groups-rings-fields

[minor gazelle]

I would rather like a nerd who doesn’t look clever and dedicated to their art of medicine to perform operations on me

Even studying literature seems to be more intellectually demanding compared to a medicine

[tardy hedge]

For sure

[cursive spindle]

I wonder why it's always a bad day when I have to visit a doctor

hmmmmmmmmmmm

[thorn jay]

I can attest to this
Source: my dad is a GP

[minor gazelle]

[knotty badger]

[minor gazelle]

It’s to express my despise for medical doctors though I use it positively

[minor gazelle]

Certainly, I mean it’s awful to need medical attention and you happen to have to encounter those very condescending people pretending moral, and fair fact human are pretty incapable fighting diseases

Well I would want money yes

But pretending morality and Hippocrates oath while you’re talking about how many girls you have played or locked door stereotyping while taking oath

That’s just fake and I would rather earn it by stochastic calculus not by faking who I’m… like I know it sounds pretty awful to generalize but they aren’t really angel they are just normal people as normal as us and seemingly very moral

[tardy hedge]

Lol

[minor gazelle]

It’s a fact

Pretty much academic people can’t imagine how their lives are

Not anymore

AI can practice medicine way more advanced and precise than any doctor

Can its chemical that heals not them another reason to invest more into chemistry and physics study

Ai for operational tasks like radiotherapy or surgery, chemist and LLM for chemical advisory

Wonderful

In 20 years

[thorn jay]

No AI will ever replace the genius of Dr House

🙂‍↕️

[minor gazelle]

Maybe protagonist

I mean they are normal people not better than anyone

It’s just very funny people think they are more moral while ignoring how awful they actually are…

[thorn jay]

Take a guess

What would the guy who makes music to express their feelings think about ai music

I am indeed not at peace with it's existence

And I will do everything in my power to not touch it at all

[minor gazelle]

You guys make music

[south patrol]

good

[minor gazelle]

Learning is hard, I kinda want to study literature and is that a good idea 🎊

[south patrol]

The only AI music i like is the silly country songs

[minor gazelle]

I loaded myself with half of literature stuff

Literature really heals and helps people to remain calm in some sense

[thorn jay]

"debate" 💔

It doesn't sound bad in the sense that AI art doesn't look "bad"

Art isn't defined as it's subjective quality, rather than the fact that there was thought and human decision making put into it

Every piece of art has a piece of the artist, and that is exactly why many people consume it, so they can take that piece and either examine it or project themselves onto it

That's my take at least

Do what you want, but try not to fuck up our already fucked up planet more

Ive said my part, I won't partake in discussions about ai further because I value my mental wellbeing

[thorn jay]

On god

[dim wagon]

is there any reason the theorem excludes the zero cg module? it would still hold right?

[thorn jay]

By convention we take the zero module not to be irreducible, partly for the same reason we do not take 1 to be a prime number

[boreal condor]

If the zero module was irreducible then no other module would be

[thorn jay]

Because we can compose a module over a semisimple algebra (like CG) into a direct sum of irreducible modules, where the amount of times each irreducible modules occurs is determined only by the module and not by the chosen decomposition

This would fail if the zero module was irreducible

[thorn jay]

in fact, if χ is the character of a finite dimensional CG module and ψ the character of some irreducible module M, then the amount of times M occurs is (χ | ψ) = 1/|G| ∑_g∈G χ(g) ψ(g^-1), if you are familiar with character theory

[dim wagon]

ah okay thank you both

[dim wagon]

unfortunately, haven’t really done character theory before, the same book i’m going through has it in the later chapters tho

[south patrol]

The psi is satisfying bro

Are you a Serre fan by any chance

[thorn jay]

Yus

Working through it rn

It's great but dense

[thorn jay]

That is sad, character theory is straight up magic

[tardy hedge]

Enpeace u know so much math bruh

Tbh like idek if i have it in me to really lock in like that

Im just doin what i can yk idk trying not to stress too much abt it but

[thorn jay]

Honestly same

[cursive spindle]

Character theory incredibly goated

[thorn jay]

I tried to lock in to properly study com alg and schemes a couple times but nO my brain refuses

[tardy hedge]

U say that but like, i just feel like ur standards are hella high

Like ik u probably know a lot of com alg

What would be in atiyah macdonald or w/e

[cursive spindle]

Just study algebraic number theory. You'll get both and will lock in just great

[thorn jay]

Uhuh

Not falling for that propaganda

[cursive spindle]

Too bad

[tardy hedge]

I did a bit of alg number theory last semester. It was an easy professor though

I didnt really “learn” it but ig im familiar with a bit of it now

It was quite a beginner course tho like the course ended with definition of class group

[cursive spindle]

bruh

[fading acorn]

ANT

[tardy hedge]

And like finiteness of class group

Yea was just an intro class

[cursive spindle]

Or not even algebraic number theory tbh

Enumerative geometry has nice stuff

It motivates you to use both + combinatorics in this case

[tardy hedge]

Idk like he defined number fields and rings of integers early on but idk what we did in the course for so long

[fading acorn]

wanna learn number theory pilled things but I'm busy with doing other algbra pilled ones

[thorn jay]

Idk, wasn't properly able to start with primary decomposition and krull dimension and such, integral dependence, descent and I barely know homological methods

[tardy hedge]

Primary decomposition just feels boring to learn

[cursive spindle]

I don't care about tensors
sees fiber product of schemes
wow time to lock in now this is very cool

[tardy hedge]

I actually enjoy learning about tensor product stuff i dunno y

[thorn jay]

You can take advantage of krull dimension for computations though

[cursive spindle]

Cruel dimension

[tardy hedge]

For what kind of computations?

I havent really seen dimension theory in action too much yet

[thorn jay]

Honestly with how lattice pilled I am, krull dimension and stuff isn't that hard to visualize, it's just the primary decomposition

[cursive spindle]

As an extension you can look at things like Spec(Z[x,y]/(f(x,y))) snd intersect it with all the things you want and see what do you get

Very fun exercises

[tardy hedge]

That is cool i dont know how to think of the prime ideals of that ring though

I guess if f(x,y) is irreducible its prime right

I mean as ideal in Z[x,y]

[thorn jay]

I found that seeing commutative algebra through the lense of universal algebra is easier than doing the actual thing

Maybe it's because I'm baiting my brain into thinking I'm doing something useful, somehow

[tardy hedge]

Haha yea no youre probably on to something with that

Sometimes these days i just feel that math is so much for me that i kind of feel like ill never really get there

[cursive spindle]

Well if f(x,y) is not something you recognize immediately I'd just play around

It helps looking at C if things get complicated in Z

C then Q then Z

[thorn jay]

Honestly just localise, my guy

[cursive spindle]

yes localization as well

[thorn jay]

There is a nice analogue of localisation for generalised commutative rings where a similar correspondence theorem of "prime" congruences holds

[tardy hedge]

Meaning like if u have a ring u want to understand its ideal structure localize at a prime and it could be easier to understand?

[south patrol]

Who pung me...

[south patrol]

Ah this

[south patrol]

very cool

You know what makes me ragequit about character theory

[thorn jay]

I'm sorry for replying to you I guess

[south patrol]

Why do they do it over C instead of Q bar

(or better yet, Q(roots of unity), once you know a bit)

[thorn jay]

Really? Is Q(roots of unity) enough?

[next obsidian]

I’m a Chmonkey don’t you know

[south patrol]

The point is that the matrices involved all satisfy x^n - 1 for some n (since everything is torsion), so they are diagonalisable with eigenvalues given by roots of unity

[delicate orchid]

If n is the exponent of ur group Q adjoin the nth root of unity is enough

[south patrol]

If and only if F is a field

[thorn jay]

Omg of course

[south patrol]

In particular it is not true that if R is a Euclidean domain then R[x] is a Euclidean domain (since F[x] isn't a field but is a euclidean domain)

Wew can we think about this funny problem with Fibonacci numbers

[thorn jay]

Man, math is pretty cool

[south patrol]

As a bit of entertainment

[cursive spindle]

I'll be back for more disturbing facts
-Prismatic Potato

[south patrol]

Let $F_n$ be the $n$th Fibonacci. By pigeonhole, for any $m \ge 1$ the sequence $F_n \mod m$ is periodic; call the period $\pi(m)$. I conjecture that $\pi(p^k) = p^{k-1} \pi(p)$

[cloud walrus]

Prismatic Potato

[south patrol]

Presumably this is very well-known but I am curious about it

[thorn jay]

How is it periodic mod m by pigeonhole

It could be no periodic

[south patrol]

It is

So the thing is that like the nth only depends on (n-2, n-1)th

And by pigeonhole those two must repeat at some point

(in particular we see pi(m) <= m^2)

[thorn jay]

Mm I see

[thorn jay]

Uhm smt smt matrices over Z/nZ

[sonic coral]

i learned about the pisano period just the other day

[south patrol]

An interesting first step is just showing that lol the periods are even distinct

[south patrol]

Very cool

[sonic coral]

have you seen that the period mod m is the order of the matrix [[1,1],[1,0]] in Z/mZ

i wonder if that would help

[south patrol]

Ig this is sort of tautological though right, like this is just cause this matrix gives the next one right

But actually yeah maybe you can think of invariants of this matrix

[thorn jay]

Would reducing every matrix entry mod n give a homomorphism wrt multiplication?

