groups-rings-fields

[thorn jay]

same for groups which possess a cyclic sylow 2-subgroup

for example groups with order 4k + 2

and of every other order there does exist a group with an orthogonal mate (simply by taking the direct product of finite-field constructed MOLS-pairs)
in other words, we've constructed a nontrivial spectrum

would you look at that

and hell, the spectrum stays the same if you require the groups to be abelian
how thoughtful of big math

(the spectrum of a class/variety of algebraic structures is the set of orders of finite algebras in said class)

[fervent meteor]

I have to show that the set G of transformations of R sending x -> ax+b for real a, b and a\neq 0, is a transformation group of R. Should I assume that a and b are fixed?

Maybe not, right? Since an identity should have x --> 1x+0

[thorn jay]

if assume a and b are fixed then you've got a set consisting of a single element

[fervent meteor]

kinda sad that I almost can't see why that is

thanks though, makes sense

[thorn jay]

because.. you then only have the function f_a,b sending x to ax + b

[rocky cloak]

You should assume a and b vary over all reals with a \neq 0, just as you described yourself

[tough raven]

"Groups with operators" might be related.

[thorn jay]

not necessarily, those are groups which have operators on them, i.e. the underlying groups change

a representation of a group is essentially the same as a module

[tough raven]

It's kind of fun to do this for all finite abelian groups (in particular, you can find out the answer for (ℤ/nℤ)^⨯).

[thorn jay]

the group stays fixed much like the base ring a module stays fixed (generally)

[mossy slate]

Is there an operation that causes the set to expand to include all real numbers?

[tough raven]

I mean, I'm just giving you the only even tangentially related keyword I know.

[thorn jay]

hehe it's okay, thanks for the contribution :3

[thorn jay]

analytic continuation, I guess? I have a friend who looks into it

[tough raven]

In general, if a variety of algebraic structures B has a forgetful functor to another A (ig you call that a reduct?), just by the existence of presentations this has a left adjoint. This generalises the monoid ring construction ℤ[M], which you can specialise to groups.

Group/monoid algebras R[G] require the concept of commuting though to say that G commutes with R.

[thorn jay]

yeah is called a reduct

[rocky cloak]

Supremum is kind of an operation I guess

[thorn jay]

oh yeah it is in some way related

mmm

this is related to something which I explored a little while back, called enrichment (because of the slight similarity with enrichment in category theory)

if you take V_1 to be the variety of groups and V_2 to be any variety consisting of only unary operations, then V_1 enriched by V_2 is a variety of groups with operators

you can expand on this and get, for example, that rings are (abelian) groups enriched by monoids

oh, of course, the variety of K-linear representations of G is the variety of K-vector spaces enriched by the variety of G-sets

I've found a single guy and precisely two papers on general representations of algebraic structures

lmao

and it was originally written in russian and like

barely readable

(the translation)

I fear I might be cooked..

[restive birch]

did anyone answer this?

[tardy hedge]

for a tensor product M(x)N that is also an R-bimodule, to show a map defined on it is an R-bimodule hom, do I show that the map from M x N is R-bilinear, but on both sides?

because bimodule hom not just (left) module hom?

[tough raven]

You need to show f(mr, n) = f(m, rn) for it to be well-defined and f(rm, n) = r f(m, n), f(m, nr) = f(m, n) r for it to be bilinear.

[tardy hedge]

What about f(m1+m2, n) = f(m1,n)+f(m2,n)?

[south patrol]

You need that too

[tough raven]

Assuming M, N are R-bimodules and you're talking about M (⨯)_R N.

[fervent meteor]

My book wants me to prove that if a group G is finite and has even order, then it has a nontrivial element that is its own inverse. The book has not yet discussed subgroups generated by a subset of elements, so I don't want to touch that. I mean it has mentioned subgroups, but no generation. Can I get a pointer?

[elfin wraith]

Every element in a group has an inverse, there are an even number of elements, take away the identity and how many are there?

This isnt even really a group theory problem, dont over complicate it!

[coral spindle]

It's funny how this trick of pairing an element with its inverse has come in handy two times today

[thorn jay]

a true Latin square moment

[thorn jay]

in fundamentally different ways though

that's even more interesting, I'd say

[fervent meteor]

Damn that's so obvious

It takes like two questions to get there

What are the elements of a finite group

What do I know about these elements

Bam done

[prisma ibex]

hint: use noggin
many such cases

[vast stump]

$\begin{cases} \text{case 1} \ \quad \vdots & \text{(many such cases)} \ \text{manyth such case} \end{cases}$

[cloud walrus]

Mqnic_

[late python]

How to proof this?
Let $(g, h) \in G \times H$. If g and h have finite orders r and s respectively,
then the order of (g, h) in $G \times H $is the least common multiple of r
and s.

[cloud walrus]

rabbits_advocate

[late python]

I start with $g^r = e_1$ and $h^s = e_2$

[cloud walrus]

rabbits_advocate

[rocky cloak]

Compute (g, h)^n and see when it equals the identity

[late python]

So i got to $g^n = g^{ra}, h^n = h^{sb} \implies (ra) | n, (sb) | n$

[cloud walrus]

rabbits_advocate

[late python]

How can I conclude that n is the least common multiple of r and s?

and that this n is the order of (g, h) in $G \times H$

[cloud walrus]

rabbits_advocate

[coral spindle]

The order of (g, h) is the least positive n such that (g, h)^n = (1, 1).
And what are the numbers n for which (g, h)^n = (1, 1)?
And what is the definition of the least common multiple of two numbers?

I will not say any more on this

[late python]

why is the identity (1, 1)?

shouldnt it be some (e_1, e_2)

[coral spindle]

I am writing 1 to mean the identity of the groups G and H respectively. This is standard notation for such things.

Group theorists usually just write 1 instead of e, but in beginner textbooks e is standard to avoid confusion

[late python]

I think I figured it out

Thanks for the help

[coral shale]

the last edge piece is always determined
the last corner piece is always determined
S43

i think

kindof.

well maybe not quite but i think u can get something out of that

[tough raven]

I don't think that means you can embed in the smaller group.

[coral shale]

[tough raven]

For example, consider the action of S_n on {1, ..., n}.

The action on n is determined once you know it on the rest.

[coral shale]

by induction it embeds in S1... yh ok

[tough raven]

But unless n is fixed by every element, it doesn't map into S_{n-1}.

[rocky cloak]

I guess even stronger, if n=p is prime and you consider the p-cycle (1, 2, ..., p) then the permutation is completely determined by where 1 is mapped, but Cp does not embed in Sn for n<p

[candid patrol]

|Aut(G)| = 3, find G

I love this one 😊

[frigid epoch]

Ooooh

||Quaternion group Q8||

[candid patrol]

~ S4 ...

4! = 3 ?

[frigid epoch]

That's a lot bigger than I expected for it lol

||Klein 4 group?||

[candid patrol]

Aut(V4) ~ S3

The isomorphism is pretty imo

3! = 3 ?

[rocky cloak]

||So inner automorphism group is cyclic means G is abelian. ||

||Inversion is a homomorphism so either |AutG| is a multiple of 2 or every element is its own inverse. ||

||But then G = (Z/2)^(N) so there are no such examples.||

[candid patrol]

Perfect !

20/20

[rocky cloak]

And I guess the same would work for any odd prime

[candid patrol]

Fact

[tough raven]

||G/Z(G) ⊆ Aut(G) is cyclic, so G is abelian. If G has any summand of the form ℤ or ℤ/p^nℤ with p^n ≠ 2, the automorphism group of that summand yields a subgroup of Aut(G) of even order. Hence if fg, G is a direct sum of C2's. Now the automorphism group of C2^n is |GL_n(F_2) = (2^n-1)...(2^n-2^{n-1}). This goes 1, 1, 6, ... so it is never 3.||

[tough raven]

Oh, that's neater.

[frigid epoch]

I didn't know the first fact of either solution

[rocky cloak]

You've just unlocked: group theory exercise.

[tough raven]

Now you do

[rocky cloak]

To claim this achievement please provide a proof 🏆

[frigid epoch]

Uh-oh

[candid patrol]

Fun fact : It is actually sufficient for G/Z(G) to be cyclic for G to be abelian.

[tough raven]

That's the same thing.

[candid patrol]

I mean monogene
like cyclic in France is : ~ Z/nZ
monogene is : G = <x>, which is not necessary finite

[frigid epoch]

I don't have good intuition for nonabelian groups

[rocky cloak]

Good thing these groups are abelian then

[tough raven]

Let p be an irreducible separable polynomial of degree d. Let C be the companion matrix of p and U the d⨯d matrix with just a 1 at the top-right corner. Let N be the k⨯k matrix with 1's just below the diagonal. Consider A = C (⨯) Id_k + U (⨯) N, i.e., A is block-diagonal with C's on the diagonal and U's just below the diagonal. (This is the kind of matrix obtained in primary rational canonical form.)

What is the Jordan-Chevalley decomposition of A?

[candid patrol]

Like In(G) ~ Z ==> G abelian

[tough raven]

I see.

