groups-rings-fields

[celest furnace]

If E contains F and is degree 1 then since F is also dimension 1 over F we must have E = F by vector space reasoning

Like if E contained something not in F then that thing would be linearly independent of a generator for F so E would be dimension at least 2

[slim kayak]

Yeah, so you've shown that finite field extensions are "noetherian" in this sense

You will run out of elements to add

[celest furnace]

Woah thats awesome

[slim kayak]

Oh no, wdym?

[celest furnace]

No just like cool connection

[slim kayak]

I think the "finite" part here ought to make it less wow inducing

Actually nvm that, "noetherian" goes too hard to not use it

[cobalt heath]

Ah like, when F is finite extension of E, then F is Noetherian module (vector space) over E

In G = (Z/nZ)^×, consider subgroup H generated by some primes p1, .., pk. For integer c, What is a minimal representative of c H, when projected back into Z?

If possible, I would like to reformulate this because "minimal representative when projected back into Z" part is not particularly clear.

[slim kayak]

I have no idea what that means either

Oh wait, a minimal n s.t. nH=cH?

[cobalt heath]

Yes

But, minimal in the sense of absolute value of Z

Which.. is not ideal, I admit.

[slim kayak]

Idk what else it would mean

I'd almost go so far and ask for a canonical representative from Z. Like how you usually think about Z/pZ as 0,1,...,p-1 (in my case we put a bar over it)

[cobalt heath]

Hmm

Anyway I would like a way to estimate such minimal n

Given some criteria to (c, n)

[slim kayak]

Uh... so what exactly is H here

[rotund aurora]

isn't it c

I didnt and still don't understand the question

[cobalt heath]

H is a subgroup of G generated by p1, .., pk.

[rotund aurora]

So it's c

H contains 1

[cobalt heath]

Oh, I mean the generators are p_1, ..., p_k, certain numbers

Also G = (Z/nZ)^×, the multiplicative group

So H contains 1 vacuously.

[slim kayak]

Vacously?

[cobalt heath]

(Sorry that the question is not well-structured, I am still figuring out right statements for this)

[cobalt heath]

1 is identity of the group

[slim kayak]

I know but usually that refers to making universal quantifications over the empty set

[cobalt heath]

True, I am.. idk, bad at English(?).

[slim kayak]

Trivially seems appropriate there ig

Shouldn't it be some relatively prime part of your c?

[cobalt heath]

What do you mean by "relatively prime part"?

[slim kayak]

Like, if the p's were say primes you'd keep dividing c with them and the remainder is minimal

And analogously for more general p's

[cobalt heath]

Btw let me give some examples, maybe it might help
For n = 13 and p_1 = 3,
G = (Z/13Z)^×,
H = {3, 9, 1}.
So c = 1 then minimal m = 1
For c = 5 the minimal m = 2

[cobalt heath]

How would you get m = 2 from c = 5, given n = 13?

[slim kayak]

How did you get 2?

Nvm I see it now

[cobalt heath]

Because 2 * 9 = 5 on (Z/13Z)^×

[slim kayak]

9*5=35 was lowkey a bit off on my end

Okay yeah, in this case you get it by taking your generator times your c mod p

[cobalt heath]

Yep (it's mod n in my notation, btw)

[slim kayak]

Right

Have you checked this for more generators?

Perhaps mod the ideal generated by your generators

[cobalt heath]

When n = 13, any subgroup of (Z/13Z)^x is cyclic

So you won't get much difference

I guess n = 15 will be different?

Since (Z/15Z)^x is isomorphic to (Z/2Z) * (Z/4Z)

H = <4, -1> <= G
Then there are two left-cosets H, 2H.

[cobalt heath]

Anyway, thank you Kerr! I will try to refine this further.

[slim kayak]

You're welcome! (What did I do)

[grizzled crane]

Are there zero divisors in set of all 2 by 2 integer matrices ?

To show it’s not an integeral domain, i suppose I need to choose a non zero element ‘a’ and show the existence of a non zero element b, such that ab=0

The ring M_2(Z) of 2 x 2 matrices over the integers is not an integral domain.

I’m trying to understand this ^

[obtuse bear]

you should be able to find a simple example using matrices that have only one nonzero element

[grizzled crane]

For A= \begin{pmatrix}
0 & 1 \
0 & 0
\end{pmatrix}, i can choose B= \begin{pmatrix}
1 & 0\
0 & 0
\end{pmatrix}, so that AB=0

what’s TeX error in this?

[chilly ocean]

I'm not sure if anyone minds if I ask this here or not, I asked in "advanced-algebra" earler. I'm studing Knot theory and was wondering if Knot's form a specific group name, other than just "Knot group"?

Like it's classification among other groups, like dihedrial and such.

[cloud walrus]

.doc
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[rotund aurora]

@chilly ocean the groups associated to different knots needn't be isomorphic, so your question doesn't make a lot of sense

It would make sense to ask what groups occur as knot groups, but I have no idea tbh

https://en.m.wikipedia.org/wiki/Knot_group

Key word: fundamental group

This is brtter suited for #point-set-topology I think, ask there if you have questions

[sour spear]

How do I know if Z/aZ x Z/bZ is isomorphic to Z/abZ?? It only happens when a and b are coprimes??

[dense void]

use chinese remainder theorem

$x \bmod a b \mapsto (x \bmod a, x \bmod b)$ is an isomorphism if $a$ and $b$ are co-prime (by CRT).

[cloud walrus]

klmn

[next obsidian]

That provides one half, to show that they aren’t isomorphic when an and b aren’t coprime is done most easily by showing Z/aZ x Z/bZ doesn’t have an element of order ab

[glossy crag]

If L/K is P, where P\in {finite, algebraic, separable, Galois, etc.}, then L(x)/K(x) (function fields) is P as well, right?

[north tusk]

Hey I'm trying to understand the proof for the division algorithm for polynomials.

nvm

Someone responded to my help channel

[slim kayak]

Hey, what's the argument for why R_p (the localization of R) is flat as a R-module?

[glossy crag]

Probably use that R_p\otimes_RM\cong M_p?

[slim kayak]

Oh right, use the exactness of localizing

[cloud walrus]

HausdorffT1

[sour spear]

Oh god, I didnt realise about that, thank u so much

[rocky cloak]

I believe preserving lines and cross ratios is enough. And in the case k>1 preserving lines should be enough.

In any case, lines and cross ratios is structure to preserve

@long obsidian

[chilly ocean]

@rotund aurora Thanks, that makes scnse, certain knots form certain groups.

[hidden kite]

If anyone is free and wants to help, could you help me check my proof and give me some feedback?

I have checked it a couple of times, so hopefully there are no typos or anything

[mighty kiln]

Looks good

[hidden kite]

Alright, thank you! :D

[tough raven]

The Gram-Schmidt algorithm works and can be used to construct an orthogonal basis for any anisotropic symmetric bilinear form of a finite-dimensional vector space over a field, right?

[alpine island]

yeah that looks good

[hidden kite]

thanks

[alpine island]

I wonder what the generator of <S> is

[celest furnace]

Infinite product in the denominator? What is q_i for i > n?

Also super clean love it other than that

[alpine island]

oh, yeah good point

[crystal turtle]

yeah I would just write \frac{1}{q_1 \cdots q_n}

(Another fun exercise: prove there are uncountably many additive subgroups of Q)

(Another exercise: Classify all subgroups of Q. Prove that your classification is complete)

[hidden kite]

Ahh sorry! I meant to n 😅

Okie thanks ryx for the exercises

Thanks everyone for your help!

[hidden kite]

Yayy thank you!

[tough raven]

The "gcd" of the elements.
More precisely, if the elements of S are written with a common denominator, then the gcd of the numerators with the same denominator (this doesn't depend on the common denominator because gcd(k a_1, …, k a_n) = k gcd(a_1, …, a_n)).

[alpine island]

yeah that was my guess

[crystal turtle]

(not insanely difficult, but hard and definitely quite involved. It took like 3-6 pages handwriten for me to do? and I think that was kinda just a sketch? and I think it was correct perhaps even???)

[tough raven]

||Scalings of localisations of Z||, right?

[cobalt heath]

Nvm

[crystal turtle]

Oh yeah I guess that should work lol

You can do it more explicitly with just elementary number theory and basic group theory which is what I had in mind

[tough raven]

I stumbled upon (roughly) this while trying to classify intermediate rings/R-modules between an integral domain R and its field of fractions (you get the best answers for R a PID).

[crystal turtle]

/that's what I did because I was a wee lad who didn't know much about rings lol

[crystal turtle]

I imagine it should be similar for an arbitrary PID?

Or the same or something

Or uhh wait that might not be right, how would you get something like {a/2 | a in Z}?

[tough raven]

That's why I had to add scaling to the statement.

OK, here's what I got more precisely:

1. If R is a UFD, any subring of Frac(R) containing R is a localisation of R obtained by inverting some set of irreducible elements of R (and the set is uniquely determined as the set of irreducibles whose inverses are in the subring).
   (IIRC existence of factorisations gives existence and uniqueness of factorisations gives uniqueness.)
2. If R is a Bezout domain (any two elements have a GCD which is a linear combination of them - i.e., a Bezout identity), then any R-module of Frac(R) containing R is generated by elements of the form 1/r, r in R.
   We may assume that the set of r used is closed under reverse divisibility, i.e., if s | r and we use 1/r, use 1/s. Once we do this, the set of r used is uniquely determined (namely as {r | 1/r in the R-submodule}).

Now, for a PID, the non-zero elements are all products of prime powers, and by CRT we can just look at which prime powers have inverses in the R-submodule.
For each prime p, either p^{-k} is there for all k (in which localise at p), or there is a largest k such that p^{-k} is in the R-submodule (in which case scale by p^{-k}).

IG 1. isn't needed for this.

[tough raven]

Also, I might be misremembering a few details here (especially the uniqueness parts), so you might want to check it. Broadly, it should be correct though.

[crystal turtle]

Okay now I think I agree with 2 here... From what I remember that works out to be what I got for Q and that was using nothing more than Bézout's identity and I guess maybe something about PIDs, so that should hold in that generally

Been a while since I did it lol

But it's a fun exercise to work out on your own

And then the rescaling bit is to correct from "submodule of Frac(R) containing R" to all submodules

[tough raven]

Well, not exactly.