Let f : R → S be a ring hom, and M = (m_ij) be an nxn matrix over R. Denote with F(M) the matrix (f(m_ij)). nxn matrix multiplication is given by polynomials in Z[x_ij | 1 ≤ i, j ≤ n], which are respected by the homomorphism f, and therefore F is a monoid homomorphism. The restriction to invertible matrices over R maps to the invertible matrices in S too, naturally.

If is f : Z/p^k → Z/p this might help

[south patrol]

Yes

But i think all it says here is that fibonaccis mod m satisfy the usual recurrence relation

Lol

[tall igloo]

this would in fact imply your statement

[south patrol]

I can believe this aha

I think it is probably not hard to provide this as a bound

[tall igloo]

uh wait lemme test how to use the latex bot, i can never get it to do inline math. is $ this $ correct? how about (this) or (this())?

[south patrol]

yes $ $ works well, though i think you do not want spaces before/after the math(s)

[tall igloo]

$bruh$ thats annoying

[cloud walrus]

hatfuel

[south patrol]

Yeah

lol

[tall igloo]

lovely, so

[south patrol]

Actually wait now i think i see like you basically get multplication by p after pi(p), or something close lol

and so you would expect stuff to work out after repeating that like p times each time you go from p^k to p^{k+1}?

[tall igloo]

interesting, can you show why?

[south patrol]

oh i mean is this what you mean aha

I mean this is very false as stated but heuristically what is happening in my mind anyway

😭

I have not thought about this properly

[tall igloo]

proof: let us assume $\pi(p^k) \ne \pi(p^{k-1})$.

now, let $F = \begin{bmatrix} 0 & 1 \ 1 & 1 \end{bmatrix}$. Then, by definition, $\pi(p^k)$ is the smallest value such that $F^{\pi(p^k)} = Id + p A$. If we raise to the power of p:
$$ F^{p \pi(p^k)} = Id + \sum_{i = 1}^{p - 1} {p \choose i} p^i A^i + p^n A^n $$
$p \choose i $ vanishes mod $p$ when $i\ne 0, p$, so this vanishes mod $p^n$. Therefore, $\pi(p^{k+1})$ is either $\pi(p^k)$ or $p \pi(p^k)$.

[cloud walrus]

hatfuel

[south patrol]

Lol that is super cool

Yeah okay classic argument, very nice

Okay so the matrix thing is useful aha

[sonic coral]

i’m glad i could contribute.

[south patrol]

Sorry for my scepticism lol

[sonic coral]

i also just had the matrix wrong lol but we don’t worry about that

i should’ve known the form given that it generates the next one but i didn’t think about it that way and just wrote down what i thought i remembered

[tall igloo]

i don't know of a way to show pi must increase though. you might need to get your hands a bit dirty. some calculations i did but was too lazy to finish that might be useful:
$$ F = Id + M, \text{where} M \coloneqq \begin{bmatrix} -1 & 1 \ 1 & 0 \end{bmatrix} $$
Exercise:
$$ M^n = (-1)^n \begin{bmatrix} F_{n+ 1} & -F_n \ -F_n & F_{n - 1} \end{bmatrix} $$
Note that:
$$ F^{\pi(p^k)} = Id + \sum_{i > 0} {\pi(p^k) \choose i} M^i $$
You want to understand the decomposition of $\pi(p^k)$ in terms of $\pi(p)$ and powers of $p$ by induction. You might get somewhere by explicitly computing some of these sums. I guess the best case scenario is that each of the $M^i$'s is already divisible by $p$ (and this gives you that the sum is $0 mod p^k$). So in principle you might just need to understand that the prime decomposition of $\pi(p)$ and that of those annoying alternating binomial sums of fib numbers

[south patrol]

I think i have another idea too

Yeah okay yes what i had works i think

[cloud walrus]

hatfuel

[tall igloo]

thank you @hidden raven lol

[tall igloo]

ph?

oh*

also about the last sentence: this is a bit tricky since for example, \pi(5) = 20 which is divisible by 5

proof by python script

[thorn jay]

Community effort 💥

[covert cliff]

Given a short exact sequence of groups 1 -> H -> G -> K -> 1 is it always true that H can be seen as a normal subgroup of G? From first isomorphism I know that H is isomorphic to im(H -> G) but is it also normal?

[south patrol]

The image of H is equal to what? (By exactness)

[covert cliff]

its true cuz its the kernal of G -> K

[south patrol]

Yes

Gg

[covert cliff]

these are op wtf

Oh and so it follows that K is isomorphic to G/H from there

[south patrol]

Lol it just ended up being what you did with matrices tbh

[tall igloo]

huh really?

[south patrol]

I should say i mean smth new i had in mind

[tall igloo]

like the upshot was that you needed A to not be divisible by p already, how did you prove this?

[south patrol]

Not the original vibes based thing

Hmmm

[tall igloo]

also -- the thing i said works for any linear recurrence relation in any # of variables. i dont have a great intuition for this but i would be surprised if the periods are distinct for any such relation. i wonder how one would go about showing it is specifically true for Fib

[south patrol]

Sorry as in you expect the period to be independent of the relation?

Im confused what u mean sorry

[tall igloo]

ah sorry i phrased it badly. you are conjecturing that \pi(p^k) must all be distinct for all values of k, for the fib series (and this implies your specific conjectured form). I am saying that I am not convinced this would be true for all recurrence relations

[dull ginkgo]

when yall have a second may I ask a probably stupid question, if you hit me with the "just ask" sticker I am going to turn into a fine paste or perhaps a slime

[south patrol]

Ah yes agreed

[thorn jay]

Ask just

[tall igloo]

ya, so you might have to actually use something about the fib relation 💀

[dull ginkgo]

So like in the category of modules over a comm ring,
L_A(M) = Hom(A,M) is contravariant and R_A(M) = Hom(M,A) is covariant right, and does this same idea work when the Hom endofunctor is substituted for it's adjoint Tensor endofunctor?

[tall igloo]

- \otimes M and M \otimes - are both covariant in -

is this what you mean?

[dull ginkgo]

I see, what about Hom( - , M) and Hom(M, -)

[thorn jay]

Fixing A, then Hom(A, -) is covariant

[tall igloo]

Hom is always contravariant in the first variable and covariant in the second

[you can also view it as a functor C^op \times C --> Set (or whatever you want)]

[dull ginkgo]

what the fuck

wait is Hom(-,M) not adjoint to - (x) M

oh wait

mixed variance functors

truly the bruh of all time

Can we have adjunctions of mixed variance

[tall igloo]

btw @south patrol in case it helps:

pi(2) = 3
pi(3) = 8
pi(5) = 20
pi(7) = 16
pi(11) = 10
pi(13) = 28
pi(17) = 36
pi(19) = 18
pi(23) = 48
pi(29) = 14
pi(31) = 30
pi(37) = 76
pi(41) = 40
pi(43) = 88
pi(47) = 32
pi(53) = 108
pi(59) = 58
pi(61) = 60
pi(67) = 136
pi(71) = 70
pi(73) = 148
pi(79) = 78
pi(83) = 168
pi(89) = 44
pi(97) = 196
pi(101) = 50
pi(103) = 208
pi(107) = 72
pi(109) = 108
pi(113) = 76

[rocky cloak]

Fun fact Hom(-, M) is adjoint to itself.

[dull ginkgo]

I need to learn more shat a gory theory

[rocky cloak]

Viewed as a functor Set -> Set^op and Set^op -> Set respectively

[tall igloo]

i mean, C^op \times C is a category, so you can have functors F: C^op \times C \to D that have an adjoint yes

[dull ginkgo]

• (x) M is adjoint to Hom(M, -) right

[tall igloo]

idk if this is what you meant morally tho

[white oxide]

Silly question, but is p[x] the set of all polynomials with coefficients in p (just making sure)

[dull ginkgo]

yeah

or the generated ideal (p, x)

[thorn jay]

How can the eilenberg Moore category of algebras of the resulting monad be interpreted?

[dull ginkgo]

Interesting. I was going to see if you can characterize tensor transformation laws as constructing a functor from R-modules to R-algebras as:
F : M -> T(M) (x)_R T(Hom(M,R))

R being a comm ring or perhaps a PID or field

[rocky cloak]

Well the monad would be
T(X) = Hom(Hom(X, M), M)
at least
the unit would be evaluation
not entirely sure what the multiplication is, would need to unwrap the definition a bit more

[south patrol]

Thanks, v cool

[thorn jay]

Set^op is the category of, like, complete atomic Boolean lattices right or something

[rocky cloak]

It is equivalent to that sure

It's also just the opposite category of sets

[thorn jay]

Yeah yeah

[dull ginkgo]

preimage moment (i think)

[thorn jay]

But maybe it is a similar situation to varieties of algebraic structures, that Hom_Set(-, M) : Set^op → Set is monadic?

