groups-rings-fields

[steel light]

How did you get that implication

Or rather why do you need the i < m and j > l part

[maiden ocean]

if i >= m or j >= l then c y^i z^j is in a or b

im claiming that i >= m or j >= l for every term in the sum (y + z)^(m+l)

by the binomial theorem

either i >= m, or, if i < m, j >= l

[steel light]

I think I'm on the cusp of understanding, I don't think I'll get it properly until I write it out myself

Which I cannot do right now

[steel light]

I understand this at least lol

Just not that verrryyy last implication (x^n)^(m+l) is in a + b

[maiden ocean]

you are writing it as a sum of terms that are either in a or b

hence the sum is in a + b

[steel light]

Don't we need one term to be in a and the other to be in b

[next obsidian]

x = x + 0

y = 0 + y

[steel light]

I think I see

If for a sum x + y we have x, y in a, then just write x + y = x + y + 0, 0 is in b so we now have a sum of elements in a and b and 0, and obviously 0 is in the radical of b, so the sum is in r(a + b)

Why do we keep explicitly stating that the elements are sums? Isn't this implicit in an ideal being an additive subgroup?

If y1f(x1) and y2f(x2) are in a^e then so is their sum

[next obsidian]

Because if you don’t write the word “sums” then it won’t be closed under sums

There’s a difference between {yf(x)} and {Sum y_if(x_i)}

[steel light]

Oh it's a sum of sums

What

[next obsidian]

What

[steel light]

LMAO

[crystal turtle]

what

[steel light]

but why not

Doesn't closure under sums come right from additive subgroup?

[next obsidian]

Try an example lol

If you don’t close under sums it won’t be a subgroup lmfao

That’s what I’m saying

[steel light]

But it's an ideal

[next obsidian]

No it isn’t

It’s an ideal after you close under sums

[steel light]

Oh

I'm an idiot

[next obsidian]

Prove to me right now that if you take the set {yf(x)}

That this is closed under sums

[steel light]

I forgot f(a) isn't necessarily an ideal lol

Even though they literally just said it

ahah

[next obsidian]

f(a) is very far from being an ideal, it won’t even be absorptive for multiplication

It is closed under sums though

I feel like you are more confused

f(a) = {f(x) | x in a}

[steel light]

Class just ended so I haven't gotten the opportunity to process your latest messages

[next obsidian]

Closed under sums, not closed under multiplication by B

{yf(x)| x in a} closed under multiplication by B, no longer closed under sums

{Sum y_if(x_i) | x_i in a} an ideal

[white oxide]

aight thanks boss

[steel light]

Chm I still don’t understand but I sorta came up with my own justification? Does this work?

Let’s say a has elements a1, a2, …then Bf(a) looks like {Bf(a1), Bf(a2), …}

Each of these elements themselves are sets

I can't figure out what the product of Bf(ai)Bf(aj) is

Never mind I don't understand actually lol

If we take the sum Bf(ai) + Bf(aj)

Then you're just adding up bf(ai) + b'f(aj) for all b, b' in B

Rings are closed under multiplication so both summands are in B, rings are closed under sums so the sum is in B

Ah wait the goal is to make it an ideal

[coral shale]

a to indicate a set

[steel light]

I think it's supposed to be the set of all bb'f(ai)f(aj)? Lol

[coral shale]

ok so ideal axioms

R is a ring

[steel light]

I'm not making sense to myself

I'm going to restart

Yes

[coral shale]

I is ideal

[steel light]

Yes

[coral shale]

closed under products and sums

(correct me me idiot btw)

rx is in I for any r in R, x in I

[steel light]

Yes

[coral shale]

a + b is in I for any a, b in I

i think these 2 make your ideal axioms

[steel light]

Yes

[coral shale]

Then you want to think about this:

I give you a set

[steel light]

Oh so I need to show why one isn't true

[coral shale]

You want to generate the smallest ideal

and that is the ideal generated by that set

[coral shale]

i didnt necessarily follow your convo so far

[steel light]

Good

LMAO

Don't

[coral shale]

You mean for why f(I)

is not necessarily an ideal

Ur right - u need to pick an example and show one axiom no hold

But about generating ideals --- i will use the notation say
(S) to mean ideal generated by set S.

(S) = {...} ?

[steel light]

No, why Bf(I) is not necessarily an ideal until you say that it's the set of sums b_if(x_i) for b_i in B and x_i in I

[coral shale]

f(I) is a set

what ur after is the ideal generated by this set

[coral shale]

So this boils down to you figuring what this has to be in general

[steel light]

Yes

[coral shale]

So use each axiom one at a time

rx is in I for any r in R, x in I
x + y is in I for any x, y in I

[steel light]

Okay

[coral shale]

(we can use an explicit example and say S = {3, 5} in Z)

so like if s is in S

you know that ??? has to be in your ideal

[steel light]

So the claim is that Bf(I) is an ideal

So I need to show that for any r in B and x in Bf(I), rx is in Bf(I)
And for any x, y in Bf(I), x + y is in Bf(I)

[coral shale]

Well it isnt

[steel light]

Yes

One of these will be false

[coral shale]

Well the only thing u can do is provide an explicit counterexample

[steel light]

Oh

[coral shale]

But im trying to tell you that this problem is more general

And it fundamentally boils down to figuring out the smallest ideal generated by a set. Smallest in the sense that its minimal. The intersection of all ideals containing that set

[coral shale]

ZS for example is not an ideal

to give the appropriate analogy to your problem

[night onyx]

Well abstractly Bf(a) is the smallest ideal containing f(x) for all x in a (by definition), but concretely the set of all sums \sum y_i f(x_i) is an ideal, it contains each f(x) for x in a, and must be contained in any ideal containing each such f(x)

[next obsidian]

Literally try to prove that you can factor a sum

You will have different coefficients on the f(something in I)

And you can’t in general find a single element to squish that into

You’re trying to go from b1f(x1) + b2f(x2) = bf(x) and you can’t do it

If you want an explicit example, consider the inclusion of k[x,y] into k[x,y,z,w] and the image of (x,y). If you look at zx + wy you can’t possibly turn this into a monomial

[coral shale]

if b1, b2, have preimages or something it might be more possible

[steel light]

I thought it would be if x had a preimage

[next obsidian]

Then everything is happening in the image of A and it’s a done deal

how could x have a preimage, it’s a part of the domain 😭

[steel light]

sorry f(x)

LOL

[next obsidian]

Bro, x

Is the preimage

????

