groups-rings-fields

[mighty kiln]

No

Well yes if you're just looking for a choice of representatives (via axiom of choice)

But not a homomorphism

[limber bramble]

Thanks , could you explain a bit more on the choice of representatives ?

[mighty kiln]

Suppose we have a quotient q: G → G/kerφ

For each coset g ∈ G/kerφ, pick some f(g) ∈ G whose coset in the quotient is g

i.e. q(f(g)) = g

Then there's your function f: G/kerφ → G

[limber bramble]

Yes but would this not led to one input being sent to multiple outputs?

[mighty kiln]

We pick an output among the possible ones

[topaz solar]

To see it’s not gonna be a homomorphism, Z -> Z/pZ very much so won’t have homomorphism inverses

[mighty kiln]

Which is where axiom of choice is involved

Say q: Z → Z/3Z, then I could pick f([0]) = 3, f([1]) = 7, f([2]) = 2

[limber bramble]

Now I get it . Thanks 🙂

[untold basalt]

Let ( X ) be a set and ( \Sigma(X) := { f : X \to X \mid f \text{ is bijective} } ).
( (\Sigma(X), \circ) ) is a group (the operation is the usual composition of functions).

Now define the operation ( g \cdot h := h \circ g ).
I have to find a isomorphism of ( (\Sigma(X), \cdot) ) onto ( (\Sigma(X), \circ) ).

[cloud walrus]

Heywood Jablome

[untold basalt]

$\phi(g)=g^{-1}$ doesn't work

[cloud walrus]

Heywood Jablome

[untold basalt]

well the identity does though wtf

[coral spindle]

Are you sure?

[untold basalt]

oh no yeah I see my mistake now

[coral spindle]

By the way, there is nothing special about the group that they show you. Let $(G, \cdot)$ be a group and define a group $(G, \odot)$ by $g \odot h = h \cdot g$. It may be worth thinking about why these two groups are isomorphic.

\bigskip
If you are seeing this before you know what groups really are, I apologise.

[cloud walrus]

Boytjie

[spice whale]

(tbf every group is isomorphic to one of the given form)

wait

no nevermind

ignore me

that's Sym(X)

[coral spindle]

Every group is a subgroup of one like that :P

[spice whale]

yeah

[young thorn]

does anyone have the topics in algebra by herstein solutions

i cant find them

[coral spindle]

Often solution manuals don't exist. I would suggest just asking about questions here

[twin sand]

Can someone help me with item d)?

I dont know how to prove that zeta_n is in Q(i,sen(2pi/n))

[dim widget]

what is zeta_n

as a complex number

as a + bi with a, b real numbers

[twin sand]

e^2pi*i/n

[dim widget]

please provide your answer in the form suggested above

[twin sand]

cos(2pi/n)+i*sen(2pi/n)

[dim widget]

yes

okay so now you just need to show that the extension contains cos(2pi/n)

do you see how to use the hint?

[twin sand]

No

[dim widget]

ok

[twin sand]

Maybe proving that sen(4pi/n) belongs to the extension, it is proved

[dim widget]

yes that would be enough

[twin sand]

But i dont know how yo prove that sen(4pi/n) belongs

[dim widget]

well cos^2(2pi/n) belongs right?

[twin sand]

Why?

[dim widget]

because it's 1 - sin^2(2pi/n)

[twin sand]

Yes yes

Okey

And then?

[dim widget]

so then cos(4pi/n) belongs

so then sin(4pi/n) belongs

so then cos(2pi/n) belongs

now you can fill in the reasoning yourself

[twin sand]

Why cos(4pi/n) belongs?

[dim widget]

you should think about it! But every step only uses standard angle identities

in fact just the double angle identities

[twin sand]

And why that cos(4pi/n) belongs implies that sen(4pi/n) belongs?

[teal vessel]

if the cosine value belongs, doesn't its square also belong?

[twin sand]

This is my problem

[dim widget]

i would pull up a list of double angle identities for cos, sin, tan and stare at it

if i were you

[twin sand]

You are using sin²+cos²=1?

[tardy hedge]

For each element of generating set, take the r in R that kills each element, then the product of each ri is an element of R that kills every element in M

Is that the idea?

[twin sand]

You cant use this propery

[tardy hedge]

Integral domains are commutative right ? So this would work

[dim widget]

we already used this for the first step, but afterwards we don't need it

[twin sand]

I dont know why this is true

[dim widget]

do you know about a relation between sin and cosine

coming from translation

hint hint

[twin sand]

I dont see it

Following that cos(4pi/n) belongs

You obtain that sin(pi/2 - 4pi/n) belongs

[twin sand]

Why it implies that sin(4pi/n) belongs?

[tardy hedge]

Alright dudes its time for me to learn measure theory. Been on modules for too long

I think i will learn a bit of tensor products first and then go to measure theory

I am trying to prepare for all my classes this september is the goal

Prepare as best i can

[twin sand]

Can someome explain me why this is true?

[chilly radish]

Yes

[dim widget]

cos(2A) = 1/(2sin(A))(sin(3A) - sin(A)) by the angle addition formula for sin

when A is 2pi/n

[thorn torrent]

Hello, I am very new to group theory and not working from a curriculum so I have a question which is maybe kinda silly:
I am trying to number the 4 edges of a playing card 1-4, and understand all the different ways of doing that as a group. One way to do it is to say that we number them clockwise from the top edge and it’s the permutations of (1, 2, 3, 4). But this doesn’t account for the fact that the physical object has 180° rotational symmetry, so (1,2,3,4) is congruent to (3,4,1,2). I don’t want to have 24 elements, I want to have 12.
My thought is that I want to treat the permutation (3,4,1,2) as a kind of pseudo-identity
Can someone explain how to do this or give me some keywords or videos to learn more?

[dim widget]

sorry is your playing card a square?

[thorn torrent]

Rectangle

[dim widget]

okay then the symmetries are not all permutations of the edges

it's not an issue of there being relations between the permutations, its that some permutations are not symmetries of the card

[chilly radish]

I hope they meant vertices

[dim widget]

Ah I see

maybe you meant vertices

[thorn torrent]

I might have meant vertices. Like I said I don’t know anything

[dim widget]

anyway its still not true

[twin sand]

I donk know this formula but you will obtain that sin(3A) belongs

[dim widget]

okay so number your cards corners (1, 2, 3, 4) then the rotations that are allowed are only 1 \to 3, 2 \to 4 or in permutation notation (13)(24)

[dim widget]

sin(3A) is just -4sin^3(A) + 3sin(A)

so it's obvious that it "belongs

[twin sand]

But

I want to prove that sin(2A) belongs

You say that is was easy? Many People dont know this formulas

Why did you say that use the translation?

[dim widget]

because it's easy if you use the translation formula

the identity i showed you follows from the translation formula

you can't just berate people into solving your homework for you

[twin sand]

My problem was to prove that if cos(4pi/n) belongs to Q(i,sen(2pi/n)) then sin(4pi/n) belongs to Q(i,sen(2pi/n))

[dim widget]

ye

and it's sin not sen

[twin sand]

And why belongs?

You look like ChatGPT, you say things that have nothing to do

If you werent going to help, dont respond, but dont leave ideas to waste

[dim widget]

all ideas i could provide to you would be wasted

you've made that pretty clear

[twin sand]

Can someone who knows math help me?

[dense forum]

i wouldnt assume he is trying to get his homework done, maybe he just wants to understand the solution to his problem properly

[dim widget]

without doing any work

[dense forum]

we dont know how many time he has spent on his own thinking about the problem, you are just assuming :/

[dim widget]

suit yourself really

[main tiger]

I wouldn't say that if cos(2pi/n) belongs to a field then sen(2pi/n) belongs

[dim widget]

here n is odd

and the problem is to prove the reverse not that it makes a big difference

[main tiger]

You have that the degree of the extension of the cosine over Q ir the Euler function divided by 2, and you can see by item e) that the degree of the extension of the sine over Q is the euler function at n

[main tiger]

Yes i know

I'm not seeing then why u can say that sen(4pi/n) belongs to the extension using that cos(4pi/n) and sen(2pi/n) belongs

[dim widget]

because n is odd

are you the same person?

[main tiger]

chill

I would like if anybody else agrees with that and could explain to me how you get to that

[twin sand]

You dont know that it is our question? Why if n odd, implies that if cos(4pi/n) belongs then sin(4pi/n) belongs?

