groups-rings-fields

[steep hull]

Well we can have division in a ring

So this would be a counterexample

Like nothing is breaking ring axioms here

[topaz solar]

well of course, i just like to stay away from division as possible since
you know
doesn't always play well

[steep hull]

Doesn’t this mean the result is false if the sequences aren’t bounded

[topaz solar]

what's this a counterexample against, btw?

like, which questionable statement?

sorry, smoothbrain

[steep hull]

n+1/n~n, n+1/n~n, but (n+1/n)^2-n^2 tends to 2

[topaz solar]

yeah, hm
suppose it does

makes sense

well, good to know

[steep hull]

Anyway, if boundedness is added, then the result holds true for the reasons of my old proof

[topaz solar]

yeah, just makes it more obvious where a mistake is in something I'm still working on

yeah i definitely missed something there, not a hard mistake to fix (i hope)

yeah i put something on the wrong side of a descending chain

fuck

welp, helpful at least to know I f'd up

on the plus side, thanks to this, I now should (assuming no other mistakes, which there likely are a couple, I'll have to comb back through later) have an l-group to work with, problem now is extending multiplication since clearly pointwise is messy with unbounded sequences (or maybe i can restrict to bounded sequences without losing anything, I haven't figured out yet, since i'd probably need a weaker boundedness thing when throwing down x^+ anyway)

Things work nicely when adding an element which is bounded above by a positive element at least

[steep hull]

Anyway, with the restraint of boundedness, the proof isn’t that messy. We have $a’_n=a_n+r_n$ and $b’_n=b_n+s_n$. Then $a’_nb’_n-a_nb_n=a_ns_n+b_nr_n+r_ns_n$. We show that these also converge to $0$. Since $s_n$ tends to $0$, it is bounded above and below by $s’_n$ and $s’’_n$, and $a_n$ is bounded above and below. We know that $s’’_n<0$ because it is monotonically increasing and a positive value would block $\inf s’_n$ from being $0$. Letting $\sup |a_n|=a’$, we get $a’s’’_n<a_ns_n<a’s’_n$, so it tends to $0$, and these are monotonically increasing/decreasing for because $a’>0$ and because $s’_n, s’’_n$ are monotonically increasing/decreasing. The same argument works for any pair of sequences where one sequence is bounded in absolute value and the other tends to $0$, so the other two tend to $0$ as well. Hence the whole thing tends to $0$ and we’re done.

[topaz solar]

Anyway, with the restraint of boundedness, the proof isn’t that messy. We have $a’_n=a_n+r_n$ and $b’_n=b_n+s_n$. Then $a’_nb’_n-a_nb_n=a_ns_n+b_nr_n+r_ns_n$. We show that these also converge to $0$. Since $s_n$ tends to $0$, it is bounded above and below by $s’_n$ and $s’’_n$, and $a_n$ is bounded above and below. We know that $s’’_n<0$ because it is monotonically increasing and a positive value would block $\inf s’_n$ from being 0. Letting $\sup |a_n|=a’$, we get $a’s’’_n<a_ns_n<a’s’_n$, so it tends to $0$, and these are monotonically increasing/decreasing for because a’>0 and because s’_n, s’’_n are monotonically increasing/decreasing. The same argument works for any pair of sequences where one sequence is bounded in absolute value and the other tends to 0, so the other two tend to 0 as well. Hence the whole thing tends to 0 and we’re done.

[cloud walrus]

Darkrifts:
Compile Error! Click the reaction for details. (You may edit your message)

tree3:
Compile Error! Click the reaction for details. (You may edit your message)

[topaz solar]

yeah, with some slight tweaking that might work

[steep hull]

Rigor is hard

Oops

I think this is pretty rigorous though

With some minor informalities

[topaz solar]

there's a few rigor things

like using |x| and sup

which, ya know
might not exist

[steep hull]

|x| definitely exists

[topaz solar]

since |x| = sup { x, -x }

might not exist

this is only a partially ordered ring, not lattice-y

[steep hull]

It’s the element of x,-x that is greater than or equal to 0

[topaz solar]

maybe neither is

partially ordered, not totally ordered

makes thigns messier

but like, can you say whether or not X in Z[X] is less than or greater than 0?

but you can still treat it as a normal element by using other laws and having $\dots \leq x - 1 \leq x \leq x+1 \leq \dots$

[cloud walrus]

Darkrifts:

[topaz solar]

just can't necessarily compare to 0

think something like the star game * in CGT

(not the best example, but i'm tired)

[steep hull]

x<=0 implies x+(-x)<=0+(-x) or that 0<=-x.

[topaz solar]

well, what if neither $x\leq 0$ or $0 \leq x$ hold?

[cloud walrus]

Darkrifts:

[topaz solar]

you can't compare {a} and {b} under the \subset partial order

[steep hull]

Oops I don’t know definitions yikes

[topaz solar]

yeah

Partial order just has antisym, transitive, and reflexive

Total has a <= b or b <= a for any pair

Well order has a least element in each nonempty subset

naturals are well ordered, integers (in the usual order) aren't

[steep hull]

We do have an abs value in an I-ring

[topaz solar]

We do, yes

but we aren't in an l-ring unfortunately

which is v annoying

but

you can still bound it

just not with |x|

i'll work on it

Nows the fun part: finding a suitable multiplication

pointwise just works fine enough* if bounded, so good enough there [*pending more rigorous formulation of argument]

when unbounded idk i'll just shoot myself in the foot later

hm

[steep hull]

Yeah rip my formalism

I’ll have to change stuff a little

[topaz solar]

definitely

but it's a start

so, just multiplying the things out and because r_n, s_n go to 0, as well as boundedness of a, b, you can say a good bit

[steep hull]

What does boundedness mean then? There exists an r that is greater than all the a_n?

[topaz solar]

yeah

[steep hull]

So that there is an relation

[topaz solar]

by a simple monotonicity argument you can kinda rig the whole damn thing up

thanks

[steep hull]

I guess things aren’t as bad as I thought

Yeah

Np

[topaz solar]

problem is it doesn't always work, but works well enough in the nice case

[steep hull]

It definitely doesn’t work in the unbounded case

So yeah

[topaz solar]

at least not in this sense anyhow

sequences gotta be bounded somewhat

i'll look into more precise "when"'s later

[steep hull]

Okay

What is this for

[topaz solar]

So, say $R$ is an l-ring, and $R\subset R^\prime$ are po-rings (restriction of your order giving you a lattice)

if you took an element $\phi \in R^\prime\setminus R$ and appended it to $R$ (closing as necessary, etc etc whatever), you don't get an l-ring back out

[cloud walrus]

Darkrifts:

[topaz solar]

so, i'm working on creating an l-ring extension in the small collection of cases where it behaves nice enough to allow it

[steep hull]

Huh well I get all the definitions now, which is nice

[topaz solar]

i mean, the whole boundedness thing here is obviously a tad bit weaker than at first glance, since there are some cases where it would work

but bounded is sufficient

[steep hull]

Yeah, it should be sufficient

[topaz solar]

if your \phi isn't bounded above by some positive element, then you get some problems with boundedness

so this construction i have atm doesn't work

since you cant do a downward monotonic thing without adding infinitely large things (a la adding \omega) for positivity, and that really messes things up since bounding things gets a bit messier

like the n+1/n example you gave doesn't work even if you bound above with shoddily added infinites

anyhow, works nice enough for some cases now

still can't make the lattice extension for R[X]_\ell tho

[tribal pasture]

Let H be a normal subgroup of prime order p in a finite group G. Suppose that p is the smallest prime that divides the order of G. Prove that H is in the center Z(G).

Take an x ∈ H-{1} and consider C(x). Since H is normal we have that C(x) ⊂ H => |C(x)| ≤ |H| = p.
Moreover C(x) | G thus we have that C(x) = 1 or p since the smallest prime that divides G is p.
Assume |C(x)| =p. Then we have that there exists a g such that gxg⁻¹ = 1 => x = 1 thus contradiction. Hence we have that |C(x)| =1 <=> x ∈ Z(G).

Can somebody tell if this is correct?

[quiet cave]

What's C(x)

[tribal pasture]

Conjugacy class of x = { gxg⁻¹ | ∀g ∈ G}

[quiet cave]

How can a conjugacy class equal a number

[tribal pasture]

Thanks edited.

[wind steeple]

Why there exists g st gxg^-1 = 1 ?

