groups-rings-fields

[thorn jay]

In fact, I is prime iff R/I is an integral domain iff (0) is a prime ideal in R/I

[glad osprey]

So what's the relation with reduced rings?

[thorn jay]

A ring is reduced if the intersection of all prime ideals is (0)

Aka, if x^n = 0 for some n => x = 0

Oh i didnt see this lmao

[mighty kiln]

Oop I should go back to my algebra course

[thorn jay]

Ring theory has a lotta definitions lol

[glad osprey]

Lol, got confused for a second, but thanks

[thorn jay]

[dull tiger]

Then is there a quick way to see if $(\bar{m})$ is a prime ideal in $\mathbb{Z}/n\mathbb{Z}$ ? For now i've always checked the elements in the ideal but that gets quite tedious if you start to look at bigger $n$

[cloud walrus]

mathrie

[arctic trail]

what is bar m?

the congruence class of an integer m?

[dull tiger]

yes right. sorry, i should have specified. I meant an element in the ring $\mathbb{Z}/n\mathbb{Z}$ that generates that ideal

[cloud walrus]

mathrie

[quiet pelican]

I think it's iff m = p mod n for some prime p dividing n

[next obsidian]

Prime ideals of a quotient are in bijection with prime ideals of the original ring containing the quotiented ideal

So this happens if and only if m is a prime factor of n

[arctic trail]

You can use the chinese remainder theorem

to find them

[arctic trail]

^

[arctic trail]

an ideal I of a ring R is prime iff R/I is an integral domain, or in this case a field since it's finite.
Decomposition into the product of rings using the CRT makes it easy to see what happens

[next obsidian]

I guess if you are working off of when m is such that (m-bar) is prime you have to ask when m mod n is a prime factor of n

So you just apply division algorithm to reduce m to between 0 and n-1

[next obsidian]

Which is this

[arctic trail]

Two steps of Euclid's algorithm is enough

[dull tiger]

So lets say i have $(\bar{5})$ in $\mathbb{Z}/21\mathbb{Z}$. Then that is not a prime ideal since 21=3*7. The prime ideals would be $(\bar{7})$ and $(\bar{3})$

[cloud walrus]

mathrie

[next obsidian]

Yeah

[arctic trail]

(5) is (1)

[next obsidian]

(5-bar) is actually the entire ring

[arctic trail]

^

[next obsidian]

Cuz it’s coprime to 21

[arctic trail]

though the entire ring is prime depending on the definition.
I think some include not being the entire ring in the axioms, while some don't

[next obsidian]

No

The entire ring should not be prime

This is bad

[dull tiger]

yes, we included that in the definition. The ideal can't equal the ring

[arctic trail]

?

depends on the place you read it

not sure why you argued with me I'm just stating some inconsistencies I've seen in the definition

[next obsidian]

I can think of a million places this entirely destroys statements of things, and have never once seen a prime be allowed to be everything

[arctic trail]

Why're you writing that with such passion

[next obsidian]

I think definitions can be bad and letting (1) be prime is a bad one

[arctic trail]

Okay?

I never said that I consider it to be prime

[next obsidian]

I never insinuated you did

[arctic trail]

In Wolfram Math World this definition includes the whole ring

[next obsidian]

I’m saying that definition is wrong

And should be corrected

[arctic trail]

Okay?

That's your belief

[next obsidian]

It doesn’t reflect its usage in anything else

[spark veldt]

i think i need more clue idk what to do 😭 how do we send quotients into K? Like the countability of the rationals argument, the zig zagy thing?

[next obsidian]

I study commutative algebra which concerns itself heavily with prime ideals and am stating that one shouldn’t ever consider (1) to be prime, that’s it

[arctic trail]

Okay

I am quite aware of the importance of prime ideals in commutative algebra

But at the end of the day the author can choose their own definition as long as they're consistent and clear on what they're writing

since you study algebra, surely you've seen examples of non-equivalent definitions with the same name

[next obsidian]

That’s silly, and one shouldn’t do that when the entire rest of the literature doesn’t do that. I could say that a continuous function is what everyone else refers to as a differentiable function but that would be bad

[arctic trail]

I've seen that to be specially true for things that generalize commutative ideas to non-commutative settings

[next obsidian]

I said what I did to clarify that if you see prime in virtually any context, you should know that it is going to mean proper, unless you can show me that in noncommutative settings it’s preferable to let it be equal to (1)

[coral spindle]

Where do you think you send the formal quotient a / b to in K?

Remember a and b are in K.

[next obsidian]

In that case I learned something and I’ll concede, but a statement like this I think just leads to confusion

If it isn’t actually used anywhere

[spark veldt]

Does this mapping need to be bijective?

[arctic trail]

🤷 I don't know what their professor has in their definition

it's not completely absurd that they don't have that in it

in which case I found it important to state

[next obsidian]

I just don’t think this is worth even considering, like I don’t worry if the professor defined a function as something non-standard

[arctic trail]

womp womp, I don't want to continue discussing this. Have a great rest of your day

[next obsidian]

Kk

[rocky cloak]

remember that if a field contains a and b, it contains a/b

[next obsidian]

Oh yeah? What’s 2/0?

[spark veldt]

do we send like a/b to ab?

[next obsidian]

This is no longer additive

[spark veldt]

but this is not a bijection i think is it ok?

[next obsidian]

It doesn’t do what you want it to do

[rocky cloak]

I guess youre missing some intuiton for how F is constructed

[coral spindle]

No

[coral spindle]

So you really want to send e.g. 1/b to b?

[rocky cloak]

Maybe think about the case when D = Z is the integers

[spark veldt]

In that case F is constucted by choosing soe integer a and b and forming their quotient a/b right

then you have the rationals

[coral spindle]

Let's say K = C, the field of complex numbers

so can you think of a map F -> K in this case

[rocky cloak]

Where would say [2, 4] get sendt?

[spark veldt]

to a + bi?

[rocky cloak]

What is a+bi ?

[coral spindle]

Have fun yall

[arctic trail]

dawg if K is a field it's closed under fractions with non-zero denominator.
What is the field of fractions of D

[rocky cloak]

Maybe back up even more.

Do you know how D sits inside F?
@spark veldt

[spark veldt]

D is the integral domain right

[rocky cloak]

Yeah

[spark veldt]

D is a subset of F i think

[arctic trail]

the field of fractions of D is the set of fractions with numerator and denominator in D with non-zero denominator

[rocky cloak]

Yeah, at least isomorphic to one in a canonical way.

[arctic trail]

K contains D, what do you conclude

[rocky cloak]

So F consists of these equivalence classes of pairs [a, b].

And if D = Z you said earlier that F should be the rational numbers.

Okay, so which pair corresponds to for example 1? What about 2? What about 1/2?

[spark veldt]

[1,1], [2 1], and [1 2]

[rocky cloak]

That's right. So in general
[a, b] is which rational number?

[spark veldt]

a/b? oh..

[rocky cloak]

Yup

So you see how to solve the exercise?

[spark veldt]

so this mapping sends [a,b] to a/b, now I just show that these ‘fractions’ form a field?

[rocky cloak]

Yeah, or that this mapping is a homomorphism

[spark veldt]

oo right, wow ok thanks. I think the mistake I had was that I already viewed [a,b] as a/b, then got really confused as to how I was supposed to send a/b to K when I should’ve been looking at the equivalence class itself.

