groups-rings-fields

[chilly radish]

In a commutative ring, if xy is a unit then it has an inverse (xy)^-1, then
x*y(xy)^-1=1 and similarly for y

[tardy hedge]

ty

[sharp ice]

still no. S_5 is generated by (1 2 3 4 5) and (1 2 3 4) with orders 5 and 5-1 resp. S_5 has order 120 and p(p-1) is 20

[south patrol]

Nice

[inner steppe]

Guys can someone explain to me what group action is like I am 5 years old. I have read the book and watched YT and gone thru the slides and I still don't grasp anything.

[tender wharf]

if this is for a class you should check out some examples of group actions

are you familiar with the usual ones

like left translations

conjugation

integers acting on R

S_n acting in {1,...,n}

you can sort of think of it as the following

if G acts on S, each element of g defines a permutation on S

for example say with S_n

every element of it is actually already just a permutation of {1,...,n}

[charred iris]

A group action associates a group with bijections on a set X, in such a way that if we have a product $gh$ in the group it doesn't matter if we first calculate $g'=gh$ and then find the map associated to $g'$, or if we first find the maps associated to $g$ and $h$ and then compose them.

[cloud walrus]

Edward II

[tender wharf]

With D_n every element in there can be thought of as a rigid motion on a regular n-gon

[charred iris]

It's a way of associating (by homomorphism) a group G with a sugroup of the group of bijections on X (compositions are associative, bijective functions have inverses, and the identity map f(x)=x is a bijection)

[tender wharf]

rigorously what I mean here is that there is a homomorphism of G into Sym(S), Sym(S) is the permutations of S

[coral spindle]

Groups are supposed to be symmetries. But the data of a group doesn't see any symmetries. Group actions are giving a group an interpretation as a collection of symmetries.

[inner steppe]

what does it mean that G acts on S_n?

[coral spindle]

No they didn't say that

S_n is the symmetric group but S here is just some set.

The example was not S = S_n but G = S_n

[inner steppe]

So given like a triangle with the corners A, B and C. We can rotate or flip them, given that my actions are r, r^2, f, rf then these functions or elements form my group? Am I correct so far?

[coral spindle]

No this isn't what a group action is

Imagine we already have a group

A group in itself does not describe symmetries of a thing

It is just a collection of objects

We have forgotten entirely what it means in terms of symmetries

[inner steppe]

this is what is confusing me

[coral spindle]

A group action is reinterpreting a group as a collection of symmetries of a thing.

A group action is not elements of a group. It is something defined on the whole group.

Now back to your example of a triangle

We can take the group of symmetries of a triangle

And this has a natural action on, as you say, the set of corners of the triangle.

That's a perfectly good example of a group action.

Where are you seeing the definition of a group action that is confusing you? It surprises me because the definition is quite straightforward and imo intuitive. What text are you working from?

[inner steppe]

What does natural action man exactly?

[coral spindle]

I am saying that it has an action, and it was a very natural thing to define.

It was not a strange thing, it was quite natural.

Natural here is an informal word.

[inner steppe]

What is perfectly good example? Where is the group action in the case of the triangle (I feel so stupid lol)?

[coral spindle]

Again this is just an English phrase

it's a fine example

it's perfectly good

just means it's fine

[inner steppe]

But didn't you tell me that the actions of rotation and flips were not group action just now?

[coral spindle]

The group action is permuting the vertices in the way dictated by the symmetries of the triangle

so you look at the triangle, look where the symmetry you chose sends the vertices, and permute the elements accordingly.

[coral spindle]

The symmetries and flips are the elements of the group in this case, because you defined it as such. The group action is information that tells us how to get an element of the group and interpret it as a permutation on {A, B, C}.

What definition of a group action have you seen.

It bothers me that you're asking this question because group actions have a very precise definition so there shouldn't be confusion regarding this

[coral spindle]

OK great there you go

[inner steppe]

I am a special breed. Seriously tho I just wanna grasp it, I feel like my brain doesn't understand it fully. 2 days ago it felt striaghtforward on the lecture but today while going thru notes I have no clue WTF it is or what I thought it was 2 days ago.

I even wrote a note to myself saying: "Intution: process by which group elements move or transform elements in the set it acts on."

[old hollow]

I personally dislike the G x X —> X definition, i feel it is confusing

I think it’s much easier to understand as a homomorphism G —> Perm(X)

That is, every group element g is associated to a permutation of X

[south patrol]

Though it is certainly worth being familiar with both, in particular since the G x X -> X will be what you usually use

[old hollow]

Sure

Usually actions are defined like “the group G acts on X via the rule gx = some stuff…”

which is intrinsically a map G x X —> X

[south patrol]

But yeah the form you did is more elegant and generalisable ofc

[small tendon]

Its what a group "does". For example, the group D3 is just the set {e,r,r^2,f,rf,r^2f} with the rules that r^3 = e, f^2 = e, and fr = r^2f.

The group action is what the group D3 actually does to the vertices of a triangle

[white oxide]

could we just say that a group homomorphism preserves identity since f(e)f(a) = f(a) = e'f(a) implies f(e) = e' by cancellation

[vapid vale]

yes

[somber turret]

Note that you assumed f(e) = e' out of nowhere and then used it to substitute going from lhs to rhs, so this is sort of a circular argument

[white oxide]

sorry where did i do that?

[somber turret]

I don't think preservation of identity can be derived from other axioms like additivity; you should include it in the definition

[somber turret]

actually lemme think, i might have made a mistake

[white oxide]

ur good

cuz f(a) = e'f(a) by definitoin of e' in H

[somber turret]

I think you're right

[white oxide]

ye

[somber turret]

f(e) = f(e)e' = f(e)f(e)f(e)^-1 = f(ee)f(e)^-1 = f(e)f(e)^-1 = e'

[knotty badger]

yep, they’re essentially related by currying/uncurrying

[limber sequoia]

@mental venture can you send the rest of excersizes?

i tried it for a good while and im stuck

[mental venture]

I think I have solved it

But I wrote my answer in Chinese

If you want to have a look, I can translate it into English

[limber sequoia]

yeah id like to

it kept me up

[mental venture]

@limber sequoia

As we have already known how “3/4” works, the second question is easier.

[warped bramble]

[limber sequoia]

super neat!

[cloud lynx]

are all intermidiate fields Fix fields?

[rocky cloak]

If the extension is Galois, then yes. That's the Galois correspondence.

[cloud lynx]

ah okk yes right ty!!

[brisk oak]

Hello guys 🤠 I often see people here reading atiyah macdonald book. What if we start a group chat for reading this book? 🤔

[surreal dagger]

Id be generally interested. Im taking a commutative algebra class next semester. One way could be just to open a thread in this channel

[warped bramble]

[wild jasper]

Let k be a generator of (Z/pZ)^* \cong Z/(p-1)Z for a prime p. Let w be a p'th root of unity, how is w^k = (w^k)^(p-1)=w^(-k) ?

[rocky cloak]

So
(w^k)^(p-1) = (w^k)^p * (w^k)^(-1) = (w^p)^k * w^(-k) = 1*w^(-k) = w^(-k)

[wild jasper]

indeed

[tardy hedge]

I am currently working through that book. I just started

[sonic coral]

i am starting this book in the spring

[hidden wind]

same

[rocky cloak]

So that's why (w^k)^(p-1) = w^(-k), and the other equality just isn't true

[wild jasper]

ok

[waxen moth]

The order of a group is equivalent to the cardinality of the underlying set right? Is there a reason why we have diffferent notations for both of them?

[tender wharf]

well both Z and R have infinite order but Z has a different cardinality

[waxen moth]

Oh yeah right eh

[tender wharf]

we just call them both infinite groups

as a more annoying example take any set X, and consider P(X). make this into a group by defining a+b to be the symmetric difference of a and b

so you can make loads of infinite groups that have different cardinalities

i guess order is more interesting if its finite

[waxen moth]

yeah

Now this sentence makes sense.

[waxen moth]

Oh so order isn't defined as the cardinality of the set of cosets of the trivial subgroup of a group?

[tender wharf]

oh i haven't seen this one

like

with the cardinal stuff

[wild jasper]

that's because I completely misunderstood what I was doing, what I should have been asking was how come k^(p-1) = 1 when k was a generator for the multiplcative group Z_{p-1} and that is obvious (even to me :P)

[tardy hedge]

Why is it that if you have a maximal ideal m, and x not in m, then m+(x) generate the whole ring?

Lol wait i might have gotten something wrong'

im tired and shii

[wild jasper]

@tardy hedge because all the non-units will be in m

[tardy hedge]

shhhhhiiit

oops sorry maybe no swear

[tardy hedge]

yea i gotta review that fac

fact

Bu

But

[wild jasper]

wait

[tardy hedge]

yea

something weird

[wild jasper]

It's "every non-unit is in some maximal ideal"

[tardy hedge]

yea

[tardy hedge]

i think maybe this isnt true in general, it was part of a bigger proof so lemme check it

about jacobson radical

[sharp ice]

its because m+(x) is an ideal. by maximality, m+(x) is either m or the ring itself. since x is not in m, m+(x) must be the ring

[tardy hedge]

Ohhhhh

cause it contains m

and m maximal

Thanks a lot

I like how atiyah macdonald doesnt give every detail like that

how does the jacobson radical relate to nilpotent elemetns

nilradical was all nilpotent elements, and it can also be viewed as intersection of all prime ideals

jacobson radical was just intersection of all maximal ideals

[tardy hedge]

Havent done the exercises yet but i got them assigned

Nice

And thank you for your reply

Ok yes so Jacobson radical contains nilradical

you know, a lot of set inclusion stuff still gives me trouble

[hidden wind]

i’ve been stuck on finding a proof for the very first proposition in a&m for a bit too long now

[tardy hedge]

Which one was it?

