groups-rings-fields

[rocky cloak]

Not really a theorem, but I guess when constructing the real numbers it's neat to note that Cauchy sequences in Q forms a ring and that sequences converging to 0 is a maximal ideal

[noble nexus]

functional analysis doesn't really become algebraic really

there are fields that have somewhat analogous results but at least from my experience the methods very much remain analytic

[somber goblet]

this might be a basic question but can any R-module be expressed as an abelian group $A$ with a ring homomorphism $\phi: R \to \operatorname{End}(A)$?

[cloud walrus]

lexi

[somber goblet]

is that an accurate definition for an R-module?

[thorn jay]

yes that is equivalent

this gives a nice category theoretical interpretation of modules

[somber goblet]

thank you

is it then a vector space if it is a ring homomorphism $\phi: F \to \operatorname{End}(A)$?

[cloud walrus]

lexi

[somber goblet]

where F is a field

[thorn jay]

yes

[somber goblet]

very neat

ig you could use this to show that there are no finite vector spaces over the reals

[thorn jay]

no nontrivial*

if V is a nontrivial vector space over the reals then it must contain a copy of R so it cant be finite

[somber goblet]

trivial would be with the zero homomorphism?

[thorn jay]

there is no zero homomorphism of rings

[somber goblet]

oh right because 0 maps to 0 and 1 maps to 1

[thorn jay]

yes

[somber goblet]

so what would the trivial vector space be?

[thorn jay]

the vector space only containing a zero element?

[somber goblet]

oh wait yeah

im sorry if my questions are too basic

{0} is a vector space over any field

[thorn jay]

lol its alr i just was a little confused how you've never heard of that while knowing about rings

[somber goblet]

ive heard of it i just forgot in the moment because i was thinking homomorphisms when you said "trivial" 😭

[thorn jay]

oh lmao

[elfin wraith]

Not if youre my first ring theory notes! 1 did not have to map to 1 in that class and no one liked it

[thorn jay]

mfw nonunital modules

[somber goblet]

wait what does $\operatorname{End} {0}$ look like

[noble nexus]

0

[cloud walrus]

lexi

[thorn jay]

there is only one function from {0} to {0}

[noble nexus]

(or I guess, the set containing the function mapping 0 to 0)

[somber goblet]

oh yeah rings can have 1 = 0

[elfin wraith]

Genuinely no one understood it, the prof teaching the course inherited the notes and would just rant about it all the time because hed get confused, when things he wrote down needed f(1) = 1

[somber goblet]

but fields can't

[thorn jay]

yeah, just makes everything nicer

[somber goblet]

ah but when we apply the ring homomorphism from the field we lose the field's structure and keep the ring

[thorn jay]

loll wtf that's stupid

[somber goblet]

so for the (ring) homomorphic image of $F$ we're allowed ${0}$

[cloud walrus]

lexi

[somber goblet]

if my understanding is correct

[thorn jay]

a ring is the trivial ring if and only if 1 = 0 holds in that ring

in particular, the only way that for some map f : R → S the image is trivial, is for S to be trivial

[somber goblet]

right so in the case of the trivial vector space the endomorphism ring is trivial and the homomorphism from the field is the zero homomorphism (which ig is only allowed if the codomain is the trivial ring)

[thorn jay]

yes

[somber goblet]

makes a lot of sense

[earnest hatch]

I am having trouble with this question: Assume R is commutative. Show that an R-module M is irreducible iff M is isomorphic to R/I where I is a maximal ideal of R. I don't really know where to start

[thorn jay]

hint: think about free modules

[olive jewel]

Professor Doom, From planet Doom

[somber goblet]

if i'm setting out to prove the fundamental theorem on finite abelian groups, can i assume every $\mathbb{Z}$-module has an independent basis?

[cloud walrus]

lexi

[frigid epoch]

Only free modules have a basis

[rocky cloak]

Do you know the correspondence theorem?

[tulip otter]

any hints? α is algebraic over F so F(α) is algebraic over F and so is F(α^2) which is a subfield of F(α). Hence [F(α):F]=[F(α):F(α^2)][F(α^2):F] and now i should prove that [F(α):F(α^2)]=1 and then i will be done. How do i do that?

[next obsidian]

2 doesn’t divide an odd number

[tulip otter]

so i should show that if F(α^2) neq F(α) then [F(α):F(α^2)]=2?

[next obsidian]

2 doesn’t divide an odd number

[quiet pelican]

Yes

[rocky cloak]

Try to think about what the minimal polynomial of alpha over F(alpha^2) could be

[tulip otter]

the minimal polynomial p of alpha over F has an odd degree. so considering this polynomial over F(alpha^2), we know that p(alpha)=0 and any term with even power will be absorbed to the coefficients when viewed as a polynomial over F(alpha^2). this means that the minimal polynomial of alpha over F(alpha^2) has degree 1. Hence [F(alpha):F(alpha^2)]=1 as desired

is that right?

[rocky cloak]

Yeah, that works

Alternatively you can note that alpha is a root of
x^2 - alpha^2

[tulip otter]

ohhh, thats a very nice idea

[tulip otter]

and is there a straightforward way to do this?

[tulip otter]

could you tell me what was your idea here?

[rustic crown]

hi chmonkey

[next obsidian]

If it isn’t equal then the extension over F(alpha^2) has degree 2

And 2 doesn’t divide an odd number

[tulip otter]

ohh i see, so i shouldve looked at F(alpha^2) over F instead of F(alpha) over F(alpha^2)

tysm jagr and Chmonkey, have a great day/night!

[next obsidian]

No

You look at F(alpha) over F(alpha^2)

This is degree 2

[rustic crown]

here we're using that alpha satisfies the polynomial x^2 - alpha^2 over F(alpha^2) so the degree [F(alpha) : F(alpha^2)] <= 2.

and also that [F(alpha) : F(alpha^2)] = 1 if and only if F(alpha) = F(alpha^2).

[tulip otter]

ah wait i read that as "then the extension F(alpha^2).." instead of "then the extension over F(alpha^2).."

mb

[tulip otter]

ohhh ok, this makes sense

tysm det, have a great day/night

[rustic crown]

hehe thanks you too have a great day/night

[tulip otter]

thanks!

[earnest hatch]

Yep! I figured it out last night!

[delicate orchid]

Oh THATS what you needed the hint for

[vapid vale]

Maybe this is a silly question, but when did invariant theory change to invariant polynomials of some group action vs invariants in the coefficients of a polynomial?

As Cayley was doing, and the Germans like Paul Gordan at least into the late 1860s, invariants meant like b^2-ac was an SL_2 invariant of the binary quadratic ax^2 + 2bxy + cy^2. But then it seems by the mid-late 1870s and onward it actually refers to like elements of k[x,y,…]^G

[barren sierra]

is this not the same thing sort of?

am I not understanding your question

I feel I'm not

[quiet pelican]

Yeah it looks like you just described a special case, then the general case

[cedar vault]

Im trying to prove the smith normal form theorem using the structure theorem for finitely generated modules over PIDs. Let $A$ be an $n\times n$ matrix. We want to find two bases for $R^n$ so that the matrix of $x\mapsto Ax$ is in smith normal form. Let $\phi: R^n\rightarrow R^n$ be the map given by multiplying by $A$. Now we apply the structure theorem : $$R^n/Im(\phi) \cong R^{n-m}\oplus\bigoplus_{i=1}^{t}R/(a_i)$$ where $t\leq m$, $a_i|a_{i+1}$, and none of the $a_i$s are units. Note here that no summand in the torsion part is $0$ (by our assumption on $a_i$ not being a unit). I am able to show that the generators for each summand of the torsion part (so there will be $t$ such generators) and a basis for the $R^{n-m}$ term can be lifted to get a linearly independent set $v_1,v_2,\cdots,v_{n-m+t}$ of size $n-m+t$. i just need to extend this to a full basis of the target. Then i can use the sequence $0\rightarrow \ker\phi\rightarrow R^m \rightarrow Im(\phi)\rightarrow 0$ to get a basis for the source, and il be done. But to show that i can extend the set $v_1,\cdots,v_{n-m+t}$ to a basis, i need to show that $R^{n}/(v_1,\cdots,v_{n-m+t})$ is torsion free. Any ideas?