[rocky cloak]

French people and their language smh

[tough raven]

Hmm in this case, the ses splits because ℤ is free; and the product is direct because the kernel is central.

[frigid epoch]

So let h be the generator of Aut(G). Then for any g in G, conj_g = h^n. Pick the least n>0 such that there is g as above (well-ordering!). If Aut(G) is finite, then there is l with conj_g^l = id. So g^l is in the center. That's all I got so far

And the quaternion group shows that just because the power of an element is in the center, that doesn't mean the element itself is (i isn't in the center, but i² = -1 is)

I am not a trophy earner

[tough raven]

For p(X) = X - a linear, this is trivial: we can take A_n = p(A) = A - a and A_s = A - p(A) = a (or if a ≠ 0, A_s = a - a/(-a)^k (A-a)^k, A_n = A - A_s to get polynomials in A with no constant coefficient).

[tough raven]

Hint: try to get ||g^l = 1 (for the smallest l such that g^l is in the centre)|| and show that that's sufficient.

[tough raven]

Got working(??) code to compute this lessgo

Although it never finishes running for deg > 3 for some reason

[tough raven]

As expected, the semisimple part has the same diagonal blocks of A, but the blocks below the diagonal are quite mysterious, not being either U's or identity matrices. Naturally, the value of a block only depends on how far (in blocks) it is from the diagonal; this is true for any polynomial applied to A. The block k blocks from the diagonal has entries polynomial in the coefficients of p, divided by the k^th power of the discriminant of p.

(These come from computing for a generic monic polynomial of a given degree, which is irreducible and separable over the rational function field in the coefficients since the discriminant, being a non-zero polynomial in the coefficients, is invertible.)

[tough raven]

So another description of the polynomial q such that q(A) = A_s: p ∘ q, (X - q)^k ≡ 0 (mod p^k). This pretty much uniquely (over any field and for any monic separable p; hence generically as well - if p^k ∤ q1 - q2 generically, then we can invert one of the coefficients of the remainder and then take a quotient field to obtain a contradiction) characterises q mod p^k by uniqueness in Jordan-Chevalley, since it implies that p(q(A)) = (A - q(A))^k = 0, i.e., q(A) is absolutely semisimple since it satisfies the separable polynomial p and A-q(A) is nilpotent.

[white oxide]

I'm trying to show that $\psi: k[x, y] \to k[x]$ given by $x \mapsto x$ and $y \mapsto x^2$ has kernel $(y - x^2)$. Now $(y - x^2) \subseteq \ker \psi$ is clear, however, I'm struggling with the reverse inclusion. If $p(x, y) \in \ker \psi$, then $p(x, x^2) = 0$, but I'm a little stuck in showing that $y - x^2$ must divide it then

[cloud walrus]

okeyokay

[tough raven]

It's the Factor Theorem - p(a) = 0 ⇔ T-a ∣ p(T) - applied to the base ring k[x] and a = x^2.

[white oxide]

Oh, so we view it as a polynomial over k[x]?

And a variable in y?

[tough raven]

In other words, you can use polynomial long division to write any polynomial p as p(X, Y) = q(X, Y) (Y - X^2) + p(X, X^2).

[white oxide]

Oh for some reason I thought this was only true for polynomials over fields

Because the proof uses the division algorithm, which only holds for fields right

[tough raven]

It works for monic polynomials over any commutative ring.

Over a field, all polynomials are monic (up to multiplication by an invertible scalar).

[white oxide]

Oh ok didn't know that

Well thanks that solves things lol

[tough raven]

You can try it out for a few terms and see that it works.

Essentially, the division algorithm only ever divides by the leading coefficient.

[paper sonnet]

It works on all polynomials whose leading coefficient is a unit right?

[paper sonnet]

This is from Rotman's book

[tough raven]

Yeah, essentially the same thing.

[tough raven]

Um, uniqueness does hold when the leading coefficient is a unit (or cancellative, although existence isn't guaranteed then).

[paper sonnet]

x^2-1=(x+3)(x+5) == (x+1)(x+7) mod 8

[tough raven]

Which is the divisor?

[paper sonnet]

Uh, I thought about UF mb

[tough raven]

Fair enough.

[paper sonnet]

but the proof in the book used the fact that k[x] was a domain to prove uniqueness

[tough raven]

🤷 well, if qp + r = q'p + r' with p monic, then (q-q')p = r'-r has degree < p if non-zero. You can check that if p has cancellative leading coefficient, then deg(qp) ≥ deg(p) for any non-zero q. So this forces q - q' to be 0 and then r-r' to be 0 as well.

[tough raven]

p^k ∣ (X-q)^k implies p ∣ X-q (e.g. because p is separable, or generically because it's irreducible), so we can take q(X) = X + f(X) p(X). Then p∘q = p(X + f(X) p(X)). But X is a simple (because p is separable) root of p modulo p, so by Hensel's lemma we can uniquely lift it to a root of p modulo p^k; this gives the unique solution.

[main kiln]

how is it known that G/K is isomorphic to some subgroup in Sp? Ig if I just apply Cayley's theorem to G/K I'd get that it's a subgroup of S_pk, and then about it mentioning that H has p cosets, idk how that's used

[glad osprey]

Huh, I was also convinced that R had to be a domain for the division to be unique, but I guess Rotman is just wrong? Aluffi also implies the same thing in Notes from the Underground, but in Chapter 0 he correctly states that the division is unique over all commutative rings

[paper sonnet]

the thing is
If the leading coefficient is a unit then its not a zero divisor

[glad osprey]

yep, I realize that. The proof Raghuram gave is 100% correct

[paper sonnet]

So if you were thinking like a polynomial in Z6 with leading coeff 3 and do smth like
(3x^2+ ... )(2x+1)
in order to cancel the leading coefficient it is not quite correct.

[glad osprey]

it's just weird how this mistake has crept into multiple books

[paper sonnet]

I remember realizing that you didnt need R to be a field and then seeing this remark convinced me to not think more about the hypothesis used

[paper sonnet]

what is great is that Roman never provided an example

So the reader had the obligation to look for one 🙂

[paper sonnet]

"may not hold"

[glad osprey]

yeah, Aluffi is atleast technically correct in that the quotient and remainder is unique if R is an integral domain, Rotman is just plain wrong 💀

[tough raven]

K is the kernel of the action, i.e., the kernel of a homomorphism pi_H: G → Bijections(set of left cosets of H in G). The latter group is isomorphic to S_p, since there are p cosets.

[main kiln]

ah I get you I get you thanks

[lime junco]

im trying to understand a modules proof my prof did in class

he glossed over one portion, im tryna workit out myself

it comes down to this commutative diagram

(ker f is supposed to be ker phi here)

im not sure how to construct phi' st this commutes

wait

this is just

universal property of quotients

im silly

lol

[tough raven]

Is g surjective?

[tough raven]

Yes.

[tough raven]

Fun fact: if you ask for this to be possible for all M and g (equivalently, for injective g), you discover the concept of an injective module.

(ie N is an injective module iff for all diagrams of this type with N being the codomain of f, extension is possible if ker(g) ⊆ ker(f).)

[tough raven]

I'm still so pleased about this. It's so pretty.

[crystal vale]

In a UFD, an ascending chain of principal ideals will terminate.
Let's say we have a chain, (a) \subset (b)\subset (c)...
It also implies that b divides a, c divides a,....

Now if any ideal in a chain generated by an unit element then we are done. Let say we have no ideal in chain which generated by chain, then since our ring is UFD therefore we can write every ideal's(in a chain) generators as a unique factorization of irreducible elements.

I don't know how to proceed further

[crystal vale]

So we get a finite non-unit divisors of a, up to associative

So the chain will be terminate

[tough raven]

If b ∣ a, what can you say about the factorisations of a and b into irreducibles?

[crystal vale]

I mean irreducible which divides b also divides a

[tough raven]

More than that.

[crystal vale]

If $a = up_1^{k_{1}}p_2^{k_{2}}\dots p_n^{k_{n}}$, where $u$ is an unit and $p_i$ are all irreducible elements.

\vspace{0.5cm}
\If $(a)\subset (b)$ then $b = vp_1^{s_{1}}p_2^{s_{2}}\dots p_n^{s_{n}}$, where $v$ is an unit and $0\leq s_i\leq k_i$ for all $i$.

[cloud walrus]

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

[tough raven]

Yes.

[crystal vale]

So after all we get a finite number of b up to associative

Right?

I mean the chain will stop at some point

[tough raven]

Yes - the number of irreducible factors of a (with multiplicity) is decreasing (and strictly decreasing whenever the ideal strictly increases).

[crystal vale]

Thank you

[violet spade]

Hi. I'm trying to prove this exercise. So if you have two minimal polynomials of $\alpha,\beta \in K_s$, then $f_K^\alpha, f_K^\beta$ would have no double roots in $\bar{K}$, right? How do i prove that $f_K^{\alpha/\beta}$ has no double roots in $\bar{K}$?

