groups-rings-fields

[wraith cargo]

A cool thing to do is if you ever encounter some really fucked definition is to come back a while later and see how many words you understand!

[white oxide]

I'm having trouble checking the distributivity property of $\text{Hom}_\mathbb{Z}(R, J)$ as a unitary left $R$-module, with the action of $R$ defined by $(r\phi)(x) = \phi(xr)$, any hints?

[cloud walrus]

okeyokay

[south patrol]

it should be p clear if you just evalute on an element

[white oxide]

like

[south patrol]

like what is ((r + s)\phi)(x)

[white oxide]

oh i was talking about the other distributive property but yea that's probably also instructive

[south patrol]

tbf there are actually two distributivity things

i find this funny cause like

ig there are least two meanings of homotopy category lol

[wraith cargo]

this is the HOMOTOPIC category

[south patrol]

lol

[white oxide]

yea i'm just struggling to use the definitions lol, i'm just confused about what $r(\varphi \circ \phi(x))$ is, it would be $\varphi\phi(rx)$ but I'm unsure as to how that equals $\varphi(rx) \circ \phi(rx)$ lol

[south patrol]

i assumed that was just a typo

or nonstandard

[cloud walrus]

okeyokay

[south patrol]

this notation is kinda wrong

well

[white oxide]

i don't know maybe i have to use something about homogeneity

is that the term for extracting scalars out of a r-module homomorphism

homogeneity and additivity right

[south patrol]

i assume you mean (r \varphi \phi)(x) but in any case Hom_Z(R,J) is just a module

like

There isn't a multiplication in it

you can't compose the elements unless R = J

[white oxide]

ye

[wraith cargo]

here the homotopic category is that all the objects are the same but you quotient out all the homotopies from the morphisms because the category is assumed to be enriched over Ab so the morphisms homotopic to 0 form a subgroup that you quotient out

[white oxide]

wait wdym i thought by definition it's an abelian group under composition

[south patrol]

no

Under addition of homomorphisms

[south patrol]

Yeah I understand it to be that one just not heard the term "Homotopic category" used for this ever

But fair yeah seems some people do that

fair nuff

[white oxide]

ohh

[south patrol]

I mean like

[white oxide]

okay that makes a lot more sense

[south patrol]

Composing homomorphisms is basically never gonna give you a group (unless they're all automorphisms by definition) let alone an abelian one

Here it wouldn't make sense to compose them

[white oxide]

oh wait yea composition wouldn't even be defined

[south patrol]

Ye

[white oxide]

oops

okay got it ty!

[south patrol]

n

p

[rocky cloak]

If your abelian category has enough protectives, the derived category is equivalent to the homotopy category of protectives. (So you don't need to worry about quotients or localizations)

Bounded derived category that is

[warm wyvern]

googles quotient category
googles homotopy category
googles chain of a category
googles abelian category

[south patrol]

lol

[chilly radish]

Category of chains*

Chain complexes to be exact

[rocky cloak]

Don't bring model categories into this. Filthy homotopy theorists

[south patrol]

Isn't that what you're talking about basically lol

[rocky cloak]

Shhh

[south patrol]

Homological algebra is just a type of homotopy theory anyway

[rocky cloak]

We don't speak of the model structure

[south patrol]

what kinda math(s) do you do jagr

[wraith cargo]

no lol

[south patrol]

wdym irony

[rocky cloak]

Homological algebra for finite dimensional algebras

[south patrol]

well when you say projectives what you're doing is taking a model structure and using that to localise wrt weak equivalences in a nice way

[south patrol]

inch resting

[rocky cloak]

Yeah the protectives should be the cofibrant or fibrant or whatever word homotopy theorists use

I'd rather think of them as the image of the adjoint to verdier localization

[south patrol]

cofibrant ye

[rocky cloak]

Or just you know... Projective modules

[south patrol]

but i think it's nicely defined here given how like projective objects are defined in terms of lifting properties already

i need to learn more about the like

derived category for algebra though lollike the verdier stuff

probs just weibel chap 10

[rocky cloak]

Also don't you need like your category to be complete or something to define model structure and stuff?

[delicate orchid]

any category that isn't complete isn't worth thinking about

[rocky cloak]

(I'm just trying to come up with reasons to hate on homotopy theory)

[south patrol]

yeah usually they are required to have all small limits and colimits

[rocky cloak]

Finite dimensional modules would like a word

[south patrol]

i think originally Quillen just required them to have all finite limits and colimits

but ye

[delicate orchid]

gulp

[rocky cloak]

I think maybe Neemann is the usual reference for localization of triangulated categories, kind of technical though.

[south patrol]

Oh okay thank

[rocky cloak]

I don't know what Weibel does

[south patrol]

Me neither yet but defines triangulated cats at least lol

[wraith cargo]

how category theorethic do you want your explanation of derived cats

[topaz solar]

Neemann

Time to add to the list of books

[wraith cargo]

https://www.math.utah.edu/~milicic/Eprints/dercat.pdf
You might be interested in this potat

it works up from localizations to derived cats

and proves way too much lol

[topaz solar]

238 pages, we got a lot of material

[white oxide]

here they mean f + g for fg right

having trouble seeing homogeneity f

fbar(rg) = f(rg) but like you can't just pull the r out to get rf(g)

[topaz solar]

No it’s composition

[white oxide]

oh right

[topaz solar]

f: A->J, g: R -> A

So fg: R -> J

[white oxide]

got it

thx

[coral steeple]

Am I interpreting this wrong or are the last two sentences not actually contributing to the proposition as it's stated?

Ah wait I think aH=Ha is stronger than "left cosets are right cosets" and that's why those sentences are there...

[warm wyvern]

no it's weaker

"left cosets are right cosets" implies H is normal

aH=Ha doesn't

[coral steeple]

But if aH=Ha then for any ah there is an h'a so that ah=h'a, meaning aha^{-1}=h', so aHa^-1=H, right?

[wraith cargo]

Do you meant this for a specific a or for any a in G

[warm wyvern]

oh wait is that how you define normal groups?

yea, that looks correct sorry

[coral steeple]

Ah I see

[coral steeple]

Also as I understand, normality is a property had by a subgroup so technically aren't there no "normal groups"?

[warm wyvern]

yea

[formal ermine]

every group is normal in itself

conjugation is an automorphism

[coral steeple]

Ah oops yes

[coral shale]

"This group is isomorphic"

[north sand]

let $\mathbb F_q$ be a field of odd order $q$ and consider the polynomial $\frac12(1+x^{(q+1)/2} + (1-x)^{(q+1)/2)})$. I want to show that this polynomial is the square of another polynomial (this is exercise 3.47 in lidl and niederreiter). I have shown so far that if $z\in \mathbb F_q$ is a root, then it is at least a double root and it satisfies that both $z$ and $1-z$ are not squares in $\mathbb F_q$. But that doesnt really seem to help me cause I dont even know if there are enough of these elements so that the poly splits in $\mathbb F_q$ and not in some extension (for which my argument breaks down). in the book the surrounding questions are all about irreducible polynomials which really stumps me cause clearly this poly cannot be irreducible. but maybe I am just on the completely wrong track?

[cloud walrus]

denascite

[warm wyvern]

[molten viper]

So, Im not really understanding what an algebra is. Would someone mind providing me something I can read that could help me get an intuitive understanding of them

[delicate orchid]

it's a vector space but you can multiply the vectors

[rocky cloak]

The definition of algebra varies quite a bit depending on the field youre in. Typically the definition is just a module over a commutative ring (or vector space over a field) equipped with a bilinear operation usually called multiplication. In many context there are extra assumptions, such as associativity or unitality

In which case an alternate definition would be a ring with a commutative ring (or field) in it's center.