[tardy hedge]

Whats going on here lol

[rotund aurora]

By symmetries I just meant elements of G. I am looking at the action of G on the vertices and using the orbit-stabilizer formula

[kind temple]

how do you see that the exponents must all be the same if the reduced words are the same?

this is from D & F section 6.3

[cloud walrus]

c squared

[coral spindle]

You're right I don't think it's as simple as they're saying, I can't see an immediate argument at all.

I think you have to argue by induction on the length of the word s, by which I mean the smallest k for which s_{k+1} = 1

Oh no wait hold on

Hold on a frikking second

[kind temple]

lul

[coral spindle]

These are literally just sequences of letters

Like the last letters must be the same, etc etc

[coral spindle]

On the contrary, I don't see how this is valid at all

[kind temple]

could you clarify?

[coral spindle]

I would like you to clarify!

[kind temple]

no, like, i don't know what you meant by this lol

i would also like myself to clarify haha

[coral spindle]

What a coincidence, I don't know what you mean by that!

[kind temple]

uh

am confused

[coral spindle]

What is the claim here.

[kind temple]

is you have two reduced words which are equal but there is an index where their exponents are opposing, then we reach a contradiction

[coral spindle]

This is just by definition

Remember they are literally just sequences of elements

And sequences are equal iff their indices are equal

N.b. this doesn't require the definition of reducedness, just the definition of this notation (a, b, c, ...)

[kind temple]

right, so you get s_i^{ep}_i = r_i^d_i

[coral spindle]

what's p_i

[kind temple]

but this doesn't restrict me from writing
$$(s_1,s_2,s_3,\dots,1,1,1,\dots)=((s_1^{-1})^{-1},(s_2^{-1})^{-1},(s_3^{-1})^{-1},\dots,1,1,1,\dots)$$

[cloud walrus]

c squared

[coral spindle]

And?

[kind temple]

the exponents are not the same

[coral spindle]

Because this is not dealt with by the claim

s_i^-1 is not in S

[kind temple]

oh oh

[coral spindle]

They did say specifically, in the paragraph where they simplify the notation, that s_i in S

[kind temple]

i did not see that

[coral spindle]

OK that would explain it

[kind temple]

so if you did have s_i = r_i^{-1}, then S and S^{-1} would have non-empty intersection

[coral spindle]

Yeah

[kind temple]

thanks. reading is hard

[kind temple]

is this also justified by a modified version of eckmann-hilton?

no wait

this operation isn't commutative

nvm

[kind temple]

the free product of two groups is just F(G U H), right?

[rocky cloak]

Well, you also have to impose relations so that G and H become subgroups

[kind temple]

why?

[rocky cloak]

Because otherwise it wouldn't be the free product

[kind temple]

so you need presentations of G and H

and then quotient by the relations?

[rocky cloak]

Like F(G u H) only depends on the underlying sets, not the groups

[rocky cloak]

Yes you can do that. Or just impose the relations that
g * h = gh
whenever g, h, gh are in G, and similarly for H

[kind temple]

am confused

why do i need to impose relations on F(G U H) then

oh wait

the F's

[dim widget]

what the F

[rocky cloak]

Yeah, free functor plays nice with free product. As one might expect from the name

[kind temple]

okay, so G * H = (F(G) * F(H)) modulo all reduced words consisting of elements only in G or only in H

[kind temple]

fr

[kind temple]

well

wait. is this the normal closure of G and H?

[dim widget]

no normal closure is with respect to an overgroup that contains both

but G, H will not be normal in G * H

[dim widget]

no not modulo the reduced words involving only elements of G, just modulo the relations in G

[kind temple]

but jagr also mentioned this

[dim widget]

or if you like just modulo $g_1g_2 = m_G(g_1, g_2)$ and the same for H

[cloud walrus]

Math_Discord_Final_Girl

[dim widget]

where $m_G: G \times G \to G$ is multiplication

[cloud walrus]

Math_Discord_Final_Girl

[kind temple]

o

[kind temple]

why are those not the same

[dim widget]

one way to think of it is that its words in H and G but consecutive strings of elements of either H or G can be evaluated as if they were in H or G

[dim widget]

if you mod out by all reduced words in H and all reduced words in G then you have nothing left

since every string is a conglomerate of blocks which are either in H or G

[kind temple]

its still not right even if you make an exception for strings of length one

okay

cool

thank you 🙂

[dim widget]

yeah even then it would get you something weird

[kind temple]

oh, and for this, when we mod out by these guys, how do you get the normal subgroup that you mod out by?

[dim widget]

I would think of it as like the simplest group containing both H and G, where "simplest" means with the minimal possible generators and relations.

[kind temple]

sorry that was wrong

[dim widget]

just to be clear

the relations are like

if * means composing strings

then g_1*g_2 = (g_1g_2)

so the word of length 2 g_1*g_2 is equivalent to the word of length one (g_1g_2) which is some element of G

[kind temple]

artin nor dummit and foote discuss this. do you know of any other books which go into more depth?

[dim widget]

i do not know

[kind temple]

okay. was worth an ask. i think i need some more detail on how this process works. I found a book that i think goes into this more

thanks for the explanations tho

[barren sierra]

is there any good notation for indicating the field we are talking about when talking about minimal polynomials?

Since of course the minimal polynomial of $\sqrt(2)$ differs over $\mathbb{Q}$ vs $\mathbb{Q}(\sqrt(3))$

[cloud walrus]

Spamakin🎷

[barren sierra]

but $m_{\sqrt{2}}^{\mathbb{Q}}(x) \in \mathbb{Q}[x]$ is ugly

[cloud walrus]

Spamakin🎷

[sly frost]

because otherwise you don't have a valid inclusion map i: G → G * H into the coproduct

let's say you have a free groups F(C₂ ∐ C₃) without any relations

then if you have a map i: C₂ → F(C₂ ∐ C₃) having i(g) = g then gg = i(g)i(g) ≠ i(gg) = i(e) = e from

so it's not a homomorphism

but the universal property for coproducts requires it

so you need to quotient it with the relations g² = e in C₂ and h³ = e in C₃

[sly frost]

I guess you have to prove it yourself

[kind temple]

what book is this from? allufi?

[kind temple]

will try to do this exercise when i get some time

[chilly ocean]

I have a question about where to peek at current, pure or applied, problems in Algebra? I'm an amateur and don't have access to other than a local Collge Library.

[coral spindle]

There is a big pfd somewhere on google titled “unsolved problems in algebra” that you can find

pdf*

[chilly ocean]

okay, I'll google it, thanks.

[barren sierra]

How do I prove identity (ii)? I've proven everything before. I feel like there's gonna be some clever choice of x for binomial expansion but I got nothing.

[chilly ocean]

Has anyone heard of this "Unsolved Problems in Group Theory. The Kourovka Notebook"?

[barren sierra]

Yea I have. Link for those curious: https://kourovka-notebook.org/

It's cool, wish something like that existed for algebraic combinatorics and related stuff

Maybe one day I'll start such a thing

[coral spindle]

Even just Wikipedia has good lists of open problems. E.g.: https://en.wikipedia.org/wiki/Homological_conjectures_in_commutative_algebra

[chilly ocean]

Yeah, I was wanting to look at something that wasn't "Mellinial Problems"-like, lol. I'll check out that link.

[dim widget]

Millennials are all very concerned about the Hodge conjecture

[prisma umbra]

This klein group4?

[barren sierra]

what do you think?

and why?

[prisma umbra]

Every element is self inverse

And identity is 1

[prisma ibex]

yeah I agree

[barren sierra]

😭

[prisma ibex]

you can just write f(x) like a normal person

[delicate bloom]

you sure about that?

[barren sierra]

Ok this was a bad example

The minimal polynomial of i over C vs R

(that 2 should have been a 3)

Anyways notation hard

[prisma umbra]

I am waiting for your answer sir

[prisma ibex]

just use succinct notation like f(x) and explain it separately

[prisma umbra]

Here

[prisma ibex]

use noggin

[vocal pebble]

And also, the third non identity element is the product of the other two non identity elements

[prisma ibex]

waow there are so many groups of order 4

[prisma umbra]

This is the necessary condition?

[vocal pebble]

Yeah, or else the order of the group might not be 4. If e,u,v,w are in the group and
w=/=uv,then wu, wv might also be in the group

https://groupprops.subwiki.org/wiki/Klein_four-group the verbal definitions section contains more info

More simply, to see why it is necessary, the multiplication tables wouldnt match up if the product of two non identity elements isnt the third non identity element

[rapid junco]

What in peoples opinion is the best comprehensive book for groups alone?