You need the scaling even if the module contains R, to get something like (1/r) R.
That requires only scaling by inverses of elements of R.

Adding scaling by elements of R as well (i.e., by any element of K) is precisely enough to get all non-zero R-submodules (proof: any non-zero submodule M contains some r, then M/r contains 1, so it is S^{-1} R / s for some set of primes S, so M = (r/s) S^{-1} R).

Also

[tough raven]

bump

[glossy crag]

kek, how'd you make that emote into the end of proof sign?

[glossy crag]

I think the characteristic had to be \neq2, but iirc that's enough. I don't think you even need the form to be anisotropic (what even is an anisotropic form, don't you mean non-degenerate?).

[hidden kite]

\newbox\eeveebox \setbox\eeveebox\hbox{% \raisebox{-5pt}{\includegraphics[height=3ex]{./figures/iibui.png}}% } \def\eeveeKawaii{\copy\eeveebox}
Full credit goes to detto

this was also drawn by them

[glossy crag]

You guys usually overdo it with that emote spam, but I gotta admit that's cute.

[hidden kite]

You're goddamn right (the cute part :hehe:)

[rustic crown]

[tough raven]

Anistropic: <v, v> is non-zero unless v = 0.
You do need this for Gram-Schmidt as I know it.

(I do believe that it is true that symmetric (even degenerate) bilinear forms have orthogonal bases even if they're not anisotropic in this sense, but Gram-Schmidt doesn't always suffice.)

And for all of it, I either never read up on the proofs or don't want to check the details to see what fields and conditions on the form are required.

I should probably just do that work though.

[glossy crag]

Oh wait, did you ask about orthonormal bases? Because what I said was about orthogonal ones. You need anisotropicity for the vectors to have norm 1, if you just want them to be pairwise orthogonal you don't need this.

[tough raven]

Orthogonal

[glossy crag]

Then you need absolutely nothing beyond the characteristic not being 2.

[tough raven]

Oh?
Let <v1, v1> = 0.
How do you compute v2 - <v2, v1>/<v1, v1> v1?

[tough raven]

Any idea what characteristic not 2 is needed for?

(Or is it superstition against bilinear forms working well in characteristic 2? That's very relatable.)

[glossy crag]

Ah, wait, you want to compute them with Gram-Schmidt, my bad. An orthogonal basis exists, but by induction on dimension, not GS.

[tough raven]

Yes
Otherwise I have to look up/figure out the more general procedure

Which again I should just do

[cobalt heath]

How do you do this, my LA class was too flimsy to cover this.

[glossy crag]

Assume the form is non-zero (nothing to prove otherwise), by the polarization identity (here's where you need the characteristic to be \neq2) there must exist a vector x such that (x,x)\neq0, then V=Kx+(Kx)^c (this is the hardest step of the proof), where (Kx)^c is the orthogonal complement. By induction there is an orthogonal basis in (Kx)^c, add x to that.

[glossy crag]

Nothing hard about it, see above.

[slim kayak]

Never seen that way of doing GS before

[cobalt heath]

Ah, so having one x s.t. <x, x> /= 0 is enough

[slim kayak]

Take some random vector v, normalize it. Then take another vector w and subtract <v,w> times w from v. Then normalize the result. Repeat this by subtracting the two weighted vectors

[cobalt heath]

Well, this is Gramm Schmitt, isn't it? I think this fails when <v, v> = 0 (which was the point of the question of Raghuram above)

[slim kayak]

Isn't that the case when v is the zero vector

[glossy crag]

It's not "enough", this isn't some special assumption you have to make, every non-zero symmetric form in characteristic \neq2 has such a vector.

I mean it is enough ofc.

It's just not some special restrictive requirement.

[cobalt heath]

Do not leave characteristics 2 off the picture, then it is reasonable to say 'enough' >.>

[glossy crag]

It actually is reasonable to say "enough" because afaik characteristic 2 bilinear/quadratic forms is a special and thorny subject completely apart of all else.

[cobalt heath]

Ok

[cobalt heath]

Consider x = (1, 1, 1, 1, 1) over F_5. <x, x> = 0

[tough raven]

I see. Thanks.

[tough raven]

Not really, you also need it to hold on the orthogonal complement.
So the condition would be more complicated.

[cloud solar]

Let R be a ring with n^2 elements and n zero divisors. Show n is a prime power.

[rocky cloak]

Let m be the characteristic of R. Since R is a Z/m module, it must have order dividing m^r. So if n is not a prime power, then neither is m, so m = ab for relatively prime a and b. Since Z/m = Z/a x Z/b, R has central idempotents so R = AxB where A has characteristic a and B has characteristic b.

Thus |A| and |B| are relatively prime with |A|*|B| = n^2. Then it must be that either |A| or |B| is bigger than n, say |A| > n.

But AxB has at least |A| + |B| - 1 zero divisors given by (x, 0) and (0, y), and |A| + |B| - 1 > n

[dull marsh]

Is there a term for a set of generators S such that there is no s in S such that <S\{s}> contains s? Similar to linearly independent sets for a vector space

[coral spindle]

Minimal-by-inclusion

(As opposed to minimal-by-cardinality)

[dull marsh]

I see, thanks

[celest furnace]

In my class we called this a minimal set of generators

Exercise: show that every minimal set of generators need not have the same cardinality

Second exercise: show this is true for p-groups (VERY HARD)

[dull marsh]

<1/2, 1/3> = <1/6> in the set of rationals

[dull marsh]

Wait wdym exactly by that

[rocky cloak]

If G is a p-group, then every minimal generating set for G has the same cardinality.

[dull marsh]

Ah

[celest furnace]

Yeah this shit is rough to prove

It is very cool though

[rocky cloak]

The proof feels a lot like ring theory. Dealing with maximal subgroups and chains of subgroups.

[dreamy granite]

can someone real quick give an example and explanation for a question on groups and rings that may come on an exam? normally i would have enough time but i legit had 2 exams today and 2 tomorrows so there are parts i literally can't understand

like for example:

Assume that the equation xyz = 1 holds in a group G. Does
it follow that yzx = 1? That yxz = 1?

[crystal vale]

Yes it imply yzx=1

[glossy crag]

Yes to the first (multiply both sides by x' from the left, then by x from the right), no to the second (take any two non-commuting x,y in some group, take z=(xy)', then if yxz too it would follow yx=xy).

[dreamy granite]

thank you!

so kinda like theortical computer science in a sense got it

my mind isn't even a mind anymore sorry if i don't make sense since i legit have no way to explain my thought process but it makes sense in my corrupt brain

[glossy crag]

You need some bro

[dreamy granite]

i do, but i can't lol

[celest furnace]

Hello everyone, today in class my instructor said that (t) is a prime ideal of F_p(t). I’m not quite sure how to show this.

Hmm now I’m confused

Hold on

[coral spindle]

Are you sure they didn't say that it was a prime ideal of F_p[t]? F_p(t) is a field so has no nontrivial ideals.

[celest furnace]

He was trying to show us that not all field extensions are separable by showing that X^p - t is irreducible in F^p(t)[x]

By applying Eisenstein

[rocky cloak]

Yes, (t) is a prime ideal in Fp[t] so the polynomial is irreducible over Fp[t], and since Fp[t] is integrally closed (like all UFDs), the polynomial is also irreducible over Fp(t)

Alternatively, if you don't want to deal with all that, it's not hard to see that the polynomial has only a single root. So the minimal polynomial of said root can't be seperable.

[celest furnace]

Thanks ! He didnt mention anything like this at all lol

[south patrol]

Hm random. Let A be an R-algebra and everything be commutative. Is there a nice description of Hom_R(R[[x]], A)?

well pointed out lol

But yeah it's interesting, like we can view this set as a subset of Hom_R(R[x] ,A) = A (since any map here is determined by its value at x)

[mighty kiln]

Is it left-adjoint to the map from R-algebra to group of units

[south patrol]

I don't get what you mean

I've not mentioned any functors, at least by name

[mighty kiln]

G → R[[G]]
S → S*

So this probably Ab(Z, A*) or sth?

[south patrol]

Uhh are you thinking of group rings

[mighty kiln]

Is this be group ring

[south patrol]

I'm just confused like I don't have any functors out of groups here

So they can't be adjoint to the group of units functor

[mighty kiln]

Oh it be group ring

[south patrol]

Your thing ig

[mighty kiln]

R[[x]] = R[Z] no?

[south patrol]

Huh

No

Power series with R coefficients

[mighty kiln]

Oh wait I read this as Laurent polynomials

[south patrol]

Oh

For that I agree with you that there is a bijection between Hom_R(R[x,x^-1], S) and set of units of S

But that isn't an adjunction

Or do you mean smth else

[mighty kiln]

Hm I wonder if there is something similar for "monoids where each element has finite factors" → R-Alg

[mighty kiln]

Tho you probably need some proper map condition for morphisms in the first category

[slim kayak]

Left adjoint? As a functor taking R to R[x,x^-1]?

[mighty kiln]

No like G → R[G]

[mighty kiln]

Say all fibers are finite

Is M → R[[M]] left-adjoint to S →

[slim kayak]

M a module?

[mighty kiln]

Monoids where each element has finitely many factors and morphisms have finite fibers → R-Alg

Nvm infinite sequences be bad

[slim kayak]

Uh huh

What are factors in the context of a monoid? Factorizations?

[cobalt heath]

Hmm, what is 'S →' ? Hom(S, -)?

[slim kayak]

It assigns to every S

[wild solar]

Hi, I'm wondering could I get some suggestion on this problem? I'm not sure how I should think about this.

[outer valve]

When you take a group presentation, is there any way to show that it will get "reduced" to a trivial group?

Like, how can I definitely say "No, you cannot shuffle these relations around anymore to reduce this group any further"?

[rocky cloak]

(a) Consider an element in both K1 and I1. Then it is the image of some element, what does that say about K2?

(b) The equality is saying that for any x there exists a y such that f(x) = f(f(y)). What does that say about x - f(y)?

