groups-rings-fields

[thorn jay]

in a sense

the explicit isomorphism is given by (m1, ..., mn) ↦m1 + ... + mn

[noble nexus]

Yeah often you write elements of the direct sum as "formal sums" just because it's easier to keep track

[coral spindle]

what would the multiplication be?

But I don't see a way to compose M → N with M → N

seems I have things in N that I can't shove into a function taking things in M

In general it's a J-module as well, but we only get a ring structure like that if we're talking about M = N

in which case we tend to call that End(M)

I don't know what that means I'm afraid

[glad osprey]

the right hand equation seems to be just a matter of writing out the definitions, dunno what sigma is though, or what the _ij subscript means

yeah, that's the same as End(M)

but that doesn't seem useful for this case, since you also have maps to and from M_j

yeah, but M_i -> M isn't an element of End(M), so you can't use any ring properties there

[thorn jay]

Hom_R(M, M) is a ring because more generally the composition map
μ: Hom_R(M, N) x Hom_R(N, Z) → Hom_R(M, Z)
is bilinear and associative

[glad osprey]

linear in each parameter, so f(g + h) = fg + fh and (f + g)h = fh + gh

(ø + psi)(pi(x)) = ø(pi(x)) + psi(pi(x)) is correct, this is just from the definition of addition of homomorphisms

dunno what the rest of your argument is, you should write it out in more detail

[glad osprey]

this isn't the distributive property btw

you know Hom(A, B) is an abelian group for modules A, B, right? How is it defined?

I'm sure they give it a group structure later in the notes

wtf

this is the closest they get to defining the group structure of Hom(M, N)

here is from D&F, part (2) is what you're interested in

if they gave you that exercise without even defining what phi + pi means for homomorphisms, then your lecturer sucks 💀

what you're trying to prove follows directly from this definition

[cloud walrus]

mq

[glad osprey]

what third bit?

(3) from D&F?

No. What is lambda?

[cloud walrus]

mq

[glad osprey]

yeah, what does it mean for lambda to be in Hom(M, M_j)?

yep, and a homomorphism is?

in terms of definition, you don't any advanced theorems here

yep, that's all you need

yes

the hard part of this exercise is paying attention to what is addition of homomorphisms and what is addition of elements in M. It's confusing because we use + for both

nooo

it's not, I promise

just write down what you want to prove, and see if you can expand the definitions

I gotta go now, but someone else will help

this follows from the definition of a homomorphism

[tardy hedge]

elements of that are homomorphisms

[glad osprey]

@tardy hedge help this guy out, I'm passing you the baton

[tardy hedge]

whats the original question?

[thorn jay]

can you give the full context

[tardy hedge]

yea like i have no idea what the question is tbh

[lime badge]

Sorry, I forgot to follow-up on the discussion that followed this question of mine

The question was motivated by the following proof sketch that (c) \cap (d) is contained in (cd):

Let p in (c) \cap (d). Then p = r_1 c = r_2 d for some r_1, r_2 in R = k[x,y]/I

*Since c and d are univariate in different variables, r_2 = c r_3. Then p = r_3 cd, so p \in (cd)

Now the step labeled * doesn't work in general, for example if I = (x -y). So that's what I was hoping to get a handle on, what simple conditions I could put on I to ensure the step * goes through.

I think the ideal I being of the form I = (p(x), s(y)) works, unless I'm mistaken?

[tardy hedge]

i feel like u just kinda have to do some definition unpacking for that one

it looks annoying ngl

[elfin wraith]

Yeah this is sorta just “how matrices work”

[dim wagon]

the solution uses the regular character on $G=C_2$. I realised that the regular character on any group of even order will work since $\chi(1)=|G|$ and $\chi(g)=0$ for everything else.

does anyone know if there are more interesting examples?

[cloud walrus]

somethingwrong

[thorn jay]

why does it work for any even ordered group?

whats stopping the regular representation to split into a direct sum of two reps with character 1 ↦|G|/2 and the rest 0

[dim wagon]

ah okay so for example the regular character on C_4 and the regular character on C_2 right? the former is twice the latter.

ohh then does this mean that the only abelian group it works on is actually C_2?

[thorn jay]

yeah i think lol

cuz of the classification theorem, any even-ordered abelian group must surject onto C2, and thus the regular character can be written as a sum of regular characters of C2

that is a strange thing to think abt

that cant be true though can it? that would mean the regular character can be written in terms of the irreducible characters of C2, which in turn are irreducible characters of your abelian group

which would mean that there would only be 2 irreducible characters

[dim wagon]

im not sure what the classification theorem is unfortunately.

[thorn jay]

every finitely generated abelian group is a direct product of cyclic groups

[dim wagon]

i was thinking that if we had an odd abelian grp, the condition would not be satisfied so we are done. If we had an even abelian group $G$, then the character of $G$ would be twice the character of $C_{|G|/2}$

[cloud walrus]

somethingwrong

[thorn jay]

thats not how characters work

[dim wagon]

hmmm why not?

i mean the regular character*

ah okay, i get it, i got confused

thanks alot

[thorn jay]

because you cant just use a character from another group

[dim wagon]

i cant compare between two different groups like that

[thorn jay]

i really want someone to correct me on this

ohh lmao

i see my fault

im stupid

[thorn jay]

this is false

like for the odd order abelian group thing too

[thorn jay]

your initial idea was correct though, huh

i remembered an exercise from Serre's book: every character such that χ(g) = 0 for all g ≠ 1 must be an integral multiple of the regular character

so if G has even order, its regular character cannot be written as twice some other character of G

proof is fairly simple: let χ be such a character and ψ any character. We compute:

(χ|ψ) = 1/|G| • χ(1)ψ(1) = Χ(1)/|G| • ψ(1)

Then taking ψ(g) = 1, i.e. the trivial character, one sees that (χ|ψ) = χ(1)/|G|. But the inner product with an irreducible character must be an integer, so χ(1) = k • |G|, and we are done

[kind kindle]

Alright I'm a bit of a novice to group theory. I'm working on a cryptography problem. In particular, trying to find a finite nonabelian group G with the following properties.

1. |G|>=2^128
2. There is no known polynomial time algorithm to determine if two elements of the group are conjugate, and the problem is thoug,ht to be hard on average for G
3. There exists a Low dimensional nonlinear representation of G, a(I know every finite group has to have a linear representation. This just means that there is some map a that applies to elements of g such that a preserves group structure, but makes findling a linear representation of g given only a(g) hard.) that is easy to construct and compute for a sane computer. It has low complexity. But finding a linear representation of g is hard given only a(g)

Thinking about it, this is based moreso on finding that nonlinear representation. Rather than the group itself. You could probably do this over almost any non ablian group of sufficient size if you had the right nonlinear representation.

Basically, Eliptic curve cryptography and RSA use these sorts of nonlinear representations over abelian groups. In particular, one or two dimensional nonlinear representations.

The groups i've thought about would be... basically just a trivial collection. the multiplicitve group of F_p[Q_8] that obeys the normal quaternion rules over R. Where F_p is a finite field of prime order, Q_8 is the quaternion group, and we set i^2=j^2=k^2 = p-1.

or in particular the group of them that have nonzero norm over the field

But uh... that has a low dimensional linear representation and figuring out conjugacy would be trivial

4. This group shouldn't be nilpotent

My main thought would be semidirect products of large cyclic groups and symmetric groups. Where we somehow eek out a low dimensional nonlinear representation

Also researching finite groups of the lie type, from what i've been able to parse they sound similar to eliptic curve cyclic groups(but eliptic curve groups are for the degenerate case where the dimension of the matrix is 1x1)

[gilded saffron]

gonna need more context

rewrite it as a matrix and two vectors

[topaz solar]

For part 2, it’s finite, so there’s an algorithm

[kind kindle]

Yeah. Doesn't mean it has to be practical

you can factor an rsa 4096 key eventually but that doesn't mean it's a practical approach to break the encryption

[quiet pelican]

It’s the set {f(a + b) | a \in I, b \in J}

[thorn jay]

do you know what I + J is

[tough raven]

Reading your proof, it amounts to the fact that mult(trivial in regular) = 1, so regular can't be a multiple of any character of a representation, right?

[tough raven]

What is a nonlinear representation? I can't understand your explanation. Could you give a definition directly?

[vocal pebble]

Polynomial time in what? Order of the group? I can give you a Theta(|G|) algorithm for any finite group in that case. conjugate by each element.

[thorn jay]

aha thats much simpler lol but yeah

inner product of characters is always a positive integer

[limber tapir]

In cryptography one usually means polynomial in the bit-length of the order of the group

[kind kindle]

I misspoke when i said polynomial time. I guess I met that "the average case complexity of conjugation in this group has complexity greater than 2^128 operations"

"While being viable to use for cryptography

[kind kindle]

yeah. We don't want a polytime algo like this here

[glad osprey]

what's a non-linear representation?

[kind kindle]

look up eliptic curve point addition. That's an example of a nonlinear representation of a cyclic group

[glad osprey]

that sounds like an oxymoron aren't representations by definition maps from G to the automorphism group of a vector space?

