groups-rings-fields

[dull ginkgo]

if f is a polynomial in F[X], and there exists an element such that f(x) = 0, then then does that imply that g = (X - x) divides f in F[X]?

"Better" question

For any commutative ring [afaik] we have the partial ordering of "x divides y" which means what you think, aka (R, |)

[devout crow]

yes

[dull ginkgo]

alright

[devout crow]

comm ring is enough

[hidden haven]

Preordering because 2 unequal things can divide each other

[chilly radish]

Imagine your category not being preadditive or having a 0

So cringe

I bet @delicate orchid likes non preadditive categories

[chilly radish]

Oh my god they were associates

[delicate orchid]

Word salad

[chilly radish]

Preadditive = homsets are abelian groups such that composition is bilinear

And 0 object = object that is both initial and terminal

[delicate orchid]

How are these functions

[chilly radish]

What

What do you mean

[delicate orchid]

How are these individuals

[chilly radish]

This is true in e.g. R-Mod

[delicate orchid]

I’ve seen it for modules yeah but I’ve forgor why they’re groups

[chilly radish]

Well for modules the inverse is obvious

It's just the inverse al

And composition is bilinear

A general category in which this is true is just called preadditive or an Ab-category

For example the category of CW-Spectra with addition given by the pinch map

[delicate orchid]

I just can’t remember why composition is even well defined

[chilly radish]

Cuz you define an internal operation

[delicate orchid]

See this

[chilly radish]

So f+g means something

So h o(f+g) also means smth

[chilly radish]

What

[delicate orchid]

If it’s like Hom(M, M) or Hom(M, R) then yeah

That’s just the dual space

I can see that working

[chilly radish]

Composition doesn't have to be internal

Like

If f,g:A->B and h:B->D then
h(f+g)= hf + hg

Where hf, hg are in Hom(A,D) and so is h(f+g)

[delicate orchid]

Right so the operation in the homset isn’t composition

[chilly radish]

No

It's a separate addition operation

Which distributes over composition whenever defined

[delicate orchid]

Then what did you mean by this

[chilly radish]

This

[delicate orchid]

Because calling this composition really fuckin confused me , even if it is composition I read it more akin to scalar multiplication (in the case of working in cat of R modules)

[chilly radish]

Technically these are different compositions

[delicate orchid]

I think I’ve only seen this as scalar multiplication which is the problem

[chilly radish]

Hm?

Wdym scalar multiplication

[delicate orchid]

Hom_{R}(M, N) as an R-module

So the only “composition” I’ve seen is multiplication by elements of R

But this just seems like a more general version

[dull ginkgo]

Is there another way to phrase algebraic completeness using homomorphism properties

[chilly radish]

We always mean composition of morphisms

If scalar multiplication is defined e.g. as in R mod

Then composition is usually bilinear w.r.t it too

[delicate orchid]

Yeah, exactly

[chilly radish]

Like (af) o g = a(f o g)

[delicate orchid]

So when I hear “composition” I don’t assume you’re talking about, as far as I knew, had to be scalars

[chilly radish]

What do you mean

[delicate orchid]

Hence my confusion

[chilly radish]

I see

Composition is always a little circle to me

[delicate orchid]

It’s cleared up now though

[chilly radish]

Anyways preadditive categories are nice

Even moreso when they have a 0 object and all finite biproducts

[delicate orchid]

Now I’m going to try and reverse engineer why the fuck my com-alg lecture specified that they had to be scalars

[chilly radish]

Idk what you're referring to

[gilded gull]

Uh idk if you guys are done I don't wnna interrupt but

How can I show if a morphism is well-defined?

[delicate orchid]

Show that if x = y then f(x) = f(y)

[gilded gull]

So like injectivity but backwards?

[delicate orchid]

Yeah basically

[gilded gull]

o

ok ty

[dull ginkgo]

like, is there something beyond roots of polynomials that make algebraicly complete fields important

[chilly radish]

Uh, all the properties that follow from algebraic completeness?

Why does it need to be phrased in terms of homomorphisms to be important

[dull ginkgo]

fair enough

[delicate orchid]

representation theory is much nicer over algebraically closed fields because you have guaranteed existence of every eigenvalue

[chilly radish]

Polynomials winning again

[agile burrow]

Feeling like a solvable lie algebra today

Might have to apply Cartan's criterion to be sure

[tough raven]

Given a ring R and (say, left) R-modules M and N, does Hom_R(M,N) have any more structure than “abelian group” in general?

[agile burrow]

In general, no

However, if R is commutative or if one of M or N is an (R, R)-bimodule then you get an R-module structure on the Hom set

[tough raven]

If L and R are a left and right ideal respectively of a ring A, then LR is a two-sided ideal. OTOH, any right ideal containing L should contain LA, being the right ideal generated by L and also a two-sided ideal. Hence LR = LA.
Have I made a mistake somewhere here? (I'm guessing it would be that LR is a two-sided ideal.) The conclusion sounds wrong.
(Note: for X, Y subgroups of A, XY := subgroup generated by {xy|x in X, y in Y} = {Σ_{i=1}^n x_i y_i | n >= 0, x_i in X, y_i in Y}.)

[hidden haven]

LR need not contain L?

[chilly radish]

You also always have a Z(R)-module structure

[tough raven]

Let R be a ring, I the set of isomorphism types of simple left R-modules, M a left R-module, and for each i in I, M_i the sum of all simple submodules of M of isomorphism type i.
Is it true that M_i for i in I are “linearly independent” i.e. their sum is their internal direct sum?

By semisimple module properties, a sum of simple submodules is a direct sum of some of those submodules.
So given m_i in M_i for i=1,…,n s.t. Σ m_i = 0, we can decompose m_i into a sum in such a direct sum and end up with the goal
if L_{11},…,L_{1i_1}, … ,L_{n1},…,L_{ni_n} are simple submodules with L_{i1j1}, L_{i2j2} isomorphic iff i1 = i2, and the sums L_{i1} + … + L_{i j_i} are internal direct sums, then the whole sum L_{11} + … + L_{nj_n} is a direct sum.

[rustic crown]

you can show that M_i and sum of M_j for j != i intersect trivially. if it were non-trivial then find a simple sub module of the intersection. it should be in the class of i as well as not in i

[tough raven]

Why would that be?

I don't know how to prove a simple submodule of M_i must have type i.

[rustic crown]

oh lemme think, i don't remember the details very well

ideally one would prove the equivalence conditions for being semi-simple
(1) it's a sum of simples
(2) it's a direct sum of simples
(3) complement property holds
(4) every short exact sequence splits

[tough raven]

123 I know

4 give me a minute

[rustic crown]

oh 3 and 4 are the same

[tough raven]

Oh the semisimple one is in the middle?

Cool.

[rustic crown]

oh oops, yea

so this is like the "existence of basis" somewhere you would need to use AoC

[tough raven]

Yes there was so much Zorn's Lemma in this equivalence.

[rustic crown]

similarly you can prove the "uniqueness of basis" with similar ideas

[tough raven]

What is uniqueness of basis?

Basis cardinality?

[rustic crown]

so if you write a semi-simple module as a direct sum in two ways, then for a given isomorphism class of simples, both sides have the same cardinality of number of copies of it

[rustic crown]

so yea, pretty much this, but since there could be many isomorphism classes of simples, it says the uniqueness of "dimension" for each

there was a direct proof using WOP

but as far as i rememeber the idea was pretty much same as the "replacement" for usual vectorspaces

[tough raven]

In finite case this is the uniqueness of tower thing

I will take it for granted for infinite for now

[rustic crown]

i think we should be able to prove that using these two

[tough raven]

👀

Ah.

[rustic crown]

if M is a sum of simples, then we know it's a direct sum of a subset of them

[tough raven]

Sure.

Take a wrong simple module in M_i; write some complement of it in M_i as a direct sum.

[rustic crown]

now if N is any other simple submod, then by complement property M = N o+ something else

by uniquness N would appear in the earlier decomposition

i think this works

[tough raven]

So that gets us to M_i, M_j disjoint.

[rustic crown]

yea, they only intersect at 0 >.<

[tough raven]

But the same trick works finitely many times.

[rustic crown]

i think this would make sense now

the simple submod in the intersection would has type i as well as type j for some j != i

this is bad, so the intersection was trivial to begin with

[tough raven]

Oh you said sum of M_j

I thought it was just M_j

[rustic crown]

semi-simples are nice

[tough raven]

Hmmm

Maybe I should look for the infinite-case proof after all.

Or maybe restrict to a special case of presence/absence?