[hidden wind]

it looks really easy and probably is so i’m very stressed out by not having managed to set it up yet

[tardy hedge]

Ur working thru that text rn too?

[hidden wind]

mhm

[tardy hedge]

Nice

We will have a journey together

[hidden wind]

[hidden wind]

…will have a look at it tmrw

[tardy hedge]

What was the statement

[hidden wind]

there is a one-to-one order-preserving correspondence between ideals of R containing some ideal I, and ideals of R/I

iirc

very possibly i don’t

[brisk oak]

Well it is an analogue of the same theorem for groups

And the correspondence is basically just the canonical homomorphism I think

[hidden wind]

yeah it’s just some work checking every detail… i’ve been on a drought for the past month barely done any problems at all

just gotta get back into the flow somehow

have some bad habits i have to get rid of

[brisk oak]

Atiyah-Macdonald book discussion club

[chilly radish]

what part are you stuck on

[hidden wind]

it’s really just sitting down and doing it that i struggle with

but i will try tmrw

and ask here if i get stuck

[inner steppe]

A finite group G acts on a set X and two of the orbits of G have 5 and 21 elements, respectively. What can be said about the order of G, ∣G∣?

Is the only thing we can say that |G| = 5x21xk where k is an integer>0 ?
Or can we say something else?

[quiet pelican]

That’s the only thing we can say generically, yes

[arctic trail]

that's essentially all you can say

I don't think you can say anything else, that's not immediate from the group having order multiple of 5*21

[south patrol]

Yeah e.g. note you can take the group G = Z/5 x Z/21 x H for any finite group H and let this act on the set {1,...5} disjoint union {1,...,21} in the obvious way to ensure you have one orbit of size 5 and one of size 21

So you can get any multiple of 5 * 21

[old hollow]

So this is the converse

[south patrol]

Sure I guess. I'm just saying this shows you cannot sharpen the statement at all beyond "it must be divisible by 5*21" yeah

[cloud walrus]

wordslinger

[tranquil wagon]

I've been thinking about this for a while; in my mind it seems like its not true but I cant come up with a counter. any hints are appreciated!

[hidden wind]

phew finally managed to check all the details (which i severely needed, seems i am not quite sufficiently used to working with either cosets or ring/group homomorphisms) for the bijection between the two sets of ideals, but fsr brain noped out at checking the order-preserving, though i'd expect this to be the easiest part... (the direction via the inverse mapping is clear)

took me a bit almost an hour writing everything out, but at least now i am officially out of the problem drought (which has lasted a full month now), just have to keep the ball rolling somehow

[rocky cloak]

Shouldn't be true no.

Take ||L = N = Z/2||, and have M = ||L(+)N||. Then take M1 ||spanned by (1, 1)|| and M2 ||spanned by (0, 1).||

||Then neither intersects L and both map onto N.||

The main issue is that alpha^-1 measures whether M1 / M2 intersects L, but it don't really say much about whether they intersect each other

[tranquil wagon]

Ah yeah ty

[elfin wraith]

Any recommendations for reading on graded and filtered rings? Didn’t seem like Artin or D&F had anything on them and the notes from my lecturer are kinda lost on me

I think seeing some examples would be nice, preferably non commutative, thanks!

[kindred vector]

Are you specifically working with just rings and not algebras?

For algebras Schedler's notes on deformation theory have a good introduction in the first section

Not sure about just rings though

[wraith cargo]

probably read a book on commutative algebra

stuff like matsumura or eisenbud talk about this stuff in detail

[elfin wraith]

That section of the notes specifically deals with rings, but we’re working with plenty of algebras in the class so I’ll take a look there, thanks for the recommendations I’ll come back if I’m still a bit lost

I do enjoy that from my first look at these notes, he mentions the stared exercises are quite difficult and can be skipped, but the first stared exercise is question 1 of 8 on my homework lol, guess these will be good to look at though

[rustic rapids]

I am not too sure how to approach this in general. I think i can find a few examples. Namely, for the injective maps, we can have perhaps a "swap" in which f(x) = -x. Since 0 is the identity it would stay at 0 and the kernel would be trivial. I think we can extend that to be f(x)=-ax for any a. Similarly, I think we can have a surjective map that 'shifts' f(x) = x+a, but im still not sure how to generalize any of this. I know the homomorphism is determined by its kernel, but I'm not sure how it pertains to isomorphic or surjective homomorphisms, only that its kernel is trivial iff injective

[hidden cairn]

for part b. Let $a, g_1 \cdot a, \dots, g_{r-1} \cdot a$ be representatives of left cosets of $HG_a$. Consider the map $$ g_i HG_a \mapsto { k \cdot (g_i \cdot a) : ~ k \in H } = O_i $$

[cloud walrus]

pink_panther

[hidden cairn]

is this the right isomorphism that proves $r = |G : HG_a|$

[cloud walrus]

pink_panther

[tardy hedge]

Kind of silly question incoming:

A free R-module with basis A means every m in M can be written uniquely as R linear combo of elements in A. What about 0? I am just trying to remember the relevance of 0 to uniqueness and all this

Basically im trying to remember linear algebra stuff

[coral spindle]

0 = 0a_1 + 0a_2 + ...

Just think of a vector space.

It's the same

[tardy hedge]

Its been a while. I was just confused cause like also 0 = 0a_1 so 0 cant be written uniquely ok ok silly

But i noticed now in df they specify nonzero elements for the uniqueness thing

[coral spindle]

Since only finitely many entries are nonzero you can see it as an infinite sum

There's no problem there

[tardy hedge]

Finitey many are nonzero?

[coral spindle]

Yes, only finitely many coefficients of a linear combination are nonzero

This is why I was saying you can treat it as an infinite sum

[coral spindle]

Then you don't have to worry about this nonzero business

[coral spindle]

You can really truly just see it as an infinite sum as I did here.

[tardy hedge]

Ok

Also, Ra1 is not isomorphic to R as R-modules in general right. If a1 has uhh … zero torsion or something? Then its isomorphic ?

[coral spindle]

Like are you saying that if we have some R-module M and element m of M then Rm is not typically going to be iso to R?

Yes. Find an example.

[tardy hedge]

Yeah i was just going to clarify

[coral spindle]

Idk what you mean by zero torsion

It can have torsion, yes, but R can also have torsion

Try R = Z.

[tardy hedge]

Ok yes i will need to think through it, i was asking because in the free module context Ra1 iso to R for each a1 in the free generating set

And i saw its injective because (r1-r2)a1 = 0 implies r1=r2

[coral spindle]

Yes that's right

[tardy hedge]

Since its a free module every element is written uniquely so you cant have nonzero thing times a1 = 0 cause also 0a1+… = 0

[coral spindle]

Uh-huh

[cloud lynx]

is M_\infinite(R) a ring?

or is it only a ring if the dimension is finite

[tardy hedge]

R can also have torsion as an R-module? Like if R has zero divisors

Why try R = Z?

Oh like a Z-module

Yeah it a group has element with finite order

If

Or?

Yeah for a Z-module then the submodule Za for some a in module wont be iso to Z yes

If a has finite order in the group

Makes sense

[rocky cloak]

Depends what M_\infty means. If you just naively consider a matrix as an infinite grid of elements of R, then this won't have a well defined matrix multiplication.

If however you require that for each column only finitely many elements are non-zero you get the endomorphism ring of R^(N) (free module on N)

[cloud lynx]

:((

do u know a infinite non communative ring with char p>0?

hö

but isnt it finite?

I wanne cry

[next obsidian]

What about 2x2 matrices over an infinite

Fuck you tteppa

Smd

Okay how about the endomorphism ring of an infinite dimensional F_p vector space

Chmowned

I went infinite on yo ass in a different way

[cloud lynx]

mmh this isomorphic to M_\infinite(F_P)

and multiplication is not well difined

[next obsidian]

Yeah but you need to be careful about what an infinite matrix is

[rocky cloak]

Take F the algebraic closure of Fp, G the absolute Galois group and R = F#G the skew group algebra.

For example

[cloud lynx]

what is F#G

lol

[rocky cloak]

The skew group algebra

[next obsidian]

The skew group algebra

[cloud lynx]

another example? because we havent that lol

[cloud lynx]

and its well defined?