Thank youu sorry for taking a loong time to understand 😭

[glad osprey]

I guess a/b is unambiguous in a field, but I really prefer writing ab^(-1), not sure what other people think

and just to be sure, it's a good idea to check that the map is actually well-defined, for example f([a, b]) = ab is not just not a homomorphism, it's not even well-defined @spark veldt

[amber burrow]

I think i might be misunderstanding this problem: Prove that there exists a constant c>0 such that in every nontrivial finite group G there
exists a sequence of length at most cln |G|with the property that each element of G equals the product of
some subsequence. (The elements of G in the sequence are not required to be distinct. A subsequence of a
sequence is obtained by selecting some of the terms , nont necessarily consecutive, without reordering them;
for example, 4, 4, 2 is a subsequence of 2, 4, 6, 4, 2 but 2, 2, 4 is n

(Putnam 2008 A6)

cant you just pick your list to be e,g_1, g_1^2,..g_1^n,g_2, g_2^2,...g_2^n,...g_n,...g_n^n

where g_k is the kth generator

and n = |G|

the list has n^2 + 1 elements

cln(n) = n^2 means c = n^2/ln(n)

so you can always find a c that works

oh its because it might not be abelian

[lusty marlin]

Seems to me that you want a c that works for all nontrivial finite groups G

[narrow wagon]

Suppose K be a field extension of F, and let B= {b1,b2,b3} be the basis of K over F

is it correct to say K = F(b1,b2,b3)?

the smallest field containing b1,b2,b3

I’m trying to prove this

[thorn jay]

Yes

I wouldnt see why not

[spark veldt]

Let $I$ be an ideal that consists of all polynomials in $\mathbb{Z}[x]$ with constant divisible by $7$. Show that $I$ is a maximal ideal using the definition.

[cloud walrus]

bluepianist

[spark veldt]

I'm not really sure how to approach this

[barren sierra]

what is the definition of a maximal ideal?

[spark veldt]

$M$ maximal if for $I$ s.t. $M\subseteq I\subset R$ then $I=M$ or $I=R$. I tried letting $I$ be some arbitrary ideal s.t. $M\subseteq I\subset R$, not sure where to go from here though...

[cloud walrus]

bluepianist

[barren sierra]

Well if I = M, you're done there's nothing to show

so suppose I is not M and show that I = R

[spark veldt]

do I need to do the reverse?

[barren sierra]

what reverse?

[spark veldt]

assume I=R then show I=M?

[barren sierra]

no

how do you prove statements of the form P implies (A or B)?

[spark veldt]

P implies A or P implies B

[barren sierra]

exactly

but we can do better actually

One really common way is to of course assume P is true, then say "If A is also true. Then we are done. So suppose A is false. Then we want to show B is true"

you've surely seen this before

essentially using the fact that P implies (A or B) is equivalent to (not(A) and P) implies B

[spark veldt]

got it

thank you spamakin

[barren sierra]

yea

[barren sierra]

so do you see why this suffices?

[spark veldt]

yeah i do

[barren sierra]

🔥

[spark veldt]

tysm,

[spark veldt]

I'm wondering why this is so?

How did they make the multiplicative inverse appear? It's not a field?

[storm kiln]

alpha is a real number, and they assumed it isn't 0

[spark veldt]

ohh i see

thanks!

This is my attempt I think it's not fruitful. How should I repair it?

[spark veldt]

I wanted to imitate the proof on Ex. 3.5.2 (post just above this) but since the polynomials have coefficients in Z I can't get the inverse and conclude 1 is in I, hence I=R.

[storm kiln]

Inverses do exist in Z/7Z though (for nonzero elements)

[void cosmos]

a quick y/n

every group generrated by 2 elements a,b of order 2 is solvable by modding out by <b|b^2=1> and using that if N and G/N are solvable then G is solvable

correct?

[coral spindle]

Why must <b> be a solvable subgroup?

sorry whoops, I mean a normal subgroup?

I.e. why can you mod out by <b>?

E.g. in the group with presentation G = <a, b | a^2 = b^2 = 1>, the subgroup <b> does not contain aba (though I may be mistaken about this -- hopefully someone can confirm)

[kind temple]

<b> = {1,b} right?

[coral spindle]

Yeah

Did you mean something else when you wrote <b | b^2 = 1>? I can't see what it would be.

[kind temple]

no, i was misremembering a fact about normal and abelian subgroups

[coral spindle]

I was referring to this lol

[kind temple]

yea, i was trying to think about it too

[coral spindle]

Also new username and pfp, interesting

Yeah I’m pretty sure this proof is wrong as written, but I’m not sure if every group generated by two involutions is solvable or not

I think the derived subgroup is <ab>, whence it is solvable, but I have not convinced myself of this yet

[void cosmos]

yes i meant <b> ={1,b}

[void cosmos]

i actually dk ur right

i think ur right yes

<ab> is abelian

[void cosmos]

^

and the quotient is abelian too

yeah thank you!

[coral spindle]

I am not sure this is true, to be clear

[void cosmos]

doesn't the quotient have index 2

[coral spindle]

Oh, you’re right. Nice

[void cosmos]

ty

for ur help

[spark veldt]

Suppose $M$ is a maximal ideal and $ab\in M$. Can you say that $M=ab+M=(a+M)(b+M)=MM$ hence $a+M=M$ or $b+M=M$, which gives $a\in M$ or $b\in M$?

[cloud walrus]

bluepianist

[spark veldt]

I'm particularly wondering if the MM to a+M=M step is legal

[quiet pelican]

Assuming what you’re trying to prove is “maximal implies prime”, that step is either circular, or wrong (leaning more towards wrong, based on this comment)

[trim wind]

No, just look at M = (2) and a=b=2

[south patrol]

(a+M)(b+M) = abM + aM + bM + MM

there is no reason that should MM

Also uh off the bat, m = m^2 is an equation which one would usually expect not to hold

e.g. if m is f.g. then this is doesn't hold by nakayama unless m = 0

[coral spindle]

psst potat you added an extra M

abM

[rocky cloak]

k^2 would like a word

Though Nakayamas lemma implies that m is generated by an idempotent.

[south patrol]

sorry lol ye

in a domain i guess

but yeah

[rocky cloak]

Could be (1) as well. Though I guess m was assumed to be a maximal ideal or something

[south patrol]

i guess what i had in mind is that in p-adic hodge theory there is a common local domain where m^2 = m

and then you're like okay that's super non-noetherian

(or a field)

[chilly ocean]

I have read a lot online about the significance of the finite simple groups as "building blocks" for all finite groups much in the same way primes are the building blocks of the naturals. I'm now reaching the end of my second course in abstract algebra and I feel no closer to understanding why this is. In the case for primes and naturals I understand that any natural number can be uniquely factored into a product of primes. For a given finite group G, what would the equivalent statement be in terms of simple groups? I've tried searching online and not found many satisfying answers

This particular topic is one that interests me maybe more than anything else I've encountered in math so far, I'm endlessly fascinated by the classification of finite simple groups and just how complicated it is, I would just like to understand its significance a bit better out of curiosity

[rocky cloak]

The Jordan-Hölder theorem: every finite group has a composition series, and the composition factors are uniquely determined by the group.

(i.e. doesn't depend on choice of composition series)

This means that every group can be built as iterated extensions of finite simple groups, and which simple groups make up a given group is unique. Similar to how the primes that make up a prime factorization are unique.

[chilly ocean]

I see, I had encountered this theorem during class but I didn't realize that all groups had such composition series (in my class they were called complete subnormal series which I think may have added to the confusion for me)

Thank you

[rocky cloak]

The difference between groups and numbers is that groups can be combined in more interesting ways than simply "multiplying"

You have the direct product, but also semidirect product, central extensions, and other types of extensions, and there is interesting theory understanding these ways to build groups

[young thorn]

let's say you prove a map is a linear map

what would be a good concise concluding sentence to ur proof

[bitter rover]

Hoopity doopity dap, and that's why we have a linear map.

[solemn garden]

“by the above, the map is additive and homogeneous. QED”

Is the union of every chain of completely reducible modules (ordered by inclusion) also completely reducible? If so, how can this be proved? I got stuck when finding the simple direct summands of the union

[rocky cloak]

Yes, this is a result about semi-simple modules:
If a module is a sum of simple modules it's a direct sum of simple modules.

You can prove it by Zorn's lemma. Consider the set of all simple submodules, and then consider the poset of subsets for which their sum is direct.

[rocky cloak]

"And thus, as we have meticulously expounded through an extensive array of dialectical reasoning and corroborative attestations, the cartographic representation under consideration can be incontrovertibly ascertained to possess linearity in its constitution."

