groups-rings-fields

[dim widget]

I think n > 6

[twilit wraith]

oh we are getting to symmetric groups soon

i know thats essentially the heart of abstract algebra

[next obsidian]

Sounds right

[twilit wraith]

group theory i mean

since it started with analyzing permutation groups

[dim widget]

Ah wait that’s just outer automorphisms

To have no automorphisms at all you have to be Abelian

[twilit wraith]

man ive been kinda struggling with this class so far but this stuff is so cool

[dim widget]

So then probably you do have to be Z/2

[next obsidian]

Oh yeah lol

Wait Aut(S_n) is usually S_n

Lol

What about if you’re infinite abelian tho

I guess also like

Non-finitely generated

[dim widget]

I think you still have the problem of inversion

x \to x^-1

[next obsidian]

Oh yah

True true

[dim widget]

So you have to be all 2 torsion

[next obsidian]

Yup

[dim widget]

Then i think it’s easy to see you can only be Z/2

[twilit wraith]

wait

[next obsidian]

Then vector space swap basis elements

[twilit wraith]

theres something to do with counting the number of proper divisors that has to do with having a trivial automorphism group i think

[glad osprey]

Nice, I think you just reproduced the proof here: https://math.stackexchange.com/questions/8379/g2-implies-g-has-non-trivial-automorphism

[twilit wraith]

i think its just that if the order of a group equals the number of proper divisors of its order then its automorphism group is trivial

maybe not even proper

[next obsidian]

Well TTEG and I just showed that it’s only e or Z/2Z

So idk

[twilit wraith]

ah i see

[next obsidian]

If anything this result just sounds like it can tell you something that is not possible for most groups

[twilit wraith]

so no number greater than two is equal to the number of its divisors

[twilit wraith]

yeah

i would imagine for n > 2 that its always less

[velvet hull]

The implicit assumption is that it’s finitely generated, no?

[next obsidian]

No

[velvet hull]

Because yall just wrote out the exact same thought process I had

[next obsidian]

It’s abelian because of conjugation

[dim widget]

no this argument has nothing to do with being finitely generated

[next obsidian]

And then inversion has to be trivial so everything is order 2

So it’s a vector space over F_2

Then you swap basis elements

[velvet hull]

Mmmh okay

[twilit wraith]

oh btw, to get in A in the algebra class im taking, you have to do some sort of project on a topic related to abstract algebra, and im wondering if you all know anything cool that could reasonably be covered in a first course

be presented rather

[next obsidian]

Only group theory?

[twilit wraith]

i think rings are at the very end

but i think groups are the most reasonable

if we cover enough about rings i might be inclined to do smth about galois theory

since i know the basic idea of the fundamental theorem of galois theory

[next obsidian]

Maybe you can talk about Lie groups or something

If you have the background to do so

You have the problem of needing to get people on board with what a manifold is but like

You could talk about group actions and quotients of them for topological spaves

This is pretty cool because it’s a pretty central topic in algebra, and being able to kind of nicely do that in topology using these algebraic gadgets is nice

And also very useful in practice

[twilit wraith]

oh yeah

i mean i dont know much about diffgeo

[next obsidian]

You don’t really need to know much I think

And depending on what “presentation” means

You don’t need to explain why something works

Like to say “projective space is obtained by quotienting K^n+1 \{0} by an action of K*”

Where K is R or C

And then maybe say why quotients in topology can be whack, like uhhh

[velvet hull]

Ping pong lemma, banach tarski paradox

[next obsidian]

A manifold locally needs to look like R^n, but once you do a quotient this can easily be broken

But hey, in this nice situation it still looks like R^n

You could also talk about homology very very vaguely, but this again requires some background on your end

But I think even the question of

“Take a donut and a donut with two holes, these are clearly different with our eyes, but how can we make this precise?”

Well you can associate something called homology which spits out groups and in one case you get Z and in the other you get Z^2 or whatever

[dim widget]

Or you could talk about stuff that has nothing to do with topology

Since its an algebra class

[twilit wraith]

yeah but like

algebraic topology is super cool

[next obsidian]

I just find it hard to think about a topic to present on that isn’t Galois theory or rep theory lol

Unless you want to be like “hey we didn’t talk about semi direct products” or something

[kind temple]

what about the jordan-holder program and composition series?
i think it stays within the realm of group theory. i also thought it was cool to see for the first time

[dim widget]

one cool purely group theoretic theorem that has a very beautiful proof which uses only basic group theory (but in a fiendishly clever way) is the following

if G is finite and Z(G) = 1 then the chain of finite groups Aut(Aut(…(Aut(G)))…) eventually stabilizes

an easier version is that if G is simple nonabelian then Aut(Aut(G)) = G

[twilit wraith]

oh that is pretty cool

[twilit wraith]

if Z(G) isnt trivial then does that chain eventually just become isomorphic to Z(G)?

nothing is really motivating that claim just a guess

[dim widget]

The chain doesn’t become trivial it just stabilizes

So like at some point applying Aut n times is the same as applying it n+1 times

I think if the center is nontrivial the result can fail

Though I don’t see how right now

[twilit wraith]

oh i see

[rocky cloak]

Krull-Schmidt theorem and Jordan-Hölder are nice classification type theorems gir finite groups

[chilly ocean]

Unipotent matrices is maximal subgroup of invertible upper triangular matrices

[rocky cloak]

Not maximal. But it is a subgroup

[chilly ocean]

If I allow to vary first diagonal entry to be non zero real then , it contains unipotent subgroup

[rocky cloak]

For example

[chilly ocean]

i maybe wrong can you tell me the intermediate subgroup

[rocky cloak]

What do you mean? You already gave one example, do you just want more?

[chilly ocean]

okay thanks one is enough for now.

[rotund aurora]

do you know an example?

[thorn jay]

There is a conjecture that it will at least always fall into a loop

[rotund aurora]

There are two notions of "stabilizes". If you ask that the inner homomorphism G-->Aut(G) be an isomorphism at some level, then G=D_8 never stabilizes. However, Aut(D_8) is abstractly isomorphic to D_8. So the notion of "stabilizes" that I'm asking for is whether G is abstractly isomorphic to Aut(G) at some level

[thorn jay]

I doubt that being the case; I think there must be a group where it's periodic

[rotund aurora]

is this true? People have asked whether this is the case, but is there any evidence supporting this conjecture?

[rotund aurora]

it seems to be unkown

[thorn jay]

We know basically nothing about automorphism towers

Besides theorems about centerless groups

[crystal vale]

I have to show that if G is a group of odd order and H is a normal subgroup of order 5, then H \subset Z(G).

Any hint?

I know if I can show that any non-identity element of H in Z(G), then we are done.

But I don't know where I have to use the hypothesis that G has odd order

[chilly ocean]

well please check it We have a map from G to aut(H) taking an element g to an conjugation map by g , check if this is surjective and order of aut(H) is 4 so 4 divides |G| which cant happen similarly if image is of order 2 , 2 will divide |G| now remaining is trivial map hence which means conjugation by any element fixes H, implies for any h in H and g in G ,ghg^-1=h or in other words every element of h commutes with any element of G , so H is subset of Z(G)

[crystal vale]

Great, how did you come up with this idea?

[chilly ocean]

I am dealing with group actions nowadays, similar things are in dummit foote group actions chapter .

[crystal vale]

I see, can I DM you?

[valid fox]

If two elements alpha and beta have the same minimal polynomial over a field K, is K(alpha) isomorphic to K(beta)?

say like $i\sqrt[4]{5}$ and $\sqrt[4]{5}$ over $\mathbb{Q}$

[cloud walrus]

Tiessie

[glad osprey]

Yes, they would both be isomorphic to K[x]/(p(x)) where p(x) is the min poly of alpha and beta

[valid fox]

thanks thats what I thought as well

I felt like the fields in the example I gave would be very different but I guess not

[glad osprey]

Interesting is this also true if H has order p for an odd prime p?

[chilly ocean]

In that case p-1 can have odd factor

[glad osprey]

Oops, right, thanks

[glad osprey]

I guess it's kinda confusing that Q(5^(1/4)) is a subset of R while Q(i5^(1/4)) isn't, but they actually are isomorphic, they're just embedded in C differently

[chilly ocean]

r u doing galois theory recently

[glad osprey]

Yep, did it this spring! One of the most fun courses I have done so far

[chilly ocean]

undergrad or grad

[glad osprey]

undergrad

[chilly ocean]

nice ,

[glad osprey]

So not too difficult, we followed Bhattacharya for the most part

[chilly ocean]

my prof made that difficult

asking to calculate lattice structure

[crystal vale]

You are doing a PhD?

[chilly ocean]

yes, in phd we have to do a freshman course to recall everything

whole algebra were covered

[glad osprey]

You mean like the lattice of subfields?

[chilly ocean]

hmm

[crystal vale]

And your research area?

May I get to know which university you are in?

[chilly ocean]

ergodic theory

[chilly ocean]

no

[crystal vale]

[chilly ocean]

it must be private , otherwise other people from my department are also here they will recognize me

[crystal vale]

You can tell me in my DM, if you want

[chilly ocean]

let it be bro, it wont affect anything

my wife checks everything so i got instructed to keep everything hidden

[crystal vale]

[chilly ocean]

from where u r getting these problems

[crystal vale]

Gallian

[chilly ocean]

okay nice

[crystal vale]

Now i am doing Dummit

I have an entrance exam for master( HRI)

[crystal vale]

21, 22, How can we do this without using binomial, maybe induction will help, but how can I prove that (1+p)^p^(n-2) ≠ 1 mod p^n?

