of the form $a \otimes b$ with $(a, b) \in F$? sure i understand, the typical element is of the form $\sum_{i = 1}^r n_i(a_i, b_i)$ but then don't we get that this element is in $F$ by closure?

[cloud walrus]

okeyokay

[coral spindle]

A consequence is that we understand the quotients of quotients there ya go

[delicate orchid]

you literally just said "sure I understand" and then stated the answer to the question you're asking

[white oxide]

LMAO

nah i'm wondering why $\sum_{i = 1}^r n_i(a_i, b_i) \neq a \otimes b$

[cloud walrus]

okeyokay

[white oxide]

or is not of the form $a \otimes b$

[cloud walrus]

okeyokay

[delicate orchid]

get 2 (x)_Z 1 + 1 (x)_Z 3 into an elementary tensor and i'll pay pal you 10 quid

[chilly ocean]

for example the second Isomorphism theorem is used to prove that any subgroup of a metabelian group is metabelian, I'm looking for similar use cases of the third one for proving other facts

[white oxide]

what's the conversion rate to dollars again

[delicate orchid]

lemme just make it explicit

[karmic moat]

,w 10 quid to usd

[cloud walrus]

Results provided by WolframAlpha

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

Requested by aanzhuu

[karmic moat]

oh bug off

[delicate orchid]

thank you wolfram alpha, very cool!

[karmic moat]

now all i need to know is pence to usd

then i will know quid to usd

😎

[crystal turtle]

I saw what you deleted.

[delicate orchid]

https://tenor.com/view/hello-neighbor-like-subscribe-bell-hello-neighbour-hello-neighbor-2-hello-neighbour-2-gif-2484330488843852459

anyway

[karmic moat]

ok guys i just got 2 (x)_Z_1 + 1 (x) _z_3 into elementary tensor

[delicate orchid]

point is you can't write shit as elementary tensors all of the time

[white oxide]

no way! it's almost like i'm asking why that's true! 🤯

[delicate orchid]

what

"why isn't every vector in the space also in the basis"

[white oxide]

elegantly put. correct!

[rocky cloak]

I'm not entirely sure what you're doing in that example, but 6 is an integer. You'd want to look for a ring R where not all the elements can be expressed as some combination of integers

[delicate orchid]

alternatively think about a vector space tensored with it's dual

even more specific, R^2 (x) R^2*

[rocky cloak]

In general if S -> R is an epimorphism of rings, then tensoring over R gives the same as tensoring over S

[delicate orchid]

you can't get something like
(1, 2)
(3, 4) as a multiple of an elementary tensor

[white oxide]

hmm ok

yea maybe something like Hom_S(R, S) (x) Z_3 or somes hit would work

[lime junco]

for group theory, is there an example of a non cyclic group G such that Aut(G) is cylic?

[delicate orchid]

if Aut(G) is cyclic then so is Inn(G), meaning Inn(G) must be trivial so G is abelian. Then we use the structure theorem to conclude that G must also be cyclic

important to note that the other implication doesn't hold, there are cyclic groups with Aut(G) non-cyclic

[lime junco]

wdym the structure theorem?

[delicate orchid]

the structure theorem of finitely generated abelian groups

there might be a way to do it without but I'm not going to think that hard

[lime junco]

oh ok

just we havent covered that

and one of the practice problems is to show it is impossible or give an example

can you argue that if one element generated the automorphism group, then every automorphism is of the form g maps to g^a, so the group is cyclic?

[rocky cloak]

I guess this still leaves the case of G abelian infinitely generated

[delicate orchid]

what if Inn(G) is trivial (which it is if Aut(G) is cyclic as I said before)

[delicate orchid]

fair point

maybe it doesn't hold for infinite groups actually

focusing on finite groups for now, if G is abelian non-cyclic then it has to have some C_p x C_p in it for some prime p, meaning GL_2(F_p) is a subgroup of Aut(G), hence Aut(G) isn't abelian

that's what I was getting at using the structure theorem

[delicate orchid]

apparently it need not even be infinitely generated

[rocky cloak]

I think this is a pretty challenging problem. Are you sure there isn't more assumptions on G, like it being finite abelian or something?

[delicate orchid]

or just finite

[rocky cloak]

I guess just ||ZxZ/2||

[crystal turtle]

If G is finite and Aut(G) is cyclic then G is cyclic

[lime junco]

No, just G

[rocky cloak]

Then I guess maybe they want you to look for an example like this

[crystal turtle]

I actually figured this out a few weeks ago in response to another person's questions, it's basically what Wew noted. But yeah for infinite groups it can fail

[rocky cloak]

Damn, didn't notice that was Wew, thought there was a new person here

[crystal turtle]

(Here is the full solution, but spoilers ofc if you want to do it yourself)

[white oxide]

are they not equal because for instance, (x^2x, x^3) - (x^2, xx^3) = 0 in R tensor R over R but is not equal to zero in the tensor product over Z

[delicate orchid]

are you sure?, ||we have the (1,0) -> (-1,0) map in Z but also the map (-1, 0) -> (1, 1) right||?

[rocky cloak]

Hmm, yeah I guess the automorphism group is C2xC2.

I'm confused actually. If the group is rank 2, surely the endomorphism ring has nontrivial idempotents, so can't be Z

[delicate orchid]

what I'm picturing in my head is two copies of Z ontop of eachother two copies of Z on opposite sides of a cylinder

we can swap the copies over, or spin one of them, or both of them, around

[lime junco]

Thank you so much!

I really appreciate it

[delicate orchid]

b-b-b-b-but this is what I said

[crystal turtle]

I had it lying around glad it came in handy a second time

[crystal turtle]

I said it first :^)

[delicate orchid]

:o)

[crystal turtle]

Maybe some day I will have answered every mathcord question and can just reference back to previous answers for everything

[delicate orchid]

this does happen eventually I'm sure of it

[white oxide]

can somebody give me a hint for b pls

[rocky cloak]

Seems rank 2 doesn't mean finitely generated

[delicate orchid]

strange but good. Because it should absolutely be true for finitely generated

[delicate orchid]

can I pick A, B, R?

like you just want some dudes right

[white oxide]

i think i'm still confused on the notion of having a free abelian group generated by the product of two modules. maybe i need to review but i don't see why it would just be equal to A x B by closure

oh wel

[white oxide]

no homo but yea

[delicate orchid]

half homo

[chilly radish]

It's like that joke:
A time traveler walks into an open mic. He sits down and hears the person on the stage, "13!" Everyone erupts into laughter. He continues, "32!", again everyone laughs. He leans over to the person next to him and asks what's going on, the person replies "we've heard the same jokes so many times we decided to just give them numbers". The time traveller decides to try it out and goes up to the stage. He yells into the mic "72!" And the crowd erupts into laughter lasting minutes. After he gets off the stage and the laughter dies down, he asks the person what the joke was that it was so funny. He goes "we:ve never heard that one before!"

[delicate orchid]

so just let R be your favourite field, A = B = R^2 with basis {e_1, e_2}, then {e_i (x) e_j : i, j = 1 or 2} is a basis of A (x)_R B, so just take e_1 (x) e_2 + e_2 (x) e_1

if this was an elementary tensor it would imply a linear relation between elements of a basis which is a contradiction

[chilly radish]

That's a stupid joke tho the world building makes no sense

[karmic moat]

i dont get the joke at all

[white oxide]

oh so the free abelian group on A x B would just be A x B again

[karmic moat]

what does having to be a time traveler have anything to do with that

[delicate orchid]

no?

[white oxide]

i'm confused

[crystal turtle]

Nothing in that joke makes any sense

[white oxide]

so we're considering A = R^2 and B = R^2

[karmic moat]

someone said72 and they laughed so then they went back in time and made that joke?

[delicate orchid]

the free abelian group is all formal sums of elements of (A, B)

[white oxide]

as R-modules

[delicate orchid]

do you not have the theorem about the basis or something

[chilly radish]

No?

The joke is that no one ever said that number before

[white oxide]

yeah but then wouldn't that give you back elements of (A, B)

[delicate orchid]

formal sums

[white oxide]

i probably need to review free abelian groups lol

[karmic moat]

well ok so like someone said "72" for the first time and they laughed so then the time traveller heard that, went back in time, then said "72" and the crowd burst into tears because he was the first one to say 72

[chilly radish]

The dissonance in the payoff against the established rules of the universe is part of the joke

[delicate orchid]

we're not actually adding them in AxB, the "+"s are just a formality
you could write it as an infinite vector of coefficients if you wanted

[chilly radish]

No the time traveller is in the future

[karmic moat]

oh

[chilly radish]

I thought I said that, oops

[karmic moat]

oh ok it's starting to make sense

[delicate orchid]

chat this joke isn't really that hard to get

[karmic moat]

basically every joke was made

so now they're just stating numbers

no this joke is mind boggling

[delicate orchid]

reading comprehension issue

[chilly radish]

Ok we can keep discussing my impeccable humour in one of the discussion channels

[tardy hedge]

Does this kind of make sense?