Or well, a

[steel light]

💀

brain so dumb sometimes

Sorry I think I meant that if b1(f1) + b2(f2) = bz, there may not be an x such that z = f(x)

[next obsidian]

That doesn’t matter?

If you don’t put a condition on z and b and f are both allowed to be just “exist in B” then that equation is pointless

z = b1(f1) + b2(f2) and b = 1

[coral shale]

I wanted to say earlier that
(S) = { sum r_i s_i }
which is all (finite) linear combinations of things in S, coefficients in R.

And then you could apply this to your problem by attempting to show set equality (f(I)) = {bf(i) : b in B, i in I} - showing both inclusions. One inclusion holds, but as chmonkey said, ull be stuck in the other (and once u are, this is instructive of how to construct a counter)

[steel light]

I cannot understand this and I don't want you guys to waste any more time trying to help me 😭

[night onyx]

lol it's not a waste of time

[steel light]

It is with me

Maybe with someone who has a proper background in ring theory it wouldn't be

I'm trying to just visualize what a^e looks like right now

[night onyx]

the important thing to understand is that by definition Bf(a) is the smallest ideal containing each f(x) for x in a. It's a purely abstract definition

now consider {all sums of r_i f(x_i) for x_i in a and r_i in R }

[steel light]

Yes

The thing I don't get is what if B is bigger than a? By the time you reach the last index for all of the x_i, you won't get every element in B in your linear combination (for one element)

[night onyx]

proving that this is Bf(a) is actually super easy:

1. the set of all sums is an ideal (prove this)
2. the set of all sums contains each f(x) for x in a (prove this)
3. if I is another ideal containing each f(x) for x in a, it must contain all sums of r_i f(x_i) (prove this)

that's really all there is to it

[steel light]

So do you just reindex B and then redo the sum and call that another element

[coral shale]

its all possible linear combinations, you can choose whatever you want to put in it

if S is the set youre generating ideals from, youre allowed to pick any finite subset of S

then attach them with any coefficients from your ring

And then add.

[steel light]

I think I understand

Any linear combination can, at most, have |a| terms in it though right

[coral shale]

yes, but theres also nothing wrong with having more

because

3a + 2a = 5a

if there are any repeated things from your set

you can combine (coefficients live in R)

[steel light]

Yes

Okay then let me rephrase, once you combine all the like terms then your linear combination will, at most, have |a| terms

[coral shale]

yes

[steel light]

Okay

[coral shale]

ive lost slight thread of what a is

[steel light]

I'm gonna say something maybe silly

[coral shale]

isnt that an ideal from your domain?

[coral shale]

your homomorphism need not be injective, so even less.

[steel light]

Yes

[steel light]

Yeah it could be less

[steel light]

This process sounds a lot like linear algebra

[coral shale]

and if f(a) is infinite. finite combinations only

it is

modules

[steel light]

Considering f(a) as a basis

and B as the "field"

but it's a ring

[coral shale]

it is a bit like that, except no guarantee of "independence"

Do you know what a module is

[coral shale]

what we are looking for is the span

span of S

or span of f(a)

[steel light]

Not formally, I know it's like a vector space over a ring instead of a field though

[coral shale]

yes. thats all it is.

thats why it looks like linear algebra to u

the ideal generated by f(a) is the span of f(a)

[steel light]

that's so much easier to understand bruh

😭

[coral shale]

i believe youre actually looking at the module

of a ring over itself

and the span in that context

[night onyx]

yeah, and in that sense ideals are like vector subspaces, closed under addition and scalar multiplication (by elements of R)

[steel light]

Yes

What the fuck?

That's pretty cool lol

[coral shale]

oh right of course

[steel light]

Good motivation for why modules even exist

[coral shale]

ideals are additive subgroups

[steel light]

Yes

[coral shale]

ideal over a ring is therefore a module

[topaz solar]

You can turn B into an A module I think too

[coral shale]

module is abelian group over a ring

[steel light]

Is there an analogue to Hamel bases?

[coral shale]

idk what that is

R over Q?

[steel light]

It's a type of basis for an inf dim VS

[night onyx]

no, there are modules with no R-basis

[coral shale]

In linear alg, all linear combs have to be finite

cus there is no analysis

dunno what an infinite sum is

[night onyx]

that's one of the first differences you see between linear algebra and module theory, the existence of bases

[topaz solar]

Also, consider, {2, 3} span Z as a Z module

[coral shale]

thats how direct sums differ from cartesian products. Direct sum - can only be non-zero on finite number of coords

[topaz solar]

There’s a basis for Z here even

You just can’t reduce that spanning set to it

You can have modules where na = 0 for a not equal to 0 too ofc

Even when your ring doesn’t have zero divisors

[night onyx]

lol you can think of module theory as "lets see what happens when you have a vector space over a general ring instead of a field" and you discover all the easy stuff from linear algebra isn't as easy anymore

[coral shale]

its interesting there isn't a "trivial" example to demonstrate your point of confusion here. I don't think

your rings have to be a bit 'bad' in some sense

If your ideal is principal, pretty sure it doesn't come up

so not PID

[topaz solar]

Agonizing

[steel light]

I have reached the end of the chapter

[next obsidian]

It’s a functional analysis thing where you require a topology

Hamel bases

It lets you avoid the annoyance that there’s no countable basis for an infinite product of the field

[ashen heron]

chapter? what topic? what book?

[steel light]

Chapter 1 AM

[coral shale]

a Hamel basis of a vector space V over a field F is a linearly independent subset of V that spans it.

[ashen heron]

algebraic basis

[coral shale]

i see, its just the generalisation in the sense basis is at first defined only for finite ig?

[solar vessel]

now it's time for the exercises

[ashen heron]

it is the one defined for where linear combinations have all scalars 0 except finitely many i think

[coral shale]

oh to differentiate between topological and la basis

[ashen heron]

the other is schauder when you have norm or sumthin

[coral shale]

right ok

So like your 'typical' la basis is hamel

[next obsidian]

Oh wait yeah it’s Schauder that’s infinite? I forgot

Hamel is normal I guess

[solar vessel]

yeah Hamel is algebraic

Schauder is some FA thing

[coral shale]

so aoc refers to hamel right

[solar vessel]

then there's Hilbert/orthonormal basis

[ashen heron]

exercises? what book?