Actually I can get away with finite case

[rustic crown]

oh really?

[tough raven]

Later after all.

[tough raven]

This started with me trying to understand why the M_i for a semisimple ring are simple.

Since a semisimple ring is a finite direct sum of simple submodules, it's enough.

[rustic crown]

oh i see

wait

[tough raven]

Actually wait

[rustic crown]

so one definition of a semisimple ring was that all modules over it are semi simple

[tough raven]

Did I even need this direct sum stuff

It was enough to show M_i had no characteristic submodules

which I did

[rustic crown]

idk what a characteristic submodule is 🙈

[tough raven]

I learnt it when trying to decode a one-line hint that was all I found to prove this

[rustic crown]

ah

[tough raven]

Characteristic left ideals are precisely two-sided ideals. Thus two-sided ideals are sums of isotypic components. Conversely, isotypic components being characteristic are two-sided ideals.

It means preserved by all endomorphisms

I'm assuming.

[viscid pewter]

ok i want to run some of my solutions for some d&f through you guys

i'm just gonna post a bunch of solutions and then you guys can chip in if anything seems wrong, if anyone's around

[viscid pewter]

let X be {(aK, bH)} where a can be any in G but b must be in K, so this corresponds to |G: K| |K: H|

and let Y be {cH} where c can be any in G, so this corresponds to |G: H|

so consider the map f: X -> Y, where f(aK, bH) = abH

then if f is a bijection, we have the desired result

first, surjectivity:

any element g of G is in some coset of K, so g is in some aK, ie. g = ak

any element k of K is in some coset of H, so k is in some bH, ie. k = bh

so any element g of G is some abh where a is in G, b is in K and h is in H, so f is surely surjective

now injectivity: show abH = cdH iff aK = cK and bH = dH

abH = cdH iff:

(c^-1)a(bH) = (dH)

but bH and dH are in K

so (c^-1)aK = K

so aK = cK as required

now what? i actually realised my original proof doesn't cover this enough, this sucks

no wait i'm dumb

any element g of G is some abh where a is in G, b is in K and h is in H. suppose there are two different ways to write g. then:

if abh1 = cdh2:

(c^-1)a bh1 = cdh2, but bh1 and dh2 are in K so (c^-1)a is in K, so aK = cK

then:

(d^-1)(c^-1)ab = h2(h1^-1)

no clue

jesus i hate this

next

if there is h in H such that h is not in N:

consider the subgroup in G/N generated by hN. let the order of hN be x. then x must divide the order of G/N, as <hN> is a subgroup of G/N. but x must also divide the order of <h>, which in turn divides the order of H.

so x is a common factor between the order of G/N and the order of H. so if |G: N| and H are coprime, x is 1.

i think that one's fine

[rustic crown]

second one looks fine. another way to do it would be to use second iso theorem,
HN/N = H/(H n N)
the left is a subgroup of G/N. Thus the size of H/(H n N) divides both |H| and |G/N| which means it's 1 and so H n N = H

.
the first one has a little problem i think

the map f you define, f(aK, bH) = abH isn't really well defined

a better strategy would be to define a map f : G/H --> G/K and show that this is surjective and each fiber f^-1(gK) is in bijection with K/H

give it a try, and let me know if you get it

[agile burrow]

I think I'm being silly, but why is the intersection of ideals zero?

methinks maybe the author has already assumed the ideals are simple? In which case the intersection is an ideal in I_i and I_j, so by simplicity it must be zero

[oblique river]

Hmm that is strange

[rustic crown]

i think they're trying to say L is direct sum of ideals if it's direct sum as vector spaces

so we're given the intersections are 0

[oblique river]

That was my first thought too, but the wording is a little strange

[agile burrow]

This is to guarantee elements of L are written as a unique sum, right?

[rustic crown]

yep

[agile burrow]

Woke up feeling like a semisimple Lie algebra

computing my Killing form to be sure

[wooden ember]

Every galois field extension comes with associated automorphisms. In particular if we consider an algebraically closed field as a galois extension of some other field (like C with R) can we further extend this field to a sort of “super algebraically closed field” by requiring that every polynomial equation involving the automorphisms of the galois group admit solutions?

So with the example of C over R we might require that z * conj(z) = -1 have a solution

I’m imagining you can’t but curious to know why

[oblique river]

If you wanted every equation to have a solution youre going to run into a problem with something like z * conj(z) = i

If you take the complex conjugate of both sides you get conj(z) * z = -i so now your “field” is no longer commutative

[tribal moss]

It's not even clear what conj(z) ought to mean in your extension. Who's to say that complex conjugation extends uniquely to your new extension field?

[wooden ember]

Both fair points, thanks

[tough raven]

Actually if it might not be commutative you might want to make conj an antiautomorphism, in which case you get z*conj(z) = -i, i.e. everything collapses.

[tribal moss]

I suppose one could still speak about whether a field extension L/K is "super-algebraically closed" with respect to whichever K-automorphisms L happens to have. However, Buncho's argument essentially shows this is not possible unless the group of K-automorphisms is torsion free.

(And it doesn't help to drop the requirement of fields to be commutative, because we could make the same argument for the sum of z and its conjugate(s) as for the product).

[severe cedar]

is it true that: if $(R, +, \cdot), (S, +, \cdot)$ are rings and $\phi\colon (R, +) \to (S, +)$ is a group homomorphism, then it is also a ring homomorphism? i think it must be, but i don't really see what structure there is to take advantage of apart from the distributive law

[cloud walrus]

capitalsigma

[delicate orchid]

hmm

[chilly ocean]

if you're working with rings with unity, you usually ask that a ring homomorphism takes 1 to 1. so you can send everything to 0 for a counterexample

but you probably only care about whether or not f(rs) = f(r)f(s)

[severe cedar]

yeah, f(rs) = f(r)f(s) is what i'm hoping to show, i hadn't thought about whether the rings have 1

[chilly ocean]

it's not true, f: (x, y) -> (x, x + y) from Z^2 to itself is a group homomorphism, and f(1, 0)f(0, 1) = (1, 1)(0, 1) = (0, 1) is not equal to f((1, 0)(0, 1)) = f(0, 0) = (0, 0)

@severe cedar

[severe cedar]

hmm, i guess i am misunderstanding what i have to show for this question, then. thanks!

[opaque pilot]

Since $\langle g \rangle \preceq G$, then the possible values of $k = 0,1,2$. That is, possible $|g| = 1,5,25$ by Lagrange. If $|g| = 1$, then g is identity and obviously is in H.

[cloud walrus]

KN

[opaque pilot]

If $|g| = 5$, then call $K = \langle g \rangle = {e, g, g^2, g^3, g^4}$.

[cloud walrus]

KN

[opaque pilot]

Im not sure where to go now. Maybe if assume g not in H and get a contradiction?

oh I think I got it

[terse crystal]

The order of g in G is 5^k

Therefore the order of gH in G/H divides this 5^k

It also divides [G:H]=4

So order of gH in G/H must be 1

[keen sparrow]

Unless I'm missing something obvious, but shouldn't the quotient in theses maps be the other way around

A is a commutative ring with unit and q is an ideal of A

and d should be a unit

[agile burrow]

could someone help me see that L = Z(L) + [L, L]? I've shown that L is completely reducible since ad L is isomorphic to L/Z(L) = L / Rad L, which is semisimple.

[opaque pilot]

I did this, I think yours is a lot faster..

[terse crystal]

I see, although I don’t how you obtained g=e in the end, a counterexample could be when G=Z/25Z direct sum Z/4Z, H=Z/25Z, g=(1,0) ,or (5,0),clearly doesn’t equal e=(0,0)

[opaque pilot]

Are those quotient groups?

[terse crystal]

Yeah

[opaque pilot]

we'll learn that next week, but yeah I dont think g = e should be there

thanks for pointing that out. I thought I proved that k != 1 or 2, so k = 1 and that means g= e

[terse crystal]

I see, you just made a logical error

When g doesn’t belong to H, contradiction

So it contradicts g not being contained in H, doesn’t contradict g not being e

[opaque pilot]

That makes sense

Thanks again

[glossy widget]

We know thanks to Abel and Galois that there's no quintic formula, more precisely, no quintic formula with finitely many nested radicals. What about an infinite quintic formula? One that's infinitely long or one that has infinitely many nested radicals?