[mighty kiln]

Surely this doesn't require choice

[rocky cloak]

It wouldn't surprise me if it did, but idk

I mean the fact that every vector space has a basis is a special case

[mighty kiln]

Wait how

[rocky cloak]

A vector space is a sum of simple modules

And a basis is precisely writing your space as a direct sum of simple modules

[mighty kiln]

Oh you take the sum over all non zero vectors

[rocky cloak]

Sure

[mighty kiln]

Okie

[rocky cloak]

Now what soup_norm said is sightly weaker, but feels like it should be choicy as well. At least like countable choicy

[next obsidian]

Arki red fox

[mighty kiln]

Chmonkey chair monkey

[rocky cloak]

Alright, 4am. Time for everybody to post other people's names, make sense

[cloud walrus]

nastasya

nastasya

[serene dune]

cant show the equality

how is 2x=0

[rocky cloak]

There should be a minus sign somewhere.

Either v(x) = (x, -x) or pi(a, b) = a-b (or negatives of these)

[serene dune]

oh the question was...

[solemn garden]

I see, thanks, I’ll try that

[serene dune]

and what is that do not work

was he consuming alcohol at late night while constructing this problem

[rocky cloak]

I mean the hint says quite clearly that using those maps will not work, and that you have to introduce a sign somewhere

[serene dune]

yeah now i get it, but why is it in a hint, i feel the problem statement could be better

that is no hint ig

AxB \to A+B

is that not canonical ?

maybe im worng

[rocky cloak]

Yes, there are canonical (i.e. obvious) maps

[serene dune]

[rocky cloak]

But the problem isn't saying "prove that this canonical sequence is exact"

They're asking you to find one

[serene dune]

i looked at the definition of 'establish'

sorry i was drunk

[serene dune]

is that a film reference ?

[rocky cloak]

Yeah, classic Jackie Chan movie

[serene dune]

now i feel why i need to be a linguistic wizard before doing anything serious with maths

[rocky cloak]

Sequel to Drunken bachelor ||jk||

Prequel to Drunken PhD

[serene dune]

jagr, can i get some book recommendation ?

idk what im trying to do

im not in any school

nobody told be to do maths, im about to be 22

[rocky cloak]

Well I'm terrible with books, but you can ask I guess

[serene dune]

oh, i dint have a clue with what im tryna do with algebra

if not book atleast tell me what is out there

topic wise

trying to learn module atm ig

[rocky cloak]

I see, well I'm in representation theory of finite dimensional algebras. So it's like modules, but with more linear algebra I guess.

Galois theory is pretty clean.

From commutative algebra you can branch into like number theory stuff or algebraic geometry

Representation theory of finite groups is also pretty nice, like character theory and stuff

[serene dune]

well alg geo is hot topic nowadays, but as far as i have done maths i think i like algebra more than anything

yeah i tried to get into some basic character theory, but i couldnt keep myself motivated enough

[rocky cloak]

Could always learn some category theory and homological algebra stuff. People usually recommend Algebra Chapter 0

[serene dune]

whats this homological algebra, is that something that we need in alg geo too ?

[rocky cloak]

Yeah, it's essentially the study of exact sequences

Useful for alggeo, algtop and rep theory

[serene dune]

alr

whats the pre req ?

idk, i feel lost because i dont have a school or a prof to give me work

[rocky cloak]

Well, you want to know the basic of modules I guess. Then pick up some cat theory on the way

[serene dune]

ok, noted

[rocky cloak]

Here's an archive of bachelor and master thesises in algebra
Algebra | Oppgavearkiv https://search.app/vTcqdcXwKaeUnDcC7

[serene dune]

can i tackle commutative algebra too ?

[rocky cloak]

I'm sure you can

But I guess you can't do everything at once

[serene dune]

yeah, takes time

tryna to do topo too at this moment

but havent touched in a week

and words for topo ?

trying to solve this

feels hallucinating

[serene dune]

https://media.discordapp.net/attachments/1054780485408133170/1327127712967557230/attachment.png?ex=6781ef7f&is=67809dff&hm=925a18739376d754417a7b1c7a1ce92b70402e2c57372b2de3b160aee74a2ecb&

[woeful sage]

would that work?

[serene dune]

well take the root of x^4=1

and show that x^k \neq 1 for all k<4

mind if the order is w
any integer multiple of w satisfies the condition

thats a topic closely related to the proof of existence of infinite prime number through furstenburg topology right ?

[lusty marlin]

Yes

[thorn jay]

Algebra chapter 0 my beloved

[woeful sage]

what

[woeful sage]

are you saying yes to my proof that R has no element of order 4?

[serene dune]

someone posted about profinite integers and it vanished

[woeful sage]

I saw a topology related number theory exercise in the Niven, Zuckerman and Montgomery, I should try to do it sometime

[sturdy spear]

I Love Algebra !
(Haven't studied AA yet but I'm badly curious to stdu

[lusty marlin]

Yes

[woeful sage]

Have you finished Abbott and Rudin

[sturdy spear]

No

[serene dune]

does anybody do zorich here ?

[woeful sage]

This shows us that the direct product of groups is abelian and associative

[sturdy spear]

Zorich analysis?

[woeful sage]

does that mean w.r.t the direct product, the category of groups can be turned into a group (not an actual group since the underlying collection isn't a set, but a group like object)

[serene dune]

yeah as both A,B are normal subgroup of the mother group

[serene dune]

yep

[sturdy spear]

Maybe i use it for some topics. But using Abbott and rudin.
Hbu

[serene dune]

bartle and rudin mainly

[sturdy spear]

Cool

And how about AA

[serene dune]

but i would like to solve zorich

oh idk in AA, mainly all over the internet

[woeful sage]

Dummit and Foote is the best

[serene dune]

currently doing modules from bylth

[woeful sage]

(objectively true statement)

[serene dune]

bible like book

[sturdy spear]

I see. A famous book

[serene dune]

i like when problem set does it for me

not only the proof but also the vision, ehh its hard to word sensantions

[sturdy spear]

Category of groups of categories of groups

[sturdy spear]

I see, i got it.

[hidden wind]

i’ve been meaning to!

[serene dune]

should we consider creating a thread, count me in

[hidden wind]

from experience i know i wouldn’t engage much in that, but feel free to dm me with or about zorich things

anyone got any suggestion for some rings to think about in looking for how to find a proof of this claim: the set of prime ideals in any cring has a minimal element

[woeful sage]

what is a cring?

[hidden wind]

the domain and finite cases are obvious but i dunno how to to do an argument for the general case

commutative ring with identity

[woeful sage]

ah okay

[serene dune]

[woeful sage]

ring with identity

[serene dune]

i used to say CRU

[woeful sage]

...you don't include identity in your rings

[serene dune]

analysis...

[woeful sage]

that's an interesting proposition

wait how is the minimal element defined here?

minimal as in cardinality?

[hidden wind]

wrt inclusion

[woeful sage]

ah okay, so alternatively there is a prime ideal which is contained in every prime ideal of the commutative ring?

[hidden wind]

no

that’s only in the uninteresting integral domain case

where (0) is prime

actually no i don’t know that is the only such case just a such case

as in, where there is only one minimal element

[storm kiln]

What about looking at the intersection of all prime ideals?

also if (0) is prime, then ab = 0 => a = 0 or b = 0, so it's an integral domain

[hidden wind]

thanks i’ll try that, i gather this is an important construction which i haven’t at all looked at yet

but first i’m having a little break mwahahaja bye

[storm kiln]

nvm actually, you can have $ab\in I, J$, but $a\in I$ and $b\in J$

[cloud walrus]

Jelle

[storm kiln]

so neither a or b have to both be in $I \cap J$

[cloud walrus]

Jelle

[storm kiln]

Perhaps the intersection of all prime ideals is prime though, maybe look at ||nilpotent elements||

[boreal inlet]

we say they are rngs.... (I'm sorry please don't be offended)

[thorn jay]

I think they were about they weren't calling rings with identity simply "rings"

[serene dune]

oh no its not Ralgebra its unitary associative Ralgebra

[south patrol]

this is my main criticism of dummit and foote

[thorn jay]

:(

I only accept nonassociative algebras over a field

[south patrol]

a field like H

[serene dune]

i still do not understand why we need that definition of the module in your name

[south patrol]

(we don't)

[woeful sage]

wait... Dummit and Foote doesn't require rings to have identity???