[balmy prairie]

Tough question. I don't know many books that are purely group theory and no rings and fields. But in my mind Fraleigh's book has a group theory section that covers more than enough. If you skip rings only and read fields, then you also get to read some Galois Theory(which uses groups a lot). But if you want something comprehensive as in "teaches all the group theory you will need for grad school and get into research," Lang's Algebra is the holy grail.

Some books I haven't personally read and have heard good things about: Dummit and Foote, Artin

[delicate bloom]

why do you want groups alone

[dim widget]

I think they mean the best book to learn the material of groups, regardless of how it treats other topics

not the best book that only mentions groups

[topaz solar]

Rotman has a groups specific book

Which I’ve been pointed to by an old group theorist (since retired, I believe)

But I’m not a group theorist to comment on the quality of its exposition and choice of topics covered

[surreal sluice]

I am trying to understand the statement that a discrete normal subgroup of a group is closed as written in this answer https://math.stackexchange.com/a/511612/343701

I don't see why it is the case.

Please let me know if I should ask in the Alg Top channel. Sorry if it is case.

[untold basalt]

Let $F$ be a field. A non-constant polynomial $f \in F[x]$ has a root with multiplicity greater than $1$ if and only if $\text{gcd}(f, f') \neq 1$.

Is this meant to be up to associated elements to $1$? I remember seeing it with $\iff \text{gcd}(f, f')$ is not invertible, but I'm having trouble prooving it:

if $f$ has a root $r$ with multiplicity $\geq 2$ I get that $r$ is a root of both $f$ and $f'$, so $x-r$ divides $\text{gcd}(f, f')$; this however doesn't necessarily mean that $\text{gcd}(f, f')$
is not invertible, right? It could be $\text{gcd}(f, f')=(x-r)h(x)$ with $(x-r)h(x)$ invertible...?

[cloud walrus]

Heywood Jablome

[surreal sluice]

(x-r)h(x) cannot be an invertible element in the ring F[x] as it will have degree at least 1.

[untold basalt]

Can't there be invertible elements of degree greater than 1?

[surreal sluice]

Nope. In fields, or more generally in integral domains, degree of p(x).q(x) is sum of their degrees, and 1 has degree 0.

[untold basalt]

oh I see, so this could happen for example in F_4[x] but not F_p[x]

no nvm F_4 is a field, brain fart

Z_6[x]

[surreal sluice]

Think so, although I don't remember an example rn

[rotund aurora]

is units of Z[G] isomorphic to G?

[lusty marlin]

No

[rotund aurora]

right I guess Z[Z]=Z[t, t^{-1}] and the units are +- t^k, so Z/2Z x Z I think?

[untold basalt]

Prove that there exists no rational fraction $f \in \mathbb{Z}_p(t)$ such that $t=f(t)^p$.

[cloud walrus]

Heywood Jablome

[untold basalt]

writing $tq^p(t)=h^p(t)$ is supposed to give a contradiction modulo $p$...

[cloud walrus]

Heywood Jablome

[south patrol]

Presumably Z_p is meant to be integers mod p?

[untold basalt]

yes

[south patrol]

This should be true over any field

[south patrol]

Yes, for example use unique factorisation

And the fact t is is prime

[rotund aurora]

you can just count degrees

[untold basalt]

thanks

allright, so I get deg(LHS) =1 mod p and deg(RHS) =0 mod p

[rocky cloak]

I'm not sure it's true.

Certainly if H is just the trivial subgroup it's discrete, but only closed if G is Hausdorff. And I guess this example can be expanded by taking the product of a non-Hausdorff group and a discrete group.

[south patrol]

True

[rocky cloak]

gcd is only defined up to associates yes

[south patrol]

Well

okay sure like np = mp + 1

Lol

[long obsidian]

If I have a ring R[x,y] and an ideal I=<x-2> can I show that R[x,y]/I=R[y]?

[south patrol]

Yes

[long obsidian]

Is it basically obvious that since I is generated by elements in R[x,y]\R[y] that the ideal I is in R[x,y]\R[y]? Does this have to do with I being a subring

[south patrol]

Well you can apply the first iso theorem to an appropriate homomorphism R[x,y] -> R[y]

It is probably easiest to see once you note we can replace x-2 with x

[cloud walrus]

clubsoda14

[nimble folio]

Is this fine? Seems easier than showing a map between Z[x] and Q[x] is not surjective.

[languid trellis]

I don't think Z[x] and Z are isomorphic

[vapid vale]

Z[x] is not ismormophic to Z

[nimble folio]

Oops

[vapid vale]

Z[x} is not a PID

[dim widget]

Phi_1 has not been defined, but I have a strong suspicion it’s not an isomorphism…

[nimble folio]

Yea just realized that

LOL

[languid trellis]

Triple whammy

[vapid vale]

😭

[dim widget]

anyway you can just show there’s no surjection Z[x] \to Q

that’s the way to go it feels

[nimble folio]

https://math.stackexchange.com/questions/382654/proving-that-mathbbzx-and-mathbbqx-are-not-isomorphic

I don't really understand the first answer

[dim widget]

If two rings are isomorphic they have the same units

but these rings do not have the same units

[vapid vale]

the ^X refers to the units (multiplicative invertible elements)

[nimble folio]

Yea ik

[dim widget]

a polynomial over a ring is invertible if and only if its constant and that constant is a unit in the ring

[nimble folio]

How do you prove this

I don't think it was mentioned in dummit and foote

[dim widget]

it’s automatic from the fact that a homomorphism preserves the identity and respects multiplication

ab = 1 iff F(ab) = 1 iff F(a)F(b) = 1 iff F(a), F(b) are both inverses of each other

for F an injective map of rings

[nimble folio]

ok that makes sense

Thank you!

[dim widget]

then if F is surjective this shows the map on units is surjective

[languid trellis]

I have no idea where they're trying to lead me when they bring in exercise 12. For reference, this is $$R^{n}/I R^{n} \cong R / IR \times \cdots \times R / IR$$, where $I$ is an ideal of $R$, $IR = { \sum_{\text{finite}} i_j r_j, i \in I, r \in R }$

[cloud walrus]

swifteeee

[languid trellis]

Any push in the right direction would be appreciated (:

R^n an R-module btw

R/I is a field iff I is maximal....

So R/IR is a submodule of R/I. We already have (R/I)^n = (R/I)^m iff n = m.

hm

from (12), we immediately get $R^{n} / IR^{n} \cong R^{m} / IR^{m}$

[cloud walrus]

swifteeee

[tardy hedge]

Bruh i just did that one

Im so mad i have to go to measure theory now. I wish i could just study algebra tbh

[rotund aurora]

but you can study sigma-algebras

[tardy hedge]

Not the same 😡

[languid trellis]

I'm getting lost in the sauce. For the forward direction, suppose Rn and Rm are isomorphic. I want to show that n=m. Suppose that n != m. Then this contradicts (12), as both R^n/IR^n and R^m/IR^m are isomorphic to each other and to the cartesian product R/IR n or m times. For the opposite direction, suppose that n=m. I want to show that R^n and R^m are isomorphic. What from here?

[tardy hedge]

if n = m then its trivial isnt it

[languid trellis]

Yeah you're right actually

then I'm questioning what I wrote for the forward direction

I have not invoked maximality of I once

[tardy hedge]

I dont think doing it by contradiction is the way

[languid trellis]

But then again, I feel like the forward direction is fine

hm

[rocky cloak]

R/I being a field is equivalent to I being maximal. So that's where it's evoked

[languid trellis]

I understand that, could you elaborate on where specifically we're using it?

[wraith cargo]

you want to show that your statement is equivalent to the statement that F^n = F^m if and only if n = m for F a field

[rocky cloak]

Say R^n is isomorphic to R^m, that implies (R/I)^n is isomorphic to (R/I)^m . By linear algebra dimension is well defined so m=n

[wraith cargo]

because F^m and F^n are vector spaces over F and it's known that two vector spaces are isomorphic to each other if and only if their dimension is the same

[languid trellis]

Oh right I see

[coral spindle]

Welcome back swift. Exams went well I hope?

[languid trellis]

I actually worked for them and I am very pleased with how they went

Quite a nice experience actually

[coral spindle]

Congratulations!