[hidden haven]

https://m.youtube.com/watch?reload=9&v=BSHv9Elk1MU
This gives a proof that there is no continuous function on the coefficients which solves the quintic

[glossy widget]

I actually watched that yesterday and that's where I got my question from lol

[hidden haven]

Oh lol

Yeah so I think an infinite nesting of radicals, if uniformly convergent, would be continuous, so that should be the answer

If I recall correctly, the statement that's proven in the video is something along the lines of "there do not exist continuous functions f_1,...,f_5: ℂ⁵ → ℂ, such that for a list of coefficients c = (c_0, ..., c_4), the elements f_i(c) are precisely the roots of the quintic with c as its coefficients"

[chilly ocean]

45 minutes long

[glossy widget]

it's honestly worth the time

You know how you can have "2 answers" to an infinite fraction? this way you can kinda have a quadratic formula without roots ig? I thought that maybe a similar argument could be made for a general quintic, but this time with infinite radicals

[south storm]

I don’t think the LHS can be understood in a way to be well defined here, one can‘t define it as the limit of "finite versions" of the fraction as they largely include 1/0.

This is evidenced by x^2-x+1 having no real roots.

[hidden haven]

This is not the statement, it's almost immediately true. The one in the video is about multivalued functions, and I was trying to mimic that using multiple functions but that makes the statement weaker

I'm not sure how to talk about continuity for multivalued functions, maybe as functions into a quotient of ℂ⁵

Quotienting by the action of S_5

[tribal moss]

Define a distance between (a1,a2,a3,a4,a5) and (b1,b2,b3,b4,b5) as the minimum of sum_i |ai-b_sigma(i))| for sigma ranging over S_5?

[hidden haven]

Hmm is that any different from the quotient topology?

Seems very similar

[tribal moss]

I'm not sure if it is -- I just find metric spaces more concrete to think of.

[hidden haven]

Understandable

[tribal moss]

Intuitively I would expect the function that takes the coefficients of a quintic to its multiset of roots to be continuous under this metric.

[hidden haven]

Yeah fair, I'm thinking the same about the one I mentioned

[hidden haven]

I went through the proof again and figured out that we need one more condition on the multivalued function to make the proof work

If the multifunction is F: C → C with n values, then we can construct the following topological space Z: Its elements are pairs (x, y) where y is an image of x. Give this the subspace topology from the product topology. If the fundamental group of Z is abelian then we cannot build a continuous function involving only standard arithmetic functions and F that solve the degree n polynomial

I think the Z I have constructed is the Riemann surface associated to a multivalued function? I am not sure idk much about Riemann surfaces

The condition makes sense because the proof proceeds by looking at paths given by commutators of permutations of roots

And by saying that the fundamental group is abelian we are saying exactly that those act trivially on F(a function in the coefficients) but non trivially on the roots

[chilly radish]

You forgot ruffini, rude

[glossy widget]

I apologize 😢 btw was Ruffini's proof actually valid? I heard that it was incomplete or something

[chilly radish]

Wiki says he completed his proof in 1813

[glossy widget]

looks like a cool proof, sadly I still need like a year or so to understand it, I don't have enough algebraic topology knowledge (not enough algebra and not enough topology knowledge actually lol)

[hidden haven]

dw I didn't really write a proof, just the general condition that makes the proof in the video work

Your question was still more general, so it doesn't really answer it

[glossy widget]

hmm I see, thanks for the effort anyways

[narrow marsh]

Does (a+b,a) mean the standard inner product in euclidian space or the killing form K(a+b,a)

the book defined both the killing form and inner product to have the notation (.,.) so im confused

[narrow marsh]

humphreys?

[agile burrow]

yeah :(

[narrow marsh]

do you know how to solve my problem 👀

[agile burrow]

haven't gotten to root systems yet lol

[narrow marsh]

:((

oh i think its inner product im being dumb

[sweet echo]

Here $\alpha$ should be acting on $V := V(1) \times V(2)$ and then its action on an element of V should yield another element of $V$. But here $V(\alpha)(v_1) \in V(2)$ and not in $V$. Are they viewing $V(2)$ as a subspace of $V$ and if so should the action of $\alpha$ on $(v_1,v_2)$ technically be $(0,V(\alpha)(v_1))$?

[cloud walrus]

ΣAC

[sweet echo]

ah the answer was yes I should have just read on

[tough raven]

Yes there's some theorem about this

which I think has come up in this server before 🤔

Continuity of roots

[delicate bloom]

if you take f(x)=1/(1-x) and iterate f(f(f(...))), it doesn't converge to anything because f(f(f(x)))=x. It's somewhat similar to defining x=1-1+1-1+1-1+... = 1-x and getting x=1/2. If your original function had not been periodic it's possible there would be a way for it to have 2 attracting fixed points and possibly prove that without too much effort with the banach contraction theorem or something from analysis

[glossy widget]

yeah right this doesn't work, but look at this

[delicate bloom]

sure, that's what I'm trying to tell you, different functions can have attracting fixed points under iteration and the one you picked happened to not

[glossy widget]

I see

[gilded gull]

How would i prove the converse of if gcd(m, n) = 1, $C_{mn} = C_{m} \times C_{n}$

[cloud walrus]

Rishi

[tribal moss]

In general, to get the Riemann surface of a multivalued function, you need to pair each x up not only with a value of the function at x, but with an entire germ (or an entire Taylor series or whatever). Otherwise what you get might not even be a manifold.
Consider for example e^sqrt(z) as a multi-valued function. Its value at z=-pi² is -1 in both branches, but there's nevertheless two distinct points on the Riemann surface corresponding to it.

(And even that description doesn't fully give justice to all the subtleties, because you do want a single point corresponding to f(0)=1).

[hidden haven]

Hmm I see, that's cool. Clerk has been telling me to read this one book on Riemann surfaces for some time, I should get to it at some point

[tribal moss]

Or even just (z-1)sqrt(z), to show that we can have the same phenomenon without a transcendental function in play.

In that case, we can identify the Riemann surface with the variety { (z,t,y) in C³ | t²=z, y=(z-1)t }, but if we try to project away the t coordinate as irrelevant, we get a spurious self-intersection.

[tight flare]

Which book did Clerk recommend? I've also been planning to read some stuff about Riemann surfaces

[hidden haven]

Otto Forster, Lectures on Riemann surfaces

I read a little bit and it seemed quite good, and clerk is a big fanboy

[tight flare]

Oh even a german one, very good

Maybe it'd make more sense to read it in english anyway tho

[south patrol]

Deutsche Sprache, beste Sprache

[tight flare]

Indeed

[tribal moss]

The assumption that the fundamental group must be abelian sounds like it makes sense in the context of making the proof work -- and radicals satisfy it, so we do get the Abel-Ruffini theorem as a corollary. But if we're contemplating making use of more base functions than radicals, then it doesn't feel like a particularly generous condition. For example, sqrt(z³-z) doesn't satisfy it (despite being a composition of functions that do); its Riemann surface is homeomorphic to a torus minus one point.

[hidden haven]

Yeah it can probably be weakened, I didn't want to spend any more time on this today

[sour plume]

holy SHIT i always thought it relied on expressing the formula via roots and polynomials, that's insane

how did i go through ten years of mathematics with this misunderstanding

[chilly radish]

Well that's what abel-ruffini proves

Well via repeated roots

And all other field operations

This is what it means to be an element of a radical extension

Am I misunderstanding what your misunderstanding was

[south patrol]

Is the understanding of the category of k-linear reps of a given group G as just the cat of functors from the delooping BG (i.e. the corresponding groupoid w one object) into Vect_k (and more general things besides) often a useful pov? I find it pretty and I guess you can use it to motivate certain constructions (like taking a functor Vect_k -> Vect_k (say a Schur functor?) to produce new reps of the group) but i'm not sure if it's more useful in itself

Eh maybe the way to create new constructions gives a cool perspective

[south patrol]

"See star" algebra

[delicate orchid]

banned

[south patrol]

At least that message was more on topic

[round portal]

one day I wanna be just like Wew Lads Tbh

[gilded gull]

If G is a subgroup of $S_n$ and G contains an odd permutation, prove that
G contains a normal subgroup of index 2.