(c) Kn is an ascending sequence, now think about (a).

(d) Use (c) to conclude something about Kn and f^n

[wild solar]

sorry why is (b)? Why does it imply that

[rocky cloak]

if I1 equals I2, then any element of I1 is contained in I2

So f(x) is contained in I2

[wild solar]

O right

I was looking at the conclusion xd

Thank you so much!

[hidden kite]

If anyone has time and wants to help, could you help me see if these (relatively similar) proofs are alright, and provide some feedback?

[mighty kiln]

https://en.m.wikipedia.org/wiki/Word_problem_for_groups This may be of interest

[outer valve]

Thanks, I'll return to it once I have a bit more experience under my belt, when I can understand a bit more.

[slim kayak]

You can't

[narrow wagon]

Consider the cyclic group Zp under multiplication modulo p, is it easy to find the generator of this group? (aka primitive root)

[slim kayak]

Okay, you can in a lot of circumstances. But we have mathematically proven there exists no "algorithm" to solve the problem. You will need tricks and strategies that only work for some

[outer valve]

Alright, thanks. I re-read through D&f, and missed a part where it said that representing D2n as the changes in a polygon is what gave it a lower bound. So I guess the stretegies and stuff mean these kinds of stuff.

[south patrol]

That isn't a group

Well

Unless you are using odd notation

[narrow wagon]

yes, i know the notion is a bit off

The set is residues modulo p

[south patrol]

With what operation?

[narrow wagon]

multiplication modulo n

[south patrol]

If it is multiplication mod p then it isn't a group

0 has no inverse

[narrow wagon]

Unitary group

[narrow wagon]

mb

[south patrol]

I assume you mean like

[narrow wagon]

U(n)

[south patrol]

Yeah

Okay cool

[narrow wagon]

Specifically for n is a prime

I’m sort of looking for ismorphisms to Zp

[south patrol]

Hm it is sorta nontrivial that primitive roots even exist but often trial and error will work quite quickly

I'm unsure if there is an algorithm or anything, I'm sure there is

[narrow wagon]

Are there isomorphisms from multiplicative group to additive group?

[south patrol]

Yes, multiplication and addition are just notation

[south patrol]

Well that is impossible

I guess you mean Z_(p-1)

[narrow wagon]

Not really

I’m trying to see, if there a multiplicative group with prime number of elements

so it’s going to cyclic by lagranges theorem

nahh it’s impossible

Euler totient function is always even

[south patrol]

Yes

[narrow wagon]

It’s the cardinality of U(n)

[south patrol]

Do you mean a group of units with prime number of elements

[narrow wagon]

yes

[south patrol]

There are multiplicative groups more generally

[narrow wagon]

Impossible right?

[south patrol]

Well

there is one exception

There is one even prime

[narrow wagon]

Yeah

Suppose U(n) has a primitive element

then U(n) is cyclic right?

which makes it isomorphic to Z_phi(n)?

2 is a primitive root mod 5, so is U(5) isomorphic to Z4?

@south patrol

[south patrol]

Yes you are right

And yes

Everything you'vfe just said is good

[teal vessel]

I assume by "isomorphism type" they mean the archetypal group, like Z8 for the rotation subgroup?

[south patrol]

Yeah basically this means describing the "equivalence class" of the group under isos, so yes try to give the simplest possible description

[teal vessel]

been burned by formal language and natural language mixups too many times, and had to check 😅

cool, so <r> is congruent to Z_8, and both <s,r^2> and <sr,r^2> are congruent to D_8. working with lattices makes this so much easier.

(yes I did do my proper diligence and ensure that it wasn't just lattice matching, there was already a problem about non-isomorphic groups with identical lattices)

[delicate orchid]

C_p for two different primes :uponthewitnessing:

[teal vessel]

in this case it was Z_2 x Z_8 vs M_16

[viral mountain]

Can we prove that for the binary operation subtraction on $\mathbb{R}$, there exists no identity element?

This is my attempt at a proof, but I'm not sure if it holds

Assume, for the sake of contradiction, that $\exists e \in \mathbb{R}$ s.t. $\forall a \in \mathbb{R}, a - e = e - a = a$. Then, $2a - e = e$, and hence $a = e$. But, $e$ is a fixed element in $\mathbb{R}$, and so cannot depend on $a$. Hence, there exists no identity for subtraction

There are 2 problems with this proof I feel. One, I've used addition on real numbers, and even though it is a valid operation, I'm not sure if I'm even allowed to use it since I'm just considering the binary operation subtraction on $\mathbb{R}$. Two, the logic "$e$ is a fixed element" doesn't sound very rigorous, so there might be a flaw here

[cloud walrus]

BlazingSaber

[coral spindle]

The first problem is not a problem at all, I don't see why you think it is. The second problem is resolved by seeing that you mean that this would imply e = 1 = 2, for example.

Shorter -- but fundamentally equivalent -- proof: if $e$ is an identity, then $1 - e = 1$, so $e = 0$. But then $e - 1 = -1 \neq 1$, contradiction.

[cloud walrus]

Boytjie

[viral mountain]

Right, this is a nice proof
The reason why I thought the first problem is a problem is becasue I assumed that I wouldn't be allowed to use any other binary operation when I'm considering just one (subtraction), but you're right, there shouldn't be any reason why I can't use another operation

This also now makes me curious about what addition on real numbers fundamentally even means

[coral spindle]

One way to construct the real numbers is as equivalence classes of Cauchy sequences of rationals, and addition is just pointwise addition on these sequences.

But this is besides the point.

[viral mountain]

Like, per definition, I'd define

$+:\mathbb{R} \to \mathbb{R}$ as $+((a, b)) = a + b \forall x \in \mathbb{R}$, but that'd essentially be using the definition before I even define the concept lol

[cloud walrus]

BlazingSaber

[viral mountain]

interesting, so implicit in that would be the definition of addition on rationals, right?

And I'm assuming that could potentially be based on the definition of addition on integers, which is further more fundamentally defined?

[coral spindle]

Yes, and we can define addition on the rationals in terms of the integers, then the naturals, where we define addition purely in set-theoretic terms. This is really quite besides the point though.

[viral mountain]

I remember seeing some S(0) = 1, S(S(0)) notation in a video some time ago

[coral spindle]

If you want a book that guides you through this, see Tao's Analysis I.

[viral mountain]

This is super interesting stuff!

[coral spindle]

There's no need to speculate either. Here is the wikipedia page on the construction of the reals.

[viral mountain]

Thanks for the link
But yeah you're right, it's way besides the point now 😅

[coral spindle]

A remark that is more on-brand for this channel: the reals are just a quotient of the ring of Cauchy sequences by the ideal of sequences that converge to 0.

[cloud walrus]

Absta

[cloud solar]

I have a problem

And I thing it is wrong

A finite ring, the number of invertible elements is equal to the number of nilpotent elements. Show that x is invertible iff 1+x^m is nilpotent for every m non zero natural number

I think they wanted us to show that x is nilpotent iff 1+x^m is invertible for every m non zero natural number

[daring nova]

that would make more sense
It may already be known (or a similar statement, I don't quite remember) from linalg

[cloud solar]

And the condition that the number of invertible elements is equal to the number of nilpotent elements is to show the "<--" case I think

[cloud solar]

Like for det(A+I)=1 A nilpotent?

[daring nova]

that if N is nilpotent, then I + N is invertible

[cloud solar]

Yes and that

[daring nova]

is quite easy to see

[cloud solar]

This can happen for F2

[daring nova]

yeah that's the one case I had in mind

in which case the statement is remarkably trivial

[cloud solar]

For n>=3

We have at least one zero divisor which is not nilpotent

Idk if this problem is wrong

Or not

2 is not invertible

Because phi:N(A)->U(A) bijective phi(x)=x+1

And if 2 is in U(A) then there is a nilpotent with a+1=2 a=1 false

Interesting

[cloud walrus]

MasakaBakana

[stuck fiber]

So I was working a problem that asked me to prove that $$\mathbb{R}[y]\cong \mathbb{R}[x,y]/(x).$$ I ended up doing this by having $$\varphi:S[x]\rightarrow S$$ defined by $$f(x)\mapsto a_{0}$$ where $S$ is a (commutative) ring and then proving this was a surjective homomorphism with kernel $(x).$ Then when you use $R[y]$ as your ring you get the desired isomorphism. I'm thinking with this I should be able to generalize out to prove results such as $\mathbb{R}[z]\cong \mathbb{R}[x,y,z]/(x,y).$ But for some reason I have a hard time wrapping my head around this I feel like it should be relatively simple.

[cloud walrus]

Kenshin

[stark helm]

I have a question, if i is a subring, can I say since it is abelian under addition, then it satisfy subgroup?

[celest furnace]

What

[stark helm]

I heard someone get this conclusion, but I don't know why it can be true

[celest furnace]

When you take a quotient you should be imaging it’s like setting the thing quotienting by to 0

[stuck fiber]

f(x) is a polynomial in S[x]

[stuck fiber]

What?

[celest furnace]

Like quotienting out by (x-1) you should think of that quotient is setting x = 1

[stuck fiber]

Oh I see yeah I get that but I don't see how that helps me generalize this out to n variables

[celest furnace]

The same map doesnt work?

sending f(x, y, z) to f(0, 0, z)?