[south patrol]

maybe i am confusing it w smth else but i'm pretty sure

[woeful sage]

[south patrol]

cause many times on this server i have been confused when helping people only to realise that that was the case

lol

[serene dune]

pulling a centrist stunt to not offend the analysts

[south patrol]

Amusingly i now have use for non-unital rings

[woeful sage]

[south patrol]

the main thing is that if you consider augmented rings (rings with a map R -> Z) then like

[woeful sage]

Friendship ended with D&F, Artin is my new best friend

[south patrol]

those end up being equivalent to non-unital rings

[serene dune]

i like the flow of artin

[south patrol]

and similarly if you consider augmented k-algebras and non-unital k-algebras

[thorn jay]

How non UA of them to not mention the additive identity or inverse

[south patrol]

so if you're interested in augmentations then you start thinking about nonunital stuff ig

[serene dune]

oh sorry i meant to say serge, lang

[woeful sage]

UA?

un-assistive

[thorn jay]

Also, heaps and trusses

[woeful sage]

wait those are actual algebraic structures?

[thorn jay]

Universal algebra

[woeful sage]

but that comes under (R, +) being an abelian group innit?

[serene dune]

that is a meme for me

[thorn jay]

Heaps are abstractions of the term xy^-1z for groups

[thorn jay]

Hey! That so happens to be my favorite branch of math

[woeful sage]

who would've known!

[thorn jay]

😔

[woeful sage]

wait I had a question

to ask a universal algebraist

[woeful sage]

[thorn jay]

It does, and they are uniquely determined by the operation (incidentally something im looking into rn), but universal algebra has permanently corrupted my brain

[serene dune]

3x3 lemma
can someone motivate me to learn all this ?

commutative diagram 🗑️

[thorn jay]

Well, you can make a monoid object in Cat out of it

Probably

But not a group object

[woeful sage]

oh right yea

[thorn jay]

Though i suppose you can take the grothendieck completion

[woeful sage]

since inverses aren't a thing there

[thorn jay]

incidentally something in looking to generalise

UA brainrot is real

[thorn jay]

Grothendieck forces inverses to be a thing!

[serene dune]

i love how people outside of maths are so hyped about cat

[woeful sage]

Grothendieck is too OP

[thorn jay]

Im inside of math and hyped about cat

[serene dune]

since you are module person(by name) im gonna ask u for suggesting me a book

i think my brain shut down, how am i constructing those sentences, its so embarassing

[thorn jay]

Lmao, i dont know many books, but im finally properly learning modules from "a course on homological algebra"

[serene dune]

bruv

[thorn jay]

Learnt modules very sporadically at first, to be honest, so im not the one youbshould ask for recs

[serene dune]

trying to do this

currently at chapter 3 exercise

well does this have the content ?
one might need(loosely speaking)

[thorn jay]

It depends, but I'd assume so

What do you want to do with it?

[serene dune]

idk

i dont have a school or anyone to guide me

[thorn jay]

Yeah then it's hard to really know when to finish

Perhaps choose some maximal chapter/concept you want to get to or understand before you move on

[serene dune]

whats next

CA ?

HA

[thorn jay]

What's CA/HA?

Oh

Commutative/homological algebra

[serene dune]

yeah

[thorn jay]

Hmm, you could do commutative algebra -> AG, or just HA and then into AT or the like

[serene dune]

anyother thing i might need for AT ?

[thorn jay]

Topology

:p

But not much more

[serene dune]

haha

[thorn jay]

Group theory

[serene dune]

will spend 10 hours a day on desk to be more of a loser

that is the quest

[rocky cloak]

Something you could do is find some university which keeps their teaching plan or whatever available online.

Then try to vaugely follow that.

[serene dune]

that is a noteworthy advice

[thorn jay]

Agreed

As long as you eventually do universal algebra

We need more peoplee

[serene dune]

can i get a job under jagr ?

[storm kiln]

My university uses Dummit and Foote for modules

(and groups, rings, fields and galois theory lol)

(and representations)

[sturdy spear]

so how are you studying

[serene dune]

broadband, i7 9700, 1TB HDD, 16gigs DDR4

should i continue ?

[sturdy spear]

what is this

[serene dune]

just a pc

[sturdy spear]

oh lol

bruh

what subjects are you studying nowadays?

oh notknown is our matual friend Epic

[serene dune]

modules, a bit of ring, and toplogy(not in last 2weeks)

[thorn jay]

Determination and desire to write cool looking symbols

[serene dune]

he has shown me a lot of cool books, which im very thankful for

[cloud walrus]

mcdnldsmngr

[tardy hedge]

For R->S ring hom, can we make R into an S-module by s.r = sf(r)?

I know I usually see R->S making S into an R-module by r.s=f(r)s

I was also thinking this in the context of field extensions, with K/F we can say K is a F-vector space but was wondering if we could say F is a K-vector space

[rocky cloak]

This would make S into a (right) R-module. Not the other way around

Could be beneficial to think about the inclusion Z -> Q

(it should be clear Z cannot become a Q-module)

[thorn jay]

s.f(r) is an element of S, not of R

[tardy hedge]

oops

[thorn jay]

:p

[thorn jay]

Ah that is what you meant here

[crystal vale]

Yeah

[sturdy spear]

Cool yeah
He told me about nice books as well

[crystal vale]

Carother 😂

[sturdy spear]

Yeah lol

[serene dune]

i just know the cool quote from that book

[sturdy spear]

Which one

[rocky cloak]

I thought the devil was in the details

[serene dune]

same thing upto sentiment

[rotund aurora]

Is that the evolutionary biologist? I wonder what was the context for him to say that

[serene dune]

haha, but it goes well with a lot of contexts i believe

[tardy hedge]

In the quotient field this equation implies …

I dont understand that part

[vast stump]

if you pass the identity given to the quotient ring then (x^3 - 2) vanishes and x becomes theta, so a(theta) * (1 + theta) = 1 which is why the text says they are multiplicative inverses

[tardy hedge]

Oh

The quotient Q[x]/(x^3-2)

I thought they were talking about field of fractions

Which is not relevant here in any way… right ? (Sanity check)

Ty

[south patrol]

tbh that is normally what quotient field means so like dw

However, the field field of Q[x]/(x^3 - 2) is very relevant to this question

since it is Q[x]/(x^3 - 2) again

lol

[tardy hedge]

Thank u 😆

Stating the “obvious” is nice sometimes

[candid patrol]

What is the dual of Z to you ?

[potent condor]

Z^*

[rocky cloak]

SO(2) I guess

[candid patrol]

u sure ?

[rocky cloak]

That's what it is to me, sure

Why?

[candid patrol]

No u right

[rocky cloak]

Not really a right or wrong without some context

[candid patrol]

But why do you think of SO(2) ?

To me it was the the unit circle

[rocky cloak]

SO(2) is the unit circle

[candid patrol]

But there are isomorph so it"s ok but why

Oh really ?

[rocky cloak]

Like you don't just want the unit circle in R^2 you need some group structure on it.

[candid patrol]

Ok i don't get it :/

[rocky cloak]

So SO(2) or U(1), but same thing, they're isomorphic

[candid patrol]

How do you proove that the dual of Z is isomorphic to SO(2) ?

[south patrol]

It is immediate from SO(2) = S^1

Oh lol

this is amusing to me jagr

[rocky cloak]

Well, first you have to define what dual means.

You define it as Hom(-, SO(2)), then it's pretty clear

[candid patrol]

To me it's Hom(Z, C*)

[south patrol]

to me S^1 is also common notation for the group

[rocky cloak]

Well, then you shouldn't be getting the unit circle as your answer

[south patrol]

What can you say about homomorphisms out of Z?

[candid patrol]

but it is

[south patrol]

I guess yeah C^x is probably better than S^1 / U(1) etc in general

[south patrol]

Well it just depends on how you are defining dual

[rocky cloak]

Hom(Z, C^*) = C^*

[south patrol]

But usually ig Pontryagin duality means S^1 not C^x right

[candid patrol]

what is C^x lmao ?