[next obsidian]

Yes

[solar vessel]

uh well AoC is also used in the proof of Hilbert basis existence

it's pretty similar

[next obsidian]

Are they not equivalent?

[coral shale]

i mean is hamel the strongest

in the sense if hamel exists, the rest do?

or do we have other defns kindof independent

[solar vessel]

hmmmm

[ashen heron]

hamel always exists

[next obsidian]

With axiom of choice

[solar vessel]

well existence of Hamel for every vector space is equivalent to choice in ZF right

[coral shale]

yh

[ashen heron]

which is a truth

[coral shale]

🙄

[next obsidian]

Based, but we’re talking about aoc rn

[coral shale]

oh ok, i take it the rest need not always exist

[solar vessel]

I'm not sure if existence of Hilbert in every Hilbert space implies choice in ZF if that's what you mean

[next obsidian]

Afaict

[steel light]

I'm seeing so many terms I recognize this is exciting and scary

[solar vessel]

[ashen heron]

wait why are we in #groups-rings-fields

[coral shale]

modules

[solar vessel]

idk something about bases

[coral shale]

#groups-rings-modules-fields

[steel light]

Boolean rings, algebraic closure, spectrum of a ring, …

[coral shale]

convo took a curve

oh shit

https://math.stackexchange.com/questions/951846/over-zf-does-every-hilbert-space-have-a-basis-imply-ac

apparently this is open

no idea what a hilbert basis is

[next obsidian]

Imagine talking about the axiom of choice and Schauder based in #linear-algebra lol

[next obsidian]

It defined it

Lol

[coral shale]

right schauder is fairly quick to get

topological vector space, we have limits

[solar vessel]

a Hilbert basis is an orthonormal set with dense span

schauder is a lot more

[next obsidian]

Ummmmm he didn’t say linearly independent huehuehuehue

[solar vessel]

who

[next obsidian]

You. You didn’t say linearly independent set

[solar vessel]

I said orthonormal

[coral shale]

exdee

[next obsidian]

The joke

Your head

[solar vessel]

ratio

[next obsidian]

[ashen heron]

what am i missing here what the joke

[next obsidian]

The joke is orthonormal implies linearly independent

[coral shale]

i totally didnt get module shit 3 yrs ago. Can confirm doing nothing teaches u math

[void cosmos]

yo 9 is trivial right?

cuz its a pid so i get maximal ideals

if R is not a field

the -> direction

and for <- u cant have them intersect

cuz they would be associates ig

[summer path]

I didn't understand modules 2 years ago either, can confirm reading about modules makes you learn modules

[white oxide]

Dumb/trivial question: "By an $A$-algebra we mean a module together with a bilinear map $g: E \times E \to E$. Does this imply that $E$ is an $A$-module?

[cloud walrus]

okeyokay

[white oxide]

or rather E is an A-algebra if and only if E is an A-module equipped with a bilinear map

[south patrol]

Isn't that what they are saying

Tbh there are various different definitions of an A-algebra

This is like what lol nonunital nonassociative

[topaz solar]

What else would it be a module of tbh

[south patrol]

Z

[void cosmos]

anyone?

[sly crescent]

Is the group ring Z[C_3] isomorphic to the Eisenstein integers?

[night onyx]

yeah E is an A-module (when they say "a module together with" it should really say "an A-module E together with...")

[coral spindle]

No, but Z[C_6] is. Remember that the Eisenstein integers adjoin a primitive sixth root of unity to Z, rather than merely a primitive 3rd one.

[sly crescent]

Then what is Z[C_3]?

[coral spindle]

That's a vague question

ZC_3 is ZC_3

[void cosmos]

math is math

[sly crescent]

But can’t you represent a sixth root as a third root times -1?

[coral spindle]

Oh hm yes I see what you're saying

[sly crescent]

BUT

[coral spindle]

We don't have the same relations in any case

But now that I mention it, we don't have the same relations for ZC_6, so I spoke too early

[sly crescent]

Adding together the elements of the base group doesn’t result in 0 in Z[C_3].

[coral spindle]

Yes exactly

ZC_3 is iso to Z[x]/(x^3-1) whereas the Eisenstein integers Z[omega] is iso to Z[x]/(x^2 + x + 1) or equally x^2 - x + 1.

I think it should be clear the two are not isomorphic just from the fact that one is a domain and the other is not, in that case

I thought that ZC_6 would be isomorphic because of something I know about the relationship between the group ring and the group of units, but I made a mistake in the logic there that was clumsy.

[white oxide]

can somebody help me understand this part of the easy proof? it's kinda on the tip of my brain but i'm too smooth brained to spit it out

[topaz solar]

If your group has torsion, do you have zero divisors?

[delicate orchid]

Wtf are the eisenstein integers

[south patrol]

Z[w] ig

[topaz solar]

This?

[south patrol]

ZC_3 is a rank 3 Z-module, whilst Z[omega] is a rank 2 one

actually are there more meme arguments

[steel light]

Is this rank similar to tensor rank?

[steel light]

anyone up

o well let me know about these as a problem set

they’re the problems that interested me

Too many interesting problems wtaf

I need to pick out of these 😭

If I had to pick even fewer out of those then I’d say

1 - 4
7
13
15
19
26

What do we think of those

That’s still 10 problems yikes

[steel light]

If I had to pick even fewer from these I would take out 7…? But that’s it…the rest of them are just too damn interesting lol

[summer path]

i'd just say just do enough to feel reasonably comfortable with the material, then continue on

[steel light]

wait I just noticed commalg belongs in #advanced-algebra

LMAO

[summer path]

because otherwise you could spend a good bit of time here

lol

i don't think it really matters that much which channel you put this in

i think it's more so a the intro material belongs here and not there

[steel light]

That’s why I’m asking y’all, idk how much is that much

and also bc idk which of the exercises I like are actually good

for developing your skills and testing your knowledge

[summer path]

i think the statements in 2 are nice to know, but proving some of them are pretty tedious lol

[steel light]

I’m studying commalg to hopefully move to ag so I feel like 2 is important

now 3 sounds hella tedious

but unfortunately also really important too

[summer path]

i think there are definitely statements where you can just use them without having proven them prior, (e.g. 3 if you have already done 2)

i think it'd be a better use of time if you just did say like 5-6 problems here you aren't sure how to do immediately and then continue onwards

10 problems is a bit much

[steel light]

I agree

26 isn’t even really a problem by the looks of it…?