[chilly ocean]

binomial is best way, for 21

[crystal vale]

But it computational

[south patrol]

You can also yeah just do induction and it follows nicely

[crystal vale]

But, does induction works if I want to prove that (1+p)^p^(n-2) ≠ 1 mod p^n ?

[dim widget]

That’s the easy part

(1 + p)^{p^{n-3}} is not 1 mod p^{n-1} by inductive hypothesis, so then if you multiply by (1 + p)…

[chilly ocean]

i did it in 1 line in mind

[crystal vale]

But when you have to write it cannot be written in one line, right?

[chilly ocean]

rest terms getting 0 module so if i need to write i need two lines

[crystal vale]

But we have to show that the rest terms are getting zero

[chilly ocean]

thats easy part for u, p^{n-1}(p^{n-1}-1)/2 * p is zero mod p^n as 2 divides p^{n-1}-1

similarly work with rest

[crystal vale]

Yes I know, I am struggling with other terms

[chilly ocean]

binomial coefficient is an integer , use this

also use p is prime

[crystal vale]

Sorry, I don't get it how it helps me by multiple by (1+p)

[south patrol]

Well you can split it into a sum of two terms

[crystal vale]

Oh

Is it tricky? I am not getting it

[velvet cave]

is this from D&F?

[crystal vale]

Yes

[white oxide]

If f: A --> B is a surjective ring homomorphism and p is a prime ideal of A, is f(p) a prime ideal of B

I got yes but for some reason it seems wrong

Wait nvm

My logic was wrong

[dim widget]

I assume you mean f(p)

[white oxide]

I think

[dim widget]

and not necessarily

[white oxide]

Yeah I meant f(p)

[dim widget]

for instance p = (x^2 + 1) in Q[x] = A mapping to B = Q[x]/x^2

but it is true if p contains the kernel or something

for trivial reasons

I mean the condition of f(p) being prime is literally that <p, ker(f)> is prime

by definition

[white oxide]

Oops in this case p does contain the kernel 😂

Okay yeah I got it

Thanks

[dim widget]

no worries

[crystal vale]

I have to prove that D_2n / < r^k > isomorphic to D_2k, where k divides n, so I am trying to show r<r^k> has order k in D_2k/< r^k>, so it is clear that order of r<r^k> divides k

[candid patrol]

Find the correct generators

[crystal vale]

I think r<r^k>, s<r^k> will be the generators

And if r<r^k> has order m, then m divides k, so m = kx, for some x in N, so it implies m has to be k

So r<r^k> has order k

[candid patrol]

Yes you just have to choose $\bar{r}$ and $\bar{s}$

[cloud walrus]

UGOBEL

[cedar vault]

Alternatively, by the first isomorphism theorem, there is a surjective homomorphism D_2n -> D_2k defined by r^{x} -> r^{x mod k}, s -> s whose kernel is <r^k>

k divides n ensures that this is map exists

[cedar vault]

Is there an example of a ring (commutative, with multiplicative identity) where:\

1)there is an element without unique factorization and factorization does not terminate\

2)there is an element with unique factorization but factorization does not terminate\

Are both of these possible? Im aware of examples with non terminating factorization, like the monoid algebra $\mathbb{F}2[M]$ where $M ={x \in \bR | x \geq 1}$ . Here we have for any nontrivial non-unit element $n$, the following $$n = \sum{m\in M}c_m\cdot m$$ where $c_m=0$ almost always and $c_m=1$ for atleast one $m\in M, m\neq 1$. note that $$\left(\sum_{m\in M} c_m\sqrt{m}\right)^2$$ is a proper factorization for $n$. I am trying to see if factorization is unique or not here. I suppose for 1), the answer would be yes, and the same ring is an example (by picking two infinite subsets $A,B$ with $\prod_{a\in A} a = \prod_{b\in B}b$ but $A\neq B$)

[cloud walrus]

Herzog

[cedar vault]

If both arent possible in same ring, what about a ring where 2) occurs?

I suppose these "infinite products" make sense only in some rings(those with topologies), so my question is restricted to those

[toxic zephyr]

trying to prove that distinct field automorphisms (over lets say F) are F-linearly independent (dedekind's lemma i think). working with the case of two, it's pretty simple. if they're linearly dependent, one is a scalar multiple of the other, but they have to fix 1, so if sigma2=ksigma1, then 1=k. but i'm struggling to find what goes wrong if we take
sigma3=c1sigma1+c2sigma2
(i.e. proof by contradiction) like i'm trying to show that sigma3 can't be "distinct". i'm not sure what that precisely means in this context. would that be to show that sigma3=sigma1 or sigma3=sigma2? just want to make sure i know what i'm trying to show i don't really need a hint beyond that

[velvet hull]

by symmetry, it shouldnt really matter if you show that sigma3 is equal to sigma 1 or 2

but I'm not exactly sure that this is the right approach

[toxic zephyr]

i'm more just trying to understand what this actually means. like is it going to be one of the others (sigma1 or sigma2) or something else (what does distinct mean in this context)? why do we care about linear combinations of automorphisms which i think shouldn't usually result in another automorphisms? or is that kind of the point? that this shows that field automorphism linear combinations can't result in a new distinct automorphism?

i guess that one of the ci's has to be 1 and the others need to be 0, if sigma_n=sum c_i sigma_i

but, then again, why do we care that linear combinations can't lead to new automorphisms? idk it just sort of seems to come out of nowhere and i'm not sure what the goal or use of this result it, beyond just being cool.

[velvet hull]

this was taught to me as an intermediate result for proving the existence and uniqueness of canonical forms

[velvet hull]

For instance, suppose that E/F is a finite Galois extension such that Gal(E/F) is cyclically generated by T. Let the order of T be n.
Then the set {id, T, T^2,.. T^n-1} is F-linearly independent. Hence there exists an x in E such that {x, T(x), T^2(x), ..., T^n-1(x)} is an F-basis for E, and so by change of basis we get that the matrix representing T is similar to a matrix in rational canonical form

[toxic zephyr]

hm interesting

[dim wagon]

We start off with a unital ring $S$ and a unital subring $R$ where both share the same identity.

Let $N$ be an $R$-Module. It is generally not possible to make $N$ an $S$-Module, so instead we try to embed $N$ as an $R$-submodule of some $S$-module or equivalently to find an injective $R$-module homomorphism between $N$ and an $S$-Module. It turns out this is also not generally possible but below gives a "best choice" of the $S$-module where best means an injective $R$-module homomorphism is possible.

We consider the free abelian group with the basis $S \times N$. For this group to satisfy the $S$-module axioms as well as be compatible with the $R$-Module on $N$, we quotient out by the minimal relations. This quotient group is called the tensor product $S \otimes_R N$. We then somehow define the action of $S$ on the tensor product making it an $S$-module.

Finally, there is a natural way to define an $R$-module homomorphism from $N$ into $S \otimes_R N$. We note that this homomorphism is not necessarily injective but is the best choice since we quotiented out only minimal relations to construct the tensor product earlier.

[cloud walrus]

somethingwrong

[dim wagon]

I am trying to understand the basic motivation/construction of the tensor product for modules(just for extending scalars) given in dummit and Foote chapter 10. Does this sound right?

[tribal moss]

Looks right to me. Not sure it should be considered the motivation for tensor products, but it's certainly one motivation.

[kind temple]

other than this and making two modules multiplicatively compatible, what would you say a good motivation for the tensor product?

[dim wagon]

Okay thanks alot for checking

[dim wagon]

Yes wanted to ask that too, is there another preferred way to motivate the tensor product? I will try to look at this multiplicatively compatible thing too

[tribal moss]

I'm not immediately sure; it is not my area of core expertise. It's more that tensor products of rings do so many different things at once that it would surprise me to have only one motivation for modules. (For example, the usual "representation of bilinear maps" motivation for tensor products of vector spaces ought to apply to the module case too, as far as I can see).

[kind temple]

loosely what i mean by this is that the tensor product of two R-modules can be thought of as the smallest module structure where you can multiply two module elements as if they were elements of a ring. these multiplication relations are exactly the relations that you quotient out by to obtain the tensor product

i dont think i have enough familiarity with them to give better exposition

[velvet hull]

To give my two cents, the tensor product of two modules is the largest module that there can still exist a surjective bilinear map onto it

So it’s the largest thing the two modules can “agree” on, in some sense

[rocky cloak]

Being adjoint to the hom functor is pretty important, the adjunction -(x)M, Hom(M, -) between ModR and ModEnd(M) comes up a few places (see Morita theory or tilting theory).

Any functor ModR -> ModS that is right exact and preserves coproducts is given by tensor product by a bimodule. So it can help to understand such functors by studying the modules (see for example Tame-Wild dichotomy).

The tensor product has a nice universal property classifying multilinear maps, which is very useful for constructing multilinear structures (see the tensor algebra, exterior algebra, symmetric algebra).

The tensor product and Tor can be used to measure torsion.

Localization is given by a tensor product (though this might fall under extension of scalars).