[tribal pasture]

If |C(x)| =p and we know that C(x) ⊂ H where |H| = p we have that C(x) = H. Since H is a subgroup it contains 1. Thus C(x) also contains 1. By defintion, there exists a g such that gxg⁻¹ =1

[quiet cave]

Probably add this line to the proof

[tribal pasture]

Oh okay will do! Do you think the rest is okay?

[wind steeple]

No it doesn't work

Uh wait

Mh yes indeed lel

[quiet cave]

[wind steeple]

It works x)

[tribal pasture]

Yeshhuuu. Thanks for checking it!

[quiet cave]

Be careful with notations tho

I know that algebraists like to shorthand things a lot, but it might lead to some confusion

[tribal pasture]

Can somebody also tell me when its possible for N(H) = H where N is the normalizer and H is a subgroup?

[fading wagon]

so normaliser is set of x such that xH=Hx

[tribal pasture]

If H is normal we have that N(H) =G where G is the parent group.

[fading wagon]

https://groupprops.subwiki.org/wiki/Self-normalizing_subgroup

[tribal pasture]

Thanks!

[elder valley]

i've seen homogeneous polynomials plenty before, but i'm trying to grok Hartshorne's definition of a homogeneous ideal. He writes the ring $S = k[x_0, \ldots, x_n]$ as a graded ring $S = \oplus_{d \geq 0} S_d$, where the abelian groups $S_d$ consist of all polynomials of total degree $d$. i understand this.

he then defines a homogeneous ideal $\mathfrak{a}$ as one which satisfies $\mathfrak{a} = \oplus_{d \geq 0} ( \mathfrak{a} \cap S_d)$. i'm guessing this is more general than just saying a homogeneous ideal is one in which every element has total degree d for some fixed d, but i'm just struggling to take it in

[cloud walrus]

Auvera:

[elder valley]

it seems this definition has the advantage of working in any graded ring. but it doesn't seem to coincide with my previous understanding of homogeneous polynomials

[outer solar]

What does this group mean?

(Z/mZ, +)

Its in some lecture notes I found but isn't formally introduced

[magic owl]

Addition of integers mod m

Do you know what that means?

[outer solar]

yeah kinda

[magic owl]

Thats all itnis

Convince yourself this forms a group

Or just look up “cyclic groups”

[feral crystal]

@chilly ocean uh, those numbers are the sequence of i, i^2, i^3, ..?

[feral crystal]

well, k^2 = -1 if i recall correctly,
so if i multiply every one of them by -k,
i get -k^2, -k^2*i, k^2, and k^2i

--> 1, i, -1, -i

@chilly ocean

i^2 = k^2 = j^2 = ijk = -1, right?

[final gulch]

what you just did was multiply each of the four numbers by 1

[feral crystal]

@final gulch -k is 1?

[final gulch]

for one thing, the only solutions of k^2 = -1 in C are k = i and k = -i so I will assume that k is one of these and ignore the long equation you wrote at the bottom which is not possible to solve in C
but my main point is that you said you multiplied by -k but really you multiplied by -k^2 = 1

[feral crystal]

@final gulch isnt k^2 = -1 because
1 is a matrix of
1 0
0 1

and k is a matrix of
0 i
i 0

so, k^2 =
-1 0
0 -1

which is the same as -1?

this is about quaternions

[final gulch]

https://cdn.discordapp.com/attachments/496785752936546304/649188067277799425/unknown.png

I thought this was your problem

it has nothing to do with quarternions

[feral crystal]

oh hm, the whole chapter had to do with quaternions, that's why

[final gulch]

you are just looking for four complex numbers

[feral crystal]

maybe im stepping into the wrong direction

yeah

well, if i remember correctly,

wait, a rotation that map the square onto itself, implies that these four vertices end up having the same vertices?

in no particular order?

[final gulch]

the order is not completely arbitrary

for example you can't swap 2 vertices and leave the other 2 fixed with a rotation

[feral crystal]

what if i multiplied each vertex by its opposite?

1 by -1, i by -i, -1 by 1, and -i by i?

[wild sapphire]

I have a question. Given the continuous function f mapping R to R, where f(r) = r^2 for all r in Q. What is f(sqrt(2))?

I don't know whether to use the definition of continuity, or the fact that Q and I are dense in R, or both. I just don't really know where to start

[mild laurel]

Use all of those

I mean, you know what the answer should be right

[wild sapphire]

yeah, of course. 2

[cloud walrus]

QuickMaffs:

[delicate bloom]

I'd probably start by listing a bunch of simple random facts and juggling stuff around until I saw pieces that fit

[tribal pasture]

Tried that for a 30 minutes......

[delicate bloom]

did you find anything that felt potentially good but you're not sure?

[tribal pasture]

No I mean I derived thousand of new identities but none of them let me proved that b^2 = 1

[golden pasture]

Try to get something like $b=a^xb^{-1}a^{-x}$

[cloud walrus]

Ariana:

[tribal pasture]

Not sure if one can derive that for arbitrary x, since it would imply that b^2 =1 for any order of a given aba⁻¹ = b⁻¹ which seems to be a very stronger statement

[golden pasture]

<x here is odd>

b=something
b^-1=something

So just plug one into another and observe

[random quartz]

I have a question about Boole Algebra, is it here I have to ask it?

[spark bear]

[random quartz]

I have to simplify this equation using Boole Algebra, I dont know where I could start

[golden pasture]

boole algebra as in boolean algebra?

[spark bear]

@random quartz
First of all for the first two terms you could use distributive property

And yes Ariana I think so

I will write ˜B instead of B with a bar
(and same for the other elements)

So you have
˜A˜B˜C + A˜B˜C
= ˜B˜C [ ˜A + A]
= ˜B˜C [1]
= ˜B˜C

[random quartz]

Alright ! I see now, You changed [~A+A] to [1] since it act as a wire?

[spark bear]

Well it is known that for any ˜a + a = 1 for any element

Since one of them must be 1

[random quartz]

Alright thanks! 🙏

[tribal pasture]

tried that

Yep I know the square goes inside. But it didnt work here

[latent anvil]

Happy Thanksgiving everybody! I ran off to do atiyah macdonald problems after my family got to be too much and noticed something interesting: let R be a ring

• Every module over R is flat iff all finitely generated ideals of R are direct summands of R
• Every module over R is projective iff all ideals of R are direct summands of R

The first one might require commutativity

But I thought this was neat

[golden pasture]

@tribal pasture You can get b in terms of b^-1 and b^-1 in terms of b
Try to plug one into the other

[bleak abyss]

@latent anvil wait are direct summands of R???

[latent anvil]

Yes

[bleak abyss]

Oh all ideals

Oh oh oh

[latent anvil]

Lol

Did you read modules?

[bleak abyss]

Yeah

And I was like fammmmmm lay off the cocaine

[latent anvil]

I don't think that's possible for R ≠ 0

[bleak abyss]

Yeah pretty sure R^{massive fucking cardinal} kills it

[latent anvil]

Lol yeah

but what if R has an ideal with cardinality > |R|?

[bleak abyss]

[fickle brook]

[eager willow]

hey has anyone seen theory on symmetric polynomials being applied to differential operators? My idea is that you can get a sort of 'symmetric partial differential operator' by taking a taking a symmetric polynomial $f(x_1, x_2, ..., x_n)$ and naively plugging in $f(\partial_1, \partial_2, ..., \partial_n)$.

[cloud walrus]

datorangeguy:

[eager willow]

One simple result is that such an operator will always map symmetric functions to symmetric functions

and beyond that I've looked at some simple PDEs with very aesthetic solutions, and I'd like to begin generalizing

but googling for this stuff has been useless, there's so many uses of the word symmetric even in PDEs that aren't relevant

[outer solar]

If I did a group table with Z4 under addition, how does one get 0 for 2 mod 4 + 2 mod 4

[mild laurel]

What's 2 + 2

[stone fulcrum]

In Z4, 2 + 2 = 0

@outer solar
That's just kinda how it is

You can think of a clock with 4 numbers on it. If you wait 2 hours, then 2 more hours, the clock goes all the way around

[tribal pasture]

https://math.stackexchange.com/questions/205641/is-o-n-isomorphic-to-so-n-times-pm-i

The first answer is voted four times but a comment by a very well-reputed user claims that the technique is not valid. Is the answer valid or not?

