groups-rings-fields

[chilly ocean]

Like... am I supposed to assume subrings contain 1. Or homomorphisms preserve it. Sometimes you got to figure it out

[somber sleet]

I noticed that my prof defined a ring with commutativity

[chilly ocean]

I'm kind of complaining because the author I'm reading rn never explained it lol

[somber sleet]

I don't know if you understand german, but I think this is what he means

like in my case the difference between a ring and a field is just that it doesn't have to hold 1!=0 and the inverse of an element under multiplication doesn't always have to exist

[rustic crown]

fields are usually assumed to be commutative, the non-commutative rings where every non-zero in invertible are called division rings

[chilly ocean]

Yeah seems so

It's not that non-commutative rings are not important, but commutative ones are easier to work with (and have a richer theory?)

@rustic crown yes?

You're the algebra expert :p

[rustic crown]

prime ideals are fun, but hard to deal in non-commutative rings >.<

i've only worked with non-commutative rings while doing some representation theory

everywhere else the world is commutative for me

[chilly ocean]

Do you have some duality between ideals of a ring and filters on some space

Or something analogous

Sorry, random question

[rustic crown]

was looking at the definition of filters lol

[chilly ocean]

Can you study non-commutative rings geometrically to some extent like you do in commutative algebra?

[rustic crown]

if i had to guess, then probably the answer is yes

it's hard to keep mathematicians from not generalizing things :p

[lavish spoke]

analysis has predominantly non-commutative algebras unfortunately

[chilly ocean]

I wonder if Boytjie knows more

[rustic crown]

only non-commutative geometry i've heard have more stuff to do with C*-algebras

[chilly ocean]

The rings I'm studying in topology rn are all commutative and nice

with lots of zero divisors

@coral spindle hey, good morning. Did you get anything about this lattice question of mine from yesterday?

[coral spindle]

I’m afraid not :( nobody seems to know

Sorry for not @chilly ocean you, I forgot to

Whoops, that’s a person

Sorry

[chilly ocean]

Lol

[formal ermine]

wouldn't have expected linear algebra to be used in the classification of cyclic extensions

[delicate orchid]

unaware

[hidden haven]

Exact sequences of commutative non-unital rings show up in K-theory

[rustic crown]

exact sequence of groups are bad enough already

[delicate orchid]

all SES of groups split so it's not that bad :trollface:

[formal ermine]

explain

[lethal dune]

all groups are projective ℤ-module anyway

[smoky ivy]

Can I prove that if m is a maximal ideal in R, and u \in R\m, then there is some v \in R such that uv-1 \in m?

[formal ermine]

let f in Q[x] with degree >= 3. why does f(x) = c(x^n - a) imply that its galois group has degree <= n^2?

where a = alpha^n for a root alpha of f

do I just like

calculate the degree of its splitting field manually

then use that |gal| <= [extension]

[chilly ocean]

the ideal (m, u) contains m strictly so it must equal all of R. so you can find r in m and v in R such that r + uv = 1, ie uv - 1 in m

[warm wyvern]

not for the fr*nch, I have learned

[warm wyvern]

LOL

[vagrant zinc]

Guys I need help, it's my first time applying something to show that it's a group, I don't know how to go about it.
I would like you to give me some ideas, and how I could demonstrate that it is an abelian group.

I have done it like this

[next obsidian]

I didn’t look too carefully but it seems correct

I don’t think you showed the operation is associative though

[vagrant zinc]

I don't know if this is the right way to do it

[lavish nexus]

Yes

You still need to show it is assoc

which is boring

[vagrant zinc]

how do I do that, I don't understand 😦

[delicate orchid]

show that (a*b)*c = a*(b*c) for all a, b, c

it's the most boring one I agree

[vagrant zinc]

look

[formal ermine]

if |G| = 2^(r - 1) why does there exist a subgroup of order 2^(r - 2)

[rustic crown]

sylow

[agile burrow]

i don't even think you need Sylow for this, I've always proven this by induction

It's a more general fact about p-groups, that if |G| = p^r then G contains normal subgroups of order p^k for all k <= r

[rustic crown]

ah i see

sylow only gives a normal series right

here you get that everything is normal :3

[agile burrow]

but I guess Sylow 1 lets you conclude that if p^k divides |G| then you have a subgroup of order p^k which is cool in its own right

[rustic crown]

ig the only difference between the two proofs is exactly that for p-groups you have a non-trivial center so you can choose a nice element of order p than some arbitrary one

[next obsidian]

for p-groups the proof is easy tho without any Sylow nonsense

use useful fact #1 about p-groups and mod out by Z(G)

I guess you need to handle the abelian case which could be difficult unless u have structure thm

[formal ermine]

what's useful fact #1 about p-groups

[agile burrow]

non-trivial center

[next obsidian]

that Z(G) is non-trivial

[formal ermine]

ah

[rustic crown]

you can prove abelian cauchy also by simple induction

[next obsidian]

Actually not true just like

take an element

of non-maximal order

exists by like, cauchy, or by whatever else you want

lots of ways to do

just like, take some multiple of an element of maximal order and you see you get an element with order not p^n and also not 1

if |G| = p^n

[rustic crown]

i was thinking more like this, ig G is a p-group, by induction find a normal subgroup of order |H|/p, then find some nice p thing in G/H

[next obsidian]

I mean, that's the point of Z(G) right?

as long as it isn't all of G you can by induction find something of index p in G/Z(G)

then pull that back

[delicate orchid]

I could really "kill a fly with an atom bomb this" I think

[rustic crown]

makes sense. i was using Z(G/H) non-trivial to find the normal subgroup H'/H in G/H of order p >.<

ig in your argument, instead of taking all of Z(G) just take a subgroup of that of order p to avoid dealing the abelian case separately.

[next obsidian]

why is that normal in G?

wait thonk

[rustic crown]

because Z(G) center

[delicate orchid]

all subgroups of Z(G) are normal lel

[next obsidian]

Okay but

Det

actually your proof doesn't help at all

cuz how do you now you can do that in Z(G)?

the abelian case is only when G = Z(G)

but then you can't do that in Z(G) cuz no induction go through

[rustic crown]

finding elment of order p you mean?

.<

[next obsidian]

but

[rustic crown]

i had cauchy in mind, but ig if you avoid that then it increases the owrk a little

[next obsidian]

I mean

finding an element of order p in Z(G)

is equivalent to handling the abelian case

[rustic crown]

true

permutation of inductions

[rustic crown]

i wonder what all things people consider nonsense :p

like category theory is abstract nonsense

are there more concrete nonsense examples?

[quaint minnow]

Is category theory really that looked down upon ?

[rustic crown]

nope

people affectionately call it that

[quaint minnow]

Sounds like a love-hate thing

[rustic crown]

it's nonsense in the sense that it talks purely about structure of things and not about meaning

[quaint minnow]

That’s the premise of it right? To study about its relationships and not to care about the inner details

[wind locust]

can someone help me prove/disprove this statement? If two actions of G on some set X are both sharply transitive, and they agree on an element of X, then they are the same

ah i think that's easy

[coral spindle]

Yes, you factor through this element they agree on.

