groups-rings-fields

[viscid pewter]

your wording could be clearer but yeah i think that works

[crystal vale]

Yes

[barren sierra]

yea I presume if this is for HW you'll write up up clearer not too worried about that here

[viscid pewter]

so that's like the group G^X with the operation induced by the operation of G

fun

[crystal vale]

No this is not HW

[crystal vale]

Yes

[lime junco]

question,

given a field extension over k, k(a)

let a be algebraic

we know that like

k(a) is isomorphic to k[x]/(f(x)) (f minimal poly of a)

and we know the isomorphism

i dont understand why it implies this

wait ohhhh

it just clicked im dumb

one of those - properly phrase the question then it hits moments

[chilly ocean]

🦆 moment

[vivid tiger]

🦆

is my lord

[crystal vale]

For Vector space V we take V as a group under + operation so this + operation is notation? It can be multiplication, composition? Right?

[chilly ocean]

It can be anything as long as it satisfies the vector space axioms

In wikipedia it says a Dedeking domain is an integral domain in which every nonzero proper ideal factors into a product of prime ideals.

Why do they say "nonzero"?

Doesn't (0) factor as (0) which is prime in an integral domain?

[coral spindle]

You don't have uniqueness if you permit the zero ideal

[chilly ocean]

I know

[crystal vale]

Thank you

[coral spindle]

See the very next sentence:

It can be shown that such a factorization is then necessarily unique up to the order of the factors.
So we need the ideals to be nonzero :)

[chilly ocean]

Is there a generalization of Dedekind domains to non integral domains? So a commutative ring where every ideal factors (not necessarily uniquely) as a product of prime ideals?

[charred iris]

a relatively common example is (0, \infty) with multiplication as the 'addition', and exponentiation as the scalar 'multiplication', which forms a real vector space

where by 'common' I actually mean it is the only one I can remember ever seeing and not common

[crystal vale]

Yes I saw this therefore I asked, thank you

[pliant forge]

Let A be a free module and P be a projective module. Then if we have a map f: A \to P that is onto, by definition there is a map g: P \to A such that gf = id_P. Now I am asked to show that A is iso to im f \oplus ker g. How can I approach this

To me it seems like ker g is just 0, otherwise how could the composition be the identity?

[delicate orchid]

do you know about short exact sequences

[pliant forge]

yes

[delicate orchid]

ok so we have a sequence 0 -> ker f -> A -> P -> 0 with the last -> being split as you correctly identified

[rocky cloak]

That should be ker f (+) im g, since those are the things in A

[pliant forge]

hmm ic

[delicate orchid]

now excuse me as I try and remember how this proof goes

ah yeah

define a map h : ker f (+) im g -> A with h(k, p) = i(k)+g(p) where i is the map ker f -> A
h is injective as h(k, p) = 0 => i(k)+g(p) = 0 => f(i(k))+f(g(p)) = 0 => f(i(k)) + p = 0 => p = 0 (as img i = ker f). So we have to also have that i(k) = 0 => k = 0 by injectivity of i.
I've forgotten how surjectivity goes

damn it

[delicate orchid]

jagr if you could fill in the surjectivity proof I'd much apprechiate it. I have to go to a seminar now lol

[languid trellis]

Nerd

[pliant forge]

slightly confusing. So Im g lies in A, and then you are sending elements in A with g, but g's domain is P

[rocky cloak]

For x in A

f(x - gf(x)) = f(x) - fgf(x) = f(x) - f(x) = 0

So x - gf(x) is in ker f, and gf(x) is in the image of g hence
x = (x - gf(x)) + gf(x)

[pliant forge]

alright i see. also found this nice

[rocky cloak]

Splitting lemma my beloved

[delicate orchid]

Yeah we were just proving 2 => 3

[south patrol]

I guess that proves 1 => 3 by dualising right

Well okay

Not quite

If we prove it in a general abelian category then yes

[delicate orchid]

you're so stinky it's unreal

[south patrol]

Why

:*(

Cat brained

[delicate orchid]

and I presume you mean 1 => 3? I don't see 1 => 2

[south patrol]

Yes typo sorry

[delicate orchid]

fair play

[tough raven]

Do directed unions and directed intersections of ideals of a ring commute?

More precisely, suppose $D_1, D_2$ are directed sets and $I_{s,t}$ is an ideal of a fixed ring $R$ for every $s \in D_1, t \in D_2$ such that $I_{s_1,t} \subseteq I_{s_2,t}$ whenever $s_1 \geq s_2$, while $I_{s,t_1} \subseteq I_{s,t_2}$ whenever $t_1 \leq t_2$.

[cloud walrus]

Raghuram

[south patrol]

I guess these are like codirected intersections

[delicate orchid]

are "directed unions" a weird name for colimits

[south patrol]

Directed colimits

[delicate orchid]

so colimits but over (N, \leq) got it

[tough raven]

For subobjects of algebraic structures, this is the usual “not category theory” terminology .

[south patrol]

This should be false

[tough raven]

Then we have the ideals
$$ I = \bigcup_{t \in D_2} \bigcap_{s \in D_1} I_{s,t},\qquad J = \bigcap_{s \in D_1} \bigcup_{t \in D_2} I_{s,t}. $$

[cloud walrus]

Raghuram

[tough raven]

Pure set theory shows that $I \subseteq J$ (in fact, this is true without any directedness assumption, replacing $\bigcup$ with $\sum$).
My question is about the converse.

[cloud walrus]

Raghuram

[south patrol]

but tring to come up with an explicit counterexample lol

[tough raven]

Hmmm.

[south patrol]

if you take I_{a,b} = 2^{a-b} Z

Then uh

$\bigcup_b \bigcap_a I_{a,b} = \bigcup_b 0 = 0$

but

[cloud walrus]

Süßkartoffel

[south patrol]

$\bigcap_a \bigcup_b I_{a,b} = \bigcap_a \mathbb Z = \mathbb Z$

[cloud walrus]

Süßkartoffel

[south patrol]

I think this is a valid example

[tough raven]

IG even though we have gotten to unions rather than sums, x in I requires a single t to work for all s, whereas x in J allows t to depend on s.

[south patrol]

Oh and for mine

[tough raven]

Oh exactly what you wrote.

Sad.

[south patrol]

use the convention that 2^{n} Z = Z for n <= 0

lol

But then it is fine

[tough raven]

Yes, realised.

[south patrol]

Cool

[tough raven]

Hmm.

[south patrol]

But yeah this is like

[tough raven]

Sad.

[south patrol]

finite limits commute with filtered colimits (in particular directed ones)

But when you get to the infinite case on both sides, it breaks down

[tough raven]

Oh directed commutes with finite.

Well, a finite (co)directed thing is pointless.

[tough raven]

Does it hold if $D_1$ is the set of finite subsets of some set $X$ and $I_{{x_1, \dots, x_n}, t} = I_{{x_1}, t} \dotsm I_{{x_n}, t}$ (i.e., the directed intersection is of finite products of a family of ideals for each $t \in D_2$)?

[cloud walrus]

Raghuram

[tough raven]

(Probably not, but let me try to break this too.)

Ooh, this might actually work (for Z?).

[tough raven]

Let's say X = N and I_{s, {t}} is independent of t; so that the inner intersections are \cap_n I_s^n.

Then, in a Noetherian integral domain, the inner intersections are R if I_s = R and 0 otherwise (Krull intersection theorem IIRC).
Thus I = R if some I_s = R and 0 if all I_s are proper.

OTOH, the inner union \cup_s I_s^n is equal to (\cup_s I_s)^n (using directedness). So again in the intersection, we get R if some I_s = R and 0 otherwise.

[tough raven]

Oh, we can still achieve this if we let I_{{m}, b} = Z for m <= b and 2Z otherwise (X = D_2 = non-negative integers).

Again inner intersections are 0 but inner unions are Z.

So that's quite thoroughly false IG.

[hollow mica]

Sorry

Lol

[delicate orchid]

R/I isn't an ideal of R so how we doing this quoti- and there they go

[hollow mica]

SORRY

[delicate orchid]

apology.... accepted.... :thecosmoshumswithatunemostsweet:

[hollow mica]

If R is a ring, I, J ideals with J containing I, is the image of J under the quotient map R -> R/I isomorphic to J/I ?

[delicate orchid]

should be

try and construct the obvious map here and then show it's an isomorphism

[coral spindle]

I wish you'd use tex instead of monospace 😭 it's so much harder to read

It's not even hard in this case, you'd just write $ whenever you're currently writing `, and the only change is \to instead of ->

[rocky cloak]

I find the monospace very readable actually

[coral spindle]

in other cases

[hollow mica]

I'll meet you halfway and use half tex half monospace

[molten rivet]

I'm stuck on this problem, I just cannot figure out an approach. It seems like you just need to do the obvious thing and map the simple tensors to the simple tensors, but I don't know if that's right or if I am completely misunderstanding something.

[coral spindle]

Have you tried it?

[molten rivet]

Yes but I'm not sure how to show that doing that actually makes a homomorphism, am I just overthinking it?

I mean it seems obvious

[coral spindle]

Are you maybe forgetting that a map out of N (x) M is determined by a bilinear map out of N x M?

You know, the universal property

[molten rivet]

Oh

Ok I might be able to do it now, I'll message again if I need help, thank you!