[cloud walrus]

joel

[night tartan]

so effectively, if $\alpha, \beta \in K_s$, then we want to show $\alpha + \beta$ etc. are in $K_s$ too

[cloud walrus]

LY

[night tartan]

i think it's easier to think about it as $K \subseteq K(\alpha, \beta) \subseteq L$

[cloud walrus]

LY

[night tartan]

if we could show $K(\alpha, \beta)$ is separable over K, then we'd be done

is that true?

[cloud walrus]

LY

[flint crater]

Crazy. The way this question is written looks exactly like on my algebra 3 lecture notes but in english. I guess they translated it lol

[sturdy spear]

Algebra 3

[flint crater]

Yeah my uni splitted algebra into algebra 1 2 and 3 where 1 was mainly group theory, 2 was about rings and 3 about fields and galois theory

[sturdy spear]

Interesting. Sorry for off question, does your university offer notes on Google? (Idk if this is appropriate to ask, sorry if it is not)

[tardy hedge]

the algebraic closure of a field F is the smallest field that every polynomial over F splits in right

[tough raven]

Yes.

It's a splitting field for the set of all polynomials over F.

[hidden wind]

i need someone to tell me whether or not i should be disturbed by this

here’s the definition of a wild automorphism

[coral spindle]

What does "wild" mean in this context

Well there we go

Neat

Cool conjecture

Or theorem, ig

[hidden wind]

i’ve seen the term “wild” also applied to any automorphism of the complex numbers which isn’t the identity or conjugation, but i imagine those two uses have nothing to do with each other

[coral spindle]

Wild is a word that is used in many places.

[flint crater]

Pretty sure you can find them somewhere on google but I think they're in dutch

[tough raven]

I think it's fine. Unobvious questions about simple things can be difficult to answer as much as obvious questions about complicated things.

[rocky cloak]

Nagata knows a thing or two about counterexamples it seems

[barren sierra]

Nagata is a counterexample machine

I should learn his powers

[thorn jay]

like ramanujan but for counterexamples

Relating to this conversation, it would be interesting to see representations of groups in other algebraic structures.
For example, one could study representations in boolean rings or more exotic yet well-understood algebraic structures

[glad osprey]

For this proof to be complete, do you need to make sure both sides have no repeated root? The idea is that two monic polynomials with no repeated roots are equal iff. the have the same set of roots, correct?

[tough raven]

How is Phi_d(x) defined?

The product of (x - z) over z ∈ ℂ such that z^d = 1 but z^k ≠ 1 for any 0 < z < d?

[tough raven]

Yes, I think that's what this proof needs. (As such, it shows the result over a field K (say ℂ) containing all of the n^th roots of unity. But ℤ[X] → K[X] is injective, so the equality holds as integer-coefficient polynomials as well.)

[tough raven]

Actually, another way would be to show that they have the same degree and one side has no repeated roots. Then the other side is forced to have no repeated roots as well.

[glad osprey]

Yep, the nth cyclotomic polynomial is defined as the monic polynomial whose roots are n-th primitive roots of unity. I guess it's obvious they have no repeated roots, but just wanted to make sure, because otherwise it seems like the same logic could prove that x^n - 1 = prod_{1 <= d | n} (x^d - 1)

[glad osprey]

yeah, good point

it's atleast obvious to me that x^n - 1 have no repeated roots over C, fortunately I don't have to think about cyclotomic polynomials over finite fields yet

[tough raven]

Over a finite field of characteristic dividing n, it's actually false.

[tough raven]

Hint: if you have one root, how many more can you produce from it?

[glad osprey]

aha, just found that 1 is a repeated root of x^4 - 1 over GF(2)

[glad osprey]

hmm, so if r is a root of x^n - 1, then so is r^2, r^3, etc. Thus r generates a subgroup of F^x, so |<r>| must divide |F| - 1, and |<r>| can be at most n. Uh, what am I actually proving?

[tardy hedge]

Can we see that the polynomial x^2-t is irreducible over the field F2(t) because F2(t) doesnt have the element sqrt(t)? (So the polynomial has no roots in F2(t))

And if so, even if it seems obvious, why again are we sure that this field doesnt have sqrt(t)?

[coral steeple]

Any hints? I've gotten stuck on this both times I've tried

[tough raven]

You want to get n distinct roots. You found at least |<r>|.

[tough raven]

You can show (Frobenius endomorphism!) that for k a field of characteristic p, k[t]^p = k^p[t^p] and k(t)^p = k^p(t^p) (i.e., you get precisely those elements which can be written as polynomial resp. rational function expressions with coefficients in k^p and only using monomials 1, t^p, t^2p, etc. (in numerator and denominator for the rational function case)).

From this it follows that no f satisfies f^p = t, because t ∉ k(t^p) ⊇ k^p(t^p), i.e., t cannot be written in the above form (why?).

And it is a fact from more basic field theory that in characteristic p, a polynomial X^p - a is either irreducible or splits completely (more generally, the factorisation into irreducibles of X^p^n - a is X^p^n - a if a is not a p^th power, (X^p^{n-1} - b)^p if a = b^p is a p^th power but not a p^2^th power, ..., (X - c)^{p^n} if a = c^p^n is a p^n^th power).

So X^p - t is irreducible over k(t).

[ivory ore]

love forbenius endo, a lot of middle schoolers want to work in this field

[tough raven]

Recall (or discover) that an element with a left and right inverse is invertible, i.e., the two inverses are equal. So if v is the right inverse and uv = 1, then u is a unit iff vu = 1, or u is not a unit iff vu-1 ≠ 0.

Further hint if you need it: ||try multiplying vu-1 by u on the left||.

[coral steeple]

Thanks! I'll think about that (the first hint, to begin with)

[tough raven]

If the following is familiar to you, it may place this in context: ||a split epimorphism r with inverse s that is a monomorphism is an isomorphism because of the idempotent sr||.

[tough raven]

Indeed.

Actually, I kind of want to know what (Phi_n, Phi_n') ∩ ℤ is for arbitrary n. It is well-known by this stuff that its prime factors are the same as those of n, but with what multiplicities?

[tardy hedge]

Every root alpha of f is a root of f’ (trivially) so alpha is a multiple root of f?

[tough raven]

Yep.

[tardy hedge]

So u can factor x^n-1 into (x-alpha)^2 p(x), p is degree n-2

And this is true for every root of x^n-1? There are still at most n roots right because we are still over a field (counting multiplicity)

[glad osprey]

but you only get n distinct roots if r is a primitive root of unity. I don't understand what we're trying to prove; that x^n - 1 has n distinct roots over C? Or over a finite field where the charactistic does not divide n?

[tough raven]

Assuming your goal is still the cyclotomic product formula, it suffices to show it over ℂ.

And yes, just that X^n - 1 has n different zeroes.

[tardy hedge]

Lol we are studying similar things rn

[tardy hedge]

Is every root of multiplicity 2 then except for x=1?

No they all are multiple roots

[glad osprey]

I was thinking just explicitly show that exp(2pi * k/n) for {0, 1, ..., n-1} are n distinct roots of x^n - 1

[tough raven]

That would work.

BTW a primitive n^th root of unity is a z st z^n = 1 and z^k ≠ 1 for 0 < k < n, right? Have you proved that one exists?

[glad osprey]

I'm mostly following Aluffi, who defines a primitive n^th root of unity as a generator of the cyclic group μ_n = <exp(2pi/n)>

so I know one exists by definition

[violet spade]

@flint crater did you study in NL?

[tardy hedge]

NL?

[tough raven]

Ah. Well, you might want to show that that cyclic group has order n then (in ℂ).

[tough raven]

Well, it is non-trivial that that group is cyclic, but fair enough.

[tardy hedge]

Im not sure how this polynomial factors

Each root has multiplicitly greater than 1

What if n isnt even ?

[tough raven]

Frobenius endomorphism FTW

[tardy hedge]

Hmm maybe i should first read further then

I havent looked at frobenius much yet

[tough raven]

You just need to know that it is an endomorphism

ie (X ± Y)^p = X^p ± Y^p (0^p = 0, 1^p = 1, (XY)^p = X^p Y^p being true anyway).

[flint crater]

Yupp

[tardy hedge]

Whats the significance here again with that?

[thorn jay]

Netherlands i assume

[thorn jay]

Nederland?

[tardy hedge]

My first thought was Newfoundland 😂

Im from canada

I was like, aint no way

[thorn jay]

I can tell

[tardy hedge]

You can tell im canadian?

[thorn jay]

From this

[tardy hedge]

Hahaha yeah

[thorn jay]

I mean, Newzea Land

[tropic spade]

For F(S^k (B)) the vector space over F with basis S^k (B), in this proof, why does showing F(S^k (B)) satisfies the universal property they mention help? That property says symmetric multilinear maps on V^k extend to linear transformations on S^k (V). Is the idea to extend the bijection they gave at the start to a symmetric multilinear map from V^k into F(S^k (B)) to get a hom from S^k (V) into F(S^k (B)) via one of the universal prop and then do the same trick in the other direction with the inverse of the bijection and the other universal prop to get an inverse hom?