[kind kindle]

i'm using a wierd definition

[glad osprey]

you don't mean representation in the sense of representation theory?

[kind kindle]

Just some way of representing group elements as something where the group operation isn't a bilinear map on the elements. But the group operation given 2 elements is still efficiently computable

[kind kindle]

No sorry for using weird definitions

[glad osprey]

hmm, ok

[kind kindle]

I was thinking some weird generalization of an eliptic curve addition group could maybe be non abelian

[thorn jay]

functor from group as category with one object to some category that is not K-Vect, or A-Mod for a K-algebra A

:P

[kind kindle]

I've looked at the multiplicitve group of elements of nonzero norm in F_p[Q], with the additional rules that:
{i^2=j^2=k^2 = p-1, -i = (p-1)i, -j = (p-1)j, -k = (p-1)k}

but i suspect that this isn't much better for cryptography than a normal abelian group

due to the quaternion group being nilpotent

while i'm here some resources to self study abstract algebra would be useful

Also, I suspect that eliptic curve additon groups are a special case that reduces to an abelian group of some family of groups related to finite lie variety groups but idk

[quiet pelican]

The typical maths recommendation would be something like Dummit and Foote, but I’m not sure if there’s something better specifically for cryptographic applications

[kind kindle]

I'm interested in abstract algebra in general but i haven't quite gotten there in college yet

i get distracted from my calc 3 homework doing abstract algebra nonsense

[quiet pelican]

In which case Dummit and Foote is probably a good idea

[kind kindle]

i also might be doing an independent study of category theory with my physics advisor next semester

(with it's applications to physics in mind)

I suspect that will make abstract algebra easier when i get there

[quiet pelican]

Otoh it would be very useful to know some abstract algebra before category theory, because a lot of your examples are categories of algebraic objects

[kind kindle]

I mean i'm familiar with the basics of abstract algebra. I have the definitions of field, group, ring, and vector space memorized

and i'm vaguely familiar with lattices and modules

and yeah it makes sense that category theory would be useful for describing those things

[quiet pelican]

That’s probably sufficient
Like if you’re following a book you may need to take an example or two on faith but like

That doesn’t matter that much

[kind kindle]

I'm just not super familiar with the more advanced theorems and stuff

[quiet pelican]

What are you counting as more advanced theorems?

[kind kindle]

basicaly anything nontrivial :p

[thorn jay]

what kind of lattices

the cool ones or the subgroup of R^n one

[kind kindle]

over finite fields mainly because most of my study of abstract algebra thus far has been motivated by cryptography

[thorn jay]

aeeuah and here I was excited

[kind kindle]

in particular the hobby i have of hyperfixating on finding a fast classical factoring algorithm somethimes(because i'm 90% sure one exists)

[thorn jay]

but yeah fair, for cryptography

[kind kindle]

I know i probably won't find one, but it will teach me a lot.

And if I ever do find one, all I'll do is publish the factors for all the RSA challenge numbers anonymously on pastebin with a vaguely ominous message. And do nothign more(not releasing the algorithm even or using it to hack)

but i'm almost sure i won't find one with my current state of knowledge. that's fine. trying has taught me a lot.

I rmemeber i more or less independently discovered a version of fermat's factorization algorithm in high shcool

Also, my goal here is to find a post quantum crypto algorithm that's similar in speed and keysize to current algorithms(ECC basically)

[glad osprey]

This looks like a good intro to abstract algebra if you're interested in cryptography: https://shoup.net/ntb/

[kind kindle]

The most promising canidate for that seems to be a natural generalization of current cryptographic primitives to non ableian groups(basically, all of the crypto we use today is based on hard representations of finite abelian groups.)

There has seemingly been some research on infinite non-abelian groups, but not a whole lot on finite ones.

[kind kindle]

probably because conjugacy over finite groups is seemingly reducable to linear algebra a lot of time

but if we had a representation that is hard to convert to a linear representation, and was low dimensional over some big (sub)group, than that'd essentially be a holy grail.

Assuming conjugacy is hard over this representation

[limber tapir]

Did you look at isogeny based cryptography yet?

Specifically CSIDH is a key exchange algorithm that does not have massive keys and has not yet been broken

Although it has not been around for particularly long and the somewhat related scheme SIDH was broken a couple years back

Either way it's extremely interesting

[tardy hedge]

for m a maximal ideal and M an R-module, if M = mM then M = 0 right

M fg

[karmic moat]

For R a local ring yes, that’s nakayama’s lemma

If R is not local then idt it works

[tardy hedge]

i was talking to chmonkey about this before. Oh yeah I think we were trying to show Mm = 0.

we can say Mm = 0 here

[brisk adder]

P25920.!

[thorn jay]

type shit

[thorn jay]

wdym, you want the conclusion M_m = 0 or have that also as assumption

[tardy hedge]

The conclusion

[thorn jay]

hm, well im sure you can pass to the localisation

you just have to prove that (mM)_m = m M_m

because then 0 = (M/mM)_m = M_m / m M_m, so m M_m = M_m and by Nakayama you are done

[thorn jay]

and this should be pretty immediate i think

[rapid cave]

If I have polynomials $f_1, f_2, f_3,..., f_m \in \bQ[x_1,...,x_n]$ such that there exist polynomials $g_1, g_2, g_3,..., g_m \in \bC[x_1,...,x_n]$ such that $$f_1 g_1 + f_2 g_2 + ... + f_m g_m = 1$$ How can I show that $g_i \in \bQ[x_1,...,x_n]$ or at least there are polyonomials like $g_i$ that are with rational coefficients.

[cloud walrus]

ExpertEsquieESQUIE

[tough raven]

R = ℤ, M = ℚ, m = (p)?

[tardy hedge]

Ya right

So def not M = 0

[knotty badger]

Consider $\sum_i (f_i)$ in the rational polynomial ring

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[rapid cave]

it can be anything

[knotty badger]

What’s the sum in the complex polynomial ring?

[rapid cave]

the same?

[knotty badger]

Same as what

[rapid cave]

there is no difference to this sum if we work in C or in Q

[knotty badger]

I don’t think that’s true

The rings are different after all

So the ideals must be different too

[rapid cave]

oh its a sum of ideals

[knotty badger]

Oh

Yeah sorry

Should’ve made that clear

I was using (f_i) to denote the ideal generated by f_i

[rapid cave]

I though it was the sum of f_i, didn't see the parenthesis

I need to show this sum is (1)

[knotty badger]

Yep

[rapid cave]

and I know this sum over C is (1)

[kind kindle]

I have not. I shall take a look at that sometime

[rapid cave]

something along the lines of f1Q + f2Q + .... + fmQ = (f1C + f2C + .... + fmC) \cup Q[x1,...,xn]?

[kind kindle]

matrix conjugacy over finite fields can be easily reduced to a linear system right

as in conjugacy for matrix groups over finite fields

yup, it does

[knotty badger]

Yep

My suggestion would be to argue that the sum over Q is (1) as well

[kind kindle]

ahh i need papers on nonlinear representation theory

[rapid cave]

yeah how would I do that

I have a hint of linear algebra

[knotty badger]

Hm maybe using that the rational polynomial ring is a PID helps

[rapid cave]

is it?

in multiple variables

[knotty badger]

Wait

Ah yeah

[tough raven]

Are you familiar with Grobner bases?

[knotty badger]

Forgot that only works for single variables

[rapid cave]

no

[tough raven]

Hmm. They're essentially like a multiple-variables version of e.g. the Euclidean gcd algorithm in one variable. You could have used that, because the algorithm (say, to determine whether 1 lies in the ideal) uses nothing except field operations and so has to give the same answer over any field.

Let's see if we can't unfold it into more elementary words.

[kind kindle]

definiton. A polynomial representation of a group is a function a from G -> F^n (where F is a field) such that the following holds:

1. for all n, m in G, n*m = q -> a(n) * a(m) = a(q)
2. a(n * q) = P(a(n), a(q)), where P is a polynomial from (F^n)^2 -> F^n
3. a(n^-1) = Q(a(n)) where Q is a polynomial from F^n -> F^n

Definition:
A quadradic representation is a polynomial representation of a group G such that Deg(P) = 2 and Deg(Q) = 2

[tough raven]

This looks pretty algebro-geometric in nature. You might have luck looking at linear algebraic groups over finite fields (elliptic curves are projective algebraic groups AKA abelian varieties, linear algebraic groups are another class of algebraic groups which need not be abelian unlike abelian varieties).

In particular, "finite groups of Lie type".

[kind kindle]

Conjecture, there exists a finite non abelian group G such that conjugacy is exponentially harder than the dimension of the polynomial respresentation

[kind kindle]

Yup, i've made that connection too. My attempts here are to generalize the eliptic curve group operation to be non abelian, to construct a group wth this sort of representation

[thorn jay]

what do you mean "harder than the dimension"

[kind kindle]

i was inprecise

[thorn jay]

is * here just some product you assign to the set of polynomials?

[kind kindle]

And I have a feeling that Eliptic curve groups are degenerate cases of these sorts of groups.