[coral spindle]

This should be the first thing you think of whenever you want to work with tensor products btw

[molten rivet]

ok

[wooden rain]

$V_i, X \in \mathrm{Obj}(\mathrm{Vect}_\mathbb{K})$. Let's assume I is infinite as otherwise the results are clear to me. Could someone explain the intuition behind the first isomorphism to me? On the LHS we have morphisms where the domain is the external direct sum of vector spaces, so the vectors we take from it are zero at almost every coordinate. On the RHS we have vectors of linear maps that can be non-zero at every coordinate. It's super counter-intuitive to me

[cloud walrus]

Thingoln

[coral spindle]

If I have a bunch of maps all going into the same thing, I can take finite sums of them if I have some finite number of things in the domains of the maps.

That's the intuition of the first isomorphism to me.

Remember, you can't take infinite sums in a typical module!

[rocky cloak]

On the left Vi is a subspace of the direct sum for each i (by just having all other coordinates be 0). So to specify a map you necessarily have to specify it on all the Vi.

And since you can take finite sums, you don't have to specify it more than that.

For example let Vi = V be some fixed space and let X also be V.

Then we can define a map that just sums all the coordinates together. This map is nonzero on each Vi, so corporesponds to an infinite product of nonzero maps in ProdHom(Vi, V)

[wooden rain]

I'm afraid I don't understand this construction. Let's say that $I = V_i = \mathbb{K} = \mathbb{R}$. Let $\varphi \in \prod_{i \in I} \mathrm{Hom}(V_i, X)$ satisfy $\varphi(i) = \mathrm{id}\mathbb{R}$ for all $i \in I$. Which morphism from $\mathrm{Hom}(\bigoplus{i \in I} V_i, X)$ would correspond to it?

[cloud walrus]

Thingoln

[wooden rain]

Oh, would it be simply $V_i \mapsto \mathrm{id}_\mathbb{R}$?

[cloud walrus]

Thingoln

[wooden rain]

Wait, no. That doesn't make sense. I need to map Vi to something from the field

[rocky cloak]

So given functions fi: Vi -> X for all i. What you do is apply fi to each coordinate, then sum everything together.

The reason you are able to sum the coordinates together is that almost all of them are 0.

For example if Vi = X = R as you said and all the maps are identity then

(1, 0, 0, ...) maps to 1
(0, 1, 0, ...) also maps to 1
(0, 0, 1, 0, ...) also maps to 1 and so on.
(1, 1, 0, ...) maps to 1+1 = 2
(1, 0, 2, 0, ...) maps to 1+2 = 3

[wooden rain]

What if I have two morphisms from the RHS: one that is the identity at the first coordinate, the zero map at the second; and another one, that is the zero map at the first coordinate and the identity map at the second. Other than that, those two morphisms are the same. Won't they correspond to the same morphism from the LHS?

I feel like what I've said is stupid lol Because on the LHS we could "specify the first coordinate" by just checking what happens at the argument V_1

[rocky cloak]

No, one will map (1, 0, 0,...) to 0, while the other would map (0, 1, 0, ...) to 0.

[coral spindle]

Consider the following:

• Every field of characteristic p contains F_p as a subfield
• Q is of characteristic 0
• Q = F_0

Thank you all for your time

[rocky cloak]

I'm down with that convention. Now we just need a nice canonical choice of F_(0^2)

[chilly ocean]

What is F_n when n is not a prime?

[wooden rain]

Ring

But I guess then you should use the notation Z/nZ

[rocky cloak]

Let's go with the product of F_p^k where p^k apears in the decomposition of n

[chilly ocean]

Oh I see

thanks

[rocky cloak]

I mean, that's not something someone actually defines, just while we're extending the F_n notation...

[chilly ocean]

Yes but it makes sense to extend it like this, at least in the lecture notes I'm reading

[wooden rain]

I think I'm still confused. Elements from the RHS can have an infinite number of non-zero coordinates because we're taking the Cartesian product of the spaces of morphisms, so we can't sum them

I'm so sorry. I guess it's pretty straightforward and I just can't see it

[rocky cloak]

Your not summing the maps, your summing the values you get when you apply the maps to each coordinate

[coral spindle]

We want F_n to be a field of order n, so it most certainly doesn't make sense to extend it to all n since there is usually no such field.

My joke above isn't consistent with this, it's just a joke after all

[rocky cloak]

Going back to the example, you agree
(1, 0, 0, ...) can map to anything right?
Similarly
(0, 1, 0, ...) You should be able to map wherever
and also (0, 0, 1, ...)

Would it be reasonable that at some point you're forced to map all the remaining basis vectors to 0? Or could you specify somewhere for each to be maped?

[wooden rain]

Those are elements of $\bigoplus_{i \in I} V_i$, right?

[cloud walrus]

Thingoln

[rocky cloak]

Yes

(where Vi = R for all i)

[wooden rain]

I want to say yes because vectors from this space are zero almost everywhere, but at the same time I feel like it's more subtle than I think. For example, each V_i can be of infinite dimension and we need to map every vector from its basis to something

But just by looking at the construction of the direct sum, yes. We take the zero vector from almost all vector spaces we sum

[rocky cloak]

So two things to think about:

If Vi=R=X, is the function where we just sum all the coordinates a linear map?

Is there some i for which this maps (0, 0, ..., 0, 1, 0, ...) (1 in position i) to 0?

If the you answer yes to the first and no to the second, then you see that this is specifies a nonzero map for each Vi

[wooden rain]

1. Yes, it is
2. Not necessarily

Oh, if we assume that specific map, then 2 is no ofc

[rocky cloak]

So is it clear then?

[wooden rain]

The construction yes

Okay, riiiight. All that matters is the values of the map at (0, ..., 1, ..., 0, ...) because if the domain consists of only finite linear combinations of these vectors, then they make for the basis

Well, maps Vi -> X

Or less chaotically. Let $B_i$ be a basis of $V_i$. Then $\prod_{i \in I} (B_i \to X)$ is a basis $\mathrm{Hom}(\bigoplus_{i \in I} V_i, X)$

[cloud walrus]

Thingoln

[wooden rain]

If we sum the coordinates*

If we don't, then that's a basis of the other space of morphisms

Or maybe not, lol. I need to think about it

[hollow mica]

let q: R -> R/I be the quotient map. Then the map f: q(J) -> J/I that sends j + I to j + I the bijection? That feels so weird to write

[delicate orchid]

maybe...

[hollow mica]

Ok yea that guy is obviously an iso

[coral spindle]

It's q restricted to J

N.b., what kind of isomorphism do you mean?

It is equal to J/I, so you don't need to discuss isomorphism in any case

[hollow mica]

LMAO

ok

I've officially lost it

whats a good way to show something isn't a field

[delicate orchid]

y^2 = 0 so you have zero divisors

More generally R/I is a field iff I is a maximal idea

[coral spindle]

By using the definition

[delicate orchid]

Assuming R is commutative and then you just do a one line proof using the correspondence theorem

[coral spindle]

But yeah like, the usual observation is the presence of zero divisors

[hollow mica]

You're saying that y^2 + (y^2, x^3 - 17) = (y^2, x^3 - 17) ?

Ok yeah

[coral spindle]

The important part of that is that y is not zero in the quotient

[hollow mica]

Right, cuz y is in y + (y^2, x^3 - 17) but isn't in (y^2, x^3 - 17)

The actual problem I was solving asked me to determine if (x^3 - 17, y^2) is maximal in C[x, y]/(y^2 + x^3 - 17), and I reduced it to this

[coral spindle]

You reduced then un-reduced it...

This is equivalent to seeing that (y, y^2, x^3 - 17) is a larger nontrivial ideal

[hollow mica]

Ohhh

Yeah I basically did that part out loud

in showing y^2 = 0

[coral spindle]

y^2 is in an ideal generated by, amongst other things, y^2

[hollow mica]

wait is thta the batman chess guy

[coral spindle]

[hollow mica]

yea he looks familiar

pretty sure it's him

[coral spindle]

It's him

[hollow mica]

Trying to think of an element not in (x - 1, y - 4) so that after adding it the ideal stays nontrivial

[coral spindle]

What ring are 1 and 4 from

The ambient ring matters...

[delicate orchid]

They’re from the “set of integers”

[hollow mica]

This is also in C[x, y]/(y^2 + x^3 - 17)

[coral spindle]

Get clowned Wew

they were complex numbers the whole time

[delicate orchid]

Egads… the quotient structure….

[delicate orchid]

Wait is it a comma or a +

[coral spindle]

Uhhhhhhhh

[hollow mica]

• I typoed the first time

[coral spindle]

So you need to redo the previous argument.

I'm gonna sleep so I leave this to you.

[hollow mica]

My previous argument went like this:

• let R = C[x, y]/(y^2 + x^3 - 17).
• (y^2, x^3 - 17) in R is just (y^2, x^3 - 17)/(y^2 + x^3 - 17) since (y^2, x^3 - 17) contains (y^2 + x^3 - 17)
• so R/((y^2, x^3 - 17)/(y^2 + x^3 - 17)) = C[x, y]/(y^2, x^3 - 17)
• then I argued C[x, y]/(y^2, x^3 - 17) is not a field by showing y^2 as nonzero with y being nonzero

[hollow mica]

Ok, thanks for the help today

[eager willow]

Ideal correspondence theorem tell you that it is equivalent to look for whether (y^2 + x^3 - 17, x-1, y-4) (y^2 + x^3 - 17, x+1, y+4), (y^2 + x^3 - 17, x^3 - 17, y^2) are maximal in C[x, y]. Is this ok?