[topaz solar]

A field extension is an example

I believe 3d real vectors with the cross product is too, for one that isn’t commutative or associative?

[rocky cloak]

In which context are you learning about them?

[delicate orchid]

it is yeah, a very spooky space indeed

it's a lie algebra as well for a nice bonus

[topaz solar]

Idk any spooky other algebras that are fairly often used except like obvious things

like R[X] if rings are allowed

[delicate orchid]

heisenberg algebra?

[topaz solar]

I don’t know that offhand so maybe

[delicate orchid]

take the commutators from the presentation of the heisenberg group and just pretend they're lie brackets

[topaz solar]

[rocky cloak]

The power set with addition given by symmetric difference and multiplication given by intersection.

[frigid lark]

Witt vectors

[molten viper]

@delicate orchid @topaz solar so Im working on a research problem centered around polynomial rings, which Im told are algebras (specifically they’re k algebras if we have R = k[vars] iirc) but I dont really understand what that means

[delicate orchid]

yeah they're algebras

vector addition is given by polynomial addition, the multiplication is polynomial multiplication, and scalar multiplication is just multiplying by some element of k

[topaz solar]

Well R = k[x, y] might not be finitely generated as a vector space, but it is an algebra I believe

[delicate orchid]

true, is that meant to mean something I said was wrong?

confused by the phrasing

I also don't think k[x] is finitely generated as a vector space

[topaz solar]

I’m not saying that, I’m saying the vector space generators aren’t the same as the algebra generators

[wraith cargo]

An algebra is a ring that's also a module
So k[x] is a ring but its also a k-module, so it's a k-algebra

[wraith cargo]

it's not finite

[topaz solar]

Certainly not as a vector space (or cardinality)

Should be generated by x and y though?

[wraith cargo]

It's generated by x and y as an algebra
But algebras are usually called finite when they're finitely generated as vector spaces

[open sluice]

an algebra is a σ-algebra but you only require closure under finite unions, not countable ones

[topaz solar]

Finitely generated, not finite

[wraith cargo]

Made me pull out my copy of Eisenbud smh

[topaz solar]

Yes

[wraith cargo]

Finitely generated - finitely generated as an algebra
Finite - finitely generated as a module

[topaz solar]

I hate algebra terminology sometimes but yep

[wraith cargo]

The difference is very important

[lime badge]

I've also seen "finitely generated" vs "finite type"

The latter being used in the context of R-algebras and the former for modules

[rocky cloak]

# all fields are F2

[molten viper]

But every ring is a module, isnt that kind of trivial?

[delicate orchid]

yeah, every ring is an algebra over itself

but there are others

[molten viper]

I see

Meant for that to be a reply sorry lol

[cloud walrus]

Irony Incarnate

Irony Incarnate

[coral steeple]

Is the negation of "left cosets are right cosets" the statement "the partition of the group into left cosets is not the same as the partition into right cosets, even up to relabeling of the cosets"?

Put another way, at least one of the left cosets is not equal to any right coset

[warm wyvern]

they often occur appear in analysis

for instance, the set of real continuous functions over [0, 1] is an algebra

there's this famous result called stone-weierstrass

beyond this I got no clue either lmfao

and I would love to hear an answer

[static yew]

Okay if I understand this stuff correctly

Z/pZ is a homomorphism Z -> Fp (which is interesting that it produces a field with nonzero characteristic from a non field)

But Z/qZ where q=p^k, k>1, would not be a homomorphism

Because there doesnt seem to be a way to preserve + and × while keeping a field characteristic less than q

[formal ermine]

how do you define a field being a homomorphism

[ivory trail]

and I've mostly seen finite vs finite type for modules and algebras respectively

[formal ermine]

btw your Z/qZ isn't a field lol

[static yew]

Well if a homomorphism h(x) on R -> R' is a mapping such that + and × are preserved

I.e. h(a) + h(b) = h(a+b)
h(a)h(b) / h(ab)

[formal ermine]

What

you said "Z/p is a homomorphism"

wtf is that supposed to mean

[static yew]

"Mappings of rings that preserve + and × are called homomorphisms"

[ivory trail]

do you mean Z -> Z/p is a homomorphism

[formal ermine]

but how can A RING be A MORPHISM?

[static yew]

Fine. Z/nZ refers to the ring produced by a homomorphism

If n is prime, the result is a field

[wraith cargo]

have you heard of flat morphisms

flat rings are a thing
flat morphisms are a thing
ergo morphisms are a ring

[formal ermine]

so true

[ivory trail]

i haven't heard of this theorem

big if true

[elder wave]

[wraith cargo]

[static yew]

Its based on the flat earth proof

anyway

The ring Z/nZ is can be produced by a homomorphism from Z

If n is prime then the resulting ring is also a field

[wraith cargo]

tf is going on

[formal ermine]

the only time I've used flat morphisms was in the context of toric degenerations

[static yew]

If n is a greater prime power the result is not the field F_p^k. There is no homomorphism from Z to F_p^k with k>1

[formal ermine]

ok bye guys I'm going to sf now

I won't have internet for the next 5 hours

[ivory trail]

most people would say "Z/nZ is a quotient ring of Z" since asserting B is a quotient ring of A is the same as giving a surjective homomorphism A -> B

[static yew]

Okay those are some words I can work with

[ivory trail]

first sentence is correct, second is not

[white oxide]

help why is $\bar{f}$ a monomorphism 💀 $\bar{f}(g_1) = \bar{f}(g_2) \rightarrow f \circ g_1 = f \circ g_2$ but the condition that $f$ is a monomorphism only works with elements of $A$ right

[ivory trail]

there is precisely one homomorphism from Z to any other (unital) ring

[cloud walrus]

okeyokay

[white oxide]

make sure you hit land's end trail

super cool stuff

[ivory trail]

hint: if f: Z -> R for R any ring, and i know f(1), what does that say about f(n)? what does f(1) have to be?

[wraith cargo]

since $f$ is a monomorphism it has the left cancellation property, i.e. $f\circ g_1 = f\circ g_2 \implies g_1 = g_2$

[cloud walrus]

Irony Incarnate

[ivory trail]

maybe you mean that F_p^k isn't a quotient of Z for k>1

[white oxide]

oh i didnt know that i guess that's a basic fact about functions

[wraith cargo]

it's the definition of a monomorphism

[white oxide]

o

[wraith cargo]

a monomorphism is defined to have this property
That's why it's more general than an injective function tho a lot of the time they're the same

[white oxide]

hungerford just defined it as an injective map of sets which is a homomorphism

[wraith cargo]

tho you can also prove this holds for injective function
It's the other way around for surjective/epic morphisms

[white oxide]

or rather a homomorphism that is injective as a map of sets

[wraith cargo]

I think on Set the concept of a monomorphism and injective homomorphism are the same

[static yew]

What would a homomorphism from Z to GF(4) look like?