I was talking with friend about how to interpret cosets, normal subgroups, quotient groups

[white oxide]

ok sorry i'm still a bit confused; (1, 0) (x) (0, 1) + (0, 1) (x) (1, 0) = (1, 1) (x) (1, 1) which is a elementary tensor, where am i off

[coral spindle]

A quotient subgroup is one of the factors 'on top' of the group

You can only obtain some of the factors in the composition series via quotients

[white oxide]

damn this shit's hard

[tardy hedge]

Right , “on top” does make more sense

[coral spindle]

:)

To justify the terminology a bit, if we think of subgroups as a lattice, we can see the structure of quotient groups as upper portions of the lattice

[tardy hedge]

Normal subgroups are inside tho so can u sort of think of them as being factors in some sense?

[coral spindle]

This is the subgroup correspondence theorem, at least it might be called something like that to you

[tardy hedge]

Can u get G’s overall structure by looking at all its normal subgrps?

[coral spindle]

No, normal subgroups are typically not factors.

And no, we definitely cannot.

[tardy hedge]

Okay

[coral spindle]

There are plenty of groups with no nontrivial normal subgroups

[tardy hedge]

Oh yeah

[delicate orchid]

https://math.stackexchange.com/questions/2314697/showing-a-tensor-is-not-a-simple-tensor

[white oxide]

thanks

[coral spindle]

Anyway this justifies why we might think of certain factors being 'on top' vs 'on the bottom' of the group

They're almost literally on the top or on the bottom of the lattice structure

[tardy hedge]

Wow this sounds weird

[coral spindle]

Quotients view the top, and subgroups view the bottom

[tardy hedge]

Dont know tbat stuff yet lol

[coral spindle]

OK well

[tardy hedge]

Sounds fun though

Im new to group theory

[coral spindle]

You know how subgroups can contain each other? If we draw all the subgroups with arrows from one to another if the first contains the second, we get a picture of what the subgroup structure of the group looks like

We'll put bigger subgroups near the top, and smaller subgroups near the bottom.

[tardy hedge]

Oh yeah i have seen that

[coral spindle]

If you choose a subgroup, then the subgroups of that subgroup just looks like the same picture, but where we forget about all the points that don't have an arrow going into the subgroup we chose.

Similarly if you take a quotient by a normal subgroup, the picture we get there looks like everything above the normal subgroup we quotient by.

This would be easier with drawings but hey ho.

[delicate orchid]

yeah this is the corrispondence theorem

I'll do D_8

[tardy hedge]

Lol ur pic

[delicate orchid]

Q_8

[coral spindle]

Wew's grand artistry 🙏

[delicate orchid]

Q_8

<-1> lol

[tardy hedge]

Is that quaternions or someth

[delicate orchid]

I tried to show how you can see the subgroup lattice for C_2 x C_2 in here as well

I've never thought of the correspondence theorem as taking upper/lower closures in the subgroup lattice

ah closures isn't the right word

I think I want to say ideals/filters

[crystal turtle]

filter specifically

since it's upward closed, nonempty, and closed under intersections (since if A and B contain N, then so does AnB)

[delicate orchid]

just chop everything underneath off nerds

[crystal turtle]

or uhh i guess it's kinda trivially a filter. It's the upward closure of {N}

[delicate orchid]

oh so I did want to say closures

I should stop doubting myself

[white oxide]

to expand $v_1e_1 + v_2e_2 \otimes w_1e_1 + w_2e_2$, would you just "multiply them" together to get something like $(v_1e_1 \otimes w_1e_1 + w_2e_2) + v_2e_2 \otimes w_1e_1 + w_2e_2$?

[cloud walrus]

okeyokay

[south patrol]

You are missing some important brackets lol

[lime junco]

what helped me get them

[south patrol]

$(v_1 e_1 + v_2 e_2) \otimes (w_1 e_1 + w_2 e_2)$

[cloud walrus]

potato

[lime junco]

is idk if youve done the 1st isomorphism theorem yet

[south patrol]

and then yes, just use linearity

[lime junco]

if not dw you will sson

but normal subgroups come from the idea of kernels of homorphisms

and the quotients are an extension of this

becuase one of the best uses of quotiengts

is they let you reduce a surjection into a isomorphism

you can think of it as surjections have a 1 to 1 correlation to quotiengts

(ofc surjective homorphisms not just set theoretic ones)

thats how I think of it

sorta

[white oxide]

i'm not really sure what system to get to tho. i mean i expanded $(v_1e_1 + v_2e_2) \otimes (w_1e_1 + w_2e_2) = (v_1e_1 \otimes w_1e_1) + (v_2e_2 \otimes w_1e_1) + (v_1e_1 \otimes w_2e_2) + (v_2e_2 \otimes w_2e_2)$ all the way

[cloud walrus]

okeyokay

[south patrol]

well you can take out coefficients

the vectors e1 (x) e1, e2 (x) e2, e1 (x) e2 etc form a basis

so put it as a linear combiantion of those

[white oxide]

oh so like for example $v_1e_1 \otimes w_1e_1 = (v_1 \otimes w_1)e_1$

[cloud walrus]

okeyokay

[delicate orchid]

no

[white oxide]

wait no

it would be equal to (v_1 + w_1) \otimes e_1

[delicate orchid]

v_1e_1 (x) w_1e_1 = v_1w_1 (e_1 (x) e_1)

[white oxide]

oh

[delicate orchid]

? why are you trying to tensor coefficients

[white oxide]

idk

ok now i'm confused as to why for example $v_1e_1 \otimes w_2e_2 = v_1w_2(e_1 \otimes e_2)$

[cloud walrus]

okeyokay

[coral spindle]

$va \otimes b = v(a \otimes b)$, agreed?

[delicate orchid]

review the basic properties of the tensor product

[cloud walrus]

SCARY Boytjie

[delicate orchid]

or even bilinear maps

[white oxide]

oh

i didn't know you could pull out entire scalars

[night onyx]

tensoring is bilinear

[white oxide]

these were the properties my book listed:

[delicate orchid]

they're the only things you can pull out?

[delicate orchid]

well that's idiotic

[rapid junco]

Does there exist a non cyclic group G such that Aut(G) is cyclic?

[coral spindle]

Are you having R be a non-commutative ring?

[white oxide]

they didn't specify it's commutative

[coral spindle]

It just depends on the definition of the module structure right – it makes sense

OK

[delicate orchid]

are you in the same class as the other guy from earlier

[delicate orchid]

no. f(ra, b) = rf(a, b) for all bilinear maps, that's the definition

[coral spindle]

Yes but we don't necessarily say it's R-bilinear

[white oxide]

oh this is in the context of middle linear maps

[coral spindle]

if we're working with noncommutative stuff it would be different

[delicate orchid]

it's the tensor product? what else could possibly be factoring through it

[coral spindle]

Remember the hom-tensor adjunction there is different:

[delicate orchid]

incomprehensible cringe

[coral spindle]

ugh I'm not gonna write it out let me just find it

Anyway, to get back to what okey was struggling with.

[white oxide]

ok i'm lost how does $v_1w_2e_1 \otimes v_1w_2e_2 = v_1e_1 \otimes w_2e_2$

[cloud walrus]

okeyokay

[coral spindle]

If R is a commutative ring, we typically define A (x) B to have an R-module structure given by r(a (x) b) = ra (x) b.

Got it?

[white oxide]

yeah

[coral spindle]

So without a definition, we don't have an R-module structure necessarily and it's not meaningful to talk about pulling out scalars like so.

This is what wew presumed you were doing, but because you didn't mention noncommutativity this wasn't clear

[coral spindle]

Where'd you get this idea

Who says that's true

[white oxide]

oh nvm i misread

[coral spindle]

Cool

[white oxide]

so $v_1w_2(e_1 \otimes e_2) = v_1w_2e_1 \otimes e_2$ correct?