[chilly ocean]

how localization is given by tensor product

we modded out the equivalence relation

[rocky cloak]

S^-1 M = S^-1R (x) M

[cloud walrus]

Couldn't find an attached image in the last 10 messages.

Jun 8, 2025, 10:06 AM

Couldn't find an attached image in the last 10 messages.

Jun 8, 2025, 10:07 AM

[empty kernel]

,rotate

[empty kernel]

ok so the question 5 asks for which n in the natural numbers exccluding 0 and 1 is the following statement true

a,b were pretty easy
i did c and i believe that it isnt true for any n becayse 2z is always even thus meaning u will only be able to get the even/uneven numbers depending on if your n is even or uneven

but on d im completely stuck idk what to do
i tried for a few numbers and i believe it is true for all n that have gcd(n,3)=1

but well idk how to proof it

[candid patrol]

Let $f : \mbb Z/n \rightarrow \mbb Z/n$ with $f([z]) = [2z]$. Then, $\forall a \in \mbb Z/n, \exists z \in \mbb Z : a = [2z]$ iff $f$ is surjective

[cloud walrus]

UGOBEL

[empty kernel]

so u say i gotta define a function?

[candid patrol]

i.e. $f \in Aut(\mbb Z)$, so iff $f([1]) = [2] \in \mbb Z/n^{\times}$, and so $pcd(2,n) = 1$

[empty kernel]

wait am i right on c that it isnt true for any n?

[cloud walrus]

UGOBEL

[empty kernel]

ok here u lost me

[kind temple]

(d) is asking for which n is the group homomorphism Z/nZ —> Z/nZ taking [z] to [3z] surjective

[empty kernel]

the hint of the function helped tho im doing something now

[candid patrol]

Aut(Z/n)*

[empty kernel]

ok i think i did it

i proved that f: Z_n to Z_n is a surjective function

if n isnt a multiple of three

thanks!!

[candid patrol]

👌

[candid patrol]

I guess f(k) = 3k

[glad osprey]

No, c) is true for odd n

[thorn jay]

Fellow Nederlander

[empty kernel]

oh yeah you are right well the oproof is pretty much the same tho

[empty kernel]

hi

[candid patrol]

I wrote it 😵

[chilly ocean]

ITS FINE

[candid patrol]

😓

[solid prawn]

Are vector spaces a way to show how rings (V,+,∙) works geometrically?

[knotty badger]

Could you expand on this? Vector spaces aren’t usually rings

[rocky cloak]

I guess a you can interpret a vector space as how fields "work geometrically", in that it's something acted upon by a field.

Though it feels like stretching the use of the word "geometrically".

[knotty badger]

I mean mathematicians stretch the word “geometry” all the time right

Look at AG

[rocky cloak]

There are different levels of stretch though

[knotty badger]

AG seems like a much further stretch to me

[wraith cargo]

lol

why do you think AG is such a stretch to be called geometry?

[rocky cloak]

Well, AG actually involves topology and smoothness and geometric ideas.

Just saying vector spaces are geometric without any justification seems like a big stretch.

[cedar vault]

Well, a 1D vector space is more than enough to capture the geometric behaviour of a field , but a 1D v.s over a field is just the field, so like, i think trying to interpret it that way is not really the way to go

[rocky cloak]

If you already have some idea of geometry in your field, and then you just slap that onto a 1d space, then the space hasn't really done anything

[knotty badger]

To be fair I know very little about it but

It’s all algebra

Which isn’t exactly known for having pictures

[rocky cloak]

Not known for having pictures?

[solid prawn]

As I'm wondering why vector spaces can be defined only by those 8 axioms, which I think will be useful to connect it to groups stuff(? , I haven't start learning groups' properties, though I was interested about it, hope I'm not presenting like a dumb.

[knotty badger]

Yeah it’s symbol-pushing right

[wraith cargo]

I mean if ur in high school that's true

[rocky cloak]

You can argue that all of math is symbol pushing if you want I guess

[knotty badger]

I’m not tho

I’m talking about algebra specifically

[rocky cloak]

Then I have no idea what you're talking about

[knotty badger]

Algebra

[rocky cloak]

Yeah, but as I recall you're definition of algebra is something close to "math I don't like" right?

[knotty badger]

No…?

[rocky cloak]

I just remember you have this rant about how category theory can't be algebra because you like category theory and don't like algebra.

[knotty badger]

That’s part of it sure

If they really were the same then I’d expect to like them the same amount

If x = y then f(x) = f(y)

So if f(x) is not f(y) I can conclude x is not y

[chilly ocean]

algebraist

[solid prawn]

some relations can do that but not mappings

[knotty badger]

What do you mean?

[solid prawn]

I remember there are relations that take one input to two output but they're not mapping and functions, maybe I mixed up

[knotty badger]

Here f is a function, apologies if that’s unclear

[solid prawn]

anyways, I know what you mean by that

[rocky cloak]

Sure, but the conclusion that you don't like algebra would hinge on the assumption that category theory isn't algebra.

Anyway, I don't really care how were categorizing the different fields or what counts as geometric or not.

The main point was just figuring out what living meant I guess. Which I'm not sure I'm closer to...

[knotty badger]

I mean I think my argument is very clear

If x is category theory and y is algebra

You’re claiming x = y or something to that effect

This would imply Like(x) = Like(y)

But that’s false

Hence x is not y

I did not use the assumption that x is not y in deriving this

[solid prawn]

or maybe could anyone introduce this subject to me, as I want to learn it rapidly and deeper later.

[rocky cloak]

You used two unstated assumptions here, namely Like(x) and NotLike(y)

[knotty badger]

Yes but that’s my own subjective experience

It’d be pretty presumptuous for you to argue you know that better than me, don’t you think?

[solid prawn]

chill

could anyone answer me

[solid prawn]

this

[rocky cloak]

Well, if you don't like super Mario, but you do like legend of Zelda. Then you claim to not like video games, people will find that weird. I don't think that would be presumptuous

[knotty badger]

Then please do enlighten me with your mind-reading powers

[solid prawn]

As I want to derive these axioms of vector spaces

[rocky cloak]

Then you go on to argue that Zelda can't be a video game because you like it

[knotty badger]

I’m sure logically convincing me that I like algebra will work out great

[chilly ocean]

https://tenor.com/view/spock-dazzling-display-logic-star-trek-gif-16343803

[rocky cloak]

I mean it's not an argument about what you like, just what algebra is

Anyway, we've had this discussion before, so no point in doing it again

[knotty badger]

Yet for some reason you always take issue with me saying I don’t like algebra

As if I can’t have my preferences in peace

Somehow it’s morally wrong to dislike algebra or something

[solid prawn]

sorry for saying that, but could anyone answer me: Why the group-like structure of those 8 axioms can expaned to a vector space uniquely?

or it is not relevant to this subject?

[knotty badger]

It’s hard to give a proper answer because in a sense these 8 axioms define what it means to be a vector space

[rocky cloak]

I have no problem with anyone disliking algebra, so then you have misunderstood

[knotty badger]

In the sense that, at least as far as I’m aware, there’s no separate definition of vector space which these 8 axioms are meant to capture

[knotty badger]

Sure sure, it’s always me who’s misunderstanding

Whether it’s you or my own thoughts

[solid prawn]

CHILL, though I can tell you two are about the end of 'discussion'

[rocky cloak]

No one is claiming you're misunderstanding you're thoughts.

Just that you're defining "algebra" differently than other people.

That's it, that's the only claim being made against you.

[knotty badger]

Then why on earth do you care so much about that

[solid prawn]

I want to derive them, like, why I don't need more axioms, or why all other properties are non-trivial (means they can be derived by other axioms)

[knotty badger]

What we call algebra is a social construct anyway

Why be so insistently prescriptivist about what is or isn’t algebra

[solid prawn]

hey bro, answer me first

[knotty badger]

I think studying linear maps is a good motivation for a lot of them?

[rocky cloak]

I don't care. That just was the topic of discussion right now.

What does geometry mean vs what does algebra mean

[knotty badger]

also if you’re referring to me then I’m a sis :)

[glad osprey]

Lol, you're not entitled to an answer. There's nothing really to answer, a vector space is defined by those axioms, that's it. If you had different axioms, it wouldn't be a vector space

[solid prawn]

I mean, some geometrical properties can be translate into linear algebra, and why we can ensure that we have "enough" axioms

[thorn jay]

I can argue from a logic point of view.

We have some collection of things we call "vector spaces". We make an educated guess at what axioms should define that collection, and we can show (albeit using some advanced math at times) that the only objects satisfying these axioms must be these things we call vector spaces

[knotty badger]

To be fair there’s an argument to be made that there aren’t enough axioms

E.g. nothing about continuity enters the picture

But we can still give a name to whatever objects satisfy those 8 axioms

[solid prawn]

no Godel's stuffs pls

[knotty badger]

And currently we call them vector spaces

[knotty badger]

As far as I’m aware this is unrelated to Gödel stuff

[thorn jay]

There's nothing inherently special about these axioms, they just happen to work, and are nice to work with

[knotty badger]

There are ways to repackage these axioms in a nicer way

[solid prawn]

I mean, there must be a reason to define things

[knotty badger]

For example, a vector space is a field acting on an abelian group

[knotty badger]

Indeed but this reason isn’t necessarily understandable

Sometimes it’s just luck

[thorn jay]

We noticed that these were the properties of vector spaces we used, so we abstracted them

[knotty badger]

Sometimes it’s complicated social forces

Sometimes it’s because the definition leads to very interesting math later on

[solid prawn]

but why we choose those axioms, and not others?