[wraith cargo]

x mod 4

or sorry no

[ivory trail]

well there's a copy of F_2 inside F_4, so it's the map Z -> F_2 followed by the inclusion F_2 -> F_4

if you write F_4 as F_2[x]/(x^2+1), then you can see the inclusion here as the elements 0, 1

[static yew]

Ohhhhh I see

I guess technically you could say that

(Cue futurama reference)

You're just taking the homomorphism to a subring (subfield)

That doesn't seem very useful though

[ivory trail]

what makes it useful or not

homomorphisms are often not surjective

[static yew]

There doesnt seem to be a way to have a ring automorphism that doesnt preserve 1 and 0

[topaz solar]

that's kinda

the point?

[static yew]

Just checking

[ivory trail]

well maybe you're looking for homomorphisms of not rings

[topaz solar]

rng homomorphism

[ivory trail]

or Z-module homomorphisms

[topaz solar]

(Abelian group homomorphism)

[static yew]

No. I've been given a new concept here (homomorphisms, automorphisms) and I'm trying to construct scenarios to check my understanding

[white oxide]

Also doesn't the map $\varphi: A \to \text{Hom}R(R, A)$ require that R is commutative, because $\varphi(r_1a) = f{r_1a}$ and for any $r \in R$, $f_{r_1a}(r) = r(r_1a)$ but $r_1\varphi(a) = r_1f_a$ and for any $r \in R$, $r_1f_a(r) = r_1(ra) \neq r(r_1a)$ if R is non-commutative

[static yew]

The book glosses over the fact that the homomorphism might not use the entire target ring

[cloud walrus]

okeyokay

[static yew]

Btw which letter is a v with a vertical bar

This book uses that one

[topaz solar]

$\psi$?

[cloud walrus]

The great Sharp

[static yew]

An automorphism on a group doesnt need to preserve the identity

In the additive group of Z/nZ, psi(x) = x + k would be an automorphism

There would be a similar thing for the coprime multiplicative group, but it's not as simple

Multiplicative would actually preserve identity thoug

So forget that one

[topaz solar]

that is not an automorphism

[static yew]

Oh

See? Obviously I missed something

But I gotta get back to work. Meh

[white oxide]

any heroes?

[void cosmos]

yeah

[white oxide]

exhibit your heroic nature.

[round hull]

by the way, a composition-preserving map of a group -> group necessarily preserves the identity

which is kind of neat

[delicate orchid]

$\phi(e) = \phi(ee) = \phi(e)\phi(e)$ so $\phi(e) = e$ tada

[cloud walrus]

wew

[round hull]

or e'

[delicate orchid]

yur

[void cosmos]

eeeeeee

my fav number was e

2.71

how do you guys decompose ideals

can someone do an example

like for example (x^2,xy,2)

?

over Z[x,y]?

[delicate orchid]

(6) = (2) \cap (3)

[void cosmos]

yea

[delicate orchid]

ok u get it then

[void cosmos]

over Z or any UFD i can just factorize the elements

but

idk how to factorize x for example

[delicate orchid]

Z[x,y] is a UFD

[void cosmos]

cuz x is prime lmfao

yea

[delicate orchid]

why are you trying to factorise a prime?

[void cosmos]

cuz i wanted to do it like this

u factorized 6

[delicate orchid]

6 isn't prime

[void cosmos]

ik

how would you decompose (x^2,xy,2)

[ivory trail]

the "fundamental theorem on homomorphisms" says that the image is canonically isomorphic to the quotient of the domain by the kernel

so if you haven't gotten to that it's worth it

[coral spindle]

It is more commonly known as the first isomorphism theorem

[wraith cargo]

are you trying to do primary decomposition?

[void cosmos]

yes

[rocky cloak]

No the module structure on (R, A) is defined as (rf)(x) = f(xr). Remember R is a bimodule

[white oxide]

what is a bimodule?

[white oxide]

ah ok

thank u

o

left and right module

[night onyx]

both a left and right module, where (rm)s = r(ms) for r,s in the ring, m in the module

[glossy crag]

Context is crossed products (i.e. f is a 2-cocyle and (L,G,f) is an algebra generated by L and elements satisfying relations in the 1st image), the statement is about how if f and g are cohomologous, then A=(L,G,f) and B=(L,G,g) are isomorphic. Is it just me or is the calculation in the 4th image wrong?

By definition $\varphi(1_A)=\varphi(f(1,1)^{-1}u_1)=f(1,1)^{-1}\lambda(1)v_1$, but by definition of coboundaries $g(1,1)^{-1}=\sigma(\lambda(1))f(1,1)^{-1}$, not $\lambda(1)f(1,1)^{-1}$.

[cloud walrus]

Ocean Man

[glossy crag]

Maybe the definition of $\varphi$ should've been $\varphi(xu_\sigma)=x\sigma(\lambda(\sigma))v_\sigma$?

[cloud walrus]

Ocean Man

[glossy crag]

No wait, that doesn't make sense either, because then it won't be \sigma(\lambda(1)). What do?

Lol, nvm, I forgot to set the outside sigma to 1. Bamboozled again.

[molten viper]

I'm attempting this problem:
Suppose A is a ring such that, if an ideal I is not contained in its nilradical (N), then I contains some non-zero idempotent element. Prove that the Jacobson radical (J) = N in A

I'm specifically trying to prove that J \subseteq N, and doing so by contradiction. So supposing j \in J but \notin N, there exists some x such that x = aj and x idempotent. Since j \in J, 1-jy is a unit for all y \in A, specifically 1-x is a unit. I've used this to show 1 - x = 1 + x, and now I'm kind of stuck

I'd like a hint

[topaz solar]

Is it commutative

if 1 = 1 + 2x, that’s not great for x

[molten viper]

it is commutative yeah

But that doesn't necessarily mean x = 0

which is what I'd like to show in a perfect world

[topaz solar]

Right, but it’s certainly not good for something like (1+x)^2

[molten viper]

I guess like, 1 + x = 1 + 9x also

er

1 + 3x = 1 + 9x

[topaz solar]

Funny characteristic 2 lookin thing

[molten viper]

oh well, I can just keep adding 2s

hm

[topaz solar]

-x = x

[molten viper]

yeah

but like, I'm not sure what I can conclude from that

[topaz solar]

(1 + x) = 1 + 2x + x^2

= 1 + x^2

and similar on the 1-x side

Something using that x is idempotent?

[molten viper]

well it's the specific combination of 1-x being a unite and x idempotent

unit

Well, more generally 1-jy is a unit for any y

(where x = aj)

so not only is 1-x, but also 1 - 2x, and 1 - jx, etc

[topaz solar]

[molten viper]

This is similar to some other problems I've done, but I'm not seeing a connection

[topaz solar]

If it’s idempotent we should be able to get like, x^n - x zero right?

[molten viper]

yes

x^n - x = 0 for all n > 0

[topaz solar]

For all n?

[molten viper]

yeah as x^n = (x^2)^(n-2)

er

[topaz solar]

Uhh that would be 2n-4

[molten viper]

hush

x^2*x^n-2

but yeah for all n

[rocky cloak]

If x is idempotent then so is 1-x, hence not a unit

[molten viper]

why is that?

[topaz solar]

(1+x)^2 = 1 + x^2

Well if it’s idempotent that’s hard to divide

[delicate orchid]

(1-x)^2 = 1-x^2 = 1-x lol?