[wind locust]

yeah lol i guess im a bit tired i thought about this too hard

also we don't even use the freeness of the action

i was about to type 'freedom of the action'

thanks though

[coral spindle]

That's not true, we do indeed use freeness of the action

or am I misremembering what that means

if your point is that we don't need to use the fact that the transitivity is sharp, this isn't right

[wind locust]

no, no you're right I got it wrong. one sec

say both actions fix $x_0$. Let $g\in G$, and $x \in X$. Let $h \in G$ be such that $h \circ x_0 = x$. Then
$$g\circ x = g \circ (h \circ x_0) = gh \circ x_0 = gh \cdot x_0 = g \cdot (h \cdot x_0) = g \cdot x $$

[cloud walrus]

I Abhor Hatcher

[wind locust]

I don't think we used sharpness at all?

[coral spindle]

You're quite right, my mistake

[wind locust]

alright, thanks for helping out 🙏

[molten viper]

So, my text uses the notation for the sum of 2 ideals. Is that defined as
I + J = {x + y| x in I, y in J }?

Rather, is that a standard operation I should know

[delicate orchid]

Yes

[chilly ocean]

yes to both

[molten viper]

Thanks!

[chilly ocean]

now what about the product

[delicate bloom]

it's also worth mentioning it's kind of like the gcd of ideals, since you can do the euclidean algorithm by addition and subtraction. Look at ideals of Z for examples.

[molten viper]

Interesting

[next obsidian]

It is quite literally the same as “the smallest ideal containing both” so within the lattice of ideals it is the meet!

Err, join

I think

[woeful sand]

You can define about direct sum too

in some cases

it might be useful

[toxic zephyr]

why does the semi/quasi-dihedral group need to have an order that is a power of 2? does it collapse into some other dihedral group otherwise?

like why is this not a quasidihedral group?

$\gen{r,s\mid r^{12}=s^2=e,s\inv rs=r^{6-1}}$

[cloud walrus]

nilpotent nix

[toxic zephyr]

is it perhaps that the order isn't actually going to be 24? like r^6=e and this is actually just D12 or something?

how could I theoretically show that that's the case? that r^6=e i mean

[chilly ocean]

This is for ppl in the numerated help channels

[toxic zephyr]

ah my bad i didn't see that in the rules anywhere

[tender wharf]

I don't see how order of r would be 6

[chilly ocean]

#info

[vague granite]

Is it just me or is this proposition written weirdly?

Also I believe it's missing the fact that R has to be non-trivial

[south patrol]

Is this Rings and Modules by any chance

[vague granite]

It is indeed

[south patrol]

Yeah okay the notes are weirdly written at times oofies

[vague granite]

But importantly... R has to be non-trivial right??

[south patrol]

Good point

[toxic zephyr]

yeah me neither. but quasidihedral groups apparently need order of 2^k and i don't see why.

[south patrol]

I'd definitely write it differently cause I just managed to confuse myself thinking about it lol

[toxic zephyr]

if it isn't quasidihedral, i guess it has to be dihedral ? but idk how to find it's order. unless maybe it's neither, and is actually an entirely different group

[south patrol]

But yeah it should be like "suppose R is a non-trivial ring which is also a f.d. F-vector space, such that left/right ring multiplication is F-linear. Then if R has no non-zero zero divisors, U(R) = R^x" and that long condition in my first sentence can be summarised as being a non-trivial f.d. F-algebra

[vague granite]

Suppose that $R$ is a non-trivial ring with no non-zero zero divisors which is also a finite dimensional vector space in the which left and right multiplication maps are linear. Then $U(R)=R^*$, and in particular if $R$ is a commutative ring then $R$ is also a field.

[south patrol]

Yeah I mean sure for the last bit, but the point he's trying to make in the last line is that like

finite integral domains are fields

[chilly radish]

Also saying U(R)=R^* is just a fancy way of saying R is a division algebra

[south patrol]

Indeed

[cloud walrus]

Philka

[south patrol]

Well, or a less fancy way I suppose lol

the guy doesn't define division algebras etc iirc

[chilly radish]

It's phrased very weirdly tho, why would they not just say "if R is also commutative then it's a field"

[chilly saddle]

It is a vector space over what field?

[south patrol]

any field

[chilly saddle]

Like, F2 too?

[south patrol]

Why would that make a difference?

[vague granite]

Yes

The field plays no relevance to the proof since you appeal to the rank nullity theorem

[chilly saddle]

Okay, that makes sense

oh, given no non-zero divisors the field must by of characteristic 0 i think

[south patrol]

No, because e.g. F_p has no non-zero zero divisors and is a f.d. F_p vector space

[chilly saddle]

The one-dimensional vector space, sure

time to sleep

[cloud walrus]

potato

[ember hull]

I would like to show that the action of the special linear group SL(E) on the projective space P(E) defined by $u.\overline{x} = \overline{u(x)}$, with $\overline{x} \in P(E)$ is doubly transitive. Can someone help me to show that ?

[south patrol]

The point of writing it this way is to show that actually the important thing in showing r has a left inverse is showing that the sequences R \supseteq rR \supseteq r^2 R \supseteq ... must eventually "stabilise"; if this is the case for any collection of ideals in R then we call R (left i guess) artinian

[cloud walrus]

Hmz

[toxic zephyr]

do i just have to do a multiplication table or something to brute force what it is/is isomorphic to
groups really confuse me

[ruby sundial]

no

Dihedral group has two generators

One is order n

the other is order 2

there is also a relation between the generators

[toxic zephyr]

srs=r^-1

[ruby sundial]

have you learned about lagrange theorem or quotient groups yet?

or normal subgroups?

[toxic zephyr]

learned: yes. understand super well: not so much. which is the Lagrange theorem? the subgroup order divides the group order?

[toxic zephyr]

so with the example, r^12=e, s has order 2, and there is a relationship where srs=r^5, that sounds like it falls under the category.
but then r^5 should be r^-1 implying that r^6=e?

unless that last condition somehow breaks something else and it's not even a group. I'm not sure...

[formal ermine]

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

[ruby sundial]

oh yeah my bad for afk

use lagrange theorem and thats it

and let reflections be the normal subgroup

order of normal subgroup (order 2) divides the group

|N| is order of subgroup, |G| is order of group

so |G|=k*|N|

and remember we know this because of cosets

cosets have equal order

and they partition the group

so from these two facts we fet lagrange theorem

[toxic zephyr]

the subgroup generated by s {e,s} you mean?

is that a normal subgroup? rsr^-1=sr^-2 doesn't it?

[chilly saddle]

$$ rs = sr^5, r^2 s = rsr^5 = sr^{10}, r^2 s r^2 = s, r^2 s = r^{10} s, r^2 = r^{10}, r^8 = 1 $$

[toxic zephyr]

so that makes sense that two divides the group. but i don't see why <r,s|r^(2n)=s^2=e, srs=r^(n-1)> necessarily implies it needs order that is a power of 2

[cloud walrus]

mzg147

[toxic zephyr]

oh interesting

[chilly saddle]

sorry I'm just trying to learn how to use latex here

and also wake up

if I didn't mess up then r^4 = 1 so rs = sr

Did you come up with this presentation by yourself?

[chilly saddle]

Oh, this is the problem

[toxic zephyr]

how did you go from r^2 s r^2 = s to r^2 s = r^{10} s?

[chilly saddle]

i multiplied by r^{10} from the left

oops

that's not right...