[kind kindle]

i made a mistake. corrected

[thorn jay]

what, i see no difference in your messages

[kind kindle]

* is the group operation

[thorn jay]

that doesnt answer my question

the * in a(n) * a(m), is that just any group operation you put on polynomials? is it multiplication of polynomials?

[kind kindle]

i made a typo. i met a(n * m), where n and m are elements of G

[thorn jay]

???

im talking about point 1.

[kind kindle]

that was inprecise ignore it pls

[tough raven]

OK, I can give an answer without invoking Grobner bases but it's much actually only slightly less explicit.
Consider the ring R = ℚ[x1,...,n]/(f1,...,m) and S = ℂ[x1,...,n]/(f1,...,m). We want to show that if S = 0 then R = 0. But R is a ℚ-vector space and S is a ℂ-vector space and they have the same dimension as a ℚ- and ℂ- vector space respectively[*]. So obviously one is zero iff the other is.

The easiest way to justify * is that S = ℂ (⨯)_ℚ R, but I don't know whether you're familiar with base-change / tensor products.

[rapid cave]

I am familiar with tensor products.

I don't think I have the best intuiton for it but I learned it in commutitive algebra

[tough raven]

Great.

Can you see that my argument works provided S = ℂ (⨯)_ℚ R?

[rapid cave]

yes

wait

isn't C not f.g. as a Q-vector space?

[tough raven]

Correct.

Neither is R in general.

It's a tensor product of a possibly infinite-dimensional vector space with an infinite-degree field extension.

[rapid cave]

so its dimension should be infinite no?

[tough raven]

We are equating a ℚ-dimension (of R) with a ℂ-dimension (of ℂ (⨯)_ℚ R).

[rapid cave]

ok I see

wrote it down

now why is S = C \otimes_Q R....

[kind kindle]

Lets generalize my definition a bit. P and Q can be rational functions as opposed to just polynomials. I've looked into linear algebreic groups, but I can't think of any examples of nontrivial ones(i.e, ones who's polynomials include addition at all.)

because a matrix sum could have determinant zero even if the two matricies do not

and in genral, the problem of "given a polynomial and a finite matrix group, does the polynomial define a linear algebreic group" feels co-np hard

[tough raven]

If you saw tensor products in commutative algebra, you might be familiar with using their universal property?

In particular, for S, T R-algebras (all commutative rings), a ring map out of S (⨯)_R T is the same as ring maps out of S and T that agree on R.

[rapid cave]

yes

[tough raven]

How about GL_n? There are also unipotent groups (such as the group of upper-triangular matrices with 1's on the diagonal), though I think those will only give you p-groups.

[rapid cave]

now what maps do we have here?

[tough raven]

Try to use this to show that ring maps out of S and ℂ (⨯)_ℚ R are the same.

(Using the universal properties of polynomial rings and quotients as well.)

[rapid cave]

what does this mean?

[kind kindle]

alright so reading up on this, a rational point in a finite field means that you have some curve in the algebreic closure of F, such that the point's elements are in F right

[tough raven]

Formally, that there is a natural isomorphism Hom(S, T) = Hom(ℂ (⨯)_ℚ R, T) for all commutative rings ℂ-algebras T.

[tough raven]

Informally, the same thing, but just verbally argue that a map S to T and a map ℂ (⨯)_ℚ R to T consist of "the same data".

[rapid cave]

I would prefer to do this formally

after showing this by the universal property S isomorphic to C \otimes_Q R right?

[tough raven]

Have you seen that any two things satisfying the same universal property are (uniquely) isomorphic?

[rapid cave]

I rememeber this

[tough raven]

Then what I am asking you to do is show that S has the universal property that ℂ (⨯)_ℚ R has by definition, and therefore get an isomorphism between them.

[rapid cave]

ahhhh

would like some help

if f: C x R --> T is bilinear, I want to show f = g o (x) for a unique homomorphism g: S --> T?

[kind temple]

could you give more context? what are C,R,T,S?

assuming (x) is the tensor product

[rapid cave]

.

need more information?

[kind temple]

this is just the universal property of S

or is that what you are trying to show, that S has this universal property

[rapid cave]

Yeah I don't know how to show that

[knotty badger]

oh neat universal properties

[rapid cave]

I think tensor here is
(z, r mod (f1,...,fm)_Q) -> zr mod (f1,...,fm)_C

[knotty badger]

yeah

[rapid cave]

But how would I show every bilnear map f: C x R --> T factors through it

[elfin wraith]

Do the words universal property summon you in the same way that bifunctor summons wew?

[knotty badger]

i didn't know bifunctor summons wew

[knotty badger]

well i think - given a bilinear map C x R -> T, how do you make a linear map S -> T?

dw about factoring for now

and really it's not just $\textit{any}$ bilinear map, it's one where $B(z r, p) = B(z, rp)$ for every $z \in \mathbb{C}, r \in \mathbb{Q}, p \in R$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[rapid cave]

I don't know

[knotty badger]

i think the universal property of the quotient may be helpful...?

[tough raven]

Not just bilinear, but an algebra homomorphism (so respects 1 and multiplication).

[rapid cave]

I get it....

[rapid cave]

Thinking about this

[knotty badger]

ah right yeah

[rapid cave]

I am lost

[rapid cave]

How would I do this?

[knotty badger]

i think you can use the algebra hom $C[x_1, ..., x_n] \to T$ by sending $\sum_J a_J x^{(J)}$ to $\sum_J B(a_J, q(x^{(J)}) )$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

here J is a multi-index $J = (j_1, \dots, j_n)$ and $x^{(J)} = x_1^{j_1} \dots x_n^{j_n}$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[tough raven]

To recap, you have maps a: ℂ → S = ℂ[x]/(f) and b: R = ℚ[x]/(f) → S = ℂ[x]/(f) (can you write down what they are?) and you need to show that for any (commutative ring T and) pair of maps a': ℂ → T and b': R → T that agree on ℚ, there exists a unique map f: S → T such that a = f ∘ a', b = f ∘ b'.

(Where the word "map" means "commutative ring homomorphism" every time.)

[rapid cave]

a: z --> z mod f
b: r mod fQ[x] --> r mod fC[x]

Let me try writing it down

[rapid cave]

linear here i.e. ring homomorphism? Q-module homomorphism? which one?

[knotty badger]

-# sorry maybe ask raghuram, can’t really help rn :c

[fickle dirge]

Can someone verify this proof? If we have a string of elements from the dihedral group and we have an even number of occurences for each symbol then its a rotation.

Proof: We define an equivalence relation where $a \sim b$ iff $ab$ is a rotation. Then, $ D_n/ \sim $ will be the two-element group ${e,R}$ where $R^2 = e$. We can see then, that $s$ is a rotation iff it's image $\pi(s) = e$. Then, given our string s, it's image under the quotient map is the product of images $\Pi_{i}^n \pi(s_i)$. Since our quotient group is commutative, and by assumption, we have an even number of elements for each occurrence, we can commute each element and its pair to be multiplied. Then, we obtain the identity for each pair. Therefore, our image is $e$. Hence, we get that $s$ is a rotation.

[cloud walrus]

XDStar

[thorn jay]

in the same way that anything tangentially related to universal algebra summons me

[rapid cave]

will try that tomorrow

[thorn jay]

🔥

[rapid cave]

I won't bother you dw

[thorn jay]

if its about universal algebra i will never be bothered

[rapid cave]

I don't know any

[thorn jay]

not that i have a life, anyways

[rapid cave]

so I would just be shitposting to summon you

[thorn jay]

good enough

[rapid cave]

my lecturer for CA and ANT was really good and talked about some history. But the recordings aren't public and its not in english :/

[tardy hedge]

the worst videos ever are those "socratica" ones or whatever

just made me thought of that

i mean ok for like basic basic intro sure

[glad osprey]

If we can just figure out the second worst, then the third worst, etc, then we can eventually give mq the best youtube recommendation 💪

[tardy hedge]

but even for that i wouldnt recommend them

[tardy hedge]

so true

Yea i agree

Yeah

[rapid cave]

@tough raven sorry for the ping

can you check my work?

[tough raven]

I don't understand the notation you have used for B.

Which universal property are you using?

[rapid cave]

B should be the tensor map (z,r) -> z \otimes r

[tough raven]

This one?

[tough raven]

Do you want to use the universal property of tensor products of vector spaces?

[rapid cave]

its the same one no?

just a different formulation

[tough raven]

In that case, you'll get a ℚ-vector space isomorphism and you'll later have to show that it is a ℂ-algebra homomorphism (i.e. ℂ-linear and multiplicative (including mapping 1 to 1)).

[tough raven]

No, the universal properties are quite distinct.

One is purely about linear/bilinear maps; the other about algebra maps (which preserve multiplication, of the things being tensored and not just scalar multiplication).

[rapid cave]

the problem will be showing the isomoprphism (which is not explictly given) satisfies these properties

[rapid cave]

but if I replace all linear maps with homomorphisms it should be the same as your version

[tough raven]

Not exactly. The bilinear one is about a function of two arguments; the algebra one is about two functions each of one variable.

[rapid cave]

ok

I will change some things

this will probably require little changes

something like this?