[rocky cloak]

(1-x)(1-x) = 1 - 2x + x^2 = 1 - 2x + x = 1-x

[topaz solar]

also they said they had x = -x or smth

[molten viper]

Is there a result that says idempotent elements aren't units

[topaz solar]

If a was idempotent and b was the inverse

[rocky cloak]

x(1-x) = 0

[molten viper]

ah

[topaz solar]

ab = 1 but aab = a

[delicate orchid]

the 1 in question:

[molten viper]

ah

[delicate orchid]

1 is the only unit that's idempotent

[molten viper]

that's the connection to the problem I've done before

Because that other problem used that exact expression x(1-x)

Welp, I appreciate the help

I think I uh, might be a bit tired lol

[topaz solar]

I may be stupid

[molten viper]

tbf last time I did this problem, it took me forever to find that x(1-x) expression

not this problem, the last problem I did

[glossy crag]

If A is a K-algebra of dimension n and M is an A-module of K-dimension m, what is the dimension of End_A(M)?

[round hull]

what about n

[wraith cargo]

I think only bounds are known?

at least as far as I've searched

[glossy crag]

Nvm, I forgot I had this in my notes: if A is simple (forgot to mention this) and B=End_A(M), then (dim_K(M))^2=|A:K||B:K|.

[chilly radish]

Double centraliser moment

[formal ermine]

what part of this is algebra

@molten orchid

[wooden ember]

Take it to #real-complex-analysis

[molten orchid]

Ok

[coral spindle]

There is no formula for this, unfortunately. It depends on the structure of the module and algebra deeply.

There are nice examples in rep theory of such things

[rocky cloak]

I guess an interesting question is what possible values can be achieved. Maybe add the condition that M be indecomposable.

Getting 1 is pretty simple, and there is an obvious upper bound of m^2 .

[coral spindle]

I only know interesting restrictions in the case of the rep theory of groups

When A = KG, There is an integer m dividing dim M such that dim End(M) = [K(chi):K] m^2, where K(chi) is the field adjoin all elements of the character of M

Well ok, addendum: this is when M is simple

[rocky cloak]

That's neat. But I guess in general [K(chi):K] could be pretty much anything, so maybe doesn't help so much

[coral spindle]

Yes but there are also some nice bounda for that in terms of G

Plus it’s relatively easy to calculate

On the other hand, this mysterious value m is considerably harder to calculate

Search term: Schur index

[vocal estuary]

Any tips for someone heading into abstract algebra this fall? I've already started reading/studying Judson's Abstract Algebra.

[delicate orchid]

first things first, fantastic pfp

secondly, if you can understand that book you'll be fine in a first course of AA

wait nvm this book is weird haha why does it have sage exercises

nvm this is pretty good

[vocal estuary]

Many thanks!

[glossy crag]

Anyone else get off to writing morphisms without parentheses, e.g. fx instead of f(x)? Makes me feel strangely powerful.

[void cosmos]

hernstein moment\

or no

other textbooks write it in reverse

like xf

lmfao

[delicate orchid]

xf is more common than fx

so composition xfg goes left to right

[void cosmos]

yea

sucks

[glossy crag]

Group theory moment

[round hull]

get off

[south patrol]

Sometimes for functors lol but not so much functions unless we're in a linear algebra context or smth

[formal ermine]

rep theory moment

[crystal turtle]

Always for functors and (pre)images of sets, but I like parentheses for image of an element

[topaz solar]

I hate reverse polish

[formal ermine]

are you a nazi????

[rocky cloak]

The trick is to never evaluate morphisms, then you can just write f

[formal ermine]

if you hate the polish

[topaz solar]

My algebra professor asking if we can still call it that or if it’s racist

[topaz solar]

Idk the Soviets weren’t fans either

[solemn dew]

when is this lemma used?

[cloud walrus]

Galstaff, Sorcerer of Light

[night onyx]

I think once you know that x^{n-1} - 1 \in I that's enough, because x^{n-1} + I = 1 + I, implying x + I is invertible in R/I. If n = 2 then x + I is the identity in R/I which is fine, so in any case R/I is a field and I is maximal.

[molten viper]

Oh I see

[night onyx]

I took a class on banach algebras where the prof wrote (f x) for linear functionals f, which was really nice when embedding X into it's double dual, cause you could write (x* f) = (f x), had a nice symmetry to it

[lavish spoke]

i have only ever seen <x, f>, which I don't mind

[molten viper]

That does seem easier, thanks! This was the first problem I was able to do on my own so it seems Im improving

[glossy crag]

The notational idea (bilinear pairing of X and X*) is pretty standard in FA afaik, if not the exact notation your prof used.

[white oxide]

Is there any easy way to find the subgroups of Gal$\bigl(\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})/\mathbb{Q}\bigl)$? I've included the table of the permutations of the group as follows, but through computations I'm beginning to suspect that the group is abelian (is that even possible? for a permutation group of order 8 to be abelian)?

[cloud walrus]

okeyokay

[white oxide]

I guess it would have to be isomorphic to either Z8, Z4 x Z2 or Z2 x Z2 x Z2 if that's the case

well it would have to be isomorphic to Z2 x Z2 x Z2 since every auto is of order 2 if I'm correct

[molten viper]

If memory serves correctly you're gonna want to look at permutations which preserve specific elements

but my galois is very very weak

[white oxide]

hm okay ye i'll just look at the cyclic subgroup generated by each, shouldn't be too bad

i suspect that this is gonna be the lamest lattice ever lol

[molten viper]

also consider like

[white oxide]

oh wait i'm dumb all elements except the identity are of order 2

so obviously it's gonna be a lame ass lattice

[molten viper]

if it were isomorphic to Z8, consider where you might map 3

but yeah you get it

[white oxide]

still pretty sick tho

[molten viper]

It gets more complex tho, like what are the subgroups of like, Gal(Q(sqrt2, cuberoot3)/Q) for example

[white oxide]

oh yea i should think about that

[molten viper]

Also you haven't found all the subgroups yet

there's a few which are like {p_0, p_1, p_2, \mu_1}

Which is isomorphic to uh, Z2 x Z2

[white oxide]

oh wait ur right

isn't there one also iso to the klein V group

[molten viper]

I have no clue what that is

[white oxide]

oh wait Z2 x Z2 is isomorphic to the klein V group my bad

smallest non cyclic group of order 4

[molten viper]

ah

[white oxide]

ugh now i have to destroy my lattice

[molten viper]

It might be easier to ask "hey what are the subgroups of the thing this is isomorphic to"

but you also know by the uh

[white oxide]

ye but i don't know those lmfao

or rather

i did an exercise for that

[molten viper]

Lebesgue theorem?

[white oxide]

but i forgot

nah

lagrange you meant?

[molten viper]

yes

[white oxide]

ye i know there are subgroups of order 4

2

and 1

or rather

potential ones but i know they exist

since we were just talking about them

so there can be at most two subgroups of order 4

if i'm not trippin and i remember correctly

ugh but i don't want to identify each auto with an element of z2 x z2 x z2

[molten viper]

No there's 3 actually I believe

or more

Yeah there's lots more than 2

[white oxide]

oh right oops

[molten viper]

For another example, $p_0, p_1, \delta_2, and \mu_2$ are a subgroup

[white oxide]

this is annoying

[cloud walrus]

Galstaff, Sorcerer of Light

[white oxide]

how were you able to figure that out so quickly

or are you just fast and i'm slow

[molten viper]

pick any 2 elements you like

the 3rd and 4th element of the subgroup are the identity and the product of the first elements you picked

Like, we're just making more copies of Z2 x Z2

[white oxide]

ah that makes sense

[molten viper]

I'm not sure how many there are tho....