Now argue that each of the quotients are finite dimensional over C

Finite algebras over C can only be fields if they are one dimensional over C, since C is algebraically closed.

[dull ginkgo]

This is a kind of a conversational question, but I’m unsure how it’s supposed to be with exercises and abstract alg

For context I’m going off of Jacobson. Most of the exercises take me a comparatively long time to do. They take around 10-15 minutes each to process sometimes, and don’t come as they should immediately. Is this how it’s supposed to be going off of a textbook? Sometimes I have to look them up and the answers rather obvious

[coral spindle]

That's normal

[dull ginkgo]

I haven’t really “struggled” with math related concepts before, so I’m wondering if it’s just not a good route for me, or it’s a good thing to be challenged

I enjoy it nonetheless

[coral spindle]

15 minutes is not a long time to do an exercise lol

[delicate orchid]

no they're saying it takes 15 minutes to just understand what they're asking you to do

[dull ginkgo]

Sometimes

[coral spindle]

Oooh 👻 math is hard oooh

[delicate orchid]

there's a particular exercise in Fulton Harris which I'm pretty sure I still haven't fully proceed 2 years later

[dull ginkgo]

It takes a bit to process though 90% of the time it’s a difference in terminology in D&F versus Jacobson

Usually I get what it’s trying to say, just can’t immediately think of a route to do it

[coral spindle]

As per my last email,

[dull ginkgo]

Versus analysis where I usually can even for problems ahead of where I’m supposed to be

But I wonder if that’s a good thing, to be challenged in that way

[coral spindle]

No you should never try to do anything you find hard it's bad for your constitution

Do you have any questions about math?

[dull ginkgo]

like I was stuck for a while last night trying to think of a formal proof that finitely generated fields with each element being torsion is finite

Mainly because trying to show every element is some sum over Z like a module

But like, I wonder if that’s a too “high power” approached to it, since I already am a bit ahead since I went off another textbook, not what the exercise intended

I do sometimes wonder how the exercise wants me to approach it without me going nuts and using theorems that we haven’t learned yet

Like decomposition of finite abelian groups into cyclic groups

I think my insecurity is bleeding through

lol

I think I’m too focused on extreme rigor

[delicate orchid]

oh this sounds fun uhhh. If every element is torsion then 1 is torsion so ur field has a prime field F_p inside of it for some p and since it's finitely generated it has to be a finite dimensional vector space over this finite prime field ergo is also finite

[dull ginkgo]

Especially for a book like Jacobson lmao

[delicate orchid]

does this work chat? @coral spindle chat does this work chat?

[dull ginkgo]

I meant group

Fields that works omao

[delicate orchid]

https://tenor.com/view/mario-sad-sad-mario-mario-look-phone-sad-look-sad-gif-17200741509957979964

[coral spindle]

Yeah, every element must be p-torsion, but this doesn't preclude something like F_p(x)

[dull ginkgo]

this still works in the group case lmao

What do you mean by preclude

[delicate orchid]

x isn't torsion in that u stinker

[coral spindle]

Get a dictionary lmao

[delicate orchid]

oh wait torsion like Z-module ok yeah

no it is

[coral spindle]

Um um x + ... + x p times 🤓

[dull ginkgo]

[coral spindle]

Ok google look up "preclude definition" thank u

The issue is that the multiplication doesn't play ball

[dull ginkgo]

he ain’t ballin

[coral spindle]

If every nonzero element is torsion under multiplication I think we have something

[delicate orchid]

I guess the problem here is that I used "additively generated as a F_p (or Z)-module" but ur using algebraically generated

[coral spindle]

Ye

[delicate orchid]

in which case I'm right you're wrong. Smell you later stinky

[dull ginkgo]

The field itself IS finite so obviously it’s multiplicative torsion lmao

[delicate orchid]

F_p(x) isn't finite though

[coral spindle]

F_p(x) is not finite.

[dull ginkgo]

I mean the context of

The field itself

Which means our statement doesn’t include it

[delicate orchid]

https://tenor.com/view/huh-huh-whuh-huh-huh-whuh-crab-scared-crab-gif-23792302

[coral spindle]

:shovel_emoji: :hole_emoji:

[dull ginkgo]

hand hand hand

[delicate orchid]

anyway can someone help me with my group theory homework

[coral spindle]

Every element is torsion under multiplication => 1+1 is 0 or torsion under multiplication => 2^n = 1 for some n

So we have positive characteristic for free

Then uh

Take all the finite generators x_1, ..., x_n

F_p[x_1, ..., x_n] is finite extension

[delicate orchid]

yur

[dull ginkgo]

True

[coral spindle]

So done

Kinda cute

[dull ginkgo]

Last time I did field theory I keep getting derailed trying to prove Artin’s lemma with Noether normalization

[coral spindle]

Yuh bc they're roots of x_1^m - 1 for some m, ezpz

[delicate orchid]

you are nitpicking and biased

[dull ginkgo]

Mainly because I love Noether normalization

[coral spindle]

That's nice

[delicate orchid]

I don't even know what that shit is

sets were only supposed to have ONE operation as GOD INTENDED

[dull ginkgo]

Take a field: Algebraicly finite generation -> Module finite generation

yes module in this case is vector space but I don’t want to type out two words

[coral spindle]

You're such a group theorist I love it

Slay, Wew

Slay

[dull ginkgo]

Wew when automorphism group of a group

[coral spindle]

What

[dull ginkgo]

hur dur group action that works with group hur dur curry rice

[delicate orchid]

if u mean abelian group then yeah that's an upon the wit. (as the kids say)

[coral spindle]

What

[dull ginkgo]

Implied

[delicate orchid]

god and I thought I was a proffesional bullshitter

[coral spindle]

You're being outplayed

[dull ginkgo]

I learn from the best

[languid trellis]

Why do we care about orbits?

[delicate orchid]

conjugacy classes are orbits

so there's one good example

[languid trellis]

Are cosets just a particular form of orbit defined by the left/right translation map

[low wyvern]

atoms, satellites?

[delicate orchid]

there's another

[languid trellis]

Right because jacobson introduces orbits, gives half an example, then says 'wow look they're cosets' and proves some stuff about cosets

Also what is a conjugacy class

[delicate orchid]

the more I hear about this book the less I like it

[languid trellis]

I don't like you

[delicate orchid]

orbit under the action g.h -> ghg^-1

[languid trellis]

Action?

[delicate orchid]

sorry how the ever loving fuck are you discussing orbits without knowing what a group action is

what is this wack ass book

no wonder you feel like they're completely unmotivated

I'll give you an exposition if you want

[languid trellis]

Let S be a set, G a subgroup of sym S. Consider the equivalence relation x = y if there exists an a in G s.t a(x) = y. The equivalence class of x is called the G-orbit of x

[languid trellis]

Do you have a reference or some supplementary reading

[delicate orchid]

right ok so that's how you do it without defining actions ig lol

[delicate orchid]

section 6

I knew these stupid notes would come in handy one day

[coral spindle]

FINAL_FINAL_Final

[languid trellis]

Lol

He does talk about group actions in 4 subchapters time

I think the reason he did it now was just to show look cosets are one of these things

[cursive spindle]

When is FINAL FINAL FINAL FINAL

[delicate orchid]

soon

[cursive spindle]

PhD thesis be like

[languid trellis]

Then I imagine we can apply our discussion of group actions and whatever to costs

[coral spindle]

Groups describe symmetries, but without an action we can't interpret the elements as symmetries. Orbits are one tool for analysing actions, whose properties reflect back on the group.

[languid trellis]

Wait those were notes for a class wew?

[delicate orchid]

wow I was dog shit at writing back in 2022

[delicate orchid]

these were notes in case I ever needed to teach group theory

so they're not really meant to be used for anyone else lol

but the examples at the start of section 6 are the main thing here that's useful

[languid trellis]

Ah okay I was wondering how it was possible to cover like 3 things in 12 weeks of teaching

I'll also read chap 4 I think

Thanks cutie

[delicate orchid]

expect numerous typos and me randomly switching variable names without warning because I am insane

[coral spindle]

Um, did you just call Wew a cutie? He's not a cutie, he's adorable. Get it right

[delicate orchid]

thanks babes

[coral spindle]

Um did you just call me babes? I am an adult not a baby smh

[delicate orchid]

silence

[languid trellis]

Nvm I'm not reading chapter 4 because I hate permutations and all cyclic notation

[delicate orchid]

https://tenor.com/view/baby-gangnam-style-gif-3560468

[delicate orchid]

this is quite frankely the worst opinion anyone has ever had about anything that's ever existed

[coral spindle]

Permutations are very nice indeed and I suggest you give them a second chance

Especially since they're going to provide the canonical examples for group actions

[languid trellis]

I spent 2 hours yesterday trying to read this proof in jacobson that sign isn't changed by composing with a transposition

[coral spindle]

Yes that's the one really tricky proof

[languid trellis]

Then I looked for a different proof on MSE and it made sense immediately

[delicate orchid]

sgn(t) = -1
sgn(tst^-1) = sgn(t)sgn(t^-1)sgn(s) = (-1)^2sgn(s) = sgn(s)
problem??