[cloud walrus]

okeyokay

[coral spindle]

By the way we define the R-action, yes

assuming R is commutative here I presume

[white oxide]

yes

and now i'm trying to see how that's equal to $v_1e_1 \otimes w_2e_2$

[cloud walrus]

okeyokay

[white oxide]

because doesn't $w_2e_1 = 0$

[cloud walrus]

okeyokay

[delicate orchid]

why would a multiple, of a basis vector of R^2, be 0

[coral spindle]

Why would this be true

[white oxide]

i think i'm losing track of what things are

[coral spindle]

But anyway

in any case it does not matter

[white oxide]

oh

we would just move w_2 to the other side

bruh

[coral spindle]

use (8)

[white oxide]

ok wait

let me get things straight

so here our module is R^2

[coral spindle]

No

[white oxide]

and we're considering it over itself

[coral spindle]

Our module is R (x) R which is not R^2

Oh my bad you are talking about the right thing

I got ahead of myself

[white oxide]

okay

and we're considering it as a module over itself

right?

[coral spindle]

What

[delicate orchid]

fuck it

[white oxide]

never mind

[night onyx]

You'd consider R x R as a module over R, not over itself

R x R is the module, R is the ring

[coral spindle]

R^2 (x) R^2 in fact but nvm

[white oxide]

ohhh i was considering w_2 to be an element of R^2

like

[coral spindle]

Wh

[white oxide]

as in (0, w_2)

[coral spindle]

but didn't we agree these were scalars

[white oxide]

wait but doesn't this imply that it's a scalar

multiple = scalar

[delicate orchid]

WHAT???

[coral spindle]

What no

[white oxide]

huh

[delicate orchid]

w_2 is a scalar you JUST SAID THAT

[coral spindle]

Hey if I have twice the vector (1, 0) it's a scalar

2(1,0) = (2, 0)

[white oxide]

oh wait i misread boytjie's comment

[delicate orchid]

https://tenor.com/view/hello-neighbor-like-subscribe-bell-hello-neighbour-hello-neighbor-2-hello-neighbour-2-gif-2484330488843852459

[coral spindle]

You know what I'm just gonna stop talking

[delicate orchid]

gimme a min

assume towards a contradiction that $e_1 \otimes e_2 + e_2 \otimes e_1 = v \otimes w$, we can decompose $w, v \in \bR^2$ into our basis vectors $e_1, e_2$ giving $w = w_1e_1+w_2e_2$ and likewise for $v$. Hence
\begin{align*}
v \otimes w &= (v_1e_1+v_2e_2) \otimes (w_1e_1+w_2e_2) \
&= (v_1e_1) \otimes (w_1e_1+w_2e_2) + (v_2e_2) \otimes (w_1e_1+w_2e_2) \
&= (v_1e_1) \otimes (w_1e_1) + (v_1e_1) \otimes (w_2e_2) + (v_2e_2) \otimes (w_1e_1) + (v_2e_2) \otimes (w_2e_2) \
&= v_1w_1 (e_1 \otimes e_1) + v_1w_2 (e_1 \otimes e_2) + v_2w_1 (e_2 \otimes e_1) + v_2w_2 (e_2 \otimes e_2)
\end{align*}
we want this last line to be $(e_1 \otimes e_2) + (e_2 \otimes e_1) $, hence we have to have $v_1w_1 = v_2w_2 = 0$ and $v_1w_2 = v_2w_1 = 1$, but this system is inconsitent so it's impossible

[cloud walrus]

WewGhostTbh

[delicate orchid]

I still don't buy this non-commutative tensor-hom adjunction bullshit btw u get right back here

[coral spindle]

[delicate orchid]

you tellign me that r(a (x) b) isn't ar (x) b

[coral spindle]

So because we need to worry about sidedness, when R is noncommutative we define the tensor product on a left R-module A and a right R-module B

[delicate orchid]

yeah, that's fine

so we can still move scalars inside we just have to be careful about sidedness

[coral spindle]

Yeah

[delicate orchid]

ok, maybe say that instead of posting THIS

[coral spindle]

So now we need to think about the R-module structure right

And so we need to define that carefully

[delicate orchid]

mfw all modules are (R,R)-bimodules :thecosmoshumswithatunemostsweet:

[coral spindle]

bc we can't define a left structure by r(a (x) b) = ar (x) b bc that would need the ring to be commutative

so that's what we need to think about

[delicate orchid]

wait give me a minute

[coral spindle]

Sure, I'll write something else quickly

[delicate orchid]

sorry I had to hold my hands up in "L" shapes to figure out why that broke

[coral spindle]

LOL

[delicate orchid]

we'd need r(a (x) b) = (ra) (x) b

[coral spindle]

Yeah, or ig we could equally well define r(a (x) b) = a (x) rb

I think that works

[delicate orchid]

br surely

[coral spindle]

Oh wait actually that doesn't work anyway

but nah B is a left R-module

[delicate orchid]

_ _

[coral spindle]

Whoops! I wrote it the wrong way around

[delicate orchid]

ok we good then

[coral spindle]

I want it to look like ar (x) b = a (x) rb

[delicate orchid]

yurrrr

[delicate orchid]

I'm so glad this is always the case in my research

[coral spindle]

mb I think that does work lmao lol even

But yeah you get the gist

[delicate orchid]

anyway

[delicate orchid]

@white oxide does this make sense?

[white oxide]

yes

i took like

a 5 minute break

came back

regrouped

[delicate orchid]

remoduled

[white oxide]

yes

thank hyou

[delicate orchid]

no worries

eventually we stop caring about tensor products like this and just think of them as regular products in R-Alg, because they are

ok they're coproducts but w/e

[white oxide]

i don't know what that is but i look forward to when i can struggle with that

thank god for exam replacement policy in my class

[delicate orchid]

actually that perspective made them way easier for me to work with

[white oxide]

second midterm today on tensor products flat modules and category theory and i don't know shit about em

😹

[delicate orchid]

all torsion free modules are flat and I won't hear anyone tell me otherwise

[coral spindle]

hey wew guess what

||otherwise||

[delicate orchid]

MODS

what's the requirement for a ring for it to actually be true though

[coral spindle]

Not sure. I would guess dedekind domains work but that's just a hunch

[delicate orchid]

like I know PID is projective => free

actually wha's the one for flat => projective

[elder wave]

local?

[coral spindle]

I mean this isn't even true for Z is it

Seems like a v strong requirement

[delicate orchid]

googling...

I bet the definition of perfect ring is flat => projective

close!

[coral spindle]

Look at Bass' theorem, that's sick

(Bass' Theorem P) R satisfies the descending chain condition on principal right ideals. (There is no mistake; this condition on right principal ideals is equivalent to the ring being left perfect.)

wild

[delicate orchid]

I hate rings so much

[rocky cloak]

I guess it makes sense if a ring is perfect iff flat implies projective. Since wheter a left module is flat is about how it interacts with right modules

[coral spindle]

Ooh cool observation

[last spoke]

Really

So how do you find the discriminant of (\mathbb{Z}[x]) when (x^3 + ax + b = 0)

[cloud walrus]

StarvinPig

[white oxide]

I just found out what a colimit exists today on my midterm

Previously I’ve never heard of that word

I think I got the solution tho, here’s my work:

[void cosmos]

is this

supposed to be a definition

or am i stupid

[delicate orchid]

homeslice that is the question

[void cosmos]

i meant his answer

where is the definition in the answer

[white oxide]

Lol

I meant that I didn’t put anything down

I didn’t turn in the test

[void cosmos]

oh i was so confused lmfoa

yeha

hahaha my bad

[white oxide]

We have an option to not turn it in

[void cosmos]

fuck it

[white oxide]

So it doesn’t count

[delicate orchid]

fun exercises

[crystal turtle]

What

[lime badge]

Suppose M is a free R-module of finite rank and let phi be a R-bilinear map phi: M x M --> R. Let A = (a_ij) be the matrix corresponding to phi with respect to a basis (e_1,...,e_n) of M, so a_ij = phi(e_i, e_j).

I'm supposed to show that phi is non-degenerate (here, my book defines this as when the induced maps M --> M* are injective) iff det A is non-zero.

But my book has shown that a matrix with coefficients in some ring has non-zero determinant iff the corresponding linear transformation is injective. So could I just cheat and use the iso M to M* to make my argument?

[white oxide]

this is our prof's policy

You do not have to take any of these tests except for the Final; and if you take the test, you do not have to turn it in. If you turn it in, it counts. If you do not turn in a midterm, it's weight will be added to the weight of the final exam.