[toxic zephyr]

it came up in a paper i was using for a project. it just seems odd that n apparently has to be a power of 2 for the third one. like why can't we choose n=6 for example (which is where mine came from)

[toxic zephyr]

i wonder if this general principle could work though. like maybe the fact that it's 5 and not 2^k-1 will cause some kind of issue?

but i can't think of where that issue could arise

what does 3,7,15,31,etc. have that 5,9,11,13,etc. don't?

[chilly saddle]

groups that have the size of the power of 2 are important in group theory, they are called 2-groups

I would guess that's why semi-dihedral groups are defined for the powers of 2, they are next in the sequence of some 2-groups

Your example would be a semidirect product of C_12 and C_2 with a twist of 5, too general for the name of "semidihedral"

[chilly saddle]

huh

[toxic zephyr]

interesting. what exactly is "a twist of 5"?

oh

[chilly saddle]

yeah, r^3 commutes with everything

So {1, r^3 , r^6, r^9} is normal

and it's C4

dividing we get a group of order 6 which is not abelian, hence it's D3, aka S3

and something something splits

[toxic zephyr]

so S3 is the r part and C4 is the s part?

[chilly saddle]

it's neither

C4 is from {1, r^3, r^6, r^9}

[toxic zephyr]

oh

so this is saying the group is isomorphic to the product of the cyclic group of order 4 and D3?

[chilly saddle]

Yes

[toxic zephyr]

interesting

what is this function you used? is it something i could experiment with?

[chilly saddle]

I used SageMath on https://cocalc.com

I didn't sign up so I cannot share you the workbook, it would've been easier

[gilded osprey]

I have a reduction map [-] that sends an integer to its equivalence class.
Addition is then a map from ([a],[b]) -> [a+b]. I need to show addition is well-defined and unital.

I have already shown that it is well-defined, but unsure what to do for unital. Do I just show that there exists an equivalence class E s.t. [a]+E=[a]? Does it just suffice to show that E can be defined as [0], which since addition is well defined gives us:

[a]+E=[a]+[0]=[a+0]=[a]

Or do I need to show more?

[rustic crown]

strictly speaking, depending on your axioms you would need to show [a] + [0] = [a] = [0] + [a], but i think it's clear from what you have done.

[gilded osprey]

The problem says I don't have to show symmetry, I can just assume it to be true

[toxic zephyr]

for a semidirect product
$$G=\bZ\ltimes\paren{\bZ/p\bZ}^k=\gen{s^av\mid v\in\paren{\bZ/p\bZ}^k,a\in\bZ,s^{-\a} v s^\a=M^\a v}$$
would you denote the cosets
$$G/\paren{\bZ/p\bZ}^k=\left{s^a\gen{\paren{\bZ/p\bZ}^k}\right}$$
like this?

[cloud walrus]

nilpotent nix

[toxic zephyr]

i feel like i'm completely messing up the notation

[rustic crown]

then it's good :3

[rustic crown]

you don't need to put the extra <.> at the end, it's already a normal subgroup

tbh i would just denote it by \overline{s}^a

instead of carrying the whole subgroup around

[toxic zephyr]

so $s^a\paren{\bZ/p\bZ}^k\defeq\overline{s}^a$?

[rustic crown]

yep

[cloud walrus]

nilpotent nix

[toxic zephyr]

great thank you!

[rustic crown]

strictly speaking it is =:

overline{s} is (by definition) the coset of s inside G/...

so overline{s^a} = overline{s}^a is the coset of s^a inside G/...

whenever there is a single group G and a single normal subgroup N in sight, anyone would understand that \overline{g} means the coset gN in G/N

[toxic zephyr]

thank you @rustic crown, @chilly saddle @ruby sundial @formal ermine for help with my earlier question, and the #groups-rings-fields channel in general. you have all helped me so much ❤️

[coral shale]

illu is no help, lies.

[long nebula]

quick question, is (phi) meant to be coker(phi) here?

I assume that's what it is but I don't wanna go about proving something wrong lol

[agile burrow]

yes, it should be the cokernel

[long nebula]

cool 👍

[long nebula]

wait wtf how am I even supposed to prove this, "this just follows from the definitions"????

LOL

I guess I'll just write out the definition of exact and call it a day

[vivid tiger]

i guess doesn't that technically require the first isomorphism theorem?

[chilly ocean]

No why

All you're checking is if ker/im's are the same

First isomorphism theorem would be like, if you were given exact sequence 0 -> K -> L -> M -> 0 and wanted to replace M with a quotient for example

[chilly ocean]

K is a subgroup of G
Why is this true

[rustic crown]

show it's closed under the operation and inverses

[tender wharf]

the normality of H comes very much into play here

[warm wyvern]

this isn't true in general

oh wait

H is normal, mb

in that case, it is

[rotund aurora]

x^8-2 irreducible in Q(i) simply follows from Eisenstein's criterion right? Since 1+i is prime in Z[i]

[formal ermine]

yes

[warm wyvern]

omg

[formal ermine]

omg

[warm wyvern]

are they called prime ideals coz prime numbers generate the prime ideals of Z?

[formal ermine]

don't prime elements in general generate prime ideals

[warm wyvern]

prime elements being elements that generate prime ideals?

[formal ermine]

a prime element r is a non unit s.t. if r | xy then r | y or r | x

[warm wyvern]

I see

[delicate orchid]

yes, basically by definition (but not quite!)

[coral spindle]

It should be said that while every prime element of a ring generates a prime ideal, in general a prime ideal will not be generated by a single prime element.

[formal ermine]

that's only true in pids is it?

[coral spindle]

I'm not certain

[next obsidian]

It’s true in any principal ideal ring

It’s actually equivalent to being a principal ideal ring

[coral spindle]

Very nice

[rustic crown]

Z[2^(1/2^infinity)] looks like an example of a ring with a prime ideal which isn't generated by prime elements inside it

[formal ermine]

oh wow

[rustic crown]

(that's a slightly different thing i was wondering)

[formal ermine]

what's an example of a finite ring that's not a pid

[rustic crown]

F2 x F2

that's not a domain

[next obsidian]

Any finite domain is a field

So you have to look for non-domains

[formal ermine]

is every field a pid

[rustic crown]

yep

[next obsidian]

I mean

It has two ideals

[formal ermine]

oh

right

lmao

[chilly ocean]

what's your favorite non-field PID?

[next obsidian]

Z

[rustic crown]

and k[x]

[next obsidian]

Proof that primes are all principal => PID for domains:

Finitely generated primes => Ring is Noetherian

Noetherian means Krull’s Principal Ideal theorem => ring is 1 dimensional

1 dimensional means Dedekind domain since all local rings are DVRs => ideals are product of primes

All primes principal => all ideals principal

The case of a non-domains starts the same, then you write it as a finite product of dim 1 rings with the same property, or local Artinian rings with principal maximal ideal. Either way, each component is a principal ideal ring, so the product is too.

Latter part is harder than handling the domain case

[rotund aurora]

what is this group? {x,y,z | x^2=y^2=1, z^8=1, xy=yx, xz=z^7x, yz=z^5y}

[rustic crown]

weird group of order 32

[rotund aurora]

I obtained that as the Galois group of x^8-3, I think its correct, but there should be a nicer form

The splitting field is Q(sqrt 2, i, 8rt 3), and x is the automorphism that sends i -> -i, y sends sqrt 2-> -sqrt 2 and z sends 8rt 3 -> g* 8rt 3, where g is (1+i)/sqrt 2 is an 8th root of unity.