[coral spindle]

So you're just complaining that Jacobson wrote it bad?

[delicate orchid]

is it?

[languid trellis]

Yes

[coral spindle]

Yeah people usually struggle with it

[languid trellis]

I like complaining in general I'm british

[delicate orchid]

write them as permutation matrices lol

[coral spindle]

Wow imagine being british could not be EVERYONE PRESENT

[delicate orchid]

it's really obvious then cause you can just bootstrap using the determinant

[delicate orchid]

does he know, chat?

[languid trellis]

Ur a brit?

I don't believe it

I actually sont

[delicate orchid]

wait.... ur british?

[coral spindle]

Gonna blow your mind and say that Wew is also 🤯

[delicate orchid]

did I know... chat?

[coral spindle]

Wait you didn't know

Bruh

Wew wtf

[languid trellis]

Admittedly I haven't interacted with you much boytije but you seemed too uptight to be a brit

[coral spindle]

No I'm just a prick

[languid trellis]

No banter to be found in any message of yours

Right fair

[delicate orchid]

irrelevantgardless I have now thought of 2 different proofs :thecosmoshumswithatunemostsweet:

step 1: prove conjugation preserves cycle type
step 2: it's oh so simple....

[coral spindle]

Shrimple*

[delicate orchid]

actually wait can I use something from Humphrey's to nuke this fly

[coral spindle]

There's like, 5 ways to pluck this duck

[delicate orchid]

I'm pretty sure he proves that the sign map is a group homomorphism for all Coexter systems at once in that book

[coral spindle]

Neat

[delicate orchid]

so I can absolutely obliterate this

[languid trellis]

I'm not gonna find this post but I remember the proof being nice

[dull ginkgo]

Why not, helps with group actions and symmetries

[dull ginkgo]

https://tenor.com/view/mgrr-metal-gear-rising-revengeance-armstrong-raiden-handshake-gif-25514798

[languid trellis]

Can yoy dm me this pls bbg

[dull ginkgo]

Let’s say a priori we know fuck all about Galois extensions.

Assume G is a finite subgroup of Aut(F), and F^G is the fixed point subfield. Then is there a way to prove for some x in F\F^G that F is algebraicly finitely generated by F^G through the orbit of x

I know you can show it’s finitely generated as a module, but as far as I’m aware that proof is by contradiction and doesn’t actually show the basis explicitly being the orbit of x, or if it even is

[dull ginkgo]

Probably going to try to show any valuation morphism for an element outside of F^G from the group ring F^G[G] into F is an endomorphism

Actually no, not the group ring, just F^G[orbit of x]

[dull ginkgo]

Here's the route I planned about going about it:

• For any commutative ring R, and a finite automorphism subgroup, G, the fixed point subring of R^G is integrally extended by R. Because we are working with a field, this means F^G is algebraically extended by F. Any intermediate ring between F^G and F must be a field.
• Assume G has order N, Fix an enumeration of the elements of G, s_1...s_N. Consider the polynomial ring F^G[X_1...X_N], and the valuation map for each x, S_x, that fixes F^G and maps X_n to s_n(x). For some x in F \ F^G, let the image of S_x be denoted K
• Prove K must be F

I'm trying to process this

[cloud solar]

Question

[dull ginkgo]

I know the images of x for two automorphisms, s_n and s_m must commute since F is commutative under multiplication and addition. Naievely it seems like each s_n induces an automorphism that "permutes" the X_n and X_m. The group of which would be the Abelianization of G

[cloud solar]

In a finite group the maximal order of elements is a multiple of every other order of elements?

[dull ginkgo]

the order of any element in the group divides the maximal order in a finite group

[cloud solar]

How

[dull ginkgo]

wait I think that may only be for abelian groups

Here's an idea homie:

[dull ginkgo]

Consider D_10. r has order 2, s has order 5

5 is the maximal order, coprime to 2

[delicate orchid]

sayyy whaaaa

yeah this is true for abelian groups

[dull ginkgo]

that's just bezout theorem ye

*lemma

[dull ginkgo]

Actually yeah, is this okay?

Does every automorphism $s_n \in G \subseteq \mathrm{Aut}(F), |G| = N < \infty$ induce an automorphism of $F^G[X_1...X_N]$ such that $s_n^{*} : X_a \mapsto X_b $ where $s_n \circ s_a = s_b$

[cloud walrus]

Dyfunction Executive

[fading field]

[languid trellis]

R u about to expose yourself as british too

[ebon pine]

Bro has beef with fellas init

[languid trellis]

I am a bloke, fella

[dull ginkgo]

Wait

I had a brainwave

If $R$ is a commutative ring, and $G$ is a finite automorphism subgroup, and $R^G$ is the fixed field, then $R$ is an integral extension and a finitely generated algebra over $R^G$

[cloud walrus]

Dyfunction Executive

[dull ginkgo]

To show $R$ is a finitely generated algebra over $R^G$, consider an enumeration $\sigma_1 ... \sigma_N$ of elements in $G$ where $N = |G|$. Let $n!m$ denote the index of $\sigma_n \circ \sigma_m$ for convenience. We can consider the polynomial ring $R^G[X_1...X_N]$, and it's obvious that there's an automorphism for each $\sigma_n$ that sends $X_m \mapsto X_{n!m}$. Lets fix some $x \in F \setminus F^G$, and consider the valuation morphism $v : X_n \mapsto \sigma_n(x)$. Consider the image of this morphism, the ring $P$.

It's obvious $R^G \subseteq P \subseteq R$, but it's also obvious that it is fixed under G. However, $R^G$ is the largest subring fixed by $R^G$, so either $P$ is $R$ or $R^G$. However, the identity is one of the group elements, therefore if $P = R^G$ then $x \in P \cap (R \setminus R^G) = \emptyset$, a contradiction. Thus the image is $P$, and the map is surjective. This implies $R$ is a finitely generated algebra over $R^G$

[cloud walrus]

Dyfunction Executive

[dull ginkgo]

Does this work?

I don't see why it wouldn't. The image would have to be fixed, therefore either the fixed point subring or the whole ring. it can't be the fixed point subring so it'd have to be the whole

You can use Zariski's Lemma on this lmao

and immediately get Artin's Lemma, that [F : F^G] = |G|

It's also integral too lol

[rocky cloak]

Why do you say that it must be either R^G or R?

Like if R = k[x, y] and G=C2 swaps x and y, then you can have a subring like P = k[x^2, y^2, x+y, xy] which is neither R nor R^G.

[dull ginkgo]

wait no I messed up, thanks for pointing that out

messed up a definition that I wrote down for something else

[rocky cloak]

Also, it's not true that R needs to be finitely generated. For example take R = k[x1, y1, x2, y2, ...] and let G=C2 be the group that swaps xi and yi. Then R is not finitely generated over R^G.

If on the other hand R is a finitely generated k-algebra and G acts by k-algebra automorphisms, then R is finitely generated over R^G by virtue of being finitely generated over a subring.

[south patrol]

I'm trying to remember the actually interesting statement lol

I think it is that it holds if R is a Noetherian A-algebra or smth

(And G acts by A-algebra maps)

[long obsidian]

Hey is the category of groups an abelian category?

[next obsidian]

No

The most intuitive way to see this is that cokernels “don’t exist”

Except they do, they’re the quotient by the normal closure of the image

What you end up having is an issue with the map from the image to coimage, which will translate to the inclusion of the actual image into the normal closure of the image

[delicate orchid]

also u just can't put a ring structure on the homsets

[rocky cloak]

I'd argue an easier way to see it is that Hom(C2, S3) is not a group. So in particular not an abelian group.

[next obsidian]

You don’t need that

[delicate orchid]

what

sorry

I was thinking of End(S_3)

for general homsets yeah they just need to be ab groups

[next obsidian]

This is a little chmonka tho because you have to show that there’s no way to equip one that makes sense yah?

Which requires a bit of tedious work

I guess it’s easier than making an example of what I said tho

Also that set is small

[delicate orchid]

it's a set of size 4 it's not THAT hard

[next obsidian]

So not too hard

Except uh

No it is kinda annoying

[delicate orchid]

nuh uh

[next obsidian]

I think so? Because you need to then show like

That there’s some other group such that the induced map on Homs can’t be a group homomorphism or something

Idk I might be on something

[delicate orchid]

no the more I think about it the more annoying it gets

the standard thing I know is showing that End(S_3) can't be a ring

[next obsidian]

Idk how you do that tho tbh

[delicate orchid]

you uhhh

[next obsidian]

I mean you could bash out every possible abelian group structure

[rocky cloak]

I guess there's a difference between wheter Grp is an abelian category and whether it could hypothetically be equipped with the structure of an abelian category.