Could I get help with this

[cloud walrus]

Rishi

[gilded gull]

I think I'm supposed to use the alternating group

I would consider the index of An in G but I don't know how I would show that An is normal in G

[winter token]

If $G$ contains an odd permutation it cannot be the alternating group. the alternating group only contains even permutations

[cloud walrus]

riley

[gilded gull]

No as in a way of showing the index is 2

So like there's two cosets 1 with the evens and 1 with the odds

but idk if that approach works or if it does how to do it

Like with $G:G \cap A_{n}$

[cloud walrus]

Rishi

[agile burrow]

try looking at the sign homomorphism on G

[gilded gull]

I see

I think I got it

thanks

Yep I just finished writing it out

this actually makes so much sense ty again

[past temple]

not a math question per se but

Im considering taking a grad class on algebraic number theory next semester

and Ive taken galois theory and commutative algebra

but ive never taken an undergrad number theory class

i pretty much dont know any number theory beyond like

the euclidean algorithm for gcd

is taking this class a good idea

[agile burrow]

I think you'll be fine. I feel like half the stuff covered in elementary number theory courses can be learned from a group theory course, which I'd imagine you've seen if you've done Galois stuff

[chilly radish]

I'm so sorry you feel that way

[chilly radish]

You will most likely not need anything learned in an elementary number theory class

There's really barely if any crossover

I don't think the most we did was use CRT for rings when doing ramification

[mint gulch]

Hi, I'm thinking the same as you. I found helpful the preliminary reading for the course of Algebraic Number Theory from Oxford. Supposedly it includes all that you need to see the course. It can be obtained from here for free in Course Materials

https://courses-archive.maths.ox.ac.uk/node/48862

[oblique cobalt]

Am i missing something or is it not really needed for A to be faithfully projective? Every algebra over a field is faithful and projective right?

Oh, and k is a field here

[chilly radish]

Ig this is just one way to argue that innit

Wakt nvm

I misread

Hmm

This is a weird argument

[oblique cobalt]

Right

So its just weird phrasing here on their end

[warm wyvern]

my condolences

[celest cairn]

What’s the difference between Zn and nZ

[chilly ocean]

i've never seen the first notation before. where did you see it? the second one typically means "integers modulo n"

[chilly radish]

Z is a left n module

[warm wyvern]

artin says Z/nZ and Z/Zn are the same

maybe that's related?

[celest cairn]

Oh ok.

[chilly radish]

That cuz Z is commutative

[chilly ocean]

answer the question

[chilly radish]

So nZ=Zn

[chilly ocean]

where did u see this notation

[delicate bloom]

I bet it's really written as $n\bZ$ and $\bZ_n$

[cloud walrus]

Merosity

[chilly ocean]

maybe

[celest cairn]

@chilly ocean
I belive I was looking at tensor products but I can’t remember.

[cloud walrus]

DarQ

[warm wyvern]

does this work?

[south patrol]

You seem to have done the sum, not the union?

And if it were about sums it's be false for infinite dim spaces in particular

[warm wyvern]

oh yeah

that last one should be $W_1\cup W_2\cup\dots\cup W_n$

[cloud walrus]

DarQ

[warm wyvern]

wait

hmm

yeah

[cloud walrus]

DarQ

DarQ

[undone ledge]

i think the entire proof is not correct

[warm wyvern]

why is that?

[undone ledge]

i don't think the second paragraph says anything

could be mistaken

[warm wyvern]

suppose there exists a subspace for which all of its vectors can be found in another subspace

we can then delete that subspace coz it contributes nothing

we can continue this process until at least one vector is left

[undone ledge]

what is v_i?

[warm wyvern]

v_i is a vector in B

each subspace W_i has a vector v_i that's in B which is not in its span

because otherwise B is a basis for W_i

suppose $W={W_1, W_2,\dots, W_n}$ is a finite set of proper subspaces of $V$. and let $\bold{B}$ be a basis for $W_1+W_2+\dots+W_n$

we can assume that for each subspace $W_i$ in $W$ there is a vector $w_i$ that doesn't exist in any other subspace sine otherwise we can safely delete that subspace and this process ends because the set is finite.

we can also assume that $\bold{B}$ isn't a basis for any subspace $W_i$ in $W$ because that would mean $W_1\cup W_2\cup\dots\cup W_n=W_i\neq V$ since all subspaces in $W$ are subsubpsaces of $W_i$, that means for all subspaces $W_i$ in $W$ there is a vector $v_i$ in $\bold{B}$ that's not in the span of $W_i$

since $F$ is infinite, the vectors $w_i+v_i, w_i+2v_i,\dots, w_i+(n+1)v_i$ are all unique and if two of those vectors are in a single subspace then both $w_i$ and $v_i$ are in that subspace which is impossible and since there are more vectors then subspaces at least on of those vectors isn't in $W_1\cup W_2\cup\dots\cup W_n$

[undone ledge]

hmmmm

[warm wyvern]

here is the proof with the typos corrected

[undone ledge]

where did you use infinite field?

ah

[cloud walrus]

DarQ

[undone ledge]

isn't that infinite characteristic?

[undone ledge]

or characteristic 0, yea

[warm wyvern]

no, that's not what you should focus on lol

if the order is infinite

wait, I might be misreading

[undone ledge]

i think it's implied to say "If the order of 1 is infinite"

[thorny knoll]

Finite fields have prime characteristic. Infinite fields can have characteristic 0 or prime characteristic.

[warm wyvern]

welp

we can always choose (n+1) elements of F instead

[undone ledge]

oh true

nice

second paragraph can be shortened to "since every W_i is proper, for every W_i there exists an element v_i in V not in W_i"

wait

[warm wyvern]

that doesn't guarantee v isn't in any other subspace in W tho

[undone ledge]

if you can guarantee v isn't in W, you're done

don't need third paragraph

[warm wyvern]

but v is still in W_i

which means it's still in $W_1\cup W_2\cup\dots\cup W_n$

[cloud walrus]

DarQ

[undone ledge]

for all subspaces, there exists a vector in B that's not in span of W_i

do you mean span of all W_i, or an individual W_i?

[warm wyvern]

individual W_i

[undone ledge]

actually why do you need B

isn't it just v_i in V?

[undone ledge]

i don't get what this is saying

you chose v to not be in W_i

[warm wyvern]

that's true actually

[undone ledge]

i mean span of W_i is just W_i, right

[warm wyvern]

yes

[undone ledge]

right, so don't you only need this?

[warm wyvern]

the second paragraph implies "each subspace has a vector that's unique to it"

as in

you can't find it in any other subspace in W

since W_i is proper, there exists vectors in V that aren't in W_i ofc

[undone ledge]

ok i understand

it's a huge communication thing

i meant third paragraph lmao

[warm wyvern]

yeah, well

I'm still inexperienced with proofs lol

[undone ledge]

this should've been "but w_i is still in W_i" ig

[warm wyvern]

well, I was going with your definition

[undone ledge]

ack

sorryyy

[warm wyvern]

haha

it's alright

[undone ledge]

i meant the v_i you were referring to in third paragraph lol

[warm wyvern]

yo, @delicate orchid

I solved that exercise (I think)

[delicate orchid]

which exercise lol

[warm wyvern]

this one

[delicate orchid]

wow that is

uhh

convoluted lol

go back and read my hint

[warm wyvern]

lmao

[delicate orchid]

I cannot find my hint

[warm wyvern]

found it

[warm wyvern]

doesn't that only work for finite dimentional vector spaces?

[delicate orchid]

all vector spaces are finite dimensional

and I can't answer that properly because I cannot FIND MY HINT

YOU'VE HIDDEN IT FROM ME

[warm wyvern]

lmao

#discussion message

here it is

[delicate orchid]

I found it

imagine....

[warm wyvern]

imagine wot

[chilly ocean]

finite-dimensional vector spaces 🤤

[warm wyvern]

(please not infinite dimensional vector spaces)

[delicate orchid]

every day that passes I become more of an unironic finitist

[undone ledge]

Suppose ${W_1, \ldots , W_n}$ is a finite set of proper subspaces of $V$. If $W_i \subseteq \bigcup_{j\neq i}W_j$, we may remove $W_i$ without affecting the union; this procedure guarantees the existence of a $W_i$ with a $W_i \ni w \notin \text{any }W_{j \neq i}$. \

Choose an element $V \ni v \notin W_i$, and $n + 1$ distinct $\lambda_j \in \mathbb{F}$. Then $v + \lambda_1w_i$ and $v + \lambda_2w_i$ cannot both go into a $W_{j \neq i}$. Hence by pigeonhole principle there must exist a $v + \lambda_jw_i \notin W_1 \cup \ldots \cup W_n$

[cloud walrus]

_gmz

[undone ledge]

sorry for latex spam

you don't need n + 1 distinct, because handshakes are n(n - 1)/2, but it's the same idea

(just summarized what DarQ said, don't mind me)

[warm wyvern]

that's pretty nice

[delicate orchid]

that's basically what you said DarQ

[chilly ocean]

#groups-rings-fields ?