[rocky cloak]

The elements are (formal) F-linear combinations of elements in G, and the multiplication is
(a g)(b h) = (a b^g) (gh)

Where b^g means g acting on b. (a, b in F, g, h in G)

[next obsidian]

Yeah

Multiplication is composition and addition is just addition

[cloud lynx]

mmh but Mn(R) isomorph to End(R^n)

[next obsidian]

Now ain’t that just a semi direct product

[cloud lynx]

so M_infinite(R) isomorph to End(Rînfinite)

[next obsidian]

Horrors beyond human comprehension

[cloud lynx]

so how can be multiplication well defined in End when its not well defined in M_infinite(R)

[next obsidian]

Cuz what you said isn’t true

[rocky cloak]

Yes, the group algebra of the semidirect product
F(G rtimes H)
equals the skew group algebra
FG # H

[next obsidian]

Not every matrix corresponds to a linear map

I mean this is really coming down to what an infinite matrix is

Anyway take what I said or leave it

[rocky cloak]

See this answer @cloud lynx

If you define M_\infty so that each column has finitely many nonzero entries, then it equals End(R^(N)) and matrix multiplication = composition

[cloud lynx]

what is column

[cloud lynx]

I take it

lol

I have trust in u

[rocky cloak]

A column is something stacked vertically

A matrix is a grid of elements, so has columns and rows

For example in the matrix
a b
c d

The first column is
a
c

[cloud lynx]

AHHH

Okk

Tyy

English is not my native language unfortunatly lol

Unfortunately

[sweet pendant]

How do I go about showing that “If R[x] is an integral domain then R is an integral domain”. If we take any a,b from R then since R[x] is a subset of R we can just represent a,b as polynomials i.e., f(x)=a and g(x)=b. We also know that f(x)g(x) is not equal to 0 or cannot be a zero divisor since R[x] is an integral domain. Now ab=f(x)g(x) right and ab is in R so ab also cannot be a zero divisor. Will this be the correct idea?

[sharp ice]

yes. any subring of an integral domain is an integral domain. u can embed R in R[x] through sending a to itself as a constant polynomial so R can be viewed as a subring of R[x]

[sweet pendant]

Great! Thank you.

[sly crescent]

What about a non-split extension

[rocky cloak]

What about them?

[sly crescent]

What is the group algebra of a non-split extension?

[rocky cloak]

Well FG certainly maps onto F(G/N). And the ideal is generated by (1 - n) for n in N.

Not sure what more you can say, like the extension is not particularly nice, so the resulting algebra will also be not very nice I guess

[nimble thunder]

Hi All. Is there a name for a monoid that's basically isomorphic to (-inf, 1] and equipped with a single binary operator? I'm writing a (physics related) paper where the fundamental objects turned out to essentially be three of those…

[coral spindle]

Well without specifying what operation that is, it doesn't mean very much to say that it's isomorphic to that monoid

Any monoid with a continuum number of elements satisfies those requirements.

(if by 'isomorphic to (-inf, 1]' you mean in bijection with that set)

[rocky cloak]

And what's the deal with (-inf, 1]? Such a weirdly specific set

[nimble thunder]

The binary operation isn't too crucial, tbh, it just happens to make sense. For context: it's essentially encoding the inverse of distance, but with a zero point. A good analogy would be accuracy on a dart board: missing the board maps to (-inf, 0), the rim maps to 0 and the bullseye maps to 1.

[coral spindle]

The binary operation is entirely crucial if you're talking about a monoid being isomorphic to something!

[nimble thunder]

I just used "isomorphic" so I wouldn't waste everyone's time explaining the thing if it's a well known thing… 😅

might have miscalculated xD

[coral spindle]

You know how physicists get upset when laypeople use the word "quantum" willy-nilly? I imagine I know how that feels right about now

[nimble thunder]

Hahaha

[coral spindle]

Without further details you are just talking about any monoid with continuum many elements.

[nimble thunder]

How icky would it feel if I were to describe something like that and call it "closeness"?

[rocky cloak]

Closeness seems fine, but maybe "signed distance" would be better

Not sure where the monoid structure is fitting into this though

[nimble thunder]

It's a monoid because the set of physical distances isn't closed under multiplication, so I just dropped the operation…

[rocky cloak]

You dropped it?

I'm not following

[nimble thunder]

(I also dropped addition and division, hence downgrading from a field to a monoid? I'm demonstrably still bad at this xD)

[coral spindle]

There's no use in us guessing and heming and hawing about what this could be

[rocky cloak]

Okay, but like a monoid is a set with an operation. So far you haven't said anything about an operation

[nimble thunder]

Ok, let me go and collect my thoughts a bit more. Thanks for indulging my ramblings for a bit! 😅

[tardy hedge]

Random, but are you interested at all in physics?

I’m not interested in physics

I thought I was way back when but then I realized I just liked the math

[coral spindle]

No physics doesn’t interest me personally

It was programming that made me realise the same thing

[sweet pendant]

We are given that $R$ is a domain and we need to show that the units of $R[x]$ are exactly the units of $R$.
My attempt: Let $f$ and $g \in R[x]$ where g is the multiplicative inverse of $f$. Then since $R[x]$ is also an integral domain $fg=1$. This implies that $deg(f)+deg(g)=deg(f+g)=0$. So f is indeed a constant polynomial. Now how do I connect this with the second part of the statement? Can I say that since all the coefficients of the polynomial come from $R$, then $f$ is also in $R$ and is a unit? Idk, this last statement kind of does not really conclude the proof.

[cloud walrus]

mss

[tardy hedge]

Wdym since R[x] is also an integral domain then fg = 1

Yes you basically showed that if f in R[x] is a unit then it must be constant, so its in R

So whatever the units of R[x] are, are the units of R

[sweet pendant]

Ohhhh ok ok I get it now.

[tardy hedge]

Saying this isnt relevant

At least i dont think it is

[sweet pendant]

I think I can just say let $f$ be a unit with inverse $g$ where $f,g \in R[x]$. Then $fg=1$ and move on to the degree stuff etc instead of all that.

[cloud walrus]

mss

[tardy hedge]

Yea

Oh yea cause that degree sum formula only works for R an integral domain ?

I didnt think about that before

So there could be more units

Than just constant term being unit

[sweet pendant]

Is there a faster way of finding all the solutions instead of checking every single element for $\bar{5}x^2=\bar{3} \text{ in } \mathbb{Z}/7\mathbb{Z}$? Like its fine because there are only 7 elements in $\mathbb{Z}/7\mathbb{Z}$ but what if $m$ is like 400?

[cloud walrus]

mss

[sharp ice]

there are tricks for larger numbers. for specifically 400, u note that 400 = 2^4 * 5^2
u solve the equation mod 2. lift a solution mod 2^4 using hensels lemma. then do mod 5. lift a solution to mod 5^2 the same way. then u use the CRT to find a solution mod 400

^ thats one way, im sure theres others aswell

[sweet pendant]

So this will be (x-3)(x+3)=0 mod 7? and the solutions are 3 and 4.

[tough raven]

Notice that b(M1) = b(M2) iff M1 + L = M2 + N, and a^{-1}(Mi) = Mi intersect L.
So for this to be true for all short exact sequences involving M is equivalent to the lattice of submodules of M being a distributive lattice, according to https://en.wikipedia.org/wiki/Distributive_lattice#Characteristic_properties (see the second displayed equation).
The lattice of submodules of a module is a modular lattice but not a distributive lattice, so it probably isn't in general, which should imply that there's a counterexample.

In fact, the prototypical modular non-distributive lattice is 0 <= a, b, c <= 1 (see the link above again), in which b, c have the same meet and join with a but are non-equal, and this can be realised e.g. by three distinct 1-dimensional subspaces of a 2-dimensional vector space.
So M any 2-dimensional vector space over a field and L, M1, M2 three different lines in it should be a counterexample.

[tough raven]

Oh, oops.

Oh lol more generally all complements of L in M have the same a^{-1}() (namely 0) and b() (namely all of N).

[cobalt heath]

Damn, Auslander-Reiten theory is wild

Never thought morphism versions of e.g. radical would be so frustrating..

In the class, its definition was rather glossed over. How do I understand the radical of morphisms, rad(X, Y)?

[warm ember]

when taking inverse limit of an inverse directed family of groups, why does it matter if its directed

directed poset means for every i and j there exists k such that i<=k and j<=k

[trail cave]

I need to correct part d here, the binary operation is (a,b)(c,d) = (ac, ad+b) for a,b,c,d in R. a^n implying a = 1 or -1 is obvious, but I don't see how in the a = 1 case my conclusion changes from what he said. I still have to say bn = 0 and n can't be 0, so b must be shouldn't it?

[rocky cloak]

So you can define rad(X, Y) very similar to how you would for rings.

Namely f: X -> Y is radical iff for any map g: Y -> X, 1 - gf is an automorphism.

If you're in a finite length category then it becomes easier to describe. If X and Y are indecomposable then rad(X, Y) is the set of non-isomorphisms. And rad is additive so you can just distribute it over direct sums.

See https://wiki.math.ntnu.no/ma3203/2024v/course_schedule sheet 14 exercise 3c, 3f. Also exercise 4 relates it to AR-theory

[cobalt heath]

Thanks! Looking into it rn

[rocky cloak]

You may have to use ||fittings lemma||, I don't know if you're familiar with it or not.

[cobalt heath]

Sadly, I am not familiar with it

[rocky cloak]

In short the endomorphism ring of an indecomposable finite length module is a local ring

In long: if X is a finite length module and f is an endomorphism, then there exists an n such that X is the direct sum of Im f^n and Ker f^n

[cobalt heath]

I learned about this, but did not learn about the "in long:" version

Not exactly learned about the proof tho

[rocky cloak]

Well, it's a nice exercise. Remember that finite length modules are both artinian and Noetherian

[cobalt heath]

Hmm actually, it seems like I did not learn this.. sorry

There was a similar lemma about endomorphism being a local ring, but likely was more restricted case.