[delicate orchid]

no it's smarter than that

[next obsidian]

And then show nothing works

But that’s ass

[next obsidian]

Precisely

Except actually abelian category is NOT structure

It is a property

[delicate orchid]

what's that thing where no zero divisors and finite => field? you use that

[next obsidian]

Ah

[delicate orchid]

then contradict with the fact that |End(S_3)| isn't a power of p

[next obsidian]

Yeah that’s kinda annoying tho because

The proof is hard when you don’t assume commutative I believe

You get it as a result but it’s blech

[delicate orchid]

that's cringe lil dude

also it's an ABELIAN category why wouldn't I assume COMMUTATIVE

RAAAAAAGH

[next obsidian]

Dang lil dude

Commutative on the ring operation graaaaa

[rocky cloak]

Well, that no zero divisors imply skew field is trivial, and they also have prime characteristic

So you don't need the full wedderburn stuff

[delicate orchid]

that "trivial" result is the one I'm referring to

I'll google the name

[rocky cloak]

The on a finite set injective=surjective=bijective

Does that have a name?

[next obsidian]

Pigeonhole

Lelz

[delicate orchid]

ok it's Wedderburn's little theorem apparently

[next obsidian]

Is that true?

I guess so yeah

Clear

[delicate orchid]

it's a skew filed obvs they need prime char

filed

[next obsidian]

Stfu

You know in Bourbaki field doesn’t mean commutative

[delicate orchid]

is he french per chance

[next obsidian]

😍

[delicate orchid]

that's a french thing

[rocky cloak]

Surely Bourbaki out of all people deserve the pronoun they

[delicate orchid]

I don't know anything about this individual

[next obsidian]

Has Bourbaki ever had a female

I think not

[rocky cloak]

This "individual" 😉

[delicate orchid]

https://tenor.com/view/cat-what-is-happening-dim-dim-what-confused-gif-26463317

[rocky cloak]

Some guy on quora said

There were very few female mathematicians in France during the formative years of Bourbaki (and there probably still are lamentably few today). The group of people that organized to become Bourbaki, and the ones that joined later, were all men.

So seems no women

[delicate orchid]

I have never been more on edge

[rocky cloak]

Wews mind being blown by Bourbaki

[formal ermine]

bro's edging

when does the galois group emit a ring structure?

is there anything interesting here?

the only example i can think of on the go is Gal(F_p bar/F_p) = Z hat

[delicate orchid]

I thought I had an answer here but I do not

Gal(L/K) = End_K(L)^{\times} taking L to be a K-algebra (I think) so meh there's something there

[opaque finch]

My question goes like this
Let $R = K[x,y]$, and let $\varphi: R^2 \to R^2$ be an $R$-module endomorphism of the $R$-module, $R^2 = R \oplus R$ given by $\varphi((a_1,a_2)) = (a_1 + xa_2,ya_1)$. Find a monic polynomial $p$ with coefficients in $R$ such that $p(\phi) = 0$.

[cloud walrus]

mycroftholmes1703

[opaque finch]

I understand we have to use Cayley Hamilton theorem in generalised form, but I am messed up about the computation

[formal ermine]

can u show us ur computations that u have so far

[tough raven]

Surprisingly, there isn't. If a category is abelian the abelian category structure is uniquely determined.

[pastel pollen]

is this true for exact categories? I would assume the answer to be no?

[tough raven]

IK very little about exact categories.
IIRC there are two notions of exact category, one requiring an “exact structure” and the other called a “Barr exact” category.
The former is a structure and the latter is a property.

[pastel pollen]

okay, I see. I havent heard of the latter. What I am thinking of is a structure on the category, thanks!

[rocky cloak]

You can have many different exact structures on the same category

Any additive category admits an exact structure though, and being additive is uniquely determined for the same reason being abelian is.

[random pasture]

Originally I thought I just I have to show is that any even permutation can be written as an even n number of 2-cycles, and then you can merge pairs of two cycles to form n/2 3 cycles, but I'm not sure how to prove that this will always be possible. Does the proof involve using inverse 2-cycles?

[rocky cloak]

So every even permutation can be written as the product of an even number of 2-cycles. So if you can prove that the product of any pair of 2-cycles is a product of 3-cycles you're done.

The product of overlapping 2-cycles is a 3-cycle, so then you're done. Then you just have to write a disjoint product of 2-cycles as a product of 3-cycles.

[stark helm]

how to count number of irreducibles in form of ax^2+bx+c over Zp where p is prime? I count the result when a=1. (p*(p+1)/2). I need a result when a is not 1.

[random pasture]

oh wait, even number of 2 cycles with two possible formats, (a b)(b c) => (a b c) and (a b)(c d) => (a b c) (b c d)

so you can just say that any pair of evens can be represented with only 3-cycles

[rocky cloak]

If a is nonzero then
ax^2 + bx + c = a(x^2 + b/ax + c/a) so that's just p-1 times the number of such with a=1.

If a=0, then you're just counting bx+c

[celest furnace]

Exactly

[stark helm]

Is it because monic polynomials is irreducible, so ax^2+bx+c is irreducible as well? And does it happen in all field or just in Zp?

[rocky cloak]

Yeah, that's true in all fields

[stark helm]

Another question is: if a=1. And I write x^2+bx+c=(x+s)*(x+t) to find its reducible polynomial, and I have already counted number of cases. But I am confused about if it is possible to have the case when we can't factor it into roots but we can still write it as reducible polynomials.

And must all polynomial be either reducible or irreducible, is there any possibility of being not reducible and not irreducible?

[rocky cloak]

Being reducible means that you can write it as a product of polynomials with smaller degree. Being irreducible means you can't, so that's the only two options.

If the polynomial is of degree 2, then reducible means it's the product of degree 1 polynomials, hence we can factor it completely into it's roots.

[stark helm]

OK, I got it, it should also be the same as it is of degree 3. Really appreciate that

Question: What is the difference between the field Zp and field Zp(t). Does the latter one represent rational functions?

[coral spindle]

This is a worrying question because it reveals that you don't understand the latter notation

Are you familiar with the notation R[x], where R is a ring?

If F is a field, then F(t) is similar to F[t] insofar as it contains polynomials, but as you say, yes it also contains quotients of polynomials.

Indeed F(t) is the ring of so-called rational functions on F.

So it should be clear that it's completely different.

[minor pilot]

.’

[chilly radish]

Bro really said "no bitches"

[stark helm]

so is it correct to say F(t)={q(t)/r(t): q(t), r(t) belongs to F(t)[x]}?

[random pasture]

For proving this homomorphism, what would the product be for phi(gg') = phi(g)phi(g')?

is it addition or multiplication

[quick pilot]

can someone give me a bit of a hint on this? not sure where to start

[chilly radish]

Assume that sigma transposes a and b WLOG a<b. So consider that is sgn(sigma(i),sigma(j)) for any two numbers i,j.

There's a few options:

1. If i and j are both not a and b, then sigma does nothing to them and so sgn(sigma(i),sigma(j))=sgn(i,j)=1.
2. If i=a, then for j<b we get -1, for j=>b we get 1, so we get a contribution of (-1)^(b-1)
   Can you list the other options and see what the total result of the product is?

[fading field]

bourbaki ❤️

[next obsidian]

Bourbaki ❤️

[severe tide]

You need to show $\varphi((n_1,n_2)+(m_1,m_2))=\varphi(n_1+m_1,n_2+n_2)=\varphi(n_1,n_2)+\varphi(m_1,m_2)$

[cloud walrus]

kevinhardy2

[random pasture]

i feel like it's pretty straightforward with $n_1 + n_2 +m_1+m_2$ for both of them. Am I messing something up or just overthinking?

[cloud walrus]

x496

[tough raven]

No, that's it. Do note that you used the fact that Z/2 is abelian. In general, for any abelian group G (say with the group operation written as addition), the map
G × G -> G : (g, h) -> g + h
is a group homomorphism.

Let x_1, …, x_n distinct elements of F (or more generally elements of a ring whose pairwise differences are all invertible?).
Is the n by n Vandermonde matrix
1 x_1 … x_1^{n-1}
1 x_2 … x_2^{n-1}
… … … …
1 x_n … x_n^{n-1}
right-invertible?

[toxic zephyr]

if I remember correctly, to consider a ring R as an R/I module, we need I to be contained in ann(R) (can't remember the proper notation, but ir=0 for all i in I and r in R). can someone give a nontrivial example of such an R and I? i can't think of an n such that R=Z/nZ would have a nontrivial annihilator

it's been a while since I've done any algebra so Z/nZ is my go to. but maybe PIDs only have trivial annihilators or something i honestly can't remember

[toxic zephyr]

actually that's probably true, but I can't remember that many interesting examples of rings. are there rings that have nonzero elements that zero out all elements of the ring?

or I guess since 1x=x for all x, then such an element can't exist?? i.e. there's no nontrivial way to consider R as an R/I module??
maybe I should go look back at my hw from years ago bc we had a problem on something like this

[ivory trail]

annihilators (ideals of R) are defined for subsets of R-modules

the most canonical way to treat R as an R/I module is also kind of boring: via extension of scalars along R -> R/I, in which case the R-module R turns into the R/I-module R ⊗R R/I = R/I

there's also "coextension of scalars" which replaces R by R-Mod(R/I, R), which you give an R/I-module structure by requiring linearity

but idk what you can say about this (because i haven't studied it)

[delicate orchid]

writing homsets with the name of the category

[ivory trail]

yeah let me write "by Hom(R/I, R) (the R-module of R-module homomorphisms)"

[delicate orchid]

It’s clear from just Hom(R/I, R) lol

Do Hom_R(-,-) if u must

[ivory trail]

not when we're talking about how R can be an R/I-module or whatever

R/I is an R-module through restriction of scalars anyway

[rotund aurora]

where do you use Zorn's lemma? I didnt use it, idk if I clowned

• Take x not in N(A), quotient by N(A) so you can assume N(A)=0
• If x!=0 the set (1,x,x^2,x^3,...) forms a multiplicative system and the prime ideals in A not containing x are in one-to-one correspondence with the prime ideals in A_x
• A_x has prime ideals because 1/1!=0/1, because otherwise there would exist an n such that x^n=0, which is only possible if x=0, a contradiction

mmh ok ig I need that a nontrivial ring has prime ideals

[elfin wraith]

I have a question to find the number of monic, cubic polynomials in F_3[X] which have no root. Obviously there’s 27 polynomials which may or may not have roots, and I could just evaluate the polynomials and work out which coefficients give me roots, take however many of those there are, and take it away from 27. But this feels really inelegant and kinda infeasible for any larger fields (by hand at least)

So I’m wondering, is there a nicer way to do this that I’m missing?