[rustic crown]

it has a normal subgroup isomorphic to D(8) or D(2*8) whatever you like to call it

so some weird extension of Z2 and this D(8)

it's also generated by 8rt 3 and w = primitive 8rt 1

maybe that gives a simpler presentation

(isn't galois group of x^8-3 a good enough description :p)

[rotund aurora]

no

like the exercise asks to find the galois group lmao

[rustic crown]

and you found it >.< as a presentation

[stiff karma]

${ Z_{n} } = { 0, 1, 2, \ldots, n-1 }$
Are addition and multiplication just implicitly defined for this set to be modulo n or is it that people just assume that it's used in the context of modular arithmetic?

[rotund aurora]

but it sucks

[cloud walrus]

ren_dez_oox

[next obsidian]

I think you

Should try to write it as an internal (semi)direct product

[rotund aurora]

I'll be honest

[next obsidian]

Of Galois groups of intermediary extensions

Maybe

[south patrol]

Tbf normally one would not view Z/nZ as that set lol but yes, if the operations aren't given in the standard way then it's not Z/nZ

[rotund aurora]

I don't know semidirect products, tho I could probably read the definition

[stiff karma]

what is it "normally" viewed as then, if I may ask?, are you alluding to it being the set of all integers modulo n?

[south patrol]

Yeah

[stiff karma]

is there anything more to the definition of Z n

[south patrol]

But I guess at first Z_n is sometimes introduced as you said so nvm

[stiff karma]

Could you elucidate the definition above?

[rustic crown]

a+b-c in nZ is same as saying a+b and c leave the same remainder on division by n

and there is a unique such c in that set for a given remainder

[stiff karma]

right yes I got that part , but is it a part of the definition of Z n, or is Z n just like any other normal set, and this is just so that the operation of addition and multiplication are closed?

[rustic crown]

that depends really... in the start people may explicitly write (Z_n, +, *) for this whole baggage... but as time goes on people just abuse it and refer to the whole thing by Z_n

[stiff karma]

I see ok, that was kinda confusing for me, different lecturers used it in different forms as you mentioned, thanks!

[rustic crown]

it's never a confusion unless one defines multiple operations on the same set, and when people do that they would definitely be careful about the putting all the + and +' everywhere

[stiff karma]

Yeah makes sense.

[celest cairn]

Is $\mathbb{Q}(\frac{-b\pm\sqrt{b^{2}-4ac}{2a}}) = \mathbb{Q}(\sqrt{b^{2}-4ac)$?

[cloud walrus]

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

[celest cairn]

Supposed to be divided by 2a. Idk what went wrong.

[warm crow]

$\mathbb{Q}(\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) = \mathbb{Q}(\sqrt{b^{2}-4ac)$?

[cloud walrus]

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

[celest cairn]

thanks 👍

[warm wyvern]

what's you guys' favourite k-algebra where k is a field?

[formal ermine]

k is my favorite k algebra

[warm wyvern]

cringe.

[rotund aurora]

uhm ok I think its just {x^8=y^4=1, yx=x^3y}

[woeful sage]

my favorite is a k-algebra-algebruh k is a field, algebra is a vector space and algebruh is a tensor algebra

[elder wave]

Terrible day to have eyes

[rustic crown]

so it's a semidirect product of Z/8Z and Z/4Z :3

with the map Z/4Z --> Aut(Z/8Z) = U(8) given by sending 1 to the unit 3

[rotund aurora]

let p be an irreducible polynomial and suppose that for every root a of p, the extension Q<Q(a) has no intermediate fields. What can you say of the Galois group of p or the splitting field of p ?

[formal ermine]

do algebras over rings exist

do they have a special name

[next obsidian]

Algebra

[formal ermine]

isn't that over a field

[next obsidian]

No

[chilly ocean]

Usually its over a commutative ring

[formal ermine]

oh

wikipedia was lying to me once again

[warm wyvern]

they're called R-algebras in aluffi (where R is the ring your algebra is over ofc)

I do think I heard "Algebra" is usually over a field too

[elder wave]

there's even separate articles for the two

[warm wyvern]

illu got muted?

[south patrol]

ig algebra can just mean a few things rip

i don't even assume my algebras are associative

[chilly ocean]

We have Dimension of F(a) over F is n

Can we say that a is algebraic of degree n?

[tender wharf]

wait what?

this seems to be false

[formal ermine]

I'm unmuted again

[warm wyvern]

since this section was all about how R-mod is an abelian category, I get the feeling that I'm supposed to prove M satsfies the universal property of ker p and im p

I know ker p \cap im p={0}

[warm wyvern]

how do I prove this tho?

[rotund aurora]

span is always guarranteed no?

oh yeah

bruh

dont say it

[chilly ocean]

glad you know it, illuminator

[rotund aurora]

You have this in general

[formal ermine]

o is that too much of a hint

sorry

[chilly ocean]

it's not a hint it's the whole fucking answer

[formal ermine]

oh

sorry

[rotund aurora]

how do you show that x^4-14x^2+9 is irreducible? I could show it by pretending it to factor as two quadratics, since obviously it has no roots, and modulo p it seems to be reducible always (at least for small primes). Is there an other way?

[formal ermine]

you could try eisenstein with sub

[rotund aurora]

sub?

ah

[formal ermine]

substitution

[rotund aurora]

what substitution

I don't think that works

[next obsidian]

Do you know the splitting lemma?

[formal ermine]

I think u = x - 1 works

[next obsidian]

But you don’t even need that

recall from group theory that if you have subgroups which intersect trivially, and generate the group, then G is the internal direct sum of the two subgroups

[rotund aurora]

you get this

[next obsidian]

Prove the same thing is true for modules, then prove that this is satisfies by the im and kernel of p

[formal ermine]

typo

I mean 2 not 1

wait

that doesn't work either

it's a 31

oops

[rotund aurora]

yeah lmao

[hot lake]

You could try looking at factorizations of the form (x^2+ax+b)(x^2-ax+b)

Or (x^2+a)(x^2+b)

[rotund aurora]

yeah that's what I did. But was wondering if something else could be done

[hot lake]

There should be a prime where it is irreducible I think but I can't really check anything for now

[rotund aurora]

I did check in wolframalpha. There was none

[delicate orchid]

this one's a classic

[hot lake]

Ah are the roots just +-sqrt2 +- sqrt5

[hot lake]

Ah yeah it never stays irreducible

[rotund aurora]

btw, the Galois group of that polynomial is just Z2 x Z2 right?

[hot lake]

Yeah

[rotund aurora]

Let p be a polynomial of degree n. Can you say something about the size of the Galois group (of the splitting field of p)? So of course you know that its divisible by n and it divides n!, but are there any other restrictions?

[south patrol]

hm, to be slightly picky the Galois group order needn't be divisible by n unless p is irreducible

[south patrol]

But yeah I mean you can find examples when each the galois group has order n (e.g. cyclotomics) and examples where it has order n! (e.g. the "typical" quintic has galois group S5)

So you'd need to know more about the polynomial to make any better conclusions

[rotund aurora]

oh yeah, I forgot to include that condition

[delicate bloom]

if we leave it as not necessarily irreducible, is it possible it divides a number with factors not in n!?