[dull ginkgo]

Or just power set algebras equivalently lmfao

[thorn jay]

Yeah, but I do algebra do I'd like to have a somewhat algebraic description of it

[dull ginkgo]

I'm sorry to hear about your infliction

may you recover soon

[white oxide]

Am I tripping, or why is this true? The claim is that $\bigcap_{\mathfrak{q} \in \text{Spec } S^{-1} A} \mathfrak{q} = S^{-1}\mathfrak{R} = S^{-1}\bigl(\bigcap_{\mathfrak{p} \in \text{Spec } A} \mathfrak{p}\bigl) = \bigcap_{\mathfrak{p} \in \text{Spec } A} S^{-1} \mathfrak{p}$, but aren't the prime ideals of $\text{Spec } S^{-1}A$ only in one to one correspondence with the prime ideals in $A$ which have empty intersection with $S$?

[cloud walrus]

okeyokay

[south patrol]

Hm @tall igloo i assume this is the same as what you have, but I think have reduced it to showing that pi(p) and pi(p^2) are not the same

Hmm

[tall igloo]

i would believe this

[south patrol]

I mean i guess it must be true aha so i certainly believe it but ye hm

Claim: Sps pi(p^k) is not pi(p^(k+1). Then pi(p^(k+ m)) = p^m pi(p^k) for all m <= k.

Hence it is enough to show that pi(p) is different to pi(p^2)

[Pf: let G_n = 1/p^k F(n + pi(p^k) - F(n)], which starts not divisible by p and is always an integer.

Now note F_(N pi(p^k) + n) - F(n) = p^k[ G(n + (N-1) pi(p^k)) + .... + G(n)] = p^k N G(n) mod p^(2k). So for all k <= m, this is divisible by p^(k+m) iff N is divisible by p^m. his shows that the period mod p^(k+m) is precisely p^m pi(p^k). ]

[tall igloo]

im currently beefing up my crappy python code to check your conjecture up to high values, lol

[south patrol]

Like the point here is that if period increases then it must keep increasing

[white oxide]

Actually nvm I looked this up online and I guess I have to rawdog it with the definitions

[dull ginkgo]

Another stupidly dumb question, working in the category of left R-modules, if for $n = 1 ... N$, we have the module maps $\eta_n : A_n \rightarrow B_n$ then is the map $\eta : \bigotimes_{n = 1}^{N} A_n \rightarrow \bigotimes_{n = 1}^{N} B_n, \bigotimes_{n = 1}^{N} a_n \mapsto \bigotimes_{n = 1}^{N} \eta_n(a_n)$ a well defined module map analogous to the property for biproducts?

[cloud walrus]

mizalign

[rocky cloak]

I guess it's worth asking yourself. What is S^-1 p when p does intersect S?

[dull ginkgo]

I could just inductively chain the covariance of the right and left tensor functors but idk if it commutes

[south patrol]

Oh i have done it i think @tall igloo and it is easy

Very cute actually

[tall igloo]

oh?

lets hear it

[rocky cloak]

Also, it might be easier to prove this with the nilpotency definition of the nilradical.

Otherwise you may have to justify why
S^-1 of intersection is intersection of S^-1

[south patrol]

Wait no i did a dumb dumb

😭 im sorry

[white oxide]

Note that the nilradical of $S^{-1}A$ is contained in $S^{-1} \mathfrak{R}$. Conversely, suppose that $r/s \in S^{-1} \mathfrak{R}$, so that $r \in \mathfrak{R}$. Then $r^k = 0$ for some $k \in \mathbb{Z}^+$, whence $(r/s)^k = r^k/s^k = 0$. Therefore, $S^{-1} \mathfrak{R} \subseteq S^{-1} A$. Why doesn't this work?

[tall igloo]

F

[white oxide]

This was a previous result on the page, sorry forgot to mention that

I implicitly used that in the first sentence of the proof above

[cloud walrus]

okeyokay

[rocky cloak]

I mean it works.

Though I would say that they nilradical of S^-1A being contained in S^-1 R is sort of the only nontrivial step

[white oxide]

Oh interesting, so I guess that's why it's a corollary

Because the other direction is trivial yeah

Huh so is it necessary to use the fact that S^{-1} commutes with intersections for the first direction

[rocky cloak]

I mean, I would prove it using the nilpotency definition

But you were using that in your above proof

[white oxide]

Interesting, I guess I'm wondering why they would state it as a corollary then (it immediately follows this proposition)

[dull ginkgo]

S^-1 is a lattice isomorphism as a map from ideals disjoint from localized set to the set of ideals of the localization right

[rocky cloak]

Isn't that what corollaries are supposed to do?

[white oxide]

But it didn't use any of the proposition no if you just use the definition of nilpotency

but I guess corollaries do come after propositions yes

[rocky cloak]

I mean you can prove it either way

[white oxide]

Huh most corollaries i've encountered depend on the theorem preceding them

[rocky cloak]

Like say
(a/s)^n = 0 in S^-1 A

Then there is a t in S such that t a^n = 0
therefore ta is nilpotent, so in the nilradical.

Therefore (ta)/(st) = a/s is in S^-1 R

[white oxide]

I guess I'm just confused about its placement but yeah doesn't matter

[rocky cloak]

Or you can prove it the way you did, using the proposition

Like I'm a little confused about your confusion.

You think it would be better if they had it somewhere else / didn't use the proposition?

[dull ginkgo]

Like how $\mathrm{Hom}\left( \bigoplus_{n=1}^{N} A_n, \bigoplus_{m=1}^{M} B_m \right) \cong \bigoplus_{n = 1}^{N} \bigoplus_{m = 1}^{M} \mathrm{Hom}(A_n, B_m)$ in matrix form, is there a isomorphism $\mathrm{Hom}\left( \bigotimes_{n=1}^{N} A_n, \bigotimes_{m=1}^{M} B_m \right) \cong \bigotimes_{n = 1}^{N} \bigotimes_{m = 1}^{M} \mathrm{Hom}(A_n, B_m)$

[cloud walrus]

mizalign

[tall igloo]

i would not suspect this to be true. the first equation works because R-Mod is an abelian category and Hom must commute with lims in the second variable and colims in the first one, and the direct sum is both a product and coproduct (bc R-Mod is abelian --> additive). however, tensor products are much more subtle in R-Mod

i might have my lims and colims mixed, this is my fatal flaw. but it does not matter

[south patrol]

Hom(Z (x)_Z Z, A) = Hom( Z, A) = A so no

[dull ginkgo]

ah true

[kind temple]

you can expand the LHS to like, nested homs, but idk how useful that is

[dull ginkgo]

Or alternatively, is the function sending A to A (x) A a proper functor?

[south patrol]

I would say Hom commutes with limits in the first variable (if viewed as a functor from the opposite) or sends colimits to limits

[south patrol]

What do you mean

Presumably "is there a way to turn it into a functor"?

[dull ginkgo]

Yeah

like do maps from A to B induce maps from A (x) A to B (x) B

[south patrol]

Then yes — tensor maps together

[tall igloo]

i'll believe it, but ya doesnt matter here luckily lol

[south patrol]

Like really it is the same as in the other variable up to duality i mean

[dull ginkgo]

Actually also probably just from the tensor algebra one lmfao

[south patrol]

Like send f: A -> B to f (x) f: A (x) A -> B (x) B

Sending a(x)a' to f(a) (x) f(b')

It is amusing that this map is actually linear

[dull ginkgo]

@south patrol this was the original qn

[south patrol]

Then yeah we are good

[rocky cloak]

The answer to this is yes

[south patrol]

Probably more obvious if you think on terms of multilinear maps

[dull ginkgo]

true

[south patrol]

Given multilinear B1 x ... x Bn -> C and maps A_i -> B_i, you get a map A1 x ... x An -> C which is still bilinear

This furnishes a map A1 (x)... (x) An -> B1 (x) ... (x) Bn

Idk what that notation means lol

[dull ginkgo]

goddamn it

nevermind :p

[south patrol]

Lol Hatfuel how far have you checked the conjecture btw

[dull ginkgo]

But yeah instead of this being an isomorphism it's a map from the left to the right

[south patrol]

I might try it myself as a bit of fun lol

Uhh i do not see a map ngl

[tall igloo]

debugging, lol

i had to implement ladder exponentiation or whatever its called

[south patrol]

Bruh

I keep going round circles trying to show pi(p) is not pi(p^2) lol

[tall igloo]

how convinced are you that it holds ad infinitum if pi(p) is not pi(p^2)? i did not read your proof carefully lol

[south patrol]

I wanted to have some funny contradiction like pi(p^k) being constant

[south patrol]

Quite, but i can write it out better lol to check

I am just on mobile atm

[tall igloo]

great, then it works for the first 200 primes

1000*, lol

lemme throw in some other recurrence relations and see what happens

[south patrol]

Are jusy checking pi(p) is not pi(p^2)?