Here's hoping the markers are accurate, and when innacurate they are in your favour :P

[languid trellis]

Thank you very much

We'll find out in about 6 weeks

[coral spindle]

You'll be well-prepared

(Even if you do nothing over the summer, which I encourage)

[languid trellis]

I find studying very rewarding, and I have literally nothing else to do, so I will be ignoring your advice haha

[coral spindle]

I'm glad, just don't wear yourself out

It wouldn't do if you started your degree with no motivation

[dull ginkgo]

Yooo hi swifteee

Do you want to study Nakayama lemma together

[rotund aurora]

If Z[G] iso Z[H] then G^ab iso H^ab, right?

[languid trellis]

Of course, and I would've said before that I would stop before this happens but it turns out that my burnout gauge is not very good lmao

[languid trellis]

Im good homie haha

[rocky cloak]

Do you have an argument for why?

[coral spindle]

Are you looking only at finite groups, also?

[rotund aurora]

well I was doing this exercise, here they want G and Q to be abelian in (iii), but iirc H_1(G, Z) is the abelianization of G, so it should still work I think

but

I was wondering about a more direct proof, if that is possible

[dull ginkgo]

:(

[rotund aurora]

I think it should also be true for infinite groups (?)

[coral spindle]

Maybe but the proof might be easier

[dull ginkgo]

Me

[rocky cloak]

Well, with these hints I guess it's true yeah. Since H1(G, Z) is the abelianization of G

[rotund aurora]

is there some shortcut to prove this, tho?

[rocky cloak]

Isn't the proof already pretty short?

Or what kind of a short cut are you imagining

[coral spindle]

Do you mean to ask if there's a proof without homological machinery? I think it'd be tricky, since we can't easily describe the commutator subgroup purely in terms of the ring

Or at least I can't see an easy way to do so

[rotund aurora]

well it's basically saying the co/homology is the same, then they calculate one homology group

[rotund aurora]

yeah, something like that

[rocky cloak]

I guess something that's not so obvious is that {}^\phi Z is isomorphic to Z...

[rotund aurora]

the action is trivial, isn't it then obvious?

[coral spindle]

Yeah I'm not sure I follow jagr

[rocky cloak]

Like hypothetically if phi(g) = -1, then the action of ^phi Z would not be trivial

(but then phi would not be an isomorphism of course)

[rotund aurora]

uhh good point

[rocky cloak]

In fact if G = C2 = {1, g} then phi(g) = -g should be an isomorphism for which the ^phi Z is not trivial

So then it's not as straight forward as it looks I guess...

[coral spindle]

Ah I see what you're saying now

[rotund aurora]

in this case, you could consider the "group basis" {1,-g} then this maps 1 to 1 and -g to g, and the action would be trivial (?)

[rocky cloak]

The action of g on ^phi Z would be multiplication by -1

[rotund aurora]

but the action of -g on ^phi Z would be multiplication by 1

and {1,-g} is still a group

[coral spindle]

Yes but still ^phi Z is not the same module as Z

Which is an important observation

[rotund aurora]

but it is a trivial {1,-g}-module

[coral spindle]

We would maybe want to prove that for any such isomorphism phi there is some phi' for which ^phi' Z is Z

You say 'but' but you haven't really indicated why this is fixing the issue

It's a trivial module for a different group, but we don't know if this is generally the case

[rotund aurora]

{1,-g} is isomorphic to {1,g}. Both modules are trivial C_2-modules then, regardless of the isomorphism you choose, they will obviously have the same homology/cohomology

[twin sand]

Can someone help me with item d)? I dont know how to use the hint (without using item e)

[coral spindle]

I thought the whole point of this conversation is to do this without cohomological machinery.

[twin sand]

My problem is proving that Q(i,zeta_n) is content in Q(i,sin(2pi/n))

[coral spindle]

TTEG gave you plenty of info yesterday. Think about the multiple angle formulae.

It is significantly harder than the other qns, and I think this was an oversight, but the info is there.

[twin sand]

Please can you explain me?

[rotund aurora]

well, Jagr didn't change the group G, the group rings are obviously isomorphic and G is isomorphic to itself. I believe jagr was pointing out that there are some subtleties in the outlined proof (in the hints)

[coral spindle]

I would rather you think about it yourself

Perhaps someone else will choose otherwise

[coral spindle]

Oh my bad, I thought we were going for a different proof

[rotund aurora]

oh yeah I'd still like a more "direct" proof, but I'm just saying that the conclusion of the problem is obvious in Jagr's example

[twin sand]

Sorry, i have spend many time thinking it, and i dont know how to prove using the hint

[rotund aurora]

anyway, I was wondering if from an isomorphism Z[G]-->Z[H] we can obtain an isomorphism Z[G^ab]-->Z[H^ab] ?

therefore reducing this to the abelian case

[rocky cloak]

Is ZG / [ZG, ZG] just straight up isomorphic to ZG^ab ?

Feels like yes, but I don't have the energy to think right now

[twin sand]

Some girl said yesterday that cos(4pi/n) belongs implies that sen(4pi/n) belongs to Q(i,sen(2pi/n))

But i dont know why it implies

[coral spindle]

That "some girl" has a username, and she said a lot more than that.

[twin sand]

This is the step that i dont understand

Shee said the steps before

But this step i dont know

[rocky cloak]

Airtight, so every element of ZG is a linear combination of elements of g, so a commutator is a linear combination of commutators of elements in G, which generate the kernel of ZG -> ZG^ab.

So indeed ZG = ZH implies ZG^ab = ZH^ab
@rotund aurora

[rotund aurora]

how are you defining a commutator of ZG? ZG is just a ring

[rocky cloak]

[x, y] = xy - yx

[X, Y] = ideal generated by [x, y], x in X, y in Y

[rotund aurora]

can you determine Z[G]^x (the units) ?

with G abelian

[coral spindle]

This may be crossing into extremely difficult territory because this looks a lot like Kaplansky's unit conjecture (which was disproved)

[rotund aurora]

It obviously contains +- G, so the question is if it contains anything more

[rotund aurora]

okay yeah that's what I was wondering about

uh but K is a field

[coral spindle]

Yes I'm looking at this now

[rotund aurora]

I just want K=Z

[south patrol]

I think it is still subtle

[coral spindle]

This is why I said it looks a lot like... :P

[south patrol]

This is related to my thesis, perhaps...

[coral spindle]

I'm looking at some slides by Higman and he apparently claimed that his example works in characteristic zero, which is to say it would work in QG since the example is defined over Z. But I cannot see proof of this claim, and I don't know what the inverse would be.

[rotund aurora]

btw is Z[G x H] easy to express in temrs of Z[G] and Z[H] ?

[coral spindle]

(I.e., I don't know if the inverse would also be defined over Z)

[coral spindle]

Z[G x H] is isomorphic to Z[G] (x) Z[H] where the tensor is over Z

[south patrol]

Hmm good question

[south patrol]

Isn't that with free products

[coral spindle]

No I don't believe so?

I don't see why that'd be the case, the free product would be very complicated indeed

[south patrol]

Well was because Z[]: Groups -> Rings preserves colimits

[rocky cloak]

Appearantly the units are exactly ±g, and you can use this to prove that G=H when ZG=ZH (G abelian)
https://link.springer.com/chapter/10.1007/978-3-0348-8658-1_7

[south patrol]

Would be surprised it if sent products to tensor products

But sure when G,H are abelian all these coincide

[coral spindle]

A GxH-module is a G-module and H-module such that the actions commute, so you can see why you get a map one way. The other way is also not too tricky to see

Like you can just define the maps without much of a hitch

[south patrol]

Yes

[coral spindle]

I think you'd need a more complicated thing for the free product of groups because you'd need to distinguish order very strongly

[south patrol]

Okay hmm

[rotund aurora]

yeah I guess it's obvious if you write g otimes h instead of (g,h), for (g,h) in G x H

[coral spindle]

Perhaps the tensor product isn't the coproduct for noncommutative associative R-algebras?

[south patrol]

Yeah

[rotund aurora]

ok then if G is finite maybe using the structure theorem of fintie abelian groups allows for a direct calculation of the group rings

[coral spindle]

Yeah, if you know k[x] (x) k[y] is k[x,y] from your algebraic geometry days this follows

By which I mean it's the same argument

[rocky cloak]

Tensor product is coproduct in the category of commutative rings, not in Ring

[south patrol]

Yes

[stark helm]

I am wondering if there is a normal subgroup that has the same order as An, i think it should not exist because it must exist odd permutations, but I am not sure how can I argue that

[coral spindle]

Normal subgroup of what?