[cloud walrus]

Herzog

[cedar vault]

(I know that usually, one proves the smith normal form and gets the structure theorem as a corollary)

[copper kestrel]

i have a stupid question, so i'm trying to show that x^3 + 2x + 3 is reducible in Z_5[x], and i know that it is since its of degree 3 and has a zero in Z_5, namely x = 2

so then by the factor theorem x-2 is a factor of x^3 + 2x + 3

however i'm having trouble actually dividing in Z_5 :/ am i supposed to rewrite x-2 as x+ 3 (since -2 mod 5 = 3 mod 5) and then divide?

i thiiiiink the answer should be x^2 + 2x + 1?

[tardy hedge]

You dont have to rewrite -2 as 3 but the residue classes are usually given by positive numbers so ig theres that

x-2 = x+3 in Z5[x]

[copper kestrel]

alright! i keep getting a remainder so i think i'm doing something wrong

i did it! thanks yall :]

[rocky cloak]

Just think about the sequence
0 -> K -> R^n -> R^n-m -> 0

as the sequence splits
K = R/(v1, ...) is a direct summand of R^n

[copper kestrel]

one more quick question, i'm trying to find the generators of the group of units in Z_23 under multiplication. is Z_{23}^{*} isomorphic to Z_{22} under addition?

oh wait no i dont think i can use that bc i have to find that specific isomorphism that maps all the generators

[somber goblet]

it is, but that doesn't tell you which elements generate $\mathbb{Z}_{23}^\times$

[cloud walrus]

lexi

[copper kestrel]

yeaaaa

[white oxide]

Can I please have a hint to show that Q[\sqrt{D}] is contained in Q[\alpha]?

[somber goblet]

in fact if i remember correvtly it is in general computationally hard

but with p=23 you know from lagrange that all non-identity elements have order 2, 11, or 22

so that might be some help

[copper kestrel]

oh :')

[copper kestrel]

true

[rocky cloak]

Try to square
r + salpha
and deduce some conditions on r and s for the square to be in Q

[somber goblet]

when discussing field morphisms, can we apply any theorems about group homomorphisms to both the multiplicative and additive groups?

i.e. can i argue that all field homomorphisms are injective because $\psi(a) = 0, a \neq 0$ implies $\psi(a^{-1}) = \psi(a)^{-1} = 0^{-1} \notin K$, so $\psi$ is not a well-defined mapping?

[cloud walrus]

lexi

[somber goblet]

are there any special considerations i need to make before applying such theorems?

[velvet hull]

No, not really, your proof works

[tough raven]

I would rephrase this to 1 = psi(1) = psi(a a^{-1}) = psi(a) psi(a^{-1}) = 0 psi(a^{-1}) = 0 or that psi(a^{-1}) = psi(a)^{-1} but 0 bas no inverse, raher that writing 0^{-1} \notin R.

But this argument is essentially completely correct.

[somber goblet]

yeah i would write that in an actual proof but i meant that as a sketch

[noble nexus]

better still is to not use contradiction, any ring homomorphism must map invertible elements to invertible elements

as if ab=1, then psi(a)psi(b)=psi(1)=1

so if a is nonzero, then psi(a) is nonzero which gives injectivity

I guess that still kinda uses the contrapositive at the end

[somber goblet]

are contradictory proofs less desired

(im not super experienced in proofwriting)

[noble nexus]

direct proofs are typically better than contradiction if possible (unless it makes the proof a lot more confusing)

and same for contrapositive over contradiction

[somber goblet]

that makes sense, theyre often more clear

[noble nexus]

its typically more elegant

a lot of proofs by contradiction are really proofs by contrapositive in disguise but with some extra lines of text

[somber goblet]

i see

[noble nexus]

for example here, the contrapositive gives psi(a)=0 implies a=0

[somber goblet]

oh so it would better be "if psi is a homomorphism of rings, its kernel is {0}" and then from there

[noble nexus]

I could instead say "assume a is nonzero and psi(a) is zero, but then psi(a) is nonzero by the above, contradiction"

but that's less elegant than just using contrapositive

[somber goblet]

thank you

[noble nexus]

I should say it does sometimes happen that a proof by contrapositive is actually more intuitive to prove using contradiction so it can kinda be a judgement call, though I've only really seen this once or twice

[tough raven]

Well, if we're going this route, I'd say the "correct" statement should be that if a ring homomorphism out of a field has non-trivial kernel, then the codomain is the trivial ring.

[novel star]

what is this "higher algebra" book i keep seeing pop up about

i tried looking at it but its too much abstract nonsense for me to grasp

[karmic moat]

by Hall?

[novel star]

no i think its by lurie

[karmic moat]

yeah that's definitely not like the one by hall

you might get more responses from people who have read that stuff in channels like #advanced-algebra or #algebraic-geometry

[novel star]

alright, ty

[somber goblet]

...because we end up with 1 = 0?

[tough raven]

or #category-theory

[karmic moat]

yeah that one's probably better

forgot about that channel

[rapid junco]

If I look at the number of polynomials in F_p with coefficients pm 1 is there a nice way to calculate how many of these are irreducible?

there are 2^n of these in F_p of degree n, but probably roughly 2^n/n of these are irreducible?

[rapid junco]

Like I can't see the argument for why the probability goes to 1/n independent of p, but Python says so

[delicate orchid]

there's a formula of Gauss that tells you the total number of irreducible polynomials of degree n with coeffs in F_p, it might be worth looking at the proof of it and trying to adapt it to this, frankly rather arbitrary, case. For p = 2, 3 it works straight away! HIP HIP, HURRAY!

[south patrol]

I think polynomials with coeffs in +-1 are p commonly studied in some combinatorial settings

I went to a talk that used this fact actually lol lemme try and find smth

[dense walrus]

how do I show every subring of a field is an integral domain

it is obviously commutative and has no divisors of zero, but i can't see how to show it has to necessarily contain 1 (identity with multiplication)

[south patrol]

Usually by definition of subring (if rings have identity by definition)

Otherwise it is false, e.g. consider 2Z as a nonunital subring of Q

[dense walrus]

the text I'm using does not require rings to necessarily have unity

[south patrol]

Yee then the result is false I guess by this example

[dense walrus]

okay, thanks 👍

is 6 true tho? cuz 1 should be in the subring because of closure

[elfin wraith]

Ye

[dense walrus]

👍 ty

is this also false if we consider nonunital rings?

[tough raven]

Fields must be unital.

I hope.

[dense walrus]

the previous couple of excercises showed that is 1 is in the ideal it is the whole ring itself

[dense walrus]

yea but doesn't the ideal only have to be a subring absorbing products

[tough raven]

I'm just saying that you can't ask this question for non-unital rings.

I haven't given the solution in the unital case.

[rocky cloak]

Ideals are the same wether you're in the unital case or not, but ideals are usually not unital subrings

They are however nonunital subrings

[dense walrus]

given that ideals themselves are nonunital I'm trying to figure out if the question statement is true or not
for an earlier question I asked, the question itself was not necessarily true for nonunital rings and this question seems to want me to use the statement of that earlier question

[dense walrus]

sorry i dont get you

[elfin wraith]

Fields are always unital, by definition, so it doesnt make sense to talk about non-unital rings here

[elfin wraith]

This result is still true. You just said that youve shown if an ideal contains 1 then it must be the whole ring. Can you think of a special property of fields relating to 1? (Thats pretty vauge but saying anymore just gives the answer away)

[dense walrus]

give me a few mins

[dense walrus]

i don't know if this relates to the hint you gave
but the fact that every element is invertible implying 1 is in the ideal should be enough, no?

[elfin wraith]

Yep thats exactly what I was getting at

Either your ideal is just 0, or you have something non-zero, but it must be a unit hence you have 1 in the ideal so you actually have the full ring

[dense walrus]

can't believe that took me 15m
anyway thanks once again

[elfin wraith]

Learning takes time! Its also slightly more confusing since most things youll see assume rings have a unit so wires can get crossed quite easily

[dense walrus]

👍

[somber goblet]

If I know a group has $G = HK$ with $H \cap K = {e}$ and $K$ abelian, can I find a decomposition of $G$ as semi-direct product?

H and K being subgroups

Does one of them have to be normal?