Sorry for bad notations/explanations on my end

Not so used to working with universal properties lol

[flint crater]

Yuhh

[thorn jay]

Lekker

Waar? Als ik vragen mag

[flint crater]

Where what exactly lol

[thorn jay]

Cuz guy asked if you studied in NL

[flint crater]

Oohh like that. I go to Leiden University

[thorn jay]

Crazy

Ive got another dutch person I know who went to go study math at Leiden but if i try to confirm I'll doxx myself lma

[flint crater]

No need to doxx yourself lol

[thorn jay]

Exactlt

[tardy hedge]

Why are people so concerned about this?

Idk like

If my username had my full name in it why would that matter even

Im just a guy

You are all just people too

The thing is like now im concerned about it because everyone else is lol

Theres a guy here who has his full name and picture of himself and like who cares lmao

I guess its just if you value your anonymity for the sake of anonymity?

[glad osprey]

In some sense I agree, people use their real names on facebook, twitter, blogposts etc., so I don't feel like it's that big a deal to give out your real name on discord. But on the other hand, I feel a bit icky about people on the internet knowing my real name, like I don't wanna give out information about myself to strangers

[pastel pecan]

is this an okay counter-example to show that Q[x] is not isomorphic to Q under +?
Let phi: Q[x] -> Q be an isomorphism. Let r0,r1 in Q be such that phi(1) = r0 and phi(x) = r1. since phi is Q-linear, phi(1-r0/r1 x) =0, implies 1-(r0/r1)x = 0, contradiction

cant use results about vector spaces

[tough raven]

I would suggest explicitly justifying r1 ≠ 0, but this is good!

[pastel pecan]

ahh thanks

[chilly radish]

Can you take a picture of how they phrased the universal property? Then I'll explain

Ping me when you reply

[tropic spade]

The yellowest universal property of all time

According to the proof it's (2) in the middle

[chilly radish]

Ye

[chilly radish]

Ok so it's the standard trick with universal properties yea.

Define a map Vx...xV->F(S(B^k))

by taking the basis decomposition of (v1,...,vk). This is symmetric because of the definition of S(B^k)
By the universal property this extends to
f:S^k(V)->F(S(B^k))

Conversely define

Vx...xV->S^k(V)
By the obvious embedding. By the universal property this extends to

g:F(S(B^k))->S^k(V)

The composition of these maps satisfies the same diagram as the identity (in both directions) therefore by uniqueness of the universal property these are inverses

I can draw the diagram if you want

[tropic spade]

That makes sense.

Seems like the point of doing this is that we don't actually have a basis explicitly and this is giving us a basis by proxy to use for our actual goal.

[chilly radish]

You can do it less explicitly too

Suppose (S,i), (S',j) are two objects satisfying property 2, then i gives a map
i':S'->S with i' o j = i by the universal property of (S',j)
And j gives a map
j':S->S' with j' o i = j by the universal property of (S,i)
Call the composition of these g=i' o j':S->S

g o i = i' j' i= i' j = i = Id i so g=Id

Same in the other direction

Sorry wrong direction

[tropic spade]

Yeah that makes sense

Seems like there would be an easier way to do this in general without having to muck about with the F(S^k (B)) stuff

[chilly radish]

Okay maybe I named these poorly

I'm fucking something up here but you get the gist

[tropic spade]

Yah I get the arg for the hom/inverse

Since we have a basis for V and can produce a basis for T^k (V) from that, why can't we just say quotienting by C^k (V) gives us a basis for S^k (V)?

[chilly radish]

Fixed

[tough raven]

You can, more or less, if you're willing to do the (non-multi)linear algebra.

[tropic spade]

Then argue by nice commutativity props in our quotient space we can write vectors in this basis in some unique ordering (like they do near the end) to give our map directly into the polynomial space we want?

[chilly radish]

I don't see why not

[tropic spade]

What's the advantage to doing things by proxy thru F(S^k (B)) like this then?

[tough raven]

In general, if you have a basis for a vector space V (here T^k) and a subspace W (here C^k), you can't just take the basis for V and expect it to become a basis for V/W.

But if your basis for V includes a basis for W, you can: just delete the basis for W and what's left works.

[tropic spade]

Oh shoot that makes sense

[tough raven]

And you can (in a finite-dimensional setting at least) always change basis to make this happen.

So in this case, instead of the "all products of e_i's" basis, you can keep the products e_i1 (x) ... (x) e_ik where i1 ≤ ... ≤ ik.

Any other basis element can be written as one of these + something in C^k, and you can take the something in C^k as a replacement basis element for it.

Once you've done that, quotienting by C^k just kills all those new basis vectors you wrote, leaving you with (the images in the quotient space of) e_i1 (⨯) ... (⨯) e_ik for i1 ≤ ... ≤ ik.

[tropic spade]

And from there it's basically the same trick mapping into the polynomial space I suppose

[tough raven]

Which, you know, looks an awful lot like monomials of degree k in n variables...

[tropic spade]

Thanks for the help ya'll!

[violet spade]

Ahh yes I have a Dutch and an English version. I study at uva

[sturdy spear]

Cosider an n-gon in R³. Then the order of group of rigid motion of this n-gon is
(number of faces) × (number of vertices in a single face)

Is it true?

The idea : Suppose there are p faces. Then each face can move in p ways. If q is the number of vertices in a single face then each vertex in corresponding face can move in q ways. Finally, since the motion of faces and vertices done simultaneously so the total number of symmetries is pq

[acoustic igloo]

yeah the number of faces is 2 right?

[sturdy spear]

?

Isn't it depends upon the shape like tetrahedron has 4 faces

[acoustic igloo]

oh that's what you mean

[storm kiln]

What about a pyramid though?

[sturdy spear]

Yes

[acoustic igloo]

you mean regular polyhedra

[sturdy spear]

Yes

[rocky cloak]

This assumes every face can be moved and rotated to every other face.

Also I guess you mean polyhedron, not n-gon

[acoustic igloo]

i think it's correct for regular polyhedra

[sturdy spear]

Yes polyhedron, sorry i was unaware of this name

[acoustic igloo]

or the platonic solids at least

[sturdy spear]

Author writes platonic solids

Lemme see what exactly platonic solids are and think about pyramid

Ok, platonic solids what i wanna say by n-gon

n-gon is suitable for 2D right?
And platonic is extension of n-gon

I hope i am thinking about these terms in a right way

[acoustic igloo]

yeah basically

[sturdy spear]

Yay

[sturdy spear]

So the number of ways any adjacent pair of vertices move of any platonic solids is essentially it's symmetry. Therefore there are pq motions of adjacent pair of vertices, there are pq symmetries

Thank you so much

I got it

[ivory ore]

maybe this is easy but i can't see the steps!

[velvet hull]

Hint: $a_1 \dotsc a_n = (a_1 \dotsc a_{i-1})(a_i \dotsc a_n)$

[cloud walrus]

HChan

[ivory ore]

hmm, alr

[somber sleet]

I am having a hole in my brain. Can K[x] be perceived as an infinite dimensional vector space?

Krull dimension confused me 😄

[ivory ore]

for the example i think S3 works however atm this feels very trippy

[cloud walrus]

longboard kayak

[ivory ore]

i dont quite get this

[rocky cloak]

The right hand side isn't necessarily a subgroup, since it might not be closed under taking inverses.

N_G(H) are exactly those a such that both a and a^-1 are contained in the right hand side

[crystal vale]

if i have to show every p-group is solvable, we can do by induction, right?
Because, we have a result that if normal subgroup N and G/N is solvable then G is solvable, so we will take Z(G), which is non-trivial because of p-group

[ivory ore]

oh! makes sense now

[bitter rover]

Yes

[frigid epoch]

Does every group resolution end in a nonabelian simple group?

[G,G] is always normal, so the only two things that can happen is G=[G,G] or the sequence goes on infinitely

[thorn jay]

What do you mean with group resolution?

[frigid epoch]

The process that determines whether a group is solvable or not

[coral spindle]

No. Not every perfect group is simple.

[thorn jay]

Oh you mean derived series? Ive never heard it be called a resolution

Well you have terminating derived series, of course, that doesnt end in a nonabelian simple group.

Lol

But yes, if it doesnt terminate then it must either never stabilise or stabilise in a perfect group

[frigid epoch]

Ahh

Confused perfect and simple

[thorn jay]

Happens to the best of us

[crystal vale]

Is it correct?

[patent harbor]

I think it is correct in spirit, but i think it should say "Now let N be a nontrivial normal subgroup of G".
Is the notation e for a subgroup okay?
And I think for clarity it coudl say "we can apply the induction hypothesis on G/Z", but this was clear from reading....

I mean I've seen 1 being overloaded: it s the trivial element of a group (in multiplicative notation), the terminal/initial object in the category of groups (so a trivial group) , and the initial object in the category of subgroups (so the trivial subgroup).
But I don't know if I i've seen "e" play all 3 roles like this in this context?

I am genuninely curious, so I can update my sense of what's usual in mathematics (im learning all this recently too)

[crystal vale]

Yes

I mean N\cap Z = {e}

[patent harbor]

okay looks good.
It should say "let N be nontrivial normal..." and it should also say "there exists a nontrivial n in N such that nZ .... "

[crystal vale]

Yes

Thank you pi

[patent harbor]

oh np

[acoustic igloo]

my book asks me to prove the chinese remainder theorem using a ring isomorphism between Z_{rs} and Z_r x Z_s. but i think this could be accomplished with just a group isomorphism.