@tough raven

[tough raven]

Yep.

[rapid cave]

Now how do I show uniqueness of h

[tough raven]

I'll mention that you have used that a, b are Q-algebra homomorphisms in going to the last line to move the rJ from a() to b().

[rapid cave]

yeah ofc

[tough raven]

You can use a generating set for S.

If some subset A of S generates it (as a Q-algebra, say) and two Q-algebra homomorphisms from S agree on A, they have to agree on S.

[rapid cave]

I guess {x1,...,xn} U C generate S

Tons of thanks!

[quiet pelican]

Yes

Because we can restrict the homomorphism to the domain submodule, then it just comes down to “the image of an R-module homomorphism is an R-module”

Your proof works
But a quicker way would be to go back to here, and say “so f^-1(N) = 0 or f^-1(N) = M, so N = 0 or N = R/I as f is an isomorphism”
Or actually if I’m doing the proof myself, I just need to prove R/I has no non-zero proper submodules
If N is a submodule then by correspondence theorem, N = J/I for some J containing I
So by maximality of I, J = I or J = R, ie N = R/I or N = 0

[velvet hull]

Suppose I’m given a matrix group and I want to show that it generates some kind of free product using the ping pong lemma.

Are there any tips or tricks or common constructions for me figure out what the correct action is for these kind of questions?

Cause either the solution is immediately obvious by inspecting the matrices or fucking impossible to figure out and there’s some insane mobius transformation that i would have never spotted to begin with

[thorn jay]

werent you talking a while ago about how you hated that theorem

well, youll be glad to know that statement of the theorem is really simple

[elfin wraith]

Doing algebra without the correspondence theorem sounds neigh on impossible

[quiet pelican]

Between Subgroups that contain N and subgroups of G/N

And normal subgroups correspond on each side

[elfin wraith]

The theorem states that there is an inclusion preserving bijection between subgroups G/N and subgroups containing N

This generalises to rings and ideals, and even specifically to prime ideals

(Because of the whole inclusion preserving part)

I would be willing to bet enpeace will come and tell us it generalises to whatever the UA notion of a quotentable object is

You should also try to prove the correspondence theorem yourself I think, it’s not that hard

[cloud walrus]

mq

[thorn jay]

hell ye

the nice thing is that in generalized algebraic geometry, the bijections in correspondence theorem are actually homeomorphisms

this is due to the third isomorphism theorem; (A/N)/(M/N) ≈ A/M

[cloud walrus]

mq

[quiet pelican]

This works (yeah the surjective follows by “image is a non-zero submodule of the simple module M”)
I would just assume it’s already proven that inverse of an iso is an iso though

[thorn jay]

in a sense the naturality of this isomorphism promotes a "relative" point of view where everything relative to a congruence θ can be carried into A/θ to make everything nicer

[tough raven]

Honestly, I think people wouldn't bother to consider so many things relative to a congruence if they couldn't equate them to analogous things for the quotient.

[limpid hemlock]

anyone know of a good exercise list for dummit and foote? like a list of the nice & "reachable" exercises for each section

if so, please dm me so i dont lose it

[regal mango]

What is the difference between herstein and dummit and Foote ?

In terms of new concepts and question

[thorn jay]

yeah!! makes all the proofs way nicer lol

[regal mango]

Actually I was confused in both

Earlier but I did herstein cuz many people said it have lots of good question

After that when I saw questions in math Olympiads I wasn't able to do because many question

Require concepts from dummit and Foote

Like group actions and all

With which I am unfamiliar with

[tardy hedge]

Why not

I mean tbf after reading other algebra books dummit and foote are definitely pretty verbose

Its crazy how much exposition there is in their tensor product section compared to other books

That exposition actually made it harder for me to make sense of tensor product

Cause they were talking about it in so many different ways right off the bat

Im not familiar with their group theory section

I know about chapter 7 and onwards

Their ring theory is not bad

Lots of examples

Prob next up is modules chapter

I know theyre kinda boring to read sometimes

Tbh examples are kinda goated tho cuz if u know big assortment of them then like you kinda understand the shit yk what i mean

Yea thats fair

[elfin wraith]

They define rings incorrectly so it’s not off to a good start

[tardy hedge]

Oh really? How do they define them?

[elfin wraith]

I think D&F is actually pretty good it’s just like obscenely verbose

[elfin wraith]

Non unitally

[tardy hedge]

Oh yeah thats weird

What they do that for

[elfin wraith]

Some people do, idk why, something something analysis something something semi groups

Probably because they also hate the beautiful term rng

[south patrol]

They do it to make it more annoying to answer questions in #groups-rings-fields

[tardy hedge]

Lol

[tardy hedge]

I think they do that maybe to highlight the fact we can have ring stuff without 1 but also that having a 1 makes a lot of nice theory work

Ig just highlighting the importance of 1? Idk

[elfin wraith]

I mean it’s just a different perspective, you get stuff like ideals are subrings with is nice if you decide to want to say that for some reason

[thorn jay]

oh yeah its insane how much people focused on semigroups rather than monoids

in many UA texts they'll refer to Pol1(A) as a semigroup which, like, sure?? but it is a monoid and the existence of an identity element is very important

(Pol1(A) being the set of polynomial functions in one variable)

associative magma

i.e set equipped with an associative binary operation

i guess

yeah

im too lazy to say "set equipped with binary operation"

if you forget the identity and inverse

ye

because i had to clarify

yes but groups have those as explicit operations

so you forget them and get a semigroup

yeah

nullary

[karmic moat]

Go two channels down and you’ll find farmers and agriculturalists

[white oxide]

I am trying to show that if $a \in \mathbb{Z}$ is not a power of a prime, then its radical is not prime. I have thought of the following: write $a = p_1^{\epsilon_1} \cdots p_n^{\epsilon_n}$ where $n > 1$ and $\epsilon_i \geq 1$. Then, for some $1 \leq j < n$, $p_1 p_2 \cdots p_j$, $p_{j + 1} \cdots p_n \notin \sqrt{(a)}$, but if we take $m = \max {\epsilon_i}_{1 \leq i \leq n}$, then $(p_1 p_2 \cdots p_n)^{m} \in (a)$

[cloud walrus]

okeyokay

[tardy hedge]

is this part of trying to show that the only primary ideals of Z are (p^n) for a prime p?

[white oxide]

Ye

[limber sequoia]

It's not completely wrong but you're working a bit hard there...
Try to prove it for a simple case like 6 first

A slight error in your proof is the case where all the epsilons are 1.

[white oxide]

Wait sorry why doesn't it work if all epsilons are 1?

[tardy hedge]

also, why is it only for some j? cant u choose any j?

Yeah im also unsure of why it doesnt work if all epislons are 1

if a = p1p2 then p1 and p2 are not in rad(a) but p1p2 is in it

[white oxide]

Yeah that's what I meant

[tardy hedge]

Ok

[white oxide]

Ig I should've said for any j

1 <= j < n

[tardy hedge]

idk it looks good to me but maybe we are missing something

[white oxide]

Anyways now I'm stuck on how (p^n) is "immediately" primary 💀

[tardy hedge]

cant just check directly?

[white oxide]

Yeah I'm trying to

[thorn jay]

its all just applications of the fundamental theorem of arithmetic lol

[tardy hedge]

yea makes sense

factorize it into primes right

[white oxide]

If $x = p_1^{\epsilon_1} \cdots p_n^{\epsilon_n$, $y = q_1^{\delta_1} \cdots q_m^{\delta_m}$ and $xy \in (p^n)$, then...

[cloud walrus]

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

[thorn jay]

||suppose ab ∈ (p^n), i.e. ab = k • p^n. Suppose that a isnt an element of (p^n). Then a = l • p^n1 where n1 < n and p does not divide l. By the fundamental theorem of arithmetic we then must have that p^(n-n2) | b, and therefore there is a power x of b such that p^n | b^x, thus b is contained in the radical of (p^n). Indeed, it primary.||

[tardy hedge]

compare factors i think

[limber sequoia]

because if all epsilons are 1 then p_1...p_n is a

[tardy hedge]

yea but in the proof we split that shi up

[limber sequoia]

ah i missed that

anyhow I thought it would be easier to say something like...

Say a is divided by p_1, p_2. Then p_1 * a/p_1 is in (a), but the two ideals are ...

[thorn jay]

when ab = k • p^n, then the n copies of p get distributed over a and b. This means either one of them contains n copies of p, hence must be an element of (p^n), or both contain at least one copy of p, and thus must be in the radical of (p^n)

that is the main argument

[tardy hedge]

ye some shi like dat

[thorn jay]

commutative monoids with the unique factorisation property are great

[tardy hedge]

enpeace with the Bruh moments

[white oxide]

Does the proof start like this? Suppose that $xy \in (p^n)$ where $x = p_1^{\epsilon_1} \cdots p_n^{\epsilon_n}$, $y = q_1^{\delta_1} \cdots q_m^{\delta_m}$. Assume that $x \notin (p^n)$. Since $xy \in (p^n)$, we have $p^n = p_j^{\epsilon_j}$ or $q_k^{\delta_k}$ for some $j$ and $k$. It cannot be $p_j^{\epsilon_j}$ for any $j$ since $x \notin (p^n)$

[cloud walrus]

okeyokay

[white oxide]

ONTO CHAPTER 5 OF AM BOYS

[tardy hedge]

primary decomposition was quick hey?