[crystal turtle]

Oh so now if you fuck up the final you're screwed

[white oxide]

yup

[void cosmos]

whats M*

Hom(R,M)?

if so then yes

[lime badge]

M* is the dual of M

M* = Hom(M,R)

[void cosmos]

isnt this literally the problem?

ut my book has shown that a matrix with coefficients in some ring has non-zero determinant iff the corresponding linear transformation is injective.

[lime badge]

Yeah I guess so

It seemed too easy

[fair bay]

I dont think this question makes much sense unless we assume K to be finite?

[coral spindle]

No, you only need to assume that all three indices are finite

Or in fact you can assume any two are finite, but that's kinda what you have to prove.

E.g. G = Z, K = 2Z, H = 4Z. Note K is infinite here.

[fair bay]

I mean you could have G infinite and H infinite but G : H to be infinite as well

[coral spindle]

You certainly could but the question is probably not asking you to consider cardinal arithmetic, so just assume the indices are finite like I said.

[fair bay]

sorry by indices you mean the orders?

[coral spindle]

No.

I mean |G : H|. That is the index of H in G.

If you're feeling like a big kid who can do big kid math then show there's a bijection G/H <-> (G/K) x (K/H) always.

[fair bay]

oh ok

strange how the question states no such assumption

but it makes sense why you would

[fair bay]

also one last question, do I need to assume |G : K| is finite specifically, since that's the larger group?

[coral spindle]

.

As I said, just assume they're all finite.

[fair bay]

alright sweet! thanks for the patience

[coral spindle]

Assuming that |G : K| is finite is woefully insufficient

[fair bay]

yeah I just realized now if K is larger group than the index would actually be smaller anyway 😄

[late snow]

So if I have a ring where addition is defined as x(+)y = (x+y-1) and I want to verify that the distributive laws hold, is the following the correct way to do it? ```
x(y(+)z) = x(y+z-1) = xy + xz -x?

I know I need to check the other way, as well. But I am slightly sussed out by the -x term on if I am checking correctly.

[delicate orchid]

well what’s xy (+) xz

[late snow]

regular addition, in this case

I tried to separate (+) and + as the former is the addition for the ring

[delicate orchid]

No I mean, write it out

[late snow]

OH, I see what you mean now. I think I fundamentally got the definition confused. In this case, ```
xy(+)xz = xy+xz - 1

Well, I suppose this should really be using this field's multiplication?

[delicate orchid]

Well no it’s just not a ring lol

[late snow]

Okay, I think I am just being really dumb.

[delicate orchid]

Take your original ring to be Z, x = 2, y = z = 1

[delicate orchid]

Right so you want to check \odot distributes over \oplus

Which I do believe

[late snow]

Okay so, I should have been trying to show $x \odot(y \oplus z) = x \odot y \oplus x\odot z$

[delicate orchid]

Dot the the multiplication here

[late snow]

mistyped

[cloud walrus]

statsbro

[delicate orchid]

Yeah there we go

These remind me of formal group laws for some reason

[late snow]

$x \odot (y \oplus z) = x \odot (y + z -1) = x + y + z - 1 - (xy + xz - x)$

[cloud walrus]

statsbro

[late snow]

then $x \odot y \oplus x\odot z = (x + y - xy) \oplus (x + z - xz) = (x+y - xy) + (x + z - xz) -1$

[delicate orchid]

Looks good to me

[cloud walrus]

statsbro

[late snow]

Yeah I think so too, there may be a missing x term but I want to move on. Thanks for helping me, I feel like I understand where I can go wrong a lot better now.

[delicate orchid]

No I think you nailed it, those two expressions are equal to one another

[last spoke]

I'm just wondering how you go about solving the last 3 questions/3b (This was a take home that was due 12 hours ago and already submitted, just curious)

[delicate bloom]

what'd you submit

[last spoke]

3 pages of latex with my name on it, I'll go see if I can Screencast the submission page on canvas to prove it if that's the issue

[delicate bloom]

just curious what we're working with here to start with

[last spoke]

Ahh

Well for 7, I determined what the ring of integers was. That's it

6 I something something somethinged Chinese remainder theorem though we didn't cover when there's more than 2 ideals

For 8 I definition hopped, determined it had to be 2 ideals in (\mathbb{Q}(\sqrt{6})) and then got it to like, (p = xyac + 6xybd)

[cloud walrus]

StarvinPig

[wraith cargo]

For 6 my initial idea might be to try and show that O is a UFD since a Dedekind domain is a UFD iff it's a PID

[last spoke]

Ahh

We did not discuss that

It's one of the exercises in our textbook and the hint there is CRT

[wraith cargo]

7 basically asks you to factor 7 in Q[sqrt{-3}]

[wraith cargo]

This is a problem in Neukirch as well and I don't remember what the hint there is it's something with prime elements

[last spoke]

Yea its basically what I did, and what our lecturer told us to do as well

I showed 5 was a field

[wraith cargo]

Ye that works

[last spoke]

I mean the fact you knew what textbook I was working off based on that kinda says it

Would you be surprised if our lecturer didn't define what splits completely is

[wraith cargo]

Lol
It means that your ideal factors into the most possible prime ideals
i.e. into [L:K] of them when L/K is a separable extension

[last spoke]

I learned that 1 hour before this was due

[last spoke]

The only time we covered them was the last lecture where he worked through last years exam but didn't even tell us what the question was asking for until like an hour in

[wraith cargo]

But Neukirch is a good book
It has a lot of info
Sucks that ur lecturer is kinda ass tho

[last spoke]

Kinda is a bold statement. This is also his field

Like I know for one of my classmates who he was supervising, he didn't define fuck all properly

[wraith cargo]

Dw I had a very famous topologist as my AT prof and he was a very very shit lecturer

[last spoke]

For us his definition of fractional ideal was an ideal that was a product of primes and inverses

That's pretty close to a quote

[wraith cargo]

Well
That's good intuition in NT ig lmao

[last spoke]

He also didn't define unramified

Like didn't even try

@wraith cargo I'm also probably the nicest towards this lecturer out of the bunch if you want some extra context

Here's a fun example

[next obsidian]

For 6, the way that I know how to do it is to use the Forster-Swan theorem lmao

[last spoke]

Lmao the fuck is that

[next obsidian]

Basically tells you an upper bound on the minimal # of generators for a module M based on the minimal # of generators of M_p over p in the support of M, and then the dimension of MSpec A

the finitely many primes tells you that MSpec is 0-dimensional

(Really it's this JSpec thing)

and I_p is always principal cuz the localizations of a Dedekind domain are a PID

and so 1 + 0 = 1

so it's generated by a single thing

[last spoke]

The only thing in this were meant to know is PID @next obsidian

[rocky cloak]

What was the assignment here?

[last spoke]

If a_n
X^n + ... + a_0 is irreducible, show a_0x^n + ... + a_0 is irreducible

[rocky cloak]

I see. ax would be the only exception though

[last spoke]

Yea thats what I found but like, that's not a trivial exception

[vagrant zinc]

Hi guys, does anyone know where I can find the proven correspondence theorem?

[coral spindle]

I would frankly suggest you try proving it yourself – it's a good exercise and not too long either.

[vagrant zinc]

I'm just helping a colleague find the demo.

[coral spindle]

But for a reference, if I assume you're talking about groups then this should be either proven or an exercise in any group theory textbook

I believe Fraleigh does it

[vagrant zinc]

No, it is rings

[rocky cloak]

The proof for rings and for groups are basically the same. But anyway here is the first result that comes up when you Google it
https://math.stackexchange.com/a/1290711/306319

[runic galleon]

hello, can someone explain what's the difference between mapping and relation?

[open sluice]

#proofs-and-logic but a relation is just any subset R of X x Y
a mapping f is a relation that’s also a function

[teal vessel]

a mapping is a function. We say that a relation is a function if, and only if, for all elements of the relation, the primary element of each ordered pair appears precisely once.

(at least according to ZFC set theory)

[crystal turtle]

Also, don't cross-post your question in multiple channels.

[karmic moat]

is there a standard notation for a free abelian group

like if F is a free abelian group with A as a basis

ive seen Z[a : a in A] used in algebraic topology but does this apply to just any free abliean group

[coral spindle]

That’s pretty standard yeah

[karmic moat]

sick thanks

[rocky cloak]

Z^(A) is pretty common for A-fold direct sum

[coral spindle]

You could also just say Z^n assuming your set is finite ofc

[karmic moat]

so like if i had A x B would i say Z^(A x B)

as in, free abelian group generated by ordered pairs (a,b)

[teal vessel]

"prove that if A is a subset of B, then <A> is a subgroup of <B>"

1. A being a subset of B implies A is a subset of <B>
2. for all elements a, b in A, ab^-1 is in <B> (because a,b are in <B> and <B> is a group)
3. Given any subset H of group G, <H> is a subgroup of G by the intersection definition of a generated subgroup
   Therefore, because A is a subset of group <B>, <A> must be a subgroup of <B>

right?

at least, I think so. This notation is a bit confusing to parse

"the intersection of all H such that A is a subset of H, and H is a subgroup of G" right?