[thorn jay]

Because they work

[solid prawn]

I understand what you mean

[thorn jay]

And they are nice to work with

[knotty badger]

Like there definitely is a reason for every math definition

It’s just that mathematicians are not in the habit of writing these reasons down

[solid prawn]

but the proper definition of vector spaces seems to be appear later than some linear algebra stuffs

[knotty badger]

Some do of course

But the reasons aren’t part of proofs or anything

[solid prawn]

means those axioms were chosen to followed naturally by not disturbing properties derived before or later.

[dim widget]

the point of the definition of a vector space is to abstract away a lot of theorems from linear algebra and apply them to a whole range of situations.

[knotty badger]

Yeah the history of math is very messy

Often definitions come last historically

Whereas they usually come first pedagogically

[solid prawn]

but something like determinants did have an elegant derive of definition.

and I believe everything has one

[thorn jay]

You need to understand that there is no inherent reason we chose "these axioms specifically", because they are not more special then any other set of axioms defining vector spaces

[knotty badger]

Yes but even if the reason is elegant, it might not get written down

Since it’s not part of the proof, in a sense it’s an inefficiency

At least to a lot of mathematicians

[dim widget]

what is not elegant about vector spaces

[knotty badger]

Wdym

[solid prawn]

time isn't a big problem anyways

[knotty badger]

This is more about paper length

[solid prawn]

everything in math is elegant

[knotty badger]

You can put motivation for your definitions

But you can also just, like

Not bother

And you get a shorter paper

[dim widget]

@solid prawn i don't understand what you're asking

[solid prawn]

this

[dim widget]

yeah I don't understand this

the concept could have more or fewer axioms or different axioms

[solid prawn]

why we chose those 8 axioms

[dim widget]

but it's obviously a useful concept no matter how you choose to express it

[dim widget]

because they look pretty simple and they work

[thorn jay]

That's what I've been saying qwq

[dim widget]

like there is a universal style people have agreed upon for defining algebraic structures, and these axioms are very much in that style

compare with the axioms for a ring or a module

[solid prawn]

I'm trying to derive it, not writing a poem to express how great they are

[dim widget]

You can't derive a definition

that's one of the most basic points of math

[rocky cloak]

I guess if you look at the development of linear algebra, it's starts with linear equations in R^n. So R^n is like the model vector space.

Then you get natural subspaces like kernels and images, and you can apply the same theory to things like solutions to linear differential equations.

Then you set up the minimal axioms needed to prove your theorems, and boom you got a mathematical theory

[dim widget]

if you already understand what you're trying to express, then you can sometimes easily write down axioms for it

[solid prawn]

or I'm trying to explain why thry are

for example, why determinant looks like that, which looks terrible for beginners

[dim widget]

so like Jagr says the basic point is people tried to find the minimal list of axioms that make classical linear algebra over R^n or C^n work

[solid prawn]

and we can derive it from n-volumes

[dim widget]

so they could apply it to lots of superficially unfamiliar situations like solving differential equations etc.

[dim widget]

that's one choice of derivation

[solid prawn]

why are they the minimum, say?

[dim widget]

Well a basic part of linear algebra is that you need to be able to take linear combinations

that's actually the main thing

[solid prawn]

how do I proof this set of axioms are unique, any other set of axioms are equivalence to this?

[thorn jay]

They aren't unique

[dim widget]

there isn't ever going to be a unique set of axioms for a desired object

you have to decide what you want first

then you can show that your axioms do produce an object compatible with that

[thorn jay]

Any other set of axioms defining vector spaces is equivalent to the usual chosen set of axioms because.. that's how equivalence works

[solid prawn]

or in general, how to determine is there enough axioms for an object, to fix its meaning?

[dim widget]

axioms are just sentences

[rocky cloak]

It might depend what you mean by "fix its meaning"

[dim widget]

you could have many sentences that describe the same set of mathematical objects uniquely

[thorn jay]

If you mean "we have a class K of objects and we want to determine if some set of axioms T define K" then that's, kind of very hard

[solid prawn]

the least amount of axioms, also

[thorn jay]

That's also hard

[dim widget]

there is no such thing as the least amount of axioms

you could always combine 10000 axioms into one extremely long formula

[knotty badger]

Try this for sets lol

[dim widget]

you are not thinking clearly

[solid prawn]

yes but I mean the minimum things it need to follow

[thorn jay]

Not sure I follow

I was talking in the sense of model theory / universal algebra

[knotty badger]

I was just referring how hard it was to land on the current axioms for set theory

[solid prawn]

not that hard actually, we just have to find the logic statement that expressed the whole thing

[thorn jay]

Sometimes it really feels like throwing things at the wall with educated guess and removing redundancy

[knotty badger]

Mhm mhm

Thaaaats math!

[rocky cloak]

Nothing is hard, we just have to do it. Problem solved

[solid prawn]

so, are we finally reaching the gate "Understanding the thing I want to ask"?

[dim widget]

you need to sit down and give a rigorous treatment of what you are imagining. I think you took the determinant, had one definition (in terms of permutations or something) that you didn't like. Then someone told you some other definition (the intuitive concept of oriented n-volume) which you did like, they made an attempt to mathematically formalize the properties an "oriented n-volume" should satisfy, and they showed your permutation definition was recovered from that. You are now trying to do the same with vector spaces except you're not willing to actually do the work of making a rigorous claim about what the object (here vector spaces) "should look like"

[thorn jay]

And then you have to prove that anything satisfying those axioms is actually in your class

And first you have to actually find those axioms

[solid prawn]

yes, the process is just like prooving they are equivalence definitions

[thorn jay]

Often you want those axioms to be of a certain form (equations or something), as that yields certain properties

Sometimes the class isn't even defineable using axioms

[dim widget]

that's just not how most definitions work

most/many definitions are trying to abstract away nice properties of some specific class of examples

vector spaces are like this

[solid prawn]

and I'm trying to explain why they look like this, in simple ways

[dim widget]

the important exercise is to go back and make sure that R^n or C^n actually satisfies these axioms, and that the basic theory of linear algebra still works with only these axioms

[dim widget]

that's like saying "i'm trying to explain why this painting looks like this"

well the person who invented it choose for it to look that way because they thought it was beautiful and simple

and because it successfully abstracts the key properties I mentioned above

[solid prawn]

yes, although I know it's a dumb action

[dim widget]

i.e. it gives a framework in which linear combinations make sense

[solid prawn]

just explain what the painter is thinking

[dim widget]

they were thinking this

[solid prawn]

in order to understand the painting

[dim widget]

there you go

[solid prawn]

furthermore, how beautiful and simple it is.

[dim widget]

I think the first one and arguably the second one are subjective qualities

which is part of what you're running into here

[tribal moss]

Sometimes the painter was just thinking "this looks right", and all the subsequent theorizing is not about reading the painter's mind but about figuring out why that particular thing looked right.

[atomic mesa]

Hello I have that a soluble group is simple iff it is cyclic of prime order. But I claim the group {e} is a contradiction to this?

[slim kayak]

Might either be a mistake or buried somewhere in the definitions of soluble and simple

[candid patrol]

soluable group : $\exist s \geq 1 : G^{(s)} = {e}$ \newline
simple group : $H \triangleleft G \Rightarrow H = {e}$ or $H = G$ \newline
But for every $k \geq 0$, $G^{(k)} \triangleleft G$, so $G$ is both soluble and simple if $G^{(s)} = G^1 = {e} \Rightarrow G$ is abelian etc...

[cloud walrus]

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

[south patrol]

I thought e is usually not called simple

At least to me

Lol another potato... woah

But yeah this will probably be their convention

[shell lance]

how is the symmetry operation defined in the dihedral group? do i pick one arbitrary line of symmetry and persist with it throughout all the rotation and flipping operations?

[atomic mesa]

Thanks I missed it in the def

[south patrol]

Pog

Pog

[velvet cave]

im pretty sure if u rlly wanna, you can define the dihedral group without any reference to actual geometry

using generators

similarly to how we can define e^x using a power series instead of naive definitions

[toxic zephyr]

can someone tell me if i got this right? say we're looking at the nth roots of unity in a field of characteristic p.
if n=kp^m (where (n,k)=1), then X^n-1=(x^k-1)^(p^m). so the nth roots of unity are just kth roots of unity. and so there's no primitive nth root of unity, but there are primitive kth roots of unity (if k is not 1, or maybe 1 counts as a primitive 1th root of unity idk) and any kth root of unity will also be an nth root of unity?

[crystal vale]

I think yes, any kth root of unity will be the nth root of unity, say x is the kth root of unity that means x^k = 1 implies x^n = 1, right?

[toxic zephyr]

usually we do r^a s^b or something. technically, you don't have to pick a specific one. like r could be clockwise or counterclockwise (as long as it has the right order). and s can be a reflection across any arbitrary line of symmetry. it's a cool geometric fact that you can get ANY other reflection over some line of symmetry via r^a s for some a. that is, any reflection can be obtained by whatever starting reflection you chose, and then rotating it some number of times.
so, yeah, we usually just say <r,s:r^n=e, s^2=e> or something

[toxic zephyr]

yeah just raise both sides to the p^m power cool

[crystal vale]

Yes

[rocky cloak]

I'm not entirely sure what you're asking, but the dihedral group can be defined as the symmetry group of a regular n-gon.