[cloud solar]

What could be the finite rings A with the number of idempotent elements equal the number of invertible elements equal the number of nilpotent elements

Or what property this type of rings have

Just for curiosity

I know we can make bijective maps between, U(A), N(A) and I(A)

2 must be a zero divisor

[slim kayak]

Subrings are closed under addition, so they are also subgroups

[rustic crown]

A (commutative) ring is a product of two rings if and only if it has a non-trivial idempotent. So given such a finite ring A, you can break it as A = A_1 x ... x A_n where each A_i has exactly 2 idempotents, namely 0 and 1.
also notice that units and nilpotents behave nicely in products, so it reduces to finding rings A which have exactly 2 units, nilpotents, idempotents. If ε is a non-zero nilpotent, then the nilpotents in your ring are exactly {0, ε}. But now ε^2 is also a nilpotent, so either ε^2 = 0 or ε. we can't have ε^2 = ε, as that implies ε^k = ε for each k > 1, and so ε= 0. Thus ε^2 = 0.
Now the units in the ring should be {1, 1+ε}. Since 1-ε is also a unit, 2ε=0. Take two cases, if char A = 2, then {0, 1, ε, 1+ε} is a subring which is iso to F_2[x]/(x^2). If char A != 2, then 1+ε must be -1, so ε = -2 and we have 2ε = -4 = 0. So Z/4Z is a subring.
So the ring A is either a finite Z/4Z-algebra or a finite F_2[x]/(x^2) algebra. I haven't thought more, but probably you can make more deductions from these. But anyway this already gives you lots of examples, for any pair n, m of naturals the ring
A = (Z/4Z)^n x (F_2[x]/(x^2))^m is an example of such a ring. It has 2^(m+n) units, nilpotents, idempotents.

[rocky cloak]

Pushing this a little further, since the ring is finite it's artinian, and any connected commutative artinian ring is local, with nilpotent radical. So every element is either invertible or nilpotent. Since we have two units, there must be exactly 4 elements.

So Z/4 and F2[x]/x^2 are the only examples.

[rustic crown]

[rocky cloak]

I wonder what happens if you drop the commutativity though

Hmmm, so consider the ring of upper triangular 3x3 matrices with coefficients in F2 for which the diagonal is constant.

This ring has 16 elements, 8 units, 8 nilpotents and 2 idempotents.

So taking the product of this ring with (Z/2)^2 shouldn't add any units or nilpotents, but bring us to 8 idempotents.

Wait, this also works in the commutative setting. F2[x]/x^3 has 4 units, and 4 nilpotents, but 2 idempotents.

So F2 x F2[x]/x^3 works

[rocky cloak]

So each factor satisfying this is sufficient, but not necessary

[rustic crown]

ah oops

but each factor should have same number of nilpotents and units.

[rocky cloak]

Yeah, cause you always have more units (or equal)

[rustic crown]

so each factor is a finite local ring with residue field F_2. can we say more?

(and that total units = 2^number of factors)

[rocky cloak]

So you have a local algebra with residue field F2 and 2^k elements. Then it will have 2^k-1 units and the same number of nilpotents.

After that you can just go haywire and pad the product with enough copies of F2 to get enough idempotents

[stark helm]

I know that, and it can also be verified by a-b in subrings when a and b are both in subrings

[cloud walrus]

HausdorffT1

[south patrol]

Well

You can sneakily get around it by showing if you're not in V(I) u V(J) then you're not in V(IJ)

But I don't know how direct you want lol

[slim kayak]

How does it follow from your proof by contradiction that P contains all of I or J?

[long obsidian]

I'm assuming it's not true and hence such a and b exist. But then ab in IJ but by assumption P contains IJ as a subset and so it should contain ab. But then this contradicts P being a prime. Maybe I didn't explain this enough not sure.

[long obsidian]

Actually I like the contrapositive more than a contradiction. Thank you this works

[slim kayak]

Sure, but the way you phrased it seems like you could have half of A sit in P and a complementary of B sit in P

the hypothesis you would is assuming there exists an a in A (wlog) which isnt in P and go from there

[spark veldt]

Anyone who's read Artin's Algebra:

Can you read Rings Ch.7 after reading Ch. 2 (Groups) but skipping 3-6 entirely?

[slim kayak]

idk, tell us what said chapters are

[rotund aurora]

Yes

[spark veldt]

thanks

[stuck fiber]

I'd read 6 since the symmetry stuff is pretty cool imo

[spark veldt]

Yeah I'm actually planning on reading all it's just I have a time deadline that requires me to cover rings asap if possible xD

[stuck fiber]

So I have a question, in Hungerford when they describe a category they talk about as a class of objects with a class of disjoint sets (hom(A,B)) whose elements are the morphisms between a pair of objects in the category so are these morphisms considered to be part of the category?

[slim kayak]

Wdym?

People would say that the stuff in Hom(A,B) are morphism of the category, sure

[crystal turtle]

Yes

The data of a category is a class of objects Obj and the sets (or classes, if we allow such categories) Hom(A,B) for A,B in Obj satisfying blah blah whatever laws. You can also define it as a class of objects Obj and a class of morphisms Mor, where each f in Mor has a specified domain A and codomain B (written f: A --> B) satisfying blah blah. Both are equivalent (altho there are a multitude of reasons to prefer the first one). In either case, the morphisms are essential parts of the data of a category.

[stuck fiber]

The name is very appropriate lol. What do you mean by data exactly, also is there a specific definition for class? I've been just treating it as a set of things with a shared property but wasn't sure

[slim kayak]

Its just a placeholder word more or less

[crystal turtle]

yeah both are meant fairly informally here

[slim kayak]

They are usually sets with extra structure, or categories. Sometimes data structures in CS

really whatever makes sense with the axioms

[crystal turtle]

Data = what kind of information it holds. e.g. the "data" of a group is a set + a binary operation on that set which is unital, associative, and has inverses

Class = collection of things with a shared property. What is a collection vs a set? Well idk enough foundations to give a full answer in ZFC (usual foundations in math), but think of the distinction between a collection or class of all sets {x | x is a set}. This cannot be a set by Russell's paradox, so it would be incorrect to call it a set.

[stuck fiber]

I see that's a good motivatoin behind class

[crystal turtle]

Class also includes sets though. Using this kinda informal notion, given a set S consider the class {x | x in S} = S

[slim kayak]

Sets are usually distinguished by them having the set membership relation "in"

[crystal turtle]

This is ultimately the answer and the secret is that 95+% of the time it doesn't really matter what precisely is meant by "class"

[mighty kiln]

I thought class be NBG

[slim kayak]

Classes are just something you want to put next to your quantifier and to sound serious while doing so

[crystal turtle]

That's one approach to formalize what a class is

[formal ermine]

a category is a tuple (Obj, { Hom(A, B) | A, B in Obj)

if u dont wanna use symbols u say "the data is ..."

[crystal turtle]

Calling it a tuple is bad

[formal ermine]

or whatever

[slim kayak]

Yeah, thats a pretty funny way to describe it

[crystal turtle]

Because that would in a sense force Obj to be a set

[formal ermine]

not rly

but im not advocating for that definition anyway

[slim kayak]

Ultimately it doesnt really matter what the definition is, after a little bit of toying around in 1-2 examples you ought to know what a category should be

[crystal turtle]

Not sure how common NBG actually is for working with categories (I've never seen it explicitly used). More common nowadays is the use of Grothendieck Universes when necessary. Small (/set) = a set in the universe, large(/class) = a set not in the universe

[crystal turtle]

If you want to define what a tuple is set-theoretically, I think so.

[prisma ibex]

yeah using just NBG with sets/classes causes issues in general

you want universes/large cardinals and then it's fine

[crystal turtle]

(which, again, in practice doesn't matter because you can throw the assumption of a universe at it)

[stuck fiber]

I was just trying to make sure I understand for when he introduces equivalence because says in a category C a morphism f:A->B is called an equivalence if there is in C a morphism g:B->A such that g(f)=1 so I was just trying to understand if this was abuse of notation or if we actually consider these morphism to be a part of the category

[crystal turtle]

(and fg = 1)

[slim kayak]

I mean, you would end up making a set which either all the things your mentioned or sets that contain them.

[stuck fiber]

yes

[formal ermine]

its a class

[slim kayak]

And within ZFC I dont think you can have a proper class be on the left of an in symbol

[prisma ibex]

you are considering the morphisms as part of the category yes

a category has objects and morphisms and a composition operation between morphisms satisfying certain properties

[slim kayak]

No this is allowed

if it wasnt then category seem pretty useless

[prisma ibex]

probably you want to write something different than g(f) if you mean composition

g\circ f is good

[slim kayak]

Actually, your definition of a category should talk about composition of morphisms

[stuck fiber]

he uses circ but I couldn't remember it

[prisma ibex]

okay then that's great

[stuck fiber]

It does yes

[slim kayak]

Yeah so you can freely talk about elements of the hom-sets and their compositions

[stuck fiber]

This is a very rough drawing of how I was thinking about a category

[prisma ibex]

yeah this is a silly way to draw it I think

I mean you should be thinking of morphisms as actual arrows between the objects

[slim kayak]

engage your supressed mind map creation skills from HS

[stuck fiber]

What's a mind map

[stuck fiber]

How do you think of Hom then?

[prisma ibex]

it's the set of all morphisms from one object to another

[slim kayak]

or concept map

[stuck fiber]

Right that's why I had the morphisms in the Hom circle

[prisma ibex]

I mean sure you just want to picture them as actually sitting between objects, not just some diembodied thing on the side

[stuck fiber]

Oh I see yeah

[stuck fiber]

Why did you learn this in highschool?

[slim kayak]

so that we could proof that adjoints functors preserve (co)limits and other trivial but useful results I am talking about when you were asked to make a diagram of concepts and how they interact with other things

[stuck fiber]

Right I never did that or I skipped those days LOL

[prisma ibex]

regarding size issues, the notion of "class" is kind of a silly one and you run into technical issues in general trying to think about things like this

you can really think of everything as a set here if you add a bunch of large cardinals on top of ZFC

[slim kayak]

So something like Set becomes the category of sets not on the naughty list (smaller than some large cardinal)?

[prisma ibex]

yeah exactly

[slim kayak]

fair enough

[prisma ibex]

you would have some sequence of large cardinals \kappa_1<\kappa_2<... and then you can talk about "\kappa-small" sets for \kappa one of these large cardinals

\kappa_1-small would just mean set in the usual ZFC sense

[slim kayak]

makes sense, you want your category to be somewhat nice after all

[prisma ibex]

maybe like \kappa_2-small would mean class in the usual NBG sense if you're doing things in a smart way that avoids issues

[stuck fiber]

So what is kind of the motivation behind this he gives a few examples but they don't really motivate the subject very well?