[summer path]

wall of text

im too tired to read all of that

[steel light]

1, 2, 4, 13, 15, 19

Ahhh but there’s nothing on maximal ideals

I should cover that

[steel light]

Wait yes there is

4

2 is so exhausting though and I want another problem with maximal ideals

I think I’ll skip it and replace it with 7 unless it’s one of those problems that’s strongly recommended to do

1, 4, 7, 13, 15, 19

[snow condor]

this is a typo and should be G_p, right...?

[south patrol]

ye

[snow condor]

thx

[steel light]

So yesterday John recommended I go through this German professor man dude person guy bro’s excellent algebraic geometry notes to get some motivation for learning this stuff

He said it really comes to light when you see the geometry behind the concepts

I skimmed the notes and I don’t really see what he means but I’m sure that’s because my algebra is weak and perhaps I still don’t properly understand what geometry is lol

[mighty kiln]

this German professor man dude person guy bro’s

[steel light]

In my mind geometry is the study of shapes, distances, angles, curvature, and how these properties transform via diffeo/homeomorphism or affine transformations

Both globally and locally

I did not see any of that…? Though I didn’t get far

[steel light]

yeah his name was some long name that starts with a G

[mighty kiln]

What notes are those

[steel light]

Like Gimo and Gilluminator

[coral spindle]

Gathmann

?

Stick w it

[steel light]

Yes Gathmann

[open sluice]

🤔 if you want that shouldn't you be doing diff geo

[steel light]

Oh no I’m not saying that’s what I want, just that that’s what my interpretation of what geometry is was

So is that wrong? What else is geometry?

I mean I do want that but I’m also open to more, I’ll come back to diffgeo one day

[open sluice]

all i know is hilbert geometry

[steel light]

no one appreciated my joke :sadge:

[warm wyvern]

homeo is topology

it's not geometry when a triangle is a circle

[south patrol]

Now write out an explicit homeo

[steel light]

Is the reason Gathmann uses the normal subgroup symbol for ideals because ideals play a similar role in ring theory as normal subgroups to group theory?

[coral spindle]

Most people do

[south patrol]

Yes

[steel light]

You quotient by the ideals and their multiplicative property looks similar to how a normal subgroup is defined

I see

[delicate orchid]

you would

[steel light]

What else is geometry then?

Or rather

What is geometry*

[coral spindle]

idk, sheaves

it's geometry if there's sheaves and it's sheaves if there's geometry

[south patrol]

The study of properties preserved by transformation groups

[open sluice]

https://en.wikipedia.org/wiki/Absolute_geometry

[south patrol]

Borger king

[coral spindle]

I'm at burger king with my burger queen can I get a large fries

[delicate orchid]

it's just a vibe

[steel light]

What’s a transformation group? Like S_n or Aut(G)?

[delicate orchid]

a group that transforms

[coral spindle]

Like the group of Euclidean transformations

[delicate orchid]

rather, it's a group that's acting on something

[steel light]

Hm

That’s a good definition

[coral spindle]

(This is the 'Kleinian' geometry)

[delicate orchid]

an example that comes to mind is like - all of alg geo

[steel light]

What group actions do we consider in algebraic geometry?

[delicate orchid]

depending on what vibe you get from it it can be geo or not

[steel light]

Or does that come up more in Galois theory

vibe

[delicate orchid]

like localisation can either just be "add in fractions so you only have one prime ideal" or "zoom in onto a specific neighbour hood of a point on a varieity"

[warm wyvern]

mfw all math is just vibes

[delicate orchid]

one of those interpretations is clearly geometrical

[steel light]

Mmm

Yes

[rocky cloak]

I feel like AG is born out of doing geometry (in the sense you suggested) to algebraic curves/surfaces etc.

A question like "is y^2 = x^3 smooth, or does it have a singularity?" Seems like a geometric question.

[steel light]

I agree

[mighty kiln]

Is it true that at some point the geometry starts disappearing and the algebra takes over

[crystal turtle]

it's all algebra

[coral spindle]

The whole point is that the geometric questions become reflected in algebraic ones

So yes the geometry disappears in that sense, but in fact it is just a change of perspective that occurs

[mighty kiln]

Rip intuition

[coral spindle]

No you just gain new intuition

My ring is reduced, so my space is connected. Stuff like that.

[delicate orchid]

"my category has a terminal object so it's compact"

[steel light]

but how does the latter interpretation relate to what potato said

what properties are we worried about being preserved and under what transformation group?

[delicate orchid]

my question is why do you need a definition of geometry

[steel light]

Curiosity and so I can better understand what math is

Why does anyone else study this stuff?

[delicate orchid]

grobner basis

that's why

[coral spindle]

Wew prolly knows something I don't but personally I do not see algebraic geometry as fitting in with Klein's programme

[delicate orchid]

I know next to zero alg geo dw

[coral spindle]

Defining fields of mathematics is a relatively hopeless task.

[steel light]

I see

[coral spindle]

I would say that algebraic geometry doesn't yet seem like geometry to you because you have not gotten used to it yet

[warm wyvern]

I can define analysis

[coral spindle]

Blah blah, Von Neumann quote, etc.