[white oxide]

ONLY TOOK ME 5 MONTHS TO GET HERE

[tardy hedge]

i need to review all that stuff again tho tbh i never learned it superrr well

also i got stuck on chapter 5 one time and didnt go back to it

[white oxide]

Oh nah usually I do exercises in one chapter while I read the next chapter and I can only move on once I've done 10 exercises or so or if I've been stuck on the chapter for a while

[tardy hedge]

idk why i got stuck i should look at it again

[tardy hedge]

oh thats a good way to do things

keep it interesting by reading but also do exercise

[white oxide]

Yeah, 2/3 of my time is devoted to exercises and 1/3 is reading

[tardy hedge]

realistically i should do more exercises

but i mean if im in a class i just want to do the homework and move on

the homework takes enough time already

[thorn jay]

primary decomposition is weird

[tardy hedge]

you know what yea it is a little weird

associated primes are annihilator of radicals of some shit

some weird thing

[thorn jay]

because the case of radicals is really easy; just basic lattice theory, but that it holds for general ideals with primary stuff is just weird af

[white oxide]

i've got my ba in math so I'm done with classes forever 😹

[tardy hedge]

lol yeah do u just have a lot of free time to study ?

i dont know if ill be doing that if i work a full time job

not sure if id care to

[white oxide]

well I gotta finish my CS degree too so the most I can study nowadays is like 30minutes to an hour but on weekends I have more time

once I get full time yeah will have less time to study

[tardy hedge]

oh i see yeah i mean even just a little bit of effort here and there is good and pays off over time

[south patrol]

As in 5am? That's quite early in the morning

[south patrol]

Pog...

[astral ivy]

What exactly is Euler's group and what are its properties?

Is it $F_n = \left{0 < k < n | k \not| n\right}$? The set of all numbers less than base $n$ such that they're coprime with n? Sorry if the set notation is confusing I have no idea how I'd write it.

What about properties and what is the group operation?

Let's say $n = 10$ for example. Then $F_{10} = \left{1, 3, 7, 9\right}$ but what is the operation that these elements have in common? What can I use to have $a\ast b = c$ where $a, b, c \in F_{10}$?

From this, by group properties I mean showing if it's cyclic or Abelian and that I would have if I knew and understood the group operation I guess.

Can anybody help?

[cloud walrus]

danilojonic

[limber sequoia]

You mean like the multiplicative group of n?

[astral ivy]

see my example but yeah I think it's that

but I wouldn't necessarily say that

because multiplicative group is also $Z_5^+ = {1, 2, 3, 4}$ right?

[limber sequoia]

no, for 5 the multiplicative group would be 1,2,3,4

since 0 has no inverse.

[astral ivy]

ye correct

mb

[cloud walrus]

danilojonic

[limber sequoia]

Well, im not sure what I can tell you.
The multiplicative group of n is all the numbers that have an inverse mod n, which happen to be all the coprimes.

[astral ivy]

but how would we represent 5 then?

[tardy hedge]

i said this to remind myself of it tbh

[kind temple]

i haven’t heard it under the name “Euler’s group” before (he doesn’t need any more things named after him lmao)

this is the group of units of Z/nZ, denoted (Z/nZ)^x

and it is (isomorphic to) the set of all numbers 0 <= k < n such that gcd(k,n) = 1

[lusty marlin]

n does not have a multiplicative inverse mod n

[limber sequoia]

I don't really know number theory but if we could give you a simple explanation of an arbitrary multplicative group then the field would be really simpler.

[tardy hedge]

the algebraic number theory course i took was so poopy to me

i did not understand it

[astral ivy]

this course doesn't cover fields, only groups and rings

[tardy hedge]

number theory is hard

[astral ivy]

Can you expand on the group properties, operation, is it abelian, cyclic etc?

with explanation pls

[south patrol]

Lol

[limber sequoia]

well it's always abelian.

[south patrol]

Poopy is just funny

[tardy hedge]

Lol yea

[kind temple]

-# i feel like i just got asked an AI prompt

[tardy hedge]

it works sometimes

[rapid cave]

the operation is just multiplication modulo n

[limber sequoia]

but what im saying is that there is no description of a general multiplicative group of n because it's complicated

[rapid cave]

its cyclic only for some classes of values for n

[tardy hedge]

"I is an ideal which can be generated by n elements up to radical" What does this mean?

[kind temple]

this feels like an unfocused question. is there anything more specific you want to know?

[limber sequoia]

like you want a silver bullet to understand it, but there isn't one. It's just complicated

[astral ivy]

Is it abelian because of this $gcd\left(n, k\right) = gcd\left(k, n\right) = 1$?

[cloud walrus]

danilojonic

[limber sequoia]

um
it's abelian because it's a qoutient of Z which is abelian.

[astral ivy]

subgroups of abelian groups are normal too

[limber sequoia]

or because ab = ba with whatever mod you'll like, if you want a more concrete discription

[kind temple]

-# sorry lol

[alpine plank]

The multiplicative group isn't a quotient of Z

[rapid cave]

as a ring

[astral ivy]

how does that make a group? Modulo n integers under multiplication aren't group tho

[kind temple]

the units are

and we picked specifically those elements

[astral ivy]

I see

[alpine plank]

I think this would get cleared up if you worked it out

[limber sequoia]

of Q?

yeah you're right that's the wrong description

[astral ivy]

how should I work it out?

[alpine plank]

gimme a sec

[tardy hedge]

bump

[kind temple]

just show that the operations are well-defined and is abelian

like

[rapid cave]

probably its radical is generated by n elements

[tardy hedge]

Ok. Seems like weird wording to me tbh

[kind temple]

if you take two numbers a and b that are coprime to n, show that their product is coprime to n

find the inverse for each group element

find the identity of the group

show that the operation is commutative

[alpine plank]

yeah basically this

[kind temple]

imo i think this is pretty straightforward

[delicate orchid]

this could be tricky if you don't know Bezout's

actually no it isn't

[glad osprey]

how do you do it without Bezout?

[tardy hedge]

whats bezout again

[alpine plank]

sum of coprime ideals is 1

[tardy hedge]

ya i mean u said like abstract version

[kind temple]

there are integers x and y such that ax + by = gcd(a,b)

[alpine plank]

just translate to Z

[tardy hedge]

Mb i forgot what coprime ideals are

[kind kindle]

how can i construct nontrivial bivariate polynomials which define group laws over some subset of a finite field(or generallly could be matrix valued or something)

first intuition. There has to be a nontrivial set S over which p(x, y) is associative

meaning not just solutions, but a big set S, such that I could pick any 3 elements of S and it'd still be associative and form a closed set

[white oxide]

Dumb question but in a UFD if $a = p_{\alpha_1}^{\epsilon_1} \dots p_{\alpha_n}^{\epsilon_n}$ how do we know that the only divisors are all of the form $\prod_{(\delta_1, \dots \delta_n)} p_{\alpha_1}^{\delta_1} \dots p_{\alpha_n}^{\delta_n}$ where $\delta_i \leq \epsilon_i$

[cloud walrus]

okeyokay

[white oxide]

I was thinking suppose that there was a divisor of a which is not of this form, then it has a prime factorization

Then use Euclid's lemma

[thorn jay]

that's a weird way to write it lol

[white oxide]

wdym

[thorn jay]

using \alpha_i instead of just i and that product

[white oxide]

Oh yeah I don't know why I did it

[quiet pelican]

Suppose b = [factorisation of b] divides a
Then every prime in b is one of the primes in a, as if p | [factorisation of a] then p divides some p_{a_i}
Similarly the power of p in b is at most that of a

Qed

[white oxide]

yeah that's what I was thinking

[thorn jay]

if a = p_1^k_1 ... p_n^k_n then all the divisors are of the form u p_1^l_1 ... p_n^l_n where 0 <= l_i <= p_i and u is a unit

[white oxide]

I always took such things for granted but upon closer inspection I realized I didn't know exactly why it was true lol

[quiet pelican]

Or you can just remove that p from both a and b and conclude by induction on sum of d_i

[white oxide]

Yes

That was what I was trying to see

[thorn jay]

that's what unique factorisation means

like literally the definition kinda

I guess not literally

[quiet pelican]

It’s not
The definition is that the factorisation as written exists and is unique
It’s not immediately trivial that that implies the factors are of that form

[thorn jay]

it is though? if a | b, then b = a * k. Factor a and k, and then the result basically follows immediately

[white oxide]

You're just a savant my boy

[thorn jay]

genuinely had to look up what savant means 😭

anyhow, I apologize

[white oxide]

Nah you chillen

[thorn jay]

I was writing out a way how it might be easy to see but I'm pretty sure that would make it more convoluted

skull

I guess let P0 be a set of primes such that every prime has a unique associate in P0. Then every element of your ufd can be written uniquely as a product of a unit with powers of elements in P0

[thorn jay]

then, when you've factored a, b and k then you can literally just compare the powers occuring in the primes lol

don't have to worry about all the messiness with associates

[lime badge]

If I have two matrices over F_2 such that AA^T = 0, is there anything I can conclude about A?