[urban acorn]

it's kind of also true to say that $2+2=4$ in $\frac{\bZ}{4\bZ}$ but also $4 = 0$.
You just let each integer denote its residue class rather than do it with only 0 through 3.

[cloud walrus]

Intel:

[urban acorn]

which gives motivation for computing it like $2+2=4=0$

[cloud walrus]

Intel:

[urban acorn]

first carrying out the computation in Z, and then reducing mod 4

[outer solar]

Ah I see thanks everyone i suck at modular arithmetic

[urban acorn]

[chilly ocean]

How can I show that the alternating group, $A_n$, is the commutator subgroup of the symmetric group, $S_n$? The result should involve showing that $A_n$ contains all elements of the form $[a,b] = aba^{-1}b^{-1}$; what would I put as elements $a,b$?

[cloud walrus]

Aleksandr:

[urban acorn]

Let $\epsilon : S_n \to {1,-1}$ take each permutation to its sign (even = 1, odd = -1). Observe that it's a homomorphism into the group of order 2, which is commutative, and that $\sigma \in A_n \iff \epsilon(\sigma)=1$. Then $\epsilon([a,b]) = \epsilon(a)\epsilon(b)\epsilon(a^{-1})\epsilon(b^{-1})=\epsilon(aa^{-1})\epsilon(bb^{-1})=1$, and therefore $[a,b] \in A_n$.

[cloud walrus]

Intel:

[urban acorn]

Essentially, it's because the codomain of the homomorphism is commutative, so the commutator fails to survive when reducing by the kernel of the homomorphism.

not sure how to prove the other direction, but i haven't thought of it that much

[cobalt quarry]

When you have a exact sequence of modules $0\rightarrow M \rightarrow 0$ and you fail to realize that M=0
🤦‍♂️

[cloud walrus]

Prof:
Compile Error! Click the reaction for details. (You may edit your message)

[cobalt quarry]

Wellp I need more practice

[magic owl]

You should work with

Everything from one term exact sequence to two term to three term

They each have a very specific meaning

And its good to have them memorized

@cobalt quarry

Once you get to 4term theres more variability

And at three terms you get the splitting lemma

[cobalt quarry]

I know but I was working on constructing some ridiculous counterexample
And at the end I forgot to realize that the thing is zero

[chilly ocean]

it ok u still did it :D

[cobalt quarry]

❤️ 😄

[chilly ocean]

What famous unsolved problems are there in Algebra?

[wind steeple]

Goldbach conjecture

[golden pasture]

thats algebra? feels more like nt pain

[wind steeple]

it's algebra

[golden pasture]

the weaker goldbachs proofs look a lot closer to anal nt tbh

i think a more algebraic problem that looks super simple but unsolved is like jacobian conjecture

[chilly ocean]

oo thanks

[chilly ocean]

hard to prove partial converses are really cool imo

finds all the families of cool things that are hard to prove partial converses

[mild laurel]

How do I show that localization of modules preserves ideal intersection using the fact that localization is exact?

[maiden ocean]

u dont

[magic owl]

Uh

Well ideals are kernels right

So there should be something there?

Id have to actuslly think and im at linch

[mild laurel]

Lmao

Yes ideals contained in other ideals are the kernel of the natural surjection map

[tribal pasture]

Show that char(F) = 0 or prime.

Assume char(F) = m is not 0.
Assume m is not prime. Then m = rs such that 1<r,s<m. Since 0 = m = rs => r = 0 or s =0. Since r,s< m, this contradicts that char(F) = m. Hence char(F) must be prime.

Is this correct? Seems too simplistic.

[ornate chasm]

it is

all you're doing is looking at the copy of Z/m inside your field

any subring of a domain must be a domain

[mild laurel]

How do you know that RS= 0 implies the thing?

[ornate chasm]

cuz it's a domain

[tribal pasture]

Oh... that only holds in fields of prime order?

[mild laurel]

How do you prove its a domain

[ornate chasm]

cuz it's a field

@tribal pasture what

[mild laurel]

How do you prove its a field

[ornate chasm]

by hypothesis

F is a field

[tribal pasture]

ab = 0 => a =0 or b =0 holds only in prime fields I think

[mild laurel]

Ah

[ornate chasm]

it holds in domains

because domains have no zero divisors except 0

[tribal pasture]

What is a domain?

[ornate chasm]

a ring with no zero divisors

maybe that's why you feel the proof is shaky

[tribal pasture]

Yeah so dont I have to prove that first?

which essentially amounts to proving char(F) = prime if non-zero

[ornate chasm]

no it doesn't, it's more general

you can do it directly

but yes, prove it first

field => domain

[tribal pasture]

Why does r,s < m imply r,s are in F/{0}?

[ornate chasm]

because they aren't 0

I mean

because of definition of characteristic

[tribal pasture]

And we know that how?

[ornate chasm]

it's the smallest integer that's 0

[tribal pasture]

No I mean how do we know that r,s <m = 0 imply r,s are not equal to 0. Like for example in F_6, 2 < 6 yet 2=6=0

Oh nvm, its because char(F) = m which means it is the smallest positive integer

[ornate chasm]

yes

there's no such thing as F_6

don't write that

you mean Z/6

[tribal pasture]

Oh soz

yep yep

[ornate chasm]

there's things like F_4 and F_9

they aren't Z/4 and Z/9

[tribal pasture]

Wait we have F_4?

[ornate chasm]

F_n just means "the field with n elements"

yeah F_4 has characteristic 2

it's F_2 with some more elements

[tribal pasture]

Is it Z/4Z with some additional structure?

[ornate chasm]

all finite fields are of the form F_{p^k}

no

because Z/4 has char 4

and you are proving that fields have prime characteristic

[tribal pasture]

Ah true. Thanks thanks

[mild laurel]

Okay let me repost

How do I show that localization of modules preserves ideal intersection using the fact that localization is exact?

[chilly ocean]

maybe use the fact that if $I, J \le R$, then the intersection $I \cap J $ is the kernel of the map $I \oplus J \to R$ given by $(i, j) \mapsto i-j$

and that exact functors preserve kernels

[cloud walrus]

trex:

[tribal pasture]

{a+bi | a,b in F_p} gives a field for p = 3 and p =7. Why not p =5? I understand its because while finding inverses we consider, a²+b² which is 0 for (a,b) = (2,1). But is there some algebraic explanation as in an explanation that is generalizable for other fields?

[chilly ocean]

the point is that -1 is a square mod 5, but not mod 3 nor 7

(the criterion is that -1 is a square mod p iff p = 2 or p = 1 mod 4)

therefore, if by {a+bi| a,b in F_5} you mean the algebra F_5[x]/(x^2+1), it is indeed not a field

[tribal pasture]

I am not aware of that notation. By that I mean the set {a+bi | a,b in F_5} where the multiplication and addition is inherited from complex numbers

Oh and so if -1 is a square then the iota i doesnt behave as in it does usually in complex numbers?

[chilly ocean]

okay, I see what you mean

we are talking about the same object then

the thing is that in a field, a polynomial of degree 2 has at mos two roots

but here, you have 4 square roots of -1

2, 3, i, -i

[tribal pasture]

@chilly ocean Oh I see your point. But why does that causes a problem? Is it because we lose some sort of uniqueness for the inverses because the inverse that is captured by i might also be captured by 2?

Oh okay thanks thanks

[chilly ocean]

well you actually produce some non invertibles

for example 1+2i

because (1+2i)(1-2i) = 5 = 0 mod 5

yup

this sort of ideas actually leads to a proof of the fact that a prime is a sum of two squares iff it is equal to 2 or =1 mod 4

[tribal pasture]

In prime fields, is it true that a^2 = 1 => a = 1, -1?

[mild laurel]

what does it mean to be a prime field

[tribal pasture]

As in the number of elements are prime?

[mild laurel]

Then yes that's true

But this is true for all fields

[tribal pasture]

I was just confused by an example we were discussing above where we found that -1 has 4 square roots in F_5 so I thought maybe this kinda thing also becomes true for square roots of 1

[mild laurel]

-1 does not have 4 square roots in F5 what

[tribal pasture]

Oh no sorry in F_5[x]/(1+x^2)

[mild laurel]

Read carefully what they said, and compare it with what I said

[tribal pasture]

Yep I see the difference. But that was a source of confusion for me. But thanks for clarification!