[steel light]

I have two different definitions of analysis

[warm wyvern]

it's when epsilon goes small

[coral spindle]

Sure thing bud

Schur* thing bud

[steel light]

One is directly about calculus and the other is just “approximating things”

[coral spindle]

Wat

[steel light]

kekw

[delicate orchid]

analysis is "what happens if thing goes towards other thing"

[steel light]

Yes

get close and thing go brrrr

[delicate orchid]

algerba is "what happens if thing times by thing"

[topaz solar]

A lot of variety stuff is just rando black magic imo

But sheaves are geometric

Since, ya know

[delicate orchid]

I don't know actually

[topaz solar]

Literal restriction maps

tf you mean

Bro never seen C(X, R)

[warm wyvern]

I know nothing about sheaves lmao

[warm wyvern]

I did

[topaz solar]

gg that’s a model for sheaves

[warm wyvern]

brb I'll look up the definition of a sheaf

[delicate orchid]

I have no idea what this means

the morphisms between X and R in C

[warm wyvern]

continuous functions from X to R lol

[solar vessel]

what only algebra does to a mf

[coral spindle]

Ig if Klein says that geometry is what is preserved by isometries, then sheaves say that geometry is what is described by the maps out of it

[delicate orchid]

oh so it's like the 1-simplex ok

[south patrol]

bruh

what cat theory does to a mf you mean ig ye

[delicate orchid]

just write Hom_Top you weirdos

[south patrol]

how is it like a 1 simplex

[rocky cloak]

To me it makes sense that k[x, y] is the ring of polynomials on a plane, k[x] is the ring of polynomials on a line, and k is polynomials on a single point.

Then ring maps between them corresponds to maps of geometric objects, makes perfect sense.

Then you have things like k[x]/x^2, where a map k[x, y] -> k[x]/x^2 corresponds to taking a Taylor expansion at some point in some direction. So this is also like a ring of functions on a point, except with a little bit of derivative information. It's like the ring of functions on a tangent vector!

So this relationship with rings helped find natural spaces we didn't have before

And then just extrapolate to all rings and that's affine schemes I guess

[delicate orchid]

homology

[mighty kiln]

Isn't that I → X

[delicate orchid]

it is

[south patrol]

Hm still don't get it lol

I mean ye what arki sai

ok

[mighty kiln]

Reversed arrows

The true cohomology

[delicate orchid]

brohomology

[steel light]

Can you explain informally how k[x, y] -> k[x]/x^2 is like taking a Taylor expansion? What do you mean “in some direction”; how do we get a sense of direction in a ring? Why does the y variable disappear in this map?

[mighty kiln]

Linear approximation?

[steel light]

Why wouldn’t you need partial differentiation to do that in multiple variables?

Or some notion of differential

[mighty kiln]

in some direction

[steel light]

Ah yeah oops lol

And you can define tangent vectors as derivations

Hmm

[mighty kiln]

Hm ye what do objects in alg geom look like actually

[steel light]

I don’t know

I know of varieties

And very vaguely of what schemes do

[mighty kiln]

"Solutions to polys" feels like local deformations are not a thing

[formal ermine]

a scheme is like if a manifold and a variety had a baby

and fed it some metamphetamine and ketamine

on a daily basis

for 5 years straight

[mighty kiln]

Manifolds feel almost the opposite

[formal ermine]

idk I don't know enough alg geo to have a big say in these things

but the defns are very similar

[shy fossil]

manifolds are varieties frfr

[south patrol]

french moment

[formal ermine]

beat my wife to it

[plain lava]

how do I represent the affine space n dimensional euclidean space in gruop notation (some sets, some actions)?

[warm wyvern]

you have a wife??

[plain lava]

for an affine space $A = (P, V, +)$, we know that its composed of a set of points P, a vector space V over some field F, and a scalar action +

but if i try this for eulidean space, P is the set of n tuples that is not a vector space $P = {(x_1​ ,x_2​, \dots, x_n​) \mid x_i​ \in \mathbb R} \equiv \mathbb R ^n$
and V is just the vector space $\mathbb R ^n$

How do i notate euclidean space $\mathbb E ^n = (\mathbb R ^n, \mathbb R ^n, +)$, but like how do i differentiate between the two $\mathbb R ^n$?

[cloud walrus]

TheHappyDragon

[plain lava]

or am i just misunderstnading something lol and interpreting this wrong

then bruh

[mighty kiln]

You don't differentiate

[plain lava]

what do you mean by this

hm

i mean like

[mighty kiln]

Abusing notation is fine

[plain lava]

R^n is a vector space, but i want to express n tuples that somehow arent a vector space

[mighty kiln]

Technically ((R, +, *), R^n, +, *) is a vector space

Not R^n

[plain lava]

ah i see lmao

i got used to abbreviating R^n as a vector space that i forgot about it was shorthand notation lol

[mighty kiln]

^

[plain lava]

thanks

[night onyx]

an example where you need to differentiate would be something like a a plane H that doesn't pass through the origin. This isn't a vector space, but if you fix a point on the plane as the "origin", say p, then you can think of it's associated vector space as { h - p | h in H}. Effectively these are vectors you can translate elements of the plane by

[plain lava]

well yea thats just the classic example of an affine space

[night onyx]

yeah, so in the example of P = R^n the associated vector space is just R^n again, so just abuse notation and don't differentiate lol

[coral spindle]

idk just write funny arrows above the elements of one and don't the other. Go wild

[night onyx]

or for E^n just say P = R^n, V = R^n, and differentiate that way by writing like p + v for p in P and v in V

OR you can use the notation (R^n)_p like the tangent space at p

so say (R^n)_0 for P and R^n for V

that's probably the most consistent

[sick valve]

hi, could someone tell me what d = (m,n) is supposed to mean in this context

[wraith cargo]

Gcd

[sick valve]

oh

thank you

[delicate orchid]

fun exercise

[formal ermine]

baby's first lagrange

[delicate orchid]

oh yeah you can do it with lagrange

kinda seems over kill though lmfao

[formal ermine]

yeah this is probably to introduce lagrange

like "look how cool this is! and look how we can generalize it!"

[delicate orchid]

as if the proof I just did wasn't overkill ||use bezout's lemma LMFAOOO||

[formal ermine]

||tbf I would've also used bezout||

[delicate orchid]

||fairs, there's probably a nice easy way we're missing||

[sick valve]

in the book they did it like this

[delicate orchid]

oh they did do bezout's

weird

[dim widget]

what is bezouts lemma

[mighty kiln]

Integers are a bezout domain

[dim widget]

what’s a bezout domain

[night onyx]

basically that gdc(a,b) can be written as a linear combination of a and b (with integer coefficients)

[delicate orchid]

a domain in which bezout's lemma holds

[dim widget]

I’m confused about that definition actually

Isn’t the definition of gcd the r such that (a, b) = (r)

[formal ermine]

no it's lcm(a, b) = r then r is the gdc of a and r

[mighty kiln]

E.g. Consider (2) and (x) in Z[x]

[dim widget]