[tall igloo]

yes, i also checked p^3 up to p=900 ish

[south patrol]

Nice

[dull ginkgo]

@south patrol My final dumb question:

So I've seen a bit of the tensor transformation law shit, and I'm wondering if that boils down to this in the algebraic context:
The tensor alg functor T(-) from R-mod to R-alg is covariant, and the tensor alg functor composed on the "dual" functor, T(Hom(-,R)) from R-mod to R-alg is contravariant

[tall igloo]

running into a fun subtlety: the period of mod m powers of the transition matrix is not the same as the period of the mod m terms in the sequence. this works for the fibannoci numbers since the powers of the transition matrix are in terms of the fib numbers

[dull ginkgo]

Isn’t that due to the Fibonacci numbers being the eigenvalues of the Fibonacci recurrence matrix

If I am not mistaken

[tall igloo]

so my initial claim that this worked for all recurrence relations was wrong

[tall igloo]

yeah something like this. funky diagonalization stuff

[south patrol]

Oh interest8ng

[dull ginkgo]

not really

[south patrol]

Hmm

Ok i will gain computer access in a sec hm

[tall igloo]

previously i was skeptical since nothing in the proof seemed to rely on the fib sequence being the one studied, but now i am more convinced

[dull ginkgo]

The powers of the matrix kill off the eigenvalues with absval < 1 since their eigenspan tends to 0, while what’s left over is the largest one

Which leaves issues with other recurrences if they like have complex eigenvalues or especially if they have eigenvalues 1 or -1 I would think

I’m not too well versed at all in matrix dynamics :p

[tall igloo]

me neither lol, this is two topologists going back and fourth about a dynamics problem lol

i learned this fibannoci matrix stuff from a math teacher in high school,,, i know not much past that LOL

[dull ginkgo]

Real p much the same for me learning eigenvalues in HS for no reason

[tall igloo]

i have never taken a linear algebra class and so far this decision has caused precisely 0 problems in my life

[dull ginkgo]

Dynamics is like all my Uni does

Fucking big dynamics has my uni at gunpoint

[tall igloo]

interesting, i have heard similar things about the uni im about to enter but cannot find any evidence of it online 💀

[dull ginkgo]

I go to Penn State it’s ALL DYNAMICS

NIGHTMARE NIGHTMARE NIGHTMARE

[tall igloo]

lmao wild. mine is not penn state lol

i guess big dynamics has a chokehold on multiple unis

[covert cliff]

How is this different from the product group?

[white oxide]

Yeah ig

[dull ginkgo]

Hm doesn’t look different

[thorn jay]

Free product of two finite groups is infinite, for one

[dull ginkgo]

Free product is the coproduct for groups

Misread

[covert cliff]

ah I goofed the direction of arrows

[dull ginkgo]

Could call it the coproduct group

[covert cliff]

I somehow misread that h: G -> H and not h: H -> G

thank you

[tall igloo]

you can embed g from any G_\alpha into the product via g at alpha and 1 otherwise for each coordinate. in the boring product these embeddings commute with each other when they are at different alphas

the free product does not do this

[covert cliff]

I know this is the wrong place to ask this but if a product exists in a category then does a coproduct also exist?

[dull ginkgo]

You can construct it via like words of different elements of the groups

[tall igloo]

not necessarily

[south patrol]

So my idea hatfuel was like suppose $\pi(p^k) \ne \pi(p^{k+1})$. Let $G_n = \frac{1}{p^k}( F_{\pi(p^k) + n} - F_n$, so this is an integer-valued sequence which isn't always divisible by $p$. Note that this sequence also satisfies the Fibonacci recurrence, so it is also periodic mod $p^k$ with period $\pi(p^k)$. Hence $F_{n + N \pi(p^k)} - F_n = p^k ( G_{n + (N-1) \pi(p^k)} + \dots + G_n) = p^{k} N G_n \mod p^{2k}$, since all the in teh 2nd bracket are the same mod $p$. Hence $N \pi(p^k)$ is a period mod $p^{k+m}$ for $0 \le m \le k$ iff $p^m \mid N G_n$ for all $n$, i.e. iff $p^m \mid N$, and hence $\pi(p^{k+m}) = p^m \pi(p^k)$. Note that this in particular shows that $\pi(p^{k+1}) \ne \pi(p^{k+2})$, so we can keep going

[kind temple]

think about lattices

[tall igloo]

notation clarification, what do you mean by F_{...}?

[south patrol]

nth Fibonacci lol sorry

[tall igloo]

ohh so true. i was using F for my transition matrix so i got confused haha

[south patrol]

Ah sorry lol

[tall igloo]

you dont have to apologize for me using bad notation in my own handwritten work lol

[south patrol]

I weakened my claim and then realised it was not good enough lol so the stronger is needed

[cloud walrus]

Prismatic Potato

[south patrol]

Okay only slightly dodgy thing might be why the period of G_n is exactly the same as that of F_n

aagh

[tall igloo]

and p^m | N is implied by p^m | NG_n for all n because not all G_n are divisible by p?

[south patrol]

yeah exactly

I think it is quite believable though that all Fibonacci seqeunces have the same period mod p^k as long as you aren't always divisible by p lol

[tall igloo]

can you elaborate why G_n is not always divisible by p?

[south patrol]

Oh i mean like if G_n is always divisible by p, then F_{pi(p^k) + n) - F_n is always divisible by p^{k+1}, i.e. pi(p^k) is a period mod p^{k+1} (and hence "the" period)

[tall igloo]

is this not assuming a priori that pi(p^k) is not pi(p^k+1)?

[south patrol]

that was my assumption yes

[tall igloo]

ohhh wait i see, you are saying that if it is true once it is true forever afterwards?

[south patrol]

sorry like this is just more details on the proof that it's enough to show pi(p) not pi(p^2)

yeah exactly

[dull ginkgo]

What is pi here

[south patrol]

pi(n) is pisano period of n

(period of fibonaccis mod n)

[dull ginkgo]

I see

[south patrol]

Oh okay

So hatfuel has shared that what we have been doing is an open problem

lol

[dull ginkgo]

Lmfao

Open as in it’s important but too tuff to tackle or like no one gives a fuck to solve it

[south patrol]

So apparently like lol

A counterexample to a specific case of FLT would give an example of a p violating my conjecture

so people tried to find such p befre FLT was proven

lol

but this also means like it has been around for like decades as an open problem

but the conjecture has been verified for all p < 10^18 or something lmao

[tall igloo]

i could've done p = next prime after 10^18 if i left my python code running smh my head

[dull ginkgo]

You can count higher than 10^18 bro I believe you just keep checking

[south patrol]

I am curious though that like they said it is conjectured there are indeed counterexamples

[dull ginkgo]

Number theory Sisyphus

[thorn jay]

I think I have been able to reduce the question of π(p) = π(p^2) to checking that powers of the Fibonacci matrix in the group of invertible matrices with coefficients in Z/p^2 never hit a matrix of the form I + pA

To the ring Z/p^2 you can just adjoin the golden ratio an work with the diagonalized form

Which is if and only if we can find some n≠0 where F_n = 0 mod p and F_n+1 = 1 mod p

[thorn jay]

Holon I think I made a mistake somewhere 😔

[thorn jay]

What I have found, by Cayley-Hamilton, is that π(p) is the smallest n such that x^2 - x - 1 | x^n - 1 in F_p[x]

Lmao

[covert cliff]

Is there a group G with a non trivial normal subgroup N such that G/N is isomorphic to G?

idek if thats a silly question to ask tbh it definitely cannot be true in the finite case

[next obsidian]

Yes

[covert cliff]

I think I found one

S^1 with ker(z -> z^n)

I guess the stronger question is if this happens with every infinite group?

[next obsidian]

I don’t know

Are there infinite simple groups? Probably

Maybe S_infinity

[cedar vault]

With Z, quotient by any nontrivial subgroup is finite thus nonisomorphic to Z

[covert cliff]

oh

rip

[covert cliff]

is this cuz the only subgroups of Z are nZ?

[next obsidian]

Oh duh

[south patrol]

A_oo and SO(3) for example

[covert cliff]

like how one might view R as the set of complex numbers with imaginary part 0

so you are looking at R as a subset of something else and you have an isomorphism to that subset

K contains a subset that is field isomorphic to F

So C contains the subset {a + bi | b = 0} which is isomorphic to R

like R x 0 is not technically R but there is an isomorphism of this field into F

yeah

because p(pi(x)) = pi(p(x)) = p(x) + (p(x)) = 0 + (p(x))

its so that the domain of pi doesn't really matter

thats the point

cuz pi maps F into K

sorry pi maps F[x] -> K

ong

So identifying F with a subfield of K is only required to show that there is a field extension for F

it has nothing to do with roots

F contains 1 so 1*x is in F[x]

the theorem states that there is a field that contains an isomorphic copy of F and has a root to p

So you have found the first part

now you want to show the second part

$p(x) = \sum_{i=0}^{n}a_ix^i$ then $\pi(a_ix^i) = \pi(a_i)\pi(x)^i$ since homomorphism then the sum also splits

[cloud walrus]

Khush

[covert cliff]

cuz homomorphism

that follows from the identification of F with pi(F)

well it is exactly the identitfication

yes but F is not a subset of K

I think its best if you work out the example with R and p(x) = x^2 + 1

because it is

they are familiar sets

I don't think 2Z is a quotient field of Z by some polynomial

so im not sure how you're doing that here

in this case it is

Because $F \cong \pi(F) \subset K$

[cloud walrus]

Khush

[kind temple]

K is a quotient (of F[x] by an irreducible poly), so it is a quotient field.
F is contained in K because it is preserved under the quotient map F[x] -> K

so K is an extension field for F

[covert cliff]

yeah cuz its a quotient of a ring by a maximal ideal

[kind temple]

there is a natural map F -> F[x] taking an element f of F to the constant polynomial f

there is a quotient map F[x] -> K

the composition of these maps is an embedding

its a isomorphism (of fields) onto its image

the quotient map doesn't disturb the field properties of F

so F is kind of cannonically a subfield of K

F is isomorphic to a subfield of K

im out of practice with this stuff

[covert cliff]

its not a subfield

its just isomorphic to one

its like how a finite dim vector space is isomorphic to Fn and we can write it as a column vector

the vector is not actuall the list of numebrs

the list of numbers are just a representation of a vector

yes

[kind temple]

https://en.wikipedia.org/wiki/Field_extension#Caveats

i think the wiki page here does a good job at explaining why you can, for most cases i think, just treat naturally embedded fields as actual subfields

you use the isomorphism

if you wanted to, you could just give the isomorphic copy of F inside of K a new name, say F', and then just prove things about F'

but every property of F', as far as fields are concerned, will be exactly the same as F

[nimble folio]

if $G$ is a group, what does it mean for $\Aut(G)$ to be trivial?