[white oxide]

oh yea and i guess each element is its own inverse

[white oxide]

yeah ugh

[molten viper]

(identity omitted)
||p1 p2 m1
p1 p3 d1
p2 p3 m2
p3 m1 d2
m1 d1 m2
d1 d2 p2||

I think this is all of them?

[white oxide]

my question how did you know that the product of the first two elements (omitting the identity) guarantees that the subgroup is closed

like how do you not have to check that the subgroup itself is closed

[molten viper]

So

because each element is its own inverse, you basically get closure for free

[white oxide]

ohhh

right

[molten viper]

And these are abelian

if they weren't abelian you might have a problem

[white oxide]

but like how do you know that the product of distinct elements in the subgroup is in the subgroup

like say for example e, p1, p2, m1

so obviously we know p1 x p2 is in the subgroup

but why do you not have to check p2 x m1

or p1 x m1

[molten viper]

well, p2 * m1 = p2 * p1 * p2 = p1

[white oxide]

o

right.

[molten viper]

There's also like only 2 non trivial groups of order 4

Z_4 and Z_2 x Z_2

But this wouldn't necessarily work if we wanted the subgroups of D_8

jagr is about to pwn me

[rocky cloak]

(Z2)^3 has 16 subgroups

So that will be a pretty big drawing

[white oxide]

oh f

nvm then

[molten viper]

Don't they correspond 1 to 1 with uh

the subfields of the original field

[rocky cloak]

Intermediate extension, yes

[white oxide]

ye the intermediate fields

but i thought it was generally easier to find the subgroups

of the permutations

then the intermediate field

s

[molten viper]

In this case, it's definitely easier to just find those

[white oxide]

like that's the whole entire point of the fundamental theorem of galois theory

[molten viper]

Least for me

[white oxide]

oh rlly

hm

[molten viper]

Well yeah

just take like

But like

[white oxide]

i mean it's just the 6 subgroups of order 4 that galstaff listed plus the trivial ones, the cyclic groups of each non-identity element of order 2 right

plus the identity of course

[molten viper]

for a more complicated extension it absolutely won't be super easy

[rocky cloak]

Pretty easy to find the subgroups of (Z/2)^3 if you just think of it as a vector space over F2

[molten viper]

it will be very very hard

So like, in my earlier example of Q(sqrt2, cbrt3), the subfields might be harder to find

[rocky cloak]

Yes, though I only count 7 subgroups of order 4

[molten viper]

Wait did I miss one

I got 6 earlier

[white oxide]

nvm i can't count

ye

p0 p1 d2 m2

[molten viper]

ah thank you

[white oxide]

mhm

so the whole group = 1 + 7 + 7 + 1

ah

[molten viper]

I think the galois group here is gonna be Z2 x Z3 but don't quote me on that

Probably not tbh

¯_(ツ)_/¯

I think it's iso to Q[x]/(x^2-2,x^3 - 3)

O might be Z2 x S3 then, if you start permuting roots of unity

Like I said, my galois is weak

[white oxide]

UGH I DREW IT IN A WAY THAT I CAN'T DRAW THE SUBGROUPS OF ORDER 2 DIRECTLY UNDERNEATH EACH SUBGROUP OF ORDER 4 I'M SUCH AN IDIOT

[white oxide]

stronger than mine fs

[molten viper]

I mean, I passed the course

I'm much better with rings

because that's what I am actually looking at on a daily basis

and if you read back in this chat, you'll see I'm not the best lol

[topaz solar]

Better than me at least

[rocky cloak]

I'm actually gonna be examinator for a Galois theory exam in two weeks

[molten viper]

Slay

I might get the pleasure of grading group theory psets

[white oxide]

no surprise there

W

lol

good enough

looks pretty cool

now time for the field lattice!

[molten viper]

wish I could draw the problems I was solving lol

I mean, you can ig but I don't think they'd be super enlightening

especially considering it'd need to be n-dimensional

[south patrol]

Lol

[molten viper]

and n = 3 is TRIVIAL

[south patrol]

Imagining drawing like homotopy theory now

[molten viper]

Isn't that what all those arrows are for

[south patrol]

Gonna draw an infinite sequence of topological spaces real quick

[topaz solar]

(X_n)

[white oxide]

help what's the easiest way to show that this series is abelian i don't want to give a million names to a million things

i guess i can use the third isomorphism theorem but that requires me to show that HiN is a normal subgroup of Hi+1N and that Hi+1N is abelian

nvm

i'm dumb

each one is obviously iso to Hi and the quotient groups are abelian

oops

[chilly radish]

Wdym iso to H_i

The key point here is the 3rd isomorphism theorem

[white oxide]

nvm im a moron

yeah you're right

what am i doing

LOL

[white oxide]

so basically i have to show this then

wait doesn't HiN normal in Hi+1N follow from butterfly or am i trippin

hmmm

[chilly radish]

It follows from the correspondence theorem yes

Well

HiN/N in Hi+1N/N

[white oxide]

Oh right that’s better

Agh that’s also annoying

[coral spindle]

Type theorists would write f x, so you're in good company

I personally like that style but it's not popular in the least

I remember hearing some great advice along the lines of convention being preferable to personal preference in notation

[void cosmos]

let R be commutative with identity , P be a prime ideal in R , A be a noetherian R-module and C be a P-primary (rad (C) = P) submodule of A

then there exists m int >0 such that P^mA is contained in C

any hints?

[silver scaffold]

It would be really appreciated if someone can check if my proof is correct

[formal ermine]

can you send it as text

[silver scaffold]

I tried multiple times in #latex-testing but the syntax keeps on getting messed up. I'll try again

[formal ermine]

that image is too small for me to read on my phone

[wooden tapir]

Hi i need some hepl with advanced university math 🙂 differential equation

[unkempt stream]

#odes-and-pdes

[wooden tapir]

coul someone help me please:)

[formal ermine]

this isn't the correct channel mate

[unkempt stream]

Assume R is Noetherian.
Let I_n be a sequence of strictly increasing ideals

1. If I_n is strictly in R, then contradiction with Noetherian-ness (lol)
2. If I_n is strictly outside R*, then they're contained in the ideal (X), which R[X]/(X) is iso to R, so by the isomorphism theorem there's a contradiction

[wooden tapir]

Oh sorry i didnt find it but thank man!

[formal ermine]

mizalign linked you it 🙂

[unkempt stream]

Assume I_n are all prime, and after some N, R intersect I_n is non-trivial

However, for any prime idea P, P is maximal amongst ideals J such that P and J have the same intersection with R

So the chain R intersect I_n is strictly increasing in R, which violates Noetherian-ness

However, idk how to extend this proof beyond prime ideals WITHOUT invoking Choice to assume every ideal is contained in a prime (maximal) ideal via Zorn's Lemma

Because the normal proof R[X] is noetherian uses the leading coefficient map (I'm avoiding because I want to try to understand the Ideal structure) DOESN'T need the full power of choice

is there any way to somehow extend this proof for prime sequences to all ideal sequences without invoking choice?

[topaz solar]

Why would you have ideals not subset of R

[unkempt stream]

It's basically to split the proof into 3 cases

really I can get rid of the second argument

It was a byproduct of me trying to find an original method

Afaik if ideal A is contained in B, then A intersect (X) need not be strictly contained in B intersect (X)

i.e B\A can be disjoint from (X)

BUT if A and B are prime, then B\A CANNOT be disjoint from R

I really want to avoid the leading coefficients map because it isn't an actual ring morphism

[formal ermine]

outside R?