Each line of symmetry will give you a reflection.

[shell lance]

yeah so i was asking about the line of symmetry of s in the generator, whether i am just picking an arbitrary line at the start and generating the rest of the group using s corresponding to the same line of symmetry or not.

for example if i had an equilateral triangle line this
A
B C

and my LOS passes through A and midpoint of BC. then after performing a rotation the triangle now looks
B
C A
now my LOS is passing through B and the midpoint of CA right?

[toxic zephyr]

yes you pick an arbitrary one, and the rest of the reflections can be achieved through composing it with some number of rotations.

[shell lance]

heres how my confusion arose: i read somewhere that traditionally the LOS is the one passing through a specific vertex and the center of the polygon. i assumed this to remain invariant even after rotations and so initially my LOS rotation along with my rotations. i found i couldnt verify the basic relations this way.

[toxic zephyr]

if you pick s to be the one that flips B and C but fixes A in your example, then the one which flips A and B and fixes C is just sr. the first rotation you did, and then the same reflection (flipping the bottom two nodes) gives
B
A C

[toxic zephyr]

if i understand what you're saying correctly (idk if i am) then you do need to keep the same exact line of symmetry (i.e. the reflection you use) the whole time. imo it's simpler to think of the dihedral group as being a composition of just one reflection and just one rotation (i mean you don't have to, but i find that easier to keep in my head). i kind of struggle with the shape manipulation in my head, personally, so an algebraic definition in r and s is easier fo rme.

[shell lance]

yes you understood it right, thanks

[cinder onyx]

a good way to think about it is as a permutation of the vertices of a polygon and how you work with products of permutations

[twilit wraith]

is it two that if two groups are isomorphic then they have the same number of generators of the same order?

i know the converse doesnt hold

it makes sense in my head that this is true but im not sure if theres a strange counterexample

[echo vine]

Consider the ring ( A = \mathbb{Z}[i] ).

\begin{itemize}
\item[(a)] Prove that ( (2 - i) ) is an ideal of ( A ).
\item[(b)] Prove that ( A/I ) is a commutative field with 5 elements.
\end{itemize}

\textit{How could I begin with b)? I don't even know what I'm looking for or how to start.}

[cloud walrus]

Aguacate

[wraith cargo]

every finite integral domain is a field

so prove (2-i) is prime and the quotient only has 5 elements

[echo vine]

[tardy hedge]

youre annoying lol

[errant wedge]

You could try constructing an isomorphism from A/I to Z/5Z

@echo vine

[echo vine]

the thing is I don't really understand how A/I even works

[fresh bison]

Yes, isomorphisms preserve order of the elements

[echo vine]

Its like Zn but with Z[i] right?

[errant wedge]

or sorry I think FIT is easier here

[echo vine]

I don't know what that is

[errant wedge]

first iso theorem

construct a surjection Z[i] -> Z/5 with kernel (2-i)

[cinder onyx]

I'm not sure what you mean? I mean, you could certainly construct differing generating sets for the same group so in that sense, no. But if you have an isomorphism φ:G->H and (S) is a generating set of G then φ(S) is a generating set of H, so

[thorn jay]

Isomorphisms preserve anything that isn't the precise elements

[cinder onyx]

^

[thorn jay]

So order, generating sets, minimal generating sets, etc.

You can see this abstractly as isomorphisms preserving any (possibly infinitary) proposition obtained from equations in the language of groups

[errant wedge]

uhh if it helps, for any equiv class [a+bi], we have a^2+b^2 = (a-bi)(a+bi), so [a+bi] = [a^2+b^2] and every class has an integer representative. Then note 5 = (2+i)(2-i) so [5] = [2-i] = [0], so this is like ur sign to just think of it as Z mod 5 (but u already have this since its the unique finite field of 5 elements)

im not sure if thats wym

[echo vine]

hmmm

[cinder onyx]

useful way to think about this yes

[echo vine]

What does the fact that a^2+b^2 = (a-bi)(a+bi) have to do with the rest? Or with the class

I'm sorry this subject is very new to me and I dont understand a lot yet

[thorn jay]

What doesn't fit there is stuff like orderability, but I am sure you can just add a relation to your signature to fix that

[errant wedge]

Sorry, I was going about it wrong. This should help

Note that (a+bi) - (a+2b) = bi - 2b = -b(2-i), so [a+bi] = [a+2b]. You can then use this to construct ur surjection

[errant wedge]

im just trying to explain how "A/I works" and like how you'd think of it being Z/5Z. We have 5 as a multiple of (2-i) as above, so 5 is 0. Since 2-i is 0, 2 is i, so i^2 = -1 is 4, do you kinda see

[echo vine]

I kind of see this

I don't really understand how I can build the surjection from this though. Thank you for all your help yet but I think im missing something important

[errant wedge]

Think about how you can construct a map from Z[i] to Z/5Z that "enforces" this relation

Basically the answer: || Let ur map be [a+bi] -> [a+2b], and verify that (2-i) is the kernel ||

[echo vine]

hm...

thank you so much for your help

im going to try make sense around it

[toxic zephyr]

apparently we can assume that the r_i's are prime here, but i'm not sure why? say we have like a_1=2^(1/6), then a_1^6 is in Q. is the idea like. we can instead choose a_1=2^1/2, and then a_2=2^1/6? then a_2^3 is in K1=K(a_1)=Q(2^1/2)?

[twilit wraith]

I see

I had to prove that Z under addition isn't isomorphic to Q under addition

I ended up just showing that Q isn't cyclic but I was wanting to show that Q is generated by more than two elements

[white crown]

Anybody know how to solve this in here

I’m in need of a help is a linear algebra problem

[toxic zephyr]

can you just post a screenshot

[white crown]

Gotcha u

This is just a little bit of the work that got to be done

[glad osprey]

Yeah, that sounds correct 👍

[white oxide]

Uh stupid question but if R and S are commutative rings are the prime ideals of R x S of the form p_1 x p_2, with p_1 and p_2 prime ideals of R and S respectively

[thorn jay]

R x S / I x J ≈ R/I x S/J which is an integral domain if and only if either I=R or J=S, and the other is prime

[white oxide]

So the answer is not in general right, because prime ideals are proper

I guess that makes sense, we could have R x J be a prime ideal since it could be proper

but R wouldn't be a prime ideal of R since it's the whole ring

[thorn jay]

Just not ever, as prime ideals by definition aren't the whole ring :3

[white oxide]

Wait wdym, aren't cartesian products of prime ideals still prime?

[thorn jay]

I just explained why not

[thorn jay]

The product of two integral domains is never an integral domain

[white oxide]

oh something like (1, 0)(0, 1) = (0, 0) right

[thorn jay]

Yup

[white oxide]

mm makes sense

thanks

[knotty badger]

thank you!

[cobalt heath]

So on cyclotomic field, the ring of integers is R = Z[zeta].

Considering identification Z^n = Z[zeta], which operations will permute it?

For simplicity, let's assume n is power of two

Also, you can think of a map R^n -> R^n as Z^(n * n) -> Z^(n * n)

How could I express the transpose map in terms of R?

This should be trivial but I dunno how to simplify this

[rocky cloak]

One key property Q has that Z does not have is that it's divisible.

Meaning for every x in Q, and positive integer n there is a y in Q with x = ny (repeated addition of y n times).

It might be easier to think in terms of a specific isomorphism. Like say f:Z -> Q was an isomorphism. Then we can define x to be f(1)/2. What could then f^-1(x) be?

[crystal vale]

If you want to show Q is generated by more than two elements then just take a generating set Q

[rocky cloak]

So assuming zeta is prime-root of unity, and your basis is 1, zeta, zeta^2, ... Then multiplication by zeta would permute the basis.

In general I'm not sure that you can find a basis where permuting them corresponds to something particularly natural.

As for your second question I'm not sure what "express in terms of R" should mean necessarily. I guess you can do some fourier transform trick to extract the coefficients of the matrix and do something from there...

[cobalt heath]

Oh, Fourier transform?

Hm, can one express arbitrary permutation of Z^n in terms of R = R[zeta]?

[rocky cloak]

Well, I'm still not sure what "in terms of" should mean. But I'm guessing the answer is that you can't

[cobalt heath]

By "in terms of R", I mean using arithmetic operations in R.

[rocky cloak]

Then I'm leaning harder towards no

[chilly ocean]

hilarious pattern

[lime badge]

Very dumb question, but there's the usual theorem that R[x]/f(x) --> R^d sending a polynomial in R[x] to its remainder by division with f(x) is an isomorphism of abelian groups.

Does this generalize to polynomial rings in multiple variables nicely? For example, let's say R[x,y]/(x^l + 1, y^m + 1). I'd like to say there's an iso to R^(lm) and it seems plausible since the x^i y^j monomials form a basis, but is there an easy set of quotient ring manipulations I can do to show this?

[velvet hull]

as long as all the leading terms of the generating polynomials are units. this is probably not a necessary condition though

[lime badge]

I guess there's the isomorphism that R/I (x)_R R/J = R/(I+J).