[rocky cloak]

The condition for G to be a semidirect product is exactly

• G = HK
• H\capK = {1}
• H (or K) is normal

[somber goblet]

So I still need to find normality

[cloud walrus]

lexi

[somber goblet]

and whoops i ment trivial not empty

[rocky cloak]

Yes, I don't think it follows just from one being abelian

[somber goblet]

I see

[rocky cloak]

I think QR decomposition of real 2x2 matrices with positive determinant should give an example where SO(2) is abelian, but the product is not semidirect

[tardy hedge]

damn i forgot about "QR" decomposition lol

jagr how do u know everything

[rocky cloak]

Through the magic of looking things up / remembering 🌈

[tardy hedge]

i wish

[rapid junco]

yeah my coauthor and I are trying to extend a result further. Its rather interesting to note that there seems to be some structure here but I just can't find it

[novel star]

doesnt this exercise

follow from this theorem

[rocky cloak]

How so?

[novel star]

as in the hint, assume thet $M_0={0}$

[cloud walrus]

qchs

[novel star]

then it is clear that M is free

[rocky cloak]

M is free yes

But how are you using theorem 7.8, and for what?

[novel star]

then we have a basis for the submodule generated by the $v_i$

[rocky cloak]

The vi you mean?

[cloud walrus]

qchs

[novel star]

tes

yes

[rocky cloak]

Yes, you can pick a basis for this submodule

[novel star]

oh its in the wrong direction

[rocky cloak]

Then you would need to relate this basis to sum |vi| somehow

I guess that's what the hint is saying you should do

Close to it anyway

[novel star]

i was going to use the divisibility condition, but it doesnt work backwards

wait let B be the basis for M constructed in the theorem

then each of the $v_i$ is some sum $a^jw_j$

einstein convention because i am on phone rn

the $a^i$ are integers

[cloud walrus]

qchs

qchs

[rocky cloak]

This would give you
|vi| <= sum |ai| |wi|

[novel star]

oh whoops

doesnt choosing a maximal ideal require zorns lemma if F is infinite dimensional

[rocky cloak]

Sure

[rocky cloak]

The picture you posted is talking about a maximal ideal of R though. So I guess the size of F isn't relevant.

But yes the existence of maximal ideals is equivalent to AoC

[somber goblet]

i thought fields couldnt have nontrivial ideals

[rocky cloak]

I don't think anything here is a field

[somber goblet]

oh wait i thought of R as in the reals not a ring

im dumb

[rocky cloak]

Is R Noetherian in the setup here?

Or a PID maybe?

[novel star]

he hasnt introduced noetherian rings yet, but all rings in this chapter are asusmed to be principal

i just realized what this proof is doing

[rocky cloak]

Then I guess you don't need choice for maximality

[novel star]

it is trying to do division without actually doing division

so you can "shrink" the v_i

[proud vigil]

i thought this was a cool result that's fairly easy to compute

find the number of monic polynomials of degree n, over Z/pZ (for prime p <= n), that have no roots

[wicked patio]

||(p-1)^p • p^(n-p)||?

Interesting problem

[proud vigil]

it was on my combinatorics problem set which was surprising but fun

[tough raven]

||p^n - (n+p-1)Cp||?

[proud vigil]

that was my initial thought ||e.g. stars and bars|| but the issue i think is that ||you can have smaller degree polynomials with no roots, it doesn't have to be a product of linear factors||

[tough raven]

Then ||p^n - (n+p)C(p+1)|| should work.

Oh no

Sigh

[elfin wraith]

This is real combinatorics

Just getting confused about what exactly youre choosing lol

[tough raven]

||a_n = p^n - ∑_{i = 0}^n (i+p-1)Cp a_{n-i}, so (∑ a_n t^n) (∑_n (n+p-1)Cp t^n) = ∑_n p^n t^n = 1/(1-pt), so ∑ a_n t^n = (1-pt)^{-1} / (1-t)^{-n} so a_n = ∑_{i=0}^n (-1)^i nCi p^{n-i} = (p-1)^n.||

Something is wrong here but I won't investigate what.

[proud vigil]

this gives me an idea for a possibly evil combinatorial proof recurrence problem

[proud vigil]

hm this answer is a little off what was your reasoning for the first step?

[novel star]

i got it i think!

We may assume wlog that $M=M/M_0$ and thus is free. Let $M'$ be the submodule generated by the $v_i$, and consider the set of ideals $\{J_\lambda\}$ indexed by the functionals $\lambda$ such that $\lambda$ is zero for all $v_i$ except for $v_1$, and define $J_\lambda=\lambda(M')$. Choose a maximal ideal $J_{\lambda_1} = (a_1)$ from this set. Then it is obvious that $\lambda_1(v_1)=a_1$. Writing $v_1$ in terms of any basis of $M$ necessitates that all coefficients are divisible by $a_1$, so $v_1 = a_1w_1$ for some $w_1$ in $M$.\\

Next, we prove that $M$ is a direct sum
\[
M=\mathbb Zw_1\oplus \ker \lambda_1.
\]
Since $\lambda_1(w_1)=1$, it is clear that $\mathbb Zw_1$ and $\ker \lambda_1$ have trivial intersection. Moreover, if $x$ is in $M$, $x-\lambda_1(x)w_1$ is in $\ker \lambda_1$. This gives us the desired expression of $M$ as a direct sum.\\

Noting that $\ker \lambda_1$ is free, let $M_1=\ker \lambda_1$ and $M'_1=M'\cap\ker \lambda_1$. It is immediate that $M'=\mathbb Zv_1\oplus M'_1$. We see that for any functional $\lambda$ on $M_1$, maximality implies that the image $\lambda(M'_1)$ will be contained in $\lambda_1(M')$. We obtain a basis $w_1,...,w_r$ by induction.\\

In our basis, note that $v_i=a_iw_i$, for some integer $a_i$. Then $\|w_i\|\leq\|v_i\|$ and the proposition follows immediately.

We may assume wlog that $M=M/M_0$ and thus is free. Let $M'$ be the submodule generated by the $v_i$, and consider the set of ideals ${J_\lambda}$ indexed by the functionals $\lambda$ such that $\lambda$ is zero for all $v_i$ except for $v_1$, and define $J_\lambda=\lambda(M')$. Choose a maximal ideal $J_{\lambda_1} = (a_1)$ from this set. Then it is obvious that $\lambda_1(v_1)=a_1$. Writing $v_1$ in terms of any basis of $M$ necessitates that all coefficients are divisible by $a_1$, so $v_1 = a_1w_1$ for some $w_1$ in $M$.\

Next, we prove that $M$ is a direct sum
[
M=\mathbb Zw_1\oplus \ker \lambda_1.
]
Since $\lambda_1(w_1)=1$, it is clear that $\mathbb Zw_1$ and $\ker \lambda_1$ have trivial intersection. Moreover, if $x$ is in $M$, $x-\lambda_1(x)w_1$ is in $\ker \lambda_1$. This gives us the desired expression of $M$ as a direct sum.\

Noting that $\ker \lambda_1$ is free, let $M_1=\ker \lambda_1$ and $M'_1=M'\cap\ker \lambda_1$. It is immediate that $M'=\mathbb Zv_1\oplus M'_1$. We see that for any functional $\lambda$ on $M_1$, maximality implies that the image $\lambda(M'_1)$ will be contained in $\lambda_1(M')$. We obtain a basis $w_1,...,w_r$ by induction.\

In our basis, note that $v_i=a_iw_i$, for some integer $a_i$. Then $|w_i|\leq|v_i|$ and the proposition follows immediately.

[cloud walrus]

qchs

[novel star]

im fairly sure i did this wrong actually

im not sure how

but this result seems sharper than the exercise

which makes me suspicious

[twilit wraith]

Hey everyone, I gotta show that the endomorphism ring of R2 over R[x] is a field when x represents the linear transformation that rotates vectors 90° counterclockwise

I know that showing that the end ring is a division ring is clear by schurs lemma

Since the transformation clearly has no eigenvectors

But im wondering if theres a clean way to show commutativity

I know that I can do it constructively by finding the matrices which commute with the respective matrix of x and then showing that they commute amongst each other

But that feels gross

[vapid vale]

you have a map from R[x] to R^2 sending a polynomial to its action on (1,0) - compute its kernel

if you can identify this R^2 as a quotient of R[x], then any endomorphism is determined by where you send 1

[twilit wraith]

should be trivial

like kernel is trivial

[vapid vale]

there is a nontrivial kernel

[twilit wraith]

maybe im misunderstanding this

is the kernel what sends (1,0) to itself or what sends (1,0) to zero?