[warm wyvern]

yeah, but they made it prettyyyy

[delicate orchid]

#real-complex-analysis ig lol

[young hollow]

Im not sure why 1.3 is needed. Can’t we just use some point r such that f(r) /= r snd a single fixed point, a?

[chilly ocean]

?

[young hollow]

Are you saying this is the wrong place?

[chilly ocean]

i am. what you posted is not really algebra, but basic analysis

[warm wyvern]

algebra doesn't study distances afaik

[young hollow]

Oh, sure then. It’s just buildup of motivation for introducing groups but I can post it there too

[delicate orchid]

wait really?

odd direction to go in for motivation

[young hollow]

Yeah this is from a nice book I found that approaches abstract algebra as how the need for it arose

[chilly ocean]

what is the book? interesting

[young hollow]

So it starts with concrete geometry and gets abstract

[sand cradle]

I guess Euclidean geometry to symmetry or smth?

[young hollow]

Exactly

Nice little gem I came across

[prisma thunder]

Anyone know what Spec(C[x, 1/x]) looks like?

[warm wyvern]

that's algebraic topology, i think?

[oblique river]

no

[warm wyvern]

wait, I'm in the wrong channel

[oblique river]

it's not lol

[warm wyvern]

lol

[prisma thunder]

I was trying to write down some prime ideals and I'm like ??? Uhhh is there a better way of understanding this?

[sand cradle]

Go to alg-geo

[prisma thunder]

Ah okay

[oblique river]

it's the open subset of Spec(C[x]) where you've removed the point 0

i.e. you've removed the ideal (x)

[chilly ocean]

algebraic geometry, which can work in this channel

[delicate orchid]

wait isn't C[x, 1/x] just C(x) lol

[oblique river]

no

[delicate orchid]

where does the equality fail

[oblique river]

C(x) is a field but C[x, 1/x] is not

C(x) contains 1/(1-x)

[delicate orchid]

got it

[oblique river]

@prisma thunder in general if you have a ring R and you invert a set of elements S = {r1, ..., rn} -- call that ring R_S -- Spec(R_S) is homeomorphic to an open subset of Spec(R)

namely, the one where you've removed all of the prime ideals which contain one of r1, ..., rn

as a special case, if each of the ideals (r1), ..., (rn) is maximal, then Spec(R_S) is just Spec(R) but you've removed the points corresponding to those n maximal ideals

[prisma thunder]

Ah, so that's why you said removed at the point 0, i.e., removed the ideal (x)

That makes sense now

[oblique river]

great!

[keen sparrow]

Anyone knows if there's an errata for "Algebraic K-theory by Hyman Bass"

google not showing anything

[chilly radish]

Maybe @scarlet estuary ?

[scarlet estuary]

i know nothing about that book unfortunately

[chilly radish]

Rip

[keen sparrow]

https://tenor.com/view/po-dies-dies-of-death-teletubbies-gif-23885910

[prisma shuttle]

guys can someone explain wut these weird bracket thingies mean

is this like a common notation or smth

the same notation is used ehre too:

for context these appear on pages 6 and 8 of this papr

https://arxiv.org/pdf/2111.14573.pdf

[chilly ocean]

at first glance it just seems like weird notation for the ideal generated by those elements

[prisma shuttle]

do u get why in the first pic the relation u^4=u^2 would hold then

why isn't it just u^2=1

[chilly ocean]

the image of x_1 in R doesn't even satisfy this

[prisma shuttle]

i feel like i am missing smth

oh wait x_1^2=0

right

[keen sparrow]

Has anyone seen this notation before? the confusion is in (4)

P is a maximal ideal

and b is an element of a dedekind ring

reading this book is fustrating

[pulsar sorrel]

Im guessing that p divides b?

so px=b for some x or maybe this translates to b in p?

[keen sparrow]

p is a maximal ideal

[prisma shuttle]

guys wut is the def of the orbit of an element in a ring

i've seen it used in group theory for group actions and stuff but ima not too sure wut it means in the context of ring theor

[keen sparrow]

This book 's notation is all over the place. Denoting x with the ideal (x)

[balmy vale]

how does the first condition imply acts freely

[hot lake]

if the action is not free then there is a point p and a nontrivial group element g such that p.g = p. then the first condition can't be true because for any neighbourhood N of p you can pick q1 = q2 = p, gamma = g, and you have g.q1 = q and g <> 1

[balmy vale]

oh that makes sense

made me realize i was thinking of the wrong definition for free action

thank you for the help

[woven obsidian]

Is there some neat way to show that $PGL_2(K)$ acts triply transitive on $P^1(K)$? Without coordinate bashing and case distinctions?

[cloud walrus]

AoiKunie

[chilly ocean]

Are there some videos online that explain how to solve polynomial equations using special functions etc.?

Quintics can't be solved by radicals but that doesn't mean they can't be solved

Or book/article but I'd prefer a video in this case

[hidden haven]

Did you also see that video on the insolubility of the quintic without galois theory

[chilly ocean]

Oh, do they solve them there?

[hidden haven]

Nope

They give a different proof that also talks about continuous functions rather than just algebraic

Continuous functions + radicals

And you can generalize radicals to some well behaved multivalued functions

[fickle brook]

@hidden haven link?

[hidden haven]

What I said there is not entirely correct, scroll down for more discussion on it, but the result in the video is not as general as the one I stated

[chilly ocean]

Huh. I'm pretty sure there should a formula for quintics in terms of elliptic integrals though

I remember someone mentioning it to me once

I'm just looking for solutions to those kinds of problems in an analytic way

Curious

https://math.stackexchange.com/a/541027/476484

I found something that I wanted

[hidden haven]

Interesting

[stuck cosmos]

What exactly is an orthogonal representation of a group on an inner product space?

[delicate orchid]

Id define an orthogonal representation as a homomorphism from G -> O(n, F), so it’s probably just a generalisation of this to a general inner product space

[cloud walrus]

SK2099

SK2099

SK2099

[agile burrow]

It should be the second one. Orthogonal matrices preserve inner products so applying \rho(g^-1) to both arguments preserves the value and yields the expression on the right

[chilly radish]

The second one is also just true by the definition of an orthogonal transformtion

And the fact that rho is a hom

[west heart]

Hiii guys

Why if $A\in GL(2, R)$ with $\vert A\vert$ is finite then $\det(A)=\pm 1$ ?

[woven obsidian]

What do you mean by |A|?

[west heart]

Order ?

[cloud walrus]

Potato

[chilly ocean]

take the determinant of both sides of A^(order) = I

use the fact that you're working in R

[west heart]

Ohhhhh

That's actually simple

[prisma shuttle]

how do u show s(R) = 4, where R is the following ring

the s function is defined as follows, and s(R) is a shorthand for s(R,R)

[chilly radish]

Find a polynomial of degree 3 satisfying this and show it's not.true for any polynomials of degree <3?

You could probably argue quite explicitly

Maybe there's asmarter way idk

[barren oasis]

help

[tight flare]

Im not too sure if this is abstract algebra

[delicate bloom]

since your ring R is characteristic 2 you have x^3=x^4 so you know p(x)=x^3=x^4 works. To show this m=4 is minimal you can use the fact that because p(x) maps all elements of R, then p(x) must map x to x^m, but deg(p) < m so you can't do it.

I might retract that answer tbh lol

[tough raven]

Still confused about this TBH

[tough raven]

Given this

[tough raven]

I think there is a similar result for subgroups of a group of infinite index from which this follows.

[rose fiber]

does it make sense to talk about an irreducible representation in a different field than the groups field?

[tough raven]

What is the group's field?

Typically a group does not come associated with a specific field.

[rose fiber]

maybe I phrased that awkwardly. I was asking someone why if SO(2) is isomorphic to U(1) that the former has a 2d irred rep but the latter doesn't, and my impression was that it had to do with the implied field in both cases.

[chilly radish]

Last part seems....sus

[delicate bloom]

the first part is sus

p(x)=x^3=x^4 doesn't actually imply for any element x in R that we have p(x)=x^4 lol

[chilly radish]

That's also true

I thought you said something you didn't

[dapper nebula]

How can SO2 have a 2d irrep when it is abelian

I guess if you're considering it over the reals then it is true

And it is indeed true that representation theory over different fields are different

[rose fiber]

yeah, my confusions comes from people telling me slightly different things... so does the phrasing in my first question make sense, or how would you rephrase it?

[dapper nebula]

Well the group itself doesn't have a field

But the representation is a vector space over some field so it makes sense to ask about that

[rose fiber]

I see. Do you know any resources which discuss this (not like book length)

[warm wyvern]

concerning (b)

is there a clever way of proving the existence of a multiplicative inverse?