[crystal turtle]

Sure that works. You can also just note that <A> is a subset of <B> through the intersection formula

Since <B> contains B, <B> contains A, hence is part of the intersection

[teal vessel]

yeh

seems like that's saying the same thing I did but starting at the largest object

[scarlet field]

i know there exists advanced technique to prove this but we haven't learn, so my way is denote |<a>|=|H|=15, |<b>|=|K|=2, b\notin H, and try to show |ab|=30, by langrange the possiblities of |ab| is 1,2,3,5,6,10,15,30, by the condition given we can rule out 1,2,15, and by |<ab>H|=(|<ab>||H|/|<ab>|\cap H|) i can rule out 3,5,6, because |<ab>|\cap H| need to divide both |<ab>| and |H| and <ab> cannot be contained in H, but for 10, the above equation fails, because we could have |<ab|\cap H|=5 which divides both 10 and 15, any idea about how to rule out |ab|=10?

[teal vessel]

so since this definition doesn't really distinguish between things like <s,r,sr> and just <s,r> do we have a term for something to the effect of the minimal generators of a group?

[rocky cloak]

Notice (ab)^10 = a^10 b^10 = a^10 which isn't the identity.

[scarlet field]

we don't know G is abelian

[rocky cloak]

I see

It's not that hard to prove it's abelian though

[rocky cloak]

Just think about what could aba^(-1) possibly equal

[scarlet field]

it should equal b right bc it's obviously abelian but i couldn't find how to show it based on the given condition

[rocky cloak]

What is the order of aba^(-1) ?
Hint: ||multiply it by itself||

[scarlet field]

oh it's 2 so aba^(-1) must equal to b cuz only one subgroup of order 2

[rocky cloak]

Bingo

[scarlet field]

thank you so much

[teal vessel]

gotta find an example of some abelian subgroup H of some group G such that <H,C_G(H)> is not abelian. This is hard XD

[rocky cloak]

Can't you just pick H to be the trivial subgroup?

[teal vessel]

.... you ever see a solution staring you in the face and laughing at your stupidity for twenty minutes?

like, that seems a valid solution given the parameters of the problem

seems a bit cheeky, but otherwise yeah

[rocky cloak]

Seems the problem is trying to trick you into thinking it's harder than it is

So I'd say the problem is the cheeky one, not the solution

[teal vessel]

trivial groups are groups. trivial groups are groups. trivial groups are groups.

[open sluice]

no they’re not

the group theorists are lying to you

[teal vessel]

"prove that if H is a subgroup of G, <H-{1}>=H"

uh... H-{1} is a group with the identity removed, which means that for all elements a,b which are not inverses, ab is still there. the group generated by that set must provide closure, and since inverses are present, 1 must then be part of that group, but since no other new elements are generated, that's just H without 1, with 1, i.e. H.

these questions be weirding me out with how easy they are.

[delicate orchid]

you mean H is a subgroup of G, I hope

[teal vessel]

I do, brain going faster than my fingies

[noble lynx]

I was wondering if anyone knew any applications to group theory that uses the second isomorphism theorem? I'm having a hard time gaining intuition for why we even care about the "product" of subgroups like AB

[lilac mango]

How would you prove that if phi is a surjective homomorphism then the image under phi of any Sylow p-subgroup is a Sylow p-subgroup?

[void cosmos]

conjugate

[delicate orchid]

thanks moamen, very cool

actual answer, the homomorphic image of a p-element is a p-element as the order of the image must be a divisor of the order of whatever
thus, phi maps sylow subgroups into sylow subgroups, which then in turn must be onto because surjective

[lilac mango]

Why does it send sylow subgroups to sylow subgroups?

Why cant it send Sylow subgroups to p-subgroups

[delicate orchid]

because every p-element is contained in a sylow subgroup by definition

I said into, not onto

onto comes from the fact that it's surjective

lets start naming things

ok nevermind my proof requires knowing that you can split group elements into p' and p-parts

might be a bit too advanced

do you have the result that S in Syl_p(G) => Sker(f)/ker(f) in Syl_p(f(G))?

[chilly ocean]

just conjugate dude

[warm urchin]

One small application is to show for a normal solvable subgroup N and a solvable subgroup H we have H.N is solvable

[delicate orchid]

I'll assume you don't: Let f : G -> H be surjective with kernel K, S in Syl_p(G), then f(S) = {sK : s \in S} = SK/K. Since |S| = |G|_p, ([G : S], p) = 1, and so we can conclude that ([G : SK], p) = 1 as well because SK contains S. Therefore [H : SK/K] = [G/K : SK/K] = [G : SK] is also coprime to p, so SK/K must be in Syl_p(G/K) = Syl_p(H), thus f(S) is in Syl_p(H)

[delicate orchid]

oh and f(S) is a p-group by this discussion

put that at the start

[tardy hedge]

What are solvable groups

[abstract rock]

solvable means different things to different people, but in most contexts it means you can decompose the group with abelian groups

this'll matter when you learn about jordan holder

[abstract rock]

good examples outside of the prior one about solvable groups having a nice decomposition are like the one wikipedia mentions, where 2nd iso can let you identify two groups where one might contain the other obviously, but an isomorphism would seem unattainable until you find a good normal subgroup

[void cosmos]

i meant conjugate , the powerlifting program developed by louie simmons

[delicate orchid]

thanks moamen, very cool

[tardy hedge]

Sorry im high again and was thinking about this stuff

And got mind blown

Thinking about how group theory is the study of symmetries and how that makes sense

And i was thinking about some results in group theory and how its really saying something about symmetry

Has anyone else had moments like that when first studying this stuff?

Anyone

[delicate orchid]

so based

[topaz solar]

Group theory is the study of how you can permute structures, because all the cool questions are about actions

[delicate orchid]

group theory is representation theory etc. etc.

[tardy hedge]

Yeaah

And the group in its entirety is like the “structure” u are preserving

That corresponds to some real life structure or pattern or something

Ok well how im thinking of it makes sense to me but verbalizing things is weird

[topaz solar]

All Polish Roelcke precompact groups are Aut(M) for some separably categorical structure

With the topology lifted from pointwise convergence stuff

[tardy hedge]

Bro how deep does math get

Like what the actual fuck

[topaz solar]

It goes deep

[tardy hedge]

I dont think im ready its like too much

[topaz solar]

This + some stable independence stuff & Hilbert space independence -> all Polish Roelcke precompact groups have Kazhdan property T

[open sluice]

baby steps

[topaz solar]

Which this is like zigzagging from a dynamics-y topological group property back to a dynamics-y topological group property

[last spoke]

j o i n u s

[topaz solar]

You don’t have to jump straight into the abyss like I did

[tardy hedge]

Yeah this course has convinced me

Ive never been this engaged with thinking about the concepts in a math course before this

[delicate orchid]

we start forming groups out of groups

[tardy hedge]

I want to

I need to stop comparing myself to others though

I do it too much and get motivated to demotivated back and forth

I think im just always trying to gauge how im "doing" and if im able to do it or no

not

hopefully what matters most is just the desire to keep trying to learn and understand

[cobalt heath]

Wasn't group theory originally the study of permutations

[sly crescent]

Infinitely deep

[cobalt heath]

How did it become representation

[open sluice]

just do what’s fun

[delicate orchid]

permutation matices

[cobalt heath]

Ah everything is matrices somehow

[delicate orchid]

everything is matrices except matrices, they're an additive category.

[rocky cloak]

I think groups first arose in Galois theory as permutation groups of roots yes

[cobalt heath]

You mean permutation was not among the focus of research before Galois theory?

[rocky cloak]

Well depends how sharply you define Galois theory. But the origins of groups is as permutations of roots of a polynomial to attack problems like the (in)solvability of the quintic

[cobalt heath]

People loved to solve polynomials, didn't they

[tardy hedge]

Its cool how in like {0,1,2} integers mod 3 the elements themselves are literally symmetries

Because you can think of them corresponding to a function (permutation)

[cobalt heath]

I personally understand that as giving numbers to rotations

[tardy hedge]

Like you have a circle and 1 represents the rotation by 360 by 120 deg?