[rocky cloak]

If you can count the number of polynomials with 1, 2 and 3 roots, then you can just subtract that from the total number of polynomials

[dire siren]

I think it's a bit faster with PIE.

[elfin wraith]

This is what I was trying to say sorry if it wasn’t clear, I was just wondering if there’s a more direct way than this because it feels like it’d suck for anything much larger than F_3, even even for F_3 it’s not exactly elegant

[elfin wraith]

PIE?

[dire siren]

principle of inclusion and exclusion

the advantage is that you can find the number without classifying polynomials

and it works fine on anything larger than F_3

[elfin wraith]

I’m not sure that I see how I would use that in this situation, could you elaborate?

[dire siren]

let A={polynomials that have 0 as a root}, B={polynomials that have 1 as a root}, C={polynomials that have 2 as a root}
you want to find 27-|AuBuC|

|A|=|B|=|C|=9 (because you can choose the first two coefficients arbitrarily, but the last one is fixed)

AnB consists of polynomials of the form x(x-1)(x-a), where a is 0,1 or 2, so |AnB|=3, and similarly |BnC|=|AnC|=3

and finally |AnBnC|=1 (the only element is x(x-1)(x-2))

[elfin wraith]

Ohhhh yeah that’s way nicer!

[dire siren]

this should work over any finite field K

[elfin wraith]

Thank you very much, thats definitely a much cleaner approach than having to just blindly calculate

The step I wasn’t seeing was how to deal with the intersections but that feels obvious now I’ve seen it

[dire siren]

@elfin wraith what's interesting is that with this approach we can find that if |K|=q, then the number is (q-1)^3, which makes me think there might be an even faster way to count

or maybe it's just a coincidence, idk

[crystal vale]

In question 3, I showed that I have (1 k) for all k=1,2,......,n , so I can generate any permutation.
Is it correct?

And for question 4, I showed that there is (i i+1) for all i=2,......,n-1).

[elfin wraith]

That is quite interesting, there’s almost certainly something there but I don’t see it right away. I wonder if it generalises and for a monic nth degree polynomial you have (q-1)^n

I might play around with that on sage later if I have the time (and remember to)

[dire siren]

I think the case n>3 is different. The argument I used for counting |AnB| will no longer work, because unlike in the previous case, the third factor is no longer linear.

or maybe it will still work

yes, I guess that last factor should be chosen arbitrarily, so we can count

nike, for n=4: (x-a)(x-b)(x^2+cx+d), with c,d arbitrary

[elfin wraith]

Yeah it would definitely require some thought, but there could be something there

Not sure it has much use other than general interest but still

[rocky cloak]

It doesn't get worse with the size of the field in any way.

If the field has size q, then there are q(q-1)(q-2)/3! polynomials with exactly 3 roots. And q(q-1) with exactly 2 roots. The ones with exactly 3 roots are the q ones with a triple root, and those that are a multiple of an irreducible quadratic polynomial. So you just have to count the quadratic polynomials in a similar way.

[elfin wraith]

Oh yeah, I definitely didn’t see that, i don’t think I was saying what you were saying after all.

That is probably even easier for larger fields than using inclusion exclusion

Either way my takeaway here is that I should think more carefully about how we I can write the polynomial to make the roots appear more clearly, so thank you both

[rocky cloak]

You can even write the whole thing out as a polynomial in q without every specifying what q is

[dire siren]

I just remembered that there was this question in an Olympiad in my country

for a finite field with q elements, it was asking for the probability that a polynomial of degree n>=q doesn't have any root

[rocky cloak]

A cubic polynomial is irreducible iff it has no roots, so if f is such a polynomial then f is the minimal polynomial of an element in the field of order q^3.

So if x is an element in the field with q^3 elements either x is in K or the minimal polynomial has degree 3. And in that case x has 3 conjugates, so the answer should be (q^3 - q)/3

So same argument would work if you replace 3 by any prime (to count the number of irreducible polynomials)

[dire siren]

that's a very nice argument

[rotund aurora]

are you from Romania lol

[dire siren]

yes

[rotund aurora]

Romania's olympiads are so unique

[dire siren]

yes, that works

[dire siren]

hmm, I think you are missing cases in which the polynomial is irreducible even in the field with q^3 elements

the solution of the problem I mentioned find the number of polynomial with degree n>=q to be (q-1)^(q+1)*q^(n-q)

if we only want the monic ones, we should divide this by q-1, and for n=q=3 this becomes (q-1)^3

[glad osprey]

It's probably correct, but it's hard to say without seeing your work

[rocky cloak]

I'm not directly counting polynomials, im counting elements that have minimal polynomials of degree 3.

For q=3, (q^3-q)/3 = 8 which indeed matches (q-1)^3, but it shouldn't be (q-1)^3 in general.

[stark helm]

Find the roots of f(x)=x3 + x2 − 1 ∈ Q[x]. I realize that there should be no rational roots by rational roots theorem. Hence I determine that f(x) is irreducible and find its extension field Q[x]/(x^3+x^2-1). But it seems to me that there will be too heavy if we find elements in this extension by supposing u is a root of f(x) and u^3=1-u^2 and find all elements and plug these values into f(x)? So I am doubting if there is some easier method to find its roots in extension?

[dire siren]

Ah, I understand now. Also, I see what the problem with my (q-1)^3 is. In the case q=3 there were 3 sets that I used to apply PIE. But in general there are q, so the answer should be q^3 - Binom(q,1)q^2 + Binom(q,2)q - Binom(q,3), which is indeed (q^3-q)/3.

[dire siren]

use Cardano

[stark helm]

can you describe what it is?

since this is the first time that I heard this name

[dire siren]

@stark helm here is a step by step explanation: https://www.math.ucdavis.edu/~kkreith/tutorials/sample.lesson/cardano.html

and here are some worked out examples: https://brilliant.org/wiki/cardano-method/

[rocky cloak]

On what form do you want the roots? Like as radicals? Or do you want the two other roots described in terms of one of the roots, or how?

[stark helm]

just the roots as we construct extension , the diagram is the original question

[rocky cloak]

I'm not sure what you mean by that. Like the polynomial has only one real root, so the extension Q[u]/(u^3 + u^2 - 1) contains only one of the roots.

You can describe the minimal polynomial of the other roots in Q(u) I guess...

"Find" isn't really a technical term. Maybe it's clear from the greater context of your exercise set, but on it's own it doesn't really mean anything.

[lethal cipher]

So i was trying to prove the second isomorphism theorem, and I came across something a little strange. Let $S$ be the normalizer of the subgroup $N$.
For $(SN)/N:={ xN, x\in SN},$ for any $x\in SN$, $\exists s\in S$ and $n\in N$ s.t. $x=sn$. So $xN=snN=s(nN)=sN$. So this suggests that every element in $(SN)/N$ can be represented as $sN$ for some $s\in S$. But, it that were true, then it suggests $(SN)/N$ is isomorphic to the quotient group $S/N$, which doesn't sesm right.
So...I think my logic is flawed somewhere and I could appreciate some help identifying the issue

[cloud walrus]

dackid

[shrewd sandal]

You can't quotient S wrt N since N is not necessarily a subgroup of S.
If it were then you'd just have your result via the second iso theorem

[lethal cipher]

Ah, I see. So although each element in (SN)/N can be represented as sN for some s, which kinda looks like what S/N would, S/N doesn't always exist since N need not be a subgroup of N

[shrewd sandal]

Ah you said normalizer. In any case what I said stands, NnS=N so this just results from the second iso theorem

[shrewd sandal]

Right. If N is normal in S, then what you have shown is correct, it just happens to be the result of second iso

[lethal cipher]

Well, I was working on proving the second iso, and that sN thing came up, so I just wanna make sure my thoughts are justified.
But yeah, if N is a subgroup of S, then (SnN)=N so (SN)/N iso to S/N due to the second isomorphism theorem.

Thank you for the clarification sqeral. That helped a lot

[shrewd sandal]

Ofc 👍

[minor wraith]

wait

can I say, $D_6$ is isomorphic to $C_{12}$ cyclic group

[cloud walrus]

Kakaka

[minor wraith]

which is isomorphic to $\mathbb Z_{12}$

[cloud walrus]

Kakaka

[wraith cargo]

This isn't true

[coral spindle]

You can certainly try to say false things

But they will be wrong

[minor wraith]

can someone please give me a more intuitive explanation of congruence relations

can u relate it to congruence equivalence relations in modular arithmetic/number theory?