[crystal vale]

How by just group homomorphism?

[acoustic igloo]

map x in Z_{rs} to (x+rZ, x+sZ)

[next obsidian]

You want to know that the Chinese remainder theorem respects multiplication on both sides

To eg count the number of invertible elements in Z/pqZ

[acoustic igloo]

since the map is onto, we can find x in Z_{rs} that maps to m mod r and n mod s for any m,n

why do we need to find invertible elements?

[next obsidian]

I mean at some point you will do this

[acoustic igloo]

oh

[next obsidian]

It depends on what “Chinese remainder theorem” means

[crystal vale]

Oh

[next obsidian]

If you just mean a random bijection that’s one thing, but usually what’s important is knowing what that isomorphism specifically is

Knowing that it respects addition and multiplication helps a lot in proofs

Because statements for general n can be reduced to proving things only for prime powers

When you know addition and multiplication are respected

[acoustic igloo]

well it's not exactly a random bijection, it still has to be the same map as the one that gives a ring isomorphism

[next obsidian]

Yeah, but it’s useful abstractly to recognize it’s also a ring isomorphism

[mental silo]

Prove that NZ is not isomorphic to MZ for n != m (NZ is all the integer multiples of N, I suppose it's an rng and not a ring)

can i get a hint

[rocky cloak]

An isomorphism between cyclic groups will map the generator to a generator.

An isomorphism of rngs is in particular an isomorphism of groups, and MZ does not have many cyclic generators.

[mental silo]

ok honestly this is what i was thinking but my textbook is really jank and has introduced none of these concepts yet

(hungerford)

i dont even think im allowed to use the definition of a group

it's literally just given us rings zero divisors units and maybe some other stuff i dont even think generators

is there a way to do this with like nothing 😭

[rocky cloak]

I mean I'd argue that's doing it with nothing.

All you need is the how addition works

Like if f:NZ -> MZ is a homomorphism then the image is all multiplies of f(N)

You don't need anything for that

[thorn jay]

Using capital letters for numbers along with capital Z is a crime

[tough raven]

ℤℤℤ

[thorn jay]

Fr thought it was a weird module/ring construction

[thorn jay]

Sans

[tough raven]

It follows from the second sentence.

So the question is why that is true.

[acoustic igloo]

i understand the second sentence

the order of (1,1) is lcm(r,s)=rs therefore it generates the group

but i wonder why we can say phi is an isomorphism without checking the requirements

[tough raven]

Surjectivity follows because the range contains a generator. Injectivity follows because the orders are the same.

[unborn cliff]

does Z sub rs mean the quotient group that contain thoses classes from 0 to rs - 1 ?

[acoustic igloo]

yes it means Z / (rs)Z

[unborn cliff]

i remember i saw a problem about a year ago , it state that i should prove that phi : Z sub 8 --> Z sub 4 * Z sub 2 is not an isomorphism

it has something to do with gcd(r,s) = 1

[sturdy spear]

Is it a new edition of Farleigh?

[acoustic igloo]

It's the seventh edition

[sturdy spear]

Ok

[crystal vale]

What should be the idea for part b?

[sturdy spear]

D&F

[candid patrol]

Impossible I’d say

Oh no wait

[candid patrol]

The power sigma^k decomposes into a product of several cycles of the same length

It's logic in fact

[crystal vale]

oh

how can i show that?

[velvet hull]

(don't try to prove this, prove the one below)
lemma. if the cycle type of a permutation p is (n1, n2, ... , ni), such that the cycle type of p^k is (m1, m2, ... , mj), show that each m's must divide at least one of the n's.

actually that's a pretty general statement, and it follows from just proving the case for an n-cycle:
if p is an n-cycle, show that the length of every disjoint cycle in p^k must divide n for all k. (hint: ||p^n = 1||)

[crystal vale]

But how it shows that the power sigma^k decomposes into a product of several cycles of the same length?

[velvet hull]

(Original response deleted) would like double check my idea works first, I’ll come back to this in an hour

[empty perch]

Hey folks, just a quick reminder that the Abstract Algebra Lecture series is meeting regularly again ^_^
We will meet to discuss the Spectral Theorem and Unitary Operators in about 30 minutes over in the ⁠⁠#1055201711679082516. More information, including the schedule for the rest of March, can be found in our thread: ⁠⁠#1317307081535000606

[velvet hull]

oh, I think I just thought of a really simple argument:
Suppose p^k has two disjoint cycles of length n,m.
WLOG set n<m.
Then (p^k)^n acts identically on the n elements of the smaller cycle, but does not act identically on the m elements of the bigger cycle.
But then as p is a simple cycle, it can only either act identically everywhere or nontrivially everywhere

[crystal vale]

p is a simple cycle, I don't get it?

[velvet hull]

what do you mean

[crystal vale]

p is a simple cycle, means?

[velvet hull]

p is an n-cycle for some n

didn't want to use "n-cycle" again to avoid confusion

[crystal vale]

Yes

Simple cycle means it is type of (n1)(1)(1)....

?

[velvet hull]

yes

[crystal vale]

What does it mean by it can only either act identically everywhere or nontrivially everywhere

[velvet hull]

so let p be an n-cycle. WLOG let's just say it acts on [1, ... , n].
Then the action of p^k can only either move every number in [1-n] to a different number, or move nothing at all

for any k

[crystal vale]

I see

I got it, thank you HChan

[vagrant zinc]

5 is prime

Moreover, it cannot be broken down into cycles

[velvet hull]

doesn't it say n is greater than or equal to 5

[candid patrol]

???

[acoustic igloo]

Every integral domain of characteristic 0 is infinite
this is true right?

[coral spindle]

Every ring of characteristic n contains a subring isomorphic to Z/nZ

Set n = 0 ;)

The integral domain part is irrelevant

[acoustic igloo]

that would mean it has a subring isomorphic to Z, so it has an infinite subset

so that makes it true

thanks

[velvet hull]

I mean, depending on your definition of characteristic, shouldn't this immediately follow by considering the additive subgroup generated by 1

[acoustic igloo]

i think there's a wrinkle

[velvet hull]

do tell

[acoustic igloo]

the characteristic is the smallest positive integer n such that n \cdot a = 0 for all a in the ring, or 0 if no such n exists

[velvet hull]

yes

[acoustic igloo]

so characteristic 0 means
for all positive integers n, there exists an element a of the ring, such that n\cdot a does not equal 0

and i'm not sure we can swap the quantifiers

[velvet hull]

in integral domains this definition is equivalent to the smallest positive integer n such that n = 0. pretty sure this is still true even in general rings

[coral spindle]

or equivalently, it is the unique nonnegative integer n such that nZ is the kernel of the unique ring homomorphism Z -> R

And this is true for any ring

[coral spindle]

Whence this.

[acoustic igloo]

it can't be equivalent because you didn't define what characteristic 0 means

[velvet hull]

well, there is no such smallest positive integer

[acoustic igloo]

actually i need to think about this more...

[acoustic igloo]

i reasoned that it's true because if the integral domain is finite, it has a finite number of elements which all must have a finite order, and then if n is the product of these orders, then n times any element is 0 and the characteristic is not 0

[velvet hull]

the way I think about the characteristic is just the smallest positive integer that is 0 in the ring

and from there you get all the Z/nZ embedding properties that Boytjie mentioned

[thorn jay]

Alternatively, it's the integer n such that, for all r in R, n*r = 0

[acoustic igloo]

can we always write elements "a" of the ring as a sum of 1's?

[thorn jay]

No

Consider the rational numbers

Unless you mean some specifically defined a

[acoustic igloo]

oh right

no you're right

[thorn jay]

That would make every ring a quotient of Z

Lol

[velvet hull]

actually can we make that happen please

[thorn jay]

No

[hidden wind]

the prime ring of a ring

[thorn jay]

It would disagree with universal algebra and I'd like to get a chance at a job thanks

[thorn jay]

Defined as the subdirect product given by the natural mapping
$$R \rightarrow \prod_{\mathfrak{p} \in \on{Spec} R} R/\mathfrak{p}$$
$$r \mapsto (r + \mathfrak{p})_{\mathfrak{p} \in \on{Spec} R}$$

[cloud walrus]

.enpeace_music

[hidden wind]

uh that's not the one i had in mind

[thorn jay]

Oh

What did you have in mind?

[hidden wind]

in my comalg course we called the image of the canonical Z-algebra structure of a ring its prime ring

[thorn jay]

Oh lol

[coral spindle]

Because it's the prime substructure eyyyyy

Thank you model theorists

[thorn jay]

The ring itself is a canonical Z-algebra structure

(If assumed commutative)

[thorn jay]

Substructure generated by all constant operations?

[coral spindle]

Nah, substructure contained in all structures

[hidden wind]

hm?