[rocky cloak]

I mean you can replace finite length with finite dimensional I guess. Otherwise I'm not sure of any similar lemma.

There is the Krull-Remack-Schmidt theorem, that says that if a module is a finite direct sum of modules with local endomorphism rings, then it can be written as a direct sum of indecomposables in a unique way.

[cobalt heath]

Ah, yep, it was proved for finite dimension case

So it does not use artinian/noetherian machinery.

[rocky cloak]

Well, I mean the proof is the same. Whether you use the word "artinian" or not. Your just saying that a decreasing sequence of subspaces eventually stabilizes.

[cobalt heath]

Anyway, thanks a lot!! The exercises would hopefully help me with understanding.

[elfin wraith]

Would appreciate if anyone could give me a pointer of where to start with this problem, not quite sure how to actually approach it (not looking for a full solution or anything, genuinely just an idea to get going)

Let $k$ be a field and $q\in k^\times$. Let $\tau_q$ be the unique automorphism of $k[y]$ such that $\tau_q=qy$. Show that $k[y][x;\tau_q]\cong k\langle x,y\rangle/(xy-qxy)$

[cloud walrus]

Nope

[elfin wraith]

I’m guessing it’s some application of the universal property, and unpacking the notation for the ore extension it does certainly look like this should be true, but I’m not quite sure what the details are

We obviously have xy-\tau(y)x in the kernel of the left, but then I’m guessing we use the fact that the image of tau is qy and that’ll somehow give us what we need to use the universal property, but I’m struggling to work out what any of the actual details are here

[kindred vector]

Recall that given an endomorphism (\sigma:R\to R) the Ore extension of (R) is just [R[x,\sigma]=R[x]/\langle xr-\sigma(r)x\rangle]

[cloud walrus]

vir783

[cloud lynx]

how do i show that f in Z[X] is irreducible?

[south patrol]

What is f

In general this is hard

[cloud lynx]

because reduction criteria and eisenstein is only that f is then over Quot(R)

X^3+10x-5

in Z[X]

[south patrol]

Well Eisenstein is very useful

Oh lol

I mean yeah Eisenstein applies to this guy

[cloud lynx]

yeah but eisenstein states that its in Q[X] irreduicble

but were looking at Z[X]

[coral spindle]

Well if it were reducible in Z[X] then certainly it would be reducible in Q[X] -- can you see why?

[cloud lynx]

yes

[coral spindle]

OK then you're done

[cloud lynx]

why?

take 2X for example

[coral spindle]

Think about this again.

[cloud lynx]

its reducible in Z[X] but irreducible in Q[X]

[coral spindle]

Yes because there is something about 2X that makes this untrue that your other polynomial does not share

Hint: leading coefficient

[cloud lynx]

leading coefficient is a unit

[unkempt stream]

I forget the name of the type of ring

[south patrol]

Relevant theorem: Gauß' lemma

Lol

[unkempt stream]

Where everything has a (not unique) prime decomposition

[south patrol]

I have seen factorisation domain I think

[unkempt stream]

That’s it

Yeah it’s a statement for the field of fractions and their polynomial rings of a factorization domain

[cloud lynx]

ah ok thank you!!

do u know a way how to determine how many possible homomorphism there exists for a homomorphism

?

[coral spindle]

There's no single way to do this, but a good method is to choose a nice generating set of whatever object you're looking at, since any homomorphism will be entirely determined by where it sends those elements.

[cloud lynx]

ah ok ty!

[warm ember]

when taking inverse limit of an inverse directed family of groups, why does it matter if its directed

directed poset means for every i and j there exists k such that i<=k and j<=k

[south patrol]

Well I mean you can still take the inverse limit anyway, but it behaves particularly nicely when you have directed things

Very often one losens this to being "cofiltered" which is essentially the same but slightly more flexible

[trail cave]

bumping this pls

[glad osprey]

I think you'll have more responses if you write it up in latex. I can't even tell what the original problem is

[cloud lynx]

how does S_3 ->Z/2Z even look like?

something with signum?

[south patrol]

It is the signum

[cloud lynx]

okk

tyy

[trail cave]

I went to my tutoring center before class and ended up answering the question myself, I don't know why I was struggling with it

but thank you

I'll be more clear next time

[grave sedge]

Every element in the limit is "represented" by an element coming from one of the groups

And if you are doing some computation involving finitely many elements you can always find one of the groups containing all those elements

Which means that a bunch of the properties that are true for each of the groups are also true for the limit

[cloud lynx]

has anyone an Idea how to solve there exists a bijection between {U<G with Index 2} and {G->Z/2Z Epimorphism}?

[coral spindle]

Hint: all subgroups of index two are normal.

[cloud lynx]

Yes I know so doing something 1.isosomorohism theorom?

[coral spindle]

Try it and see

[cloud lynx]

But I dont know how to construct it

[coral spindle]

Construct what

[cloud lynx]

Do they mean U->Z/2Z?

[coral spindle]

No

They mean G -> Z/2Z, they made no mistake.

[cloud lynx]

Because i dont even understand how it look like

ok

G->G/U

but we know that for every U there exists a Epi G-> Z/2Z

so also G->Z/2Z

so U isomorphic to Z/2Z

and how does it show that set is bijective?

to the set of epis

[rocky cloak]

This is direct limits, not inverse limits though.

Direct limits are very nice, much nicer than general colimits.

[grave sedge]

You're right, i read "directed set" and forgot to read "inverse"

[rocky cloak]

I'm not sure there is a big advantage.

For direct limits vs colimits, direct limits are exact and commute with the forgetful functor. And the latter allows a nice construction of them.

But for inverse limits vs limits, the forgetful functor already commutes with limits, and inverse limits are not exact. So I'm not sure what the gain is.

But inverse limits of a system of surjective maps is exact though, so that's a slight advantage.

[south patrol]

I'm confused what you mean by colimits "allowing a nice construction of them"

[rocky cloak]

Commuting with the forgetful functor allows a nice construction

Sorry if that was unclear

I.e. you just take the direct limit of the underlying set, and then equip operations in a nice way

[south patrol]

Oh sorry I see what you mean, misread your post

Lol

Thought "the latter" was referring to general colimits and was like huh

[vapid vale]

atiyah-macdonald

[gilded talon]

I have a stupid question regarding quotient group with AoC
Say we have a function from G to G/N defined by

$\phi:G\rightarrow G/N$ where $\phi(g):=gN$.

[cloud walrus]

AlexATG

[gilded talon]

and now i want to show \phi is surjective, so for each C\in G/N, i want to find a g such that \phi(g)=C

[warm ember]

oh

i guess i havent seen anything nontrivial with this

[gilded talon]

this can be done trivially by choosing some representative g\in C, and thus \phi(g)=gN=C

but i think the action of choosing g\in C involves using the Axiom of Choice, or does it not. And if yes, is there a way to bypass using AoC

i tried to explain by for each C\in G/N, C is non empty and there exists a g\in C, for the same g, we have \phi(g)=C, but im not sure if this is rigorous enough

[coral spindle]

It's rigorous enough.

I think it does require AoC but you'd have to ask someone who actually knows about foundational stuff. I would say don't waste your time on it if you just wanna learn group theory. AoC is very necessary for algebra.

[rocky cloak]

You don't need AoC, since there is no reason to make all the choices simultaneously.

The statement for each coset X in G/N there is a g such that X = gN is true without needing to define a function G/N -> G making such a universal choice.

[gilded talon]

Thanks a lot! So when I do it "one by one" for each X\in G/N I don't need AoC, but when i do it "at the same time", say writing out a set of representative S, or repeating the process of picking each one by one indefinitely, this is when i need AoC

[sweet pendant]

Any hints on how to get started for this question?

[gilded talon]

bijective means surjective and injective. So by surjectivity, for any y in B, y=f(x) for some x in A. Using injectivity, what can be told about such x?

[sweet pendant]

So $\forall x \in A \exists y \in B | f(x)=y$ ? So $y=f(x)=y$? So $y$ has the same image as itself?

[cloud walrus]

mss

[gilded talon]

think about what injectivity means, would there be two different a,b \in A such that f(a)=y and f(b)=y

[topaz solar]

Yes

In another manner of speaking, you can uniformly pick representatives

Since choice guarantees a function which picks them out

Whereas without choice, you can’t necessarily pick out a rule describing how to pick a representative

[gilded talon]

Thanks!

[chilly radish]

this is not the channel for this problem. Use #proofs-and-logic or #discrete-math

[old sphinx]

Can somebody help me with this.

[glad osprey]

If g is a Lie algebra, is End(g) also a Lie algebra? What is the Lie bracket on End(g)?

[wraith cargo]

it's just [f,g] = fg - gf I believe

[south patrol]

yes

[glad osprey]

Alright, thanks 🙏

[south patrol]

Well really I suppose this is a property of the underlying vector space

But a map g -> End(V) of Lie algs for a vector space V is equivalent to a rep of g on V, so this is saying ad is a representation on (the underyling vector space of) itself

(well change vector space for module as appropriate :))

[rotund aurora]

For 33, do you see why these operators are GF(2)-linear? Then you just have to show that the zero function is represented iff all the coefficients are zero (Euler would have had problems proving this, btw, but I think you shouldn't have problems if you know a little bit), and then use a dimension argument.