[coral spindle]

Yes

Let G = Z and H = nZ = {nz | z in Z} where n is some integer

Then congruence on H is congruence mod n

That's really the best intuition you're going to get.

[timid latch]

how do i prove a group is acting on a set

[coral spindle]

Using the definition of a group action

[sour depot]

congruence relations let you take quotients

if you have a congruence relation on a group it partitions the elements of the group into equivalence classes

the equivalence class of the identity defines a normal subgroup, and the quotient by the congruence relation is the same as the quotient by that normal subgroup

[coral spindle]

No, they do not.

This only happens if H is a normal subgroup.

But you may note there is no such assumption here.

[sour depot]

H should be assumed to be a normal subgroup here

[coral spindle]

I don't see why that would be the case. There is no barrier to defining this equivalence relation, which defines the cosets.

[sour depot]

the reasonable notion of a congruence relation is an equivalence relation on the underlying set which is compatible with the group structure in the sense that a~a' and b~b' implies ab~a'b'

[coral spindle]

I am aware in universal algebra that the word congruence—yes well there you go.

[sour depot]

in which case such a thing corresponds uniquely to some normal subgroup

[coral spindle]

But this is not saying that.

[sour depot]

right okay

sure I guess more generally you can talk about congruences where a~b means a^{-1}b just lands in some subgroup H, rather than a^{-1}b being the identity

you can still talk about quotients, they just won't be groups anymore, they'll be sets of cosets

taking quotients by subgroups and talking about sets of cosets is still a good thing to talk about, so that gives some intuition in this more general setting

it's just a bit weaker of a notion since you're not quite demanding that the equivalence relation is compatible with your group structure as is the case for the more restricted notion

[random pasture]

I completed part a, what exactly is part b asking me to do?

[coral spindle]

What part of the question confuses you

[random pasture]

like can i just say that theres no surjective group homomorphism because sym is not abelian?

[coral spindle]

No

Why would that follow?

You should consider the hint, and other facts you know about the symmetric group.

[random pasture]

I know there's sign homomorphism from Sym-> z/2

oh i see

we showed phi z/2 x z/2 -> z/2 and we know there exists surjective from Sym-> z/2

so we assume there exists surjective homomorphism Sym-> z/2 x z/2, and use h1 (a,b)=a and h2 (a, b) = b to contradict that?

is that sequence of steps logical

[coral spindle]

Just try to prove it

You don't need to seek approval for your plan

[stark helm]

how about saying finding complex roots?

[crystal vale]

Both question?

[crystal vale]

I am not sure how to write in correct way

[dull ginkgo]

Simple exercise: Let $G$ be a group with subgroups $H_1$ and $H_2$, both with finite index, then $H_1 \cap H_2$ must have finite index. Also let $G/H$ be the set of equivalence classes / cosets, doesn’t need to be a group for this lol.

Immediately the Cartesian product $ G/H_1 \times G/H_2 $ has cardinality leq $[G : H_1] [G : H_2]$. Consider the “diagonal” subset $D$ of pairs $(xH_1, xH_2)$

It’s easy to see that $x(H_1 \cap H_2) = xH_1 \cap xH_2$. So therefore $[G : H_1 \cap H_2]$ is the cardinality of the image of the following map $D \twoheadrightarrow G/(H_1 \cup H_2)$, so therefore it must be less than or equal to $[G : H_1] [G : H_2]$ by transitivity and the domain being finite. Is this a valid proof?

[cloud walrus]

Dyfunction Executive

[dull ginkgo]

Next is this god awful exercise that for two subgroups of a finite group $|HxK| = |H| [K : x^{-1}Hx \cap K]$

[cloud walrus]

Dyfunction Executive

[dull ginkgo]

which is scary but I’ll try it in a bit. I assume you have to stratify HxK, into hxK and show |hxK| = [K : x^-1Hx \cap K] or something

[minor wraith]

can I get a hint for this question please

solution is apparently 6

[vivid tiger]

Well

What's the kernel?

[minor wraith]

do I need to first count the number of homomorphisms

[vivid tiger]

Actually

[minor wraith]

note, both of them are additive groups

[vivid tiger]

Maybe we need to look at the cokernel?

[minor wraith]

do not know what that is

[vivid tiger]

codomain/img

[minor wraith]

current attempt, although, I am not sure whether this the right approach:

Let $f(x):=x$. We require $f(63)=64x\equiv 0 \mod 147$; that is, $63x=147k\iff 7\mid x$. So we have 21 distinct homomorphisms.

If we consider a few, say,

$f_7(a)=7a\in \mathbb Z_{147}\implies |img(f_7)|=\lfloor \frac{147}{7}\rfloor =21$,

$f_{14}(a)=14a\in \mathbb Z_{147}\implies |img(f_{14})|=\lfloor \frac{147}{14}\rfloor =10$,

$\vdots$

$f_{7t}(a)=7ta \implies |img(f_{7t})|=\lfloor \frac{21}{t}\rfloor$ for $t\in{0,...,20}$(verify inductively).

Thus, setting $\lfloor \frac{21}{t}\rfloor=7\implies t=3$ as the unique $t$ within our specified range. What other 5 homomorphisms with images of size 7 am I missing?

[cloud walrus]

Kakaka

[vivid tiger]

img is a subgroup and it has index 21

Why must f(63)=64x=0 mod 147?

why does this imply 63x = 147k?

[minor wraith]

because 1 is the generator

for Z_63

[vivid tiger]

Ah

[minor wraith]

so by homomorphism additivity

[vivid tiger]

f(1)=x

[minor wraith]

yeah

wait

[vivid tiger]

63x=64x

[minor wraith]

f(63)=f(1)+...+f(1)

63 times

[vivid tiger]

I don't think 63x=64x

.

[minor wraith]

and also $63\equiv 0 \pmod {63}$

I'm trying to ensure that f(0)=0

[cloud walrus]

Kakaka

[minor wraith]

mapping identity to identity

[vivid tiger]

f(63)=0, I get that.

Why does this equal 64x?

[minor wraith]

typo

63x

[vivid tiger]

Anyways, image is size 7, so index of image is 147/7=21.

[minor wraith]

what good does index do me

it's the number of cosets right

[vivid tiger]

I don't know, actually.

I feel like it should be helpful.

[vivid tiger]

yup. you can also take the quotient cuz abelian

Wait

Duh

What am I doing

[minor wraith]

wat

[vivid tiger]

Yes, it's generated by 1

So, we need to look at where 1 goes, as you say

A prior we have 147 possibilities

We can narrow this down.

7f(1)=0

[minor wraith]

huh

[vivid tiger]

Well, the image is of size 7

so, f(1) generates the image, and the image has size 7, so 7*f(1)=identity=0

=147 in Z/147Z

[minor wraith]

why does f(1) generate the image

[vivid tiger]

well, 1's a generator

f(1+1+1+1...)=f(1)+...

[minor wraith]

ok

[vivid tiger]

so f(x)=nf(1)

[cobalt heath]

For group of 147 elements, wouldn't it good to inspect the 7-group?

Or are you guys doing something else

[vivid tiger]

So, we need to look at all numbers x mod 147 such that 7x = 0

[cobalt heath]

Ah sorry, so you guys are working with Z/147Z.

[vivid tiger]

e.g. x=0, x = 21

[minor wraith]

im trying to find the number of homos with this restriction

[vivid tiger]

x=42, x= 63, x = 84, x = 105, x = 126

Where have I fucked up?

I count 7 homo's.

Oh

Yeah, this is a necessary, not sufficient, condition.

e.g. f(1)=0 leads to an image of size 0

so we need to check the cases f(1)=21,42,63,84,105,126

Well

We have a problem

If we kept going we'd get a lot more cases

nvm

126+21=0

duh

[minor wraith]

would euler's totient function help

[vivid tiger]

because 21 | 147

[vivid tiger]

where do we need it?

[minor wraith]

idk

[vivid tiger]

This is additive group

[minor wraith]

friend did similar question used it apparently

yah

[vivid tiger]

Point is that f(1) = multiple of 21 mod 147

So, besides f(1)=0, there are at most 6.

We need to check that the rest actually have images of size 7.

[minor wraith]

hold on, f(1)=0 \implies f(a)=0, mapping everything to zero

[vivid tiger]

so, 21,42,63,84,105,126,0 has the right size.
42,84,126,21,63,105,0 has the right size

[minor wraith]

so image is size 1?

[vivid tiger]

this is why f(1)=0 doesn't work, the image has size <7

f(1)=21 works, as it generates the entire group {0,21,42,63,84,105,126}

This group is isomorphic to Z/7Z

so everything but 0 generates it.

Therefore f(1)=42, or f(1)=63, or etc., also work.

And thus we have 6 homomorphisms.

[vivid tiger]

So...is this the 7-group?

{0,21,42,63,84,105,126}?

I don't think we need Sylow here?

That is, in full, here's the cleaned argument:
f(1) generates the image. image of size 7 => 7f(1)=0=147. thus f(1) is a multiple of 147/7=21 mod 147. the multiples are
{0,21,42,63,84,105,126}, which itself is a subgroup of order 7, isomorphic to Z/7Z. So, we need to find the generators of this subgroup. There are 7-1=6 of them

[tacit hemlock]

Oh did u use Lagrange’s theorem

[tacit hemlock]

Otherwise what is 21?