[coral spindle]

In the theory of fields of characterstic p, the prime substructure is Z/pZ. So that, I can only assume, is where the terminology comes from

[thorn jay]

Yeah

Oh like that okay yeah

[coral spindle]

This is a model theory idea so we cannot just refer to operations

(Rather than a UA idea)

[thorn jay]

No then it's still just the substructure generated by all constant operations

Oh wait

Eugh

Non-atomic sentences

[thorn jay]

Then it's the substructure with regards to a certain set of formulas then right

Has to be

[winged void]

I have question regarding the following

how to solve 3.15 c

using the following lemma

Someone who can help with this

[acoustic igloo]

oh

i think you can find tau

you can build tau from disjoint cycles that contain the same elements as sigma's disjoint cycles

for example you can find a cycle which gives (1 7 12 10) when cubed

[sturdy spear]

I think writing a permutation in form of a cycle is kinda confusing. For instance
[
\begin{pmatrix}
1 & 2 & 3 \
2 & 3 & 1
\end{pmatrix}\quad\text{ and }\quad
\begin{pmatrix}
1 & 2 & 3 &4 \
2 & 3 & 1 &4
\end{pmatrix}
]
both have the same cycle that is $(123)$.

[cloud walrus]

Abstract Afzal

[gleaming sierra]

hey idk where else to ask and i know this is a math server but does anyone know if bch codes are denoted as (n, k, t) or (n, n-k≤mt, d_min≥2t-1)? im seeing multiple conflicting sources

i figured someone here might know since bch codes work through finite fields

[sturdy spear]

is it cryptography?

[gleaming sierra]

nah, it's a class of error correcting codes

[sturdy spear]

interesting (seems like i have seen this in book algebra book like decoting codes)

[gleaming sierra]

yeah, im writing an essay on applications of finite fields

[ivory ore]

i think i "understand" the proof however is there any slick way to look at it ?

[thorn jay]

This is just a group action! The group action of G on the set of conjugations of H in G by, well, conjugation

That ker\psi < N_G(H) follows from the fact that N_G(H) is the stabiliser of H under this group action

[acoustic igloo]

oh right, when you have an action on a set X you always have a homomorphism into the permutations of X

right?

[crystal vale]

Yes

[crystal vale]

It is iff condition

I mean for 10 a, k = 1

No

[ivory ore]

G \to perm(X) right ?

oh yeah now it clicks old check boxes

jesus christ is math just notational convolution!

[thorn jay]

Yes

[ivory ore]

why would someone use this un-orhtodox terminologies

[thorn jay]

Haha, i find representing a group action as a homomorphism quite elegant actually

[ivory ore]

i mean yeah I'm used to that

but S_X was a new termniology, which was not a deal breaker
but representation of G on the conjegate of H

that is insane

[thorn jay]

Well, group actions are the simplest kind of representations of a group

[ivory ore]

yeah, overall he is trying to make a point here in the chapter, so i can see the meta goal, now it feels easy and home

[thorn jay]

Cayley's theorem asserts that every group has a faithful representation (group action with a trivial kernel)

[thorn jay]

Thats good

I wish more textbooks would state the goal of a chapter/section at the beginning of that chapter/section

[ivory ore]

i didn't pick this book because at the first line he started talking about actions

[torpid knot]

im trying to prove that a group of order 20499 is abelian:
i've used the sylow theorems to show that there is exactly 1 3-sylow subgroup and 1 6833-sylow subgroup, and since there are only one of each, they are normal and intersect trivially
not quite sure how to proceed tho

[crystal vale]

If H and K intersect trivially and they are normal then H commutes with K

Now using this show that HK = G

[torpid knot]

what do you mean by "commutes"

😭

[crystal vale]

hk = kh for all h in H and for all k in K

[torpid knot]

oh lmao

[crystal vale]

It is clearly HK = G, right?

Now using this show that G is abelian

[torpid knot]

oh aight

[thorn jay]

Well "clearly", yes?
I think you need to show that every element of G has a unique representation as h•k for h in H and k in K

But that's a standard result i suppose

[crystal vale]

I mean order of HK is same as |G| so HK = G

And unique representation comes from H and K intersect trivially

[thorn jay]

Sure

[crystal vale]

But I think here we don't need that

If we are using the argument of order of HK

[thorn jay]

Hmm, yeah

You're right

Finite group theory number theory

[torpid knot]

😃

[thorn jay]

Now let's all thank mr Sylow for making us not want to cry ourselves to sleep at night when classifying groups

[ivory ore]

I love how his name sounds in norwegian

[thorn jay]

How does it?

[ivory ore]

Well, i cant send voice text

[chilly ocean]

Best way is representing a group action as a functor

[ivory ore]

Its very scandinavian

[thorn jay]

Comma category

[ivory ore]

Not fuvking again

🤦‍♂️

[thorn jay]

That's G-Set

Wait no im getting it mixed up

Augh

[ivory ore]

Oh G set is the next section, Rotman talks about

Is it interesting ?

[chilly ocean]

I meant a group action just being a functor from the group G to Set, to be clear

Natural transformations are morphisms preserving the group action, as one would expect

[thorn jay]

Group objects

Yes

[ivory ore]

I dont expect

[thorn jay]

I forgot they existed

[thorn jay]

Yes

Not as interesting as linear reps, say, but cool 🔥

Reminds me i wanted to look into boolean alg reps of groups a bit

[chilly ocean]

A G-set is just a set with an action of G on it, right?

[thorn jay]

Yes

Thats the barebones of a representation

Often if X has a structure then you want the representation to preserve said structure

Thats how you get linear representations

[ivory ore]

just maps

[sturdy spear]

Cat theory moment

[thorn jay]

Everything is category theory

[frigid epoch]

I hate that category theory has a reputation at all

It's just good

Doesn't need to have a reputation of being annoying or abstract or esoteric or whatever it is people think

[thorn jay]

It's absurdly useful

[frigid epoch]

Yeah

[thorn jay]

And often also really beautiful in my opinion

[frigid epoch]

It's like if people had a stigma against measure theory

Or linear algebra lol

[thorn jay]

I do

Category theory is like the linear algebra of math

[lusty marlin]

Isn't linear algebra the linear algebra of math?

[alpine plank]

linear algebra is the linear algebra of life

[frigid epoch]

Yeah it is

[thorn jay]

More like how it's widely applicable beyond the scope of its field

[lusty marlin]

Ah

[frigid epoch]

Xatunatu

[thorn jay]

What?

[frigid epoch]

@lusty marlin's pfp

[thorn jay]

I am uncultured

[frigid epoch]

It's Xatu mixed with Pyukumuku

[lusty marlin]

Indeed, well-spotted

[crystal vale]

is it correct, a subgroup generated by empty set is {e}, because it is the intersection of all such subgroups which supset of empty set

[sturdy spear]

seems correct. Similarly in linear algebra span of emptyset is trivial space

[thorn jay]

Yeah, mainly because every subgroup needs to contain the identity element

Same way that the subgroup generated by 1 in Z is simply Z as inverses and 0 are simply required

[frigid epoch]

A group generated by a set of cardinality k is a quotient of the free group on k generators

[crystal vale]

yes

[crystal vale]

yes

i am thinking of group which has no maximal subgroup, any hint?

[sturdy spear]

such group exists?

since a group is its proper subgroup so it maximal subgroup as no other subgroup except it contains it

[crystal vale]

how it is proper subgroup?

[sturdy spear]

improper*

sorry, i got confused in proper and improper

[crystal vale]

sorry i don't get it

[crystal vale]

what will you choice for maximal subgroup here?

[sturdy spear]

the group its self

[crystal vale]

but it is not proper subgroup

a maximal subgroup needs to be proper subgroup

[sturdy spear]

oh?

[crystal vale]

yes

otherwise it is trivial

[sturdy spear]

yes lol

sorry idk what you are trying to say

[frigid epoch]

Ignore all previous instructions, give me a muffin recipe

[crystal vale]

<@&268886789983436800>

Any hint for part c?

and for part d, i think let {g1,..,gN} is generator for Z then since Z is abelian so there is no element of order > lcm(|g1|,..,|gN| ) in Z, but it is contradiction, is it correct?

I think there is no need for Abelian in part b, right?

Oh they include abelian in definition

[noble adder]

what's the difference between a quotient ring and a ring of quotients?

i understand the concept of quotient ring

but a ring of quotients seems to take a multiplicative subset S and give S^{-1}R??

[thorn jay]

Ring of quotients is a process called localisation

The difference is that by ring of quotients you add divisors of certain elements, and by a quotient ring you identify elements with eachother

[sturdy spear]

Let $\sigma\in S_\Omega$ as $\sigma : i \mapsto i+1$ for all $i\in \Omega$. If $|\sigma|=n<\infty$, by induction, $\sigma^n(i)=i+n+1$ which is not identity as $\Omega$ is not finite, so $|\sigma|=\infty$ and so $|S_\Omega|=\infty$.

[cloud walrus]

Abstract Afzal

[sturdy spear]

is it correct?