Ah in a UFD

Not a PID

Ok

[mighty kiln]

Bezout ufd is pid
Bezout noetherian domain is pid

[dim widget]

Yes that is obvious

I was confused how you make a definition which is not tautological

[night onyx]

You mean (a,b) as in the ideal of Z generated by a and b? Then yeah, (a,b) = (gcd(a,b)), that's one of the motivations for abusing notation and using (a,b) instead of gcd(a,b)

[dim widget]

I was wondering how you define gcd, I was under the impression for some reason that it only makes sense in a PID, but I see what you’re meant to do in general

[night onyx]

it's the largest d such that d | a and d | b

but in terms of ideals if d = gcd(a,b) then (a,b) = (d)

lol it's definitely an abuse of notation, it just works well so we do it

[coral spindle]

Considering there is no notion of 'large' in a general ring, it's clear this definition isn't it

Furthermore, in the ring Q[x,y] it's fair to say that x and y are coprime, yet (x, y) is not (1)

[dim widget]

I guess the correct definition is (a, b) = (c)

For bezout domain

Then you don’t need to introduce the gcd in an ad hoc way or assume ufd at all

[coral spindle]

We can define the GCD up to associates as the largest in the divisibility preorder, but typically this won't exist

[night onyx]

oh well yeah, in a general ring you'd say d = gcd(a,b) if

1. d | a and d | b
2. if d' | a and d' | b then d' | d

[dim widget]

And you get interesting examples like holomorphic functions on C and the Robba ring

[dim widget]

This doesn’t have to exist

[hollow mica]

this is a nitpick but isn’t it antisymmetric hence a partial order too

[coral spindle]

-1 | 1 | -1

Not a partial order

[hollow mica]

oh negatives

forgot about them

[dim widget]

Just units in general

[hollow mica]

right, you guys are in a field?

[coral spindle]

No

[dim widget]

No then there’s no point in any of this

[hollow mica]

why not

[dim widget]

Because everything divides everything

[hollow mica]

oh

[dim widget]

Except 0

[hollow mica]

fields have a lot of units ig

forgot about that one too

[night onyx]

true, not all rings are GCD domains (https://en.wikipedia.org/wiki/GCD_domain)

[dim widget]

that is one of the main things about a field

besides being commutative

[summer path]

I can't tell if tteg is still trolling anymore

[dim widget]

I actually didn’t know any of this lol

I’ve never thought about “gcd domains” in my life

[summer path]

I feel like I'm being gas lit again

[steel light]

What's so special about PIDs? Any element in your PID can be written as a product of principal ideals?

[south patrol]

uh wdym

[steel light]

Or rather a unique product

[south patrol]

writing an element as a product of ideals oop

But uh I mean that's a UFD

[steel light]

Well you know what I mean lols

[south patrol]

Well I suppose yes being a UFD is one of the important bits but also we wish to study ideals generally and this is one of the easiest cases

But then some of studying ideals just comes from wanting weaker notions of factorisation too so hm

[delicate orchid]

all ideals are generated by a single element

[steel light]

Yes

Just to be clear you mean each ideal right?

is generated by a unique single element

[delicate orchid]

what is the difference between that and what I said

oh

then yes

[steel light]

😎

[delicate orchid]

for all (ideals) there exists (an element) that blah blah blah

[steel light]

Yes

[delicate orchid]

so divisibility works exactly how it does in Z

[south patrol]

not unique

[rocky cloak]

unique up to unit

[delicate orchid]

unique up to a unit

aka unique

[steel light]

That's what I meant, any ideal (element) can be written as a product of principal ideals (elements in those ideals)

[delicate orchid]

intersection rather than product no?

[steel light]

Uh

Idk

[south patrol]

same thing

[delicate orchid]

same thing if ur a nerd perhaps

[south patrol]

i would normally write a product

[steel light]

Isn't it only the same thing if the generators are coprime

[south patrol]

because it's nicer and more tangible in this case

well okay they aren't the same thing in general

but the statement is invariant under changing intersection to product

[steel light]

also

isn't it unique in the sense of how equivalence classes are unique (up to representative)

that's what you guys mean by up to unit?

[delicate orchid]

yeah I realise that now - my apologies

[south patrol]

that'd be kinda tautological ig

[south patrol]

no dww

[void cosmos]

yo can someone just check my reasoning here

[delicate orchid]

we're saying that if (a) = (b) then a = ub for some unit u

[steel light]

yes

Oh

[south patrol]

In fact this is one of the nice things about factorisation of ideals lol

units are immediately taken care of

[steel light]

Ahhh there's the "divisibility"

I see it now

[south patrol]

So yeah an ideal has a unique factorisation as a product of prime ideals in a PID

And that's actually unique (up to permutation of the ideals lol)

[steel light]

Ah yes I should've said prime

[south patrol]

Have you done any / are you doing any algebraic number theory

[steel light]

No to both

[delicate orchid]

what's the definition of two elements being associates?

[steel light]

I've not touched number theory at all

[void cosmos]

if they differ by a unit

a = ub for some unit u

--> a and b are associates

[delicate orchid]

oh that's surprisingly relevant to the convo happening atm

[steel light]

Yeah

LOL

[south patrol]

Basically feather like

[delicate orchid]

I agree wiith -> being obvious

and field -> semi simple also being obvious

[void cosmos]

yea once i see X is a field ik everything is juist obvious

so i just skipped it

like i just assumed its not a field instantly

lol

but for the <-?

[delicate orchid]

an infinte number? that surprises me

[south patrol]

Why I mention number theory feather is that lots of (at least basic) stuff in alg nt is about factorisation of ideals lol

like you can do some stuff with more general rings than PIDs

which is nice

[delicate orchid]

I'm probably thinking about this wrong anyway - in my head I'm attempting a block-decomposition esque thing lol

[south patrol]

(the relevant notion i mean is that of dedekind domains)

[rocky cloak]

Isnt this like the original motivation for ideals? I.e. ideals are "ideal numbers" in the sense that they satisfy unique factorization in rings of integers even when "numbers" dont

[south patrol]

yeah lol

[void cosmos]

who are you talking to right now

who is it you think you see?

[delicate orchid]

could you explain how not being associates implies they must have trivial intersection?