[rotund aurora]

Let me reformulate the question: Let m be an integer that is divisible by n and divides n!. Does there exist an irreducible polynomial p of degree n whose Galois group has size m ?

[south patrol]

regardless of the properties of the polynomial, you will still get a faithful action of the Galois group on the roots of the polynomial

(so that G embeds in S_n and hence |G| divides n! anyway)

[subtle flame]

in vector space R[X], is this true: (p+q)(x) = p(x) + q(x) ?

[formal ermine]

yes, that's how we define a vector space structure on R[x]

[subtle flame]

thanks

[long nebula]

can a submodule of a finitely generated module be non-finitely generated?

I can't see any obvious reason why the submodule would have to be finitely generated

[rotund aurora]

yeah

[chilly ocean]

look at a unital ring as a finitely-generated (by 1) module over itself

easy examples follow

[long nebula]

grrrr

[chilly ocean]

the submodules are ideals btw

[long nebula]

okay yeah I see, I just need to figure out how to prove that submodules of a finitely generated module are finitely generated in the case where R is noetherian then

[rotund aurora]

there are many stupid examples like F[x1,x2,x3,...], and look at the submodule generated by x1,x2,x3,x4... (over F[x1,x2,x3,...])

[long nebula]

Oh I see

That makes sense

So the key is that I have to use the fact that R is noetherian somewhere hmmmmmm

oh wait this is in dummit-foote

[rotund aurora]

its not hard to prove

[long nebula]

okay I set up a list of conditions that should be equivalent and I proved that 2=>3 and 3=>4, time to prove the other implications..

[rotund aurora]

How do I show that the Galois group of x^4-2x^2-2 is D4 ?

I figured it out I think

[delicate bloom]

I guess if r is a root, then -r is also a (distinct) root, so you have the roots {r, s, -r, -s}, then if sigma(r)=s I think this basically fixes a bunch of stuff

[rotund aurora]

yes

[warm wyvern]

I saw the proof for that not too long ago actually

[long nebula]

I think you need to do something w quotients of free modules

[rotund aurora]

that way is easier, yes

[warm wyvern]

yea

the proof relies on the fact that ||if R is noetherian then so is the free finitely generate module||

[long nebula]

okay so now this reduces to proving the acc for free modules of rank n hMMM

I guess we just use the fact that the rank is at most n

[rotund aurora]

look at R^n

[long nebula]

yeah

if the chain isn't eventually constant then the rank would continue increasing forever

I think

okay yeah

finished my hw 🥳

[rotund aurora]

cool

you don't need to use the acc, you can just ||look at the first coordinate of the vectors in R^n and realize its an ideal, then blah blah induction ||

[warm wyvern]

what's the acc?

[rotund aurora]

ascending chain condition

[long nebula]

oh yeah, I was proving that four conditions were equivalent

that makes sense though

(if R is noetherian)

[warm wyvern]

I'm not sure if your book mentions this but if every submodule of M is finitely generated then it's called a noetherian module fwiw

which I find cute

[rustic crown]

(what's fwiw?)

[long nebula]

for what it's worth

[warm wyvern]

for what it's worth

sniped

[rustic crown]

[long nebula]

fwiw d+f defines noetherian modules using the ACC lol

ugh I wrote a proof down but I can't find where I used the fact that R is noetherian

uh oh

[rustic crown]

first 3 are equiv for any ring

[long nebula]

oh I see what I did wrong

accidentally assumed that R is a field 😭

[rustic crown]

[rotund aurora]

[rustic crown]

that's a huge accident

[rotund aurora]

what is a field?

lmao

[warm wyvern]

lmfao

[long nebula]

Lol I wrote that if m_1, m_2, ..., m_n, m' is linearly dependent, then m' is a linear combo of the others

Oops

[rustic crown]

//headpat

[warm wyvern]

M can just be Z, right?

[rustic crown]

over what ring

[warm wyvern]

over Z?

lol

[rustic crown]

then it's just free

[long nebula]

Lol

Prove a contradiction and then use principle of explosion

[rustic crown]

🤯

[long nebula]

hmm wait is it true that if (M \subsetneq N) and the rank of (N) is finite, then the rank of (M) is strictly less than the rank of (N) over a Noetherian ring?

[cloud walrus]

lebesgueric (he)

[long nebula]

clearly this isn't true for a general commutative ring

[rustic crown]

is N free?

[long nebula]

no

[rustic crown]

else consider M = 0, N = Z/2Z over R = Z

[long nebula]

submodule of a free module tho

oh wait

that is free nvm

[warm wyvern]

I thought I was working in Ring and thought Z is initial

[long nebula]

okay just gonna erase this and start over lol

[rustic crown]

still not true 2Z < Z for R = Z

[long nebula]

pain

[rustic crown]

//patpat

[agile burrow]

are you trying to say Hom_{Z-Mod}(Z, Z) = 0?

[rotund aurora]

you can consider Z and 2Z, both over Z. The inclusion 2Z in Z is proper and 2Z and Z have both the same rank

[rustic crown]

hewwo walter

[agile burrow]

hi det

[rustic crown]

[rotund aurora]

(lmao its what det said, didnt read)

[warm wyvern]

yes lmfao

[agile burrow]

that is not true

[warm wyvern]

I know haha

brain fart

[long nebula]

okay just gonna show the ACC via induction on n for R^n then

like croqueta said

[rustic crown]

nuuu

.<

[long nebula]

😭

[delicate orchid]

yus

[rustic crown]

oh i mean you can do that

.<

but...

.<

[delicate orchid]

we showing R^n artinian or something of the sort?

[long nebula]

noetherian

if R is

[rustic crown]

do this instead... if M is a submodule of N such that both M and N/M are notherian, then N is noetherian

[warm wyvern]

oh, I know

[long nebula]

isn't this still just induction on n

[warm wyvern]

M=Z/nZ and R=Z

[long nebula]

R^(n+1)/R = R^n

[rustic crown]

yee >.<

[warm wyvern]

:)

[rustic crown]

but... didn't have to do the argument specifically for R^n with all the messy n+1 and n's >.<

[delicate orchid]

:)

[rustic crown]

🙂

this emoji looks like it's in pain

[agile burrow]

what lies beneath those eyes

[warm wyvern]

in several of my friends groups there's this inside joke that 🙂 is a psychopath

[next obsidian]

Use Hilbert basis theorem and that Noetherian is closed under quotients

[warm wyvern]

lmao

[rustic crown]

does that work tho?

you only get noe as an R[x] module or something

[next obsidian]

If you’re a finite algebra you’re finite type too

[rustic crown]

[next obsidian]

Or are you trying to show it’s Noeth as an R-module lol

It probably still follows

[rustic crown]

(only problem being hilbert basis theorem uses R^n is noe inside its proof >.<)

[rotund aurora]

you would quotient in R[x] by x^n right?

[next obsidian]

I don’t think it does?