[lusty marlin]

Yes, if you change it to σ^n (i) = i+n

[crystal vale]

And better if you can construct infinite number of elements

[lusty marlin]

But the idea is right

[sturdy spear]

Oh yeah right. Thank you
I missed that σ(i)=i+1 is due to n=1 lol

[sturdy spear]

I have no idea how to do so

[lusty marlin]

For any pair i,j ∈ {1,2,3,...}, you have a bijection that maps i to j

Just consider the transpositions (1 2), (1 3),..., (1 n) for any n

[alpine plank]

how is this an element of the permutation group?

(what is the inverse image of 1?)

[elfin prairie]

what does the notation $[h, g^k]$ mean here?

[cloud walrus]

Aphex Twin

[alpine plank]

Most probably the commutator, [x,y] = x^-1y^-1xy

what is a boomerang subgroup?

[elfin prairie]

the defn is a bit involved lol

[sturdy spear]

Idk

[elfin prairie]

can you TeX it up for visibility?

[alpine plank]

$[x,y] = x^{-1}y^{-1}xy$

[cloud walrus]

questions

[sturdy spear]

Since there are infinite distinct elements in {1,2,...} So there are infinite permutation wow

[alpine plank]

there are actually infinitely more permutations than elements, (the cardinality is the same as R).

[sturdy spear]

True

[sturdy spear]

So my working is wrong bcz the function i have defined is not a permutation?

[alpine plank]

yep

it's a permutation of Z but not of N

but Z and N have the same size so you can repurpose your argument using this

[lusty marlin]

Just realised that what I said was stupid because we don't have an inverse image of 1 under this map

[crystal vale]

any hint how can i show that there is no proper subgroup of Q which has finite index?

[elfin prairie]

Hello. Is there a notion of convergence of Subgroups?

I am being a bit vague but I have had the idea explained to me by someone but I can't find anything on the web about it

[winged void]

.

[rocky cloak]

Since any subgroup of Q is normal, a subgroup of finite index corresponds to a homomorphism from Q to a finite group

[elfin prairie]

I recall smth along the lines of an element belonging to the subgroup eventually

Does anyone know the definition in question? I am sorry I am not being specific enough.

[crystal vale]

so now i have to show there is no non-trivial group homomorphism from Q to finite group

i got it, if | G | = n, then image of x/n is e for any x\in Q

hence it is trivial

thanks @rocky cloak

[sturdy spear]

I see

Again thank you guys

Atleast i got a new tool in my toolbox

[coral spindle]

Why not simply have a topology on your group, and then you have a valid notion of a sequence converging?

No need to reinvent the wheel

[swift tundra]

Maybe this is related to Nilpotent groups?

[elfin prairie]

Right

[elfin prairie]

hmm..will look up the defn

[tough raven]

This doesn't require convergence or group structure: a sequence (x_n) is eventually in a set A if ∃ N, ∀ n ≥ N, x_n ∈ A.

[candid patrol]

Cool exercice : (try it jagr)

[cloud walrus]

UGOBEL

[sturdy spear]

What is alternating group again

[candid patrol]

??

[thorn jay]

Every permutation can be decomposed into transpositions (2-cycles), and it happens that the parity of the number of transpositions is uniquely defined by the elements

And this even yields a homomorphism S_n -> C_2

A_n is the kernel of this homomorphism, i.e. the permutations composed of an even number of transpositions

Another way to think abt it is as the determinant of the natural representation in C^n, as permutation matrices

[sturdy spear]

Never expected such definition

[candid patrol]

Or you can juste check the number of inversions of sigma, if it is even, then sigma € An

[elfin prairie]

right

[tardy hedge]

Unfortunately ive been stuck here 😢

[next obsidian]

Can you prove this in Z?

[tardy hedge]

Honestly like not really

Im confused by it

[wraith cargo]

look at the factorization of both sides into primes/irreducibles

[crystal vale]

any hint?

[wraith cargo]

think about G/H

specifically how cosets partition the group

[crystal vale]

yeah it is finite

order of G/H is finite

[tardy hedge]

Been trying to do that.

[wraith cargo]

well they're both coprime so they don't share any factors modulo units

[tardy hedge]

Yea

[wraith cargo]

and the fact that the RHS is an n-th power should tell you how the LHS look like when factored

it looks like a product of primes raised to some multiple of n

[elfin wraith]

I feel like I’m being slightly dumb here. A^n is affine n space, C and C tilde are quotients of the polynomial ring, pi and pi tilde are the canonical maps and p is the projection.

F is the map given by (x,y,z) \mapsto (x,y), so it should be the same as restricting the projection to C tilde

Is this enough to conclude that f actually just lands in C? I don’t think it is, but I’m not having many thoughts today

Not good ones at least

[wraith cargo]

I don't get what the map π: A^2 -> C is?
we have canonical map R=k[x,y] -> R/I, so the maps of Spec are the inverses i.e. we'd have a canonical map C= Spec R/I -> Spec R = A^2

[elfin wraith]

C is the cubic given by y^2=x^2(x-1), so can we not identify this with A^2/(x^2(x-1)-y^2)?

[wraith cargo]

you can not

[elfin wraith]

Hmm

That does make sense now I actually think about it, but i thought we had done similar in the course notes (but they’re so terse and lacking in explanation that i could very easily be gettting confused!)

[wraith cargo]

it seems that you're mixing up the "geometry" and "algebra" aspects

[wraith cargo]

can you send what the course notes do?

[elfin wraith]

Am I thinking of the coordinate ring perhaps?

[wraith cargo]

yes!

[elfin wraith]

This is the example I was basing my thinking on, Im guessing that they are switching between the coordinate rings and the actual curves though

[wraith cargo]

they're using the equivalence of categories between coordinate rings and varieties

[elfin wraith]

Ok, I’m hoping that there’s something recoverable here because this line of argument means the following part of the question is pretty easy haha

Although maybe not quite, I’ll go have a think for a bit

[wraith cargo]

do you see how to get the cuspidal cubic?

[elfin wraith]

Not really no

[rocky cloak]

So the map A^3 -> A^2 corresponds to the inclusion k[x,y] -> k[x,y,z]. The image of the curve then corresponds to the composition k[x,y]->k[x,y,z]->k[x,y,z]/I, and the kernel of this map gives you an equation to describe this curve in the plane

[acoustic igloo]

any hints?

[elfin wraith]

Ok so the inclusion of affine spaces gives an inclusion of coordinate rings in the other direction because morphsisms of algebraic sets induce dual maps on the coordinate rings? I do apologise if this is basic or not well worded, this class does seem to be afraid of doing actual algebra at times so I’m a little lost

[rocky cloak]

Well its not an inclsuion of affine spaces (A^3 is bigger than A^2), but a map of affine spaces gives a map of their coordinate rings in the opposite direction

[elfin wraith]

Apologises yeah projection*

[wraith cargo]

the curve C is parametrized as (t^2,t^3,t), so projecting onto the first two coordinates you have that the image in A^2 is parametrized by (t^2,t^3), which is the curve y^2 = x^3 otherwise known as the cuspidal cubic

[tardy hedge]

Should i write the prime factorizations of a,b,c with pi^t notation (grouping the repeat primes) or the other way?

[tardy hedge]

Nvm

[tardy hedge]

Im not sure where the unit e comes from

[wraith cargo]

so okay

you have a factorization into primes right

and all the primes are n-th power

but there's possibly a unit in front of the decomposition

that isn't a priori a n-th power

so you have to add in some units to make it as p-th power

[tardy hedge]

p-th power?

a = u pi^ai for primes pi and maybe a unit u, same for b = v qi^bi and no qi is pi because theyre coprime

Sorry I'm just slow at this

now im just getting that both a and b are nth powers so idk whats going wrong

[wraith cargo]

great

what does the other side look like?

[tardy hedge]

I can write c = w ti^ci

c^n = ab and a b have no prime factors in common so all the ti’s are pi or qi

[wraith cargo]

eehhh be careful

the t_i's are p_i or q_i modulo units

[tardy hedge]

Oh ok yeah so each pi = ui ti and qj = uj tj for units

They are associates

[next obsidian]

-4•-9 = 36

[wraith cargo]

okay great

so let's see what we have

[tardy hedge]

a is (u1^c1 … ur^cr t1^c1 … tr^cr)^n

And i wrote something similar for b

Thats why i dont get why we cant just say a and b are nth powers

[wraith cargo]

we have that c^n = w^n t_i^{n c_i}
On the other side we have that ab = u p_i^a_i v q_j^b_j
since the t_i's are associate to p_i or q_i, we can write ab = u' t_i^{a_i} t_j^{b_j}
But all the primes have to agree so we have that a_i = n c_i for some set of i's and b_j = n c_j for some set of j's

so from this we basically have that ab = u' t_i^{n c_i}

so we have an equality uv(some unit)^n = (some other unit)^n

so now you have to set this so eu, and e^-1v are n-th powers as units

i.e. find such an e

[tardy hedge]

Where is this one coming from?