[delicate bloom]

I think since x^3-2 and cx^2+bx+a are relatively prime so you can do the euclidean algorithm to get polynomials f, g such that f(x)(cx^2+bx+a) + g(x)(x^3-2) = 1

[warm wyvern]

wait

suppose you found f and g

how do they relate to the multiplicative inverse?

[delicate bloom]

oh when you plug in alpha the second term goes away since alpha^3-2=0, leaving f(alpha) as the inverse

[warm wyvern]

oh

[prisma shuttle]

coudl u explain a bit more? I don't really follow your logic here

wut do u mean that "p(x)=x^3=x^4" works?

and how does the ring R having chracteristic 2 help u with the rest of the proof

ima just having a ibt of hard time linking up the details in your expalnaiotn

[delicate bloom]

my next comment was me retracting my anwer haha, it doesn't work

the characteristic 2 bit was just to say x^3=x^4

[chilly ocean]

❤️

[chilly radish]

?

[chilly ocean]

<3 looks like a heart

[chilly radish]

Oh

Lmao

[prisma shuttle]

oh i thought u were talking about smth else lol

k thats fine thx for hleping anyways

[chilly ocean]

It's probably helpful that (ax^3+bx^2+cx+d)^4 = (a+b+c)x^3+d

I think p(X) = (x+1)X^2+xX should work

Note that (ax^3+bx^2+cx+d)^2 = (a+b)x^3+cx^2+d

@prisma shuttle

[prisma thunder]

Since ideals in Dedekind domains have a unique prime ideal decomposition, is there a notion of a generalized Dedekind zeta function whose norm on the ideal class group of such a Dedekind domain is compatible with the original Dedekind zeta function?

I found something called an ideal norm but I’m not sure if this is the direction I should take.

[somber thorn]

Is there always an isomorphism between two groups of the same cardinality?

[chilly ocean]

Z_4 and Z_2 x Z_2

[somber thorn]

ah

good point

thanks

[prisma shuttle]

remeber an isomorphism is a MUCH MUCH stronnger condition

\phi(ab)\phi(a)\phi(b)

and \phi has to be bijective

same cardinality is a consequence of isomorphism

but the otehr way around doesn't always hold

[woven obsidian]

I am having some trouble understanding this

First, what is meant by a cubic ring over another ring (e.g $Z_p$), is it a ring that is also a $Z_p$ algebra, and isomorphic to $Z_p^3$ as a module?

[cloud walrus]

AoiKunie

[dull ginkgo]

I guess given a cubic ring, it's maximal if it cannot be embedded into a strictly larger cubic ring. IE, there is no non-isomorphic ring-monomorphism to another ring {as the image would have to be a subring and therefore it's not maximal}

[chilly radish]

You just repeated what the text said in a more convoluted way, but that's not what aoikunie was confused about

[dull ginkgo]

Not sure what a "Ring over a ring is" unless it's just an extension

[chilly ocean]

I know how to show this, but it’s not very intuitive, can anyone provide a solution?

[terse crystal]

ab(ba)^-1=aba^-1b^-1=abab=(ab)^2=1

[acoustic island]

This should be obvious but I cant' see it, how do you show that $a \cup 1 = 1$ in a Heyting algebra?

[cloud walrus]

Dudu

[acoustic island]

I'm using this definition of a heyting algebra

[fierce verge]

I literally started abstract algebra today. I was given the exercise to come up with the cayley tables for a group of order 4. I was told that I would find 4 tables but 3 of them would be identical and so there would be only 2 unique cayley tables. Would someone please point out which 3 of these are identical and explain how?

[delicate bloom]

might help to look down the diagonal, the 4th table all elements square to the identity, the first 3 tables you have only one element that squares to the identity, while the other two elements square to the same element in the group. If you relabel these, they're functionally the same

since this is your first day, I doubt you've learned what a homomorphism is yet, but they will make it more clear how to relabel

[fierce verge]

i've learned abt homomorphisms in linear algebra.. would you mind pointing out how it comes into play in identifying identical cayley tables?

[fierce verge]

i see the distinguishing factor now. but is this the only one or there are other ways to distinguish functionally identical cayley tables?

[woeful flint]

is there any difference between a quaternion algebra and a division algebra. I need to understand the ramification of division algebras but whenever I look online all I can find are references to quaternion algebras

[chilly radish]

A quaternion Algebra is a specific type of division Algebra

Actually

I'm not sure it always has to be division

But it's always a CSA

Yea no it doesn't have to be a division Algebra. It's either a division Algebra or 2x2 matrices

But not all division Algebra are quaternion Algebras

[delicate bloom]

if they're functionally the same you can find an isomorphism between them, which is a bijective homomorphism, so for starters you can start learning about some properties of homomorphisms and isomorphisms which is probably what you'll be learning about soon enough in your course anyways

[fierce verge]

thanks!

[delicate bloom]

yup yw

[sweet echo]

If I have two modules of the path algebra of a quiver Q, and the representations of Q that correspond to these two modules are the same then are the two modules necessarily isomorphic?

[hidden haven]

Yes, the correspondence you mention is an equivalence of categories between the category of reps of Q and the category of modules of the path algebra of Q

[tough raven]

This is an interesting question actually.

Are there any efficient algorithms to determine if two Cayley tables on {1,…,n} determine isomorphic groups (given that they determine groups)?

There's “try all n! bijections”, which is O(n^2 n!) I think

which is not exactly efficient

[chilly ocean]

Since every group can be represented alternatively as a Cayley graph, and now this ends up being related to the graph isomorphism problem. A lot of analogue problems are in the same complexity class that was created for this problem (GI), apparently including digraphs

So I'm not familiar with the stuff but there's definitely literature on improved algorithms out there related to GI

I don't know if this is a good answer for you though, even though usually the Cayley table is the "legend" for creating a graph I'm not sure if it can be directly related to a corresponding matrix representing the Cayley graph

Probably tho lol idk

[tough raven]

Wouldn't the Cayley graph depend on a choice of generators?

[chilly ocean]

Oh that's cool I didn't know about that

Like they can give you two nonisomorphic digraphs?

[tough raven]

If there are a different number of generators they ought to, right?

[chilly ocean]

Was there any way that you could see that easily? I had to look up an example lol

very cool tho

[tough raven]

Isn't the degree of each vertex equal to the number of generators?

[chilly ocean]

ah ok that makes sense

[tough raven]

Or is this some other Cayley graph

[chilly ocean]

wow turns out that if we restrict our view to Cayley graphs of a given group with the property I naively assumed +relating isomorphic Cayley graphs of the same group via automorphisms you get some cool properties https://d31kydh6n6r5j5.cloudfront.net/uploads/sites/66/2019/04/michel_slides.pdf

completely unrelated to your thing sorry 😓

[tough raven]

What is the property you assumed?

[chilly ocean]

that isomorphic Cayley graphs means isomorphic groups

and vice versa

which we now know is broken in both directions

wait no its not true

the forward direction is actually true

which means what I mentioned earlier could actually help you, but even though it never gives false positives it has lots of false negatives

which is unfortunate lol

[tough raven]

Hmm

You take a graph

with a vertex labelled as 1

and recover the group?

Makes sense

[chilly ocean]

wait what

[tough raven]

Yes nvm that might only work for some kind of acyclic graph or something

How do you define whether a labelled directed graph is a Cayley graph without referring to a group?

[chilly ocean]

I guess it encodes it explicitly via the generators on the labels branching out from 1, so if there are |G| distinct vertices and everything follows the rules of a group then you're set?

[chilly ocean]

also btw my "wait what" was me thinking you were saying something snarky/making fun of me cause I'm really bad at reading tone through text 💀 sorry

don't be thrown into doubt bc of me

[tough raven]

That was because I didn't elaborate, I was just musing out loud.

Also I don't think it's as straightforward as I thought it was then

So actually nvm

[tough raven]

You can't check if it follows the group rules unless you can actually recover the group operation so you can check

which is what I'm not sure how to do now.

[chilly ocean]

you can write every element as products of powers of the generators and just keep looping around the graph can't you

found it https://en.wikipedia.org/wiki/Cayley_graph#Characterization

[tough raven]

Right, it would work for that graph.

But how would you define the product of two arbitrary elements of an arbitrary Cayley graph?

[chilly ocean]

looks like a composition of graph automorphisms works

[tough raven]

👀 I don't understand.