*by 120 degrees

Like is thay what u mean or is it more complicated

[cobalt heath]

Yes, I mean this

[tardy hedge]

Cool

[random sail]

see: cayley's thm 🙂

[tardy hedge]

Yas^

[tardy hedge]

Thats good

[last spoke]

Well we didn't do that in our galois theory course

[rocky cloak]

You didn't consider how groups can act on the roots of polynomials?!

[last spoke]

We did but we didn't get to the quintic bit

[rocky cloak]

Well sure, but problems "like" that

[last spoke]

Sure, but like that's the famous one

We stopped our galois theory at solvable iff solutions by radicals

[rocky cloak]

But then you're basically at the finish line. The last part is just fiddleing a bit to show that A5 isn't solvable, and that there is a polynomial with that Galois group

[last spoke]

Ahh, we didn't do that bit. Fair though

[cobalt heath]

You did not learn that A5 is not solvable?

[last spoke]

Nope

At least I don't think so, we might have mentioned it

This lecturer does not like proofs though

[young coral]

Can Anyone help me with the formal proof of this problem. Or tell me where can i find it

[karmic moat]

https://proofwiki.org/wiki/Chinese_Remainder_Theorem_(Commutative_Algebra)

[young coral]

Sorry but I am not very well acquainted with this part of ring theory .So as a proof to the above problem should I write the whole of what's there in your shared link. Basically , are both problems exactly the same?

oh I see ,basically I need to just proceed the same way with for the special case of n=2

[karmic moat]

yeah

it’s actually enough to just prove n = 2 case then use induction to conclude it for all n

[topaz solar]

Well, the groups I usually see are acting on a structure that isn’t necessarily a vector space, so not necessarily representation stuff insofar as G = Aut(A)

[alpine island]

Why is the symbol for this article a bike chain? lol

[karmic moat]

idk lol

[tardy hedge]

Thanks for that visual. I think i will be thinking about that a lot now when I see Zn

[abstract rock]

crt is like gear ratios

[alpine island]

so true

[clear elk]

is anyone awake

[quaint tree]

Yeah the groups you see are the groups that act on, like, models of logic, and shit like that. Which is insane.

[summer path]

[clear elk]

i just need this proof written

[open sluice]

can you ask a better question

[summer path]

that isnt even a question

[upper pivot]

anyone want to help me with this? the context is btw that r divides m

I actually think this is bs oof, like I cant think of a way to do this that doesnt fundementally depend on like

phi(m)/m

and this can get arbritarily small

so I feel like certainly with only one b I should not be able to find r with probability 1/2

(if I had 2 b's I could with 6/pi^2)

[dim widget]

Is the algorithm supposed to be probabilistic or deterministic?

If it’s deterministic it’s kind of a stupid problem

[upper pivot]

it can be probabilistic

[dim widget]

Also do you actually know that r divides m or just that they are not coprime?

[upper pivot]

yes I know r|m

[dim widget]

Presumably the former

Otherwise quite difficult to learn anything

[upper pivot]

hmm so I think actually this problem might require a quantum algorithm, I initially dismissed the idea but the more I think about it the more it makes sense oof.

[dim widget]

Are you allowed to have a quantum algorithm?

[upper pivot]

something about phase approximations ig. Ill think about it more tomorrow

yeah

[dim widget]

Aha

[upper pivot]

this is kind of a classical post-processing of a different quantum algorithm

or so i thought atleast oof

[dim widget]

I mean you can certainly look at the poset of factors d of m and say that P(d = m/r|sample) = 0 if gcd(d, b) != d for some b in the sample, and otherwise use Bayes’ theorem to compute the MLE for d. That’s the only thing you can really do

Then it’s a basic task in statistics to see how many samples you need to be sure within 5% that you’ve found d for instance

[rocky cloak]

The naive thing I guess would just be to take r to be m/gcd(m, b). Seems plausible that would have a succes probability more than 1/2

[upper pivot]

nah that would be like, phi(m)/m probability

or something like this

[upper pivot]

see thats the issue, I am only allowed 1 sample (atleast thats what the question makes me think)

[dim widget]

Oh I thought we were assuming you wanted to find the minimum number of samples

[upper pivot]

if I could for instance be allowed even 2 samples, I could guess r with 6/pi^2 probability

[dim widget]

Hmm

[upper pivot]

I think this is probably some phase estimation quantum algo stuff, cause its kinda what u said about classical algos

theres only so much that can be done

[tardy hedge]

N is a normal subgroup of G, H is a subgroup of G. Prove that H ∩ N is a normal subgroup of H.

Im gonna show it by showing that if a and b are in H, aVbV = abV, V = H ∩ N

an element in aVbV is of the form av1bv2, since v1 is in N then v1b = bv1'. Then b^(-1)v1b = v1', all of the left hand side are elements of H and since H is a subgroup then v1' is also in H, so v1' is in V

so av1bv2 = abv1'v2 so aVbV subset of abV

is that fine so far

[tribal moss]

It's probably quicker to show directly that it's closed under conjugation.

[tardy hedge]

yeah am i doing like more steps than necessary sort of thing

yeah i actually forgot that was how normal subgroups were defined in the first place lol

[chilly ocean]

When does F_p(x) have a nth root of unity?

I'm particularly interested in the case of F_p(x), where p is not 2, and a 2nd root of unity?

[rocky cloak]

A 2nd root of unity meaning -1?

[chilly ocean]

oh haha

yea

[rocky cloak]

If so, always

[chilly ocean]

Which I guess means always

Right, thank you

[rocky cloak]

In general I guess it will have a primitive nth root of unity iff n divides p-1

(assuming Fp(x) means rational functions, if x is some algebraic element it will depend on x)

[crystal turtle]

[ this is just that the multiplicative group of finite fields is cyclic btw. (F_q)^\times ~= Z/(q-1) ]

[tardy hedge]

Thanks yeah i made the solution quicker doing it that way

Also i was super stuck on this for a long time because i was trying to show H intersect N is normal in G, not H

You guys do algebra problems, how much are you thinking in an intuitive sense, like thinking about how the problem relates to a problem about symmetries etc, vs thinking in a sense of just symbol pushing and how it works in that way

[crystal turtle]

I never think of groups in terms of symmetries.

I actually kinda hate that interpretation.

[tardy hedge]

Whys that?

[tardy hedge]

I mean yeah like i guess this problem is pretty easy so you dont even really need your thinking to go further than just symbol pushing and set theory stuff

But are all problems like thay?

Or does one need a certain intuitive sense in order to make sense of problems?

[rotund aurora]

does this agree with the definition of character in representation theory?

uhm

well if you take the trace of a 1d rep then sure

but for higher dimensional representations it need not agree right?

disclaimer: I dont really know representation theory

[delicate orchid]

yes it does

[delicate orchid]

all irreducible reps of abelian groups are 1 dimensional

so there's no contention here

[rotund aurora]

ohh ok

but you could take the trace of a reducible representation, no?

[delicate orchid]

yeah but who cares

[rotund aurora]

idk

[delicate orchid]

the character of that is the sum of irreducible characters

[rotund aurora]

right

[rotund aurora]

so this agrees with \emph{irreducible characters} ?

idk why I went latex mode

[delicate orchid]

so when people say "the characters" of a group they mean either the set of irreducibles (95% chance) or the ring the characters generate (5% chance)

[rotund aurora]

thank you

[rotund aurora]

is this statement equivalent to the following?: The matrices of a finite abelian group H in GL_n(C) are mutually diagonalizable?

[delicate orchid]

if mutually diagonalizable means simulationously diagonalisable then that's actually how you prove it!

very impressive

[rotund aurora]

yes I meant that

I just reformulated the problem, idk how to prove that, but sounds fun

[delicate orchid]

||every set of diagonalisable matrices that commute with one another is simultaniously diagonalisble||

[rotund aurora]

yes, that's what I came up with

[tardy hedge]

If N is normal in G, show that the order of aN in G\N is a divisor of order of a in G, when ord(a) is finite.

Order of aN is smallest j so that (aN)^j = N

Smallest j that a^j is in N. since a has finite order, we know that a^j is e in N so aN does have finite order. Call the order j.

So now we must ask if a^j has finite order. If it does, then order of aN is a divisor of order of a.

And a^j does have finite order cause a does

So is that all?

Of course a^j has finite order , <a> generates finite cyclic group and so every element in the finite group has finite order

Good?