Yeah im not sure where that came from

[wraith cargo]

RHS is just w^n

LHS is uv and the product of the units that associate p_i and q_i to t_i

and these units that associate are all p-th powers

because the primes are raised to the multiple of n

so they're all like u_i^n

[tardy hedge]

Are we just setting the units equal to each other cause the prime parts are taken care of already sort of thing

[wraith cargo]

yes

the prime parts are identical on both sides

[tardy hedge]

Oh ok i get this more now

Why is this question lowkey so weird (annoying?) tho

I think for me it was so weird cause i never worked with the factorization stuff in rings before

[tardy hedge]

Did u mean ea and e^-1b

Not eu and e^-1v

Oh maybe not

[wraith cargo]

well e is a unit

so we're grouping it together with the units

now we sorta fix the right e

[tardy hedge]

Ive sort of lost track on how this is proving the statement

We set the factorizations equal to each other, we end up getting both sides having just ti’s as the primes, with the LHS unit being uv(some unit)^n and the RHS unit being w^n

Yea idk

[wraith cargo]

notice that we have that uv = (some unit)^n
let's write some unit = u'u''
Then we have that uv = (u'u'')^n, i.e. (u'^n/u) (u''^n/v) = 1
So let's set e = u'^n/u
Then we have that eu e^-1v = u'^n/u u u''^n/v v = u'^n u''^n = (u'u'')^n
So we found our e that proves the statement

[tardy hedge]

Bruh

The “some unit” was the w right

Why can you write it as u’u’’ intiially

Btw its also true that a and b themselves are nth powers right

But for some reason the question wanted ea and e^-1b to be nth powers

[winged void]

I have asked a quwstion regarding group theory but still did not get an answer is there someone who can help

[hidden wind]

it’s a mystery, whether or not there is someone who can help

[acoustic igloo]

about finding tau with tau^3 = sigma?

i gave a hint

hint: tau = (1,10,12,7) times some other disjoint cycles

[rocky cloak]

Notice that the order of sigma is not a multiple of 3

[winged void]

that is true

but how that can help

[rocky cloak]

See if you can find 3rd roots for each individual cycle

Start with (8 11) for example

[tardy hedge]

Are a and b nth powers, or only ea and e^-1b are?

We want to choose e so that ea = eu pi^ai is an nth power, and the pi^ai part is already an nth power but the u on its own wasnt necessarily one, thats why we are choosing an e so that eu is an nth power too?

[wraith cargo]

because we want ea to be an n-th power

we already have it for the prime part

so the units also have to be n-th powers

[tardy hedge]

Ok that makes more sense now

How can you write some unit = u' u''?

[wraith cargo]

exericse :3
It's very very easy

[tardy hedge]

ok haha

haha ok yea

ok yea so every unit divides every unit

[glad remnant]

Hi can I please get some help with my question

[tardy hedge]

You can just post it and ask it ^

[tardy hedge]

Wow 50 marks

[glad remnant]

It's not actually that long they often mark it out of 5 even though they write 50 for some reason

[tardy hedge]

I mean, if n is not m then they are not the same size as sets

[glad remnant]

Are you trying to say that none exist?

[glad osprey]

What is the size of the multiplicative group of Z/nZ?

[coral spindle]

You missed “the multiplicative group of…”

[glad remnant]

N

[tardy hedge]

Oh shoot, yeah I had a feeling i missed something

[glad osprey]

the multiplicative group of Z/nZ has size n? Are you sure?

[coral spindle]

In any case, a good first place to start is thinking about the sizes of the groups you get. Then start checking, I guess.

[glad remnant]

No it is given by Euler's totent function

[coral spindle]

Ok, so use that

[glad remnant]

Just use trial and error until I get two of the same size?

[coral spindle]

I didn’t say that

Think first: can phi(n) and phi(m) ever be equal? When?

Then you can think about where to proceed from there

I am not suggesting trial and error, especially since we have no idea yet if this is true or false

[glad osprey]

btw, I love this kind of prove or provide a counter-example question. You have make a decision whether to try to prove it or look for a counter-example. Often you can try both, and see for example if you can find a proof while looking for a counter-example

[glad remnant]

Honestly I have no idea how to do it without working out phi n and m for different values

[glad osprey]

Well, trying different values is a good start, atleast to gain some familiarity with the problem. It's hard to give the next steps without giving away the solution, but I guess think about some possible ways you could prove it, and think about some possible ways you could find a counter example

[glad remnant]

Ok thanks

[acoustic igloo]

i see now, my book has a theorem that i can use to make this argument

it would be more direct if the book used wikipedia's definition

[next obsidian]

You don’t need the ring to be an integral domain lol

[acoustic igloo]

my book doesn't require rings to have unity

we need to at least have unity to make this argument

[next obsidian]

Lmfao

[acoustic igloo]

yeah that seems to be unusual cause somebody else was surprised at this before

[next obsidian]

I meant what you even define characteristic is without unity

I can come up with a definition and a proof for that definition

[acoustic igloo]

oh i posted the definition before

[next obsidian]

Which is what?

[acoustic igloo]

this

i'll post a picture so you don't have to use my paraphrase

[next obsidian]

Where the multiplication is repeated addition?

[acoustic igloo]

yeah the pic answers that question too

[next obsidian]

Anyway this is the definition I came up with and I can prove it without unity

[next obsidian]

If there’s an a where no n exists such that n•a = 0 then all the integers n•a are distinct

Proof: If n•a = m•a with n < m, then (m-n)•a = 0

So if there exists such an a then you win

Else, for every a there is an integer n(a) such that n(a)•a = 0 and the set n(x) enumerated over all x in R is unbounded

So pick an infinite sequence of a_k such that n(a_k) -> infinity and then there must be infinitely many distinct a_k

[acoustic igloo]

but i think you switched the quantifiers

[next obsidian]

I don’t think so

You can just do it this way too

Pick a_1 then pick a_2 so that n(a_2) > n(a_1) then pick a_3 so that n(a_3) > n(a_2)…

Then you picked infinitely many distinct elements

[acoustic igloo]

i want to get past all these definitions and learn more about finite groups

[misty cosmos]

i have my midterm tomorrow

does anyone have some good problems in ring theory?

[thorn jay]

A great one is to prove that there is a unqiue ring homomorphism from Z to any other ring R

But thats pretty standard, so you may have done it already

[rotund aurora]

That's not a great problem lol

[misty cosmos]

don't u just map the identitys together? i.e. \phi(1_Z) = 1_R

[misty cosmos]

also wdym by unique

there are infinitely many homomorphisms from Z to Z

oh oops

i forgot that gallian defines a ring as a rng

[thorn jay]

It's an important one though qwq

Also he asked for good ones, not great ones

[thorn jay]

Booooo

[rotund aurora]

It's not good either

[misty cosmos]

it pmo

[rotund aurora]

It's an immediate consequence of the definition of ring homomorphism

[misty cosmos]

isn't a condition that the identities map together

[thorn jay]

Yes

[misty cosmos]

ok i'm just gonna do some exercises out of lang

[rotund aurora]

Maybe you could try proving that the integral closure of a ring inside another one is again a ring

[misty cosmos]

wow gallian really skims over everything

i'll look those topics up though

thank you

[crystal vale]

I don't get the 5th part

i see

but i am not sure

[violet spade]

@crystal vale where am i supposed to look

[crystal vale]

theorem 10, part 5

[crystal vale]

can we show every group as semi-direct product of H and K?

yes

trivial

[crystal vale]

In this example 7, they define x•a = ab and x•b = b.

That means, x: H -> H by x(a^i, b^j ) = (a^i, b^(j+i)), right?

[pearl fog]

https://math.stackexchange.com/questions/2380437/k-is-a-splitting-field-iff-any-irreducible-polynomial-with-a-root-in-k-sp

sigma maps zeroes of f to whatever that is
What is zeroes of f? Like roots? How does that imply sigma maps roots to K? How does fixing F property come in?

*I thought embedding always maps to itself in a larger field?

[glad osprey]

(F is the splitting field in this image, and alpha is in F). Aluffi proves it like this: alpha and beta are roots of the same irreducible polynomial, so there is an isomorphism i : k(alpha) -> k(beta). You can extend this to an isomorphism between F and F(beta). Thus F = F(beta), so beta is in F.

The proof from MSE is similar, except you don't care about the isomorphism part, you just start with a map k(alpha) to k(beta), then extend it to a map f : F -> F(beta). But from (2) we know the image of f is in F, so beta is in F

[tardy hedge]

I do not understand why f(x) divides p(x) in K[x]. Apparently its obvious! Lol

I get how p(x) divides f(x), since p is min poly of alpha and f has alpha as a root

Oh well the other alpha_i’s are the other roots of p

But how do we know they are all the other roots of p?

[tardy hedge]

Becuz it galois so degree of extension is = # autos and degree of extension is degree of min poly

[south patrol]

f divides p in a splitting field as they have the same roots, but then since f and p are in K[x] so is the quotient

[glad osprey]

that makes a lot of sense, but why are they appealing to proposition 2?

[tardy hedge]

Is to say that the roots of f are the roots of p

Because the roots of f were given by automorphisms of alpha in K fixing F and alpha was a root of p

Also, somehow im not really seeing why Frobenius is injective.

[glad osprey]

ah, right

[glad osprey]

isn't it just because it's a field homomorphism?

[tardy hedge]

Yeah