The usual argument that AA^T = 0 ---> A = 0 doesn't work here unfortunately. Sorry if this is the wrong channel for this

[thorn jay]

over F2 for 2x2 matrices they must have the form

| a a |
| b b |

maybe #linear-algebra is a better channel for this? But try seeing if this generalizes to arbitrary dimensions

[lime badge]

Thanks!

[tardy hedge]

thats like classic when to one person its immediate but to someone else it takes conscious thought to make the 2-3 steps of logic that was actually automatic to you

[thorn jay]

how it feels reading anything written by McKenzie

[minor gazelle]

Let (X,•_X) be a monoid, is there a concrete example of such monoid (X, •_X)^{op} such that for all h,g in (X,•_X) we have h•X g= g•{X^op} h

Like actually example of maybe some integers or something noncommutative and nontrivial

[noble nexus]

Isnt that just the definition of the opposite monoid

or am I misunderstanding what you mean

[thorn jay]

Let M(n) be the monoid of matrices under multiplication. Then M(n)^op is isomorphic to the monoid given by A \cdot B = A^T * B^T

[ebon prairie]

ques 46

[thorn jay]

love reading sideways

[ebon prairie]

question: suppose a and b belongs to commutative ring and ab is zero divisor, to show either of a or b is zero divisor

[thorn jay]

ye that's it

nice job

[ebon prairie]

how do i ensure a and b are themselves non zero

[thorn jay]

because else ab would be zero

[ebon prairie]

oooh thanksss

[ebon prairie]

how to show characteristics of finite ring divides the number of elements in that ring

[limber tapir]

well I guess 1 generates a subgroup with charR elements of the additive group and then apply lagrange

[ebon prairie]

ring may not have unity

[limber tapir]

wait then how is the characteristic defined?

ah I see, looked it up

[ebon prairie]

it can be defined without unity too right

[limber tapir]

but yeah just replace 1 with any element

(any element s.t. a*k != 0 for k < charR)

okay I guess you have to do some more work, since there must not be an element with ord(a) = char(R), but it all comes down to the properties of exp(R)

[ebon prairie]

i was thinking claiming char r= lcm of all additive orders when unity not there

[limber tapir]

I mean thats the definition right?

[thorn jay]

well all the orders of the elements must divide the order of the ring

so their lcm must too

[limber tapir]

exactly

[minor gazelle]

Is my logic here correct?

I kinda copied and revised the logic using this free monoid concept (which is surprisingly not that hard to understand because of the structure of addition)

This is definitely correct, this is definitely one of the more abstract proof I have done in a while, 🥰 my 1 week doesn’t end up wasted

[noble nexus]

yeah looks good

[minor gazelle]

I know right I basically trying to make sense of this free groupy monoid thingy for a week 😭😭

[noble nexus]

free objects are the kinds of things that seem really complicated when you first encounter them but end up being quite simple once u get used to ti

[minor gazelle]

❤️❤️❤️

[idle girder]

The idea looks good, and yeah, free objects are nice!

I think the writing can be improved. Maybe framing it like this would make the proof structure clearer? "Let m0, m1 be given with f(m0)=f(m1). We will show m0=m1. For this purpose, take g, h : (NN, +) -> M given by .... ... ... Because f is mono, we obtain g=h. We see m0=g(1)=h(1)=m1, as desired."

[minor gazelle]

🥰

[thorn jay]

great job of getting your head around it!

once you know it you start seeing this pattern everywhere; at least in my experience

[thorn jay]

many category theory stuffs

[twilit wraith]

this is the practice exam i was given for algebra 1

does anyone else agree that the last problem is disproportionately difficult as compared to the others?

i was able to get the first five parts in 15-20 minutes and then i couldnt solve the last one even after half an hour

[obtuse reef]

Yeah 1 seems pretty easy

2 is quite easy too

Yeah 3(ii) is more difficult than the rest

[noble nexus]

Idk I think it kinda spells out what you need to do

maybe if you didn't cover group actions in much detail

Whenever your group acts on an n-element set you obtain a homomorphism into Sn

as hinted by part (a) in this case you have a 3 element set of the cosets of H, then the kernel part follows pretty easily by unwrapping definitions

but it is definitely more difficult than the rest of the problems

[hollow imp]

think about how S4 must act on cosets of D. How many elements does S4 have? Which elements send D to itself?

[twilit wraith]

i mean i know that S4 can only send D to three things

apparently if i consider the symmetry group on the set S4/D then i get that homomorphism

[twilit wraith]

didnt even spend a whole class on them

i think thats the biggest thing

[noble nexus]

Yeah that would explain it

[twilit wraith]

i know the definition and orbit-stabilizer

but nothing else

[noble nexus]

A group action of G on S is the same thing as a homomorphism from G to the symmetric group on S

understanding why is a pretty good exercise to do

[hollow imp]

well orbit-stabilizer would tell you there's indeed 3 things

[hollow imp]

and then you're done

[twilit wraith]

ah

thats pretty clear just from knowing what a group action is

but i guess i didnt bother thinking about it that way

[twilit wraith]

the problem is that i didnt know that was a thing

[noble nexus]

yeah it's a really important interpretation because it makes it clear how to generalize to groups acting on other things

[twilit wraith]

like i knew that concept existed i guess but ive never used it when solving a problem before

i actually dont know what Sym(S4/D) looks like

i mean i know its a set of three things and their permutations

but idk what the permutations look like as group actions

[noble nexus]

Well if I give you an element of S4, how might you define a permutation of the cosets of D?

[twilit wraith]

i guess just left-multiplying each element of D by it

just as the group action is defined

but how do i know which elements act nontrivially on D

[noble nexus]

Well if an element g acts trivially, it must satisfy gD = D

[twilit wraith]

true but i know nothing about D

besides that its order 8

oh wait

[noble nexus]

Yes but what does gD= D tell you about g

[twilit wraith]

that just means the kernel has to be a subgroup

duh

[noble nexus]

Yeah

[twilit wraith]

wow i cant believe how hard i overlooked this problem

hindsight is 20/20 i guess

now that im aware of this line of thinking though i should be good

[noble nexus]

bonus challenge problem: can you show there is a surjective homomorphism from S4 to S3?

[twilit wraith]

yes by first isomorphism theorem

[noble nexus]

this map in the problem isn't surjective

[twilit wraith]

oh right

[noble nexus]

which you can also see by noticing that any element that fixes D must belong to D and thus fixes all the other cosets

[twilit wraith]

and as ker phi is size 8 we must have that im phi is size 3

[noble nexus]

yeah

[twilit wraith]

so i need a normal subgroup of size 4

i guess <(12), (34)>

so S4 mod that needs to be S3

which i think is just a matter of showing that S4 mod that isnt abelian

nah because D3 is still there

[noble nexus]

Well D3=S3 so you're fine there

[quiet pelican]

That isn’t normal
(24) is conjugate to an element of it, and isn’t in it

[twilit wraith]

oh right

[noble nexus]

This subgroup also isn't quite normal

but there's something similar that is

[twilit wraith]

maybe <(13), (24)> but idk

or 14 23

kinda just making shots in the dark now

[quiet pelican]

Those are also similarly not normal

[noble nexus]

Have you learned what conjugates look like in the symmetric group

[quiet pelican]

What are the conjugacy classes in S_n?

[twilit wraith]

No

Or if I have then i dont remember

[noble nexus]

basically the conjugates of an element are just the elements of the same cycle shape

In fact conjugating an element by a permutation applies that permutation to each element of the cycle: $\tau (1 2 3)\tau^{-1}=(\tau(1)\tau(2)\tau(3))$

[cloud walrus]

Blake

[noble nexus]

however I will say there's a way of constructing a surjective homomorphism that doesn't go in this direction, but it's a bit sneaky

[twilit wraith]

Oh I see

[twilit wraith]

Its quite late so im not sure im in the mental state to motivate it myself

[noble nexus]

actually the homomorphism from S4 onto S3 is quite bizarre and nothing similar happens for larger symmetric groups, it's one of my favorite little pieces of group theory

[twilit wraith]

I am curious what it is tho

[noble nexus]

fair enough

Any permutation of 4 points also permutes the 3 different partitions of 4 points into 2 element subsets

[dull ginkgo]

Can someone check my proof of an isomorphism

I don't want to intrude on the current convo lo

[noble nexus]

thus we obtain an action of S4 on a 3 element set, and from there you can check that it is surjective

[dull ginkgo]

actually probably better suited for advanced alg

[noble nexus]

Sure go for it

[noble nexus]

alternative, perhaps less elementary proof: the group S3 can be realized as the group GL(2,2) of 2 x 2 invertible matrices over the field with 2 elements (notice that GL2 acts on the 3 nonzero elements of F2^2, which yields an isomorphism to S3)

If we allow translations, it turns out that you can achieve any permutation of the 4 points of F2^2, so the group GA(2,2) of affine transformations of F2^2 is isomorphic to S4

then since there is always a surjective homomorphism from the general affine group to the general linear group, that completes the proof

[twilit wraith]

I see

Seems like theres a ton of geometric interpretations in group theory

They're the part that fascinate me the most so far

[noble nexus]

yeah once you get used to thinking about actions and symmetries you can do a lot of cool things

But it took me a while before things clicked in that way

[twilit wraith]

Im still in that phase

I did a month of undergrad algebra in June and im now doing the grad course to have more time for more perspectives on groups and rings but I definitely haven't got the various ways to interpret groups down yet

[noble nexus]

Oh I should mention that the kernel need not be the entire subgroup D (otherwise you would be able to prove that every subgroup is normal) the reason for this is that if g is in D, then gD=D so the coset D is fixed, but gxD need not equal xD.