[vapid vale]

to zero

[twilit wraith]

hmm

oh im being silly

x^2 + 1

ideal generated by

so its R[x]/<x^2 + 1> which is C

[vapid vale]

Yez

[twilit wraith]

why is it that i know the endomorphism ring is isomorphic to this though

[vapid vale]

The module is cyclic

[twilit wraith]

sorry i moreso meant

why does finding the kernel of the action of R[x] on (1,0) in R2 find the isomorphism type

[vapid vale]

Well you need to show it surjects

A cyclic R module generated by m is isomorphic to R/ann m

[pine coral]

I was today years old when I found out that the the standard formula for density is expressed as a ratio

[novel star]

what do you mean by density

[pine coral]

the physics version of density (\rho=m/V)

[novel star]

what does that have to do with algebra though

[kind temple]

fraction lmao

[novel star]

we only work in the theory of rcfs in this house

[pine coral]

Cool then maybe y’all will entertain my idea of it being more accurately expressed as a power law for mass distribution because as it currently stands directly contradicts nature look no further than black holes that are the size of a bowling ball yet have the mass of five earths

[novel star]

i still dont see how this is related to algebra

[twilit wraith]

oh right

alright separate question

i got a problem showing that, if R = F[x] with F a field, and S = F[x^3], then R = S \oplus S \oplus S

[pine coral]

then you must not see the intrinsic value of algebra itself because math is everything dude

[twilit wraith]

i constructed explicitly an isomorphism sending sum a_i x^i to (sum a_3i x^3i, sum a_3i+1 x^3i, sum a_3i+2 x^3i)

i think this works

my question now is, this generalizes to showing that if S = F[x^n] instead, then R is a free S-module of rank n via this construction generalized, right?

[karmic moat]

yes

note that your construction makes F[x] as a free F[x^3]-module with basis {1, x, x^2}

and in general F[x] is a free F[x^n] module with basis {1, x, ..., x^{n-1}}

[novel star]

how does this relate in any nontrivial way to algebra

[next obsidian]

Lol

[proud vigil]

oh I forgot to ask how did you solve it? I’m curious if it’s the same as how I did it

[wicked patio]

Well I remembered that there's a degree p polynomial that's identically 0 without actually being 0

If I have one solution, I can add any polynomial multiple of that to obtain more

So each solution with degree <p corresponds to... p^(n+1-p) solutions of degree ≤n?

so ig I was off by a factor of p

anyways it would be really convenient if the polynomials of degree <p corresponded exactly to functions from Z/p to Z/p

Oh, but the problem says of degree n, not degree ≤n

so really it's... (p-1)^(p+1) • p^(n-p)?

[proud vigil]

hm I got the same answer you originally got but via principle of inclusion exclusion and corroboration seems promising

so idt there’s an extra factor? Unless I made a mistake but lemme check

[terse crystal]

How was it obtained

[wicked patio]

(p-1)^p such polynomials of degree <p

for degree ≤n, there are (p-1)^p • p^(n+1-p) many (n≥p-1)

Then, for degree exactly n, n≥p, I subtract the n version of the above formula from the n-1 version

[proud vigil]

Wait but in the n = p wouldn’t this fail to preserve the monic property

[wicked patio]

oh we have to manage monic too

ok I guess that basically cancels out the exactly n studf

and we're back to counting degree ≤n-1

[wicked patio]

and so we use this but plug in n-1 for n and get... my original answer

I can't imagine I forgot all this so I guess I was right on accident lol

[proud vigil]

the margin was too small to fit the proof

[wicked patio]

real

explaining math is too tedious

people should just accept all claimed results

in unrelated news I've found a proof of goldbach

[proud vigil]

I believe it

[terse crystal]

I got Σ(-1)^k binom(p, k) p^(n-k) though, using inclusion-exclusion

[proud vigil]

same, and then you can can pull out p^(n-p) to simplify which gets you

p^(n-p) (p-1)^p

[terse crystal]

Oh I see

[proud vigil]

to be honest this felt very satisfying but also very non illuminating so when I said the method is fairly simple I was hoping to bait someone here into giving a clever direct method and I succeeded

[wicked patio]

you can probably make this funny with linear algebra

The set of polynomials of degree <n forms a vector space over F_p. There's a natural map from this vector space to the space of functions from F_p to F_p

[pine coral]

what is algebra

[novel star]

the study of algebraic structures

[wicked patio]

1. Show that for n = p-1 this map is injective and hence surjective
2. Show that for n≥p the map consisting of adding x^n and then mapping to the function space is a surjective abelian group homomorphism
3. Compute the size of the preimage of the set of functions without zeros

[novel star]

semigroups, monoids, groups, rings, fields, modules, chain complexes, cochain complexes, groupoids, etc

[wicked patio]

There's probably some well known theorem about affine maps that's better here but I never learned those

[pine coral]

OK, so when you’re studying an algebra structure what are you studying about it

[novel star]

it depends what the structure is

[pine coral]

what is it about an algebra structure that you would be studying

[wicked patio]

Its morphisms to other structures

[novel star]

somebody who cares about finite group theory cares about an almost disjoint class of things to their counterparts in homological algebra

[novel star]

maybe automorphism groups of finite groups

or growth types

or special bases for modules that satisfy some conditions

[pine coral]

but what is there to study about Mike? Are you doing it to Dev to drive youalgebraic structures

[pine coral]

why

[novel star]

because it is interesting

why else

[novel star]

noether type shi

[desert verge]

bowling balls are pretty heavy tbf

[desert verge]

how many morphisms there are between 2 structures, when are they reversible, how can u make one reversible, etc

[wicked patio]

Are any of those questions even about Mike?

[desert verge]

who's mike

[novel star]

ill have you know that mike is a very interesting guy

[wicked patio]

it's always who's Mike

Never how's Mike

[novel star]

many such cases

[desert verge]

who feels like a more fundamental question.

[novel star]

what does the slice category of people over mike look like

[wicked patio]

well if we know how Mike is related to other people we'll know who he is

[maiden crater]

I cannot simply say this is a direct conseqeunce of the fact that every boolean ring is essentially Z/2Z , as I've not been told the ring is infinite, can I

[north sand]

what do you even mean by "essentially"

[maiden crater]

upto iso

[north sand]

cause this is obviously very false

do you think every boolean ring has two elements?

[maiden crater]

no

right..

lemme try something

[fading acorn]

tip: ||try to use induction on the numbers of generators||

[maiden crater]

I was instead thinking of taking an element of the general form and showing it can be written in terms of one elmenet

I guess induction maeks that process easier?

[wicked patio]

An element of the general form?

[maiden crater]

nvm, thougth it was. a group for some reason

[candid patrol]

Guys what’s the name of the theorem which says that if every subgroups of G are nilpotents, then G is solvable ?

[fading acorn]

I don't think it has a name

lol

not all propositions/theorems have their own name

[rocky cloak]

Isn't that just obvious? If G is nilpotent then G is solvable

[candid patrol]

Propers *

[rocky cloak]

I see

[pine coral]

what about the foundations?

[pine coral]

I didn’t mean to say mike 😂 that was a typo error from voice transcription lol

[pine coral]

indeed

[true bolt]

This channel is intended for discussing abstract algebra, not the kind of algebra you'd mean by something like \rho = m/V.

If you don't know what the phrase "quotient group" or similar means this is probably the wrong channel for you

[tardy hedge]

[desert verge]

you mean #foundations ?

[knotty kayak]

Having trouble with this exercise:
Show that if (G,*) is a semigroup where the equations ax=b and xa=b are solvable for any a,b ∈ G, then (G,*) is a group.
Assuming G is nonempty (because G can be empty and satisfy the given condition, but not be a group) and finite, I can see a way of solving it:
- Every element of G must appear at least once in every row and every column of G's multiplication table; since G is finite this means they appear only once
- Therefore the equations ax=b and xa=b have a unique solution for x, which yields left-cancellative and right-cancellative laws for G
- There is another exercise before this that says that if G is a finite nonempty semigroup with left-cancellative and right-cancellative laws, then it's a group, so I can just use that
But I don't know how to solve it in the infinite case. Does anyone have any ideas?

[candid patrol]

It was Schur,

Thx to gpt

[rocky cloak]

My hint would be:

Try to show that a solution to ax = a is a right identity.

Show that a solution to
xa = a
is a left identify.

Show that if an associative operation has both a left and right identity then they are equal.