[south patrol]

same as C*

[rocky cloak]

Nonzero complex numbers under multiplication

[candid patrol]

alr alr

[south patrol]

and it just happens not to matter for finite groups (or more generally torsion groups)

But Z is not torsion, so the choice of definition does matter, and you will get S^1 or C* accordingly...

[candid patrol]

You right, if we use Pontyragin duality, then it's S1, else it's C*

[south patrol]

Well, i don't buy the "else"

If you use Hom(-,C*) you get C*

There are other notions of duality

[candid patrol]

Ok ok thanks !!

[south rain]

Hi, its me (again). I'm trying to show that in every left artinian ring, every chain of leftideals can be refined into a composition series of the R-module (R, +).

For that I took any chain $L_0 \supset L_1 \supset ... \supset L_n$. Then by DCC for any $L_i \supset L_j$ I can chooe a minimal $B_0$ that contains $L_j$ and $L_i \supset B_0 \supset L_j$. Then I choose a minimal $B_1$ such that $B_0$ is contained and $L_i \supset B_1\supset B_0 \supset L_j$. But imo I would also need ACC to stop the process, or does DCC stop the process by the finiteness condition?

[cloud walrus]

dellinger

[rocky cloak]

Assuming you're rings are unital, then every left artinian ring is also left Noetherian, so you have both ACC and DCC

If your rings are not unital, then it's not true. Unless your composition series are allowed to be infinite

[south rain]

It is in general not true or is my proof just wrong?

[rocky cloak]

Not true

[south rain]

Oh, alright thanks!

[tardy hedge]

rip wew

lads tbh

[rocky cloak]

Something happen to them?

[tardy hedge]

Nah I just remember he was so active a while ago and I dont see him anymore

He helped me a lot when i was first learning group theory

Sorry didnt mean to make it sound like something bad happened

[tardy hedge]

If K is integral over F (K,F are fields) then thats the same as saying K is algebraic over F right?

I happen to be learning general integral dependence in comm alg and also field theory at the same time

and also alg number theory which ive seen has a lot of field extensions lol

this semester is all about fields for me apparently

[next obsidian]

Yes

[spark veldt]

Hm I was thinking, Dedekind cuts are strikingly similar to the field of quotients. The set of Dedekind cuts form a field, and so do the field of quotients. Can you do what you do to Dedekind cuts, the same thing you do for a field of quotient of an integral domain?

it's early in the morning and this is just a thought lol

[next obsidian]

Dedekind cuts construction involve an order so you’d need to figure that out first

But I also don’t think it’s that related

Like when you take the quotient field of Z you get Q and then do Dedekind cuts to get to R

The first process goes integral domain -> field by adding units

And the latter goes from incomplete field -> complete field by adding in all limits

[spark veldt]

oo yeah i kinda see it

thanks chmonkey

[rocky cloak]

Dedekind cuts only form a field if you start with a field.

If you just start with an arbitrary poset, then its Dedekind completion won't have any field structure.

[thorn jay]

I think the similarity is that they are both completions in a sense

One is a completion of a poset to a complete lattice, and the other is from an integral domain to a field, filling in the "gaps" where there should be a fraction

[thorn jay]

It wont have a field structure with that attitude

[acoustic igloo]

i think if a group is transitive on a set, then it acts faithfully but
if it acts faithfully it's not necessarily transitive
is that right?

[fading field]

what about Z acting on Z/2Z by m + [n] = [m+n]? that's a transitive action but it's not faithful becasue 2 + [n] = [n] for every n in Z/2Z

in general if H is a nontrivial proper normal subgroup of G then the action of G on G/H won't be faithful but it will be transitive, because any non-identity element in H will act on G/H via the identity

[acoustic igloo]

oh yeah, Z acting on Z/2Z is transitive but not faithful

but

if a group is transitive on a set, there is only one orbit, right?

oh i think i see where i went wrong

the order of X can be less than the order of G

i mean the order of the G-set can be less than the order of G

i have another question

Let $X$ be a $G$-set and let $x\in X$. Then $\lvert Gx\rvert=(G:G_x)$.

[cloud walrus]

Axe

[serene dune]

trvial automorp right ?

[acoustic igloo]

nevermind i got it

thanks for the help 🙏

[hushed grotto]

Let p be a prime number and G be Z + Z/p^2. How to count the number of subgroup of G with index p^3?

My attempt is as follows.
If H is a subgroup with index p^3. G/H must be isomorphic either Z/p^3, Z/p^2 + Z/p, or (Z/p)^3.
Each H gives rise of surjective map G -> G/H. And surjective map to the groups of these 3 types give the required kernel, but two surjectives f and g gives same kernel if they are the same up to Aut(G/H).
So what I need is to count |Surj(G -> G/H)| / |Aut(G/H)| for each type and add them up.
I guess I managed to count for cases Z/p^3 and (Z/p)^3, but struggling for Z/p^2 + Z/p.

[velvet hull]

I’m not sure what you mean by the statement “two surjectives give the same kernel if they are the same up to Aut(G/H)”

[rocky cloak]

They mean that if you compose the map G -> G/H with an isomorphism, then it still has the same kernel

[serene dune]

the surjection is not unique?

[velvet hull]

But that idea seems somewhat problematic to me, because as an example Z/pZ + Z/pZ has two ways of quotienting out to get Z/pZ and I’m not sure how you’re getting that in your method

It’s probably going to be a lot easier to think about Z and Z/p^2 each individually

Since AxB is a subgroup of CxD iff A is one of C, and B is one of D

And you can understand the index of AxB very well just by learning [C:A] and [D:B]

[rocky cloak]

So say you wanted to count the number of index p subgroups of Z/p + Z/p.

Then you start by counting the number of surjective maps to Z/p. That's p^2 - 1.

But then you have overcounted, because maps can have the same kernel. So you have to divide by |Aut(Z/p)| = p-1.

So the true answer is p^2 - 1 / p - 1 = p+1

[rocky cloak]

This is true, but CxD has subgroups not of the form AxB

So it doesn't really help you much

[velvet hull]

Wait, what? When does the projection map fail?

Oh it’s just an extension of the two projection subgroups

[rocky cloak]

I'm not sure what you're asking, but for example

(x, x + p^2Z) is a subgroup of Z + Z/p^2 not of this form

[vocal pebble]

can someone tell me what the notation G(I) means where I is an ideal? this is from the book "monomial ideals" by herzog and hibi, and they just pull this notation out of nowhere

[velvet hull]

Yeah it’s not necessarily a direct product,you’re right

[rocky cloak]

One thing to note is that a map to Z/p^2 + Z/p definitely maps p^2Z to 0.

So really you can think of the map as coming from Z/p^2 + Z/p^2.

But then all you have to do is count the subgroups of order p.

[rocky cloak]

Gröbner basis? Or monomial basis or something like that?

[rocky cloak]

It's not an extension of the two projections, but it is an extension of one projection and one intersection.

[vocal pebble]

yeah i think its the latter, and the statements there do hold for monomial bases

[rocky cloak]

Yeah, they should

[serene dune]

the the converse of the inclusion might not be true because B might be bigger than im(h) and f is defined over all of B but as the composition holds things outside of im(h) doesnt really matter

they are asking me to prove the equivalence here, but im not getting over the notational convolution to see what are they(1 and 2) truly means on their own

[chilly ocean]

If I take the subgroup generated by { 1/p | p is a prime } under (Q, + ), how can I say this is proper subgroup? How can I show that 1/4 not in that subgroup?

[empty kernel]

prove that every element in R can be written as f(X)Y+g(X)

with f,g real numbers

i have no clue what to do

i have never done an algebra with multi variables polynomials

and in our textbook its like 1 page of info on it basically saying some very obv things so idk what i can do here

ah wait maybe i can use this one theorem (i read it wrong the first time)

ye so the only theorem in the section of multiple variable polynomials is R[X1,X2,...,Xn] is isomorpg with (R[X1,X2,...,Xn-1])[Xn]

idk if i can use that

[potent condor]

Is this even true? How could you write X^2 in that form

[empty kernel]

idk man i dont know wtf is going on
idk why i cant understand algebra

im just trying to study to not give up hope before the exam even started but yeee...

not going to well

[potent condor]

well i dont think this is your fault

this just seems false

[rocky cloak]

There seems to be a typo. Should probably be f, g in R[X]

[empty kernel]

you sure?