Before, I determined that 21 was the number of homorphisms

[vivid tiger]

hmm...where did I use Lagrange?

[tacit hemlock]

But this 21 that you’re using seems to be representing something different

[vivid tiger]

Ah, to show that 7f(1)=0.

Since a^|G|=identity, let a be f(1) and we have |G|=7 for the image subgroup.

[vivid tiger]

?? 21 is the representative of the equivalence class of 21 mod 147? wdym

[vivid tiger]

shouldn't there be 147 homomorphisms, assuming no restrictions besides that we have a homomorphism?

I am unsettled a bit by the fact that I never used any detail about the domain except for the fact that it was cyclic.

[tacit hemlock]

Why did u do 147/7

[vivid tiger]

Well

7f(1)=0=n*147 for some n

so f(1)=n*21

That's why.

1 is not the identity

Additive groups.

e.g. imagine Z/2Z to Z/4Z via sending 1 to 2

You don't need to delete a message making a claim that is incorrect!

[toxic zephyr]

is this channel open?

not sure if their og question was answered

i'll just ask then. trying to prove that if $\func{f}{R}{S}$ is an onto ring homomorphism and $I$ is a maximal ideal of $S$ then $f^{-1}(I)$ is a maximal ideal of $R$ (and i'm not allowed to use the FHT)

[cloud walrus]

eigentaylor

[toxic zephyr]

i concluded that if $f^{-1}(I)$ is strictly contained in another ideal $J$, then $f(J)=S$, since $f(J)$ is an ideal bc f is onto(?)

[cloud walrus]

eigentaylor

[chilly ocean]

yeah, surjections take ideals to ideals

[toxic zephyr]

okay cool

i'm rusty on working with maximal ideals so i dont remember how you usually prove something is maximal. i was struggling to come up with some sort of contradiction when i assume that J is not R, and i wasn't able to think of a way to show that 1_R is in J somehow (those are the two ways i remember lol)

[chilly ocean]

you prove that an ideal is maximal by proving that its quotient is a field

[toxic zephyr]

yeahhhh i already did the proof the FHT way. it was much easier for me lol. trying to do a direct proof

quotient rings are so helpful it is kinda annoying to try not to use them

[chilly ocean]

they really are such a blessing

wait until you see localization

[toxic zephyr]

does this work?

if $J\neq R$ then there is some $m\in R\setminus J$ such that $f(m)=f(j)$ for $j\in J$. then $f(m-j)=0\in I$
$$\implies m-j\in f^{-1}(I)\subset J\implies m-j=j'\implies m=j+j'\in J$$
contradiction

[cloud walrus]

eigentaylor

[minor wraith]

what is this $\langle x_i \rangle$ notation?

[cloud walrus]

Kakaka

[minor wraith]

a single element inside the angle brackets?

[toxic zephyr]

isnt it just $\gen{x_i}=\bdef{x_i^k:k\in\bZ}$?

[cloud walrus]

eigentaylor

[minor wraith]

so $<1>=\mathbb Z_{99}$ since it's a generator?

[cloud walrus]

Kakaka

[minor wraith]

hence it's not a proper subgroup?

[toxic zephyr]

yeah as an additive group

[minor wraith]

is there a quick way to check

wtf

can I do better than just trying everything 2,3,...

[toxic zephyr]

for an additive group of Z_n <x> is just {kx: k in Z}

[minor wraith]

sorry, im very new to group theory

[toxic zephyr]

take an easy example of Z_6. what is <1>, <2>, <3>,..., <5>?
you already know that <1> is the whole group. so see what happens for the others

my algebra friend says Z_6 is the best goto example lol

[minor wraith]

and how am i supposed to check that it's a subgroup

just group axioms?

[toxic zephyr]

its always a subgroup

[minor wraith]

or is there a trick

[minor wraith]

Claim: subset of cyclic is always a subgroup?

proof: associative √, identity √, inverse √

right?

[toxic zephyr]

what about {1,4} as a subset of Z_6

[minor wraith]

assuming it contains the smaae identity

[minor wraith]

no identity

need 0 in it

[toxic zephyr]

then your claim is false

[minor wraith]

withj this added restrcion?

[toxic zephyr]

{0,1,4}

[minor wraith]

wait what's failing

associtive √

identity √

[toxic zephyr]

mhm. there are two other requirements

[minor wraith]

no inverses for 4

[toxic zephyr]

yeah but something more fundamental is missing

[minor wraith]

not closed

[toxic zephyr]

yeah

that's what you should check first

always

[toxic zephyr]

every subgroup of a cyclic group is cyclic though

[minor wraith]

then where did u pull this from

[toxic zephyr]

<x> is defined to be the smallest subgroup containing x

so it's already a subgroup

[minor wraith]

oops

[toxic zephyr]

that happens to be exactly this set

(all powers of the generator)

[minor wraith]

"powers"

so ,multiples here

[toxic zephyr]

groups cant tell the difference

[minor wraith]

ture

[toxic zephyr]

powers mean multiples for additive groups

[minor wraith]

so back with ur Z_6 example

what do i generalise

<2> = 2k

0,2,4

[toxic zephyr]

yes

[minor wraith]

<3> = 3k

0,3

<4> = 0,4,2

<5> = 0,5

[toxic zephyr]

not quite

yeah

[toxic zephyr]

this is also wrong

[minor wraith]

i realised

5 is -1 modulo 6

I can construct -5+6=1

so it's also a generator

[toxic zephyr]

thats true

so then we have
1 and 5 that generate all of Z_6
2,3,4 do not.
can you see a pattern?

[minor wraith]

oh so everything except 1 and its inverse

[toxic zephyr]

what relation does {1,5} vs {2,3,4} have with 6

[minor wraith]

what

set of generators vs non-generators

[toxic zephyr]

yeah but try to see the underlying pattern

what happened with 2,3,4 that meant you stopped before you got all the elements?

with 2 it was 0,2,4 and then you stopped. why?

[minor wraith]

wrap around

to 0

[toxic zephyr]

yes

[minor wraith]

so proper divisors of n and its mutiples?

[toxic zephyr]

right. another term for that is elements not coprime to n

[minor wraith]

yeah

[toxic zephyr]

so gcd not 1

and that checks with {1,5} and {2,3,4}. the first set is numbers coprime to 6, and the other has a gcd that isnt 1

[minor wraith]

so I can take ¬ euler's totient function

[toxic zephyr]

yeah that's right

[minor wraith]

ok

[toxic zephyr]

n-phi(n) specifically

[minor wraith]

ahh

why do these questinos have less to do with the group theoiry than i expect

[toxic zephyr]

id say its more that theres a lotta overlap with number theory

[minor wraith]

yeha a lot of these theorems, I like to think as generalisations of number theory concepts

like multiplicative order

pre cool

[toxic zephyr]

as someone who liked number theory but is scared of advanced group theory, i agree with this

[minor wraith]

ok but like

[toxic zephyr]

you dont need group theory to learn number theory or the other way around, but knowing one helps in learning the other imo

[minor wraith]

maybe i will like group theory

it's just that i laernt number theory already

and it's only liek week uno of me learning group theory

too early to tell

[toxic zephyr]

undergrad group theory is fine. i like ring theory and modules a lot better. but then i started learning sylow stuff and i just noped out of there and became an applied mathematician ||that's not exactly how it happened but kind of lol||

but basic group theory isn't too bad. it's pretty cool.

[minor wraith]

i think that will happen to me xD

hopefully PDEs into phd later

got scared of real math

[minor wraith]

wait

consider all the generators of $\mathbb Z_{25}$

[cloud walrus]

Kakaka

[minor wraith]

so all residues that aren't 0,5,10,15,20

so 7 is one

but if 7 were a generator, then shouldn't we be able to construct 1 using only multiples of 7?

that is, 7t+7l=1

but by the extended euclidean algorithm, no such bezout coefficients exist

oh mod 25

$(-7\equiv 18) - (-42 \equiv 17)=1$

[cloud walrus]

Kakaka

[minor wraith]

.close

[rocky cloak]

You still haven't said what "finding" means. Like do you want the decimal expansion of the real and imaginary part? Do you want a description with nested radicals? Do you want their minimal polynomial in terms of the real root?

[toxic zephyr]

7t+25l=1. and indeed, 7(-7)+25(2)=1. just as 7(-7)=1 mod 25

[toxic zephyr]

yeah sure

[minor wraith]

do u have a tip for this Q

Let $a,b$ be elements of a group $G$. If $a$ has order 15 and $b$ has order 24, what are the possible values for the order$ <a>\cap<b>$?

[cloud walrus]

Kakaka

[minor wraith]

what I have so far:

Let $H:=<a>\cap<b>$

[cloud walrus]

Kakaka

[minor wraith]

this is also a subgroup as all intersections of subgroups are also subgroups

do I need to use Lagrange's theorem somehow>

[cobalt heath]

Lagrange theorem certainly helps, I'd say

[toxic zephyr]

this is kinda out of my wheelhouse. my group theory sucks

[minor wraith]

how

how do i apply it

[cobalt heath]

You understand that Lagrange theorem is about subgroup relation and group order, right

So basically, by identifying the subgroup relation you can reduce the cases to check.

[minor wraith]

oh wait

so