[cloud walrus]

clubsoda14

[covert cliff]

identity map only

[kind temple]

the only isomorphism G -> G is the identity, but Aut(G) is trivial if and only if G is trivial or G is isomorphic to Z/2Z

[covert cliff]

I think you might just need to mull it over

I dropped my schools algebra course cuz of rings

But I started understanding when I kept at it

[tall igloo]

interesting, this also looks like some discrete laplace transform version of the problem

in a way that i am too sleepy to formalize rn since i just woke up, but this makes sense when you consider its just looking at characteristic polynomials

[rocky cloak]

S_infinity isn't simple, but A_infinity is

Also PSL(n, K) for infinite fields K will do

[next obsidian]

Oops yeah

[dim wagon]

I'm having trouble following the argument here

Why would the function be injective?

Oh wait I think I got it

[rocky cloak]

It's essentially the definition of faithful

[dim wagon]

If not injective, then there exist $z_1,z_2$ which map into the same $\lambda$, but then this will mean $v(z_1z_2^{-1})=v\lambda\lambda^{-1}$ for all $v\in V$ and this is a contradiction since $V$ is faithful and $z_1z_2^{-1}\not=1$?

[cloud walrus]

somethingwrong

[dim wagon]

Oh is there a more direct argument then what I wrote above?

[rocky cloak]

Faithful means G -> Aut(V) is injective.

So restricting this to the center it's still injective

[dim wagon]

Ah okay thanks, I get what your saying

[thorn jay]

@tall igloo @south patrol can you guys prove that, if π(p) = π(p^2), then π(p) = π(p^k)?

Because if so, then I believe there is a relatively simple proof of the conjecture for primes which do not contain a square root of 5 in their field

[south patrol]

This was my hope, since it implies the result, but given that there is no known proof of the result, I expect this is hard

[south patrol]

do you mean like "for all k"?

[thorn jay]

Yes

[south patrol]

Doesn't this quickly imply the result? Cause you can just take k big enough so that 0 < F_{pi(p)} < p^k and get a contradiction to it being 0 mod p^k

[thorn jay]

Right

[south patrol]

Or is this smth else

[thorn jay]

No it's the same

[south patrol]

Ah okay

But no i cannot prove that result unfortunately

[thorn jay]

I've gone a pretty convoluted path and just ended up going back where we started

Aarrrghhh

[south patrol]

I think I should probably give up given that this is not known

😭

I thought this would just be a cute little exercise

Lol

[thorn jay]

But it feels so closee

Honesty, taking the matrix route doesn't help because we cannot take advantage of linear algebra (we work over Z/(p^2))

[south patrol]

yeah lol

[rocky cloak]

So is the only remaining step that pi(p) is different from pi(p^2) for all p?

[south patrol]

Yes

Idk if you saw but like

This is known for all p < 10^18 or smth crazy

But still open in general

Lol

[rocky cloak]

I see

[south patrol]

Which i find amusing

[candid patrol]

Hello ! I’m Looking for a proof of this : Let p prime and odd and G be a p-group, then G has only one subgroup of order p iff G is cyclic

[thorn jay]

You only have to prove it for |G| = p^2

The rest follows from induction

The similarities to that nt problem...

If G is of order p^2 and noncyclic, then every nontrivial element is of order p, and thus we have unique subgroups of order p

(in fact every group of order p^2 is abelian)

That should be a simple enough proof

[velvet hull]

the cyclic => one subgroup direction isn't too hard

[velvet hull]

there's probably a much easier proof of the other direction that I'm not seeing, but for the reverse you can consider the lemma that if there are at most n elements of order n in a group G for all n, then G is cyclic (proving this lemma is kind of a pain though)

[rocky cloak]

I'm not sure I quite see the induction step here.

Like how are you using that p is odd?

[south patrol]

lol this is an exercise i have not done, interesting problem

[thorn jay]

Going through it now I see my fault

[rocky cloak]

Hmmm, so you have one subgroup of order p in the center so that must be the only one.

Then by induction we can assume G/Cp has CpxCp as a subgroup.

Then I guess one can classify all groups of order p^3

Feels like there must be a simpler argument though...

[south patrol]

Yeah and Q8 is the problem ig lol

[rocky cloak]

In guessing there are probably several counterexamples at p=2

[south patrol]

I guess another thing is that Cp x Cp is the only minimal non-cyclic p-group for p > 2

but that is presumably a lil hard to show idk

What I mean is like it's enough to show the result for minimal non-cyclic p-groups G, and Cp x Cp is the only example lol in which case it is obvious

but hmm

[rocky cloak]

I guess that's basically just a restatement of the result

[south patrol]

Ah I think I have a proof

maybe not

[thorn jay]

So true

[south patrol]

I just wanted to show that every minimal non-normal p-group is like Cp x Cp yeah by writing it as some semidirect product. But probabyl not easy sincey ou have to do it for p > 2 somehow - maybe Jagr's thing with groups of order p^3 is best lol

Like a hope here was you have some options for semidirect products and it's a bit simpler if p is odd because of the structure of Aut(C_p^n) being easier

[candid patrol]

I think it’s quite a nice result, yes! The full statement is actually:

Let $G$ be a $p$-group of order $p^n$. Show the following equivalence:

For odd prime $p$: There exists $1 \leq k < n$ such that $G$ has a unique subgroup of order $p^k \Leftrightarrow G$ is cyclic.

The proof isn’t very long using graded lattice arguments, but I’d like to find a more algebraic proof.

[cloud walrus]

UGOBEL

[candid patrol]

I guess that it is clear if we assume that The number of subgroups of order p^k is congruent to 1 mod p

[velvet hull]

-# hello

[candid patrol]

Didn’t see it haha sorry thx ! I’m gonna ckeck this

[south patrol]

I dont think this is enuff tho right lol

I have been using this and couldnt finish off idk

There is a fun proof of this fact i came up w tho

[velvet hull]

one thing that I did neglect is that this lemma requires G to be a finite group, but I think that this is still fixable, by considering finite subgroups of G (have not verified this fix works)

[rocky cloak]

Wait so how is this lemma helpful?

Hint could be that index m subgroups of G/N correspond to index m subgroups of G containing N

[velvet hull]

are you talking to me lmao

||If there is some p^n such that G contains more than p^n elements of order p^n, then there are two disjoint copies of Z/pZ in G by selecting the right elements of order p^n||

[rocky cloak]

"selecting the right elements"...?

[velvet hull]

||G has to contain at least 2 elements of g,g' order p^n that don't generate each other, but then g^(p^n-1) and g'^(p^n-1) both have order p such that they don't generate each other||

[rocky cloak]

What if g^p^n-1 = g'^p^n-1?

Say G is a hypothetical group of order p^3 with
p-1 elements of order p and p^3 - p elements of order p^2

Why would this be impossible?

(keeping in mind that such a group exists for p=2)

[velvet hull]

ah, I see, that's annoying

[velvet hull]

allright ignore what I said I don't think my idea is fixable

any proof probably requires some sylow fuckery then, I've forgotten basically all of it, need to reivew that shit for my quals

[candid patrol]

You can prove the following equivalences:
G has at least two subgroups of index 2
<=> K_4 is a quotient of G
<=> G has at least three subgroups of index 2

[cedar vault]

Z/2Z x Z/2Z

[cloud walrus]

UGOBEL

[white oxide]

Let $A$ be a commutative ring. Is it true that $A[x] \cong \bigoplus_{i \in \mathbb{Z}_{\geq 0}} (x^i)$?

[cloud walrus]

okeyokay

[thorn jay]

This makes it into a graded ring, yes

[velvet hull]

graded ring 🗣️🗣️🔥

[white oxide]

Haven't learned about graded rings yet but interesting

I'm trying to show that A[x] is flat and the hint said to use exercise 4 which involved showing that a direct sum is flat iff each of its summands is flat

[south patrol]

Also like how do you define A[x]

One way to define it is as A^(+)N with a funny operation

[cedar vault]

this is true

are you trying to prove there are only two groups of order 4?