[mild laurel]

There are not-field rings where 1 has as many square roots as you want

[woven delta]

It's easy to show that a polynomial of degree n has at most n roots

I think all you need is an integral domain

Or probably ufd

Yeah you need ufd

[mild laurel]

You only need integral domain I think

[boreal summit]

I have a question, Let R be a ring and suppose e1,e2 in R are idempotent elements that commute, For which integers n is e1+e2+ne1e2 necessarily idempotent? For each int n either prove that it is idempotent or which it is not.

I don't know where to start.

[steep hull]

@mild laurel Yeah, that’s all you need. We proceed inductively. Degree one polynomials work due to the integral domain restraint. For degree n, take a root r and write P(x) as (x-r)Q(x)+P(r). Since P(r)=0, so then we have P(x)=(x-r)Q(x) with Q of degree n-1, which has at most n-1 roots by the hypothesis while x-r adds at most one root, so we’re done because of the integral domain restraint.

Should be clear now

[steep hull]

Yes, after fixing the English it essentially boiled down to that

I only assumed integral domain

I’m pretty good at typoing late at night

Yes, I messed up the wording

Are you an undergrad

[mild laurel]

How do I show that localization of modules preserves ideal intersection using the fact that localization is exact?

[urban acorn]

question about the class equation. why treat the center differently? it's trivially equivalent (and more elegant since it has less special cases) to say that $|G| = \sum_{i=1}^{n}{|G : C_{a_i}|}$ with ${a_1, ..., a_n}$ being representatives of the distinct conjungacy classes.

[cloud walrus]

Intel:

[urban acorn]

(this time including those conjungacy classes that are singleton subsets of the center)

[magic owl]

it can be useful to consider them separately, e.g. the proof of the sylow theorems. Also to some extent the way its normally written describes the class equation as a measure of how far from abelian a group is

[woven delta]

Lmao "more elegant"

[chilly ocean]

Let $K$ be a field and $f: \mathbb{N}\rightarrow K$ a function in the polynomial ring over $K$ denoted by $K[x]$. The actual polynomial function for $f$ is denoted by $f_x: K \rightarrow K$. The set of all functions from $K\rightarrow K$ is denoted by $K^K$.
Let $\psi : K[x] \rightarrow K^K, f\mapsto f_x$.
Show that $\psi$ surjective is equivalent to $K$ is finite.

[cloud walrus]

N/𝔄:

[mild laurel]

What have you tried

[chilly ocean]

Declare this $g:K\rightarrow K, x \mapsto 1$ if $x = 0,$ $0$ else

[cloud walrus]

N/𝔄:

[chilly ocean]

For proving that $\psi$ is surjective implies $K$ finite

[cloud walrus]

N/𝔄:

[chilly ocean]

Then there exists a $f_x\in K[x]$ such that $\psi (f) = g$

[cloud walrus]

N/𝔄:

[chilly ocean]

Because $deg(f)<\infty$ f should have finitely many points where $f_x (x)=0$

[cloud walrus]

N/𝔄:

[chilly ocean]

But what is the next step

are you having trouble showing that if K is finite then $\psi$ is surjective?

[cloud walrus]

hochs:

[chilly ocean]

That as well

there are many ways to handle that. Dimension consideration by going thru $K[x]/(x^p - x)$ or by Lagrange interpolation to construct the polynomial directly.

[cloud walrus]

hochs:

[chilly ocean]

(That's enough of hint for now. I'll tell you more if youu're still stuck later)

$K[x]/(x^p -x)$ denotes the equivalence class modulo $x'^p-x$ right?

that $K$ must be infinite should be clear if $\psi$ were to be surjective: Otherwise you can construct functions with infinitely many zeroes

yes.

[cloud walrus]

hochs:

N/𝔄:

[chilly ocean]

yes.

Ok I will try what you said about lagrange interpolation

Thanks

you should try solving it in at least two of the ways that I mentioned above.

by the way, in the finite case the map $\psi$ descends to a bijection $\psi' : K[x]/(x^p - x) \to K^K$.

[cloud walrus]

hochs:

[chilly ocean]

err $K[x]/(x^q - x) -> K^K$ where $q = |K|$.

[cloud walrus]

hochs:

[chilly ocean]

that $p$ should be replaced by $q$ where $q = |K|$. Sorry about the typo.

[cloud walrus]

hochs:

[chilly ocean]

\rightarrow

or you can use (third method) the chinese remainder theorem

Hmm
Let $K={a_0, a_1, ..., a_n}$ where $\forall i\in {0,...,n}: a_i$ distinct and $g\in K^K$ such that $\forall i\in{0,...,n}: g(a_i) = b_i$.
Lagrange interpolation implies $\exists f\in K[x]: f(a_i)=b_i$ and $f_x(a_i)=b_i \implies g=\psi(f)$

[cloud walrus]

N/𝔄:
Compile Error! Click the reaction for details. (You may edit your message)

[chilly ocean]

yeah that's it.

Hmm now i wanna try the other proofs

@mild laurel : Also from 6 hrs and 30 mins ago, but hit the exact functor on $0 \to I \cap J \to A \to \cap A/I \oplus A/J$ to see that localization preserves intersection. This is the same way you'd show that flat base change preserves intersection.

Any reason you chose $K[x]/(x^p -x)$?

[cloud walrus]

N/𝔄:

[mild laurel]

Yep, someone just told me about this

[cloud walrus]

hochs:
Compile Error! Click the reaction for details. (You may edit your message)

[chilly ocean]

And what is p

@chilly ocean : $K[x]/(x^q - x)$ really, not $K[x]/(x^p - x)$.

[cloud walrus]

hochs:

[chilly ocean]

$q = |K|$. I made a typo earlier. I fixed it

[cloud walrus]

hochs:

[chilly ocean]

Oh ok

The reason is that two polynomials $f,g \in K[x]$ are the "same" as "functions" iff $\overline{f} = \overline{g}$ in the ring $K[x]/(x^q - x)$.

[cloud walrus]

hochs:

[chilly ocean]

Is that trivial or is there a famous theorem for this

i.e. a polynomial $f \in K[x]$ has evaluation $0$ at all points of $K$ iff $f$ is divisible by $x^q - x = \prod_{a \in K} (x - a)$.

that's trivial.

[cloud walrus]

hochs:

[chilly ocean]

$q$ is a power of the characteristic of $K$ but that's irrelevant for this problem.

[cloud walrus]

hochs:

[chilly ocean]

I will think about that

[golden pasture]

i) i could verify the cardinalities are the same(|G|/|HK|) but not exactly sure whats the 'canonical 1-1 correspondence'
iib) kinda lost at how exactly to approach this tbh
any tips for what direction to take would be q helpful

edit part i is indeed canonical so obvious once i saw it .-.

[chilly ocean]

@chilly ocean can the ring $K[x]/(x^q -x)$ be used to show an injection iff $K$ is infinite?

[cloud walrus]

N/𝔄:

[chilly ocean]

@chilly ocean : if $K$ is finite?

[cloud walrus]

hochs:

[chilly ocean]

No I mean for $K infinite \iff \psi injective$

[cloud walrus]

N/𝔄:

[chilly ocean]

Was just wondering that aside from the problem

the ring $K[x]/(x^q - x)$ doesn't make much of a sense when $q = |K| = \infty$ though

[cloud walrus]

hochs:

[chilly ocean]

but it is true that $\psi$ is an injection if $K$ is infinite.

[cloud walrus]

hochs:

[chilly ocean]

Oh you're right sorry

However, it is not an injection if $K$ is finite. The reason to consider $K[x]/(x^q - x)$ is so that $\psi$, when descended to this factor ring, is an injection $K[x]/(x^q - x) \to K^K$.

[cloud walrus]

hochs:

[chilly ocean]

(in the case of K finite)

Did you work on this exercise before or are you just that good? @chilly ocean

i've only seen this now tbh and I found it interesting.

Oh wow

Makes me wonder could you nudge me in a direction to prove that $\psi$ injective $\implies K$ infinite

[cloud walrus]

N/𝔄:

[chilly ocean]

sure, otherwise (that is, if $K$ is finite), then consider the polynomial $x^{|K|} - x$, which is the "same" as the $0$ function.

[cloud walrus]

hochs:

[chilly ocean]

or just $\prod_{a \in K} (x - a)$ is a polynomial that is the "same" if viewed as a function as the $0$ function.

[cloud walrus]

hochs:

[chilly ocean]

so $\psi$ can't be injective if $K$ were finite.