Ah, view element as graph automorphism sending 1 to element?

I'm not sure that's uniquely defined though
If there are multiple path to the element

[chilly ocean]

yeah I haven't seen the proof but the wiki source to Sabidussi's theorem links to this https://www.ams.org/journals/proc/1958-009-05/S0002-9939-1958-0097068-7/

[tough raven]

To what now

[chilly ocean]

that fucking name wtf

why god why

sabiDUSSY???

[tough raven]

Ah

This does require you to already have the group along with the graph though

[chilly ocean]

ah truee

I see the problem now

you don't know where the identity lives

so you can't just find the generators straight away

[tough raven]

No, that's not it.

If it is a Cayley graph of some group

Every vertex is “equivalent”

(technically, for every v1,v2 there is an automorphism that sends v1 to v2, so picking v2 is equivalent to picking v1 then applying that automorphism)

So you can pick any vertex and call it 1

The problem is defining the multiplication of two elements (i.e. vertices)

[chilly ocean]

yeah ok

found something also showing you can't just always look at the automorphism group either, but sometimes that works and those are called graphical representations of a group or GGR

[tough raven]

Ah

That is interesting

[chilly ocean]

https://math.stackexchange.com/questions/1098115/when-is-the-automorphism-group-of-the-cayley-graph-of-g-just-g
"If G is nilpotent and not abelian, then almost all Cayley graphs for G are GGRs (Babai and me). I proved (using some nontrivial group theory) that if G is a p-group with no homomorphism on the the wreath product of Zp by Z, and C is a connection set that is not fixed by any non-identity automorphism of G, then X(G,C) is a GRR. So in this one case we do have a characterization of the connection sets that result in GRRs."
holy crap this dude is a rockstar

I have no clue what they did but sounds cool, and they proved that you can recover the group as the automorphism group of the graph when it's nilpotent and nonabelian

or maybe they mean almost all in an actual sense of measure 0 lol fuck that

either way yeah I guess we can provide a very restricted answer to your original question now, that if you construct Cayley graphs G' and H' from your tables G and H, then Aut(G') iso Aut(H') sometimes when you are lucky but it's mostly cringe :(

that's not even taking efficiency into account

this all sounds terrible xD

[tough raven]

I think this has probably wandered very far away from being relevant to that question.

[chilly ocean]

LOL yeah

when you have the tables you have all the info

so for sure

[tough raven]

You could probably compute a bunch of stuff from the multiplication table and use that they have to be preserved to narrow down the search.

Like orders of elements or conjugacy classes

[chilly ocean]

yeah I wonder if you could wrangle presentations out of the tables via looking for power properties like you find in group presentations

[tough raven]

But I have no clue what the efficiency becomes or what would be fastest or what else you could do.

[torn niche]

is there an elegant proof that $I+J=(1)$ then $IJ= I \cap J$?

[cloud walrus]

Lizard

[torn niche]

I,J are ideals

[tough raven]

The way I know is to pick i in I, j in J st I j = 1, then write any k in the intersection as ik + kj.

[torn niche]

yeah that's what I came up with

like anything involving isomorphism theorems?

[tough raven]

IDK

Pretty sure you prove CRT using this
If you mean that kind of isomorphism theorem

[torn niche]

yeah

I remember seeing one in D&F but I forgot

[tough raven]

OK

By something similar to Lattice Isomorphism Theorem, I + J = R iff π(J) = R/I, where π : R -> R/I is the quotient map.

I think image under π preserves product of ideals

[torn niche]

it's a hom so yes

[tough raven]

Oh that doesn't quite help; nvm

[warm wyvern]

are quotients of vector spaces a thing a thing?

[chilly ocean]

yes, you can quotient by subspaces

[warm wyvern]

should I think of them the same way I think about group quotients?

[delicate orchid]

they're a tad boring though lol

yeah it's identical to group quotients

[warm wyvern]

I see

that makes sense

[chilly ocean]

their additive structure is the exact same as the quotient group structure coming from the additive structures of your vector space and its subspace

then you define multiplication in the obvious way

[delicate orchid]

they're super boring though cause all vs of the same dimension are iso to each other

do not sully me, buster

quotient MODULES are where it gets good

[chilly ocean]

so the entirety of finite-dimensional linear algebra is boring, by that reasoning

[delicate orchid]

agreed

[tight flare]

LOL

[chilly ocean]

i disagree

[warm wyvern]

lmfao

[chilly ocean]

[tough raven]

Just stop assuming Zorn's Lemma so they become interesting then. 😌

[delicate orchid]

I study representations of finite groups of course I don't actually think that

I just feel like quotients of them are a bit shallower than quotients of other structures

[oblique river]

often times vector spaces don't show up naked though, they have some kind of interpretation which they carry along with them, which makes them more than just their dimension

for example, let V be a 2D vector space with basis {x,y}. elements of V are of the form ax + by, i.e. homogeneous linear polynomials. we can carry this interpretation through other vector space operations

the spaces Sym^2(V) and V + F (F is the ground field) are both 3-dimensional and hence abstractly isomorphic. however, you shouldn't think of them as being "the same" because where they differ is the interpretation

Sym^2(V) has a basis {x^2, xy, y^2} and so elements look like ax^2 + bxy + cy^2, i.e. homogeneous quadratic polynomials

and V + F has basis {x, y, 1} and elements look like ax + by + c, i.e. non-homogeneous linear polynomials

so despite being abstractly isomorphic, they really don't represent the same thing at all, and the abstract isomorphism between them doesn't really tell you much about them

[oblique river]

in response to this

[chilly ocean]

they better not show up naked!

Good answer though

[oblique river]

naked vector spaces

[void cosmos]

horny

[chilly ocean]

many of us are

[tight flare]

[tame fractal]

Least horny math enjoyer

[keen sparrow]

lmaoo

watch mathematician naming something Horn then say an object is horny if and only if blah blah

[chilly radish]

We already have horns

In simplicial categories or smth

[spice whale]

are there any interesting properties of Aut(C)

[chilly ocean]

what's C

[devout crow]

if by C you mean the complex numbers, do you mean as a Riemann surface or as a field (or as something else)?

[spice whale]

complex numbers

as a field

[oblique river]

That’s a pretty nasty group

At least, assuming the axiom of choice

[chilly ocean]

if this were complex analysis, "Aut(C)" would be super duper nice: {az + b : a \neq 0}

[oblique river]

Basically like, pick any two transcendental complex numbers and there’s an automorphism which takes one to the other

[spice whale]

yeah i know

is there structure to the madness though

[chilly ocean]

#groups-rings-fields

[devout crow]

true but either way they are asking about properties of a group

[oblique river]

Not that I know of. Maybe you can topologize it in some nice way?

[chilly ocean]

not #groups-rings-fields

[fallen smelt]

I'm not sure where htis goes 🤔

[chilly ocean]

generic help channel, or go ask somewhere else where you should put it. not here

[fallen smelt]

sorry

[chilly ocean]

"what is the automorphism group of this manifold" is a purely geometric question, just asked in algebraic terms

[spice whale]

I've heard it's profinite and that all of the finite subgroups are order 1 or 2

but otherwise I don't know much abt it

[barren sierra]

Dumb basic question

But the generators of (Z/Zn)* are the numbers coprime to n right?

nvm it's the primative roots

[chilly radish]

The elements of that group are all numbers coprime to n

[barren sierra]

Yea

[smoky cypress]

What does a semisimple algebra mean

[smoky cypress]

For context the book claims that if K is a field with characteristic 0, G a finite group, then the group algebra K[G] is semisimple

And moreover if $\rho_i:G\to GL(W_i)$ are the irreducible representations of a finite group $G$ then $\rho_i$ extends to a homomorphism $\widetilde{\rho_i}:\bC[G]\to\text{End}\bC(W_i)$, and then the map $\bC[G]\to\prod_i\text{End}\bC(W_i)$ defined by the $\widetilde{\rho_i}$ is an isomorphism, but the proof just says ``this is a general property of semisimple algebra"

[cloud walrus]

Whoever

[smoky cypress]

It just references to Lang but Lang just has a whole section dedicated to semisimple stuff

And another reference is to Bourbaki Algebra chapter 8 which isn't translated

[wispy urchin]

in this context i believe semisimple means is a direct sum of irreducibles

[smoky cypress]

Feels like we're just in a loop

What does irreducible mean for an algebra

[wispy urchin]

mmmm

the definition i know says that it has trivial radical

as in 0 is the only element that acts as 0 in all irreducible reprs of A

[smoky cypress]

So 0 annihilator

Ok

[wispy urchin]

yea for the group algebra id assume

[smoky cypress]

And thus not an algebra for any quotient ring

[wispy urchin]

and the group algebra is semisimple because its repr is the same as a repr of the group G

[smoky cypress]

Huh

This part of the book just feels like black box magic to me right now

[wispy urchin]

its rly shuld give more reasoning beyond like

"lmao its semisimple"

the semisimple definition implies the algebra is iso to a product of matrices that each correspond to an irreducible

[hidden haven]

Yeah this is not easy to prove. You show the surjectivity of this map using Jacobson density and the injectivity using the fact that the kernel of this map into the product is the intersection of the kernels of the coordinate functions, and that intersection also happens to be the Jacobson radical, which you then show is 0

[somber thorn]

if $A,B,C$ are groups where $B \cong C$ do we have $A \times B \cong A \times C$?