[delicate orchid]

we know that a^j is e in N
no, you just know that it's in N

but I don't think you use this fact

[tardy hedge]

Ok wait

[delicate orchid]

you've shown that aN has finite order, which is obvious because ord(a) is finite, you haven't shown that it divides the order of a in G I don't think

ah no, you have - by applying lagrange to <a>

alternatively, the map a -> aN is a group homomorphism, so the order of the image aN divides the order of a

[daring nova]

could you explain ?

[tardy hedge]

Yeah nice

[delicate orchid]

if the order of aN is j, then (aN)^j = (a^jN) = N, Then because a^j is an element of <a>, o(a^j) divides |<a>| = o(a)

and (aN)^j = a^jN because... it just does ok??!! trust the science!!!

[daring nova]

have faith in the quotient

[daring nova]

why "the order of aN is the order of a^j" ?

[delicate orchid]

sorry, you're right - it just divides the order of a^j

which is still fine for our purposes

[daring nova]

because a^j is the smallest power of a in N, and e in N, so a^k = e in N => j | k right ?

I have algebra 1 midterms on Tuesday
I'm screwed

[delicate orchid]

I rephrased it to be much more readable

[daring nova]

but much less detailed

[delicate orchid]

no actually I think I have more detail now

[daring nova]

this is more detailed

[delicate orchid]

I think this also works

it's not more detailed it's just different

[daring nova]

than just stating it

[tardy hedge]

I like questions with cyclic groups and orders

[daring nova]

they're simple enough for my smooth brain to work them out in finite time

[tardy hedge]

Basically

[rotund aurora]

If you have this with K<L<M all Galois and pi the canonical projection Gal(M/K)-->Gal(L/K), when is it the case that M^ker(rho_1)=L^ker(rho_2) ?

[chilly ocean]

what do we mean here by "extension"?

[rotund aurora]

is it always true? Im unsure

[delicate orchid]

an extenstion G of a group K by H is a short exact sequence 0 -> H -> G -> K -> 0

if you don't know what this means you can think of it as "If G has a normal solvable subgroup H such that the quotient G/H is solvable, then G is solvable"

which makes it a bit clearer why we should expect it to be true as well

[rotund aurora]

so if for every x in M not in L there eixsts sigma doing something to x but not doing anything to L then it is certainly in the kernel and it's true

which follows by Galois I think

[tardy hedge]

Why are some subgroups normal and some arent

In an intuitive sense

Chat gpt is helpful with this tho lol

[warm urchin]

conjugation

[tardy hedge]

What does that mean thoufh

[open sluice]

conjugation in math refers to the relationship between two things that are opposite yet behave similarly
like x + iy and x - iy form conjugate pairs

honestly i understand "normal subgroup" as "the thing that makes quotient groups work"
my prof gave a different derivation for normal -> quotient but i don't remember how it went

[open sluice]

bit like how measurable functions just exist to make lebesgue integration work

[tardy hedge]

I understand it in that way too but like … im not satisfied or something? Lol idk

[warm urchin]

i think of them as kernels

[tardy hedge]

I think im fine for the purposes of my class rn

But yeah

[open sluice]

true

[tardy hedge]

Kernel of homo

[white oxide]

sanity (?) check: a basis of $\mathbb{Z}_3[x]/\langle x^2 + x + 2 \rangle$ would just be ${1, x}$ right?

[cloud walrus]

okeyokay

[coral spindle]

yes

that's a Z_3-basis for it

[white oxide]

thanks, going over field theory again

i forgot how fun it is holy

[south patrol]

indeed for F[x]/(irreducible polynomial of degree n) an F-basis is given by 1,x,...,x^(n-1)

it's a good exercise

[tribal moss]

That looks like it would probably be more at home in #advanced-algebra.

[ocean charm]

oops sorry

[wraith oak]

If I have questions from my algebra 1 course should I post here or in another section?

[open sluice]

here is fine

make sure to refer to the channel description

[wraith oak]

What is the channel description?

[void cosmos]

whats algebra 1

does ur question involve groups or rings or fields

[tribal moss]

Depending on what "algebra 1" means in your location it could also easily be something that belongs in #prealg-and-algebra.

[wraith oak]

Yes, I guess I should post it here then lol. Just wanted to make sure

[void cosmos]

yea post it

[open sluice]

oolong does mean #groups-rings-fields I think

[void cosmos]

u guys should make a bot

listening on this channel 24/7

if it sees 3x+2y = 2

it removes it

and dms the person the right channel

[open sluice]

tbf it doesn't really make sense to ask pre-algebra stuff in this channel anymore now that it's been renamed

wonder what the folks at #advanced-algebra are getting tho

[wraith oak]

Actually the question I have is from my algebra 1 course but it doesn't involve rings/fields/groups, it's about proving the irrationality of sqrt(n) if n =/= +-1 is not a square free integer

[void cosmos]

u tricked us

hahaha

[wraith oak]

lol sorry

[void cosmos]

jk

yeah ur stuff is correct

[wraith oak]

What I don't understand is the last sentence, I just copied it from the notes

What do they mean by "looking at the prime factorization of p on both sides"

And also, if I have more questions of this type where do I post it? Since it's not exactly ring/field/group theory, more like the precursors to it I guess

[void cosmos]

well ig u could throw this into #proofs-and-logic

since this looks like an intro to proof course

[wraith oak]

Oh right, I didn't see that one

[void cosmos]

if ur using a textbook

then the name of the textbook is prob a good indiciator

indicator*

[open sluice]

or #elementary-number-theory

[void cosmos]

unless its like basic algebra lmfao

by jacosbon

[void cosmos]

yeah

[open sluice]

hey I have that

[void cosmos]

its really a great text

u should be proud

[open sluice]

it prob is but it looks scary

[void cosmos]

it is

vol 2 is metal tho

[sly crescent]

Lots of stuff that doesn’t belong there

[void cosmos]

the only textbook i had in my entire life was brezis functional analysis

and i gave it to a friend

[open sluice]

but also the only reason i have it was because my prof was throwing away her old books and decided i could use them since i was sitting right there

[void cosmos]

yeah its great man

[open sluice]

[void cosmos]

ig just forget about it, make ur own proof

do u see how this and that implie p|t

and hence there is a common divisor > 1 which is a contradiction?

[wraith oak]

So far no, I'll try again though

[void cosmos]

its okay

do u get how p | n and p|s

[wraith oak]

yes, p | n bc we assmed it and the part where I showed p|s I wrote it on my own

[void cosmos]

yea gj

so p is a prime factor of n , and of also s. and u have s^2 = t^2*n

now

n is square-free

[tribal moss]

Once the question has been identifed as not on topic here, please take further discussion to #elementary-number-theory where Oolong re-asked it.

[void cosmos]

okay electric bogaloo

[delicate bloom]

just realized this was cross posted, or something lol

[void cosmos]

let me post a cute algebra problem

so it gets back to on topic

ugh cant find

prove that the sum of two left nil ideals is nil ( this is an open problem )

[wraith oak]

I'll be back soon with questions on rings, so don't worry lol

[gentle trellis]

A coworker and I were surprised to discover we have opposite conceptions of what t_a_b means for an isometry. Somehow our code is able to interact with each other but when he talks about isometries it just sounds like gibberish to me. I did a bit of googling and it seems like most academic literature takes his conception, not mine.
To me, the isometry represented by the matrix:

[ cos30 -sin30 3
sin30 cos30 2
0 0 1 ]

most definitely means go forward 3 steps, step to your left 2 steps and then rotate counterclockwise 30 degrees
Thus, if we call this t_a_b, then this expresses how you get from pose a to pose b
If I then were to go forward 2 steps from there, I would express this as

[ cos30 -sin30 3 [ 1 0 2
sin30 cos30 2 * 0 1 0
0 0 1 ] 0 0 1 ]

We can call this second matrix t_b_c as it takes you two steps forward from pose b to pose c

Taken altogether t_a_b * t_b_c tells you how to get from pose a to pose c

My coworker insists I have it backwards and that t_b_c is a matrix expressing how frame b somehow operates on point c when you multiply by it and that t_a_b then operates on that frame. But I have absolutely no intuition or conception of what that means and the moment we dig into details, his description sounds a lot like he's saying the exact opposite of what I'm saying in a way that, if it were true, would mean our code wouldn't work together because we're using the same notation to express different things. But somehow our code does work together and despite the gibberish that comes out of his mouth, the code he eventually writes makes sense to me when I can put it through my mental model.

I'm a smart guy. I should be able to understand his mental model too, but so far I can't. Someone please help me

[crystal turtle]

Not sure how this fits in this channel.