[cloud walrus]

hochs:

[chilly ocean]

By 0 function you mean the additive neutral element of K[x]?

yes, the point is that the in the ring $K[x]$, the two polynomials $0$ (as the additive $0$ element) and $\prod_{a \in K} (x - a)$ (which makes sense under the assumption that $K$ is finite, as this product is finite) are distinguished. However, once you hit $\psi$ to them, i.e. $\psi(0)$ and $\psi( \prod_{a \in K} (x - a) )$, the two are the "same" in $K^K$.

[cloud walrus]

hochs:

[chilly ocean]

Oh I see thanks for elaborating

here's the slightly bigger picture: Give $K^K$ also the ring structure with point-wise addition and point-wise multiplication. Then $\psi$ is actually a ring homomorphism, and the kernel is generated by $x^{|K|} - x = \prod_{a \in K} (x - a)$ in the case $K$ is finite and by $0$ (i.e., injective) when $K$ is infinite.

[cloud walrus]

hochs:

[chilly ocean]

in the $K$ finite case, you have an induced ring homomorphism $K[x]/(x^q - x) \to K^K$ that is injective (as you have modded out by the kernel, it is injective, and so the essence of this is in checking that the kernel is indeed $(x^q - x)$) and hence bijective (as the two rings are both finite).

[cloud walrus]

hochs:

[chilly ocean]

Yeah because if the kernel only has the neutral element then the homomorphism is injective, sure

I think that was said in my lecture

Thanks for the clarification and information i'll go and see how many proofs i can gather

[fringe nexus]

I'm given K= $\mathbb{Q}(\frac{-1+\sqrt{3}}{2})$ and I want to construct two non isomorphic field extensions $L_1,L_2$ of K s.t $Gal(L_1/k) \cong Gal(L_2/k) \cong Z_3$

[cloud walrus]

aaaaaaaaaaaaaaaa:

[fringe nexus]

Now that is the 3rd primitive root of unity, so it is a degree 2 extension and so L_1, L_2 are degree 6 extensions of Q

there are 2 groups of order 6, S_3 and Z_6

I get how to construct the S_3 extension, but how do i construct the Z_6 one?

The solution looked at the discriminant of cubics to find specific cubics, but I don't really know why that works?

The solution used the cubic f(x) = x^3-3x+1

and adjoined any root of this polynomial

im so lost

[wind steeple]

what's k ?

do you mean K ?

[fringe nexus]

Yes

Sorry

was he just looking for any cubic with non repeated roots

[chilly ocean]

The rigid motions of a regular polygon are rotations + reflections but the rigid motions of a platonic solid are just rotations. Why?

[chilly ocean]

(12) isn't a rigid motion even though there's a perfectly good element in the 3d euclidean group that does that.

[stone fulcrum]

@chilly ocean
I thought a bit more about that convo we had the other day. You can't allow a refection between two points of a tetrahedron because then your group is just S4

And we expect a unique structure from doing this

[chilly ocean]

So the set of rigid motions is only equal to the set of shape-preserving isometries when n=2?

[bleak abyss]

It might be an SO(n) thing? Reflections have determinant 1 in dim 2 but not in dim 3

I mean I haven't really seen any specific definition for "rigid motions" at least in general dimensions. So I'd speak of the rotation group of the tetrahedron, say

[chilly ocean]

Okay now I get it

Rigid motions = rotational symmetries

No rotation gives you (12). But rotating a tetrahedron over the edge 3-4 by a certain angle and then turning it around to back where it was gives you (12)(34)

So D_n is the set of all symmetries period but A_n is only rotational symmetry

Rigid motion... is a flexible word depending on context

[left kite]

a

What is e ?

[round lagoon]

https://cdn.discordapp.com/attachments/496785853180543027/649717286252445717/170785919962578953.png

[left kite]

you quick typer

[round lagoon]

[gleaming belfry]

Is it possible to write down a group presentation that does not describe a group?

[fickle brook]

[gleaming belfry]

Like, some generators and relations that contradict with the group axioms

[fickle brook]

[hot lake]

you can't contradict the group axioms with relations

in the worst case you get the trivial group

[fringe nexus]

Can anyone guide me on the right track here?
I tried using spectral theorem on B,C so they are diagonalizable with real eigenvalues, but the issue is that they aren't diagonalizable on the same basis unless B,C commute and there are no repeated eigenvalues
So I can't use it to find eigenvalues of A
I don't see how schur's decomposition theorem will help here unless I can split the decomposition into real and imaginary parts?
Also tried using gershgorin circle theorem but I don't think that works

[delicate bloom]

@round lagoon http://mathb.in/38567 Here's a quick little writeup for you

if you have any questions just ask

[tribal pasture]

Given a field F_p, does the element (p-1) generate the multiplicative group F_p-{0}?

[delicate bloom]

for p=2

[tribal pasture]

I think yes because the order of p-1 is p-1 and the group is also of order p-1 so it should generate but hoping if somebody can double check this reasoning

[delicate bloom]

and p=3

try it for p=5

4*4 = 1

you miss 2 adn 3

because in general (p-1)^2 = 1 you're not going to get very far

[golden pasture]

tho there will always b a generator

[tribal pasture]

Yeah that is true

[golden pasture]

for any number as long $\phi(n)=\lambda(n)$, theres a single generator

[cloud walrus]

Ariana:

[tribal pasture]

What is the notation?

[delicate bloom]

uhh maybe I should be more explicit with my response

the order of p-1 is 2 in the multiplicative group

so your reasoning is bad

[golden pasture]

basically $\phi(n)$ is the order of $\mathbb{Z}_n^\times$ and $\lambda(n)$(carmichael func) is the maximum order of any elements

[cloud walrus]

Ariana:

[golden pasture]

both have simple ways to compute using prime factorization

Could anyone help give any direction for (b) have been on it for quite some time, cant quite seem to come up with a bijective map from G orbits of X to H orbits of $\pi^{-1}(H)$
not exactly sure how to use the fact $pi$ is equivariant

[cloud walrus]

Ariana:

[magic owl]

is it talking about the preimage of the coset H?

I assume its talking about like

ker pi

or something?

if you think about G/H as a set of cosets then H is an element (identity)

If it were not my finals week I would think more about this

but i get the feeling if you write down enough definitions

there will be only one thing you can do

and that will work

[hot lake]

for a given G-orbit of X, choose an element x in it, then choose an element g in pi(x), then map the orbit to the H-orbit of g-1.x

[magic owl]

I wish I could pin messages for myself

I want to come back to this later

[hot lake]

I think you can mark some message as unread

[golden pasture]

ooo that works

thanks!

If you just want to have a note on it its
Procesi Lie groups page 8 exercise ii

[stone fulcrum]

@magic owl
Pind

[oak perch]

@gleaming belfry I mean ... you can write down a group presentation that describes more than one group?

[gleaming belfry]

Wikipedia claims otherwise:

Informally, G has the above presentation if it is the "freest group" generated by S subject only to the relations R.

[mild laurel]

How does this claim anything else than what he said?

[gleaming belfry]

Well, I assumed that the other groups would be what you get if you add a relation

[mild laurel]

Uh what

[oak perch]

hmm fair enough.

though I guess informally is kind of a cop out but I cant really complain.

[gleaming belfry]

well, it is followed by

Formally, the group G is said to have the above presentation if it is isomorphic to the quotient of a free group on S by the normal subgroup generated by the relations R.

(Though I can't parse that yet, I barely know what a homomorphism is)

[mild laurel]

Uh, then I'm not sure why you're thinking about free gropus

And I'm not sure why you think what he said is contradictory to what wikipedia says

[gleaming belfry]

I'm thinking about free groups and group presentations because that's the order the book I'm following introduces them in: https://venhance.github.io/napkin/Napkin.pdf. As for why I believe that is contradictory...

I assume that by "you can write down a group presentation that describes more than one group", you mean that, for example, $\langle x \mid x^6 = 1\rangle$ "describes" both $\mathbb Z/6 \mathbb Z$ and $\mathbb Z/3 \mathbb Z$, since both meet the relations. It depends on what you understand the verb "describes" here as...

[cloud walrus]

NieDzejkob:

[gleaming belfry]

AFAIU, though, a specific group presentation describes only one group (up to isomorphism)

If you meant something else, my bad. Would you care to show me an example?

BTW, I thought a bit about my original question, and really the answer seems quite obvious in hindsight, since the axioms don't have any $\neq$ in them...