[cloud walrus]

sean

[terse crystal]

Yeah

[somber thorn]

any hints here?

the result that they're referring to is

[undone ledge]

they gave the solution below

is there anything you want to clarify?

[somber thorn]

hm?

oh the thing in grey is my response

i haven’t finished the proof bc i don’t know where to proceed

from here

i can’t seem to find a subgroup

that does not contain R^n/2

that also satisfies the three conditions

[somber thorn]

yeah it's whether i'm on the right path or do i need to completely rethink my solution?

[undone ledge]

no, condition two is that hk = kh

H does not necessarily have to commute with itself

likewise for K

idea is to wlog assume T ∈ K, and conclude what other elements must belong to K (the ones that don't commute with T)

using this, you should be able to find n where (i) H is forced to be trivial (ii) K must not be a subgroup, or else closure of K forces H to be trivial

(i'm using T because R_(n/2) might not exist)

[chilly ocean]

What do you mean?

[undone ledge]

for two elements in H, say h₁ h₂ ∈ H, it is not true that h₁h₂ = h₂h₁

or K

basically they don't have to be abelian subgroups, because this would imply G is as well

and D_(n > 2) is clearly not abelian

[chilly ocean]

I've asked because by groups commuting people usually mean HK = KH

And not commuting for each individual member

Which is stronger

[undone ledge]

yes yes

normality

here is direct product though

[lethal dune]

just wondering if it is necessary for R to have 1 for the converse to hold

[lethal dune]

Here, if a,b are allowed to arbitrary (not a regular sequence for now), how 0 -> R -> R \oplus R is exact? like if a,b both are ZD then we may have cases when R-> R\oplus R is not injective?

[hidden haven]

Where is here? Did you post an image that isn't loading for me?

I assume R → R ⊕ R is an inclusion as one of the factors, r ↦ (r, 0)

That will always be injective

oh the image loaded

Yeah what you're saying seems right

[lethal dune]

alright

[warm wyvern]

what're linear factors?

[chilly ocean]

(t - c)

[warm wyvern]

ah

[cloud walrus]

Anton.

[south patrol]

Well F8 has a multiplicative group of order 7 so any element besides 1 will generate it

[keen sparrow]

anyone knows good sources for learning about witt rings?

[dull ginkgo]

Time for some absolutely insane shit

Let (S,+,<) be a commutative monoid with a partial order such that

1. • is idempotent
2. x ≤ x + y and y ≤ x +y
3. Every subposet of S has a supremal element
4. The identity is the minimal element

By a weaker form of Zorn's Lemma, there necessarily exists a maximal element M, of this commutative monoid.

Does there necessarily always exist a homomorphism from this complete-poset-monoid structure to the power set with inclusion and unions of some set X that maps the maximal element to X, and the identity to the empty set?

homomorphism aka
x < y => f(x) < f(y)
f(x +y) = f(x) + f(y)

I mean, maybe to P(R)

[grand cliff]

Is there a name for the area of algebra that studies the structures created by:
Getting a matrix M[R],
Calculating the characteristic polynomial p(x) of M,
Quotient R[x]/<p(x)>

[lethal dune]

prolly presentation theory

[grand cliff]

hm okay

was thinking that you could study graphs using this method

like

get the characteristic polynomial of some graph

then study the ring formed by quotienting by that polynomial

would be something kind of like algebraic geometry

[chilly ocean]

it would probably not be algebraic geometry

[oblique river]

is it not just linear algebra?

like, given any polynomial, there is a matrix with that as its char poly

[grand cliff]

I do not remember ever taking quotients in my linear algebra class

[oblique river]

R[x]/p(x) is in some sense the "universal" vector space with an M action

not in a technical sense

[grand cliff]

don't worry, i don't know what universal means in a technical sense anyway

(good thing)

I mean like

[oblique river]

what i'm saying is that this isnt really special to matrices

since you're really just studying quotients of R[x]

[grand cliff]

But you can say that about so much stuff

like

in algebraic geometry (for example).
You take polynomials such as x^2 + y^2 - 1 and quotient R[x,y] to discover stuff

I actually don't know any algebraic geometry at all, but i think this is the gist

[oblique river]

"discover stuff"?

also, tha'ts a 2 variable poly

not a 1 vareibale poly

[grand cliff]

True, does that cause a significant difference?

[oblique river]

slightly

R[x]/p is finite dimensional

[grand cliff]

I guess this idea came up because I was reading about algebraic graph theory on wikipedia and saw that a "common" thing to do is find polynomials that somehow correspond to the graph and then you can do things with those I suppose.

and characteristic polynomial of the adjacency matrix of a graph is one of these things

[oblique river]

i thought that once before too

i think there are things to say

but not much

also you want ot consider over Q

not over R

[grand cliff]

why's that

they're both fields

[oblique river]

R has only one field extension

and Q has a lot

basically matrices over R are not as interesitng as matricse over Q

because more matrices over R are conjugate to each other

[next obsidian]

You forgot about R/R :^)

[oblique river]

wait what's the other one???

;p

[tough raven]

This sounds like a complete lattice.
OTOH, I think there is a duality theorem (Priestly? Stone?) that distributive lattices can always be embedded into a “set-theoretic operations” lattice as you have mentioned. This is clearly necessary.
Distributive lattices need not be complete, but also (I'm pretty sure) complete lattices need not be distributive (consider lattices of submodules/normal subgroups), so this is not necessarily possible.
OTOH, if you just want a homomorphism, that's easy to do at least trivially: map everything to the empty set.

Oh your definition of homomorphism says x<y => f(x)<f(y), implying injectivity (in a lattice).

[tough raven]

Is this the same as a complete lattice though? 🤔

[chilly ocean]

Isn't that Boolean lattices

Can be realized as fields of sets

[tough raven]

No, it's true more generally for distributive lattices

Assuming by field of sets you mean closed under finite unions and intersections.

If it has to be closed under complements then you're probably right. But that's not a necessary lattice operation.

[chilly ocean]

I know I'm right about Boolean lattices

But didn't know it was more general than that

[tough raven]

I think it gets arbitrarily general once you allow topological spaces in the sense of pointless topology

[chilly ocean]

Maximal element isn't the same as maximum

There could be multiple maximal elements

[chilly ocean]

Yeah, I think by homomorphism he meant embedding

[dull ginkgo]

hm. good point

[dull ginkgo]

If a ring is a domain when the zero ideal is prime
and a quotient ring is a division ring when the divisor is a prime ideal
and the quotient of any ring by it's zero ideal is isomorphic to itself
that means every domain is a division ring. WTF is wrong here

oh. "and a quotient ring is a division ring when the divisor is a prime ideal" is wrong

only if it's maximal

[chilly radish]

What does 'the quotient ring is a division ring when the divisor is a prime ideal' mean. If you mean quotienting by a prime ideal, this is false, it's a domain, not a division ring

Oh okay you figured it out

[tiny jolt]

How would I go about getting a subgroup that has index 4 in the free group on 3 elements?

[vast quiver]

do you happen to know some covering space stuff? There's a theorem that says that the degree of a covering equals the index of the subgroup corresponding to the covering. For your problem in particular, you could take the wedge of 3 circles (which has fundamental group=free group on 3 elements) and a degree 4 cover $p: \tilde{X}\to X$. Then, $p_*(\pi_1(\tilde{X}))$ is an index 4 subgroup of $\pi_1(X)$

[cloud walrus]

Joseph

[lethal dune]

how do you get a subgroup of index for in Z?

[wooden ember]

this is definitely overkill. It's far easier to generate a normal subgroup of the free group that you quotient by to get a group of order 4. Ie you just want to find relations on three elements to get a group of order 4, so either Z/4Z or K_4