[velvet hull]

very loosely a graded ring is a ring that has "layers of sediments".
In particular a ring A is said to be graded if

• as an additive group, A is isomorphic to a countable direct sum of abelian groups $\bigoplus_{i \in Z} S_i$; and
• for all integers i and j, $S_i S_j \subseteq S_{i+j}$.
  the classical example of graded rings are polynomials in one variable where you grade monomials by degree; this grading basically captures all the important properties of the degree function as you have observed

[tardy hedge]

does direct sum there work? x^2 is in (x) and (x^2) so wouldn't there be two representations in the direct sum? (I may not understand direct sums correctly)

[thorn jay]

Wait yeah

Replace (x^i) by x^iA and then we're talkin

[white oxide]

Oh yeah u right

[cedar vault]

it is of order 4 and not the cyclic group, so it must be the Z/2Z X Z/2Z since there are only two nonisomorphic groups of order 4. now i will ask again are you trying to prove that there are only two groups of order 4

if yes then do you know the definition of internal direct product? try using that to find an isomorphism to Z/2Z x Z/2Z from a non cyclic group of order 4

[south patrol]

oh lol this is what i assumed kiand meant

[tall igloo]

as i told prismatic potato, i have never seen a number theory statement be true for primes up to like, a million, and be false, so in my book its true 💀

[rocky cloak]

All integers are less than 10^18

[vocal pebble]

Quite the theorem right there

[charred iris]

false; it's actually 10^19

[south patrol]

already made this joke

[woeful sage]

Physics-pilled 🗿

[tall igloo]

sin(x)=x for all arbitrarily small x

[minor gazelle]

$\int \sin(dx)\approx x+C$

[cloud walrus]

Emmaaaaa

[tall igloo]

okay tbf I've made more insane statements before. For example, anything in chromatic homotopy theory that works over an odd prime works for all larger primes

[thorn jay]

So true

[crystal vale]

Is that any proper subring of Z^3 which isomorphic to itself? I don't think so

Sorry

Subring of Z^3 isomorphic to Z^3

[rocky cloak]

Z^3 should be the only one

[crystal vale]

Yes

So can you verify one argument?

[rocky cloak]

Which you can see because there aren't that many idempotents

[crystal vale]

Yes

Someone claim there is Z[x] -> Z^3 onto homomorphism, which is not possible.

Consider the ring map F: Z[x] -> Z×Z×Z determined by x goes to (1,2,3) i.e.,

f(x) goes to (f(1), f(2), f(3))

Here, ker F =((x-1)(x-2)(x-3))

So by first isomorphism theorem, Z[x]/K isomorphic to image(F).

Using CRT, image(F) is isomorphic to Z×Z×Z.

This means image(F) is a subring of Z×Z×Z which is isomorphic to Z×Z×Z itself...

Now show that image(F) must be Z×Z×Z...

There are 8 idempotents in Z×Z×Z... Since image(F) is isomorphic to Z×Z×Z, image(F) must contain 8 idempotents as well... Now note that the idempotents generate all of Z×Z×Z...

What's wrong here?

[tall igloo]

yes there is no surjective homomorphism Z[x] --> Z^3. the choice of where Z ends up is determined by the ring structure on Z^3 (i.e. 1 --> (1, 1, 1)). x is free to land anywhere, but even just as groups it will generate a subgroup of Z^3 that looks like Z, at best

[rocky cloak]

So F looks like it should be surjective to me

[crystal vale]

But it is not

Because take preimage of (0,0,1)

There is no such preimage exists

Actually we can't use CRT here

[rocky cloak]

Hmm, let me see yeah
(x-1) + (x-3) = (2, x-1) Which is not everything. So indeed the image of F is isomorphic to
Z[x]/((x-1)(x-3)) x Z

Right so it's not everything

So I see my mistake, but guess I'm unsure about the question. Why do you say the image is isomorphic to Z^3 ?

[rocky cloak]

With a little change of variables x -> x+2 this turns into
Z[x]/(x^2 - 1) x Z
which might be more intuitive to deal with

[rocky cloak]

Right, so was that the answer to your question. That CRT doesn't apply?

[rocky cloak]

As a subgroup the image is isomorphic to Z^3

Just not as a ring

[tribal moss]

We can get "fairly close" to Z^3 by mapping x to (0,1,-1), though -- except that still only produces triples where the second and third entry has the same parity.

[crystal vale]

Yes

[crystal vale]

Not me, he said that because he thought we can use CRT here

[rocky cloak]

If only 2 was invertible, like if Z was Q or Z[1/2] it would work

[boreal condor]

Take 2Z^3 the ring of triples with even coordinates ?

The isomorphism is given by (x,y,z) -> (2x, 2y, 2z)

[thorn jay]

Look at this mf not using unity 💔

[tribal moss]

Not a ring homomorphism.

[boreal condor]

(0,0,0) is your unit element

With addition

Oh mm sure

Was working with group oops

[tribal moss]

But f(1,1,1)·f(1,1,1) = (4,4,4) which is not the same as f((1,1,1)·(1,1,1)).

[boreal condor]

Yeah yeah got it

[thorn jay]

"ring of triples" 😔

[thorn jay]

Happens to the best of us

[crystal vale]

Yes

[tribal moss]

So for which rings A does a surjective ring homomorphism A[x] -> A^3 exist?
It's a sufficient condition that 2 is a unit, but this is not necessary -- for example, the Eisenstein integers also work, sending x to (0,-1,w) where w is a nontrivial cube root of 1.

[rocky cloak]

The existence of three elements such that the pairwise differences is a unit should work

That can be achieved a few ways

There are also rings where A is isomorphic to A^3, so that would also work

Maybe it's more interesting if A is an integral domain...

[boreal condor]

Probably just having a nontrivial unit is enough

[tribal moss]

Hmm, I don't think Z^2 works, even though (1,-1) is a nontrivial unit.

[boreal condor]

Mmm yeah

Same for Z^3 anyway

[minor gazelle]

Guys is this mathematics? Or just trolling symbols

I heard a name earlier and check his name found this

[south patrol]

I mean lol

controversial

[minor gazelle]

This are random symbols right?

I just search that name yes

Like good or bad

Way

Is this math

Too weird I heard it earlier chatting and searched

Even English doesn’t make sense there

I only read isomorphism

?

I never can do it

[thorn jay]

I mean it's hard to understand a definition halfway through a paper which is 400 pages long

It's like trying to understand a plot of a book starting from the middle

[minor gazelle]

But that’s so many symbols

Kinda brilliant though I can’t even memorize many symbols

[thorn jay]

Idk, don't follow it

Don't really care either lmao

Don't know enough number theory

[minor gazelle]

Kinda feel I need probability theory to make sense of reality and my future wallet

That math is scary even by glance

[thorn jay]

Not much

Little bit of topology (but mostly algebraic)

And I'm picking up some lattice theory along the way because it's used to death in universal algebra

[minor gazelle]

Do you guys have to choose a specialization or something for math majors

I’m thinking probability

It’s kinda my stronghold

[thorn jay]

In the uni I'll be going to there's basically choice throughout

And you get to pick paths

[minor gazelle]

Is that a specialization though

Probability

[thorn jay]

Yes

[minor gazelle]

States?

[thorn jay]

Netherlands

[minor gazelle]

I kinda worry about the career aspect though I look more optimistic than I actually am, however I did enjoy probability quite a lot back then I study it as Econ student

[thorn jay]

Damn you too?

I'm going to Nijmegen

Great city and pretty good uni

[minor gazelle]

So

Yeah

[thorn jay]

Familiar?

Why Groningen

Ah that's fair

[minor gazelle]

Still is that really math

That symbol sheet

I know it sounds boring but I counted the symbols

I know none

[thorn jay]

Sure

[minor gazelle]

My brain😵‍💫

[thorn jay]

Why lol

[minor gazelle]

You guys don’t play some games or anything I mean it’s break

[thorn jay]

I do math in my free time

You can choose Wiskunde D

Which was cool but ultimately still kinda easy lol

[minor gazelle]

Isn’t high school math ultimately easy

Not really but high school math is like

Analytic geometry at maximum right?

Calculus is easy right?

[thorn jay]

We had complex numbers, basic diff eq, geometry proofs and stuff in wiskunde D

[thorn jay]

Depends on the integral or optimization problem

[minor gazelle]

Usually non calculus math is harder for high schooler

In particular proof based geometry and function-inequality

Those questions I couldn’t solve today

The more modern the method the easier it tends for high school

You can look up an author titu.. he’s coach for imo team America. Titu’s lemma is very famous for competitive mathematics and on his guide book there are many intriguing non calculus questions

So it’s to show that fraction is rational?