[potent condor]

It's definitely not true if f,g are real numbers

[empty kernel]

thats what i first thought but i kinda just assumed that i didnt understand it

[rocky cloak]

Well, that would make the statement true at least

And would be a reasonable exercise

[empty kernel]

then there is prob a typo

ok i will try it with f,g in R[X]

[rocky cloak]

f and g are also more commonly used for functions / polynomials.

Would have gone with like a, b if they were numbers

[violet spade]

hello fellow dutch person

[storm kiln]

Why would they have written f(X)Y though instead of fXY

[violet spade]

where do you study

[storm kiln]

Hello fellow dutch person

[empty kernel]

ku leuven

[violet spade]

damn belgium 💀

jk

❤️

[empty kernel]

i believe (X) is the ideal

[potent condor]

so you multiply the X first

[violet spade]

you use Lenstra right?

[storm kiln]

But wouldn't that turn the entire expression into an ideal?

[violet spade]

i'm sure i've made this exercise some time

isn't the trick to use remaindes?

[storm kiln]

oh, or is that the entire point

[violet spade]

remainders?

[empty kernel]

uuuh elements in R are of the form f(Y²-X³) with f in R[X,Y] no?

[storm kiln]

Yeah

sorta

[rocky cloak]

Could be helpful to think about it as a subgroup in Q/Z, then show it has no element of order 4

[violet spade]

Like, for any f in R[X,Y], we can have f = g*(Y^2-X^3)+r for some r, where deg_y(r) < 2

[empty kernel]

wdym sorta?

[storm kiln]

f + (Y^2 - X^3)

[empty kernel]

ah ye im dumb

ye wait now im even more confused

[empty kernel]

yuh i know that theorem

[violet spade]

Jelle where do you study haha

[empty kernel]

that might actually help

[violet spade]

Yeah and then maybe use evaluation homomorphism

[violet spade]

Actually, for 2 variables we need an extra step

but it's kind of easy

[empty kernel]

erm how do u define the degree of a multivariable polynomial?

[violet spade]

you look at the deg of y and x seperately

[empty kernel]

like the theorem we saw only works over single variable polynomials

it talks abt f,g in F[X]

[serene dune]

can we achieve a contradiction through lcm ?

[empty kernel]

so i dont think we can use this theorem unless we first prove that it works for multivariable polynomials

[violet spade]

no you can definitely use it

uhm

[empty kernel]

like what i mean is we cant iuse theorems we didnt prove in the lectures

[violet spade]

no i know

but you can still apply the theorem just in a more sneaky way

[empty kernel]

aha

how do i do that?

[violet spade]

yea i got it

alright

[rocky cloak]

F[X, Y] = F[X][Y]
so you can just think about this as the case R[Y] where R is some arbitrary ring.

The proof is the same as for R a field, as long as your polynomial is monic

[violet spade]

Suppose f in R[X,Y]. By the theorem, we can write f = gX^3 + r, where the degree with respect to x of r is less than 3

$f\in \mathbb{R}[X,Y]$. And $f=gX^3+r$, where $\deg_x (r) <3$.

[cloud walrus]

joel

[empty kernel]

aah okok

[violet spade]

Now, since $r\in \mathbb{R}[X,Y]$ again, we may say that $r=hY^2+r'$ for some $r'$ such that $\deg r' <2.$

[cloud walrus]

joel

[violet spade]

So $f=gX^3+hY^2+r'$.

[cloud walrus]

joel

[empty kernel]

waitwait

[violet spade]

i'll wait haha

[empty kernel]

i dont see how u can just despawn the y² in the first thing

[violet spade]

you mean in g or in h?

[empty kernel]

right here

[violet spade]

Like, we view the Y terms as coefficients

Like we can view any polynomial as $f = \sum_{i=0}^\infty g(Y)X^i$

[cloud walrus]

joel

[empty kernel]

yuh i understand that

but then why do u make your "constant" term go away

[violet spade]

Well, the question says nothing about $f(X)$ or $g(X)$

[cloud walrus]

joel

[violet spade]

So they didn't just go away

[empty kernel]

like shouldnt it be f=g(x³-y²)+r

[violet spade]

Well then we would just have $f=r$ right since $X^3-Y^2=0$ in this ring

[cloud walrus]

joel

[empty kernel]

huh now you completely lost me

[violet spade]

Well we divide out by the ideal $(X^3-Y^2)$

[cloud walrus]

joel

[empty kernel]

i tought we were using division algoritm over R[X][Y]

[violet spade]

Like for example, we have $\mathbb{R}[X]/(X^2+1)$, then this says that in that ring, $X^2+1=0$.

[empty kernel]

yuh i understand that

[cloud walrus]

joel

[violet spade]

Well now we are dividing out by $X^2-Y^3$ so that's obviously 0 in that ring. So it can't be just $f = (X^2-Y^3)g+r$

[cloud walrus]

joel

[empty kernel]

but over what ring are we using division algoritm?

[violet spade]

over the ring $R$ as stated in the exercise

[cloud walrus]

joel

[empty kernel]

but R is a quotientring not a poynomialring

[violet spade]

Well it is a polynomial ring, where $Y^2-X^3=0$

[cloud walrus]

joel

[violet spade]

Like elements of this ring could be $f = X^3Y^5-X^7Y+XY^3+3$ or something

[cloud walrus]

joel

[empty kernel]

but arent elements in a quotientring of the form f+(X³+Y²)

like arent the elements of R sets?

[violet spade]

And in this example, $f=(X^3Y^3+XY)Y^2-X^7Y+3$

[violet spade]

noo definitely not

[violet spade]

no not really

[empty kernel]

what

[cloud walrus]

joel

[empty kernel]

i tought that a quitientring is the ring of the nevenklasse

idk the engoish word for it

[violet spade]

i mean $R$ is a set, but elements of a set are just the elements.

[cloud walrus]

joel

[violet spade]

right, this is not the case here

[empty kernel]

what da fak

[violet spade]

Because in a subgroup $H$, we have a "nevenklasse" has the form $aH$ right

[cloud walrus]

joel

[violet spade]

for some $a$ in the group

[cloud walrus]

joel

[empty kernel]

yuh

[violet spade]

But this is just the set where some stuff is zero

it's not really the same

[empty kernel]

wait are u just viewing R as the ring of representants for those nevenklassen?

[violet spade]

Like you can view ideals for rings as the same as normal subgroups for groups, but they do not work the same in general

[storm kiln]

It is though, right, if I = (Y^2 - X^3), then X^3 + I = Y^2 + I

[empty kernel]

now thats where i guess i went wrong

[violet spade]

yeah well because X^3-Y^2=0

[empty kernel]

i tought that like ideals are basically normal subgroups but for rings

[violet spade]

i mean i guess that's right but it's just equivalent

[storm kiln]

but they are also "nevenklassen" cosets is the word I think

[empty kernel]

ye that might be it

[violet spade]

kinda true but not entirely. they work differently

[empty kernel]

uhu

but we defined quotientrings pretty much the same as quotientgroups

ye but so are we just looking at the representants?

[violet spade]

i mean it's basically just $x-y\in I \Longleftrightarrow \overline{x}=\overline{y}$ right?

[cloud walrus]

joel

[empty kernel]

yup

[violet spade]

And those $\overline{x},\overline{y}\in R/I$

[cloud walrus]

joel

[empty kernel]

yup

[violet spade]

Well, now we have $f=g \in R/I \Longleftrightarrow f-g\in (X^2-Y^3) \Longleftrightarrow f-g=0$. So any element in $(X^2-Y^3)$ is just zero. And we use this fact for calculations

[cloud walrus]

joel

[empty kernel]

yup i understand that

[violet spade]

but yea $f,g$ still live in $\mathbb{R}[X,Y]$.

[cloud walrus]

joel

[empty kernel]

ye

[violet spade]

So do you better understand now 🙂

[empty kernel]

and the f+() and g+() are in R

[empty kernel]

well i already understanded that

but like so when we are using division algoritm we arent rlly doing it in R but we are looking at the representanten for elements in R?

so like x³=y² in this ring

[violet spade]

i've not really seen the word "representant" be used in ring theory, but i think you're right yes.