[unkempt stream]

Outside R remove 0, which is the ideal (X), or ideal of polynomials with a 0 constant

[formal ermine]

wdym outside R

[unkempt stream]

R is a noetherian ring

R[X] is it's polynomial ring

https://mathoverflow.net/questions/222923/alternate-proofs-of-hilberts-basis-theorem

`"A suitable notion of a Noetherian module is given in the very fine book A Course in Constructive Algebra by Mines, Richman, and Ruitenburg: A module is Noetherian if every ascending chain of finitely generated submodules stops (think "Un=Un+1
", not "Un=Un+1=Un+2=⋯
").

This definition works fine for many purposes, but not for showing that R[X]
is Noetherian if R
is. See Chapter VIII of that book. Also it doesn't work well if dependent choice is not available, since it refers to sequences, which can be quite elusive without choice."`

violent screaming

I wonder if I can use the "every set of ideals has a maximal element"

notion

[tawny magnet]

Why cant we just stay in PIDs so we dont need ZFC 😭

[wraith cargo]

average algebraic number theorists crying at the thought of this idea

[wooden tapir]

To advanced for me here xD

Trying to understand

[crystal turtle]

Why can't we just accept choice as part of life nowadays

[tawny magnet]

why should we care about choice when all our ideals are nice and generated by one element?

[delicate bloom]

we can and we do

[wraith cargo]

Choice is the mathematical equivalent to consent
You can't do anything cool without it

[open sluice]

the illusion of choice

[open sluice]

nice save

[wraith cargo]

I will not comment on the typo lmao

[crystal turtle]

Lmaooo

[topaz solar]

you can do uhhhhh

questionable things like non-well orderable sets, but that's less cool more inconvenient

[formal ermine]

.pin

[coral steeple]

Let H, K be subgroups of a group G. Can I have a hint on how to prove that if the intersection of cosets xK \cap yH is not empty, then it is a left coset of K \cap H?

I don't see at all how we could show that some element is in K and H

I don't have any idea why this ought to be true either, cosets seem so unintuitive

[hollow mica]

This is an interesting message to pin

[ivory trail]

if xK \cap yH is not empty, then there is some element z in it. then zK = xK and...

[coral steeple]

I'll think that over thanks

[worthy sonnet]

If phi: Z30,+ -> Z30,+ is a homomorphism. And ker(phi)= {0,10,20} and phi(23)=9 what are other elements that get mapped to 9 and what is the mapping doing?

[delicate orchid]

Use the fact that it’s a homomorphism, so phi(a+b) = phi(a)+phi(b), and the fact that you know phi(23)=9

[worthy sonnet]

I tried

But 23=12+11 or 23=10+13?

So there are many possibilities

[delicate orchid]

But all of those possibilities lead to the same conclusion

[cloud walrus]

wew

wew

[worthy sonnet]

And what does the mapping do tho? So phi(10) + phi(13) = 0 + phi(13) so phi(13) has to be equal to 9

What mapping sends 13 to 9?

[delicate orchid]

An easier question might be “what sends 10 and 20 to 0”

Remember we’re working mod 30

[worthy sonnet]

I understood it now....the question was never what the mapping is, but what other elements get mapped to 9.
So 13 is also a solution and 3 is also one.

[delicate orchid]

Yeah, but the mapping is just multiplying by 3

[worthy sonnet]

I hate being under time pressure in exams. Literally was able to solve everything but this

[chilly ocean]

What does $\oplus$ mean in this context?

[cloud walrus]

Plazzi

[chilly ocean]

Just +?

[tribal moss]

Direct sum.

[chilly ocean]

Ty

[tribal moss]

Mutiplication by 21?
(Pretty sure the exercise didn't expect actually finding phi, but just "elements that get mapped to the same image by a homomorphism are exacatly the ones that differ by something in the kernel").

[solemn dew]

any hints?

what i have done so far

[chilly ocean]

-1 = (-1)^2

[solemn dew]

like this?

[wraith cargo]

or you could simplify if by instead using x-y

at the beginning of your argument

[chilly ocean]

you still have to use 1 = -1 somewhere

[wraith cargo]

hm that is true

[chilly ocean]

it works but it could be shortened a bit

when i wrote -1 = (-1)^2 i was hoping you would notice -yx = (-1)yx

and then with the hint you see this equals yx

[solemn dew]

oh

i didn't understand that 1=-1 was something that could work

[chilly ocean]

rings, man

[solemn dew]

hahah

first chapter

i'll get there

thanks

[chilly ocean]

you'll get used to it

[molten viper]

A quick question, for my exercise today the problem reads "for a ring A \neq 0, prove that the set of prime ideals of A has minimal elements with respect to inclusion". Am I right in reading that as saying there are prime ideals of A which have no subsets which are prime ideals?

[crystal turtle]

Correct

[chilly ocean]

zorn time

[molten viper]

Yeah my thought was zorn's lemma, but at least the formulation I read talks about maximal elements

[crystal turtle]

Minimal elements are "maximal" when you reverse the order of inclusion

Like taking A <= B if B is a subset of A

[molten viper]

So the proof might roughly go something like

So my ordering is let's say the "superset" relationship

rather than the subset relationship

[crystal turtle]

yeah

[molten viper]

So basically to apply zorn's lemma I want to show that every chain in the set of prime ideals has an upper bound

Now, does that upper bound need to be a prime ideal?

I think it does

[crystal turtle]

For it to be in the set of prime ideals, yes

[molten viper]

So, I notice that if A is an integral domain, (0) is a maximal element

as 0 \ge every ideal in A

and it's prime

but if A isn't an integral domain

it gets tricky

[crystal turtle]

Right

So how can you get an upper bound when the ring isn't an integral domain?

(an upper bound, since we have reversed inclusion, is a prime ideal contained in every prime ideal in the chain)

[molten viper]

Well, a principle ideal would be particularly useful

[crystal turtle]

In general, you can't get a principal ideal for this

[molten viper]

true true

[crystal turtle]

What's the largest ideal you can get as a bound for this? I.e. what's the largest ideal that can be contained in every ideal in the chain?

[molten viper]

Well, 0

but that's not necessarily prime

[crystal turtle]

Right, we can get a larger ideal though

[molten viper]

I mean, we're working under the idea that this isn't an integral domain

so I guess some zero divisor would be bigger

well no

[crystal turtle]

A general rule of thumb when applying Zorn's lemma to any poset of subsets (ordered by subset or superset) is to first take the chains intersection or union

And then go from there

[molten viper]

hmm

Ok well let's take the intersection over this chain

But I'm not sure that that's prime, so we call p the intersection over the chain. If x \in p and x = ab, it's not guaranteed that a or b is in the intersection

like a might be in half of them and b in the other half

[crystal turtle]

Well it's at least an ideal that is contained in every ideal in the chain, so it's a stat

[molten viper]

It is

So why not just remove all the elements like p

[crystal turtle]

You have to use the fact that it's a chain to show this can't happen

[molten viper]

ahhhh

[topaz solar]

Well, if a is outside of one, then it’s outside of each successive one?