[prisma ibex]

behind categories?

most objects in mathematics can be organized into categories in this sense

and then there are general constructions one can do within categories, or between categories, that reproduce lots of more concrete constructions for particular examples of categories

[stuck fiber]

Yeah he shows like you put all groups in a category, all sets, etc but he doesn't really show how or why you'd want this I feel

[slim kayak]

The real answer is functors and universal properties, you just gotta go through the groundworks for a bit until you can get to the interesting bits

[prisma ibex]

like products of groups, products of rings, products of vector spaces, products of sets, these are all the same universal construction in different categories

[stuck fiber]

Ah yeah he does that in the category chapter I believe (replying to Kerr)

Yeah the category whose objects were all morphisms from another category was pretty neat

[prisma ibex]

there are certain useful constructions in mathematics that are kind of impossible to think about without using a bit of category theory

[slim kayak]

I mean, if you I just told you for the first time what sets were you probably wouldnt think much of them. Although they are conceptually simpler at least

But you still have to learn what they are to do interesting math, unless you are part of the number theory subreddit

[crystal turtle]

functors, natural transformations, (co)limits, and adjunctions are the kinda main objects of basic category theory and as you learn more of them the motivation for categories will become clearer and clearer

[slim kayak]

Whats are non-basics cats?

[crystal turtle]

And imo a good way to view basic category theory is that it's a way of talking about universal properties as Kerr mentioned

[slim kayak]

Like, non immediately useful to most mathematicians?

[crystal turtle]

Like something you wouldn't see in a "first course" on category theory

Enriched categories, model/infinity categories, maybe monoidal cats fall somewhere between the two, ....

[prisma ibex]

idk something like the effective topos is useful for computability theory but is also pretty obscure if you're not studying that sort of thing

[crystal turtle]

Presentable/accessible cats, various types of topoi, ...

[prisma ibex]

there are all sorts of pretty obscure structures/properties for categories in the same way that there are all sorts of obscure algebraic structures you can study

[slim kayak]

My cat doesnt like visitors much, not very accessible either most of the time tbh

[crystal turtle]

sotrue

[prisma ibex]

for the most part these are all invented for some reason or another

like why on earth would anyone invent this? there are good reasons for it and they come up naturally in certain places of mathematics

[slim kayak]

this looks like physicist stuff

[prisma ibex]

indeed it is

all sorts of bizarre algebraic/categorical structures get invented to meet the demands of mathematical physics for instance

[slim kayak]

incoherent rambling-algebra is useful to study some obscure scenario, probably

[prisma ibex]

a lot of times without the proper context it's like "why on earth would anyone study this" but sometimes these things just naturally come up and it's like "oh well this is exactly the algebraic gadget I need to organize the thing I'm studying"

[stuck fiber]

I guess I'll finish up this section do some exercises and split some of my algebra time between the category chapter and the module stuff

[formal ermine]

they literally make us do mind maps ALMOST EVERY SINGLE LESSON

ITS SO ANNOYING

[slim kayak]

subconscious category theory indoctrination programme

[slim kayak]

My guess is "I expect my particle to operate in this quirky way. Let me write down the rules they seem to follow and lemme pretend this is still an algebra"

[prisma ibex]

it's literally "what structure do I need for the quantum master equation to make sense structurally"

[slim kayak]

A bit ballsy how often it seems to work out

[mighty kiln]

How does that work

Is it useful for like enabling axiom of choice or sth

[slim kayak]

Universes and large cardinals?

[prisma ibex]

no it has nothing to do with choice

[mighty kiln]

But what are the advantages as opposed to doing NBG

[prisma ibex]

I mean NBG doesn't have enough control over size

[slim kayak]

are you a secret NBG salesman 😭

[prisma ibex]

it just distinguishes between "small" things (sets) and "big" things (classes)

but for certain things to make sense you really need like, an entire hierarchy of size like this

like you can start with a "small" category and do something like taking the category of presheaves on it and end up with a "big" category

[crystal turtle]

Classes can't live in classes.

[prisma ibex]

okay well what if you want to do that again?

[crystal turtle]

That's ultimately where the problem is I think

[prisma ibex]

yeah

[slim kayak]

Idk about you, but a strongly inaccessible cardinal seems a lot easier to just accept

[prisma ibex]

yeah exactly

much less baggage than NBG

[mighty kiln]

[slim kayak]

You havent cared about anything of that size before and now you can do all the categorical constructions you desire

[prisma ibex]

a lot of these large cardinal axioms are pretty mild as far as adding axioms to ZFC goes

[crystal turtle]

"pretty milk"

[prisma ibex]

lmfao

typo

[mighty kiln]

Are most of them like provably independent

[slim kayak]

I thought that was some obscure saying

[prisma ibex]

literally any large cardinal axiom is independent of ZFC

[mighty kiln]

Oh hmm

[prisma ibex]

that's what inaccessible means, their existence is independent of ZFC, you can't construct them using the usual constructions within ZFC, they are "too large" for this to be possible

[slim kayak]

Oh, large cardinals work in consistency in a pretty funny way

[prisma ibex]

if you add a large cardinal \kappa to your model of ZFC "by hand" then you can just talk about sets of size <\kappa, that will be a model of ZFC, and then anything you do within ZFC will never leave that universe

[crystal turtle]

Yeah strongly inaccessible cardinals are like at the bottom in terms of large cardinal axiom strength (I have no clue how correct this is/how to formalize it but I think there's an ounce of truth here)

[mighty kiln]

Is ZFC be inductive

[slim kayak]

they should work out more then

[crystal turtle]

Compared to stuff like Vopenka's principle and whatnot, which are also used occasionally, this is not much more than a technicality.

[slim kayak]

Is there some notion in that the large cardinal axioms for stuff like groethendieck universe is just to make the formalism simpler? Like I could build an "open" universe up to some accessible cardinal and then just choose the cardinal appropiately large in order to perform whatever construction I wanted to do?

[stuck fiber]

I think I'm seeing the power of diagrams now

[slim kayak]

Yeah? What was it?

[stuck fiber]

If $(P,{\pi_{i}}) and (Q,{\psi_{i}}$ are both products of the family ${A_{i}:i\in I}$ of objects of a category C, then $P$ and $Q$ are equivalent

[cloud walrus]

Kenshin

[slim kayak]

your $ signs are a bit lonely

[stuck fiber]

Yeah I noticed lol should be all fixed now

[slim kayak]

oh, the classic unique to up to unique isomorphism

[crystal turtle]

I don't know what you mean.

[crystal turtle]

(Eventually you may see that this is an example of functors preserving isomorphisms)

[stuck fiber]

I think Hungerford may be my new favorite Algebra book

[prisma ibex]

assuming the existence of infinitely many inaccessibles is equivalent to assuming the existence of infinitely many Grothendieck universes. Both are needed if you want category theory to work without running into size issues

in practice you usually don't need more than just like, a few steps of this process

[crystal turtle]

i vaguely remember hearing that Hungerford is clunky (with cats) at some point in this server but i don't remember exactly what

[prisma ibex]

it's funny to me that Vopenka's principle was originally conceived as a "joke" large cardinal axiom with the intend of showing it was inconsistent with ZFC, but then the proof fell apart, and it turns out to be an important large cardinal axiom for certain parts of category theory

lmao

[crystal turtle]

lmao

[slim kayak]

I mean if there are essential uses of groethendieck universes, like, you could, if you are only working with sets with cardinality at most x many applications of power set, do all the necessary constructions reformulate the theorems to hold up to objects involving that cardinality

Like, in principle

[prisma ibex]

no, you can't really avoid this for certain constructions

[slim kayak]

Huh

Are said construction like in their own subfield or could it turn out that stuff like the weil conjectures are false in ZFC?

[prisma ibex]

take a small category C and look at the category PSh(C) of functors C->Set

In order for Set to be a nice category you can't really truncate the cardinalities you're allowed to look at, or limit the number of times you're allowed to take power sets, etc

PSh(C) is the "next size up" compared to C

constructions like this bump up side, in practice you don't ever do too many size increasing steps like this, but you certainly want to be able to do this more than once for example

[slim kayak]

Right, but that's sorta what I meant

[prisma ibex]

then okay great you have to assume the existence of like, idk 2 or 3 universes for a lot of math to work

[crystal turtle]

Well I mean ultimately you want "Set", whatever category of sets we mean here, to be bicomplete

[prisma ibex]

but like if you want category theory to work well abstractly it's easiest to just assume infinitely many universes and that's that

[slim kayak]

Like, idk what niceness criterion you are looking but something like completeness, but in practice you could just guarantee that up to a certain cardinality limits do exist

[prisma ibex]

that makes things shittier, not nicer lmao

idk what benefit you would get from doing this

[slim kayak]

The universes then just play a role in making the statement convenient

[crystal turtle]

universes are the way to get around that in a nice categorical way (along with whatever other properties you want)

[prisma ibex]

I mean ultimately all you're doing is like, shifting the issue onto keeping track of size issues within the category, rather than keeping track of size issues outside the category

which seems like a worse situation to be in

this is like trying to study groups where you're not allowed to apply the operation more than 1 million times

[crystal turtle]

I don't really think there's a reason to try to avoid universes tbh

[slim kayak]

I am just trying to understand how much it depends on said universes being true or not

They are certainly sweep all headaches under the rug, just wondering how "essential" they ultimately are

[prisma ibex]

they are essential unless you are okay with having to horribly fuck up categories/algebraic structures/whatever

can you still run a lot of proofs in this more fucked up setting? yeah probably

certain inductive arguments might break

there is almost certainly some part of "standard" mathematics that would break if you had to make limitations like this

anyways universes are such a minor thing to assume it's like, not really an issue worth worrying about

[elder wave]

i just hate the word universe for this

[prisma ibex]

yeah this is why I prefer just saying large cardinals

[elder wave]

it feels so stupid to say

[slim kayak]

I am not arguing for the fucked up setting, just interested in seeing how much you can stay out of "this result is true if you assume this large cardinal axiom" range when working with those

[crystal turtle]

MO/MSE might have some stuff on this if you go looking.