But R = F_2[x,y] here, so idk how much that helps me

[velvet hull]

over a field it's true

if the base ring is a field then polynomial division always terminates

[lime badge]

This isomorphism is true for any commutative ring due to the right-exactness of tensor product

[velvet hull]

oh, that's the tesnor product

I misread it, yeah

but quotients of polynomial rings over fields are always nice

[lime badge]

Ah, were you thinking of direct sum?

So you're saying that F_2[x,y]/(f(x)) (+) F_2[x,y]/(g(y)) = F_2[x,y]/(f(x), g(y))?

Is there a simple proof of this? In my case, f(x) and g(x) are monic

[velvet hull]

wait, that doesn't look right, let me think

[lime badge]

Yeah, it doesn't seem right to me either

I feel like it should be more like F_2^(l*m) rather than F_2^(l+m)

[velvet hull]

well the first one is an infinite dimensional vector space, only the latter is F_2^(lm)

[knotty badger]

seems like you'd want a coproduct...?

the direct sum is a product afaik

[lime badge]

Yeah, I know that when tensoring over F_2[x,y], F_2[x,y]/(f(x)) (x) F_2[x,y]/(g(y)) = F_2[x,y]/(f(x), g(y))

[knotty badger]

i see

[lime badge]

I was looking to generalize the usual iso of abelian groups R[x]/f(x) --> R^d when f(x) is monic

[queen quarry]

assuming that this "like function" is well defined

[knotty badger]

it is cause it's my preference

[queen quarry]

there are numerous things one can call algebra, and saying that it is well defined means that each representative must agree on the same value. do you also dislike homological algebra?

[knotty badger]

https://youtu.be/bD8kjpynF6A?si=sue0qpc6NzdhN2Go

$\mathbf{Cat}$ is not monadic over $\mathbf{Set}$, so in this sense it differs from the "algebra" i know

[cloud walrus]

Pseudonium

[knotty badger]

i haven't done much homological algebra, so i can't make any substantial comments there

[delicate orchid]

anything I say is algebra is algebra end of story

[chilly ocean]

catgebra

[lime badge]

Let me rephrase my original question.

Let's say I have the polynomial ring R[x,y] and consider monic polynomials f(x) and g(y) of degree l and m respectively. Is it true that as an R-module, R[x,y]/(f(x), g(y)) = R^(l*m) since a basis for this quotient ring (viewed as an R-module) consists of monomials x^i y^j where 0<= i <= l-1 and 0 <= j <= m-1?

[trail hamlet]

Have anyone seen http://arxiv.org/abs/1610.09725 before ? Its maths as well as its parent paper are really intense 😵‍💫

[knotty badger]

yeah from what i can tell, if you're viewing it as an R-module, i don't even see how the multiplication would be visible?

all you can do is add elements and multiply them by scalars from R, right

[lime badge]

Yeah for sure. But that's ok in my specific application I'm considering

[knotty badger]

mhm mhm

maybe a better way to phrase this is like

as long as f and g are monic, it doesn't seem to matter what the specific form of f and g are

in that you wouldn't be able to determine f or g from the R-module structure

like the isomorphism to R^(l*m) you suggest is independent of f or g

so long as they have the prescribed degrees that is

[lime badge]

Yeah, it just so happens that in my situation, that's all the info I need

[knotty badger]

cool cool

[lime badge]

About f and g, namely that they're monic of certain degrees

[distant summit]

Not sure on the right place to ask this.
What does it mean to say su(2) has rank 1?

su(2) is the Lie algebra

But rank refers to a matrix (presumably it can be extended to an operator)

[knotty badger]

If I remember correctly, it’s the dimension of a cartan subalgebra?

So in this case they’re taking the abelian subalgebra generated by J_3 as their cartan subalgebra

[knotty badger]

As far as I know this rank doesn’t refer to the rank of some operator, but I haven’t studied enough lie theory to say definitively

[next obsidian]

No

What if f = g?

I think you need to assume something like f and g share no irreducible component, this is assuming R is a UFD

[lime badge]

f and g are polynomials in different variables though

[next obsidian]

Oh…

Oh oh oh

Then yeah I believe so

[delicate orchid]

still correct

[rich granite]

She finitary my Eilenberg-Moore category till I variety of algebras

[thorn jay]

Categories are special cases of partial semigroups just extended to classes; algebra enough for me

[delicate orchid]

Categories are triangles and 2-categories are tetrahedrons

[thorn jay]

Alright, now define their homology

[delicate orchid]

that is how u define their homology og

[crystal vale]

I am doing Dummit and foote's exercise, i don't need any hint, i just want to confirm what does it mean by G a transitive permutation group on the finite set A, is that mean for any a and b in A there is g in G such that ga = b?

[rocky cloak]

That's correct

[crystal vale]

Then, isn't trivial because since |A | > 1, so if all permutation fix a, then there will not a permutation which maps a to b ≠ a, which contradicts transitivity, right?

Oh wait

Sigma doesn't fix any a oh

[cobalt heath]

One way to do it is to take a trace to get each coefficient, then add it back together.
That is, the field trace map Tr : K -> Q gets each coefficient, interestingly.

The issue is that this has high complexity.

[glad osprey]

This sounds like a fun problem, can you post here if you solve it?

[twilit wraith]

Suppose that φ is a homomorphism from a finite group G onto a group H, and that
H has an element of order 8. Prove that G has an element of order 8.

this problem is hella stumping me

of course if that element has a nonempty fiber then the problem is easy

but what if it doesnt

oh nvm im stupid

that element has to have a nonempty fiber bc the homomorphism is onto H

[warm dove]

in this proof, how does K_1 = F(\omega \beta) give all the roots?

[wicked patio]

What if G is trivial

H can be any group

Oh it's onto

Then yeah just take an element g that maps onto an element of order 8 in H

[warm dove]

I could see F(\omega, \beta) working, but then in this case K_1 isn't normal radical.

[wicked patio]

The order of g has to be a multiple of 8, say 8n

Then g^n has order 8

[twilit wraith]

yeah i didnt read that one word

[errant pond]

What are good book recommendations for intro to abstract algebra

I can’t find it specifically in book recommendations

[topaz solar]

Artin or Dummit & Foote are what I think of

[fading acorn]

dnf or artin

[fresh bison]

I liked undergrad Hungerford when I was starting out. It’s a lot less comprehensive than a book like Dummit and Foote though, so that’s a better option if you want a more in depth experience

[errant pond]

Thank u very much

[warm dove]

If artin or dummit and Foote are too hard, consider Judson’s open source book.

[errant pond]

Thx

[tardy hedge]

Dummit foote dumb foot

[woeful sage]

Dummit and Foote?

more like Peak Algebra

🗣️🔥

Okay so in the lattice of subgroups of a group $G,$ the join and meet of two subgroups $H, K \leq G$ are defined as $H \lor K = \langle H, K \rangle$ and $H \land K = H \cap K$

[cloud walrus]

Neamesis

[woeful sage]

so quite literally like union and intersection

except union isn't necessarily a subgroup, so we take the subgroup generated by the union

[woeful sage]

if D&F is too hard, do analysis and then come back to D&F

[warm dove]

I think algebra is a better intro subject than analysis 😦

[woeful sage]

disagree

they are both great intro subjects

🤝

[woeful sage]

so the smallest subgroup containing both subgroups and the largest subgroup contained inside both the subgroups very cool

[frigid shard]

if you want a conceptual understanding of algebra read aluffi

[crystal vale]

I need hint

Any hint? I am thinking if ga = a for all a in A, is it imply g = e, I know I have to show there exists g which doesn't not fix any a in A.

[woeful sage]

What is a transitive permutation group?

[crystal vale]

for any a and b in A, there exists g in G such that ga = b

[topaz solar]

Well, that’s not the same as saying there’s a fixed point free one

Since if you showed such a g = e, that would just rule out things with every point fixed

(And, as an aside, you won’t be able to prove g=e with these assumptions)

[topaz solar]

As in, every g would have some a with ga \neq a

[crystal vale]

I know that's a different question

How do I show there is some g in G which doesn't fix any a?

[topaz solar]

Well I’m saying your thinking above this one seems wrong

[crystal vale]

So ga = a for all a in A, doesn't imply g = e?

[topaz solar]

Well, let’s pretend A = [1, …, n] is just the list of numbers 1 to n

[topaz solar]

It does not

Unless that’s included in your definition of transitive permutation group, which might already quotient out the kernel

Basically just depends on if G is already given as a subgroup of S_n or has a homomorphism to S_n

[crystal vale]

G has homomorphism to S_n, because G acts on A

[topaz solar]

(If it’s the former, then I guess you are correct, but then that’s just by definition)

[crystal vale]

You mean injective homomorphism?

[topaz solar]

Yeah unless you already know it’s injective there may be a kernel

Which is a g with every point fixed

[crystal vale]

I see

[topaz solar]

Anyway

[crystal vale]

Yes

[topaz solar]

So what permutations in S_n have the desired property

[crystal vale]

Any n cycle

Any n cycle, doesn't fix any element

[topaz solar]

Right, but consider S_5

We could also have like, a 3 cycle and a 2 cycle

[crystal vale]

Yes (123)(45)

[topaz solar]

So it doesn’t just have to be an n cycle in S_n

[crystal vale]

Yes

So any permutation which cycle type is n1,n2,..,nk such that \sum n_i = n, n_i ≥2

[topaz solar]

I think so yep

Note however that ((123)(45))^2 isn’t as nice

[woeful sage]

How is that different from a regular permutation group S_A?