You use equiv of ACC and fg ideals

Then take a chain of ideals in R-[x]

Look at some ideals of leading coefficients, use that that stabilizes

Then conclude the chain of ideals stabilizes in R[x]

[rustic crown]

how do you conclude the final step here

[next obsidian]

You do some dumdum argument

[rustic crown]

[next obsidian]

You are able to show that at the step the ideals of coefficients stabilize

So too does the ideals in the poly ring

https://proofwiki.org/wiki/Hilbert's_Basis_Theorem

Here’s a proof I took at random

It is slightly different but

Same general idea

No use R^n is Noeth

[rustic crown]

i think that would use R^n is noetherian module, because leading coefficients will only tell you what happens beyond a certain degree

[long nebula]

😭 how is that messy

[warm wyvern]

weird that I heard about hilbert's basis theorem literally an hour ago

[rustic crown]

it's not >.<

i'm just weird >.<

[long nebula]

What was hilbert's original nonconstructive proof of the basis theorem?

I asked my algebra prof and he didn't know

[next obsidian]

It definitely doesn’t

Just go read a proof it’s never used

[rustic crown]

yea makes sense

they do the weird thign with the minimal degree

[rotund aurora]

Why is he requiring D>0 ? Isn't the same also true for D<0 or am I missing something?

[rustic crown]

the proof i had in mind was slightly different, if I is an ideal in the big ring and J is the ideal of leading coeffs, then J is fg so pick polys (f_1, .., f_n) whose leadings generate J. giving any arb poly with degree very large, you can use these to decrease the degree, and if the degree is small enough we use R^n is noe to add finitely many more thigns to do the work

good to know

[rustic crown]

yea i don't see much difference, except the obvious that you want |D| to be not a perfect square

[delicate bloom]

D<0 makes the units in the ring Z[sqrt(D)] change and now you have a fundamental solution to pell's equation along with all its powers with norm 1, I'm guessing that's the issue

[rustic crown]

maybe, but i think, since we're working with Q(sqrt(D)), it doesn't change much?

[rotund aurora]

yeah

you still have (s^2+Dt^2)^2-D(2st)^2=(s^2-Dt^2)^2

[delicate bloom]

but there are more solutions, so it's not exhaustive anymore

[rotund aurora]

you show there are not more solutions by Hilbert 90, no?

what other solutions are there?

[delicate bloom]

pell's equation gives you integer solutions to x^2-Dy^2=1

[rotund aurora]

yes, but aren't those included in that parameterization?

[delicate bloom]

I'll try to find an example real quick

[rotund aurora]

(1,0) is clearly a solution and you can just produce all solutions by drawing all lines with rational slope through (1,0). Although now its a hyperbola, not an ellipse. Maybe I'm missing something

[delicate bloom]

so let's pick x^2-2y^2=1, we have (3,2) as a solution, can we write this as that form?

[rustic crown]

s, t = 1, 1?

[delicate bloom]

a=(s^2-t^2)/(s^2+t^2) = 0 not 3

[rustic crown]

a = (s^2+Dt^2)/(s^2-Dt^2) with D = 2

[delicate bloom]

oh I'm using the wrong form

[rustic crown]

ig technically this gives (-3, -2) but shouldn't be too far off

[delicate bloom]

how does it generalize

oh I see, you wrote with -D where I was expecting +D

what's b end up being?

[rustic crown]

(i just copied croqueta's calculations 🙈)

[delicate bloom]

oh ok

something's still not smelling right lol

[rustic crown]

hilbert 90 is weird

[rotund aurora]

if you apply Hilbert 90, the parameterization is clear

and you can apply it, because the Galois group is cyclic

[rustic crown]

over Q stuff is so much simpler as compared to over Z lol

[rotund aurora]

and if you do the chord method, you can probably produce all solutions

[delicate bloom]

(s^2+Dt^2)^2-D(2st)^2=(s^2-Dt^2)^2
so we have
s^2+Dt^2=3
2st=2
s^2-Dt^2=1
with D=2, is that right?

[rustic crown]

i hope so

[rotund aurora]

yeah

[rustic crown]

so you had caculated (s + sqrt(D)t)/(s - sqrt(D) t)?

[delicate bloom]

2st=2 means s=t=+-1 and s^2-2t^2=1 becomes 1-2=-1 though

[rotund aurora]

and took the norm, yes

[rustic crown]

but st rational

[rotund aurora]

wait

[delicate bloom]

oh s=1/t ok

[rotund aurora]

you don't need s^2-Dt^2=1

[delicate bloom]

thinking

[rotund aurora]

you have to divide by that

and so the equation is equal to 1

[rustic crown]

you want (s^2+2t^2)/(s^2-2t^2) = 3

and 2st/(s^2-2t^2) = 2

[rotund aurora]

The parameterization for $x^2-Dy^2=1$ should be (I think):
[
(x,y)=\left(\frac{a^2+2b^2}{a^2-2b^2},\frac{2ab}{a^2-2b^2}\right)
]
for any $a,b\in\bQ$.

[cloud walrus]

Croqueta

[rotund aurora]

yes, sorry

you can check all these are solutions

[delicate bloom]

so now (3,2) is now on there so we should be able to solve for it

[rotund aurora]

ok let's try lol

do chords

[rustic crown]

(s,t) = (2,1)?

works i think

[delicate bloom]

yeah that works

ok good enough for me

[rustic crown]

[delicate bloom]

I think I see now in general this will work, it just won't be the cleanest form since it looks like it's gonna always be a reducible fraction that pops out

like with primitive pythagorean triples, the side lengths are all relatively prime numbers, I guess that might be part of the D>0 constraint that they have in mind, idk I gotta work on other stuff right now but good to know

[rotund aurora]

Consider the equation $x^2-Dy^2=1$. Clearly, $(1,0)$ is a solution to this equation. Consider the line that passes through $(1,0)$ with rational slope $m$: $y=m(x-1)$. We will obtain the point of intersection with this line and the hyperbola. First, solve for $x$:
[
x^2-Dm^2(x-1)^2-1=0
]
which is clearly divisible by $x-1$. This factors as
[
(x-1)(x(1-Dm^2)+1+Dm^2)=0,
]
and we obtain the solution
[
x=\frac{-1-Dm^2}{1-Dm^2}
]
and
[
y=x-1=\frac{-2}{1-Dm^2}.
]

[cloud walrus]

Croqueta

[rotund aurora]

if you write m=b/a, you obtain the above parameterization, but it's weird because I'm getting the above but with x,y multiplied by -1

uh wait this is not right

oh wait

y=m(x-1) not y=x-1

then everything is fine

but still weird because everything is multiplied by -1

[rustic crown]

so much algebra >.<

[rotund aurora]

ancient stuff

so cool

[rustic crown]

i think i lost the ability to do that

[rotund aurora]

btw, why is HIlbert 90 called "90" ?

[rustic crown]

it's the 90th theorem in somethin

[delicate bloom]

I think it was like the 90th theorem in some book he wrote lol

[rustic crown]

the 90-th theorem in Hilbert’s Zahlbericht, a report
on the state of number theory at the end of the nineteenth century commissioned
by the German Mathematical Society

imagine there are 89 theorems before it

and even more after it probably

people know so much >.<

[toxic zephyr]

so in my abstract algebra class we talked about modules, but we never talked about matrices over modules. how does the invertibility of matrices change when over a ring rather than a field? specifically, i'm thinking about what i can and cannot say about an invertible matrix A over Z/nZ instead of Z/pZ

over Z/pZ i'm pretty sure everything that's true in R or C applies... like with respect to the determinant being nonzero and being surjective/injective/etc.