[prisma ibex]

why would you want to stay out of this though

like I just don't see any argument for avoiding this

[crystal turtle]

"I'm not arguing..."
"I just don't see any argument"
Yeah you're right you don't see any argument

[prisma ibex]

that explains the problem

[slim kayak]

If your goal is studying these things in themselves, sure. But if they get used to solve some conjecture/open problem I wouldnt expect them saying somewhere they are basing this off large cardinals

Like they implicitely used it but its whatever, just for convenience instead of having assumed an additional axioms to make things work out

[prisma ibex]

I mean this isn't really an issue though

[crystal turtle]

(e.g. this one https://math.stackexchange.com/questions/375085/on-the-large-cardinals-foundations-of-categories?rq=1)

[slim kayak]

Probably. You saw something related?

[prisma ibex]

nobody really mentions this because pretty much everyone has agreed that this is a completely benign thing to add to ZFC

[slim kayak]

great reaction speed wow

[crystal turtle]

lmao

[crystal turtle]

I mean there are a decent amount of skeptics

I think minimally many in relevant fields to where universes are used, but they're still there

[prisma ibex]

okay and there are a handful of ultrafinitists who are also just like, wrong about lots of things

[crystal turtle]

Fair.

[prisma ibex]

imo not accepting universes basically makes you an ultrafinitist of a certain kind

[slim kayak]

uh

did you laugh at your own joke 😭

[prisma ibex]

yes

[slim kayak]

self-love, slay

[crystal turtle]

Also noted in a few places (i think in here too?) that it's kinda a meta-theorem that in any specific application of a universe, you probably don't need the universe

E.g. mentioned there about FLT: It shouldn't be necessary to use universes, but it's so technically convenient that we might as well.

[slim kayak]

I am happy with this then

[crystal turtle]

That's how I view Vopenka's principle (I have never used Vopenka's principle explicitly, fair warning). It's a technical tool which gives some nice properties in some categorical stuff (namely presentability results; here bounded = has a small dense subcategory)

[prisma ibex]

lmao I just opened up the latest Analytic Stacks lecture from Scholze and he is defining analytic stacks and keeps having to mutter "up to set theoretic issues..."

[slim kayak]

man, not even unique set theoretic issues

things are not looking well

[crystal turtle]

Like, in any actual application you should expect that a category satisfying (ii) or (iii) is locally presentable (if it's not, then that's also meaningful since you'd have a proof that VP is inconsistent with ZFC)

[prisma ibex]

Vopenka is equivalent to "Every subfunctor of an accessible functor is accessible"

which is like, reasonable enough I suppose

it's also a strong enough large cardinal axiom that people start to get uhhh

a little weirded out by just assuming that

[crystal turtle]

And you should expect there to be a proof in ZFC of this fact. Using VP is just a shorthand for that

[crystal turtle]

Yeah so take it with a grain of salt because it is definitely uhhh

[slim kayak]

this is too complicated

[prisma ibex]

Vopenka: obviously true for category theoretic reasons
anything stronger than Vopenka: probably straight up inconsistent with ZFC

[slim kayak]

I just assume 0=1 and ball it from now

[crystal turtle]

Oh yes

Obviously Ord cannot be fully embedded into Graph

[prisma ibex]

I mean

yeah?

I sure fucking hope not!

[south patrol]

Hm

[crystal turtle]

The fact that this is equivalent to the rest of those statements is wacky as fuck

[south patrol]

Is this a funny way to argue for universes: like

perhaps we should be maximalist with the zoo of sets lol

like be constrained only by consistency

:chad:

I'm kidding cause you could go in vrious differnet directions with that

[prisma ibex]

yeah exactly

[south patrol]

Tbf large cardinal axioms sometimes scare me

[slim kayak]

Someone start defining ordinals as certain graphs please

[south patrol]

But it seems reasonable lol

[prisma ibex]

the category of large cardinal axioms cannot be fully embedded into the category of ordinals

Hypervopenka

[slim kayak]

at that strength it might be inconsistent with itself

[south patrol]

but then it implies its own consistency, woop!

[slim kayak]

my bad you are right

but it can also prove gödel wrong. I think we owe hilbert that

[prisma ibex]

here is a helpful diagram to guide your thinking

[crystal turtle]

where is this from

[prisma ibex]

me

[sly crescent]

Vsauce

[south patrol]

Hm

[slim kayak]

lemme guess, the "huge" variety was made by a group of drunk mathematicians playing dare

[formal ermine]

veritasium

[south patrol]

I find it potentially misleading when like

wikipedia pages say "if the [...] holds" for axioms

[formal ermine]

what is Woodin

[south patrol]

like it is written as if we don't know whether it holds, when it is basically a choice

[south patrol]

don't say it

[formal ermine]

hi timo

[elder wave]

german hours again

[slim kayak]

I mean, if you think that assuming the axioms of ZFC isnt any different from assuming any other independent axioms, sure

[prisma ibex]

In the mathematical area of braid theory, the Dehornoy order is a left-invariant total order on the braid group, found by Patrick Dehornoy. Dehornoy's original discovery of the order on the braid group used huge cardinals, but there are now several more elementary constructions of it

[south patrol]

these are non-ZFC axioms

[prisma ibex]

boy I sure hope so!

[slim kayak]

Yes

[south patrol]

i mean like they are a choice right

rather than some mysterious thing

[slim kayak]

The choice is in assuming them but its kinda like faith, you may also not. No one knows

[south patrol]

Wdym

[prisma ibex]

as soon as you accept \aleph_0 you're in "magical thinking land" and anything else goes tbh

[slim kayak]

:chad :

I am gonna start megaultrafinitism

We are only allowed to count to 10

[prisma ibex]

axiom of infinity is the real large cardinal axiom

[crystal turtle]

There isn't a largest number. That's because there are no numbers!

[slim kayak]

Axiom: There are no axioms

[crystal turtle]

Large cardinals just seem wacky to us because out puny little brains cannot comprehend the eldrich horrors of anything infinite.

[slim kayak]

broskie is scared of the natural numbers smh

[ivory trail]

zeilberger would actually write this

[prisma ibex]

large cardinal axioms: I can accept the existence of infinities the mind cannot possibly comprehend
continuum hypothesis: dude I don't fucking know

[slim kayak]

I'd think most people would have some commitment that ZF is true for the most part (maybe reformulate 1-2 axioms), so in that sense you have a baseline under which independent axioms are a little bit mysterious and become increasingly mysterious if you climb up the large cardinal ladder ng posted

Wait, do large cardinals necessarily imply CH?

[prisma ibex]

no

[crystal turtle]

I don't give a shit if ZF is consistent. If it is, I get to prove stuff. If it isn't I can prove anything.

[slim kayak]

thats kinda funny

[prisma ibex]

CH and other stuff are like, completely orthogonal to this

[crystal turtle]

Some assumptions might but yeah they're very different flavors of assumptions

[ivory trail]

simply take the biggest large cardinal axiom

[prisma ibex]

it would be nice if some "reasonable" large cardinal axiom implied CH or not CH but things are not so simple

[crystal turtle]

"|R| is small" and "there are things that are huge" are pretty different, so it's no shock.

[slim kayak]

Just wanted to make sure

Knowing the existence of some cardinal that is completely unreachable by anything below and has wide consequences while meanwhile going "idk" if you ask if there is something between aleph 0 and 2^aleph 0 is a bit funny

[crystal turtle]

Does the existence of 100 imply the existence of 1/2?

[slim kayak]

Idk where you are getting at

i find it kinda funny
Oh yeah? Here is a counterexample

[crystal turtle]

idk what i'm getting at either

I've run out of funny juice for today (i never had any)

[slim kayak]

like, it sounds a little bit like the discussy HSers talking about doing the homology exercises and then complaining how hard their history essay is

[south patrol]

Real though

[crystal turtle]

I don't blame them for that

[slim kayak]

Yeah I am glad i never have to touch my HS ever again

[crystal turtle]

Out of homology and writing an essay, there's one I'd clearly rather do.

[slim kayak]

what about writing an essay about homology

[south patrol]

Computing the homology of history

[slim kayak]

Some argue that history repeats itself i.e. is cyclic. In this essay we are trying to calculate homology of history by considering the boundaries of.....

[inner acorn]

Playing with braid diagrams when writing elements of Sn in terms of standard transpositions (i i+1) in a minimal way.
I think I've hit a eurika moment and understanding, however xD I don't have any texts/references/notes - curious if someone knew of a resource (even if it's a shitty Youtube video) that just runs through the idea so I can sanity check

[wispy light]

Not sure if I should ask here or in #algebraic-geometry. Suppose you have a homogeneous polynomial f(x1,...,xn) of degree d over the complex numbers. Does there exist a linear transformation isomorphism T : C^n -> C^n such that f \circ T has the form x1^d + g(x2,...,xn) ?

[coarse stag]

Can someone point me to where I can find a proof of this

[prisma ibex]

this is literally just a case by case proof

the case of finite simple groups of Lie type follows from a result of Steinberg

but for the sporadics you literally just check by hand that all of these are 2-generated

idk that there is a satisfying proof of this lol

[ivory trail]

do you want T to be invertible and some condition on g?

[sly crescent]

If there is one, it is not yet known

[wispy light]

Ah, yes, of course T should be invertible. My bad. The only condition on g is that x1 doesn't appear in it and it's homogeneous on the remaining variables x2,...,xn.

[ivory trail]

i think you can map x_i to x_i + x_1

[wispy light]

Or x_i + w_i x_1 for some weight w_i, right?

[fast junco]

Afternoon folks! Can't wait to be studying this field

[dull marsh]

Which field

[wise hawk]

quick question from a noob in her first week of this course: if I'm given a set S and an operation, and I'm asked to determine whether some operation f is a binary operation on S, is it sufficient to check whether the operation maps any given element from S×S to S? My current understanding is that if I can show that there exists an element of S×S that maps to something outside of S under f, then it's not a binary operation

[dull marsh]

You are correct

[wise hawk]

thank you!

[fast junco]

Graph theory

Graphs and groups are similar enough

[celest furnace]

[fast junco]

Problem

[celest furnace]

Groups are quite far from Graphs

Ig until you talk about sylow subgroups of the symmetric group

But you have to go that far to find one

[static glen]

i dont have any intuition why rings are called rings and ideals are called ideals but for orbits there is a nice intuition

[mighty kiln]

The orbit of a planet around a star is more or less the positions than it can take

The orbit of an element is the set of other elements it can get sent to under the action

[static glen]

Something like this could be the situation

[mighty kiln]

You can imagine the orbit of a planet as the orbit under the action of (real numbers, +) as time

[rocky cloak]

Ring is an old word for gathering or organization, which you can see in words like 'crime ring' or 'spy ring'.