[topaz solar]

It’s not the whole thing

[woeful sage]

So what is the transitive permutation group

The smallest subgroup of S_A for which that condition is satisfied?

[topaz solar]

A subgroup of S_n with the property he said

[woeful sage]

Oh any subgroup

that makes sense

Since ga = b is at it's core a transposition applying (a b) to a would give you b

and I assume "transitive" is alluding to transposition

[topaz solar]

So if we have such group, how do you guarantee an object that moves every point at once

[topaz solar]

No

(12345) generates a transitive one on {1, 2, 3, 4, 5}

[woeful sage]

I see

[crystal vale]

See you can act A by S_n × S5, by just the first factor of S_n×S5, | A | = n

[crystal vale]

I see

So how do I prove that exercise?

[topaz solar]

So how would you go about showing such a point always exists?

What happens if you try to directly make one? What if one didn’t exist?

[crystal vale]

Then every g fix some a

For every g there exists a in A such that ga = a

[topaz solar]

But no a is fixed by all g

[crystal vale]

Yes

[rocky cloak]

Maybe you know orbit-stabilizer. What does it tell you about the proportion of elements that fix a given a?

[crystal vale]

|G : G(a) | = n, where G(a) = { g in G | ga = a} and n = | A|, since it is transitive so | G : G(a) | = n

[serene dune]

so a permutation group acts transitively on a finite set with more than one element

exploiting the definition of transitive action you get the group action only consists of a single orbit {ga} for all g \in G

as there is more than 1 such A there has to be one g which doesn't fix all the a's

[crystal vale]

I need to show there exists g such that ga ≠ a for any a in A

[rocky cloak]

So enumerating A as {1, ..., n} if we look at G(1) u G(2) u G(3) ... G(n), that's the set of all elements that fixes something.

[crystal vale]

Yes

[rocky cloak]

So is that everything?

[crystal vale]

No, every G(i) has a common element e

So at most it is n × (|G|/n -1) + 1

And it is less than < | G |

So there g such that it doesn't fix any a

Thank you jagr, Sharp ❤️

[topaz solar]

My name shouldn’t be there

[crystal vale]

You helped me

[crystal vale]

Now you can see the idea

[woeful sage]

after some thought it seems that 6 is the smallest order for which non-abelian groups exist, if you go any smaller, all the groups are abelian

1 is trivial, 2, 3, 5 are all primes so cyclic group, and there are only 2 groups of order 4 (upto iso) both of which are abelian

[chilly ocean]

liked it

[woeful sage]

Man classification theorems are so peak

[chilly ocean]

and 60 is smallest number where non abelian simple group can be found

[woeful sage]

60...

Oh yeah

uhhh A_5 right?

[chilly ocean]

all of them are A5

at the age of 60

[woeful sage]

But what about Z_2 x Z_3

Oh

it's isomorphic to Z_6

Z_m x Z_n is cyclic whenever (m, n) = 1

of course

Can't wait to properly learn sylow's theorems 🔥🔥🔥

[chilly ocean]

https://tenor.com/view/how-bro-felt-writing-that-how-bro-felt-writing-gif-2816109107878310570

[next obsidian]

All groups of order p^2 are abelian even

[woeful sage]

Ah yea you have that as well

[vast stump]

3^2 is actually odd

[twilit wraith]

Group theory seems to reveal 6 as a rather special number

[woeful sage]

And in string theory 6 dimensions are compactified

[twilit wraith]

It reminds me about how Aut(Sn) \cong Sn for all n >= 3

[woeful sage]

calabi-yau

[twilit wraith]

Except for 6

And only 6

[woeful sage]

woah what?

the automorphism group of S_n is isomorphic to S_n? for all n \geq 3 except n = 6

[woeful sage]

damn

[twilit wraith]

Yes

I think its bc 6 is the only number that is the product of two genuinely twin primes

In the sense that 2 and 3 are one apart

[woeful sage]

hiidostuff conjecture

[twilit wraith]

So true

If only I was good enough at group theory to make any meaningful conjectures

One day I will be but for now im struggling in my algebra class

[woeful sage]

Every great mathematician was once a student

[twilit wraith]

True

And i shouldn't say struggling since I am sitting comfortably with an A

But I regularly spend 3-4 hours on the homeworks

Which are daily

[woeful sage]

I mean if you aren't classifying all finite rings do you really know algebra

[twilit wraith]

So true even tho im only doing groups rn

[woeful sage]

oh wow not weekly?

[twilit wraith]

Nah bc its a summer class

[woeful sage]

as in you get homework every day?

[twilit wraith]

I do the 5 hardest questions always since I wanna improve

[woeful sage]

ah makes sense

[twilit wraith]

Yes

So daily 3-4 hours of group theory

[woeful sage]

journey to the top, one step at a time 🗣️

[twilit wraith]

Its also rough bc this is right after 5 hours of math research

[woeful sage]

what are you working on?

[twilit wraith]

So until like mid July im spending about half my day on math

[twilit wraith]

Smth to do with algebraic topology but its really not

Its basic graph theory

Which is less than i hoped but at least im doing research

The problem is the two people im working with have no clue what they're doing which is also unfortunate

One of em is a business major and was shocked that we needed to prove stuff

[woeful sage]

Huh?!?!

Why is a business major working on research with a math major?

[twilit wraith]

I mean he was curious about what math research is like

Which is fair ig

[woeful sage]

fair

Leibniz was a lawyer

[twilit wraith]

But im hard carrying which is frustrating as I was hoping on working with other passionate mathematicians

I'll do research during the year but I think that'll be an endeavor with only me and some postdoc

I suppose I am reassured by the fact that its only my first summer as a college student so I at least will eventually do more respectable work

[woeful sage]

I am sure you will

[crystal vale]

If b is in orbit of a, is that true that stabilizer of a = stabilizer of b?

I think there will be some relation between their stabilizer

Yes conjugation

Any hint for 27?

I know if G is abelian then its class equation is 1 + 1 + 1 +...+ 1

And gi \in stabilizer of g_j

Pick g1, I have to show G(g1) = G

So g1,..,gr in G(g1)

Say k is the size of orbit(g1)

Then k = |G|/|G(g1)|

[vivid kestrel]

in a noncommutative unital ring A with left ideal I, the annihilator of A/I is contained in I but is it always all of I?

[rocky cloak]

Notice that an annihilator is always a two sided ideal, but not all left ideals are two-sided

[vivid kestrel]

ah of course, thanks!

[rocky cloak]

It is in fact the largest two sided ideal contained in I.

[vivid kestrel]

how does one see that

nvm I see

ty

[quartz wind]

construct a counterexample to $G\cong H\times G$ implies $H$ trivial. does the following work? for any $n$ and odd $p$, define $\varphi : \mathbb{Z}\to \mathbb{Z}\times 2\mathbb{Z}:2^np\mapsto (n, 2p)$

[cloud walrus]

esca

[rocky cloak]

This is not a homomorphism, and it's a little unclear what happens for negative numbers, but it doesn't seem to be surjective

[quartz wind]

hmm yeah youre right i was mixing up multiplication and addition. thanks.

[near tapir]

Let ( R = \mathcal{P}(S) ) denote the set of subsets of a set ( S ).
Define two new operations on ( R ): for all ( A, B \subseteq S ), set
[
A + B := (A \setminus B) \cup (B \setminus A), \quad
A * B := A \cap B
]
Prove that R is a ring.

[cloud walrus]

hzh

[near tapir]

Does it suffice to draw diagrams to verify the ring axioms?

[delicate orchid]

depends how good your diagrams are. But I'd just do it symbolically

[true bolt]

Maybe they're proving the thing is a ring object in Set, in which case yes you can draw commutative diagrams

[delicate orchid]

I thought they meant drawing venn diagrams to prove things like distributivity

[near tapir]

That’s what I meant, I’m doing it symbolically now but stuck on proving associativity

[delicate orchid]

associativity of addition I'm assuming? Since it's just an XOR operation I'd write out a huge (8 rows) fuckin truth table personally but there's probably a cooler way of doing it

[true bolt]

Personally I'd do that using a series of venn diagrams

[delicate orchid]

well hold on, since it's 2 xors an element of A,B, or C belongs to (A+B)+C if and only if it belongs to an odd number of {A, B, C}, and same is true for A+(B+C)

so they define the same set

[rocky cloak]

I guess an alternate approach is just showing that it's isomorphic to (Z/2)^S which you know is a ring

[woeful sage]

These types of groups are interesting

A non-abelian group where every subgroup is normal

I think they are called Hamiltonian groups?

dayummm

[delicate orchid]

yeah so it's basically just Q_8

[woeful sage]

Lattices of subgroups are actually super pretty

Yes, even this monstrosity

it's a beautiful monstrosity

Thought I wonder what the consequences of the lattice of subgroups for a given group to be a planar graph is (and in general how graph theory interacts with group theory in this manner)

[chilly ocean]

Maybe in geometric group theory

[vestal girder]

is this the atlas?