[cloud walrus]

NieDzejkob:

[chilly ocean]

quick question, anyone knows what does this line above means? (in context of rings)

[gleaming belfry]

Hmm, equivalence classes?

[chilly ocean]

I don't think so, here for the full context

Or maybe

hmm yeah might be actually

[fringe nexus]

b mod n is non zero

[mild laurel]

That does not describe Z/3Z

That's why Wikipedia says the "freest group"

[gleaming belfry]

yes, but that's how I understood what @oak perch said. I really shouldn't have made such an assumption

[gleaming belfry]

so, what is such a group presentation that describes more than one group? My current understanding of groups is that there is none...

[woven delta]

So a group given by a presentation is the free group on whatever generators are specified mod the normal subgroup generated by the relations

[gleaming belfry]

I've already read such a definition...

[chilly ocean]

What would the identity be in the ring of continuous functions on [0,1] to R? Wouldn't it be f(x)=x?

[woven delta]

Yes

Obviously

That's the multiplicative identity

The additive identity is the constant 0 function

Hmm actually it depends what operation you're using

[chilly ocean]

Ok, so why isn't f(x)=x-(1/2) a unit?

cant we have g(x)=x+(1/2) then fog = x

[woven delta]

Yeah I'm wrong

[delicate bloom]

it's not closed under composition

[woven delta]

The multiplicative identity is the constant 1 function

I was thinking maps from R to itself for some reason

[chilly ocean]

Why? if we have something like g(x)=x^2 for example, then g o id (x) = g(1) for all x?

[woven delta]

Here the multiplication is just pointwise multiplication

In R

[delicate bloom]

f(x) =2 is a perfectly fine continuous function on [0,1] to R

but g(f(x)) =?

undefined cause you can't plug 2 into g(2) since it's on [0,1]

[chilly ocean]

but f(x)=x is closed under comp tho?

[woven delta]

👀

[chilly ocean]

and wouldn't it work for any g that gof = fog =g?

[delicate bloom]

I'll be honest I don't know what you're asking lol

[chilly ocean]

why isn't f(x)=x the identity in this ring

[delicate bloom]

which ring

what are your operations?

[chilly ocean]

continuous functions from [01,] to R

[woven delta]

The operation is pointwise multiplication

[chilly ocean]

f+g and compositions

[delicate bloom]

you need a set with 2 operations

[woven delta]

Composition doesn't make much sense

In this case

[delicate bloom]

so compositions distribute over addition?

interesting

[woven delta]

Lol

[chilly ocean]

Okay I don't know then, its not written in the book or I cant find it, forget it

[woven delta]

It's just pointwise multiplication

Lol

That's the standard thing

[chilly ocean]

Okay then

That owuld make sense

[woven delta]

fg(x)=f(x)g(x)

[chilly ocean]

Okay I see, thanks

[round lagoon]

@delicate bloom thank you

[delicate bloom]

😛

cool what are you doing I'm curious, seems interesting

or like where did the problem come from

[chilly ocean]

Need a hint in order to prove that a cycle in S_n needs at most n-1 transpositions.

Maybe I'm overthinking this

[mild laurel]

What do you mean?

Like the minimum amount of transposition for any cycle?

[chilly ocean]

Reposted from analysis channel, as I feel this is a better place:

Is there a "go to" book for introductory and intermediate set theory?

Something that starts from nothing, but not progress terribly slowly, if possible.

[stone fulcrum]

Intro set theory, you can learn it in many places. Fraleigh starts with a chapter in it. Discrete math books should have some

[potent lynx]

something that stars from nothing

you can use 'book of proof'

or you can use how to prove it by velleman

[stone fulcrum]

I've read some topology and analysis books that have some

[potent lynx]

^ yes

topology munkres

1st chapt

[chilly ocean]

Got it. Primarily reading this for both understanding and hopefully intuition behind ultrafilters. Am I in the right direction?

[stone fulcrum]

You might need a specialized book for something like ultrafilters

Literally everything lower than that is common and you'll find it anywhere else

[chilly ocean]

Got it. I suppose since ultrafilters are a bit specific, I probably won't find an exact recomendation here, but I'll try to find a good one myself

(if anyone can recommend one on either ultrafilters or related topics, though, please @ me.)

Thanks :)

[magic owl]

Any book on mathematical logic

Will touch on ultrafilters

[wind steeple]

Topologie generale chap 1 to 4 Bourbaki

[mild laurel]

lucky french people that can read bourbaki smh

[chilly ocean]

Did someone mention bourbaki

😍

[mild laurel]

can you read french

[wind steeple]

Bourbaki is translated in english x)

[mild laurel]

wheres ega tho

[wind steeple]

Ega isn't bourbaki

Ega is ega

[mild laurel]

yeah I know

im just sad

[wind steeple]

Just learn french

As we learn english :p

[mild laurel]

Lmao true

This is typical american

Whining that not everything is in English

[wind steeple]

Btw idk if ega is hard or not to read in french

I mean it could be basic french

But it's Grothendieck

🤷

[mild laurel]

lmao

I'm also terrible at learning languages so

we'll see

how much alg geo I want to know

[wind steeple]

You can learn french only by reading it

It's not that hard

The hardest thing in french is to speak

Because you feel that what you say is not the thing you read

But who cares when the goal is to read ?

[mild laurel]

Yeah that's true

Also on a side note

why are double cosets a thing

I hate single cosets enough wtf

[wind steeple]

What's that ?

[mild laurel]

well you take a subgroup and you have cosets of the subgropu

double cosets are when you take two subgroups and you look at cosets of the two subgroups combined

I.e, you take two subgroups H, K < G and you consider the sets {hgk | h \in H, k \in K} for all g in G

And this is denoted $H\backslash G/K$

[cloud walrus]

Zopherus:

[wind steeple]

This is very ugly

Since (h,k)->hgk isn't necessarly a bijection

[mild laurel]

yeah it's terrible

sad that its actually useful

[wind steeple]

How it is useful ?

Oh in fact this satisfies some ordered properties

[mild laurel]

It comes up in some representation theory stuff

But it also comes up in the idea of hecke operators on the space of cusps forms

[wind steeple]

In finite groups, you can see that's is an union of cosets of K or of H

In infinite groups to lel

[mild laurel]

Yeah they have some properties

But like there's no lagrange's theorem

[wind steeple]

The cardinal is divisible by both |K| and |H|

But yeah

That seems neat in fact

[mild laurel]

Yeah, but like none of the double cosets necessarily have to divide |G|

and they don't necessarily have equal sizes either

[wind steeple]

Yep

In finite group I wonder if you can study the action of G on the set of bicosets and extract some Galois correspondences

I see permutations of cosets lel

Ok thanks

I'm tired I'm going to bed

Gn

[mild laurel]

Yeah idk, don't think the action would be particularly nice

Since either right or left multiplication isn't super symmetric to the idea of double cosets

And conjugation wouldn't even be an action

[visual turret]

I'm having trouble with the following problem. Let G = Q8 (quaternion group), and H = <J> is normal in G. Show that there is no homomorphism from G onto H

I have been trying to look for a contradiction, by assuming that there is an onto homomorphism, so therefore G/ker(phi) ~ H

[topaz solar]

@visual turret is there a subgroup of Q8 with quotient of <j>?

[visual turret]

@topaz solar what do you mean by quotient of <j>?

[topaz solar]

is there a subgroup N of Q8 such that Q8/N =~ H

[visual turret]

That's where I've been trying to look for a contradiction

I've imagined that there isn't

is this correct?

[topaz solar]

I mean, you can do this by exhaustion

plus, you can also throw Langrange (butchered spelling) or whoever's theorem about the cardinality of cosets and such

[chilly ocean]

I need a hint in order to prove that if sigma is an odd number of transpositions, then so is any product of transpositions.

[visual turret]

I'm struggling with the proof of this question, Let phi:G->K be an onto homomorphism and let J be a subgroup of K. Show that phi[phi^-1[J]]=J. I have found that phi[phi^-1[J]] must be a subgroup of K, but can't figure out why it must be J itself.