So a 'number ring' is some numbers organized together.

As for ideals: in so-called rings of integers of number fields (for example Z[sqrt(-5)]) unique prime factorization can fail.

However, ideals factor uniquely into prime ideals. Kummer noticed this an introduced the word 'ideal number'. Because they behave the way numbers ideally should behave. Dedekind then defined ideals based on Kummers idea of ideal numbers.

Theory of Algebraic Integers (1996, p. 5), "Dedekind presumably chose the name 'module' because a module M is something for which 'congruence modulo M' is meaningful."

https://mathshistory.st-andrews.ac.uk/Miller/mathword/m/

Haven't heard of this one before, but makes sense

[mortal folio]

hi i'm sure a bit low level for this discussion but revising my group theory stuff from uni and i'm not sure what the notation circled means, the lecture notes assume you already know it lol

[dull marsh]

n | m reads "n divides m"

[mortal folio]

alright ty

[brittle ether]

hey i have a quick question, that I think is right, but I just want to sanity check it here since I haven't rigorously studied this subject

so suppose i have a group, drawn out like a multiplication table

like this:

does swapping rows and columns give you all possible isomorphic tables?

[dull marsh]

Yeah

[brittle ether]

thanks!

[south patrol]

Lol

[brittle ether]

https://media.discordapp.net/attachments/330160780488212480/1195455612121788426/Screenshot_20240112_145217_Chrome.jpg?ex=65b40dcc&is=65a198cc&hm=1484a0d7dd11f5af8ba1c886f19cd8e3b569fabfe5019592b855b0bb64b57eaf&=&format=webp&width=1003&height=590 full meme if you want it

[south patrol]

Nice

Where's this from lol

[brittle ether]

saw it on reddit

[south patrol]

Nice

I thought you just had an interesting teacher/prof for a second

[brittle ether]

i screencapped it to avoid spoiling when i posted it elsewhere

oh well

[dull marsh]

The third?

Wait no

[brittle ether]

I suppose I kind of am a teacher, I was thinking the exact same thing

that this is a beautiful way to rephrase and teach group theory to young kids

[dull marsh]

So 1 and 3 are crewmates

[brittle ether]

||incorrect||

[dull marsh]

||1?||

[brittle ether]

is crewmate or imposter?

[dull marsh]

||sus||

[brittle ether]

||c o r r e ct||

[rocky cloak]

I always knew there was something suspicious about the Klein 4 group

[slim kayak]

When the isomorphism class is sus

[stark helm]

Can someone tells me why <x,2> is not PID in Z[x]?

[rocky cloak]

You mean why the ideal is not principal? Because it's not generated by a single element.

Just think about what kind of polynomial could be such that both x and 2 is a multiple of it

[stark helm]

2x?

so since <x,2> can not be expressed as <a> for some a in Z[X], SO <X,2> can not be principal?

[rocky cloak]

So neither 2 nor x is a multiple of 2x, you may be mixing it with 2x being a multiple of 2

[rocky cloak]

Yeah, that's the definition

[stark helm]

You mean we can not figure out a common factor between 2 and x in Z[x] right?

[rocky cloak]

Exactly

Or rather, 1 is a common factor, but <1> doesn't equal <x, 2>

[stark helm]

so, if we have a common factor a such that <x,2> is equivalent to <a>, then <x,2> is principal?

[rocky cloak]

An ideal is called principal if it is on the form <a> for some element a.

So if <x, 2> = <a>, then it's principal. Otherwise it is not principal.

[stark helm]

oK, then in R[x], <x,2>=<1/2>, so it is principal in R[x]

[rocky cloak]

Yeah, in R[x] it's just equal to everything, i.e. <1> or <1/2> if you prefer

[south patrol]

||this < > notation hurts my eyes||

[crystal turtle]

langle rangle

[stark helm]

I found that some course use <> and others use ()

[south patrol]

Yeah

I think every mathematician I've ever met uses ()

But I've had <> in courses

[barren sierra]

I've seen a healthy mix though nowadays more ()

But usually I see ⟨ ⟩ not < >

[stark helm]

it happens in the textbook

[coral spindle]

Angled brackets is more consistent with generator notation for every other module.

[crystal turtle]

i like ⟨ ⟩, helps distinguish things from other parantheses

[barren sierra]

Same

[crystal turtle]

Same thing for inner products, using (-,-) inevitably confused me with iterated inner prods and normal parentheses mixed

[formal mulch]

any hints would be appreciated

[slim kayak]

Which part is unclear/needs hints?

If you think of A as just a finite set {1,...,n} its probably clear that such functions act as a basis, since you can identify them with {0,..,1,...,0}

[formal mulch]

i mean any on the way to solution

[slim kayak]

Pick any function from A to Z in F(A), it has only finitely x's for which it doesnt spit out 0, aye?

[formal mulch]

yes

[slim kayak]

Can you then use a Z-linear combination of f_a to create a function that spits out the same values for every x?

[formal ermine]

this sounds literally 1:1 like what ive read today

[formal mulch]

can you please share source? 🥺

[formal mulch]

i don't know, but i thought it can be somehow related to interpolation polynomials, isn't it? 😢

[formal ermine]

singular homology theory by jw vick

on one of the first 4 pages

[slim kayak]

Kind of, but I think you are overcomplicating it

Take some f in F(A). Then f is 0 outside of M. Since M is finite call it {x_1, ..., x_n}. You then start by taking f_(x_1) which assigns 1 to x_1 and multiplying it by the value of f(x_1)

actually yeah, polynomial interpolation with the lagrange basis. Same shtick you are right

[south patrol]

lol i have been BTFO'd here

[sly kernel]

Hi everyone , I have questions about direct product presentations (Free Group)
G = H x K
where H has presentation H = <X | R> and K has presentation K = <Y | R2>
I can't proof that G = <X,Y: xy = yx (where x from X and y from Y) R1, R2>
can you help me?
Thank you!

[slim kayak]

Try to define a map into the direct product and then try to see if adding the yx=xy to R1 and R2 generates the kernel from <X, Y> to G?

[slim kayak]

If you want a stronger hints sequentially, if you haven't done much with words yet and want a way how to approach it:
||Clearly anything in the relations is in the kernel by checking the relations themselves||
||You can write any word w as w_1 .... w_n where wlog w_1 only has letters in X, w_2 only in Y||
And for completeness sake...
||Take any word in the kernel, in the image the the w_i commute and after sorting they correspond to (1,1) = (w_1 w_3 ... , w_2 w_4 ...)=(u,v). So uv lies in <R1, R2>, by applying the relations yx=xy you can turn uv into w. So w lies in <R1, R2, {yx=xy}>||

[cloud solar]

If A is a ring and x^4=1 for every x in A* show A is isomorphic with Z/2Z or Z/3Z or Z/5Z

Oh the polynomial x^4-1

Im so dumb

[hollow mica]

I’m trying to show that if x is an element in a group with finite order r, then the order of x^s is lcm(r, s)/s.
Clearly (x^s)^(lcm(r, s)/s) = 1, but I must also show for any 0 < k < lcm(r, s)/s that (x^s)^k = x^(sk) is not equal to 1

Any hints

[tough raven]

What about an arbitrary product of such rings?

[cloud solar]

What about it?

Hmm I have a problem that says to find n>=2 s.t. U(Z/nZ) has all elements of order 2

Clearlry phi(n)=2^k

But what now?

[tough raven]

It should also have that property, but would not be one of Z/2Z, Z/3Z or Z/5Z.

[tough raven]

||By CRT, reduce the problem to n a prime power.||

[cloud solar]

n=p1^k1...pl^kl so phi(n)=phi(p1^k1)...phi(pl^kl). if p_i is 2 then phi(pi^ki)=2^(k_i-1). If p_i odd then k=1 and p-1 must pe a power of 2. So n is just a power of 2 multiplied by some fermat prime numbers

[hidden kite]

If anyone is free and willing, could you help me to check this short proof and give me feedback?

[obtuse bear]

for the "conversely" part, you've shown that products of an even number of disjoint 2-cycles are in <S>, but not every element of A_n is such a product, e.g. (12345) in A5 is not

[hidden kite]

Ahh right!

[hidden kite]

This should fix it right?
[If the 2-cycles are identical, then (ab)(ab)=(abc)^0.]

[cloud solar]

A ring. How can I show the number of indempotent elements is even if the set of indempotent elements is finite

[rose prism]

given a nonzero idempotent a, try constructing another

nonzero and non-1, i should say

[obtuse bear]

the idea seems promising, but it's not true that (ab)(ad) = (acb)(acd), the left hand side is (adb), and the right hand side is (ab)(cd)

[cloud solar]

1-x

[tough raven]

All that's left is to show that this always gives you a different idempotent, so that you can pair up idempotents using this map.

[hidden kite]

Ahh nuu, I made a silly mistake in the final line of my computation. But it should be correct that
(ab)(ad)=(ab)(ac)(ac)(ad)=(acd)(adc) right?
Though I somehow missed the most obvious thing you mentioned haha, that (ab)(ad)=(adb)!

Btw other than that, do you have anymore feedback you would like to share?

[chilly ocean]

What does the complement mean

I think X-V(f)
Is it right

[crystal vale]

But in question is not mention that R has unity

[rose prism]

is the statement true if R doesnt have unity

[mighty kiln]

Yes

[crystal vale]

Then how you define x-1

[rose prism]

did you reply to the wrong person

i didnt define anything

[crystal vale]

Non-1

[rose prism]

it is a hyphen

not a minus sign

[crystal vale]

Oh

Then how we construct another one

[rose prism]

idk

the statement in the problem is false if there is no unity

so it doesnt matter

[crystal vale]

Okay

[rose prism]

consider 2Z

[crystal vale]

0 is idempotent? Like only non-zero ?