[chilly ocean]

looks like some blue print of tony starc

[near tapir]

Would this table suffice? Going off that x is in A + B if and only if x is in A or B

[woeful sage]

woww

[woeful sage]

Interesting

[delicate orchid]

Open ended exercise: Compare and contrast the subgroup lattices of D_{2^n}, Q_{2^n} and SD_{2^n}. They’re very closely linked to each other in a nice inductive manner

[woeful sage]

[woeful sage]

SD? semi-dihedral group?

what is Q{2^n}?

generalized quaternion group?

[delicate orchid]

yes and yes but you can think of it as dicyclic of order 2^n if you want

[delicate orchid]

I woulda just done this

[near tapir]

Yeah that looks way better

[delicate orchid]

then cause the last two columns are equal they both define the same set

[near tapir]

Thanks

[keen badge]

Let $X,Y\subset\mathbb{C}$ such that $[\mathbb{Q}(X):\mathbb{Q}]=[\mathbb{Q}(Y):\mathbb{Q}]<\infty$ does it mean that there exists an isomorphism $\phi:\mathbb{Q}(X)\to\mathbb{Q}(Y)$ with $\phi(\mathbb{Q})=\mathbb{Q}$??

And what about if the degrees of extension are infinite

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[keen badge]

for the infinite case actually it is obviously no, because we can take $X=Q_{a\ell g}$ and $Y=R$.
But what about the finite case?

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[rocky cloak]

No, you can find examples even with |X|=|Y|=1

[delicate orchid]

Gal(-) is functorial so I highly doubt this is true, pick two Galois extenstions F/Q, G/Q satisfying your condition with non-isomorphic Galois groups, then any isomorphism f : F/Q -> G/Q would induce via functoriality an isomorphism between your galois groups, a contradiction

[rocky cloak]

I don't think Gal is functorial.

But it is true that isomorphic extensions would have isomorphic Galois group

Or at least I'm not sure what exactly the domain category for it should be when you say it's functorial

[keen badge]

I tried to find examples but like..
$\mathbb{Q}(\sqrt n)\cong \mathbb{Q}(\sqrt m)$ for $m,n$ not perfect squares and that's it actually.. I didn't try any other examples

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[rocky cloak]

Not sure why you think those are isomorphic, but that would be the easiest example yeah

[quartz wind]

which operation on Q should i assume here?

[rocky cloak]

Addition

[keen badge]

Lol, why they are not isomorphic?

OHH

WAIT

I'm dumb sry.

So $\mathbb{Q}(\alpha)\cong\mathbb{Q}(\beta)$ for $\alpha,\beta\in\mathbb{C}$ if and only if $\text{irr}(\alpha/\mathbb{Q})=\text{irr}(\beta/\mathbb{Q})$?

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[delicate orchid]

I googled it. It seems to be

[rocky cloak]

Not quite, they can be isomorphic on other cases as well

[keen badge]

Is there a "nice way" to classify all cases?

[rocky cloak]

I'm gonna guess no

[keen badge]

Not even when $|X|=|Y|=1$? Damn

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[delicate orchid]

I should've specified specifically that we're mapping out of extensions over a prime field of fixed characteristic so it's basically just the galois correspondence

[quartz wind]

thanks

[tardy hedge]

how do we know w_i not 0 in Mp

since Mp not 0 there is some m/s not 0

m is generated by wi

[rocky cloak]

If all the wi were 0, then all of Mp would be 0

[tardy hedge]

ok thats what i thought initially but then i did something that confused me i think

but no yea that makes sense thats fine

th

thcx

thx

[tardy hedge]

if x = a1m1+a2m2 is ann(x) = cap ann(mi)

or is it just one way inclusion

[rocky cloak]

I mean, what if a1=a2=0?

[tardy hedge]

then ann(x) is (1) which is probably not cap ann(mi)

[twilit wraith]

Let m > 2 be an even integer and let n > 2 be an odd integer. Find a formula for the
number of elements of order 2 in Dm × Dn

im finally getting to the point where these problems are getting cooler and im getting better at algebra enough to solve them faster

at least i hope

but the formula i got is mn + m + n + 1 and i was wanting to check with yall that this is right

mn because i can have (sr^i, sr^j) for any 0 \leq i < m and 0 \leq j < n

m because i have have (sr^i, e)

n for the opposite reason

and 1 because i can have (r^(m/2), e)

[tardy hedge]

Im trying to understand a proof - i dont get this part. M if fg A-module with A noetherian. If an ideal I contains only zero divisors of M, then there is an associated prime of M containing I?

What I know about associated prime is that P is associated prime of M if P = ann(x) for some x in M and P is prime ideal

[delicate orchid]

yeah this is right, there are n+1 involutions in D_n and it's clear that (x,y) is an involution iff x and y are

[twilit wraith]

i see

i really should get (m+1)(n+1) tho?

i guess im questioning myself because i could have used better logic

i know that if |a| = m and |b| = n then |(a,b)| = lcm(m,n)

[tardy hedge]

Adios Muchachos

[twilit wraith]

and since 2 is prime this implies that for |(a,b)| = 2 then |a| or |b| = 2

so now if i have some function that counts the order 2 elements in D_k, say f(k)

i just get f(m) + f(n) + f(m)f(n)

which is (m + 1) + n + (m+1)n

oh hey yeah it does work

phew

actually no i think i did math wrong

i should get mn + m + 2n + 1

[delicate orchid]

well no it should absolutely be symmetric in both arguments

D_m x D_n is iso to D_n x D_m but you're claiming they have different numbers of involutions

[twilit wraith]

i dont think that this is true though

even if i flip the order

it still holds that i should have f(m) + f(n) + f(m)f(n) elements with order 2

at least i think

since that expression is symmetric for f(m),f(n)

[delicate orchid]

I think my answer was off too, it's alot more complicated than I'd like actually

it depends on if m or n are even/odd

[twilit wraith]

yeah, which is why the given parities make this question more interesting

[delicate orchid]

oh you were given parities

[keen badge]

I'm trying to find a basis for $\mathbb{Q}(\sqrt{1+\sqrt{2}})$. \
I proved that $[\mathbb{Q}(\sqrt{1+\sqrt{2}}):\mathbb{Q}(\sqrt{2})]=2$ and using the tower rule we get that the basis consists of 4 elements, and it is obvious that $\left{1,\sqrt{2},\sqrt{1+\sqrt{2}}\right}$ are part of the basis, so how can I find the 4th generator?(I know it is $\sqrt{2}\cdot\sqrt{1+\sqrt{2}}$ but I don't know how to prove it)

[twilit wraith]

i think the logic i used is fairly sound but maybe i am missing something

[twilit wraith]

yes, m is odd while n is even

sorry other way around

[delicate orchid]

in that case I absolutely agree with this

[twilit wraith]

alright awesome

im happy this didnt take me hours unlike the other problems ive had in past days

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[crystal vale]

I don't need a solution, I need a hint, give me G and K

[delicate orchid]

well giving you K will give the game away immediately, so I'll just say take G to be S_4

[brittle ridge]

sqrt(2)*sqrt(1+sqrt(2)) is what you are missing

[keen badge]

have you read the brackets? lol

[brittle ridge]

i just noticed it now lol sorry

[keen badge]

np

[brittle ridge]

since yk it why don't you just prove that they're all linearly independent

[keen badge]

Maybe there is a way to understand why it's true instead of "guessing" this is it?

I mean.. the following is PROBABLY not true, but maybe a field extension of degree 4 with basis {1,α,β,γ} has γ being linear dependent with αβ

[keen badge]

(Over Q)

[brittle ridge]

in your case u have 2 extensions of degree 2 u make a basis for each then you mmultiply the sets and you get your basis

[keen badge]

yes, that's exactly what I did, so {1,sqrt(2)} is a basis for the extension Q(sqrt(2)):Q and for the Q(sqrt(1+sqrt(2))) : Q(sqrt(2)) why it gives me both sqrt(1+sqrt(2)) and sqrt(2)sqrt(1+sqrt(2)) as the generators

(sqrt(1+sqrt(2))) is obv

[brittle ridge]

Q(sqrt(2)) gives 1,sqrt(2) and Q(sqrt(1+sqrt(2)) gives 1 and sqrt(1+sqrt(2)) then multipling them gives the answer too

[keen badge]

so we are back to "guessing" the final generator.

That's fine

ok

Ill just "guess" it(and prove it is a generator)

[keen badge]

And what do you think of this?

[brittle ridge]

you didn't guess here you just get the basis of each extension by itself andmultiply them

[brittle ridge]

i do think it's correct

[keen badge]

Forreal? damn

If yes, then I get what I wanted

Question is, how to prove it lol

[brittle ridge]

wait i'll try

[keen badge]

ok.

How to even start trying and think abt this question... hmmm

[brittle ridge]

$\alpha\beta = a + b\alpha + c\beta + d\gamma \quad \Rightarrow \quad
\gamma = \frac{1}{d}(\alpha\beta - a - b\alpha - c\beta)$

[cloud walrus]

Mexn

[brittle ridge]

shouldn't this be enough?

[keen badge]

damn

[brittle ridge]

assuming d~=0

[keen badge]

yeah ofc

[crystal vale]

K4 is characteristic subgroup of S4, because it is unique normal subgroup of order 4, right?

{ 1, (12)(34), (13)(24), (14)(23) } is a characteristic subgroup of S4, but {1, (12) } is not normal in S4

{1, (12) } is a normal subgroup in that characteristic subgroup

[delicate orchid]

exactly

presuming you mean (12)(34)