[twilit wraith]

Yes but the subgroup part directly comes from how the group action is defined

At least after recognizing that you interpret it as a homomorphism

[jade gyro]

why can you treat free abelian groups as sets of multisets as doesn’t that violate the axiom of regularity?

[cedar vault]

How?

[earnest delta]

0,1 in z_2 and

1mod 2 =1

3mod2

So i don't get any roots in z

But i got B

Given answer is A

[jade gyro]

?

[cedar vault]

You are claiming that it violates the axiom of regularity. Can you give an example to prove it?

If I understand your question correctly any multiset will act as an element that satisfies the axiom of regularity

[jade gyro]

o well i mean is it not just by definition

[cedar vault]

I don't see that at all

[jade gyro]

uhh wait maybe i am being stupid

yeah i think so ignore me

[cedar vault]

If $F$ is the free abelian group over some set $\Sigma \neq \emptyset$. You are saying we can view it as a set of multisets so I'm guessing you mean that an element $$\prod_{\sigma\in\Sigma}\sigma^{\alpha_{\sigma}}$$ with all but finitely many $\alpha_{\sigma}\neq 0$, say $\sigma_{1},...,\sigma_n$ is mapped to the multiset $\bigcup_{i=1}^{n}\bigcup_{j=1}^{\alpha_{\sigma_i}}{\sigma_i}$. Then let $\sigma_1\in \Sigma$ . The set ${\sigma_1}$ does not intersect the free group which is a set of multisets of $\Sigma$ while ${\sigma_1}$ is a set with a single element from $\Sigma$

[sharp ice]

this just means these no linear factors, but it doesnt mean its irreducible. it factors as x^4 + x^2 + 1 = (x^2 + x + 1)^2 in Z/2Z [x]
its even reducible as a polynomial in Z[x]

[jade gyro]

no yh i was wrong

[cloud walrus]

Herzog

[jade gyro]

ok i guess ill just give in uh given a free magma F, and elements a and b in F, why isn’t the mapping from the pair (a,b) in F^2 to (ab) in F a bijection

i feel kinda stupid but i just cannot reason why

it seems so blatantly correct

(idk if this is the channel for this question really)

(tho obv you can’t have a bijection from a set to its cartesian product so i’ve gone severely wrong somewhere)

[cedar vault]

You can, from ZxZ -> Z for example

[jade gyro]

wait how

a bijection?

or am i being really really dumb

wait no i am aren’t i omd

yeah wow

[cedar vault]

Have you seen the argument where one draws a table of the elements in ZxZ and enumerates the off diagonal elements giving a bijection ZxZ -> Z?

[jade gyro]

i completely forgot

yeah yeah

i completely forgot lolll

[maiden drift]

There are twice as many numbers as numbers.

[jade gyro]

so is this a bijection?

(and thus an isomorphism of magmas maybe?

i mean there is a 1-1 correspondence and you satisfy the homomorphism requirement

i think

as in the magma operation on pairs of pairs is natural

yeah (ba)

well it’s a magma so it doesn’t have commutativity

but uh you can see this from the fact it is a free magma

(again i think)

[glad osprey]

if magmas were associative then it wouldn't be a bijection: let F be the free magma over {x} for example, then (x, x*x) and (x*x, x) would both map to x*x*x. But they aren't, so I'm not sure 🤔 it might be a bijection actually

[lusty marlin]

I'm trying to find the Galois group of x⁵-x+1 over ℚ.
Since this polynomial is irreducible over ℤ/3ℤ, it is irreducible over ℚ and its Galois group contains a 5-cycle. I've tried reducing modulo a few other primes but I haven't been able to prove the existence of a transposition.

[quiet pelican]

Do you know the condition for a polynomial’s galois group to be contained in A_5?
Also, consider ||complex conjugation||
Those two combined should give you what you want

[lusty marlin]

Right, conjugation can give a transposition. But how do we get the number of real roots of this polynomial?

[quiet pelican]

Turning points are at the zeros of 5x^4 - 1, so at x = 5^{-1/4}
Sub that back in and compute, then consider what the graph looks like
(It should be 1)

Complex conjugation gives a double transposition, not a transposition
That’s why you need the first part

[lusty marlin]

Ah right

Thanks

[thorn jay]

because x cannot be written as the product of two elements in a free magma

because the free magma does not satisfy any nontrivial equations

it is injective, though, for much the same reason :P

[thorn jay]

another way to see this: a free magma, is a set of binary trees with labelled leaves, where the operation is joining the two roots at a new root. From here it is obvious that the expression "x" cannot be written as the product of two other expressions

[white oxide]

Is this isomorphism induced by A -> k[y]/y^2 given by f(x, y) |-> f(y^2, y)?

[velvet hull]

no, x is contained in q

so taking the quotient by q just kills x directly

you're confusing (x,y^2) with (x-y^2)

[white oxide]

Oh okay

So it would just be induced by the projection?

[velvet hull]

wdym

[white oxide]

uh nvm I don't know what I mean lol

uhh

Would it be induced by f(x, y) |-> residue class of f(1, y)?

Maybe I can see it via composition of isomorphisms or something?

It makes intuitive sense since quotienting by (x, y^2) can be viewed as quotienting by x, which gives k[y], then by (y^2), which gives k[y]/(y^2)

[white oxide]

In that case I think this corresponds to that composition

[merry harness]

Maybe this question belongs in #recreational-math but I'll venture asking here. What is the centralizer of the Rubik's cube group in itself? I know that the identity and the super-flip are in the set, but not sure if there is anything else

[thorn jay]

wikipedia says only th identity and superflip

[velvet hull]

oooooh, the superflip is in the center? that's cool

[twilit wraith]

I think i proved successfully that if G is finite, H is a subgroup of G, N is normal in G, and |H| and |G:N| are coprime, then H is a subgroup of N

But I think i consequently proved that H is trivial and im skeptical

[velvet hull]

lmao that's the second isomorphism theorem for you

[jade gyro]

isn’t the empty word in the free magma tho

[twilit wraith]

My argument is just that for some h in H, its order k divides |H|, but since h^k N = N, we have that k is a multiple of the order of |hN| in G/N and so k is either not coprime to |G:N| or k=1

[jade gyro]

“word”

[twilit wraith]

But k has to be coprime to |G:N|

[merry harness]

Dang okay that makes some sense, I was thinking about corner twist too but I guess you can only twist 3 ccw at a time

What about checkerboard?

[white oxide]

HOLY FUCK HAVE I JUST PROVED THE SECOND ISOMORPHISM THEOREM??!?!?!

[delicate orchid]

I'd say that's more third than 2nd

[thorn jay]

no

imo not

because else the free magma would automatically be unital

and else it wouldnt have the universal mapling property

what would the empty word map to in a nonunital magma

[jade gyro]

i mean it can be unital and still be a magma no

[thorn jay]

then it wouldnt be the free magma

it would be the free unital magma

[jade gyro]

nono the free magma can be unital but not have to have been built on a unital magma i think

[thorn jay]

different signatures

it cant

[jade gyro]

why

cos isn’t it just an extra restriction

[thorn jay]

simply because of the fact not every magma is unital

if the free magmas would be unital, then every magma would be unital

[jade gyro]

yeah but the free magma could be no?

[thorn jay]

because every magma is a quotient of some free magma

[jade gyro]

wait uhh

i’m thinking that just because it can have a unital structure upon it doesn’t mean it has to be imposed upon it

[thorn jay]

what

[velvet hull]

no, because any element of the center has to fix the position of every piece

because of the existence of commutators (not the group theoretic version, the commutator algorithm for pieces)

[thorn jay]

no free magma can have an element acting as a unit

because that would automatically mean that every magma generated by that amount of elements was unital, which is false

[jade gyro]

i mean say it had the empty word as an element would that stop it from being a magma?

[thorn jay]

i can give examples of nonunital magmas generated by any amount of elements

[merry harness]

Got it yeah I thought of a counterexample to checkerboard too

[jade gyro]

it would still have a binary operation

thus it’s a magma

[thorn jay]

no, but it would stop it from being a free magma

it could be a free unital magma

[jade gyro]

it is both no?