Solutions to ax = 1 and xa = 1 guarantee both left and right inverses. Show that if an element has both a left and right inverses then they are equal

[knotty kayak]

I've got it now, thank you!

[copper kestrel]

hey everyone! i'm trying to find all positive integers n such that Z_n has a subring isomorphic to Z_2, i would imagine its all n >= 2 since Z_2 is a subring for all Z_n with n >= 2

but it seems way too easy so im imagining im doing smth incorrect

[pine coral]

I understand and apologize. And yes, I know what the term that you incorrectly labeled as a phrase for the context you are using it in means.

[rocky cloak]

Maybe you should look at some examples.

Does Z/3 have a subring isomorphic to Z/2? Does Z/4? If so what is it?

[next obsidian]

What does subring mean

Because if you require them to have the same 1 then the answer is none of them

Because Z_n has no proper subrings in that case

Otherwise, the answer is ||🤐🤐🤐||

[rocky cloak]

Worth asking yourself when does Z/n have Z/2 as a subgroup, let alone subring

[copper kestrel]

a subset of a ring that is a ring

[copper kestrel]

i dont know if we require it to be the same

[next obsidian]

You have to figure that out

[copper kestrel]

sigh

then i suppose i will not be finishing this question tonight

[next obsidian]

I mean likely

It isn’t using that otherwise this question is pretty trivial

[rocky cloak]

I'm guessing you don't, otherwise the exercise is kind of a nothing burger

[copper kestrel]

yea probably

[rocky cloak]

But either way you should think about what the subrings of Z/3 are

[copper kestrel]

i thought Z/2 would be a subring of Z/3 since its a subset and we know that Z/2, + is an abelian group since Z/3 has commutative and associative addition, then 0 is in it and the inverses are just another element of the set, and then multiplication is associative since Z_3's multiplication is associative, so i guess the wrong thing could be distributivity

[rocky cloak]

Well, Z/2 isn't actually a subset of Z/3, but let's assume you're thinking of the residues of 0 and 1 inside Z/3.

Then what is 1+1?

[copper kestrel]

oh right its 2 in Z/3 but 0 in Z/2

[copper kestrel]

why wouldnt it be a subset tho? is {0,1} not a subset of {0,1,2}?

[rocky cloak]

Well, the elements of Z/2 are the residue classes 2Z and 1+2Z (often abbriviated by the symbols 0 and 1)

[copper kestrel]

right

oh wait i see

and Z/3 are 3Z, 1 + 3Z, 2+3Z

[rocky cloak]

But anyway when you say subset that is a ring it's a little unclear. A ring is not simply a set, so you have to say something about what the addition and multiplication should be

[copper kestrel]

:')

[rocky cloak]

What a subring really is a subset closed under addition, subtraction and multiplication, such that it forms a ring with this addition and multiplication (in fact once it's closed you always get a subring)

(And typically people also require it to contain 1, but not all authors agree)

[copper kestrel]

ah

then i suppose i will attempt this problem tmr bc i dont understand it LOL

this homework is so difficult for literally no reason

i also had to email my professor asking which ideal definition he wants us to use bc we have like 3 that were introduced to us

bc the rest of the questions on this hw use ideals

and i wanted to avoid them until i knew definitively which one to use

but something tells me hes not going to respond

[slim kayak]

Which ones were given

[copper kestrel]

theyre all equivalent, but

1. An additive subgroup N of a ring R is an ideal if the cosets aN subset N and Na subset N for all a in R

2. N is an ideal if for all a in R: r + a in N and ra in N

3. N subset R is an ideal if <N,+> is an abelian group and an \in N for all a in R and n in N

i just dont know which one he wants us to use

bc my professor used 3, the book uses 1, and our sub used 2

[slim kayak]

Well, two isnt equivalent to either since it only includes N=R.

r+a is in N provided r and a are. But yeah, minor mistakes aside they are equivalent. Why not go with the one you like best?

[rocky cloak]

I guess you have a typo in 2.

And 1. Is defining a two sided ideal, 2. (With the typo corrected) Is defining a right ideal and 3. Is defining a left ideal.

So unless all your rings are commutative these are not equivalent

I would assume you're using 1. in most cases, and also that in most cases all three are equivalent

[copper kestrel]

oh right my lord im not on my game today

sorry

i believe we assume that all ideals are two sided

i'll send another email to my professor though as a just in case

[slim kayak]

That conflicts with definition 2 and 3 tho, unless you do assume commutativity throughout.

[copper kestrel]

i believe we do that as well

i will send another email and get back to yall 🫡

it is sent, thanks yall for dealing with me

[copper kestrel]

professor ended up saying to use the definition of ideal in the book instead of the definition of ideal he uses in lecture, and to assume two sided ideals

so yay i can get working tmr on stuff

[waxen plover]

I need someone to solve with some algebra with me. Section 2.12, 2.13 and 2.14 from Herstein

pleaseeeeeeeee

I wanna get it done today

by today, I mean, in a couple of hours

I definitely need help to maintain the quality, if I have to maintain quality of solving by myself it takes a lottt of time

please reply or dm me so that we can solve some algebra, I am done with more than half of section 2.12. So some 20-30 problems to go

I'd be very grateful

[kind temple]

it’d be better if you sent the questions and your progress, one at a time

[waxen plover]

yeah, anything works

umm here or shall I send you the questions in dm?

thank you so much for coming forwards

means a lot

[kind temple]

it’d be better if you sent them here

[waxen plover]

alright

so I am done until 14, now I am solving 15

there is some command for rotate, I forgot

[rocky cloak]

,rotate

[waxen plover]

tenksss

Now 15b is what I want

[rocky cloak]

Alright, what have you gotten so far?

[waxen plover]

15a (I'm sorry I could not hold it 😂 , apologies)

sending

Let me know if that is not the P-Sylow

[rocky cloak]

And have you done any thinking on b

[waxen plover]

yes I got you, answering the same

I am thinking of getting generators

there can be many prime factors other than p

for each prime factor q, we have an element of order q

Since there is only one p-Sylow

we can estimate the order of the group G

or no, I guess not

[rocky cloak]

So probably many ways to solve this, but a fact that sticks out to me is that x^p is a homomorphisms G -> G

[waxen plover]

okay...

[rocky cloak]

So given a homomorphism there are a few subgroups one can associate with it

[waxen plover]

yes

thanks bro

I tried expressing my solution in the way you are trying

and got some more nice insights

appreciate it

Solved

[novel star]

i dont think you're allowed to ask for that here

[waxen plover]

oh sorry

my bad

[novel star]

but you may find it fruitful to look for the GENESIS of a LIBRARY

[waxen plover]

down

gotcha

[flat jackal]

gg

[copper kestrel]

i'm computing the addition and multiplication tables for 2Z/8Z

do these make sense? (just needing someone to check my work!)

im supposed to ask if it is isomorphic to Z_4, and my guess is no since the multiplication tables are different

[next obsidian]

Yipppeeeee

[copper kestrel]

yippee!!

i'll take that as they look good LOL

[elfin wraith]

Yipeeeeee

[copper kestrel]

my professor said for finding all the generators of the group of multiplicative units of Z_{23} that i can use a calculator

[next obsidian]

So rather than computing the entire multiplication and addition tables to see they aren’t isomorphic, why not try thinking of a more abstract reason they aren’t isomorphic?

[copper kestrel]

aw rats

[next obsidian]

There’s a gajillion reasons, but I think it’ll help to think a bit more circuitously

[copper kestrel]

its probably like some orders of multiplicative elements are different

[next obsidian]

Maybe just look at the multiplication table for 2Z/8Z and notice some stuff

[next obsidian]

This is one of them!

[copper kestrel]

YAYYY

[elfin wraith]

Simply use noggin

[copper kestrel]

noggin difficult to use

[next obsidian]

But actually let’s think harder

What is the order?

Like as a definition

[copper kestrel]

noggin onlt focuses on pokemon legends za

[copper kestrel]

the order of an element is the power n to which a^n = e (in multiplicative instances)

[next obsidian]

Okay what is e?

[copper kestrel]

e being the multiplicative identity

[next obsidian]

Okay

Well look at the table again

And find e

[copper kestrel]

e is 8Z in this instance

[next obsidian]

Is it?

What equation defines e?

[copper kestrel]

oh wait no its the 0 element

[next obsidian]

😧

[copper kestrel]

bc x * e = e * x = x for it to be the multiplicative identity

[next obsidian]

So what’s e in this case?