I see the LORD

His light RAINNSSSS down on me

[void cosmos]

i am not in danger skyler i am the danger

[delicate orchid]

if there exists some x in both (a) and (b) that implies that x = ra = r'b but I don't see how to conclude that a = bu

[void cosmos]

its because then

over a pid

u factorize each of these elements in R

and then u divide by that unit element u get from the factorization

and by uniqueness of factorization u get that the irreudcibles are associate

does this work

thats what i had in mind

idk if it works tho im asking u ur the master here

[delicate orchid]

ah I see so, ra = u(x_1...x_n) = r'b with u a unit - with one of the x_is being a and another being b as irreducible => prime
so this means that r = ur'

very cleaver...

[void cosmos]

wait really

i did that on my own

[delicate orchid]

yeah I think that works

[void cosmos]

omg

i cna do it

never bow down never what

time to apply to my local grad school

that has 2 high school teachers

time for glory and money

||i am delusional||

[delicate orchid]

welcome to the cluub

[formal ermine]

NEVER BACK DOWN NEVER WHAT?

[void cosmos]

wait is back down

its back down

not bow down

wtf

i didnt know that

[void cosmos]

btw its not n

i think

they are infinite

[delicate orchid]

are you allowed to do that? is that legal?

[void cosmos]

wdym

[delicate orchid]

I've only seen it as a finite product

[void cosmos]

yea ur right

but then this doesnt work tho right?

cuz why must one of the x_is be a or b

[delicate orchid]

a and b are prime factors of ra and r'b

as they're irreducibles and therefore prime (they generate maximal ideals)

so they have to appear in the factorisation up to a unit - which we bring to the front to make one PHATT unit u

wait are these rings commutative

[void cosmos]

no

[delicate orchid]

UH OH!

but wait wait wait

[void cosmos]

or ig

integral domains

must be commie

[delicate orchid]

you still just have the unit at the front by definition of ufd though right?

yeah true

ok

[void cosmos]

so yeah nvm

[delicate orchid]

oh yeah I'm conflating ufd and ufr in my head

[void cosmos]

okay can u just write everything out

so i see what ur saying

i have covid

rn

[delicate orchid]

rip

[void cosmos]

and my brains not working

[delicate orchid]

if (a) intersect (b) is non-trivial then we can pick some non-zero x in the intersection
so x = ra = r'b for some r, r'
as this ring is a UFD we can therefore write ra = u(x_1...,x_n) = r'b
since a, b are irreducible they're prime - and so must appear (up to a unit) in the factorisation of ra, r'b i.e. a = x_i and b = x_j for some i, j (this can also be seen by the fact that x is in the intersection of (a) (b) and so must be a multiple of a, b)
ok but now I'm not seeing how I concluded r = ur' - from this we'd have ra = ur'b but then that doesn't imply that a = ub at all

we can definitely conclude that ra = ur' and r'b = ur

[void cosmos]

okay

[delicate orchid]

wait something has gone wrong here - ra = r'b but ra = ur'b

wait

no nvm that's fine

[rocky cloak]

So are you trying to prove that if a PID has infinitely many primes, then J(R) = 0?

[delicate orchid]

yeah

[void cosmos]

yeah

[delicate orchid]

I think

[void cosmos]

yea

it is

[delicate orchid]

I was just trying to prove that uhhh if two maximal ideals are disjoint then they can't be associate

[rocky cloak]

Then you just take the intersection though right

[void cosmos]

yeah

[void cosmos]

yea exactly

[delicate orchid]

yeah but why?

that's what I asked you like 20 mins ago boss

[rocky cloak]

But if they're associate, then clearly they generate the same ideal

[delicate orchid]

oh ok so you just contrapositive it

unsatisfying but I'll take it

[void cosmos]

its because if u put divide u

if u divide u

then u get two prime factors for the same element

[rocky cloak]

I see you want to prove that (a) = (b) implies they're associate?

[delicate orchid]

yeah actually I think that's what I was doing

[rocky cloak]

If b = ar and and a = bs, then a = ars -> rs = 1

[void cosmos]

u have u(a1a2a3...)a_i=u'(a4a5....)a_j right

so i thought i would divide by u

then get that each a_1 must be associate to each a_4 for example?

i dont think this works

tbh

cuz they are the same element :d

[delicate orchid]

ok so the dudes are disjoint or equal lemme think about how this implies semisimplicity

yeah ok got it

I think

R decomposes into a direct sum of aR

right? for a irreducible?

but those aren't simple, about as far as you can get from being simple actually

[rocky cloak]

Remember that Moamens definition of semisimple is stupid

For example Z is semisimple

[delicate orchid]

oh yeah duhh

we literally just need J(R) = 0 which we have right?

[void cosmos]

wait how did u get J(R) = 0

i kind of lost you two

[delicate orchid]

non-associate maximal ideals are disjoint

so the intersection of any two is 0

[rocky cloak]

J(R) = (x) because PID

[delicate orchid]

let alone the intersection of all of them

[rocky cloak]

x is a product of primes p1 ... pn. We have infinitely many primes, so take p not in this decomposition. Then x is not a multiple of p, thus (p) does not contain J(R), contradiction.

Hence x is 0

[delicate orchid]

oh that's way quicker

[void cosmos]

turns out im a failure all long!

hahaha

ty guys

i goti t

i got it*

[rocky cloak]

Now I'm wondering, so a field has 0 primes, Z has infinitely many, k[[x]] has 1. Are there PIDs with 2 primes, or n?

Hmmm, what about something like the subring of Q where the denominators are not divisible by either 2 nor 3?

Okay, yeah I think that should work

[void cosmos]

yo yo

just quick check

for like "rigor" and stuff

so J(R) contains no nonzero idempotents

this is true over any ring correct?

even if R is not an integral domain

cuz u go 1-e is invertible

so e^2-e = e(1-e) = 0 --> e = 0

?

[rocky cloak]

Should be true even for nonunital I think

[void cosmos]

yea but i mean

this proof that i just wrote

why did i not say either e = 0 or 1 -e = 0?

cuz 1-e is invertible

correct?

[rocky cloak]

Yes

[void cosmos]

gotchu

[glossy crag]

Wouldn't localising away from finitely many primes work? A localisation of a PID is always a PID.