[chilly ocean]

Do you know the definition of a fiber? Looks like that might a good approach to this

[visual turret]

I don't

[chilly ocean]

You know about preimages?

aka inverse image

A fiber is just the inverse image/preimage of a single element

Really important concept

[urban acorn]

@chilly ocean clarify what you're trying to prove. what is sigma exactly? this isn't true when sigma is any permutation, since there are both even and odd permutations.
it's equivalent to saying that the particular permutation you're denoting by sigma is even.

which is trivial if you have a cyclic decomposition, which is the standard format to write a permutation in

[visual turret]

I don't think my professor wants us to use fibers

[chilly ocean]

You're using inverse images right here, so using a single general element in a proof would force you to use fibers, whether you call them that or not

Your goal is to show that $\phi(\phi^{-1}(J))=J$, and you know how to show set equality (two-way containment) using a single general element, that sounds like a good proof method for this

[cloud walrus]

Deconstructed:

[chilly ocean]

@chilly ocean have a look at Hodges' A Shorter Model Theory (chap. 8)

[chilly ocean]

why is $\mathbb Z \cdot \left( \frac m{ln} + \mathbb Z \right) = \mathbb Z$ ?

[cloud walrus]

Godel:

[chilly ocean]

m/ln is some rational

I feel like it should be more than Z, for example Z(1/2 + Z), then wouldnt it also have some fractions in there?

[golden pasture]

huh where did you find this

[magic owl]

Whats the product here?

[chilly ocean]

Uhh the problem is to show you cant define a multiplication in quotient group (Q/Z, +) in a way that (Q/Z;+.*) is a ring with 1.

The solution starts with multiplying (m/n+Z)(k/l + Z) and showing it will be Z

and the thing I posted is suppsoed to be a final step

The thing that you posted can't be true, at least not in the conventional sense $Z \cdot (\frac{m}{ln} + Z) = { x(\frac{m}{ln} + y ) , | , x,y , integers }$. $(\mathbb{Q} / \mathbb{Z}, +)$ can't be turned into a ring with $1$ because otherwise if $\alpha + \mathbb{Z}$ were "1" in the supposed ring then there's a positive integer $n > 0$ such that $n(\alpha + \mathbb{Z}) = n \cdot 1 = 0$, so that $n (\mathbb{Q} / \mathbb{Z} )= 0$ entirely, but $n \cdot (\frac{1}{n+1} + \mathbb{Z}) \neq 0$ in $\mathbb{Q} / \mathbb{Z}$.

[magic owl]

Can you post the proof

[cloud walrus]

hochs:

[magic owl]

like the page of the textbook you are reading or whatever

[chilly ocean]

ok in that context it seems to make more sense, since $\mathbb{Z}$ is the "0" element in $\mathbb{Q}/\mathbb{Z}$ and $\cdot$ is the supposed "ring" multiplication.

Wait why n(a+z) = 0in your msg?

[cloud walrus]

hochs:

[chilly ocean]

@chilly ocean : Write $\alpha = m/n$ for some integers $m,n$, with $n$ positive (but this positivity is unnecessary). Then $n (\alpha + \mathbb{Z}) = n(m/n + \mathbb{Z}) = m + \mathbb{Z} = \mathbb{Z} ''='' 0$.

[cloud walrus]

hochs:

[chilly ocean]

Ahh okay, gotcha

Thanks

the main point is that $\mathbb{Q}/\mathbb{Z}$ becomes a $\mathbb{Z} / n\mathbb{Z}$ "module" if it were to be turned into a ring with $1$. The book's solution is essentially the same as what I gave you. To elaborate further on why $n (\mathbb{Q} / \mathbb{Z} ) = 0$ then, note: for any $x \in \mathbb{Q} / \mathbb{Z}$, $n \cdot x = n \cdot x \cdot 1 = x \cdot (n \cdot 1) = 0$, but take $x = \frac{1}{n+1} + \mathbb{Z}$. I'm using here of course that "1" commutes with everything.

[cloud walrus]

hochs:

[chilly ocean]

how do I show that if a permutation can be written as a product of an odd number of transpositions then it cannot be written as a product of an even number WITHOUT using homomorphisms

hint only

[hot lake]

wdym without using homomorphisms

[chilly ocean]

It's an exercise in AATA the webbook and the exercise expects you to prove it without using homomorphisms.

[little jungle]

have you learned about A_n?

That's the hint I'll give

[urban acorn]

@chilly ocean consider the parity of the amount of unordered pairs that change their order after the permutation

maybe just define A_n as the ones that can be written as a product of an even number of permutations without proving that they can't be written as an odd number of them and then show that it's normal (pretty trivial)

then it shouldn't be that hard to show that it's of index 2

this looks like a promising direction too

[chilly ocean]

What is the characacteristic of the field K if 27 * 1 = 0?

Answer is char(K)=3

Why can't it be 9 or 27?

[mild laurel]

You mean char(K) = 3

[chilly ocean]

ye

lol

[mild laurel]

What happens if the characteristic of a ring is 9?

In other words, why isn't Z/9Z a field?

[chilly ocean]

its not a group under addition?

[mild laurel]

Why not?

[chilly ocean]

no 0

hmm

ok

thx

[mild laurel]

uh

So what's your answer?

[chilly ocean]

only 3, I get it

[mild laurel]

Because Z/9Z has a zero

under addition

It's 0

[chilly ocean]

isnt Z/9Z numbers coprime with 9?

Or just all mod 9?

[mild laurel]

It's all mod 9

[chilly ocean]

idk then

should be

its abelian under addition, multiplication works, there is 1, every element has an inverse

[delicate bloom]

what's 3^{-1}?

[chilly ocean]

okay then

[mild laurel]

Are those the only axioms for a field?

[chilly ocean]

I mean only the inverse axiom doesnt work tho?

there are also those that I dont remember the english name of connected to multplying and adding but they work fine right

[mild laurel]

Sure

Z/9Z is fine as a ring

[chilly ocean]

yeah, gotcha

[fringe nexus]

How would i do this?

[fringe nexus]

nevermind i found the solution and its too hard for me to understand

[fading gulch]

Hi! I am dumb undergrad who is trying to learn category O for rep-th course I have. How would I find all objects in category O of sl2 for which the Casimir element ef+fe+h^2/2 acts as some given constant c?

it's like category O is so abstract that i dunno how to work with it

[stone fulcrum]

Here's one out of AM I'm struggling with.
If α is an ideal, V(α) is the set of prime ideals that contain α, r(α) is the radical of α,

Show that V(α) = V(r(α))

I guess I should Google it first there's probably a solution online oop

[fading gulch]

since $\alpha\subset r(\alpha)$ then every ideal containing $r(\alpha)$ contains $\alpha$, so, we have, $V(r(\alpha))\subset V(\alpha)$. We should show now another inclusion. It should follow from the fact that $r(\alpha)$ is intersection of all prime ideals containing $\alpha$

[cloud walrus]

notwhale:

[fading gulch]

I guess

but i am not sure

the fact I used is proved on wikipedia

[mild laurel]

No there's an easier idea

if α \subseteq p, then for all x \in r(α), we have that there is some n > 0 such that x^n \in α

Then, this implies that x^n \in p and since p is prime, it follows that x \in p

so that r(α) \subseteq p as well

[stone fulcrum]

Does that follow? If x^n \in p, then x \in p?

I get p is prime, so if ab \in p, then a or b \in p

@mild laurel

[mild laurel]

well

x^n = x \cdot x \cdots x \in p

and just write this as like (x \cdots x)(x \cdots x) \in p

so you get that x^m \in for some m < n

Then just keep going

[stone fulcrum]

I get that this might imply x^(n-1) \in p

Oh yeah

And if it does then it might imply x^(n-2) \in p

Duh lol. I get it now

Thx, this makes a lot of sense

[mild laurel]

it's an idea that comes up a lot

[stone fulcrum]

I can't believe it wasn't clicking before.

[gentle nymph]

Hey not trying to interrupt anyone working so if someone has a free minute please let me know...

I'm working on a proof right now for the disjoint cycles commute and this is a part of the proof:

Assuming ai is in set Alpha = {a1, a2, a3... am}

And set Beta {b1, b2, b3... bn} fixes all elements a

That means

(Alpha Beta) (ai) = Alpha (Beta(ai))

= Alpha (ai)

= ai+1

Similarly...

(Beta Alpha) (ai) = Beta (Alpha(ai))
= (Beta (ai+1))
= ai+1

What's a concrete way I can explain the step of (Beta (ai+1)) = ai+1 (which I highlighted in bold)?

I know that ai+1 can be defined as a assuming i = m but there's still something not clicking for me

(I'm just doing this since there wasn't a message before me here today yet... 😭 ) <@&286206848099549185>

[latent anvil]

Every element of the form aj is fixed by β

Since it only permutes the set {b1,...,bn}, and that set contains no aj by assumption

[gentle nymph]

@latent anvil How does that translate to beta (ai+1) = ai+1?