[empty kernel]

representant is like
a+I
then a is a representant for that coset

and obv u can have different representants for the same coset

[violet spade]

yes i think that's correct. Because anything in $I$ just equals zero in $R/I$

[cloud walrus]

joel

[violet spade]

you got it yes

[empty kernel]

okok now im with you again

so we know x³+y²=0 in this ring

[violet spade]

No $X^3=Y^2$ right

[empty kernel]

wait

[cloud walrus]

joel

[violet spade]

haha

[empty kernel]

ye mb

there is a minus

[violet spade]

yes

[empty kernel]

waitwait we dont need division thing then right?

[violet spade]

i mean no not really now that i'm thinking about it haha

[empty kernel]

like doesnt it straight follow if we jiust replace x³ with y²

or the oother way around

just substitute

[violet spade]

because any term in terms of $Y$ that is higher than $Y^2$, we have that that's equal to $g(X)Y$ or $h(X)$ right

[cloud walrus]

joel

[violet spade]

yea that's true, like if you want to do it rigorously

[empty kernel]

yuh thats when u sub it

[violet spade]

$f\in \mathbb{R}[X,Y]/(X^2-Y^3)$. We have $f = \sum_{i=0}^\infty g_i(X)Y^i = \sum_{i=0}^\infty g_{2i}(X)Y^{2i}+\sum_{i=0}^{\infty} g_{2i+1}(X)Y^{2i+1}$

[cloud walrus]

joel

[empty kernel]

the sum doesnt go to infinity i think polynomials are by definitoon finite

[violet spade]

and then $Y^{2i}=X^{3i}$ and $Y^{2i+1}=YY^{2i}=YX^{3i}$

[cloud walrus]

joel

[violet spade]

It does go to infinity, but then after some $m$ we have $g_m=0$

[cloud walrus]

joel

[empty kernel]

oh ye that also works

[violet spade]

The definitions are equivalent, but then you don't have to deal with the fact that if $\deg(f)=n,\deg(g)=m$, we have $\deg(fg)=m+n$. This is annoying because if you define the elements of the ring to have an upper bound, then surely after multiplying $f$ by some other polynomial $g$, we can exceed that lower bound which makes it not well-defined

[cloud walrus]

joel

[violet spade]

so that's why it goes to $\infty$, because then we don't have a lower bound

[cloud walrus]

joel

[empty kernel]

aaah ye that makes sense

[violet spade]

But anyway it's kind of nitpicking, it's basically the same

[empty kernel]

well doesnt matter that much

i understand how it works now lesgooo

https://tenor.com/view/gandalf-hope-kindled-gif-27484735

[violet spade]

😎

i'm glad to have helped. if you have any other questions, feel free to DM. I just took ring theory a month ago, so it's fresh in my memory

@empty kernel

We can also just talk in dutch you know, might make things easier

[empty kernel]

will do

well i will send em here and just ping u

my exam is monday

[violet spade]

Like midterm?

[empty kernel]

ermmm no clue what that means

[violet spade]

tussentoets

[empty kernel]

but here in belgium universities its like

you have one exam

if you fuck that up come back in augustus

[violet spade]

damn

at UVA, we had a "tussentoets" and a "tentamen" and we also had to hand in homework

HW is like 10%

[empty kernel]

we got "vacation" for three weeks (u just study everyday for 3 weeks) and then we get exams of all or subjects

[empty kernel]

for some classes its permanent evluation

like for differential equations we had 2 tests both counting for 4 out of 20 points and then the exam now is for 12 out of 20 points

[violet spade]

you are a year 2 student? which book did you use for linear algebra

[empty kernel]

but for algebra classica pmechanics and analysis its just one exam

[violet spade]

what 💀

[empty kernel]

this one

[violet spade]

that's crazyy

[violet spade]

OMG i used that one too hahahahah

paul igodt!!!!!

hahaha

[empty kernel]

i cant be bothereed to type commas
algebra, classical mechanics, analysis

[violet spade]

that's why i was asking because i knew our book was belgian haha

[empty kernel]

wim veys is my professor for algebra this year

[violet spade]

that's crazy

one exam, that's pretty bad...

[empty kernel]

i tought it was like that in every uni

[violet spade]

that's actually really cool

[empty kernel]

well we have some cool professors like stefaan vaes is quite famous in the math world

[violet spade]

nope, i have one exam algebra 2, one exam diff eq, one exam stochastics. thats it

[empty kernel]

ye see for some copurses u also have it that way

its not very nice but it is what it is

i dont think i ever studied this much in my life

[violet spade]

this is what our algebra 2 exam looked like

[violet spade]

well that's good right? lol

[empty kernel]

Yes very good

[violet spade]

this was the "tussentoets"

[empty kernel]

Like in my first year barely did anything

[violet spade]

hahaha

tbh the first year definitely was way easier for me

[empty kernel]

Same

[empty kernel]

Wait but how many points does it give?

For the final grade

[violet spade]

like 3-%

30%

[empty kernel]

Lucky u

[violet spade]

whatt haha

[empty kernel]

Well if it isnt all at once then i thibk it is better

I dont like how this one test will decide if i fail algebra or not

[violet spade]

yeah... very harsh....

[empty kernel]

I much rather have ways to already score some points throughout the semester

Do your subjects go an entire year?

[violet spade]

i heard the amount of work in belgium is worse

[violet spade]

Noo haha

[empty kernel]

Ah ok same here

[empty kernel]

Rip im gonna come to the netherlanfs

[violet spade]

probably not that much better hahahahahaha

idk?

you have the same book as me, so it's probbaly roughly the same?

[empty kernel]

Probably it would be weird if courses differ that much

[violet spade]

yeash true

[empty kernel]

The only difference might be in the spread of the work i guess

[violet spade]

probably yeah

[empty kernel]

For linear algebra we also had one test that determined our entire score

[violet spade]

that's not great...

[empty kernel]

I didnt fail it though

[violet spade]

gg's

[empty kernel]

Passed it first try

[violet spade]

i believe in you

haha

for ring theory

this is your second algebra course i guess?

[empty kernel]

Yuh

Wait no

Third
We had linear algebra
algebraic structures
Now algebra one

Algebraic structures was a small subject that only lasted half a semester and it was an introduction to groups and rings

And we also did like dual spaces and shit

[violet spade]

right dual spaces

we had a subject about group theory and after, we had ring theory

kinda weird that those things are combined at your uni haha

[empty kernel]

Wair i will show u what we have to learn for algebra 1

[empty kernel]

This is what we see in algebra 1

And what i have to know for the exam monday

[violet spade]

right so also some linear algebra

i guess in NL we have to know more about one specific thing, we had like 14 chapters of ring theory

[empty kernel]

yuck

[violet spade]

yeah hahahaha but yeah. i wish you luck!!!!!!

[chilly ocean]

By this presentation of QD_2n, isn't isomorphic to D_4n ?

[chilly ocean]

I think there is a mistake in presentation

[rocky cloak]

Yeah I think it should be srs = r^n+1

[candid patrol]

Hello ! It there a good way to find the subgroups of order 8 in S4 ?
Like, i quickly find out with Sylow that if H is a subgroup of order 8 in S4, then he is isomorphic to D8, but it there a better way ? To directly find H ?

[hushed grotto]

I now understand that every surjective to Z/p^2 + Z/p (Let's call this L) factor through Z/p^2 + Z/p^2. And surjective to L has kernel of order p.
I can also count the subgroups of Z/p^2 + Z/p^2 with order p (if I'm not wrong it's p+1).
But every these subgroups can be kernel of some surjection from Z/p^2 + Z/p^2 to L?

[bitter rover]

The Sylow theorems are certainly the quickest.

You can see that D_8 is a subgroup because the symmetries of a square are permutations on its 4 vertexes. It's a Sylow p-subgroup of S_4, so all other subgroups of order 8 are conjugate to it.