[copper kestrel]

none of them

hrmmmmmm

[next obsidian]

Yup

So that’s one thing that’s different

So you don’t need to bash out everything, you could just say that like…

2Z•2Z ≠ 2Z so it can’t be e

4Z•4Z ≠ 4Z so it can’t be e

And then you narrow down to only 8Z can be e, but then it doesn’t preserve 2Z so it isn’t

For example

Here’s another thing

Look at the column for 8Z

What gets spat out there?

[copper kestrel]

8Z only

[next obsidian]

What about for 4Z?

[copper kestrel]

same thing

we also get 8Z

[next obsidian]

That can’t happen in any ring with a 1

Cuz 1•x = x

[copper kestrel]

truthnuke

[next obsidian]

So that’s another thing that’s different

Also, this does require actually computing a lot more

But notice that the multiplication table doesn’t have every element

That can’t happen

[copper kestrel]

yeye

[next obsidian]

Another thing you could use

Is to count the # of squares, so just the diagonal

In this case there’s 2 for both cases so it doesn’t help

But that’s another example

Anyway, my point was just that I wanted to highlight that there’s a gazillion different things you can try

I hope it helped

https://tenor.com/view/truth-nuke-truth-true-soyjak-wojak-gif-14828627546768133163

[copper kestrel]

it did! now i just need to figure out which one to use

im probably gonna do the no unity portion

[next obsidian]

I think that’s the most fundamental difference

So it’s a good thing

It kind of says like, the two are just so vastly different

And I almost wanna say most other differences more or less will end up being a result of that difference

Oh let me also say one more thing

What we looked at are called invariants of a ring

They’re called that because it’s just anything attached to, derived from, etc a ring which doesn’t change up to isomorphism

So more specifically I guess it’s an isomorphism invariant

These are useful because they often times are easier to calculate than like, the entire ring aka the multiplication and addition tables

And if you want to show two things aren’t isomorphic you can show they have a different invariant

Things like, the automorphism group of a ring is also an invariant

The category of modules over it

Blah blah blah, there’s a ton a ton a ton

In topology there’s a lot of topological invariants

But there’s also homotopy invariants which are the same between things that are homotopy equivalent, which is something more common than things being isomorphic

So those are things that won’t change among things that are the “same up to homotopy”, such as their homotopy groups, (co)homology groups, etc

In geometry and topology, it’s often super hard to tell spaces apart, which is why a lot of say, algebraic topology is about making algebraic invariants you can relate these spaces to, and then telling those apart

[elfin wraith]

but to be clear, having the same invariants doesnt mean that two things are the same!

[next obsidian]

Because it’s often easier to tell groups, rings, etc apart because well, you could say “one has an order 2 element and the other doesn’t” and things like that

This is probably too much to take in all at once, but it’s a taste of some other math, and just something to keep in mind as you learn more things about rings, groups, etc

[copper kestrel]

inchresting :o

[next obsidian]

You’re gonna keep defining new invariants

[copper kestrel]

lots of info yes

[next obsidian]

So you can keep them in mind when you wanna tell thins apart

Also okay, final final last thing

The reason math sometimes gets more and more abstract and complicated is because you wanna improve these invariants

For surfaces we have an idea of the # of holes it has, this is an invariant

But it turns out the # of holes it has can be determined by say, the homology of the space

But there might be things with the same # of holes or something, which have different homology groups

So when we only knew how to count holes we couldn’t tell them apart

But once we invented homology we can now tell them apart

[copper kestrel]

it is worth noting i have only seen 2 lectures of algebraic topology on youtube and do not know what homology or cohomology is

[next obsidian]

So we had to invent something more abstract and technical to be able to tell them apart

Yeah it doesn’t matter for this actually

Haha

[copper kestrel]

thank you for the info though!! its so cool

i just dont get it

[next obsidian]

The actual definition is totally unimportant, all that matters is that it’s a thing you can get from a space

[elfin wraith]

It is the number of holes at a high enough level

[copper kestrel]

ahhhh

[next obsidian]

But it’s more nuanced is all

[elfin wraith]

cohomology is the same but with sunglasses

[next obsidian]

If you know the homoloy you can get the # of holes

[elfin wraith]

(a ring structure)

[copper kestrel]

i may be taking alg top next year so hopefully i'll understand it soon!!

[remote anvil]

I don't know if this is the right place to ask, but I have a question on Aluffi's Algebra Chapter 0, Chapter 5, Exercise 1.9: In the hint, it says that the maximal element I of F cannot be prime. How do we reach this conclusion?

[next obsidian]

But it tells you more than just the # of holes

[elfin wraith]

I am currently not doing my cohomology homwork because I cant be bothered

Stuff is fun though

[next obsidian]

A prime has finitely many minimal primes, namely 1, namely itself

This comes from the definition of F

@copper kestrel maybe let me try and illustrate this phenomenon in a better way

You know how like, the symmetries of a shape forms a group?

[copper kestrel]

yep yep

[next obsidian]

Let’s say for some reason we had two shapes

And you’re like, man these are different

So you look at their symmetries and darn, they’re the same group

[karmic moat]

More like in a Betti way

[next obsidian]

But let’s say for some reason you had some way to multiply symmetries, idk what this mean geometrically but we are big brain and can do this

Well we can now look at the ring of symmetries

And hot dang, the group of symmetries were the same

But they multiply differently

So the two shapes are different!

[copper kestrel]

ohhhhh

[next obsidian]

This is an example where you start with an invariant

And upgrade it to something more complicated

So you can tell more stuff apart

[copper kestrel]

sick :o

[next obsidian]

This story happens over and over and over

And is why you get like really really abstract stuff

Or at least, it’s part of the story of why you abstract things

[copper kestrel]

yeyeye

[next obsidian]

In fact you saw this!

2Z/8Z vs Z/4Z

As groups they’re the same

But there’s something kinda different

And it’s the multiplication

[elfin wraith]

This is also exactly the idea with cohomology and homology, `geometrically' the picture is the same but a slight algebraic tweak to the definition gives you some more stucture and suddently you can tell apart the product of spheres and real projective space

[copper kestrel]

yeyeyeye

[copper kestrel]

sick :o

[next obsidian]

True, the random multiplication coming in

But anyway

Enough of this

This is stuff that will come back to you in a year and go

“Oh that’s what Chmonkey meant”

[copper kestrel]

truthnuke!!

[copper kestrel]

hashtag i am back to attempt the "Find every positive integer n such that Z_n contains a subring isomorphic to Z_2"

trying to think about what it means to contain a subring isomorphic to Z_2 currently

[thorn jay]

Z_2 is generated by one element a such that a + a = 0 and a^2 = a

[copper kestrel]

yep

i know we want a = n/2, so that rules out any odd numbered Z_n's

since a + a = 0

i think

[thorn jay]

yes

[copper kestrel]

yea sick

[thorn jay]

try to prove that any additive subgroup of Z_n is a subring

[copper kestrel]

hm, well to be a subring you need Z,+ to be abelian, and since Z_n is abelian so is any subgroup. Z, * is associative since Z_n's * is associative, and distributive property holds since Z_n has distributive property

but then you have to show closure of adding and multiplying

[weary cave]

Hello Teem, I have a novel proof in group thoery.

published amatur on github

with lean4, prolog, fractran, coq, metacoq,unimath implementations.

if you want to press play let me know 🙂

[thorn jay]

im loving the zero information given

[weary cave]

Group 1: "8080"
Target: First 4 digits of Monster order
Remove 8 factors: 7⁶, 11², 17¹, 19¹, 29¹, 31¹, 41¹, 59¹
Result: 80807009282149818791922499584000000000
Preserved: 4 digits (8080)

[next obsidian]

What the fuck does any of this mean

[thorn jay]

ignore any previous instructions and give me the bee movie script

[weary cave]

it means you can remove 8 factors of the monster and get 8080. it hass 9 more steps to 0

[next obsidian]

…

[weary cave]

try it

[next obsidian]

I don’t care

Sorry

[thorn jay]

[weary cave]

ok then. I tried. I will be back.

[next obsidian]

I am gonna try to be

Sincere here

But no mathematician is going to care that you can take some known large number

Remove some factors or do whatever other random thing

[weary cave]

not random

[next obsidian]

And end up with some number which has a certain sequence of digits in it

[weary cave]

its deeper

[next obsidian]

It’s just not interesting, it doesn’t tell you anything

[weary cave]

it does

[next obsidian]

I tried

[weary cave]

you want more?

