groups-rings-fields

[steel light]

LOL

[long nebula]

Which symbols are you confused about

I'm pretty sure you know all of them

[delicate orchid]

THAT’s the example you go to?

[long nebula]

Yes

I like it

[delicate orchid]

I mean it’s nice geometrically but cmon now boss

[steel light]

C^x is supposed to complex numbers with multiplication as the operation? What even is R^>0? Wtf do you mean by “calculate them as groups”??

[long nebula]

Yes

[topaz solar]

Foul example tbh

[long nebula]

R^>0, positive real numbers

[solar vessel]

calculate R/Z as groups under addition

[long nebula]

by as groups I mean, they are groups

[delicate orchid]

Non-zero complex numbers, positive real numbers under multiplication

[long nebula]

take the quotient of them

[steel light]

Okay

[delicate orchid]

Calculate Gal(Q_p/Q)

[steel light]

um

I’ll try

[rotund aurora]

good

[long nebula]

it should be isomorphic to something familiar

[topaz solar]

Calculate Gal(\bar Q_p/Q_p)

[long nebula]

They will learn BY FORCE

[solar vessel]

anyway uh

[topaz solar]

This is just Aut(Q_p)

[noble saddle]

what are you scared of global fields??

[delicate orchid]

Oh yeah not you know C_4/C_2 nooooo

[solar vessel]

does he even know first iso

[delicate orchid]

S_3/C_3 no far too complicated

[rotund aurora]

I wonder why there are not as many "calculate this group/ring, etc" problems as there are problems about calculating limits, sums and integrals

[steel light]

I still don’t understand what calculate as groups mean

[topaz solar]

He don’t need to be reading AM if he doesn’t

[delicate orchid]

What is it isomorphic to

[steel light]

Oh

Okay

[long nebula]

I mean calculate it and then I appended that they are groups

[solar vessel]

I think we've long established bro doesn't need to be reading AM

[long nebula]

Just in case you accidentally thought they were rings or topological spaces or smth

[rotund aurora]

it means give as good of a description as possible

[steel light]

C^x is isomorphic to S^1?

[rotund aurora]

think like how you calculate integrals or infinite sums

[steel light]

Or a scaled version of S^1

[rotund aurora]

you can calculate sum 1/n^2, but not sum 1/n^3 really (tho this last assertion is kinda unsolved 😝 )

[topaz solar]

Calculate Aut(\prod_p \bar Q_p/U) for a Ramsey ultrafilter U on the primes

[long nebula]

You're cooking

(The answer is no, but keep cooking)

[steel light]

Darn

It’s not S^1 unless you only consider things with unit magnitude no?

[long nebula]

👀

[noble saddle]

hey I think you get quotients now

[long nebula]

LOLLLLLLLL

[steel light]

Idk what is similar to S^1 but scaled radius lol

[steel light]

What?

Everyone is laughing and I just feel so stupid

[topaz solar]

C^x is

[steel light]

LMFAOO

[long nebula]

It's because you're saying smart things

but you don't know they're smart yet

[solar vessel]

nah that's subthings smh

[delicate orchid]

Ring is abelian

[steel light]

But aren’t there like 49723472037409327432 different possible “circles” then

[long nebula]

What do you mean by this

[delicate orchid]

I like this example less and less as time goes on

[noble saddle]

I think it's genuinely a great example

[steel light]

Like ok if you only consider things with unit magnitude you get S^1 because complex multiplication is just rotation + scaling but you scale by 1 so it’s just the unit circle

But if you let your products have any nonzero magnitude

You can end up with circles of any radius

[long nebula]

Yes, if you ignore the magnitude then C^× just looks like S^1

[delicate orchid]

As a first example on how to construct quotients? You’re bonkers

[long nebula]

What possible values can the magnitude take on, feather?

[noble saddle]

but we did Z/6Z already :p

[steel light]

Anything no?? Except 0

[long nebula]

can you be more specific

[steel light]

Any real number

[rotund aurora]

-1

[steel light]

Positive real number sorry

[south patrol]

Und ich oope-

[steel light]

What

[long nebula]

hmmm positive real number

where have I heard that before

[delicate orchid]

Again, S_3/C_3

[steel light]

R^>0

Uh

[long nebula]

So what you're saying is...

[steel light]

Wait

That’s

S

1

[long nebula]

If I take C^×

And ignore R^>0

[delicate orchid]

Huh funny that

[long nebula]

I get S^1

[fervent dagger]

Anyone playing roblox
Joke lol

[summer path]

#chill

[long nebula]

So
[\bC^\times / \bR^{>0} \cong S^1]

[cloud walrus]

propERICly_embedded

[long nebula]

bambambambam

[topaz solar]

gg

[steel light]

ok…?