The second formula is just cubic root

You can even have more generalized formula

So theha^3 + 2theta + 2theta^2 +1 which is simplified to (1+\theta)^3

For fraction you just plug in

Easy

But I don’t know what Qtheta means

If it’s just to show the faction is rational

For fraction $\frac{1+\theta}{1+\theta+\theta^2}=\frac{1+\theta}{1+\theta(1+\theta)}=\frac{1}{\frac{1}{1+\theta}+\theta}$

[cloud walrus]

Emmaaaaa

[minor gazelle]

I know now

We have to use the previous computation

For fraction $\frac{1+\theta}{1+\theta+\theta^2}=\frac{1+\theta}{(1+\theta)^3/(1+\theta)}$

[cloud walrus]

Emmaaaaa

[minor gazelle]

After simplifying

Immediately

For fraction $\frac{1+\theta}{1+\theta+\theta^2}=\frac{1}{1+\theta}$

[cloud walrus]

Emmaaaaa

[minor gazelle]

Here you go

Easy

Though I may be wrong since I’m really working on my phone without any scratch paper

[tough raven]

Any hint for this? I can't guess what M' should be at all.

[rocky cloak]

I haven't verified this very carefully, but the immediate idea that sprung to mind for me and seemed to work in a simple example is

row reduce M so you have a column with a single unit and the rest 0s. Delete the row and column with the unit.

Any (k-1)x(k-1) minor in M' can be considered as a kxk minor in M by just using the deleted row and column.

So you just need to argue you can get the other kxk minors as linear combinations

[violet spade]

@sacred wharf have you already solved it?

[violet spade]

Well but what is (2+theta)^-1?

remember, Q(theta) is a field, meaning that any element is invertible

so you can simplify further

So we need to compute $(1+\theta+\theta^2)^{-1} \in \bQ(\theta)$

[cloud walrus]

joel

[violet spade]

Where we have $\theta$ being a root of $\theta^3-2\theta-2$. Because $X^3-2X-2$ is irreducible we have the natural isomorphism relationship given by $\bQ(\theta) \cong \bQ[X]/(X^3-2X-2)$, so in other words, we have to find $\overline{(1+X+X^2)^{-1}} \in \bQ[X]/(X^3-2X-2)$

[cloud walrus]

joel

[violet spade]

Now in this field, any polynomial is of degree $\leq 2$, because anything with degree $3$ or more can just be divided out by some division algorithm. So we set $(1+X+X^2)^{-1} = aX^2+bX+c$, and solve for $a,b,c \in \bQ$

[cloud walrus]

joel

[violet spade]

And then compare coefficients

that's one way to do it

@sacred wharf

[woeful sage]

He did not respond in 10 mins

[thorn jay]

Many such cases

[tough raven]

Indeed, elementary row and column operations don't change I_k(M) (the obvious fact I missed), and I_k(M) = ∑_{i=0}^k I_i(A) I_{k-i}(B) if M is block-diagonal with blocks A, B. If A is I_r for r ≤ k, then this gives I_k(M) = I_{k-r}(B) + I_{k-r+1}(B) + ... + I_k(B) = I_{k-r}(B).

[cloud walrus]

mqinsweden (ping on reply)

mqinsweden (ping on reply)

[thorn jay]

[vast verge]

I'm having trouble understanding this

Especially the part involving Lagrange's Theorem

[velvet hull]

is it clear why a homomorphism is determined by the image of 1?

[vast verge]

Yeah

Since that image acts as the generator

[velvet hull]

now observe that the image of a (sub)group under a group homomorphism is always another (sub)group

hence phi(Z12), whatever it is, is necessarily a subgroup of Z30, but Z30 only has so many subgroups

(this is essentially another way of stating Lagrange's theorem as done in your example)

[vast verge]

Okay so phi(Z12) will be a subgroup of Z30, specifically it will be a subgroup of Z30

Z30 only has so many (cyclic) subgroups corresponding to each of its factors

[velvet hull]

hmmm, that proof direction is not that simple actually, put that on pause

consider phi(1)

now as an application of Lagrange's theorem and property 2, you can check that any group homomorphism can only divide the order of any element

e.g. the image of an element of order 4, under any group homomorphism, can only have order 1, 2 or 4

[vast verge]

Okay I understand that, that's property 3 in the example. |φ(4)| divided |4|

[velvet hull]

so, to find all homomorphisms from Z12 to Z30, we only need to consider all the divisors of 12 (which is the order of 1)

and think about what happens when we send 1 (in Z12) to an element with order equal to each divisor (in Z30)

[vast verge]

So using property 3, $| \Phi (1) | divides |1|$ which means $| \Phi(1) | divides 12$

[cloud walrus]

Tropical Greens

[velvet hull]

yes

[vast verge]

But since we're projecting into $\mathbb{Z}_{30}$ we also get that $|\phi(1)|$ divides 30

[cloud walrus]

Tropical Greens

[vast verge]

Okay, thank you

[velvet hull]

we're not entirely done with the proof

but if you think you can figure out the rest from here go for it

[vast verge]

Here's how I did it

[velvet hull]

can you verify that they all induce well-defined (and distinct but thats trivial) group homomorphisms?

once you do that you're done

[vast verge]

Well, for the case where $a=5$ we have:
$\phi(x+y) = 5(x + y) = 5x + 5y \equiv 5x + 5y (mod 12)$

[cloud walrus]

Tropical Greens

[vast verge]

I think it should work pretty similarly for the rest of the choices of phi(1)

[velvet hull]

you're only verifying that it is a group homomorphism

are you sure they are well defined functions?

[vast verge]

So you're asking me to show that $phi(x) \in \mathbb{Z}{30}$ for all $x \in \mathbb{Z}{12}$?

[cloud walrus]

Tropical Greens

[velvet hull]

no, a function being well-defined means that if x=y then f(x) = f(y)

[vast verge]

Do I show that $\phi(x-y) \equiv 0 (mod 30)$ when $x = y$?

[cloud walrus]

Tropical Greens

[velvet hull]

otherwise to see where this reasoning might go wrong, consider the map from Z2 to Z3 given by sending 1 to 1.
Then this is a surjective map with surjective inverse, and as they respect the group strucutre this means that they are isomorphisms. Hence Z2 is isomorphic to Z3.
The mistake here is that I have implicitly assumed that if we send 1 to 1, that automatically gives a set function

[vast verge]

Because we still don't know what happens to all the other elements right?

We've only specified a part of the function

[velvet hull]

no, because if we send 1 to 1, then 1 will have to be sent to 1,2 and 3 all at the same time

and that is not how a function works

[vast verge]

Ah my bad, I forgot about that

[velvet hull]

typically we first verify that a map is a well-defined function before proving that it is a group homomoprhism

so even though in this case it does work to just prove that phi(x-y) = 0 iff x-y is 0, that's generally not a good idea in proofs

again, show that if x=y then phi(x) = phi(y)

[vast verge]

There is a section covering what you're talking about now, but I glossed over it because I had trouble fully understanding it

[velvet hull]

yes, that is describing exactly what I am talking about here

[vast verge]

I just have no idea what they mean by the correspondence $x + \langle 3 \rangle \rightarrow 3x$

[cloud walrus]

Tropical Greens

[vast verge]

Oh they're cosets

Wait, give me a minute. I got to read this another time

Okay so we want to show that if $x = y$ in $\mathbb{Z}{12}$ then $\phi(x) = \phi(y)$ in $\mathbb{Z}{30}$

[cloud walrus]

Tropical Greens

[vast verge]

If we chose $\phi(x) = 5x$ then that turns into show that if $x \equiv y (\mod 12)$ then $5x \equiv 5y (\mod 30)$

[cloud walrus]

Tropical Greens

[velvet hull]

I gtg, someone else pick this up

[vast verge]

Thanks for helping

I feel like I'm close to the answer anyway

[tough raven]

Let R be a commutative ring which is a k-algebra, k a field. Let x1, ..., xn be k-algebraically independent elements of R such that R is finite over the subring S (isomorphic to a polynomial ring) generated by x1, ..., xn. Why is it true that x1, ..., xn is a regular sequence in R iff R is a free S-module?

[karmic moat]

probably some Cohen-Macaulay argument in here

but i'm not sure about the details

[weary frost]

If I understand correctly we can assume S = k[x1, ..., xn]? Lets assume R is free as an S-module, i.e. R = S^k. It's obvious that x1 is not a zero divisor in R because it is not a zero-divisor in S. Now it is easy to see that S/(x1, ..., xm-1) = k[xm, ... xn] = S' and you see similarly R/(x1, ..., xm-1) = S'^n so xm is not a zero divisor in this module.

this proves on direction pretty easily

[karmic moat]

maybe it'd be easier to show R is projective

instead of showing R is free

yeah maybe you can use Quillen-Suslin and say it's enough to prove that R is projective

there should be some statements about Cohen-Macaulayness of a module and its projective dimension

[rocky cloak]

Hmmm, if R = k[x, y]/((x-1)y, y^2)
Then S = k[x] is a subring, x is not a zero divisor and R = S (+) S/(x-1) is not free.

Maybe I messed something up...(?)

It's somewhat close to the Auslander-Buchsbaum theorem though. If S was k[[x1, ..., xn]] (or any a regular local ring) it would be true.

[karmic moat]

yeah that's what's troubling me too, since S is not local

[vast verge]

How do I do this question?

Solutions didn't help

[charred iris]

What is G bar?

[rocky cloak]

Imagine H was equal to G-bar. Would you be able to solve the problem then?