[stark helm]

of Sn

[coral spindle]

No, there is no such thing.

[south patrol]

I just knew it sent free products to tensors for abelina groups and would assume that was what would continue to hold or whatever but ye

[stark helm]

how can I argue that? I think there must have odd permutation

and I think it relates to clousre of product

if that exist

and make contradiction

[rocky cloak]

It doesn't though. It sends direct sum to tensor product and send free product to coproduct of rings

[rotund aurora]

If I'm not mistaken, Z[Z/nZ]=Z[zeta_n] where zeta_n is a primitive nth root of unity (pretty much just by definition). Dunno if this is too helpful, tensoring these rings looks annoying

[rocky cloak]

Not quite

Z[zeta_n] is a domain

[coral spindle]

As a Z-module, Z[Z/nZ] will have rank n, whereas Z[zeta_n] will have rank phi(n)

[rotund aurora]

ah right

ok so I guess it's just Z[x]/(x^n-1) ?

[coral spindle]

Yeah

[stark helm]

so how will you argue that it is impossible to exist?

[south patrol]

Oh wait ignore my brain was dum

[coral spindle]

With great pain, I expect

[stark helm]

you mean if I want to argue it?

[coral spindle]

I found a pdf online that contains an (alleged) proof which looks just about as painful as I would've expected https://www.billcookmath.com/courses/math4720-fall2010/math4720-fall2010-An_is_simple.pdf

[rotund aurora]

Okay and $\Z[x]/(x^n-1)\otimes_\Z \Z[x]/(x^m-1)\cong \Z[x,y]/(x^n-1, y^m-1)$ ?

[cloud walrus]

croqueta3385

[coral spindle]

In general the question is can we express the units of A (x) B in terms of the units of A and B where these are all commutative rings

Even if we calculate the units of the group rings of the cyclic groups, I'm not sure this step is easy

Maybe I'm asking for something too strong here, but it is still difficult

[twin sand]

Can you help me?

[rocky cloak]

For n>=5 any normal subgroup would have to either contain An or not intersect it at all, because An is simple.

For n<5 you can check case by case

[rotund aurora]

it doesn't look like you can say something too general. But for this simple example, you are looking for polynomials P, Q such that $P(x, y)\cdot Q(x,y)=1+R(x,y)(x^n-1)+S(x,y)(y^m-1)$

[cloud walrus]

croqueta3385

[coral spindle]

Ah right, yeah I suppose if we just stick to polynomials we have nice algorithms and whatnot

[rotund aurora]

ok so if I'm not mistaken, here we want to show that P(x, y)=+- x^a y^b

mmh I mean this is a functional equation at this point, looks cute

[twin sand]

I dont know why the implication is true, i thing that the argument is wrong

[barren sierra]

First part was easy via rational root theorem. Not sure what to do for the second part. As far as I can tell there's nothing I can say about the degree of [K_a : Q]. I also tried to see if I could find some contradiction via assuming that there were only finitely many such a but no such luck

[barren sierra]

And then this one I'm still bashing my head against and getting nowhere. In fact I don't even intuitively see why it should be true?

[rotund aurora]

do you know what L^alg is? I don't think the way they are using this notation is standard, maybe that's your confusion

[barren sierra]

yea it's the algebraic closure of L

[rotund aurora]

I mean then it's false

the algebraic closure of Q(t) contains Q(t)

[coral spindle]

I mean in the first place, Q(t)^alg certainly has a square root of -1, and Q does not

[barren sierra]

wait no

I mixed up \bar{L} and L^alg

L^alg is set of elements of L which algebraic over K

which I see how to prove

ok ignore me for Problem 9 >_>

[barren sierra]

yea >_>

[rotund aurora]

maybe you can bash this. Like if $\alpha$ is a root of the polynomial $f_a(x)=x^4-ax-1$ and there exists a subfield $E$ of $K_a$ different from $\Q$ and $K_a$ then this means that you can find integers $x,y,z,w$ such that $x+y\alpha+z\alpha^2+w\alpha^3=\sqrt n$ for some integer $n$. If you square you will get some equations with $a$ as a parameter

[cloud walrus]

croqueta3385

[rotund aurora]

I would just try doing that, but there may be something smart

[rocky cloak]

This is probably overthinking it, but the resolvant of
x^4 - ax - 1
is
x^3 + 4x + a^2
If this is irreducible, the order of the Galois group is a multiple of 3, so it's either S4 or A4.

And the only index 4 subgroups in S4 and A4 are maximal, so the result would follow.

Hence the problem reduces to
x^3 + 4x + a^2
Being irreducible for infinitely many a. Not sure if that helps....

[barren sierra]

There's gotta be something simpler since technically this HW assignment comes before talking about Galois Groups or resolvants...

[kind temple]

just want to verify, the direct product of a family of groups is the quotient of their free product by its commutator subgroup, right?

[cobalt heath]

Wdym commutator subgroup here?

[kind temple]

commutator subgroup of the free product

sorry, the groups are all abelian

[cobalt heath]

Ah, ye the quotient should be same as the direct product of abelian gps.

[kind temple]

alr cool

[kind temple]

so

trying to show that the free product is the coproduct in Grp

um

the arrows from G into G * H are supposed to come from the universal property of the quotient group

but i messed up the direction originally and i am realizing now

how do you get natural maps from G and H into G * H?

where G = <S_G | R_G> and H = <S_H | R_H> are presentations of G and H

and G * H is the quotient group F(S_G U S_H) / ncl(R_G U R_H)

[mighty kiln]

There's just one way

[topaz solar]

Send g in G to g in G*H

unsurprisingly

[kind temple]

but this says i should be able to do it naturally

tbf, i have no idea what natural means

but uh

i was hoping it meant, coming from some other property or something like that

but sending g to g ker(pi) works

for g in G

but is that, in some sense natural?

like, what is an unnatural map?

[barren sierra]

They don't quite mean natural in any formal sense

but more like

consider the direct product of G x H

what is the natural map G -> G x H

natural as in "the obvious map"

natural in that sense is what they mean

[kind temple]

but it says i should use two properties to construct these maps

[kind temple]

doing this doesn't seem like the intended solution

also, how are we splitting up g if S_G isn't all of G?

ig im not seeing the 'obvious' way to construct the map

oh wait

[cobalt heath]

Either they want you to argue that it is kind of natural,
Or you are asked to prove that it is natural transformation.

Like you can prove the naturality square:
G -> G * G'
H -> H * G'

[kind temple]

i think its like this

where the map from G into F(S_G U S_H) is from the universal property of the quotient group

since G is a quotient of F(S_G)

and the natural maps are the unique maps from the quotient property composed with pi

[kind temple]

here is the full diagram

[wet zodiac]

are there groups with uncountable order with subgroups isomorphic to itself?

[mighty kiln]

Z times any uncountable group

[mighty kiln]

Are there simple groups with a proper subgroup isomorphic to themselves

[coral spindle]

I think the 'infinite alternating group,' namely the union of all A_n for n >= 5, is an example.

You just shift all numbers up by one and you get an isomorphic group

[mighty kiln]

Oh right

It feels like sequential colimits of simple groups are simple

Wait does this hold for arbitrary colimits

Surely at least it does for filtered colimits

[coral spindle]

I would expect so yeah

[true ingot]

Is $C_2 \oplus C_2 \oplus C_3$ a subgroup of $C_4 \oplus C_6$? I have a suspicion that not, but not sure how to show it ($C_k$ being a cyclic group of order $k$)

[cloud walrus]

Faputa

[coral spindle]

There is a subgroup of C_4 x C_6 which is isomorphic to C_2 x C_2 x C_3.

[true ingot]

Ahh there is? Alright then

[south patrol]

It is helpful if yu knw something abuot C2 x C3

[true ingot]

I got it, C2 x C3 is isomorphic to C6 right?

I have a new problem I am trying to solve now, quite different from the previous one.
I am given this presentation for SL(2,Z)

And I would like to calculate its abelianization

Which, to my understanding simply amounts to adding a relation for commutativity?