For 36, try factoring x^(3^(k+1))-1. It factors as f_k(x)...f_0(x)(x-1) where f_k(x)=1+x^(3^k)+x^(2*3^k). f_k(x) is the minimal polynomial of the primitive 3^(k+1)th roots of unity

[south patrol]

Also btw in general feel free to post Lie algebra stuff in #advanced-algebra

[old sphinx]

Makes sensefor that as thank you for the help with 33

[topaz solar]

But if we’re already on Lie things, what’s a good reference for Lie groups/algebras, particularly including tori and parabolic-y kinds of subgroups? :3

[glad osprey]

lol it says right there that the Lie bracket is ad_x ad_y - ad_y ad_x, should just learn to read 😅

[glad osprey]

will do 👍

[rotund aurora]

Have you considered Knapp?

[topaz solar]

Haven’t even heard of it so no

[south patrol]

Hm I should learn about lie groups properly at some point

A usual reference is Bump I think lol

Nice book

[topaz solar]

The only thing I know of offhand is uhhh Prosecci, whatever is in Fulton-Harris and Isaac or wtv, and whatever little Lang includes

As in, references I know of, not that I know the content

[topaz solar]

Buddy is bumping Bump

[south patrol]

Lol

[old sphinx]

There

[cobalt heath]

On an Artinian ring, is nil ideal nilpotent?

If so, I struggle with how to easily prove this.

[boreal inlet]

By Nil ideal do you mean Nilradical?

[cobalt heath]

Yeah, or any ideal that is made of nilpotent elements.

Nilradical is the ideal containing all nilpotent elements, right?

[boreal inlet]

Yes, in commutative unital rings

I am not sure if that's true for Noncommutative rings in general

[tough raven]

Indeed, nilpotent elements do not necessarily form an ideal in a non-commutative ring (hint for a counterexample: ||matrix ring||).

I think that's why the terminology of "nil ideal" instead of just saying "ideal contained in the nilradical" originated.

[boreal inlet]

Also, is Jacobson Radical equal to the Nilradical in case of Artinian rings?

[tough raven]

Yes for commutative Artinian rings (AFAIK nilradical is not defined for non-commutative ones).

[boreal inlet]

Oh right yeah, all non-zero prime ideals are maximal, yeah

[tough raven]

Even for a non-commutative Artinian ring, the Jacobson radical is nilpotent, which is if anything stronger than "being the nilradical" would be.

[boreal inlet]

In commutative settings

[boreal inlet]

Okay, from this, if we can show all nil ideals are contained in the Jacobson radical, aren't we done?

I'm not assuming commutativity here

[cobalt heath]

Are nil ideals contained in Jacobson radical?

[tough raven]

Done with?

[tough raven]

Oh, this.

[tough raven]

This is true for any (unital) ring.

[tough raven]

This is true for Artinian rings.

[cobalt heath]

Why though?

By nil ideals, I mean ideals which is made of nilpotent elements.

[tough raven]

Putting them together, you can show that the Jacobson radical is the unique largest nil ideal, and it is nilpotent, so that all nil ideals are nilpotent, as @boreal inlet said.

[cobalt heath]

I think it could be bigger, but I cannot pinpoint an example.

[tough raven]

Right, so that depends on your definition of the Jacobson radical.

[cobalt heath]

I would define it as intersection of right (or left) maximal ideals.

Then for Artinian rings, I think it is maximal 2-sided nilpotent ideal.

[tough raven]

Then you need to show that if I is an ideal all of whose elements are nilpotent, then every x in I lies in every maximal left ideal M.

Suppose for some x in I and maximal left ideal M that x does not lie in M. Then ...

I GTG so posting the answer:

... then ||M + Rx = R because M is maximal||, so ||1 in M + Rx <=> 1 + rx in M for some r in R||, so ||1 + rx is not (left) invertible||, but ||rx is nilpotent as it lies in I||. This is a contradiction because ||if (rx)^n = 0, then 1 + rx has (two-sided, in fact) inverse 1 - rx + (rx)^2 - ... + (-rx)^{n-1}||.

(The alternative definition of the Jacobson radical as {x | 1 + rx is (left) invertible for all r in R} is more natural for this.)

[cobalt heath]

Damn, how I never think of this line of thought

Thanks!

So it doesn't even make use of Artinian condition?

[tough raven]

Yes, this doesn't require Artinian.

[cobalt heath]

I forget ring theory so easily.. I must have forgotten this

[rocky cloak]

Usually you'd define nilradical as the sum of all nilpotent ideals. In which case nilradical equals Jacobson radical for artinian rings

[tardy hedge]

There is a prof here who is selling his humongous textbook collection

Serge Lang’s Algebra hardcover only $10!!!!!!

JACKPOT!!!!

[coral spindle]

Oh wow that's pretty good

[tardy hedge]

Yeah im so hyped dude

Im gonna get as many as i can afford lmao

[kind temple]

is ur prof okay ??

i have a certain attachment to the math books that i have, especially in paper/hardback

[coral spindle]

They're probably retiring, if I were to guess

[rocky cloak]

Typically professors have a whole office full of books, that won't do much good in their basement when they retire.

[mystic ether]

Lets say I have a group presentation with generators {a_i, b_j}, with relations that set all the {a_i} equal to all the {b_j} for any i and j, and no other relations. I know that then I will have a group presentation with all the generators equal. But I'm a bit confused because to me it seems like I should therefore have Z as my group. But I know the answer is supposed to be that this is the trivial group

so why isn't it Z?

(I know it's supposed to be the trivial group bc this comes from a van kampen calculation on a space we are told is simply connected)

[rocky cloak]

You sure you're not just missing a relation?

[mystic ether]

maybe I am. I suppose I should ask in #alg-top-geo-top about that though

[rocky cloak]

Sure thing

Anyway, such a presentation would give Z

[tardy hedge]

Yeah, his collection is insane

[tardy hedge]

Me too

[kind temple]

after an entire lifetime of research, what do you do after?

[rocky cloak]

I dunno. One emeritus at my institute runs an antique store.

Otherwise I guess you do what retired people usually do, play bridge with friends, travel to warmer places

Show up to seminars and complain about the youth corrupting math

[tardy hedge]

Lol

[grizzled spindle]

not sure i see why this is true

was taken in class and did not elaborate further on why the preimage should be a Sylow 2-subgroup

oh V is the klein four-group for reference

[coral spindle]

Well what is the size of phi^-1(A)? Every element of A will have |V| preimages, so in total we get 2*4 = 8 elements. And you can simply check that 8 is the largest power of two dividing 4!.

[grizzled spindle]

Oh right cause the kernal has to lie in the preimage

is the preimage always a group?

[coral spindle]

Try proving it!

[grizzled spindle]

ye good idea

[coral spindle]

More to-the-point, the preimage of a point is just a coset of the kernel.

this is known as the first isomorphism theorem

[grizzled spindle]

right, i know the theorem im mostly not used to using preimage instead

[coral spindle]

I was making a bit of a joke. But it does follow

[grizzled spindle]

alright yeah this wasnt too hard to show

thx

[winged void]

Hello im not really sure if somoen is awake

i have small question regarding order of an element inside the group

[south patrol]

Feel free to ask

[winged void]

the first question lets say i know that a^20 = identity element why this means not yet that its the order of element a and we should look to its divisor so in this case it might be 1 or 5 or 4 and why when a = - a then the order is 2 i mean well it makes sense for example that a + (-a) = 0 which makes its order 2 indeed but -a = a and thus a + a = 2a as well right

[winged void]

thx

[tardy hedge]

@coral spindle whats up with new user

it reminded me of charli xcx lol

[coral spindle]

I just think it's funny

[tardy hedge]

did u have a brat summer boytjie

[coral spindle]

I must confess I'm not really a fan of the album, I think it's just OK

[tardy hedge]

Hahaha fair enough. I thought it was quite fun

[south patrol]

The order of a is the smallest positive n such that a^n = identity, so if a^20 = e then the order needn't be 20

[rocky cloak]

If 2a = 0, then the order of a is either 1 or 2 yes

[winged void]

but do we know that that 2a = 0 this is my question

[rocky cloak]

a + a = a - a = 0

Just like you said

[winged void]

why is that the case this does not make sense to me

[south patrol]

Well this is mostly a definition of order

[winged void]

this is possible thus

ah i see

[south patrol]

As an extreme example, take a = identity

then a^20 = identity

[winged void]

that is true but idenity power is 1

i see

[winged void]

thanks alot

jagr 2808

and @south patrol

[south patrol]

np

[rocky cloak]

Kein Problem

[tardy hedge]

How can I show that (1,2) and (1,2,...n) generate Sn?

been a while since ive looked at this stuff

do i have to decompose stuff into 2-cycles or something

[kind temple]

shuffl-uffl-ing

[tardy hedge]

Idk

[kind temple]

every permutation is a product of disjoint cycles

so if you can get all the cycles

[tardy hedge]

yea i had a feeling it was that thing

[kind temple]

you're good

[next obsidian]

It’s clear how to get (1k) for all k

And from that u get all the transpositions

And from that you get all cycles

Okay? Very good

This is a mouse with a thumbs up

A rat

[tardy hedge]

No way

[kind temple]

https://tenor.com/view/nice-very-nice-well-done-gif-19147212

[south patrol]

Whitney and induct

You have a circular conveyer belt and a machine on the belt that can flip two adjacent objects on the belt

Just keep moving stuff under the machine whenever they are in the wrong order

😌

[tardy hedge]

thx i dont like this swapping permuting stuf

@coral spindle your research field is representation theory right? Why did you choose to go deeper into that?