[vast verge]

Using the isomorphism theorem

$G / \ker \phi \cong \phi(G)$

[cloud walrus]

Tropical Greens

[charred iris]

-# is it just some quotient of G

[vast verge]

G bar is just the image group

[rocky cloak]

I guess my point is that there's a relationship between the size of
ker phi, phi^-1(H) and phi(phi^-1(H))

And this comes from the first isomorphism theorem yes

[vast verge]

Wait hang on

I think there was a corollary I glossed over

These

[rocky cloak]

Yeah, so corollary 1 here is what I'm talking about

It essentially just is the answer to your question

[vast verge]

So I'm replacing G with H and phi with its restriction to H right?

[rocky cloak]

Almost, phi^-1(H) would be the thing that's a subgroup of G

[vast verge]

Hmm okay so

$| \phi^{-1}(H) | / | \ker \phi | = |\phi \left( \phi^{-1} (H) \right)|$

[rocky cloak]

Indeed

And the right side is equal to....

[cloud walrus]

Tropical Greens

[vast verge]

|H|

Thank you

[rocky cloak]

This is right. Hard to call it overcomplicated, don't think you can do it simpler

[cedar vault]

There is the subtlety of explaining that p(b) describes all elements in S, which is given by surjectivity ofcourse

Still worth mentioning

[karmic moat]

if S is a subring of R, then the inclusion i : S -> R is just i(a) = a

hence it literally is an inclusion

[tough raven]

Either that or the proof-free, citation-free offhand statement in the textbook was wrong (or missing some hypothesis or qualification (or I missed it)).

[karmic moat]

i think the philosophy of what you're trying to say is right but i'm having trouble reading your proof, in part due to the notation

all you have to do is look at the set $\varphi^{-1}(J) = { r \in R \mid \varphi(r) \in J }$

[tough raven]

Possibly the elements have to be homogeneous of positive degree (with R a non-negatively graded algebra with R_0 = k).

[cloud walrus]

anamono

[karmic moat]

if $r, s \in \varphi^{-1}(J)$, then $r + s \in \varphi^{-1}(J)$ since $\varphi(r) + \varphi(s) = \varphi(r + s) \in J$

[cloud walrus]

anamono

[karmic moat]

if $r \in \varphi^{-1}(J)$ and $s \in R$, then $\varphi(rs) = \varphi(s)\varphi(r) \in J$, so $rs \in \varphi^{-1}J$

[cloud walrus]

anamono

mqinsweden (ping on reply)

[karmic moat]

okay hmm i do recall some result like this for graded rings

[tough raven]

Does this theorem not prove that something is a UFD, and does that not make it redundant here since we otherwise know that polynomial rings are UFDs anyway?

[karmic moat]

but i think it still would need to use auslander-buchsbaum formula? maybe not

if you know that then you don't need to show it

[rocky cloak]

I'm not sure I follow, what is being proved is a ufd?

[karmic moat]

what's your homomorphism? write it out

okay

sure

[tough raven]

AB theorem on Wikipedia says "regular local rings are UFDs". So I assume you meant by this that if S were regular local, you would use it to conclude that S is a UFD and do something with that. But S in my setting is a polynomial ring, which is a UFD anyway, so I guessed maybe whatever you had in mind would work for it even without using AB.

[karmic moat]

yeah looks fine

[rocky cloak]

Ah, wrong theorem.

I meant this one
https://en.m.wikipedia.org/wiki/Auslander–Buchsbaum_formula

[karmic moat]

do you have a question?

[cedar vault]

I suppose you must ping on reply(as per their name)

[karmic moat]

do you have a question?

yeah

not necessarily divisible by the prime, right

(x+y)^4 has a coefficient of 6 in the middle

yeah i mean it wont be 0

because 4 does not divide 6

true

[woeful sage]

Which section is this from?

[cedar vault]

What exactly are you trying to do here? Maybe i didnt understand correctly, but to show I is not prime, you just need to give two continuous functions not in I but whose product is in I. What are you claiming those functions are?

You can take inspiration from your g(x) to find two such functions

[woeful sage]

I'm at 7.1

I know

that's why I reacted with

Help me in #1272080242004983819

[rocky cloak]

It's P contains IJ, because R containing IJ would be redundant information

(See Grice's maxims
https://en.m.wikipedia.org/wiki/Cooperative_principle )

In algebra all proofs are trivial once you understand the definitions

It's a joke, but jokes have some truth to them

I'm curious how you think of PvNP as an algebra problem, but that's just me I guess

It's a problem about complexity theory.

There are unsolved algebra problems though.

[fervent solstice]

even without leaving the millennium problems list you could still find some cool algebra ones

[rocky cloak]

I mean, would be the BSD conjecture I guess

[fervent solstice]

yeah or Hodge

[rocky cloak]

But I guess both of these fall into algebraic geometry, so if you're very picky about your labels you might call them geometry problems

[karmic moat]

there's a lot of conjectures in homological algebra

[fervent solstice]

doesn't the very existence of algebraic geometry make every geometry problem equivalent to an algebra problem?

[south patrol]

No

[rocky cloak]

Yes, the homological conjectures in rep theory are close to my heart

[karmic moat]

https://www.math.utah.edu/~roberts/homconj.pdf

[tardy hedge]

Bro even knows linguistics 😂😂😂

[south patrol]

I mean there are types of geometry further from algebra

[karmic moat]

lol i was gonna say that roberts lists a bunch in his book but i dont have the book on me

i guess he wrote an article on them too

[south patrol]

Or at least from AG

[fervent solstice]

are they beyond the reach of AG though

i've been led to believe that everything is AG if you do enough AG

[rocky cloak]

I'm slightly annoyed that commutative algebraist wants to steal the cool name 'the homological conjectures'

[karmic moat]

https://tenor.com/view/true-truth-nuke-super-truth-nova-truth-meme-gif-16889273424352737553

[karmic moat]

who would u give it to

the ones in rep theory?

[south patrol]

What even is homological algebra

[karmic moat]

letters and arrows

[woeful sage]

It's like homotopical algebra but with homology now

[south patrol]

Idk i mean like i dont see how you can say Riemannian geomstry is AG unless you widen what AG is significantly lol

[fervent solstice]

i mean GAGA is a thing

[rocky cloak]

Here's some of them
https://www.math.uni-bielefeld.de/~sek/dim2/happel2.pdf

[south patrol]

I mean yes but GAGA does not touch all of geometry

[karmic moat]

my prof said homological algebra is "the study of (co)chain complexes and their (co)homologies"

[karmic moat]

if i see auslander's name one more time today im gonna lose it

auslander auslander auslander omg leave something for the rest of us to do

anyhoo those are cool

dunno why we can't lump them in with the rest of the homological conjectures

[fervent solstice]

...yet

[south patrol]

Lol i mean i am v familiar with hom alg just i find it dunny like e.g. idk anyone who felse identifies as a "homological algebraist"

Usually when people say homological algebra it seems they mean the "basics"

[rocky cloak]

I'd go a little broader and say homological algebra is the study of categories with some notion of homology, so abelian, triangulated, exact, extriangulated and higher versions gets to join in

[south patrol]

Just eventually this eventually feels like stable homotopy thwory

[karmic moat]

group theory

[rocky cloak]

Yeah, one day your thinking about derived categories, then algebraic triangulated categories and dg-enhancment, suddenly you're thinking about stable infinity-categories and doing full blown homotopy theory

[fervent solstice]

there s a lie group joke in there somewhere

[karmic moat]

i don't really understand your question tbh

[fervent solstice]

yeah there are still people who study specifically groups

[rocky cloak]

Many people teach in the opposite direction

[fervent solstice]

the classification of finite simple groups for instance is a huge result that's been proven thanks to the contributions of many people over the last centuries

[woeful sage]

I have seen one such book

[fervent solstice]

the p sylow stuff is the very basics of it yea

[karmic moat]

why would they do this instead of groups rings fields?

like i can see how you'd start with fields and generalize to rings and then drop some operation to get to groups if you want

forgetful functor behind the scenes

[fervent solstice]

it's not immediately obvious to everyone why you'd care about groups, whereas vector spaces are something more concrete already

[woeful sage]

symmetry

[rocky cloak]

Most people have a lot experience with fields and rings from their earlier education. So you have a lot of examples at your finger tips.

[fervent solstice]

who cares about symmetry except for nerds

[karmic moat]

true but i think a lot of people have experience with groups as well

Z and GL_n being the main ones im thinking about

but actually i guess rings/fields show up more naturally than groups

hmm

[limber tapir]

But Z is a ring in peoples minds

Because you can multiply integers

[tall igloo]

We should only teach rings and just tell people abelian groups are Z-modules. No use for non-abelian groups anyways

[rocky cloak]

Z and Z/n is also a ring, and I think students have mostly seen it as that (with a multiplication and addition I mean)

I don't think a lot of people meet GLn before their algebra course, but I could be wrong about that

[karmic moat]

yea i dont think they've seen GLn called "GLn" but i'd imagine they've definitely seen invertible matrices

and they also know that matrix multiplication is not commutative

but yeah i guess Z as a ring makes more sense than Z as a group

[rocky cloak]

Yeah, that's true. GLn and the determinant would be a group example accessible

[karmic moat]

truth nuke

GLn can't be a field

[fervent solstice]

show that everything in F2[x] is in the class of one of those

[karmic moat]

a field has two operations, addition and multiplication