From which I would conclude that the abelianization is C4 x C6. However, a few results I found on the web claim that it's C12

[coral spindle]

You have seen x^4 = 1 and y^6 = 1 and concluded C4 x C6 but you have not paid attention to x^2 = y^3

[true ingot]

So the fact is they are not independent I suppose? But I am not sure how to account for that (it doesn't seem like I can, for example, explicitly express x in terms of y). Let me try a few things

Would the following be correct:
$\langle x,y | x^4=1, y^6=1, xy = yx \rangle \cong C_4 \oplus C_6$. But the group we are actually computing is $C_4 \oplus C_6$ with the additional relation $x^2 = y^3$, which is $(C_4 \oplus C_6) / \langle 2,3 \rangle$?

[cloud walrus]

Faputa

[true ingot]

Which then really is $C_{12}$

[cloud walrus]

Faputa

[true ingot]

Is there a better way of doing this?

[coral spindle]

Yup

Idk how it could get better than this method lol

This is very nice

[true ingot]

Alright, thank you

Is there any easier method to find that $C_4 \otimes C_6 / \langle (2,3) \rangle \cong C_{12}$ than the following:

[cloud walrus]

Faputa

[true ingot]

We know it's an abelian group of order 12, so the only two options are $C_2 \otimes C_2 \otimes C_3$ and $C_{12}$. Since $(1,1) + \langle (2,3) \rangle$ has order 12 (this was basically just brute force to find), we know it must be isomorphic to $C_{12}$

[sly frost]

the universal property of the quotient group gives you a canonical projection from the group being divided into the quotient, not the other way around

idk where you pulled that from

[cloud walrus]

Faputa

[coral spindle]

Are you sure you mean the tensor product \otimes?

If so we're talking about a very different problem

[sly frost]

no, it should be ⊕

like he wrote in his previous messages

[true ingot]

Yes sorry I meant \oplus

[kind temple]

my G is the G/N in this picture. my F(S_G U S_H) is the H

[sly frost]

notice what the kernel of the homomorphism has to be

[kind temple]

which one?

[sly frost]

the homomorphism from F(S_G) into F(S_G ∐ S_H) is injective

the kernel is trivial

or the homomorphism is trivial, but that case doesn't interest us

but we are assuming the possibility that G is the quotient of F(S_G) by some nontrivial subgroup of relations

[kind temple]

q : F(S_G) —> G is a quotient map. the inclusion map is a map F(S_G) into F(S_G U S_H) so by the universal property of the quotient, the inclusion descends to a unique map G —> F(S_G U S_H) for which the inclusion factors through it via q

where have i applied the universal property incorrectly?

[sly frost]

it says here clearly that f: G → H has to be a homomorphism with N ⊆ ker f

if the quotient is G/N

what is the kernel of the inclusion F(S_G) ↪ F(S_G ∐ S_H)

[kind temple]

how r u supposed to do this then

thanks for pointing this out btw

F(S_G) to G * H works if you follow the injection through pi.

then the kernel of that map contains the normal closure of R_G (by construction). then you get a map directly into G * H from G

[sly frost]

yes

[kind temple]

was just kind of blindly smashing diagrams together without reading their hypotheses

i have learned 🙃

[sly frost]

I mean in general it's pretty easy to see that if A generates G then you have a map G → F(A)/ncl(R) where you map elements of A to themselves and the rest follows the homomorphism property

you have an isomorphism G ≅ F(A)/ncl(R) by definition

[kind temple]

thats the inverse of the induced map of F(A) —> G from the universal property of the free group

[sly frost]

nope

[kind temple]

S_G —> F(S_G) —> G gives you a presentation of G with kernel ncl(R_G)

[south patrol]

Yes

[sly frost]

if you have a map φ: F(A) → G then you have isomorphism F(A)/(ker φ) ≅ G

φ doesn't have an inverse but F(A)/(ker φ) → G does

[kind temple]

that’s why i said inverse of the induced map

[south patrol]

The point is it is more natural to map out of the free group

[kind temple]

what else am i mapping out of?

[sly frost]

wdym induced map

[south patrol]

I am agreeing with you c squared

Lol

[kind temple]

o

lul

[kind temple]

this isomorphism coming from the first iso theorem

is what i mean by induced map

[sly frost]

I guess that makes sense

[kind temple]

good enough for me

i thought i phrased it well

[sly frost]

yes, I misunderstood you slightly and wanted to be a bit more precise

[kind temple]

again, thanks for pointing the initial error out to me

appreciate all the feedback from you guys

will be sure to check the conditions on uni. props. b4 using them next time

[topaz solar]

You can prove this for linearly ordered embedding things easily model theoretically

[mighty kiln]

Is this like "∀∃-axiomatizable implies union of sequence of embeddings is a model" or something along these lines

[topaz solar]

Simple iff $\forall x.\forall x’ \bigvee_n \exists_n \bar y . x’ = \prod_i^n x^{y_i}$

[cloud walrus]

Sharp-Malliaris regularity lemma

[topaz solar]

So yes

I.e. if the simple group <Inn(G)•x> = G for all x :3

This isn’t AE, because it’s not first order, of course

But it doesn’t have to be a sequence of embeddings either

[mighty kiln]

Is this infinitary logic

[topaz solar]

Yeah

But the same trick applies, pick a pair x, x’, it’s contained in some G_i, so reachable by conjugations

[topaz solar]

Oh right, add in a v for x=1 but uh

Close enough

[mighty kiln]

This looks like a useful algebra tool

Do you have recommended readings for category-ish model theory stuff

[topaz solar]

This is pretty easily doable by identifying things w/ the image in the colimit here but uhh

I know Baldwin’s Categoricity has a bit of AEC-ish stuff, I think Makkaï has a thing for accessible ones

But idk anything much about category-ish model shenanigans

[mighty kiln]

Okie, thanks for the pointers

[nimble folio]

Let $\varphi:R \rightarrow S$ be a surjective homomorphism of rings. Prove that the image of the center of $R$ is contained in the center of $S$.
\begin{proof}
Given $x \in Z(R)$, $\varphi(x) \in \varphi(Z(R))$. Since $\varphi$ is surjective, for all $s \in S$, there exists an $r \in R$ such that $\varphi(r) = s$. Observe that:
$$s\varphi(x) = \varphi(r)\varphi(x) = \varphi(rx) =\varphi(xr) = \varphi(x)s.$$
Thus $\varphi(x) \in Z(S)$ which gives $\varphi(Z(R)) \subseteq Z(S)$.
\end{proof}

[cloud walrus]

clubsoda14

[nimble folio]

Trying to figure out why x in Z(R) implies φ(x) in φ(Z(R)). Can anyone explain?

It is not obvious for me

[rocky cloak]

phi(Z(R)) means things of the form phi(x) for x in Z(R). That's just what that notation means

[nimble folio]

so theres nothing to show

[rocky cloak]

No, or you have to show phi(x) is in the center, as they do

[nimble folio]

eesh to be honest a lot of this stuff from discrete really confuses me

in general if f is a map from X to Y, then does x in X imply f(x) in f(X)?

[rocky cloak]

Yes, f(X) aka the image of X by f is exactly the set of things of the form f(x) for x in X

[nimble folio]

got it

i wasnt comfortable with the notation but thats good to know

Thank you!

[tardy hedge]

https://www.youtube.com/watch?v=nrOaFxNex7U

what do yall think about Weinstein bringing up group homomorphisms around minute 23 lmao

I dont actually think it was relevant because it seeemd like Terrence was asking a question about how addition relates to multiplication but then Weinstein's argument was using groups ... shouldnt he have responded with a ring theory POV instead since that is the structure terrence is talking about

i dont see how Weinstein saying how real numbers under X is isomorphic to some real numbers under addition is relevant

[coral spindle]

I think anyone giving Terrence Howard the time of day is very silly indeed

[tardy hedge]

Lol

[coral spindle]

Anyway yes you are right, clearly the disagreement (lol) is about ring structure (god why am I pretending this is so intellectual)

[tardy hedge]

😂

[coral spindle]

The real problem is that Howard is attention-seeking and this is his way of staying in the spotlight. \end{document}

[tardy hedge]

yeah this talk is genuinely awful

i thought it would be a bit interesting or entertaining at least

[languid trellis]

But what does it mean to multiply.... to increase, right? So how is 1x1 = 1? 1x1 is at least 2...