[gentle trellis]

SE2 is a group and I didn't see anywhere else it fit better

Some people ended up trying to answer in #help-9 but mostly asked questions that didn't seem to me to go anywhere. I'm still missing something

[eager willow]

Left multiplication by that matrix for sure is rotation then translation. Right multiplication by that matrix is for sure translation then rotation.

Really in 3 dimensions it's not an isometry, the translation is shearing. But like then you can project to the top 2 coordinates.

But yeah maybe you're confused about left vs right multiplication.

I always think of left multiplication as row operations (or column-wise action) and right multiplication as column operations (or row-wise action)

So your matrix as row operations is adding 3 times the third row to the first and 2 times the third row to the second, and then do a rotation.

[gentle trellis]

@eager willow Please see the discussion that ensued in the help channel. It's not what you think. The confusion was all about what "do a rotation" means. In my conception you rotate "about yourself" when you move which means you track your heading. In standard conception, points don't have a heading and rotations are "about the origin". It turns out that if you flip the order in which you sequence operations, these are dual operations and both are valid interpretations of multiplying isometry matrices

In other words, I was using a mental model where if I (a character standing on the fixed infinite cartesian grid) start at (x,y) = (0, 0) facing along the positive x axis, then step to the left one increment, then turn to my right 45 degrees (which is not at all the same thing as changing my coordinates by rotating myself about the origin point; in this model my coordinates stay the same, only my heading changes), then step forward 2 increments (along my new heading), I'll end at
(x, y) = (sqrt(2), sqrt(2) - 1) with a heading 45 degrees off to the right of positive x

This can be expressed as:

[ 1 0 0 [ 1/sqrt(2) 1/sqrt(2) 0 [ 1 0 2 [ 1/sqrt(2) 1/sqrt(2) sqrt(2)
0 1 1 x -1/sqrt(2) 1/sqrt(2) 0 x 0 1 0 = -1/sqrt(2) 1/sqrt(2) sqrt(2) - 1
0 0 1 ] 0 0 1 ] 0 0 1 ] 0 0 1 ]

Notice how the matrices representing each of these actions are sequenced in time from left to right, not from right to left

[delicate bloom]

@gentle trellis In a simpler case of exactly the same concept, the confusion is if someone draws the graph y=f(x) and then asks how do you make y=f(x-1)? There are two ways of interpreting this. We can either move the graph of the function to the right by 1 or we can move the whole coordinate grid to the left by 1.

Your coworker's is the first interpretation which is more common, and your interpretation is the second, since it keeps the perspective of the graph. Both are equivalent.

[tulip glacier]

Hello! Exposure to abstract algebra made me question the intuition behind a basic property that I once took for granted - associativity. Can anyone provide me a quick intuition (or way to think about it), besides its algebraic definition ? That it "allows members to be "grouped" in any way" still doesn't feel intuitive enough for me.

Thanks in advance for any help!

[thorny ferry]

Translation is an isometry, because it is distance-preserving.

[thorny ferry]

So it's transformation of a point vs. transformation of a coordinate frame. One is the inverse of the other for a given isometry.

[tardy hedge]

So any possible homomorphism out of G is already “encoded” in the normal subgroups of G?

[thorny ferry]

You can apply the operations in any order (which is not the same as saying the operations are commutative).

[tardy hedge]

Thats pretty cool

[tulip glacier]

Hmmm, I can see where this is coming from, and it is good enough. But, just out of curiosity, is there something else other than this ? It's completely fine if not.

[thorny ferry]

It's hard to think of any other conceptions of it

I guess intuitively it's just like addition or multiplication, but those have more properties than just associativity.

Note that subtraction is not associative though.

[tulip glacier]

👍

Thanks!

[indigo ridge]

Im a bit confused about this excerpt in my book on power series, it says the following "Almost always B(x) itself will be a polynomial, but its inverse B(x)^-1 is neccesarily a power series, rather than a polynomial"

I can think of a counter example, for example any given quadratic polynomial in a ring will have an inverse that is also a polynomial.

[tribal moss]

The inverse if x² is sqrt(x), which is not a polynomial.

[coral shale]

is 1/f(x) or undo f meant here....

[tribal moss]

(Unless the ring is characteristic 2, quadratic polynomials can't even have inverses unless you restrict their domain first -- as I implicitly did here).

[indigo ridge]

Sorry yeash because it doesn't have the constant term

I meant quadratics with constant terms

[tribal moss]

The inverse of x²+1 is sqrt(x-1), which is not a polynomial either.

[coral shale]

You need to clarify what "inverse" means

Surely it means the multiplicative inverse rather than the function inverse else I dont find sense in the context

[tribal moss]

But even so, the reciprocal of a quadratic will not be a polynomial.

[indigo ridge]

So Im talking about the ring F[[x]], and the inverse is just an element of the set such that given a the inverse of a aa^-1 = 1

[coral shale]

Thats the multiplicative inverse for the ring of polynomials

So x^-2 is the inverse of x^2 which is not a polynomial

[indigo ridge]

its the ring for power series*

[open sluice]

here’s another way to look at integers that you may not have noticed
you probably think of the number 5 as expressing a number of objects: 5 apples, 5 pencils, 5 bunnies
but you can also think of it as an operation, namely the act of adding 5 to something
so 5 + x may mean “add 5 bunnies to x”
if you have just 5 by itself, you can conceptualize this as “add 5 to 0”
then 10 + 5 would mean “add 5, then 10,” or in other words combining the operations of adding 5 and adding 10; this is of course the same thing as adding 15
now you can interpret the associative property differently: it’s saying that (a + b) + c is equal as an operation to a + (b + c)
for example, take 1 + 4 + 9
if you combine this as 5 + 9, you get the (single) operation of adding 9 followed by adding 5
but if you combine it as 1 + 13, the operation becomes adding 13 followed by adding 1
and associativity claims that these two functions are the same

if you compare this to commutativity, that would say that adding 2 then 3 is the same thing as adding 3 then 2

maybe this can help you think about associativity

[coral shale]

x^-2 will be an infinite series yes

[indigo ridge]

Ok so it's always the case that the inverse of a polynomial is going to be a power series?

[coral shale]

generally.

not always

Take a constant (non 0) polynomial

[tribal moss]

You can't have two polynomials whose product is 1 unless they are both degree 0.

... or unless the coefficient ring has zero divisors.

But if you say F[[x]], that seems to indicate the coefficients come from a field.

[indigo ridge]

Ok yeah that makes sense.

[indigo ridge]

mhm

[tulip glacier]

Aha, makes sense! I guess I'll have to think a little bit more about it, but it definitely helps. Thanks as well!

[coral shale]

A general note. Algebraic expressions with binary operations cannot be formally parsed without proper parenthesizing

The usual convention is a # b ? c = (a # b) ? c

[tribal moss]

Something something Polish notation.

[coral shale]

in terms of implicit notation

But, associativity means either interpretation works (they are equal)

a # b ? c = (a # b) ? c = a # (b ? c)

[tulip glacier]

So I can think about associativity as allowing the operation to take an arbitrary list (without parentheses), without leading to ambiguities ?

[coral shale]

I believe so, more or less.

[tulip glacier]

👍

[coral shale]

If you take this convention theres also no ambiguities but tbh, we don't quite take this convention in practice anyways

but also, ive forgotten --- associativity means u are referring to the same operator

# and ? will be the same

[coral shale]

but what I meant was, we read 1 + 2 * 3 as 1 + (2 * 3)

because.

[coral shale]

So if there are no conventions to follow for parenthesizing --- yes.

A useful example of associativity that always holds

thats a bit surprising to me tbh

is function composition

this is associative, generally

[open sluice]

I agree actually

[coral shale]

f o g o h = (f o g) o h = f o (g o h) = f(g(h(x)))

like, this is very general. but turns out to be associative

And to me, this is a big motivating reason to care about associativity

most nice things will be associative because of this in some way or other. I think.

[tulip glacier]

Yup.

Thanks to all the answers, I feel as if I am one step closer to finding the intuition!

Commutativity allows us to not care about the order of the elements, meanwhile associativity allows us to not care about the way the elements are grouped (parentheses). All there is left is just to make a proper difference between order and grouping, since my intuition yet cannot really do this. I mean, it works, but I feel like I can do better at this.

[coral shale]

Yeah if you have commutativity without associativity

you get. quite stuck

i would say

There's not too much you can do with this expression
a * (b * c)

Associativity is a 'default' that you will assume in almost all contexts

There are few natural things that aren't I think.

The vector cross product is one