[latent anvil]

What does it mean for an element to be fixed by a permutation?

[gentle nymph]

Ohhh I see. An element is only fixed by a permutation if permutation ( element ) = (element)

So therefore it's true by definition that beta (ai+1) = (ai+1)

[latent anvil]

I mean I guess I just rephrased your question

But the point is beta will leave every element outside the bi unchanged

[gentle nymph]

Right

What about this step though?

Alpha (ai)

= ai+1

[latent anvil]

well that's the definition of α

It cycles those elements

[gentle nymph]

I see

Okay I think that makes sense to me

Thanks @latent anvil 😄

[mild laurel]

Have you guys heard of monstrous moonshine before

[upper pivot]

nope

[mild laurel]

@maiden ocean listen

this is cool

[chilly ocean]

https://cdn.discordapp.com/attachments/585199534204780635/586853733749751818/61996144_2424153627856871_6053510095927508992_n.png

[maiden ocean]

spooky : o

[mild laurel]

yeah

but where do these numbers come from

Okay the first idea, is the idea of the monster group

In the last century, there was a big movement to classify all the finite simple groups

simple groups are those without normal subgroups

So things like cyclic groups of prime order

or the alternating groups

Are simple groups

[upper pivot]

they have been classified right

[mild laurel]

Yes we're done

So there are 18 infinite families of finite simple groups, like the two I described above

But there are 26 groups that don't fit into any of these lists

Called the sporadic groups

The biggest of these sporadic groups is called the monster group

and has order 808,017,424,794,512,875,886,459,904,961,710,757,005,754,368,000,000,000

[maiden ocean]

lmao

[upper pivot]

damn

[mild laurel]

its a big boy

chonker

Anyways, one of the tools used in classifying simple groups

is representation theory

And the idea is that you take a group G

and study homomorphisms G to GL_n(C)

These are the invertible matrices of size n by n, with complex entries

In fact, before we even knew the monster group existed, we knew a ton about its representations

We knew everything about its representations in fact

[upper pivot]

so what u researching

[mild laurel]

We're getting there

So one idea is of irreducible representations

Don't worry too much about what it means for a representation to be irreducible, but there are only finitely many irreducible representations for a group

So one thing you can study is the dimensions of the irreducible representations

And for the monster group, this list goes

1

196,883

21,296,876

Okay keep that in mind

The other end of things

There's this important complex function, the j-invariant

Which is an invariant of elliptic curves and plays a big role in the theory there

And this j-invariant has a fourier series

Because of its properties

and the coefficients of its fourier series go

1

196,884

21,493,760

and you can check that the sum of the first two dimensions of the monster 1 + 196,883 = 196,884

and further that the sum of the first three is 1 + 196,883 + 21,296,876 = 21,493,760

And in fact, we now know further, that all the coefficients of the fourier series can be written as finite sums of the dimensions of the monster group

And that's basically it, or at least, that's all I know of the story so far

It's just pretty crazy that something used in number theory to study elliptic curves is connected with something in finite group theory

[hot lake]

technically, every integer is a finite sum of ones

but yeah monstrous moonshine is super cool

[mild laurel]

okay yes you're right

I'm not sure exactly about the specifics of that statement

[hot lake]

iirc there are other modular functions that have the same relationships with some other groups

[mild laurel]

I know that the coefficients can be written in multiple ways since there are integer dependence relations between the sizes of the monster group but

Right, I think that's part of the research I'm going to be doing

[hot lake]

and like, the coefficients in the linear relations don't depend on the group / modular function

well in the case of the j function they have built an actual natural graded space on which the monster group acts, and looking at each level you get a representation of dimension the corresponding coefficient of j

[mild laurel]

Yes Arch, that's what I'm saying, but the specifics are a bit more complicated

Since yeah, you can write every integer just as 1 added a bunch of times

About your other question, about irreducible representations

If you take some representation G to GL_n(C)

Then you can consider the vector space C^n

And consider how G acts on this vector space

I.e., every element of G gives you some linear automorphism of C^n

so every element of G gives you some function from C^n to C^n

So now, a representation is called irreducible if

Since a representation is a map from G to GL_n(C)

which are the invertible matrices of size n by n with complex entries

[upper pivot]

alright, so far following

[mild laurel]

So by taking any element g of G to some invertible matrix through this map

This gives you a linear automorphism of C^n

Yes? @chilly ocean

invertible matrices are linear automorphisms

You just take some invertible matrix and multiply it by some vector in C^n

This is bijective since the matrix is invertible

So now

If you take some non-empty, proper subspace V \subset C^n

Then this subspace is called G-invariant, if, for all elements g \in G

the subspace V is fixed by all the automorphisms arising from elements of G

[maiden ocean]

i am lurking while reading

[mild laurel]

And a representation is irreducible if and only if there are no G-invariant subspaces

Since it turns out, by Mashke's theorem, if there's one G-invariant subspace, then there's an orthogonal complement of this subspace that's also fixed

I mean, take any group G

and consider the trivial map from G to GL_2(C)

I.e., send every element of G to the identity matrix

In that case, every subspace of GL_2(C) is G-invariant

The subspace doesn't necessarily need to be fixed element wise like in this example

but just that for all v \in V, the action of g on v is in V

Sure

The image of V has to be exactly V

Um, I'm not sure this question makes sense

I'm not sure, but I don't think it matters

Sure V and C^n are groups under addition of vectors

But these are abelian groups, all inner automorphisms of abelian groups are trivial, identity automorphisms

[chilly ocean]

I need help completing this proof that if x is a cycle of odd length then x^2 is also a cycle.

Here I go.

<x> = Z_|x|. Because |x| is odd x^2 generates <x>. Therefore...

[mild laurel]

I'm not sure this is how you should start

Just think about which elements go where

[chilly ocean]

Yes I know but that's hard to formalize

that IS what I'm trying to formalize!

[warped socket]

Just started on Dummit and Foote’s Abstract Algebra

One question involved looking at ‘reals modulo 1’

I’m wondering if this can be generalized to reals modulo k, for any positive integer k

[bleak abyss]

Should be isomorphic

[warped socket]

My thoughts so far is yes

Uh is that to me?

Not sure if it’d be isomorphic

Reals modulo 1 has an uncountable number of elements, while integers modulo 1 is countable

Or am i missing something

Also i dunno if this is the right channel

Nvm i found some answers online

Sorry for clutter

[oak perch]

Technically it's not isomorphic? 1x1 =1 which is not true for a general k?

like for the reals modulo pi, a(pi)/b are zero divisors where a,b are from Z. So mod pi does NOT include 1 as a zero divisor but mod 1 Does have 1 as a zero divisor? (feel free to check my logic here)

[split shard]

Hi i am french so sorry for my English, i study medicine first year i have a problem with a "theorem" i just why we did this can i post a image here?

[somber rivet]

Urm, you might want to take a better image? It's not very clear at this instance. Also, it's interesting that you're doing math as a student of medicine

[split shard]

I d like to know why we 'bring' or 'do' sh(x) on order 4 and not order 1 i hope you understand my shitty english hahahaha

[golden pasture]

uh i dont understand whats going on but this def isnt abstract algebra

i guess looks like taylor?

its like

#calculus ?

[snow bolt]

yeah thats calc

[split shard]

Oh okayyy I'll post it there sry 🙃

[chilly ocean]

How does the definition of a group action get you to "homomorphism from G to Sym A."

[mild laurel]

Every element of g

[chilly ocean]

nvm I found a paper

[mild laurel]

Gives you a bijection on elements of A

[tribal pasture]

If G/H is isomorphic to K then is G isomorphic to H x K?

[mild laurel]

No

[tribal pasture]

This is an argument as for why D_10 should not be isomorphic to D_5 x H = {1,x^5}. But almost anywhere I search, I get the answer that they should be isomorphic.

Can somebody tell which should I trust (I think they should be isomorphic since using the product isomorphism since D_5 and H are both normal in D_10)

[hot lake]

no for example, Z/4Z is not isomorphic to Z/2Z x Z/2Z

even though you can quotient Z/4Z by {0;2} to get a Z/2Z

and in your situation it doesn't make sense to say that D5 is normal in D10, it's not the same D5s

though they forgot that D5 x H has an 11th element of order 2 I think

(1, x^5)