[tardy hedge]

Yooooooo

[delicate bloom]

the thumbnail gives me too much cringe to even consider clicking it

[rotund aurora]

it's not intelectual if they don't discuss why negative times negative is positive

[untold basalt]

how can I prove that $x^{p-1}+x^{p-2}+\dots+x+1$ is irreducible over $\mathbb{Q}$?

[cloud walrus]

Heywood Jablome

[coral spindle]

Hint: if that polynomial is f(x), what is f(x+1)?

Remember the formula for a sum of powers

[barren sierra]

(rational root theorem go brrrr also)

[delicate bloom]

unfortunately I don't think that'll work unless there's some magic trick I'm not seeing, since it could factor into quadratics or cubics etc that are nonlinear

[barren sierra]

Aren't the only possibilities over Q ||-1, 1|| and it's clear that evaluating that polynomial at both is nonzero?

[delicate bloom]

try this polynomial:
x^4 + x^3 + 2*x^2 + x + 1

[coral spindle]

But this does not imply irreducibility

(x^2+1)^2 also has no rational roots

[delicate bloom]

that only rules out linear factors, not quadratics, cubics, etc

[barren sierra]

Oh wait I'm dumb

[delicate bloom]

spoiler this is ||(x^2+1)(x^2+x+1)||

[barren sierra]

idk why I got into my head that it was asking about the polynomial splitting

[delicate bloom]

all good 😎

[south patrol]

No

I always wonder what the nicest way to show that cyclotomics are irreducible is

I think I remember there being a funny slightly fancy way

[coral spindle]

By definition

But yeah these particular ones I always knew via ||Eisenstein|| spoilers

[south patrol]

Oh yeah this

https://math.stackexchange.com/a/2231360/932674

[south patrol]

Yes I think this is how any sane person does them

[south patrol]

To be fair, reasonable lol

I mean then the best way to compute the degrees of cyclotomic extensions

[coral spindle]

So wise...

[prisma ibex]

idiot's smart person versus smart person's idiot

[rapid junco]

which is which

[wanton stratus]

Terrence Howard is full of BS

Like,

he says things like "Tone of the Sine Curve"

What's tone?

[delicate bloom]

don't waste your time

[south patrol]

Lmao true

[chilly radish]

I initially read terrence howard as terrence tao

[sterile condor]

Is there a difference between a monoid and a group or is it just terminology?

[coral spindle]

There is a HUGE difference.

Every group is a monoid, but very very few monoids are groups.

What makes you think they're the same thing?

[mighty kiln]

Elements in monoids may not have inverses

[tardy hedge]

whats the context from your bio lmao

physicists

[next obsidian]

YOOO PUPPY MONKEY BABY

[sterile condor]

I may or may not have forgotten group theory, specifically the inverse property.

[leaden heart]

nice gregory edgeworth pfp

[chilly radish]

Very observant

[cloud walrus]

Spamakin🎷

[cloud walrus]

Spamakin🎷

[chilly radish]

How do you define algebraic closure?

[barren sierra]

Splitting field of k[x] over k

[chilly radish]

How is that different from all Algebraic elements over k?

Like the fact it's contained in K sure but K has all of them

[barren sierra]

Uh

Good question

Maybe this is why I felt I was going in circles?

I guess I'm done idk it feels like I proved nothing (but the statement seems very nothing)

[chilly radish]

I can't think of a reasonable definition of Algebraic closure under which this isn't immediate

[barren sierra]

Yea

Ok thanks for giving my sanity back ig

[tardy hedge]

https://mathoverflow.net/questions/32637/when-are-modules-and-representations-not-the-same-thing

How long does it take to get to this level of understanding of algebra

?

I would love to one day really see and understand these algebraic structures and how they all intimately relate to each other but being at the level that these guys in the answer are at seems soo far off!

Although I am happy that I nonetheless have seen myself learn a lot especially compared to when i first looked at group theory last year

[barren sierra]

am I going nuts or is this missing a squared for each product term

cause otherwise how is this symmetric in the r_i >_>

[delicate bloom]

I'm guessing they would have say, (r_1-r_2)(r_2-r_1) as terms

[barren sierra]

oh bruh

ok yea then that checks out

[delicate bloom]

cool

[nimble folio]

Let $S$ be a subring of $R$ and let $I$ be an ideal of $R$. If I define the map $\varphi:R \rightarrow R/I$, why is $\varphi(S) = (S+I)/I$?

[cloud walrus]

clubsoda14

[nimble folio]

I don't understand why

[mighty kiln]

Are yous struggling with φ(S) ⊂ (S+I)/I or φ(S) ⊃ (S+I)/I

[nimble folio]

I never thought about it like that

What do elements of (S+I)/I look like?

[mighty kiln]

They are cosets of I

[nimble folio]

Ok

Im supposed to show that (S+I)/I is isomorphic to S which I'm comfortable doing

[tardy hedge]

It is?

[mighty kiln]

Isomorphic to S?

[nimble folio]

Oh I left out a crucial piece of info

S ∩ I = {0}

So (S+I)/I ≅ S/(S ∩ I) ≅S/{0} ≅S

by the second isomorphism theorem

[mighty kiln]

Oh S is a subring

[tardy hedge]

Noice

[nimble folio]

But I don't understand why φ(S) = (S+I)/I

[mighty kiln]

But for any subset S of R you have φ(S) = (S+I)/I too

[tardy hedge]

Cant u just look at the elements and work it out pretty directly?

It doesnt seem that hard is it?

[nimble folio]

Oh

[tardy hedge]

Well, right?

[mighty kiln]

You could try directly bashing out both directions

[tardy hedge]

Ya thats how i thought to do it how else would u

[nimble folio]

S + I = {s + i | s ∈ S, i ∈ I}, so an element of (S+I)/I looks like (s+i) + I = s + I

which is just an element of φ(S)

[mighty kiln]

Yup

[nimble folio]

Wowwww

It was so simple I don't know why I was getting so confused

Thanks guys

[tardy hedge]

Honestly it be like that sometimes

I feel like an idiot 99% of the time and then u see it and then it seems obvious

It always feels so extreme like that lol

[nimble folio]

Yea

I'm doing problems in Dummit and Foote and the phrasing also kind of threw me off

[tardy hedge]

What chapter are you on?

7? Rings?

[nimble folio]

Let $S$ be a subring of $R$ and let $I$ be an ideal of $R$. Prove that if $S \cap I = 0$ then $\bar{S} \cong S$, where the bar denotes passage to $R/I$

[cloud walrus]

clubsoda14

[nimble folio]

Yes this is chapter 7

[tardy hedge]

Yea that wording is weird to me too

[marsh scaffold]

Why exactly is C not empty here

It looks like they're doing some sort of replacement in the process above

If AnB is non empty then we can see that C is non empty

But what if it isn't

[tribal moss]

You can always that S=Ø and get at least one element of C that way.

[kind temple]

pick any single element of A and map it to a single element of B

[marsh scaffold]

Oh we can define functions on empty sets?

[kind temple]

yes

there is only one such function whose domain is the empty set

[tribal moss]

Wait, the sentence starting with "We also impose ..." is part of the definition of C even though it follows a full stop?

If so, you need to set S = A cap B, and f the identity on A cap B in order to get at trivial element of C.

[marsh scaffold]

And they did ?

[marsh scaffold]

Wait i don't understand from here?

[kind temple]

the definition is phrased poorly

is all they were saying

[marsh scaffold]

Okay

[kind temple]

this is the important part

[marsh scaffold]

Then how does that guarantee the linear independence part of (A-S) U f( S)

[tribal moss]

If S is a subset of A and f is the identity on S, then (A-S) cup f(S) is just A, which is linearly independent by assumption.

[marsh scaffold]

I agree

I was saying about the mapping a single element part

I think the empty set map is our only option for a candidate?

[tribal moss]

Are you assuming A and B are disjoint when you say that?

[marsh scaffold]

Yes

[kind temple]

even if you map a in A to something in span(a) cap B, you still have (A - {a}) U {ca} being linearly independent

[tribal moss]

Right then. Though, of course, the third part of (5) (with the hint) asks you to prove how you can make more elements of C if you know one where S is not A.

[marsh scaffold]

I don't understand...B is any set of generators ...this is just a process of replacement right, how do we guarantee that the one that b replaces in A still results in a LI set

[tribal moss]

You do need to be a bit careful which element of B you replace.

[tardy hedge]

Somehow im confused on “the ring R can always be embedded as an R-submodule of the S-module S”.