[rotund aurora]

a matrix A is invertible if and only if det(A) is a unit

[toxic zephyr]

so gcd(det(A),n) has to be 1?

[rotund aurora]

if you are working in Z/nZ I suppose that's the case

This is true over any commutative ring

and if det(A) is a unit, you can divide

the other direction is trivial, you can try to prove it

[toxic zephyr]

wow that... is like a perfect generalization from the 1d case i was working with originally...

[rustic crown]

i'm a unit

[toxic zephyr]

that makes perfect sense!

adjugate matrix comes in clutch once again

alright so that's good. so if the inverse exists, then it's still both injective and surjective, right?

like the kernel is still trivial?

[rotund aurora]

yes

mmh

it need not be surjective

if you are looking at matrices as linear operators on a space of vectors (by left multiplication), if you are over a ring, surjectivity will be lost most of the time

for example, in Z, consider left multiplication by 2

oh wait

but 2 is not invertible what am I saying

I guess then what you said is still true

[toxic zephyr]

yeah if it's invertible then it's torsion free right? i think that's the right term that applies

[next obsidian]

What, if a matrix is invertible it defines a surjective map

[rotund aurora]

a map that is invertible is bijective always btw

yes, by brain collapsed for a sec

[next obsidian]

What does this mean

[toxic zephyr]

haha all good

[next obsidian]

A map doesn’t have torsion

In any definition of torsion I know

Do you mean like as an element of the module of matrices maybe?

[delicate bloom]

maybe they're thinking invertible matrices don't have finite order?

[toxic zephyr]

i think the matrix is acting as a scalar

[next obsidian]

Acting on the matrices via scalars?

If that’s what you mean then you’re right

[rustic crown]

[toxic zephyr]

this is the group
$$G=\bZ\ltimes\paren{\bZ/p\bZ}^k=\gen{s^av\mid v\in\paren{\bZ/p\bZ}^k,,a\in\bZ,,s v s=M v}$$
and im essentially working with a function w that does this

[next obsidian]

Like are you trying to make M_nxn(R) into an R-module?

[cloud walrus]

nilpotent nix

[toxic zephyr]

so i'm looking at the invertibility of the matrix pa(M) and how that can potentially make w a bijection

i was originally working over Z/pZ but now i'm moving on to Z/nZ

i have a proof that i think works for the former, but i wasn't sure how much i have to change my argument for the latter

there's a lot more to it... but that's the very basic gist

[delicate bloom]

decompose Z/nZ into Z/p^kZ x .... by CRT then look at inverting in Z/pZ and then go back up to Z/p^kZ by hensel's lemma

or maybe just think about solving it in Q_p since it's a field, then restrict down to Z/p^kZ again

[toxic zephyr]

googled "CRT algebra" and got "critical race theory"

chinese remainder theorem?

[delicate bloom]

cathode ray tubes

[toxic zephyr]

😆

[rotund aurora]

Let F be the splitting field of x^n-2 over Q. Let z be a primitive nth root of unity and consider the automorphism of F that sends nrt 2 -> z nrt 2 and fixes everything else. The subgroup of Gal(F/Q) generated by this automorphism is Z_n. Let K be the fixed field of this subgroup, then Q<K is Galois and Gal(K/Q)=Z_n.

[toxic zephyr]

i'm writing that down so i can ask my mentor at the next meeting if that could apply, but it's possible that having the gcd of the det with n being 1 is the if and only if i need, since M/pa(M)/a/n are all pretty arbitrary and tough to compute in general.

[rotund aurora]

Is this correct?

[delicate bloom]

yeah more than one way to do it I think, idk I didn't really understand the original problem you were doing but this seems like a rough way of vaguely splitting it up to make it simpler probably

[rustic crown]

shouldn't that be units mod n?

[delicate bloom]

it's Z/nZ cause it's addition in the exponents of z^k

[rustic crown]

the galois group is the semidirect product of Z/nZ and (Z/nZ)*

z can go to any other primitive nrt1

[rotund aurora]

uhh wait

[delicate bloom]

yeah, I see I think

[rotund aurora]

right

[rustic crown]

so Z/nZ is a normal subgroup so by fundamental theorem, gal(K/Q) would be the quotient Z/nZ*

[delicate bloom]

not an isomorphism if you're not doing units

sigma_b(z^a)=z^(ab) basically derp my bad

if b isn't invertible mod n, we're screwed

[rustic crown]

so fixed field is just Q(z) right lol

[rotund aurora]

Let F be the splitting field of x^p-2 and let z be a primitive p-th root of unity. The map prt 2-> z prt 2 is then a valid automorphism no? And this automorphism has order p, so this way we can produce Z/pZ as a galois group, no?

[rustic crown]

yea, but that won't be over Q

Gal(F/Q(z)) = Z/pZ

but you've come quite close

because Gal(Q(z)/Q) = Z/(p-1)

[rotund aurora]

oh okay I see

[formal ermine]

I could swear you helped me solve this before lol

[rotund aurora]

your exercise was showing that the galois group of x^p-2 is Z_pxZ_p

I think

[formal ermine]

Zp*

[rotund aurora]

yeah that

this lmao

[formal ermine]

yeh

[south patrol]

Aff(F_p)

😎

[celest cairn]

How do I show that $\mathbb{Q}\left(-\sqrt{2}\right) = \mathbb{Q}\left(\sqrt{2}\right)$?

[cloud walrus]

Sapphire

[next obsidian]

Lol

[chilly ocean]

show that one side is contained in the other

[oblique river]

here's a place to start

try to write down an element which is in one set but not the other

you won't be able to do that (because they are indeed equal) but that might help you realize why this is true

[rustic crown]

terra and buncho so kind >.<

meanwhile :p

[south patrol]

ok i assume this is fine here but my course in ant defines trace/norm of elements a in a number field K in terms of the images of the a under the embeddings into C. is there any point doing this when one could equivalently talk about the other roots of the min poly (or equivalently the trace/det of the map x -> ax)

i suppose the only thing is i know that in some proofs when we're actually working with it it can be helpful to use the fact that the embeddings are field homs but ye

[rustic crown]

ig transitivity becomes easier to prove with the definition using embeddings

[sonic coral]

if i have a group with 6 elements, four of which are their own inverses

then the other two are forced to be each others inverses right

[delicate bloom]

they could also be their own inverses too, right?

idk didn't think it through lol going to sleep

[sonic coral]

well they have order 3 so they aren’t

should’ve mentioned that

[tender wharf]

let's say d and f are the 2 which have order 3

e, a, b, c all are their own inverses

[tender wharf]

what is bc

bc doesn't have to necessarily be the identity

let me think

[sonic coral]

it is e

[tender wharf]

did you check

by matrix multiplication?