Strictly decreasing moment

[crystal turtle]

So you might start like "if a is in every p_i, we are done, since a is in p. Otherwise..." Then show that implies b is in every p_i, hence in p

[topaz solar]

Then invoke zorn

[crystal turtle]

"Professor, can we do this without axiom of choice??? 🥺 "

[molten viper]

Hm, so

if there's some x with a \notin x, then a isn't in any of the elements greater than x

[chilly ocean]

every time someone did this in class i wanted to throw my seat at them. it's a good thing the chairs were built into the floor in most rooms

[crystal turtle]

lmaoo

[molten viper]

so the bs are in all of these elements we're talking about, and actually they're in every element of the chain because we're working with inclusions

[crystal turtle]

Right

[molten viper]

fuck man, this is twisting my head

Doesn't help I'm very overstimulated rn but it's whatev

[crystal turtle]

b must be in all ideals after a certain point in the chain by assumption, hence b must be in every ideal by inclusion

[topaz solar]

I think asking if choice is essential is nice sometimes, but maybe not all the time in class

[chilly ocean]

you know the kid doing it in class is the kind of kid to do it all the time

[wraith cargo]

poll what is more annoying:
Asking abt choice
Asking whether a map is well defined in the proof of snake lemma

[topaz solar]

Ok true

[crystal turtle]

asking about choice

[chilly ocean]

asking about choice

[topaz solar]

Choice, the snake lemma thing is normal

[chilly ocean]

i've never seen someone ask about the axiom of choice in class when it is interesting and the answer would benefit many people in the room

it's always some really stupid shit like "do we need choice to prove that this system of equations has a solution?"

[wraith cargo]

I don't understand why ppl have a problem with choice

[molten viper]

I get the general sketch of the proof

so today we're just gonna move on lol

[topaz solar]

I mean that’s pretty much the whole thing

[coral shale]

the honorable sharp

[crystal turtle]

^^ seems like you've got the proof, last part is just "the result follows from Zorn"

[round hull]

why not

[lethal dune]

Choice is a lie

[formal ermine]

The Axiom of Choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's Lemma

[solemn dew]

Does anyone know how to think here?

Can't think of any example

[formal ermine]

algebraic closure

[warm wyvern]

coz of the existence of non-measurable sets

and results like these

also coz constructive proofs are a lot more insightful

whereas choice is such an ass pull

"just trust me bro, the function exists"

[formal ermine]

who tf even cares about analysis

[warm wyvern]

*sniff* *sniff*

do you smell that?

I smell skill issue

[delicate orchid]

Who even cares about analysis

[warm wyvern]

I do, wew

I do

[round hull]

do you know about p-adics?

[warm wyvern]

getting ganged up on in #groups-rings-fields

[formal ermine]

lol what

[lime badge]

So I'm confused about this problem I'm doing. Let $K$ be an algebraically closed field. Consider the category K-Aff of affine K-algebraic sets of some affine K-space and define $\text{Hom}(S,T) \coloneqq \text{Hom}_{\text{K-Alg}}(K[T], K[S])$, where K[T] is the coordinate ring of T and similarly for S.

If we have the covariant functor $h_s \coloneqq \text{Hom}_{\text{K-Aff}}( _, S)$ and $p$ a point in the affine K-space, I need to show there is a natural bijection between $S$ and $h_S(p)$

I'm supposed to use this result that if a point p of an algebraic set corresponds to the maximal ideal $m_p$, then the value of a function $f \in K[S]$ at $p$ equals the coset of $f$ in $K[S]/m_p \cong K$

[solemn dew]

please enlighten me🙏

[warm wyvern]

think of a finite integral domain with finite characteristic

and then use that to get something bigger

and then use that to get something infinitely bigger

[formal ermine]

@solemn dew

[warm wyvern]

how do you know the algebraic closure is infinite?

[formal ermine]

because algebraic closures need to be infinite lol?

otherwise they're not algebraically closed

[round hull]

sorry

[warm wyvern]

I'm not so sure how that's obvious

[coral shale]

this is a short one to prove (do it)

[formal ermine]

by completion theorem Qp contains an isometry of Q thus it has zero characteristic

[warm wyvern]

@lime badge can you take this to #latex-testing first?

[lime badge]

Sorry, I'm done

[warm wyvern]

[coral shale]

wait what i thought this was short to prove

maybe im missing some context

[formal ermine]

it is

[coral shale]

illumi stop trolling the analyst

[cloud walrus]

Eternal Way

[formal ermine]

darq is just having a skill issue

[coral shale]

he's not called chmonkey...

[warm wyvern]

i'm his enemy you bitch

[formal ermine]

||consider prod_{a \in K} (x - a) + 1||

[round hull]

take the countable product of integral domains, and consider the subring where everything right of a 0 is 0

[formal ermine]

why so complicated

[lime badge]

Anyway, my question is just that K-algebra homomorphisms between coordinate rings are right composing by some function that maps between our algebraic sets right? So what exactly is there to prove for my above queston?

[solemn dew]

i can barely write, my internet connecting is something else right now

[formal ermine]

eternal I think this belongs in #algebraic-geometry

[solemn dew]

but i think i got it thanks

[lime badge]

No problem, thanks!

[solemn dew]

and ill look into algebraic closures

[crystal turtle]

Not sure, but that looks a lot like Yoneda, no?

@lime badge

[round hull]

they're in the other channel

[round hull]

how is that complicated

arguably it is more elementary

[coral shale]

DarQ, it really is worth u trying to figure it out without the hint btw

[warm wyvern]

what hint?

I think I figured it out tho

just what lumin said

[coral shale]

well yeah he practically gave the answer in the spoiler

[warm wyvern]

¯_(ツ)_/¯

fam I'm not thinking about factorization and shit

that shit sounds so boring

[round hull]

darkyuuutie
obviously not thinking about factorization makes you a lesser mathematician

[warm wyvern]

now who tf taught you that name?

do I know you somehow?

also, I will sometime

but not until I have to

@formal ermine how do you know the algebraic closure exists tho?

[crystal turtle]

axiom of choice lmao

[warm wyvern]

God dang it

sigh

[wraith cargo]

There's a Zorns lemma argument

[coral shale]

zorn is something u accept bro

at the start of rings and sht

deal with it

[delicate orchid]

Why is the analysist crying about choice lol

[warm wyvern]

Yea this is definitely less complicated then infinite product go brrr

[crystal turtle]

Knowing that an infinite product of sets with cardinality at least 2 is in fact infinite requires choice as well, right?

Because an equivalence statement is that every infinite product of nonempty sets is nonempty

[warm wyvern]

That's a choice I can choose to accept

[crystal turtle]

LMAO they're equivalent

But you only need countable choice I gues

[warm wyvern]

I know lmfao

[delicate orchid]

If it’s a countable infinite product you might be able to get away with it

[delicate orchid]

Yeah

[crystal turtle]

ZF does not include countable choice by default

[round hull]

you can list the elements

in this case at least

[topaz solar]

Well ordering

[crystal turtle]

I guess since this is an explicit product you can, right?

But I feel like that still would require countable choice?

[round hull]

(1, 0 , ... ), (1, 1, 0, ... ), (1, 1, 1, 0, ... ) are all distinct and in the subring

so it is infinite

[topaz solar]

Rings you can point to specific elements in em, like 0, 1

Doesn’t need choice because it’s already a function

[glossy crag]

F_p(x), the field of rational functions with coefficients in F_p: it's a field (so definitely an integral domain), is of characteristic p (because F_p is), and is certifiably infinite (because it contains F_p[x], the polynomial ring over F_p). As @formal ermine pointed out, an algebraic closure of F_p (or any field of finite characteristic) also satisfies these requirements, but for this you need to know what an AC is and that an algebraically closed field necessarily has infinitely many elements.
Actually idk why I'm complicating things here, the simplest answer is F_p[x].