[marsh scaffold]

I see, why does such a replacement always exist

[tardy hedge]

We are assuming btw R is subring of S

Im a bit confused on saying R-submodule of the S-module

[mighty kiln]

S is an R-module

And R ⊂ S is an R-submodule

[tardy hedge]

We cant say its an S-submodule because we cant generally extend scalars right

[mighty kiln]

Yea

[tribal moss]

The hint seems to sketch an argument why there's always an element of B that we can use to replace v.
(I don't quite feel confident I can start explaining that right away, though).

[marsh scaffold]

Say A is the empty set and B is {0} and we are taking about the trivial vector space

There is nothing to replace in this case

[tribal moss]

In that particular case there's nothing to do: (Ø, empty function) is already maximal in C.

[marsh scaffold]

So as we don't know about our vector space i think we have to start from the empty function (in the case AnB is empty)?

[kind temple]

forget what i said.

[tribal moss]

We start from the identity function on A cap B -- when A cap B is empty, that is the empty function.

[marsh scaffold]

Alright

Thank you @tribal moss and @kind temple

I never knew the existence of empty functions

Now I do

[tribal moss]

They're not very interesting, and mostly just useful for avoiding a need to have pesky "unless such-and-such is the empty set" conditions in your theorems.

[tardy hedge]

And also, when we say R is a subring of S, we can also more generally just say there is a ring homomorphism from R to S and everything still works out in this way right

[mighty kiln]

Then you get a morphism of R-modules R→S

[tardy hedge]

In what way is this different

Only difference is that if its not an injection right

For R a subring of S, how is the ring action of S-modules related to the action associated with R-modules?

I know for a given ring R you can have many different R-module structures. So for S which has R as a subring …. How is the action of S related to R

Is my question specific or well defined enough? Im not even sure

[mighty kiln]

The action by R is just the restriction of the action by S

[tardy hedge]

Ohh ok

[mighty kiln]

That is, if you have f: R→S and S acts on something, then r ∈ R acts as f(r)

[tardy hedge]

Yeah

So the S action must be specified beforehand i guess

[mighty kiln]

Yea

[tardy hedge]

When I look at something like a Q-module, i mean, how do i know what actions it could have

I mean i know Q is a field so that module is a vector space so its probably limited or something

But im just not sure of the specifics

I suppose im also not sure on how the field is related to the vector space structure in general, in what way does the field matter

Because to be honest even when i took the course in “abstract” linear algebra, even still the fields we worked with were always just R or maybe sometimes C

So im not even really sure in what way is the field relevant

[spice minnow]

Im not sure if Im interrupting a previous discussion (and I apologize), but I just had a very quick question: is every simple semigroup also D-simple? I have proven the other implication, but I think my brain is starting to melt and I can't find an example where this fails.

[mighty kiln]

Many results from linear algebra carry over

Though some results require the ring to be nicer

For instance every vector space is classified by the cardinality of its basis, and more generally the finitely generated modules over a PID have a simple classification

[tardy hedge]

For this free abelian group, we do not specify what (si,ni)+(sj,nj) is, right?

Like, its just that

I didnt cover free groups before so just making sure

[tardy hedge]

I meant more specifically for a vector space how does the field interact with the elements , what kind of ring actions are allowed i guess?

[tardy hedge]

Ok sorry my question is not well posed, of course the ring action needs to satisfy the module axioms, i guess im trying to ask like, for say a Q-module, what kind of Q x N -> N maps are there?

[coral spindle]

What do you mean by that

What specific assumptions are there on this map

[nimble folio]

Let $\varphi:R \rightarrow S$ be a ring homomorphism. Prove that if $J$ is an ideal of $S$ then $\varphi^{-1}(J)$ is an ideal of $R$.

My question in this: Assuming that $J$ is an ideal of $S$, does it follow that $\varphi(x-y) \in J$ implies $x-y \in \varphi^{-1}(J)$? Wouldn't it be required that $\varphi$ is surjective? Or is this just by definition of pre-images?

[cloud walrus]

clubsoda14

[coral spindle]

That is by definition.

This has nothing to do with ideals or rings, this is just how preimages work.

[nimble folio]

Ok

So if $f:X \rightarrow Y$ is a map between two sets, then by definition $y \in Y$ implies $f^{-1}(y) \in f^{-1}(Y)$

[coral spindle]

Firstly: you mean f, not phi presumably

[nimble folio]

yes lol sorry

[coral spindle]

Secondly, no: f^-1(y) is not defined.

[nimble folio]

Uhh

[cloud walrus]

clubsoda14

[nimble folio]

Why

[coral spindle]

Consider the map f : Z → Z defined by f(x) = 0

What is f^-1(0)?

What is f^-1(1)?

[nimble folio]

Ah

[tribal moss]

f^-1({y}) will be defined, though.

[coral spindle]

In general the correct definition is $f(x) \in Y$ iff $x \in f^{-1}(Y)$.

[cloud walrus]

Boytjie

[coral spindle]

This is a definition of the set f^-1(Y)

When f is a bijection, this aligns with f^-1 applied to Y (i.e., the image of Y under f^-1)

[nimble folio]

I see

[coral spindle]

Again this has nothing to do with algebra, this is just elementary set stuff

[nimble folio]

Yes I know

My discrete is really bad

I've always had trouble with images/preimages and stuff

[coral spindle]

Well now you know the definition so you can apply that in the future

[full nymph]

What's the geometric interpretation of the conjugacy classes of D_3?

[coral spindle]

Have you calculated the conjugacy classes? What do you notice about them?

[full nymph]

This is what I got. Alpha is rotation, b is reflection

[coral spindle]

So it looks like the conjugacy classes sort things into rotations and reflections :)

[full nymph]

Ah got it. The elements in the last class are all reflections

[coral spindle]

And the elements in the second are all rotations. By the same angle, even, if you consider one to be clockwise and the other anticlockwise

[full nymph]

Hm but e there is also a^3, so why is it not in the same class as a and a^2?

[coral spindle]

1. the identity is always in its own class
2. I gave a pretty big hint just above

[full nymph]

Ah understood

[marsh scaffold]

Is the solution like consider the set of indicator functions on elements of I

They form a Li set

[marsh scaffold]

And so from this their cardinality is ....

Now all i need is to show that if a LI set has cardinality equal to that of the Basis then it is also a basis

This is where I am having difficulties

[marsh scaffold]

And I don't think that's true

[topaz solar]

Yes

[topaz solar]

This is false!

Because I can exclude a 1 dimensional subspace by removing 1 basis element easily

[marsh scaffold]

Yeah I got like {1,2,x,x^2...} And {1,x,x^2..}

[topaz solar]

That’s not linearly independent but yeah

But watch this

You have a linearly independent subset of size bigger than I

So you’re done

[marsh scaffold]

But i haven't proved that yet huh

[topaz solar]

That said, the set of indicator functions is, in fact, not linearly independent

[marsh scaffold]

Huh

[topaz solar]

Indicator for {1, 2} = {1} + {2}

Right

So all the finite subsets are spanned by (indicator functions for singletons)

[marsh scaffold]

Yea...I agree

So I would get all the kinds of functions that are non zero only on a finitely many of Is

Taking the indicator functions?

[marsh scaffold]

Why is it not LI tho ?

For any Finite subset , given a linear combination we can show that all the scalar are 0?

Just make the others vanish

[topaz solar]

Again, $1_{{1,2}} = 1_{{1}} + 1_{{2}}$

[cloud walrus]

Sharp-Malliaris regularity lemma

[topaz solar]

So those 3 already aren’t linearly independent

[marsh scaffold]

Oh so here as it's singletons so it's LI?

I am taking singletons here

[topaz solar]

Singletons give you a linearly independent thing yes

And that gives finite subsets

[marsh scaffold]

Okay but how do I show that cardinality is strictly bigger?

[topaz solar]

Now, consider a function f: I -> K

[marsh scaffold]

Okay

[topaz solar]

That’s not gonna, in general, be of finite support, but that is a dual space element of a K-vector space with basis I

[marsh scaffold]

Sure

But i still don't get where we proved the cardinality to be strictly bigger

[topaz solar]

Ah yeah that’s just the “here’s how it related to the last part”

But uh, have you seen the cantor diagonalization thing?

[marsh scaffold]

Yes

[topaz solar]

We can do something similar I think, just with a bit of tweaking to exclude things like how we can reach any finite subset by adding singletons

[marsh scaffold]

Hmmm