The only "elementary" way I can think of would be to then rule out other groups of order 8 based on various group properties. For example, the quaternion group Q_8 can't be a subgroup of S_4 because element of order 4 in Q_8 all square to -1, which isn't true of the elements of order 4 in S_4.

etc. etc.

Is that "better"? It will be much longer.

[lusty marlin]

It is good to note that for n≥3, the dihedral group of order 2n is a subgroup of S_n, for the reason outlined here.

[candid patrol]

Ohhh Yeah true ! I forgot this one haha

[candid patrol]

Tanks you 🙏

[hushed grotto]

Ah, any (Z/p^2 + Z/p^2) / (subgroup of order p) is of course iso to (Z/p^2 + Z/p)...
Thank you 🙏

[hidden cairn]

I am trying to convince myself that $$C^\times$$ is a Lie group. First, the multiplication map $$(a,b) \cdot (c,d) = (ac-bd, bc + ad)$$ is piecewise smooth because both $$ac-bd$$ and $$bc + ad$$ are polynomials of input. So multiplication is smooth. Similarly, $$(a,b) \mapsto \frac{a}{a^2 + b^2}$$ is also smooth. so inversion function is also smooth. thus $$C^\times$$ is a Lie group. Is my reasoning right? Please let me know if there is a more appropriate channel for this question.

[cloud walrus]

pink_panther

[hidden cairn]

the inverse of a + bi is (a - bi) / (a^2 + b^2) and i tried to argue both maps (a,b) --> a / a^2 + b^2 and (a,b) ---> -b / a^2 + b^2 are smooth

oki thank you!

[minor fulcrum]

I have a text on group presentations that says that a set of relators generates a word if you can create the word out of taking the relators and repeatedly either sticking a relator between two symbols or deleting a relator. Is it required that you stick the relators in the middle, is sticking them at the begining/end not enough? Like, isn't this just the normal subgroup of the free group generated by the relators?

[minor fulcrum]

like this is its definition, it feels stronger than necessary but idk why it would be?

[rocky cloak]

So this is the normal subgroup generated by the relations, so if you assume things like
w*w^-1
is a relation, then it's enough to be able to stick things in the middle of words.

Otherwise you need to also require conjugates of relations to count as relations.

[minor fulcrum]

the text assumes that ww^-1 are all relators yeah

but do we need to stick them in the middle?

[rocky cloak]

Right, then it's enough to put relations into words

[minor fulcrum]

is juxtoposition not enough?

[rocky cloak]

Say your group is the free group generated by a, b. With relations given by a.

Then bab^-1 is also a relation

The only way to produce that is to place a in the middle of bb^-1

[minor fulcrum]

ah I see.

https://en.wikipedia.org/wiki/Presentation_of_a_group
" In this perspective, we declare two words to be equivalent if it is possible to get from one to the other by a sequence of moves, where each move consists of adding or removing a consecutive pair x x − 1 {\displaystyle xx^{-1}} or x − 1 x {\displaystyle x^{-1}x} for some x in S, or by adding or removing a consecutive copy of a relator. The group elements are the equivalence classes, and the group operation is concatenation." is that just wrong then?

[rocky cloak]

"consecutive copy of a relator" sounds to me like if ab is a relation, then you can't turn c into acb.

But if you interrupt it differently, then it may be wrong sure.

It's not super clear what they mean, but is somewhat hinted at by saying "consecutive pair x x^-1", or just by which interpretation would make their statement correct.

[acoustic igloo]

i am being destroyed by group theory

[serene dune]

what does the term "inclusion preserving" mean

[Correspondence theorem] If N is a submodule of an R-module M then there is an inclusion preserving bijection from the set of submodules of M/N to the set of submodules of M that contain N.

oh its like the lattice structure doesnt get destroyed after the mapping right ?

also does the bijection means not losing information while quotienting(which happens a lot when we quotient out an intricate group structure by some nsg)

[chilly ocean]

Denote the bijection by F

If M includes N, F(M) includes F(N)

[serene dune]

right!

geeked vs locked in

[vivid birch]

Do you know of any good reference to read about Artinian groups?

[spark veldt]

Took my algebraic structures final exam yesterday, did really well and I kinda wanna learn more. I used Herstein's Topics in Algebra for it. I'm wondering where would be a natural next step after this? Commutative algebra or something? I'm not really familiar with the algebra 'pathway'. Any suggestions?

[rocky cloak]

Well, what things did you cover, which parts did you enjoy, what are you curious about?

Algebra is a fairly wide field with lots of things to explore.

[spark veldt]

The entire semester we covered groups rings and fields, really enjoyed things like the isomorphism theorems and ideals part in the rings chapter (although they were kinda frustrating).

I just found it really cool that we can construct fields from maximal ideals etc and especially had fun learning polynomial irreducibility.

[rocky cloak]

Sounds like commutative algebra can be good then.

Probably smart to learn some about modules.

[spark veldt]

oo ok

thank you sm

[patent girder]

if $V$ is a vector space over $k$ and $T\in \mathcal{L}(V)$ is given, is there a nice condition to know when there is a vector $v$ such that $V^T$ is generated by $v$ as a $k[x]$-module?

[cloud walrus]

Zander

[south patrol]

Does V^T mean the k[x] module with x.v = Tv?

(It is not really standard notation)

Such a v is called T-cyclic. I don't think there is a general way to know if such v exists (besides assumptions that are essentially the same as the conclusion) but an important thing is that the degree of the min poly of T should be the dimension of the vector space [I assume you are working with finite dim stuff...]

[patent girder]

yes. sorry about the notation, thats just how i learned it

[south patrol]

Another thing is yeah you can consider the T-annihilators of each v

Which means the set of polynomials p(x) with p(T)v = 0

For each v you get an ideal I of k[x] of stuff killing v, generated by some polynomial p_v. v is T-cyclic iff deg p_v = dim V

(Note p_v divides min poly of T)

So yeah like playing around with these can be useful but generally kinda hard as you need to check a lot of v...

[tough raven]

That's sufficient as well, actually. (Equivalently, minimal polynomial = characteristic polynomial. This is guaranteed if the characteristic polynomial has no repeated irreducible factors (or roots).)

[sweet pendant]

Can I get some help with part(e)?

[rocky cloak]

So notice that if you have an isomorphism
Q[x]/(f) -> Q[x]/(g)
then you can compose with the projection to get
Q[x] -> Q[x]/(g).

And then by choosing a preimage of x you can left this to a map Q[x] -> Q[x], that will map (f) into (g).

So the only way this can happen is if psi((f)) is a proper subset of (g).

psi((f)) = (psi(f)) and f and g have the same degree, so we would need psi to not preserve degree.

What are some easy maps that would work for psi?

How can you choose f and g to force this psi to work?

[sweet pendant]

So there cannot exist such a map? Because automorphisms of Q[x] are linear in x?

[rocky cloak]

We never said it had to be an automorphism of Q[x]

It only has to induce an isomorphism Q[x]/(f) -> Q[x]/(g)

[sweet pendant]

I'm still a bit confused... ring isomorphisms from Q[x] to Q[x] are always of the form psi(x) = a x + b I thought, so they must preserve degree?

[rocky cloak]

I'm telling you to not consider an isomorphism

So okay, let's just say for example
psi(x) = x^2

This is a pretty easy map that doesn't preserve degree.

How can we choose (f) and (g) such that psi becomes an isomorphism between Q[x]/(f) and Q[x]/(g)?

[somber sleet]

There is something I do not completely remember, what does the fact that a polynomial is irreducible over a finite field imply over its maximality/being prime?

[rocky cloak]

A polynomial is irreducible iff it is prime iff it generates a maximal ideal.

Both of these are true over any field.

[somber sleet]

Is this a theorem? I wanted to read the proof

[hollow imp]

this fact should be in any abstract algebra book, where they prove k[x] is a PID

[rocky cloak]

It is a theorem, but I don't know if it has a name or anything.

Maybe look here
https://math.stackexchange.com/q/1593553/306319

[somber sleet]

Yes I remember, that, but I don't find a proof in my notes, hence why I asked

[somber sleet]

oh that's nice, thanks a lot!

[sweet pendant]

how about x^3-a, x^3-b. Say a=2, b=16?