[void cosmos]

tysm

[void cosmos]

i love this fact it saved me in a hard ass problem in the undergrad test

[rocky cloak]

Yeah, that's the conclusion I came to aswell

[delicate orchid]

u guys say these things that really make me do double takes

I get that unit plus nilpotent is a unit but (-e)^n = e for all n > 1

block decomp R = eR+(1-e)R = eR+R = R? I mean maybe?

[rocky cloak]

But e=0

[delicate orchid]

I'll just accept I won't get this one

[void cosmos]

what

whats the problem

@delicate orchid

[rocky cloak]

For the nonunital case, choose x such that

-e -x + xe = 0

Then multiplying by e

-e -xe + xe = 0 -> e=0

[delicate orchid]

I can name tens of thousands of idempotents in commutative rings that do not have the property that 1-e is a unit

[void cosmos]

yea cuz they aint the jacobson radical tho

@rocky cloak hmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

this

characterization

of J(R)

is not

true

over nonunital rings

tho

[south patrol]

I agree with you moamen

At least for unital commutative rings lol

[void cosmos]

thanks ||daddy||

[south patrol]

Bruh

[void cosmos]

but yeha cuz

i remember proving J(R) = { x| bla bla is invertible}

[rocky cloak]

Wut, I thought that was your definition of J(R)???

[void cosmos]

one direction uses that 1-xy is invertib le for x =1

[void cosmos]

no king

my definitoin is the intersection of all left regular maximal ideals which is the intersectino of all annihlators of simple left R-modules which is the intersectino of all left primitive ideals

wake up its the first of the month

if R is unital

[delicate orchid]

I lied - I get it if the nilradical is the jacobson radical it's obvious then

[void cosmos]

then this is the same as {x |1-yx is invertible for all y}

the way to do it is

to show that a is in J(R) iff Ra is a left quasi-regular ideal

i thinnk

[rocky cloak]

Okay, so not the sum of quasiregular ideals?

[void cosmos]

it contains all quasi-regular elements

idk wdym by the sum

or yeah

yeah

[rocky cloak]

Right, so then this works

[rocky cloak]

... and all it's elements are quasiregular right

[void cosmos]

yea

[void cosmos]

yeas

its a quasi-regular ideal itself

[topaz solar]

[void cosmos]

yea ur right

what a chad u are

eren jeager

[topaz solar]

There’s some clown definitions of things

[void cosmos]

yo math is literally cool af

[mighty kiln]

^

[void cosmos]

i dont wanna learn new stuff

i wanna solve more problems

why does it have to take 3 years

[mighty kiln]

But new stuff also cool

[void cosmos]

just to get in to play

yea ig

[topaz solar]

Because mfs like Scholze and Gelfand and co. keep pushing things further and further back

[void cosmos]

wait u chose r such that r+e+re = 0?

why do u have negatives 😄

[topaz solar]

2 things are negative, one thing isn’t

[rocky cloak]

Well -e is also in the radical, so that should be okay. Just need the signs for things to cancel

[topaz solar]

r+e = re

[void cosmos]

cool

got u

much better

[steel light]

Wait I’m a little silly

[crystal turtle]

me too

[steel light]

Is (0) not always prime because we may have zero divisors

Duh

Lmao

[crystal turtle]

0 is prime iff it's an integral domain

[summer path]

Sometimes obvious statements only appear really obvious after you explicitly state them

Even though that one is just run the definitions twice in each direction

[steel light]

Yeah

[steel light]

Yeah LOL

[void cosmos]

anyone has notes

or a pdf

for semisimple rings

i dont like covering them from hungerford

too much talk

[coral shale]

Z is an ID

0 is a prime number...

[rocky cloak]

Certainly prime, but if is it a number ❓

[void cosmos]

who wants

to build the whole basic theory

of semisimple rings and modules

in the discord chat with me

i only have two things that i can do alone now

i have the definition of a semisimple R-module ( which is the direct sum of simple R-modules )

and i can prove only one fact

that sums and direct sums are the same here

that is if we have a module that is only a sum then its also a semisimple module

who can prove that this definition is equivalent to the fact that it is left artinian with zero jacobian radical

fuck

jacobson

[rocky cloak]

Proving that it's left Artinian should be easy, since an ideal is just a submodule. Then the proof that the radical is 0 I think you've gone through here before

[void cosmos]

yes eren jeager

thats what i like

to tell me if a proof is do-able by my own or not

like it would bea complete waste of time to try to prove on my own that Tor(M) = R/(a1) + R/(a2)... over a pid

so idk

i will try to do it now ty

[void cosmos]

but wait do i say that a ring R is semisimple if its semisimple as an R-module over itself?

wait

okay u went over something here for me i think

since its a direct sum of simple R-modules

a submodule would be isomoprhic to a simple R-module?

or what

wdym

[rocky cloak]

A submodule would also be isomorphic to a sum of simples, but fewer of them.

[void cosmos]

yea

so it would be semisimple aswell

[rocky cloak]

Yeah

[void cosmos]

yoo wait

i have a bad question

arent direct ssums nonzero only for finitely many indices

why cant we

like why

yea so

these things are ideals right

so they are closed by left multiplication

wait

nvm im lost

[rocky cloak]

What is the question?

[void cosmos]

yea just nvm forget it

let A be a semisimple ring

so an ideal of A would be a submodule which is again semisimple

okay wait

i prob know the answer to this

cuz i know math fucking sucks

but here is a question

is ANY sum of artinian rings artinian

or must it be finite

this guy Schur is literally the luckiest math guy in the history of the world

he just proved the most basic shit ever

and he has his name

[rocky cloak]

The sum must be finite. But if we're talking about unital rings, then an infinite sum of rings is not unital anyway

[void cosmos]

so ig

if we could somehow

turn this infinite sum into a finite one

we are dnozo

donzo

[formal ermine]

What is an infinite sum

[void cosmos]

which is impossible cuz it would have stated it in the definitoin

like why would they go in the definition " a sum of simple .. " and thne go " theorem 3.123 its actually finite lmfao "

[rocky cloak]

But is the ring unital?

If it's unital you can prove the sum is finite

[void cosmos]

okay let me try it

[rocky cloak]

If not, it's just not true that it needs to be Artinian

[void cosmos]

yea i got it instantly

its literally a vector space lmfao

cuz u go 1 is in the direct sum

but this direct sum is finite stuff + 0 garbage