I have not learned anything about that yet

[warm ember]

so for a group homomorphism f:G->H, it is called an embedding if the image of f is isomorphic to G

does that mean f is injective

i cant find any counterexamples but i also don't know how to prove it

[south patrol]

Technically this isn't correct

The following are equivalent:

1. f:G -> H is injective
2. The "corestriction" f;G -> im(f) [literally the same function but make the codomain smaller] is an isomorphism

That 2) implies 1) is clear, since they are the same function (up to changing codomain).

And 1) => 2) is also clear because after corestricting it is now surjective (as well as injective)

[south patrol]

The reason I say this is that it could happen that the image "just so happens" to be isomorphic to G

[grim hawk]

hi all, I am working on homework for a graduate course in group theory and am struggling to recall definitions of bounded and unbounded sets. We are being asked to prove that SO(2,R) is bounded but SO(2,C) is not. I was under the impression that for a set to be bounded it needed to also be well-ordered, and I am struggling to recall what that entails. Any help would be greatly appreciated!

[wraith nexus]

bounded in metric space sense I assume

there being a point x and a finite M>0 s.t. all points are within M of x

[grim hawk]

thats what I was thinking, but I am unsure, given that we don't have explicit ordering in SO(2,F or C). I know that any set can be well ordered under axiom of choice, but I am not sure how to go about solving this in that regard. Plus, technically this class has not discussed boundedness or ordering at all smh

[glad osprey]

SO(2, C) is a subset of M_2(C) which is isomorphic to C^4 as a vector space (in general M_n(K) is isomorphic to K^(n^2)). So you can use the vector space norm on C^4

There's also a notion of an operator norm on M_n(K). But every norm on a finite dimensional vector space is equivalent, so I think you can choose any norm you want and get the same result

[trail cave]

Am I correct the answers here are, respectively, (2 4 8)(3 5 6), (2 1 6 7), and (3 9 7)(5 4 1 2) ?

[warm ember]

o i see

thanks

[south patrol]

np

[lethal cipher]

In the mult. group (Z/p^kZ)^x, where p is an odd prime and k>=2, I want to show that p+1 has order p^{k-1}.

Through some binomial theorem stuff, I was able to deduce that ord(p+1) divides p^{k-1}...but I'm having a bit of trouble showing it actually is p^{k-1}.

Any help would be greatly appreciated

[lethal cipher]

Okay, so i've been able to reduce it down to trying to show $(p+1)^p-1$ is not congruent to $0 \mod p^k$, and I was able to show that
$c_1p^2+c_2p^3+\dots+c_pp^{p+1}$ where each $c_i$ is not divisible by $p$.

Though I'm not sure if this is useful in showing that this is not divisible by $p^k$ 🤔

[cloud walrus]

dackid

[lethal cipher]

Unsure if the specifics matter, but just in case,
$c_i=\frac{(p-1)!}{(p-i)!i!}.$

[cloud walrus]

dackid

[lethal cipher]

Got it! k=2 this actually isn't true, but that's fine, because p is the order of p+1 in mod p^k.

But if it's not true for k=3, then it can't possibly be true in k>3

[rustic crown]

maybe proving this separately will help you clean it up a little.

If q >= p is a power of an odd prime p and you have a number g = 1 + qx with x not divisible by p. then g^p = 1 + pqy with y not divisible by p.
(if p is 2, we would actually need q >= p^2 to do this.)

and a simple induction on this statement gives that for any q which is a power of an odd prime p, and g = 1+px with x not divisible by p. we have g^q = 1+pqy with y not divisible by p.

[brisk oak]

There are 2 lemmas that make it super easy to prove.

1. If $a \equiv b \pmod{p^k}$ then $a^p \equiv b^p \pmod{p^{k+1}}$

2. If $k \geq 2$ and $p \geq 3$ then $(p+1)^{p^{k-2}} \equiv 1 + p^{k-1} \pmod{p^k}$

[cloud walrus]

🤗🤗

[hidden wind]

i want to show that if k is a field, then (x,y) in k[x,y] is not principal, but i’m a bit lost on how to do that :S

like, intuitively it’s plenty clear, but i’m not sure what to use to set up a formal argument for this

[quiet pelican]

Whatever it is generated by can’t have degree > 1, and then we clearly couldn’t get x
So it’s generated by something of the form ax + by + c
But then that vanishes on a line, so everything in its ideal does, but x, y don’t vanish simultaneously on any line

[hidden wind]

ohh lovely

thank you

[lethal cipher]

Hehe, I think I actually did this induction proof for p+1.
Also, although the things y'all are mentioning are helpful to show it's congruent mod 1, I'm a bit unsure it addresses the core issue of order I was dealing with at the time

No worries though, I was able to figure it out

[hidden wind]

polynomials are my frens, though they have often confused me

[tardy hedge]

A maximally linearly independent subset of a vector space must be a basis. But can I see an example of a maximally linearly independent subset of a module that does not generate it?

[quiet pelican]

2 \in \bZ

[lethal cipher]

Simply put if the order is less than p^{k-1}, then (p+1)^{p^{k-2}}=1modp^k, so
(p+1)^{p^{k-1}}-(p+1)^{p^{k-2}}=0modp^k, and for k>=2, but we can actually show this is not true. For k=2, p^{k-2} is just 1...so this is a non-issue in that case

[tardy hedge]

2 in Z?

Ok yea. I forget that for example for modules just always first start by looking at groups …

Thx

Ive noticed every time i want an example of somethint, its always a way simpler example than i thought was gonna happen lol

[hidden wind]

Despite the Dostoevskian title, this is nothing more sinister than a rigorous undergraduate textbook whose subversiveness consists in presenting rings before groups.

pff

from a review of aluffi's notes from the underground

[glad osprey]

I'm not a big fan of that title, for a long time I thought it was literally just a collection of notes

[hidden wind]

i haven’t had a look at the book, but i’m a big fan of the subversive act of presenting rings before groups

[somber sleet]

Can somebody help me understand why the multiplication of two ideals is always contained in their intersection?

[brisk oak]

Just check by definition, basically

[next obsidian]

Cuz if a in I and b in R, ab in I

Big hug

[somber sleet]

it's not so bad, c'mon hahah

[next obsidian]

I’m talking about the user above’s name

They’re @ big_hug

[somber sleet]

I'm just brushing up ideals, and I'm just confused

If I take $z \in IJ$ does it mean that $z = \sum_{i=1}^n a_ib_i$ for $a_i \in I, b_i\ in J$ or can it also be an infinite sum?

[next obsidian]

Finite

[cloud walrus]

damn_guuurl

[next obsidian]

You can’t make sense of an infinite sum unless you have a notion of convergence

[somber sleet]

is this the only reason why we only look st finte sums?

[next obsidian]

I guess

The sum is a binary operation and there’s no free way to lift that to an infinite input operation

[somber sleet]

okay

So I just have to accept that we don't like innfinite sums here

[south patrol]

well they don't make sense

[somber sleet]

can you elaborate or is it really only because the notion of convergence

[south patrol]

Well like yeah infinite sums are defined in terms of limits which only makes sense when you have a topological space (or smth similar)

[somber sleet]

I feel so stupid right now, I have never really noticed that and I do algebra since like 3 years

thanks for helping me though

[hollow tartan]

I think it's a valid observation

[south patrol]

np

[grim hawk]

is GL(n,R) a metric space? If so, how would one show this?

[rocky cloak]

Its standard topology can be induced from a metric yes.

How to show that I guess would depend on what definition you use for its topology.

[next obsidian]

Idk how you’d get the topology in any other way

Tbh

[cloud lynx]

Thank you guys for helping me in the past months!!! Today I wrote algebra and it was ok!

[rocky cloak]

I mean, equip R^n^2 with the product topology I guess

[next obsidian]

Lmao true I guess

[glad osprey]

There's multiple norms you can assign to GL(n, R), and each norm induces a metric

[glad osprey]

Use the standard norm of R^(n^2) for example, as I mentioned here

[arctic trail]

if you think of a matrix as a set of n column vectors with n components you can think of the n x n matrices as R^n x R^n x ... x R^n (n terms).
This naturally gives you a norm ||(v1,...,vn)||^2 = || v1||^2 + ... + ||vn||^2

[arctic trail]

which for me makes the topology on GL(n,R) as being the same as R^n^2 more intuitive for me

[tardy hedge]

I FRIKEN HATE NUMBERS!!

The amount of times id have to review “the smallest positive linear combo of a,b is the gcd” is absurd

I just never undertsand it intuitively

[coral spindle]

I don't think it's intuitive. You just learn it and move on ig

[glad osprey]

That you can write the gcd as a linear combination of a and b follows from the extended euclidean algorithm (which I think it's pretty intuitive). I can't think of an intuitive reason that it's the smallest linear combination though, but I'm sure there is one

here's one I think: let g = gcd(x, y). Then ax + by = agk + bgm = g(ak + bm) for some k and m, so a linear combination of x and y is always a multiple of their GCD

[golden turtle]

Have I started this correctly? I'm stuck on how to "deduce that this stabilizer has order p^r for some r<=l"

[arctic trail]

If you find that unintuitive, there's a neat thing

Z is a gcd domain

so Z is principal

so aZ + bZ = cZ for some c. It is obvious that gcd(a,b) divides c.