[rose prism]

yes, in 2Z, 0 is the only idempotent

[crystal vale]

Okay

[rose prism]

so there is an odd number of them

[crystal vale]

Yes

[rose prism]

in the unital case, 0 and 1 are idempotent, and i was suggesting to prove that there is an even number of other idempotents

[crystal vale]

Oh

[obtuse bear]

(ab)(ad)=(ab)(ac)(ac)(ad) is correct, but (acd)(adc) is just the identity as (acd) and (adc) are inverses of each other.. i'm guessing there was a typo somewhere? Assuming you can fix this calculation, I don't have any other feedback - the general approach seems fine
ah i see what you were trying to do, I think you meant (acb)(adc)
but as you said, just noticing that (ab)(ad) = (adb) is sufficient

[hidden kite]

Alright, thank you so much!

[mighty kiln]

Aluffi calls an algebra over a ring finitely generated if it is finitely generated as a module and finite type if it is finitely generated as an algebra

How prevalent is this convention in general?

The usage of "finite type" and "finitely generated" seem quite arbitrary in other sources

E.g.

[rustic crown]

yea that's super weird ><

for map of algebras A --> B, i see the words "finite" and "finite type"

for modules only one make sense

this is probably confusing if F was a qcoh sheaf of O_X-algebras

but then it corresponds to an affine map Spec(F) --> X

[formal mulch]

Find the dimension of the group (SO(n)) as a linear space. The group (SO(n)) is defined as the group formed by all (linear) transformations with unit determinant that leave invariant an ( (n - 1) )-dimensional sphere with respect to some (Euclidean) metric (g_{\mu\nu}):

[ S^{n-1} = { (x_1, x_2, ..., x_n) \in \mathbb{R}^n \mid g_{\mu\nu}x^\mu x^\nu = 1 } ]

[cloud walrus]

nana

[formal mulch]

i've never seen such definition of sphere, can anybody pls explain how it came 🥺

[rustic crown]

isn't g_{mu nu} = 1 iff mu = nu?

or some rotation of that

[rocky cloak]

It's just the normal definition, x is on the sphere iff ||x|| = 1

[tribal furnace]

that's the same as [\angles{x, x} = \sum_{\mu = 1}^{n} \sum_{\nu = 1}^{n} \angles{\vec{e}{\mu}, \vec{e}{\nu}} x^{\mu} x^{\nu} = 1] in coordinates

[cloud walrus]

Feuerzangenbowle

[rustic crown]

[formal mulch]

oh, in my wikipedia it was only just this 😦

[rocky cloak]

So many ways to notate the same formula

[tribal furnace]

this is the case for an orthonormal basis with respect to the inner product

[cloud walrus]

Feuerzangenbowle

[tribal furnace]

the formula here allows for non-standard inner products

although i personally prefer a coordinate-free description rather than that

[formal mulch]

do you mean scalar product?

[cobalt heath]

Huh, what is scalar product

[tribal furnace]

yes, inner product, scalar product, Riemannian / Euclidean metric, first fundamental form, so many names

[cobalt heath]

Ah, so that is the riemannian metric but with coordinates

[formal mulch]

so, $g_{\mu\nu}x^\mu x^\nu = 1$ is just $\langle x, x \rangle$?

[tribal furnace]

sorry, custom preamble

[formal mulch]

[tribal furnace]

just use \langle x, x \rangle

[cloud walrus]

nana

Feuerzangenbowle

[hidden kite]

Oh hi zan

If anyone is free and wants to help, my proof here should be fine right?

[uncut cloud]

Yes it is

[crystal vale]

If G is a commutative then AB is a subgroup of G when A and B are subgroups of G. Is it true?

[dull marsh]

What do you think?

[cloud solar]

How to show a connected commutative artinian ring is local?

[rustic crown]

it's a standard fact that any artinian ring A is automatically noetherian. if not you can pick the minimal infinitely generated ideal I, with annihilater Ann(I) = p. you can show that this becomes a prime ideal because of the minimality. and so I is an infinitely generated module over the artinian domain A/p = k which must be a field. But this is a contradiction because I doesn't have any infinitely generated subspaces by minimality.

[crystal vale]

I think it is true

[dull marsh]

If you can prove why it is, then it's definitely so

Can you prove it?

[crystal vale]

Yeah but I am not sure therefore I asked

[dull marsh]

Have you tried already or you don't know where to start?

[crystal vale]

Yes I did it but want confirmation

[dull marsh]

Well, show your work then

[crystal vale]

I define AB={ab| a belong to A and b belong to B} now since A and B is subgroup so identity belong to AB then let any element ab belong to AB then their inverse b'a'=a'b' and A and B is subgroup so a'b' belong to AB and similar for closure

[dull marsh]

Right, or for a shorter proof you could use a characterization of subgroups

[wild solar]

Hi, I'm wondering if anyone could help me understand this problem better? I'm confused about what the isomorphism should be, and I don't know how to define IA_S (the last expression).

sorry that I cut in

[dull marsh]

A subset is a subgroup iff it's closed under the operation taking x and y to xy'

[rustic crown]

after that you show that artinian ring has finitely many primes. say p_1, p_2, ... are primes, then they're automatically maximal since A/p_i are artinian domains again. in particular these are coprime ideals and so the sequence p_1, p_1p_2, p_1p_2p_3, ... is strictly decreasing, therefore must terminate by artin-ness. this gives finitely many primes. The prouduct of all of these is the jacobson radical J, and again by artininan assumption J^(N+1) = J^N. since J^N is f.g. by noetherian ness, Nakayama tells you that J^N = 0, and thus the coprime ideals p_i^N intersect to 0. So CRT tells you that
A --> prod (A/p_i^N) is an iso.
each A/p_i^N is a local artinian with nilpotent maximal ideal given by p_i/p_i^N.

so if this was connected, then there is only one factor in that product.

[crystal vale]

Yes

[crystal vale]

Is it characterization of subgroup?

[dull marsh]

Yeah

[crystal vale]

Oh

[viral mountain]

Here, shouldn't we be more careful and verify that $(b' * a') * (a * b) = e$ as well? Since the definition of inverse requires both $x * x' = x' * x = e$

[cloud walrus]

BlazingSaber

[hidden kite]

I see, thank you!

[crystal vale]

Can you help me to solve some problems

[dull marsh]

Since you are working in a group, the other condition follows immediately

I.e., if the set you are working in is already a group, xy = e is sufficient to conclude y to be the inverse of x

Since inverses are unique

[dull marsh]

Not really at the moment

[viral mountain]

Oh, I see

is there never a case that
$x * y = e$ but $y * x \neq e$, so we don't even consider $y$ to be the inverse of $x$ in the first place?

[cloud walrus]

BlazingSaber

[dull marsh]

If x, y are element of a group with operation *, then that's never the case

Outside of groups there of course are such cases happening

[viral mountain]

I know that inverses are unique, so $x * y_1 = y_1 * x = e$ and $x * y_2 = y_2 * x = e \implies y_1 = y_2$

[cloud walrus]

BlazingSaber

[rocky cloak]

If xy = e, then multiplying both sides by y gives

yxy = y

Then multiplying by y^(-1) on the right gives

yx = e

[viral mountain]

Ok fair, that makes sense!

[rocky cloak]

The crucial thing being used here is that y has an inverse. In structures that are not groups, it could be that xy = e, but y doesn't have an inverse at all.

[viral mountain]

right

but since it's a group, every element is guaranteed to have a (unique) inverse

[static glen]

existence already implies uniqueness

[crystal vale]

Given that G is group of order (p^n)a where a and p is relatively prime then how this Lagrange theorem use here to conclude that order of AB is (p^n)b?

2.6.2 Algebra 1

[viral mountain]

Given a group with these axioms, I'm able to prove that a right identity element exists once I've established the existence of a right inverse

I was wondering if there is a way to proceed the other way, i.e. prove the existence of a right identity first somehow (wasn't able to come up with this) and then use that to prove the existence of a right inverse

Also, by symmetry, axioms about right identity and right inverse would exist, but I'm curious, can we have axioms about a left identity and a right inverse, or a right identity and a left inverse equivalent to the standard group definitions?

[coral spindle]

No, you cannot prove this. Find a counterexample.

[static glen]

Well, you need an identity element to even talk about inverses so if there's no identity it does not make sense to talk about inverses

[rocky cloak]

You can always define a' as the inverse of a iff

a'ab = b for all b.

If course then it follows that a'a is an identity element.

[vocal patrol]

So an integral domain contains no zero divisors

Doesn’t that mean if there was an integral domain subset of some ring R, that the integral domain would contain only units of R? May or may not be all of them

[hollow mica]

bumping this

[daring nova]

x^(sk) = 1 iff sk is a multiple of r, by definition of the order

[hollow mica]

mhm so sk would be a common multiple of r and s smaller than lcm(r, s), contradiction

Nice

[daring nova]

so you want ||lcm(s, r) | sk|| so ||lcm(s, r) / s | k||

[barren sierra]

No, an integral domain has no non-trivial zero divisors

[vocal patrol]

Wait I meant it contains no zero divisors

[barren sierra]

No, Z is an integral domain so let the subset be all of Z. 2 is not a unit

[slim kayak]

you can have all the zero divisors, just get yourself an integral domain

(for legal reasons I am joking)

[vocal patrol]

So in a ring there are zerodivisors units and elements that are neither?

[coral spindle]

Something not being a zero divisor does not mean it is a unit

Yes, put simply.

[vocal patrol]

Yea that’s what I’ve though before

So it’s an if then

[coral spindle]

What is an if?

[vocal patrol]

If unit then not zerodivisor

[coral spindle]

Yes.

[vocal patrol]

But not iff

Okay thanks

So the only thing being different from fields and integral domains are those elements that are neither zero divisors and units

[coral spindle]

Yes

[vocal patrol]

For Zm rings tho do those elements that are neither exist?

Because don’t I have gcd(a,m)=1 implies a is a unit

[coral spindle]

What does that mean

What is a Zm ring

[vocal patrol]

Integers mod m

Under addition and multiplication

[coral spindle]

OK, we usually write that Z_m or (much better) Z/mZ, just fyi.