[sonic coral]

yes and addition mod 2

[tender wharf]

well

then sounds like you have your answer

[sonic coral]

well i know that, but i guess i was more concerned with can i just assume that from the start

[tender wharf]

well its an order 6 group so it's isomorphic to D_3

(i mean it's obviously not isomorphic to Z_6)

[sonic coral]

knowing that 4 elements have order 2 or less, leaving the remaining 2 to be inverses of each other

we havnt covered isomorphic yet. i just know the definition bc i read ahead

[tender wharf]

oh actually yeah you are right

let r be the rotation in D_3

so {e, r, r^2} is the subgroup of rotations

yeah

[tender wharf]

you could not assume it no

but you didn't have to assume it

since all you had to do is to perform b * c

[sonic coral]

no it’s easy enough to check

[tender wharf]

yup

[sonic coral]

but the way i first thought about it was that it was forced

i don’t know enough examples of groups to think of a counter example though

[tender wharf]

that's because there isn't one

[sonic coral]

so then it can be assumed 🤣

[tender wharf]

well

no because you haven't proven that classification theorem

[sonic coral]

maybe i just have a great intuition then

[tender wharf]

although if it's Z_6

[sonic coral]

it makes sense in my head but obviously i would be better off to show it

[tender wharf]

lemme think

yeah

it's actually true for Z_6 too

[sonic coral]

i’m not familiar with the group Z

only the dihedrals

[tender wharf]

oh that's just addition modulo 6

[sonic coral]

oh yes Z mod 6

[tender wharf]

I'm using theorems you haven't learnt yet I'm assuming

[sonic coral]

yeah i’ve only sat down with my professor once

we covered all of chapter 2 of gallian

[tender wharf]

ah okay

what I said is in chapter 7

[sonic coral]

but he kinda gave me some definitions from chapter 3 too i saw

[tender wharf]

for c)

what I'd do is provide some additional justification

like actually carry out b * c

[sonic coral]

for bc

[tender wharf]

yup

[sonic coral]

i’ll ask him also about my thoughts

[tender wharf]

what thoughts

[sonic coral]

of just taking b and c to be inverses by exhausting all other possibilities

and the fact that their order is 3

[tender wharf]

for this group sure you can brute force it

[sonic coral]

what about showing it’s abelian, elements in the group must commute. which i think they do. i just don’t know how to show it unless i’m supposed to do every combination which can’t be right

[tender wharf]

ooh I don't think this is abelian

one sec

[sonic coral]

let me look harder then

[tender wharf]

yeah it isn't abelian

[sonic coral]

yup found it

[tender wharf]

I remember doing this question by brute force actually

[sonic coral]

a c

!=ca

how do you prove a group is abelian

like an arbitrary one

[tender wharf]

let x, y be from group arbitrary

then show xy=yx

[sonic coral]

this one we have that g^2 = e

and so then that implies that g^-1 = g right

then i should take a,b in g and show ab = ba

but ab = (ab)^-1 ?

which is b^-1 a^-1

or ba

[delicate bloom]

both

just change your "or" to "="

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

[sonic coral]

perfect

[delicate bloom]

cause you're saying a=a^-1 and b=b^-1 right

[sonic coral]

yes of course

singe a^2 = e anf b^2 = e as well

[long nebula]

how do you show that something isn't free, like showing that R isn't a free Z-module for example

unless it is one

showing that there exists a basis seems a lot easier than showing that there isn't one

[next obsidian]

If you can exhibit torsion that can’t exist, that’s an easy one

[long nebula]

right

[next obsidian]

But for stuff like showing Q isn’t free (projective), you might try to construct a map that you can’t lift

[long nebula]

okay I guess I should read about projective modules then

[next obsidian]

Literally doesn’t matter

Just use this property

If M is free, given any map M -> N and a surjection N’ -> N, there exists a map M -> N’ making the triangle commute

You can do this by just looking where a basis of M goes in N, then lift this to elements of N’

Send the basis to those lifted elements and the triangle commutes by construction

[long nebula]

oh huh interesting

[next obsidian]

(Projectivity is just the property that you can always lift along surjevtions like this, by definition)

With Q you can construct maps which would prove Z contains something which sums with itself to say, 1

So Z “contains 1/2”

[long nebula]

ohhh

[next obsidian]

But this is clearly bs

Do you see how?

[long nebula]

hmm

[next obsidian]

Or okay, maybe this is easier hahaha

Let’s say Q, or R was free

Free modules admit maps into any abelian group right?

Just send a basis to stuff

[long nebula]

wdym?

[next obsidian]

Free modules admit nonzero maps

[long nebula]

oh

[next obsidian]

Into every abelian group

Or let’s say module

Actually

[long nebula]

okay makes sense

[next obsidian]

Lol

But they admit tons

So

How many linear maps R -> Z exist?

[long nebula]

I feel like you can send any generator of R to any element of Z?

[next obsidian]

Well I mean by definition of free module, yeah

[long nebula]

so like Z^(rk R)?

[next obsidian]

If you have a basis B, Hom(F,M) is in bijection with B^M

You just choose where to send each basis element

Then extend linearly

The same way you do for vector spaces

[long nebula]

wait do you mean M^B

[next obsidian]

Uhh

Yes

I do

[long nebula]

okay cool

[next obsidian]

So R should admit tons of maps to Z

But…

What if n≠ 0 was in the image

[long nebula]

if phi(q) = n, then phi(q/n) = 1, so phi(q/2n) doesn't have anywhere to go

right

[next obsidian]

Yeah

[long nebula]

ohh

[next obsidian]

You’d have 1/2

[long nebula]

so that means Q can't be free

[next obsidian]

Yeah

[long nebula]

and similarly R can't be

[next obsidian]

Yes

[long nebula]

this makes sense 🤯

[next obsidian]

And over Z this is the same as projective is the same as flat is the same as torsionfree

[long nebula]

neat!

this makes total sense ty

[next obsidian]

Anyway

Once I showed something wasn’t projective by like

Being projective is the same as saying any surjection M -> P splits

So you can find a linear… right inverse?

And this says that M is isomorphic to P (+) kernel(M -> P)[specificallt you write it as an internal direct sum of the kernel and the image of P under the splitting map, which is injevtive!]

And I just found some map where you could show that it couldn’t possibly split because somehow M just couldn’t be the direct sum of those two guys

[sonic coral]

ive shown => but need help with <=

how can i manipulate (xy)^2

[next obsidian]

xyxy = xxyy

What can you do

[long nebula]

that's super cool

[sonic coral]

multiply on the left by x^-1 and on the right by y^-1?

[next obsidian]

Yeah

Then you get?

[sonic coral]

or just look and see that xy=yx in the middle by associativity

idk if thats as valid

[next obsidian]

Uh

Do what you said originally

Idk what that other thing means

[sonic coral]

lol

[next obsidian]

But yes once you do that you have yx = xy

Okay GG now you can go home to your family

[sonic coral]

like if xyxy=xxyy, then x(yx)y=x(xy)y, so yx=xy

or is that not how that works

[next obsidian]

Yeah but how do you justify that

[sonic coral]

same way i said it first

[next obsidian]

You do it by multiplying by x^-1 on the left

And y^-1 on the right

[sonic coral]

its so obvious once you point out what you said though i hate proofs

its just right there

lol

[next obsidian]

🗿

[tender wharf]

eh you learn after a while

[marsh goblet]

kind of a silly question but I was wondering if, given B normal in A, is A/B oplus B isomorphic to A?
i feel like it should be - my idea was to map (a+B,b) |-> ab but it doesn't appear to be well defined
or at least idk how to prove that it is

[formal ermine]

what's an intuition behind the direct sum

[sharp sonnet]

are you mixing additive and multiplicative notation

[marsh goblet]

Yes I am so sorry

[sharp sonnet]

maybe fixing this will help?

[marsh goblet]

I had the right notation in my own working, still don’t get it though

[sharp sonnet]

oh right