However if c = gcd(a,b)k for k != +-1 then c doesn't divide a or it doesn't divide b. But clearly cZ must contain both aZ and bZ, so cZ = gcd(a,b)Z

in particular this implies that the least positive linear combination is gcd(a,b)

[thorn ermine]

Very good explanation but to someone just starting out abstract algebra trying to get their head around gcds and bezouts that would be very confusing

[arctic trail]

fair

[warm ember]

its not intuitive?

i thoughts its intuitive

[warm ember]

smallest because factoring out gcd

[coral spindle]

If that's intuitive to you, great but like

That doesn't mean that everyone is going to feel the same

[glad osprey]

yep, every linear combination is a multiple of the GCD, I realized after I wrote it

A few years ago I made an animation to try to visualize the euclidean algorithm. It's more crappy than I remember, but I think it captures kind of how I think about it

The only thing more embarrassing than the animation itself is how much time I spent making it

[tardy hedge]

Z6 as a Z6-module has torsion elements {0,2,3,4} right? And we can see this cant be a submodule since its not a subgroup

This is just for a question about giving example of a module so that Tor(M) is not a submodule

[south patrol]

Sure

[tardy hedge]

Ya i kinda knew it was fine but i said it cuz ppl probably have better insight

Lol

[south patrol]

The torsion elements of R as an R module are just the zero divisors (and 0 if you don't count that as a zero divisor lol), and the zero divisors of Z/nZ are the things which share a factor with n

If you want to generalise the calculation

[tardy hedge]

Yea ok cool sounds good thx

[arctic trail]

I guess then that the Torsion elements of any local ring (as a module over itself) form an associative algebra

I assume, not sure

nevermind, we need the condition non unit => zero divisor which errr

maybe like a ring with a finite grading (i.e. R_n = 0 for n > N, for some fixed N)

so the maximal ideal is nilpotent

now I'm just rambling, so i guess I'll go to bed

[arctic trail]

I like that manimation, it's pretty well done

[glad osprey]

Thanks I remember struggling a lot with the euclidean algorithm, but once I got this image in my head it suddenly made sense

[golden turtle]

What does it mean "Deduce that the congruence class mod p of the number n_p,l of subgroups of G of order p^l depends only on the order of G,not on its structure"

I'm not sure what it is asking me to prove

[charred iris]

I would guess it wants you to prove that $n_{p,l}\pmod p$ is the same for all groups of order $|G|=p^km$

[cloud walrus]

Edward II

[golden turtle]

I see

[warm ember]

is the product of two objects in a category unique

up to isomorphism

is this right channel actually

[south patrol]

Yes, up to unique isomorphism

[warm ember]

o ok

[south patrol]

You can write it in #category-theory ig but this is fine

[warm ember]

ty

[south patrol]

The product of two objects has a "universal property" which distinguishes it up to unique isomorphism

[warm ember]

it is because there is 1 endomorphism that is id

[south patrol]

those are key words to google aha

[warm ember]

so any morphisms A->B and B->A have composition identities

on both sides

[south patrol]

Yes

[south patrol]

Exactly

[warm ember]

thanks

[south patrol]

Note that like

[warm ember]

o ye

[south patrol]

these A -> B and B -> A are unique too

So there is a unique isomorphism [subject to the fact that they are compatible with projections]

[south patrol]

Note yeah I should underline that this is one endomorphism which is compatible with the projections

There may be tons of other endomorphisms which don't have this property

[warm ember]

ye

[south patrol]

Nice

[dense root]

hey everyone

[celest cairn]

Hey 🙂

[tardy hedge]

This “induced map” is just phi’(m+IM) = phi(m)+IN?

How can I be sure that things are well defined and still a homomorphism?

I guess i could manually check but i am wondering if there is a higher level observation i can make

[grave sedge]

Composing phi with the projection you get a map from M to N/IN, it's easy to see IM is contained in the kernel of this map so you get a map from M/IM to N/IN

[tardy hedge]

Hm

Oh the N -> N/IN projection

[tardy hedge]

Oh ok cool so this was showing to me how we can see that this map is all well defined and still a hom

Because we can make it by just composing other homomorphisms

I suppose when things are saying “induced” maps its referring to some compositions

[tardy hedge]

I see IM is contained in kernel of pi o phi, but how does this induce a map

From M/IM to N/IN

[hidden wind]

which book is this?

[golden turtle]

I could still use help with this

I’m struggling a lot with deducing n_pl is 1 mod p.

[tardy hedge]

Im confused on this still

[wraith cargo]

this is essentially a weaker version of the first isomorphism theorem
Basically the idea is that if I is a subset of the kernel then the original morphism factors through this induced morphism

the morphism is induced how you think it is

phi-bar(a + IM) = phi(a) + IN

it's sort of just a sit down and check it's well defined sort of calculation

[tardy hedge]

Yea i thought this was the “induced” map but it made me uncomfortable cause i actually dont even “really” know whats happening yk

I just automatically thought ok its probably this

[tardy hedge]

I is a subset of kernel?

[wraith cargo]

It's not

I was using a different I

my mistake sorry

I meant that IM is a subset of the kernel

[tardy hedge]

IM subset of kernel?

Yes this is wherr i got stuck

In terms of understanding what is happening

I have this map now M -> N/IN I got from composing projection with phi

Ok that makes sense sure

The kernel of this guy contains IM, ok sure

Now i got stuck wasnt sure what to even think of really

[regal zodiac]

[wraith cargo]

Think of a diagram

here phi is your original map

phi-bar is the induced map

and pi is the projection

this is how phi-bar fits into the picture

the original morphism factors through it essentially

[tardy hedge]

Ok this is quite helpful because i am not yet thinking in terms of diagrams and factoring through and stuff

[tardy hedge]

Phi is from M to N tho?

[wraith cargo]

here I already composed it with the projection N -> IN

[thorn ermine]

Hey, anyone able to explain how to do this to me clearly?

Have no idea how to understand why it works the way it works

[tardy hedge]

Does this picture tell us how phi bar is defined?

[wraith cargo]

nope
But I already gave you the definition earlier

[tardy hedge]

Right, so this picture just illustrates how its being “induced” i suppose

[wraith cargo]

yeah basically

[tardy hedge]

Ok thanks. So now i actually have to try the question lol

I really appreciate the help btw

[wraith cargo]

the idea is to to show whether or not an input can spit out two different results

so take two classes in your group or whatever

choose some representatives for each class

and see whether or not you always just get one answer out of the function

or you're able to get two different ones

[tardy hedge]

So, i still need to justify that phi bar is 1) well defined and 2) a mod hom still right?

[wraith cargo]

Ig they sorta assume it already but if you want you can go and check yourself

[thorn ermine]

Ok so for example, if I chose these two to prove that one holds in Z4 but doesn’t hold in Z5, then it would be a correct way to prove by counterexample right:

[wraith cargo]

idk what you wrote there lol

[thorn ermine]

Basically using 0 and 4 as a and b

Which holds for Z4 but not for Z5

[wraith cargo]

anyways idk I'm gonna show you how to do it for Z_4 to get the idea
Take [0] * [0] in Z_4
If we choose a = 0 and b = 0 as representatives we get that they're clearly congruent mod 5 so we have that [0] * [0] = 1
But if we choose a = 0 and b = 4, we have that they aren't congruent to each other mod 5, so we also get that [0] * [0] = 0
Hence * isn't well defined on Z_4

[thorn ermine]

So we chose a = 0 and b = 0 as representatives to show that they’re congruent in Z5? Because 5 x 0 = 0 remainder 0?

I mean wouldn’t that work for Z4 as well because 4 x 0 = 0 remainder 0 as well?

[wraith cargo]

uhhhh

I'm not sure I fully understand what you mean

[thorn ermine]

Ok so like you said [0] * [0] = 1 for Z5 right

[wraith cargo]

No for Z_4

[thorn ermine]

I see

But wouldn’t it also work for Z5 is what I mean

Therefore they both give us 1 according to the question

[wraith cargo]

yeah in Z_5 we'd have that [0] * [0] = 1 always
But you should argue yourself why this is true

[thorn ermine]

I mean it wouldn’t really give us a counterexample then tho right

Because they’re both the same

As you said I think [0] and [4] would work

[tardy hedge]

I got that phi(a^(n-1) m) = a^(n-1)n

[thorn ermine]

But then my question is, why are allowed to use [4] even though it’s not in Z4? Z4 only has [0], [1], [2], [3] right

[tardy hedge]

Assume I nilpotent with n that makes I^n = p

0

[tardy hedge]

Can I show phi is surjective somehow from this?

[tardy hedge]

For any n in N, i know this

[tardy hedge]

[4] is in Z4, its the same as [0]

Thats kinda the point of the question, since we can represent elements of Zn with different numbers, we need to verify that functions are well-defined

[thorn ermine]

Sorry could you kind of elaborate on well defined a bit more because I’m really confused about that

Because why should we be allowed to use [4] for that even tho Z4 specifically doesn’t have it

[tardy hedge]

Z4 does have [4]

Elements of Zn are like equivalence classes

[thorn ermine]

I mean like shouldn’t we be only allowed to work with

[0], [1], [2], [3]

[tardy hedge]

Well idk cause we tend to define functions by using representatives of the equivalence class

[thorn ermine]

Ok yeah then I’ll just work with that

[tardy hedge]

So if we define functions by using representatives of [0] (like 0, 4, 8, etc) then we need to be sure that all of this maps to the same element

Or else it wouldnt be a function

[thorn ermine]

So I can provide a counterexample to prove that [0] and [4] don’t work in Z4 and Z5 by counterexample but how can I prove that Z5 is a well defined operation in the general case then?