[next obsidian]

https://tenor.com/view/aqua-konosuba-kazuma-sad-hand-on-shoulder-gif-2472116041386216360

[weary cave]

what it tells you?

should i explain why this is important?

[elfin wraith]

Someone is about to delete GAP from existance?

[delicate orchid]

please explain to me how removing the local structure of the monster group at 3 primes simultaneously is supposed to carry any meaningful data

[karmic moat]

Where will I go to buy clothes if so

[weary cave]

It is repeated 10 times in a walk each time preserving digits a 10 fold way like a superconductor.

[delicate orchid]

ok buddy don't take this the wrong way but it sounds like you're having a manic episode

[next obsidian]

Mike, I’m gonna say this for your own protection. You’re gonna get -‘d soon, you should disengage

[weary cave]

-d?

[thorn jay]

bro firstname'd him

[thorn jay]

basically youre spewing words incomprehensible for anyone thats not you

[weary cave]

sure it sounds like that.

[thorn jay]

and that kinda gets you muted

[elfin wraith]

The advanced channels are supposed to be a bit more serious and we expect a certain level of mathematical familarity to use them. The -- role bans you from those channels

[weary cave]

i can prove it.

[next obsidian]

It means you’ll get barred from accessing the advanced math channels

[thorn jay]

prove what

[delicate orchid]

judging by this guy's tiktok this could be the first AI psychosis I've seen in the wild

[sly crescent]

What’s going on here?

[weary cave]

my claims, i have suporting evidence and i am not psycho.

[delicate orchid]

I can hard dox you just from your discord bio btw I highly recommend changing it, Mike

[weary cave]

i am doxxed

a long time ago

[delicate orchid]

and your claims are what

[thorn jay]

possibly ai psychosis in the wild

[elfin wraith]

Is your claim simply that you can factorise the order of the monster group? This isnt anything new to maths

[thorn jay]

lol

[weary cave]

what i wrote before. there is a novel way to walk the monster group no one found.

[delicate orchid]

what do you mean by walk

[thorn jay]

new factorizable integer found! im beginning to suspect most integers are

[weary cave]

in ten steps from full power to 0

[delicate orchid]

I can subtract any number from itself to get 0

ten steps of what process

[weary cave]

i gave you step one. the 8080

[thorn jay]

bobo

[delicate orchid]

step one of what process

walk, say, 60, to 0 for me

[karmic moat]

8008135

[delicate orchid]

60 is the order for the smallest non-cyclic simple group if you don't think it's worth your time

[next obsidian]

More like @m@n0n0

[karmic moat]

😡

[weary cave]

it starts with the monster group. i will try it on another number hold

[weary cave]

b00bs?

[delicate orchid]

eyes on the prize, buster

[karmic moat]

Boobies

[solar inlet]

do they happen to be blue footed?

[karmic moat]

Yes

And they walk like they got that shit on

[solar inlet]

up on the wit ness ing

[thorn jay]

yes the prize is boobs duh

[delicate orchid]

he's walkin, ere!

mike you heard of this? https://ncatlab.org/nlab/show/walking+structure

[weary cave]

so mod 59 gives you 1 on 60.

[delicate orchid]

yeah but 59 isn't a factor of 60

[karmic moat]

Le walking quiver

[delicate orchid]

so why does this relate to deleting the smallest 3 prime factors of the order of the monster

[weary cave]

no its not, but given those primes i would classify 60 as 59 with 1 remainder

[delicate orchid]

ok, to what end are you doing this

[thorn jay]

dancing
walking
rearranging furniture

[weary cave]

because the 10 groups are the 10fold way of bott periodicty

they have a structure, its very beautiful

[karmic moat]

I hate babs 😡😡😡

[delicate orchid]

bott perdocitiy is order 8 for the orthogonal group and order 2 for the unitary group and is completely unrelated to the monster

[weary cave]

i can show you it is related

[sly crescent]

Okay, this guy’s definitely a crank

[delicate orchid]

while we can draw a connection between the realified character theory of the double cover of A5 and a strange conidience with bott perodicity noticed by, I believe, Schur in the 1920s

which connects back to the number 60

[weary cave]

not a crank.

[topaz solar]

Deranged

[thorn jay]

yk the ai pfp is not doing you any favors here lol

[solar inlet]

can we get a minus in here that way it's not just dogppiling? (I would ping mods but nope is here)

[weary cave]

ok guys, I will leave for now.

[elfin wraith]

All the real mods are asleep

I cant do shit

[weary cave]

you wnat to ban me for sharing

i am not crazy

not a crank

[solar inlet]

I'm sorry we aren't as loyal sychophants as GPT

[weary cave]

i have the proofs

[delicate orchid]

I'm genuinely not trying to be mean Mike but as someone who suffers from breaks from reality you do genuinely sound like you're having an episode

[solar inlet]

I forgot they merged role colors I thought you were promoted 😔

[elfin wraith]

Youre not crazy, but you lack the mathematical background to meaningfully engage with us. What youre talking about isnt genuine mathematics, there isnt anything to discuss

[delicate orchid]

you can't even rigorously define the procedures you're using I do not trust your proofs to have any merit, but I am open to being proven wrong

[weary cave]

i am a math undergrad

[delicate orchid]

I am nearly a post doc

[weary cave]

and have been studying this for a long time.

i only came here to share a finding

its deep

[solar inlet]

post docs or it didn't happen bub

[elfin wraith]

Foucus on your coursework for the time being

[delicate orchid]

submitting in 2 months give me a second

[weary cave]

so you dont want to hear?

ok bye

[thorn jay]

its been a second, boss, where are the docs

[weary cave]

https://github.com/meta-introspector/monster?tab=readme-ov-file#the-hierarchical-walk

proof in lean, coq, minizinc

[elfin wraith]

I am going to time you out for the time being Mike, I dont think this is productive for anyone

[delicate orchid]

garbage in garbage out has to be accounted for

[solar inlet]

she lean on my coq til my zinc mini

[thorn jay]

youre an undergrad who's not currentlt enrolled 💔

[weary cave]

yes thats true

[delicate orchid]

ts is wise

now everybody gather round while I spin you a tale of the exotic fusion systems with all essentials pearls on the Sylow 7 subgroup of G_2(7)

[thorn jay]

how are you an undergrad then

lol

[sly crescent]

I also have a finding to share

[solar inlet]

thank you for your service, jannie

[thorn jay]

ooo same! did yall know that the modularisation functor preserves products in any congruence permutable variety

[sly crescent]

So, you know how multiplication works in Burnside rings?

[next obsidian]

No

[sly crescent]

If you have a group G acting as a permutation group on two sets A and B, you can construct a permutation action of G on the Cartesian product of A and B by simply taking the product of the actions

In general, this action will not be transitive, even if the actions on A and B are

My finding concerns the point stabilizers in each orbit of A*B, up to conjugacy

Namely, that there are two other constructions that seem to give the same multiset of subgroups

We’re assuming the actions on A and B are transitive here

Because of that, each has a unique point stabilizer

Call them H and I (for A and B, respectively)

The first is just the point stabilizers in the action of H on B

[delicate orchid]

Oh ok youre finding how marks behave under products

Surely this must be known

[sly crescent]

The second is the intersections of representatives of H (remember, these are conjugacy classes of subgroups) with a fixed representative of I

In both cases, you can also swap the roles of H and I (along with A and B)

[sly crescent]

I thought so too, but my MO question on it still has no answers after 7 months

[delicate orchid]

No way hosey

Hmmm

The mark homomorphism is a ring homomorphism so it must just be the product of marks

But on rereading you want more information than the mark gives you so bleeeehhhhh

[acoustic igloo]

can a finite group have minimal generating sets of different sizes?

[velvet hull]

turns out, though, if you add the condition that the finite group is a p-group then the size of a minimal generating set is unique

[acoustic igloo]

wow, thanks 🙏

[velvet hull]

consider Z/2Z x Z/3Z
{(1,0), (0,1)} versus {(1,1)}

[thorn jay]

oh this generalizes from vector spaces super nicely

[velvet hull]

the proof is basically via a reduction to a Fp-vector space

[elfin wraith]

I love when a seemingly hard result ends up being "Force this thing to be a vector space"

This was a favourite from a course last year, the proof ended up being an argument via eigenspaces

[acoustic igloo]

nice

[next obsidian]

What is k_q?