[formal ermine]

facepalm

[unkempt stream]

I just realized a few days ago that Chinese Remainer theorem holds for abelian groups lol

[glossy crag]

It holds for modules in general, which includes rings and abelian groups.

[wraith cargo]

I mean it holds generally for rings too lol

[unkempt stream]

Yeah which extends up to modules

[formal ermine]

crt holds for modules too btw

[unkempt stream]

I was like wait, lol, the proof doesn't even require multiplication lol

[wraith cargo]

I don't accept this is as fact

[unkempt stream]

Why

[wraith cargo]

It isn't aligned with my world view

[unkempt stream]

:troll:

[wraith cargo]

[formal ermine]

turning 16 in 4 hours and 5 mins

[unkempt stream]

arguably thinking about it in terms of abelian groups is easier and more intuitive

[unkempt stream]

Works because submodules work nicely and analogously as abelian group quotients

And ideals are technically modules

[warm wyvern]

oh mizlang is back

@unkempt stream sup

you got into the uni you wanted?

[unkempt stream]

no

got waitlisted and rejected 😦

going to a local college

affordable

[warm wyvern]

oof

that's oki

as long as your local college has grad courses

and allows you to take them

[unkempt stream]

n o p e

They don't offer even abstract alg

[warm wyvern]

[unkempt stream]

I have dummit & foote tho, and I am doing EE as a major. Always was gonna

[open sluice]

rip

happens

[warm wyvern]

that's rough

[solar vessel]

[unkempt stream]

dummit & foote is more interesting than I think an actual abstract alg class is anyway

it's more of something I do before I go to bed nowadays, just a set of problems

[crystal turtle]

I find dummit & foote to be a bit slow/overly talkative

[void cosmos]

is there a proof of krull intersection theorem without primairy decmop?

[wraith cargo]

you can prove it with the artin-rees lemma

see Eisenbud ch. 5.3

[void cosmos]

thank you for the ref

[wraith cargo]

average mathematician when a book isn't just a list of theorems and proofs that have no correlation to one another:

[white oxide]

might be a silly question, but what does it mean to embed some algebraic object in another object? do we only require a monomorphism A --> B, or does this homomorphism have to be the inclusion map?

[crystal turtle]

Injective is sufficient

[white oxide]

oh okay thanks

[crystal turtle]

Since if it is injective, it should induce an isomorphism with it's image, which "properly" embeds into the object through the inclusion

Though some people might stick strictly to an embedding being the inclusion map

[white oxide]

ye, it seems as if it's a question if it's up to isomorphism or not

i guess algebraically there's no difference lol

or at least from an algebraist's point of view

[crystal turtle]

If you're working with anything categorical, it should be up to isomorphism almost always

[white oxide]

true

[crystal turtle]

(or up to some weaker notion like equivalence, context permitting)

[unkempt stream]

i like the problems

[unkempt stream]

equivalence is abstraction of abstraction :troll:

[white oxide]

would this simplify to the trivial homomorphism? since any $\varphi \in \text{Hom}_\mathbb{Z}(R, J)$ would just send $1_R \mapsto 1_J$ where $1_J$ is the identity in $J$

[cloud walrus]

okeyokay

[crystal turtle]

I would send 0_R to 0_J, right? Not 1? (writing J additively)

[white oxide]

oh well i was just going by their notation

$f(a) = 1_J$

[cloud walrus]

okeyokay

[unkempt stream]

D e a r g o d

[white oxide]

wut

[unkempt stream]

abstract algebra does irreprable damage to your psyche because people say shit like this and you’re like

w h a t p a r t

[crystal turtle]

R is a ring with a multiplicative and additive identity, 1_R and 0_r respectively. J is an abelian group, in the category Mod_Z of Z-modules (or abelian groups). The identity requirement for homomorphisms in this would be that it takes 0 to 0, not 1 to 1

[unkempt stream]

you’re essentially “forgetting” the ring part of it, or that monoid of multiplication on it

[white oxide]

ye i know but they wrote 1_R so i'm just going with their notation

[unkempt stream]

So the 1 part doesn’t fucking matter if it’s purely an abelian morphism

[crystal turtle]

This

[unkempt stream]

Forgetful functor goes brr

free functor go rrb

[crystal turtle]

0_R goes to 0_J, but 1_R can go anywhere

Since J is not a ring

[white oxide]

ohhh i see now

that makes sense

my b

[unkempt stream]

nothing’s special about 1 in the category of abelian groups

In rings (WITH IDENTITY) yes

[crystal turtle]

exactly

[white oxide]

i thought we were considering just the additive abelian group structure of R

[crystal turtle]

(except in the category of abeliean groups, the group 1 is the zero object)

(which is pretty important)

[white oxide]

so that Hom_Z(R, J) would make sense

[topaz solar]

g is a map L->J, but f(a): R->J, so you need to insert a value from R to get a value in J, and 1 is an easy choice since it doesn't have to map to 0

[unkempt stream]

Really R as a ring and R as an abelian group are “different objecrs” but can be mapped from a ring to an abelian group via a functor

[topaz solar]

or uh, should've replied to okey

[white oxide]

oh that makes sense thanks

also wow

ur honorable now

congrats

[topaz solar]

think polynomial evaluation

these aren't polynomials but it's similar that you want to actually evaluate the function

[crystal turtle]

mapping g(a)=f(a) is such a Yoneda-style move ngl

[unkempt stream]

for example, the complex numbers are the same as a vector space as R^2

but as rings, they are not

[crystal turtle]

C_p (completion of p-adics) and C being isomorphic as rings moment

[unkempt stream]

NO.

[unkempt stream]

i thought it’s the closure of the completion lol

[crystal turtle]

oh right, my b

[unkempt stream]

neat though how the closure-completion of basically all nontrivial metrics on Q return to the same place isomorphically (with choice)

[crystal turtle]

only as rings tho, C_p as a metric space is vastly different lol

But yeah it's quite interesting how much the p-adics and reals relate

[unkempt stream]

FORGET IT (literally)

[warm wyvern]

aluffi isn't that but it's still very very fun

[warm wyvern]

the former

[unkempt stream]

inclusions afaik are monos in SET

[warm wyvern]

the inclusion map is only apparent is only ever apparent when the algebraic structure you're trying to embed is evidently a subset of where you wanna embed it in

[crystal turtle]

right, but monomorphisms=injections in a lot (but not all) of concrete categories

[white oxide]

yea i guess that makes sense intuitively

[warm wyvern]

in that case the inclusion map is just the restriction of the identity to said subset

[white oxide]

can you explain what you mean by this? i thought monomorphisms were injective as a map of sets

[white oxide]

ye makes sense

[warm wyvern]

monomorphisms aren't always injective maps

[crystal turtle]

A monomorphism is a map f such that, if fg=fh, then g=h

I.e. left cancellable

In Set, a function is left cancellable iff it is injective, so that's why its true in Set

And in fact, in any concrete category over Set, any injective function is going to be a monomorphism

But in some categories, there are monomorphisms that aren't injective

[warm wyvern]

I actually don't know of a category where the monomorphisms aren't injective

[crystal turtle]

neither do I lol

[warm wyvern]

what's a concrete category over set?

tbh it's prolly something stupid like a poset category or smth

where elements don't even make sense for you to define injectivity

[crystal turtle]

A category C is concrete (over Set) if there is a faithful functor from C into the category (of Sets)

I.e., it means you can view the objects are "structured" sets, and morphisms as "structure-preserving" functions