[thorn jay]

no

do you know what the defining property of a free magma is

[jade gyro]

the universal property

[thorn jay]

yes

which is not satisfied if your magma is unital

[jade gyro]

how?

[thorn jay]

take the free nonunital magma F, consisting of all nonempty binary trees where the binary operation is adjoining two trees at their root.

Now, this magma is generated by x, so if any free magma F' would be unital, then F = h(F') where takes the canonical generators of F' to x. Then, as F' is unital and h is surjective, h(F') must be unital too. But this is a contradiction, so F' cannot be unital

[jade gyro]

ah i see so it’s cos the empty word could be a combination of itself any number of times

[thorn jay]

in a sense

[jade gyro]

so then ig free unital magmas are then isomorphic to products of themselves instead

[thorn jay]

like, a good thing to note is that "the empty expression" is not something that exists

every expression is a nonempty tree

[jade gyro]

is it not just the unit

[thorn jay]

no, because a unit might not exist

[jade gyro]

no isn’t the empty expression just a unit

and so it might not exist

[thorn jay]

for any n-ary operation, how would you define f(" ", p1, ..., pn)

if a unit exists then that is an expression in of itself

[jade gyro]

yeah it’s kinda semantics but you’re basically saying the empty expression might or might not exist

[thorn jay]

not an "empty expression", whatever that may mean

its not semantics

[jade gyro]

it’s still a thing when looking at say the group of words

that’s just its name

[thorn jay]

the empty word is the expression "1"

[jade gyro]

yeah

[thorn jay]

thats not the empty expression

[jade gyro]

ah that is semantics lolll

[thorn jay]

i guess?

[jade gyro]

synonyms n stuff

[jade gyro]

?

[thorn jay]

uh

no

because that would mean that every unital magma satisfied a nontrivial identity that wasnt 1x = x1 = x

[jade gyro]

sorry wdym?

[thorn jay]

well, if F ≈ F x F, then (1, t)(t, 1) = (t,1)(1,t)

which would mean there exists nontrivial expressions t, s such that ts = st was satisfied in every magma

which is false

[jade gyro]

oh uh ig then separate the units?

[thorn jay]

?

[jade gyro]

like have a left and right unit

[thorn jay]

then it wouldnt be a product of itself anymore

[jade gyro]

or does that take us back to the unit problem

[jade gyro]

oh yhyh

[thorn jay]

also fyi a magma having both a left and a right unit has a unit

cuz, let l and r be the left and right units, then r = lr = l

[jade gyro]

yeah ikik

[thorn jay]

i guess the gripe i have with "empty expression" for a unit is that it breaks down when you even have a single operation that is not binary

i guess for unary operations you could do p(" ") = " "

but something like f(" ", x, y) is meaningless

and, formally, any unit has to be added as a nullary operation anyways

[jade gyro]

i’ve realised that what i needed is when you just consider magmas ty!

[thorn jay]

[tardy hedge]

in a graded R-module, 0 cannot be written as a sum of nonzero things right

[karmic moat]

can't you just take Z/2 to be a graded Z-module with the trivial grading? then you can write 0 = 1 + 1

[wraith cargo]

They probably meant as a nonzero sum of things from different levels of the grading

[tardy hedge]

yeah

yea sorry thats what i meant

[karmic moat]

oh

[tardy hedge]

Because M = direct sum Mi so every element of M is a unique finite sum of Mi things

but does 0 mess any of that up or like

i was just trying to show why ker(f) is graded when you have f: M -> N homgeneous

[karmic moat]

yeah you can't write 0 as a sum of non-zero elements of diff gradings

[delicate orchid]

oh yeah? watch THIS

[karmic moat]

i'm watching bro i'm nervous

[tardy hedge]

He couldn't do it

[south patrol]

Yeah you can

With the power of goodwill

[karmic moat]

you're right

i'm a negative nancy </3

[kind kindle]

Question, what is known about polynomials which define a group operation for a subset of a finite field

[coral spindle]

You're describing a very particular case of algebraic groups which are groups defined by polynomials, put briefly. Have you seen those?

[kind kindle]

vaguely familiar with them

but i'm essentially trying to find compact ways to represent groups with polynomials over finite fields.(i.e, represent the group law as some relatively small easy to construct polynomial from F^n x F^n -> F^n for some finite field F)

[glad osprey]

Doesn't this just force |hN| = 1? Since |hN| divides k divides |H|, and |hN| divides |G/N|. But then you get N = G, so still much stronger than what you're asked to prove

[twilit wraith]

Why would it not force k=1

I was talking about it with someone earlier but I didnt quite understand

[glad osprey]

Can you write out your argument in more detail? I think you got the division in the wrong order, k does not necessarily divide |G:N| AFAICT

[twilit wraith]

Let h in H be arbitrary. Then |h| = k such that k divides |H|. Now consider hN in G/N. We have that (hN)^k = h^k N = N and so either k is not coprime to |G/N| = |G:N| or k = 1. However, k must be coprime to |G:N| and so k=1 and thus h in N, and therefore H is a subgroup of N.

However k=1 also implies that H is trivial

But i can come up with an example where H is not

G = S3, N = <(123)>, H = N

[glad osprey]

The problem is the step "so either k is not coprime [...]", which you haven't justified. If a and b are coprime, and k divides both a and b, then k = 1. But k doesn't necessarily divide |G:N| here

[twilit wraith]

Its not that k divides |G:N|, its just that its not coprime to it

If we let |hN| = l, then l divides k and it divides |G:N|

Oh wait l = 1 is a different third case

And we've disproven the first two

[glad osprey]

I still don't understand. We know l = |hN| divides k, and k divides |H|. Why can't k be bigger than 1, but still coprime to |G:N|?

[twilit wraith]

k cannot be bigger than 1 but coprime to |G:N| only if l > 1

I was implicitly and incorrectly assuming the premise and thats the fundamental error I think

[white oxide]

I don't really care to make this precise, but could we just say something like $k[x, y] / (x) \cong k[y]$, and so $k[x, y]/(x, y^2) \cong (k[x, y]/(x))/(y^2) \cong k[y] /(y^2)$

[cloud walrus]

okeyokay

[velvet hull]

sure

[quiet pelican]

That’s essentially precise if you add the words “by correspondence theorem” before the second iso

[white oxide]

correspondence theorem?

[quiet pelican]

If I is an ideal of R, then the ideals of R/I correspond bijectively with ideals of R containing I via J -> J/I, and (R/I)/(J/I) = R/J

[tough raven]

No, it was the empty sum.

[white oxide]

How does iii imply this highlighted sentence

We get that A[x, y] is contained in a subring D of B such that D is a f.g. A-module but I don't see how that helps

[tardy hedge]

probably consider A[x+y] and stuff

A[x+y] is contained in A[x,y] so x+y is integral over B

[quiet pelican]

By ii, A[x] is a finite A module
By ii again, A[x, y] is a finite A[x] module
By basically-linear algebra, A[x, y] is a finite A module
And A[x + y] \subseteq A[x, y], so it is a finite A module

[white oxide]

Oh yeah smart didn't think about that

[white oxide]

Ye thx

I don't really see where we use iii tho lol

We use the fact that A[xy], A[x + y] \subseteq A[x, y] is f.g.

So A[xy] and A[x + y] must be fg

[tardy hedge]

we are in condition iii) and the proposition says its equivalent to i)

[thorn jay]

we need f.g. as a module

[white oxide]

Oh yeah

I'm dumb

[thorn jay]

:P no ur not, math is hard

[white oxide]

Integral closures seem like a generalization of algebraically closed fields lol

Or algebraic closures

[thorn jay]

they are!

[white oxide]

wtf do you call them again

haven't studied field theory since 2 years ago 😹

[thorn jay]

an integral extension of fields is precisely an algebraic field extension

[white oxide]

mmm

makes sense

[thorn jay]

i.e., if a field F is an integral E-algebra, then E < F is an algebraic field extension

and commutative rings just happen to be nice enough that many nice properties translate over

ofc, not all

that would be silly

[tardy hedge]

that would be just sooo silly man

[karmic moat]

enpeace do you know if solvable groups show up in AG other than just in the context of borel subgroups/using borel fixed point theorem

i would ask in alg-geo but it doesn't seem appropriate atm

or anyone reading that might know

[thorn jay]

idk I'm not an AG guy

[karmic moat]

balls

[thorn jay]

I would love to be some day though

[karmic moat]

lol im doing a reading course w my prof and one other student on flag varieties

[thorn jay]

esp because of how much AG-esque stuff is popping up in tame congruence theory atm

[karmic moat]

but like they skip so many proofs i gotta go back and read everything

we alternate presenting every other week

[thorn jay]

for a split second I though you were talking about the UA kind of varieties

but then I snapped back to reality

[karmic moat]

are they generalizations/analogs of ag varieties

or just shits and giggles same name

[thorn jay]

it's the same concept; collections of object that satisfy equations

[karmic moat]

okay

nice

[thorn jay]

just in AG it's coordinates satisfying polynomials and in UA it's algebraic structures satisfying equational identities

and you've got like a similar Galois correspondence between varieties <-> deductively closed theories as with varieties <-> prime ideals