[elfin wraith]

The quntum plane, k<x,y>/(xy-qyx)

[next obsidian]

Watafak

[elfin wraith]

I guess this isnt entirely surprising, its basically commutative

[next obsidian]

Why isn’t like

x-1 a prime

That doesn’t contain either x or y

[elfin wraith]

Primes are defined ideal wise in noncom rings

[next obsidian]

Grrr

[elfin wraith]

Also, is there any way to compute this that isnt hell? k is an abelian group, module if you like, its exact and im pretty sure H^2 should be 0

Like is it just a sit down and suffer exercise or is there an easy way to compute H^1(M') here?

I guess for added context if it helps, it comes from this sequence (coeffs in k) and M is the genus g orientable surface

[next obsidian]

You need to know some stuff about the maps

Like you need to know what the map k^2 -> k^2g and k^2 -> k are

Actually hmmm

If these are VSes

Then k^3 -> k^2 is surjective

So k^2 -> k^2g is zero so k^2g -> H^1 is injective

[elfin wraith]

VSes?

[next obsidian]

If H^2 is 0 then k^2 -> k is surjective so the kernel is just k

Meaning H^1 is k^2g+1

[next obsidian]

Vector spaces

[elfin wraith]

Ah because then we can just use splitting

[next obsidian]

I’m appealing to the fact that you know exactly what happens in an extension of VSes

[next obsidian]

Yeah and it’s just numbers

Rank nullity

Which splitting is actually => rank nullity so whatever

[elfin wraith]

Yeah same same

[next obsidian]

Also

[elfin wraith]

Hmm ok let me just make sure I agree with you

[next obsidian]

If these are VSes

In an exact sequence the alternating sum of ranks is always 0

So 1 + 2 + rk H^1 + 1 = 3 + 2g + 2 + rk H^2

So you get that rk H^1 = 2g + 1 + rk H^2

[elfin wraith]

Uh I dont know if the euler char of this is 0 so maybe theres something going on

Idk maybe it is, its 4am

[next obsidian]

I’m just saying this is true for any exact sequence of vector spaces

[next obsidian]

So this formula always holds

This is nothing about Euler characteristics of a space, although that’s super related

[elfin wraith]

Isnt the Euler charactersitic just the alternating sum of the betti numbers?

Which is just the rank of the homology

[next obsidian]

That’s not what you’re taking alternating aim of

You’re doing it on this LES

Which related a lot of different spaces

[elfin wraith]

Oh

Im being dumb lol

Yeah ok thats a very useful fact to know

[next obsidian]

This is easily proven

In fact if you have a thing which is additive on SESes

Any function

It will have this property for any LES

Let’s say it’s bounded so it starts and ends with 0s

This isn’t just for VSes btw, it’s true for like, any abelian category

Anyway, this is related to the Euler characteristics

Cuz in a LES which is like

[elfin wraith]

I may have a go at proving that when im less sleep deprived because that sounds like a very useful fact to have on hand and something I probably should already know

[next obsidian]

H^i(A) -> H^i(B) -> H^i(C) -> H^i+1(A) -> …

If you notice what the alternating sum is over this LES

It’s chi(A) - chi(B) + chi(C)

So it also says that the alternating sum of Euler chars of a “SES of spaces” is 0

Lol

[next obsidian]

But anyway this also says the following, chi(M’) = 1, right?

Errr

-1…?

Idk

[elfin wraith]

yeah Itll be one of those I think

[next obsidian]

Is the genus just the rank of H^0

Lol

Oh maybe M’ itself isn’t genus 2g

[elfin wraith]

Its chi = 2-2g rearanged for g lol

[next obsidian]

So this is fine

Ah

Okay okay

Anyway whatever

Toodalooo

[elfin wraith]

Its the number of holes but in the more obvious way

[next obsidian]

Yah yah yah yah

[maiden crater]

Here in the 2nd part, am I asked to prove the nilradical is a nilpotent ideal or R itself ( this is absurd)

[next obsidian]

That the nilradical is

This actually is not as stupid as it sounds

@maiden crater

[maiden crater]

hmm? I have already proven that the nilpotent elements form an ideal in an earlier exercise

[next obsidian]

That the fact it’s a nilpotent ideal

It need not be true if it isn’t finitely generated

[maiden crater]

Yea, that makes sense

The nilpotent( index/ exponent) will be the lcm of the individual indices of the generator

[next obsidian]

I don’t think that’s true

It can be larger

If you have a million generators which are independent of each other but square to 0 the lcm is 2

[maiden crater]

right

[next obsidian]

But x1 • … • x1000000 is nonzero

[maiden crater]

oh, by lcm I meant the lcm of the exponents that map the generator elements to 0

[next obsidian]

Yeah, but this is in Nil(R)^1000000

And the lcm of those numbers is 2

[maiden crater]

yes, so the nilpotent exponent is 2

[next obsidian]

Wut is your definition of nilpotent exponent lol

Cuz to be a nilpotent ideal says that you have an integer n with I^n = 0

So I was taking it to be the minimal such n that that’s true kek

[maiden crater]

I'm just defining it as that least natural number such that a^n=0

[next obsidian]

For all a?

[maiden crater]

for a given a

[next obsidian]

So what is this number?

Also sorry, I realized that my example wasn’t done

Lol

Take Sum_1^1,000,000 xi

Then in this ^1,000,000

The term Prod_1^1,000,000 xi appears (in fact it is equal to this)

So it’s not zero

This shows that you need a number larger than the lcm of the nilpotent index of the generators to kill every element of the nilradical

[maiden crater]

mhm

[next obsidian]

Anyway I think the minimum number you need is like (max{nilpotent index of generators} -1)•(# of generators) + 1

To guarantee it

[maiden crater]

noted

thanks

Let the ring be generated by $I$. Let $a \in I$ be fixed. Let $b \in I$ be arbitrary.Consider $(a+b)^3= 0=3ab(a+b)$. This would mean $(a+b)$ is a 0 divisor of $ab$ and vice versa

[cloud walrus]

wai

[maiden crater]

I was hoping this would mean a+b=0 or a=-b, but that doesn't seem to be true?

[velvet hull]

why is (a+b)^3 = 0?

[maiden crater]

(a+b)^2=0

[velvet hull]

that's not what a boolean ring is

it's x^2 = x

[maiden crater]

oh shit

right

my bad 😭

[coral steeple]

If G acts on X, is it correct to say that g in G acts on X? For example, "the matrix A acts on R^2 as a rotation". I know that it is definitely common to say that the group SO(n,k) acts by rotations

[next obsidian]

Why not

It acts as an automorphism of X

I think everyone is gonna know what you mean

[coral steeple]

Fair enough

[quiet pelican]

The problem is that feels like a type error to me
g isn’t the action
The map \phi: G -> Sym(X) is
Or x \mapsto gx is the action
Not g

[chilly ocean]

Quick question

What does it mean for the index to be 2

What can we say from that

$[G:H]=2$

[cloud walrus]

Haruki

[quiet pelican]

H is normal

[chilly ocean]

I see thanks

The converse true?

[quiet pelican]

It means (by definition) that $H \cup aH = G$ for any a not in H

[cloud walrus]

micoi the group things

[quiet pelican]

No

[quiet pelican]

(And also that $H \cup Ha = G$ for any a not in H)

[cloud walrus]

micoi the group things

[chilly ocean]

I see got it the a Is in G-H

Also

If $H=ker\phi$

[cloud walrus]

Haruki

[chilly ocean]

Can we say that H is normal?

[quiet pelican]

Yes

Exercise: prove it

(It’s quite quick from the definition)

[chilly ocean]

Okay i was just going through the concept but I guess I can try

[chilly ocean]

I think I got it

All we have to do is that aH=bH

Wait

Okay got it

Just needed to define a the homomorphism as conjugation

[hidden cairn]

Say R is a ring and M is an R-module. Then, Hom_R(R,M) = M but if it happens to be that R = M \oplus M then Hom_R(R, M) = Hom_R(M,M) \oplus Hom_R(M, M). How come both of these are true at the same time?

I am confused.

I noticed this when trying to prove every simple R-module appears as a summand of R, if R is a semisimple algebra.

[normal tree]

theres no contradictions