[tardy hedge]

Trying to show phi is surjective. I took something in N/IN , n+IN and we know there is a phi(m)+IN = n+IN so phi(m)-n is in IN, so phi(m)-n = a1n1 then since I is nilpotent say for k then if u multiply by a^(k-1) you get phi(a^(k-1) m) = a^(k-1)n

Im now stuck

[glad osprey]

You need to show that if a and a' is in the same equivalence class, and b and b' is in the same equivalence class, then [a]*[b] = [a']*[b']

[golden turtle]

I still could use help with this

[thorn ermine]

Makes sense: so if I proved that 0 ≡ 0 mod 4 and 0 ≡ 4 mod 4 are both in Z4, And therefore [0] * [0] = [0] * [4] in Z4 but not in mod 5, would that suffice as a proof by counterexample?

[errant wedge]

Sanity check, Frac(Q[x]) = Q(x) right?

Or in general for any field actually

[thorn ermine]

Waterloo pmath student 😥

[coral spindle]

Yup!

[glad osprey]

I don't know what you mean by [0]*[0] = [0]*[4] in Z4 but not in mod 5. In Z4 we have [0] = [4] so we would expect [0]*[0] = [0]*[4], but [0]*[0] = [1] while [0]*[4] = [0] (since 0 != 4 mod 5)

[errant wedge]

Yippee ty

[tardy hedge]

A hint in right direction would be cool

[errant wedge]

Yes 👀 u here too?

[thorn ermine]

Makes sense, thank you!

[thorn ermine]

Nah uwo

A surprising amount of Canadians in this server

[tardy hedge]

Also canadian here

[thorn ermine]

Ooh in uni?

[tardy hedge]

Yea

[thorn ermine]

Which

[errant wedge]

Ah W

[tardy hedge]

Its a smaller school so

[thorn ermine]

Nice

[errant wedge]

u go to CUMC this year?

[thorn ermine]

Nah I didn’t even hear about it 😵‍💫

If it was truly at ubc there would be no way I would be allowed loll

[errant wedge]

Does western not fund some students?

some of my friends here didnt get funding and bought their own tickets 😭 wild to me

[thorn ermine]

No idea but I didn’t even hear about the opportunity this year

[errant wedge]

ah rip sorry

[thorn ermine]

Yeah but if it’s cool and close to Ontario next time I’ll defo try and snag a way to attend

I remember western held it online two times during Covid lmao

[errant wedge]

2021 comes up online

[errant wedge]

UW put in a bid for 2025 so it might be near 👀 (tho tbh waterloo's a kinda lame place to host it 😭)

[thorn ermine]

😯

Waterloo is kind of close to my place I’d love to attend

Plus I’ll get to tour the cheriton library thing 👀

[errant wedge]

oh DC is neat. Imo DP is more impressive as a library but ye

[thorn ermine]

👀

[tardy hedge]

Its true that IN contained in I^2N contained in … etc right?

Since phi bar is surjective, take something in N/IN (n+IN) and you have for some m, phi(m)+IN = n+IN

Then phi(m) - n is in IN so its also in I^nN = 0 (I is nilpotent)

So then phi(m) = n and phi is surjective ?

[tardy hedge]

Uhh nevermind its the other way round lol

Nevermind !

[tardy hedge]

Ok so phi(M)+IN = N

IN is a subset of I^2N+phi(M)

So then I can say I^2N+phi(M) = N?

Then repeat until ideal gets killed off then phi(M)=N?

[tardy hedge]

This part was only possible because phi bar is surjective

Im now just trying to take a step back and think about what was needed where

[golden turtle]

I could still use help on this

[lime junco]

ok,

if H and K are normal in G

and K contains H

is the map G/H -> G/K
gH -> gK well defined?

i just wrote a proof and realize

i never verified well definedness of maps

and im not sure if it true now that i think about it

wait no im silly it should be, because aH = bH implies a - b in H which implies a - b in K which implies aK = bK

nvm 😄

[static quarry]

what are some general techniques for determining the lattice of subgroups of a finite group? More concretely, how would you approach the problem of drawing a Hasse diagram for the subgroup lattice of S_5?

[grave sedge]

I guess one idea could be to find what the maximal proper subgroups are

Because if you know the subgroup lattice of the maximal subgroups then you can construct the one for the whole group

[coral spindle]

Well unfortunately doing so is still very hard. As my friend who studies maximal proper subgroups of E8 says in his talks, the general problem is "hopeless." You will probably simply need to use ad-hoc methods to determine the lattice of subgroups of any group.

[regal zodiac]

Is C[x,y,z]/(z^2024) of finite length as a C[x,y,z] module ?

[wraith cargo]

nope it's not artinian

since it has dimension greater than 0

oh wait

as a C[x,y,z] module

fml lol

I thought as a C module

[regal zodiac]

Would u have any idea about how to compute it's length?

[tough raven]

Since C[x,y,z]/(z^2024) is a cyclic module and C[x,y,z] is commutative, C[x,y,z]/(z^2024) is Artinian as a C[x,y,z]-module iff it's Artinian as a C[x,y,z]/(z^2024)-module, so your argument is valid.

[wraith cargo]

maybe there's some calculation using like grobner bases?

[rocky cloak]

In general just cooking up a composition series will do

[regal zodiac]

Yes but it seems impossible to do so it this case

[rocky cloak]

I mean you just start with some submodule, then see if you can find one that contains or is contained in it.

Like okay you got (x) for example, and it contains (x^2) which contains (x^3), ... yup that's infinite

[regal zodiac]

I got to compute the length of C[x,y,z] / ( x^3 +3x^2 + 2xy, y^2 -1, z ^2024)

So far, i tried to consider some short exact sequence to decompose the problem but i got stuck

[regal zodiac]

Hum it seems complicated in this case no?

[rocky cloak]

It's a little more complicated I guess.

You can start by taking out (z) at least, that's not gonna do a difference.

After that I'd probably look at (y+1) or (y-1)

And just chip away one element at a time

[regal zodiac]

Okay i will try that, what do u mean by taking out z?

[rocky cloak]

0 -> (z) -> M -> M/(z) -> 0
is a short exact sequence

And (z^n)/(z^n+1) = M/(z), so the length of M will just be 2024 times the length of M/(z)

And M/(z) you can just consider as a C[x, y] module, so that cuts down on the variables

[regal zodiac]

I Will take a look and Come back to u thank u very much

[sly crescent]

E8 as in the compact Lie group?

[coral spindle]

The finite group of lie type, really

[unkempt stream]

If anyone is bored and likes basic group theory: let G be a finite group, and let S be a subset of G with cardinality coprime to G’s order. Show the only element of G that sends elements of S to S under left/right multiplication is the identity

[wraith cargo]

Doesn't this just follow from orbit stabilizer?

[unkempt stream]

Yeah lol

[candid dove]

Okay but what is the action here tho

Left g multiplication on S?

And orbits are gS kind of stuff?

[pine wigeon]

Hi guys! I have no idea how to proceed on this exercise. It says "Proof that if V is a commutative K-algebra without 0 divisors, where each element is algebraic over K, then V is a field" I used the property that every element is algebraic, but then I don't know how to proceed in order to conclude that there exists inverses. Any help would be appreciated

[regal zodiac]

Hello, i'm sorry but i really struggle with length computation, i have to compute the length of A = C[x,y] / ( x^3 + 3x^2 + 2yx , y^2 -1)
So i consider the following short exact sequence :
0 -> ([y+1]) -> A -> A/([y+1]) where [y+1] is the class of y+1 in the quotient.
to get length (A) = length ( A/([y+1]) + length ( [y+1]) , i managed to compute length ( A/([y+1]) but i can't manage to compute length ([y+1]) , i can't see how to do it

[rocky cloak]

Think about the map C[x, y] ->> ([y+1]). Are you able to describe the kernel?

[rocky cloak]

An element is algebraic means that it's the root of a polynomial, see if you can use this polynomial to construct the inverse

[regal zodiac]

I'm sorry but which map are u talking about ?

[rocky cloak]

The one sending 1 to y+1

[regal zodiac]

What i tried is to find the submodules of C[x,y] I such that (x^3+3x^2+2xy, y^2-1) is included in I which is included in Q^-1( ( [y+1]) ) where Q is the quotient map
I found Q^-1( ([y+1] ) ) = ( x^3 + 3x^2 + 2xy , y+1)

But the problem is that i found an infinite number of such I , for n , take : I =
(X^3 + 3X^2 + 2XY, Y^2+1, X^n(Y+1) )

And it does not make sense , i shouldn't be able to find an infinite number right @rocky cloak ?

[regal zodiac]

I Will try to think about it

[unkempt stream]

Yeah. The natural action on the power set

[candid dove]

Okay still unclear on how it follows from orbit stabilizer.....

[unkempt stream]

What are the orbits :3

[candid dove]

Uh idk |S| size subsets?

[unkempt stream]

Well not immediately from orbit stabilizer

Better way of doing it is to realize how Stab(S) acts on S