[delicate orchid]

You have to be 13 years old to use this application

[topaz solar]

Visualized

[fervent dagger]

Lmao

[delicate orchid]

Yeah see and feather still has no idea how to construct a quotient

[steel light]

I don’t understand what I just did or what it was meant to help me understand 💀

[rotund aurora]

Joe Rogan plays roblox

[long nebula]

fine smh

[delicate orchid]

Dog shit pedagogy

[steel light]

I’m sorry Eric 😭

[rotund aurora]

I mean, just write things in polar form

thats it

[delicate orchid]

Wait just realised I’d have to explain what S_3 is to do my example uhuhunubunuubub

[steel light]

Oh

[topaz solar]

Begone from here, mortal, banished

[steel light]

Wait is it like

Wait wait pause

I think I’m boutta spit

[rotund aurora]

ok we pause queen

[long nebula]

She's cooking

[solar vessel]

let him land

[topaz solar]

He cookin

[summer path]

it sure took a long while for me to realize that C^X was \C^\times and not the set of functions f: \C to X

[long nebula]

they go by any pronouns

[delicate orchid]

Mfs u see the pink dot

[rotund aurora]

bro is color-blind

[delicate orchid]

Misuse of the pronoun role MODS MODDSOSOSODOSS

[summer path]

the pink dot takes priority

[rotund aurora]

acshually, misgendering is violence

so should be put in jail

[summer path]

muted jail

[long nebula]

this is a fair point, why do you not just have the any pronouns role lol

[topaz solar]

Bro asking feather to take things seriously

[long nebula]

dw I'm gonna be teaching high schoolers in around two weeks, so if feather doesn't get it, I'll just use this example on them

[summer path]

i think there was a saying at some point that the only people who click on every pronoun role are pre-uni people

[delicate orchid]

By the nine celestial spheres I beg you to reconsider

[long nebula]

hey it said in the description I was supposed to teach them "algebra"

what else am I supposed to do??

[summer path]

which nine?

[rotund aurora]

I can only count up to 5

[delicate orchid]

Just S_3/<(123)> = {{e, (123), (132)}, {(12), (13), (23)}} \cong C_2 it’s so swaggggg

[steel light]

Ok so you guys said some stuff about how x + I = 0 + I when x is in I
And then there was the Z/Z3 example where when I wrote it out (though I didn’t post it) I saw that when you took the coset of an element in 3Z it was just 3Z
Then there was that thing Nix said about how if a + m = a’ + m then that means a - a’ in m
Now you’re giving an example of a group quotient where we effectively “ignored” the magnitude (because that’s what we were quotienting by, we sorta “likened together” all of the unique magnitude into a single thing)

So is the idea for a quotient ring is that you basically identify all of the elements that when added together are in your ideal as one thing?

Idk if I just cooked sorry if I ruined the barbecue guys

[delicate orchid]

You can even directly see how everything multiplies by bashing cosets together

[long nebula]

For a ring quotient, you identify r and s iff r-s is in the ideal

(Prove that this is an equivalence relation)

(Prove that this is equivalent to your usual notion of Z/3Z)

[delicate orchid]

proves it

[rotund aurora]

when you have an identity, you can just tell which things you kill

[long nebula]

⚔️

[rotund aurora]

but you have to make sure your equivalence relation is a congruence

[steel light]

Wait this is just modular arithmetic

Oh fuck

[rotund aurora]

so that it respects the operations

[steel light]

LOL

[crystal turtle]

yes

[rotund aurora]

thats why you consider ideals, and normal subgroups (and not just an subgroup or smthing)

[steel light]

Why did no one say this before man 😭

[toxic zephyr]

I DID

[long nebula]

Exercise

[steel light]

YOU DID I WAS JUST TOO STUPID SORRY NIX

[crystal turtle]

[steel light]

LOL

[steel light]

Oh

Okay

[long nebula]

If your textbook doesn't leave definitions up to exercises, why are you even reading it??

[delicate orchid]

Whoops nearly spoilt the exercise

[solar vessel]

yeah I also think you should exercise

[steel light]

Let me try R[x]/(x^2 + 1) again

[solar vessel]

ok

you may try

[long nebula]

Wait I better delete that before Sharp sees it

Carry on

[solar vessel]

huh

[summer path]

i can only count to 4

[steel light]

Wait (x^2 + 1) means the ideal generated by x^2 + 1 right

[delicate orchid]

Yus

[steel light]

Gonna have to get used to that lol

[cloud walrus]

tutor named finger (nix)

[toxic zephyr]

i dont remember when you use each one

[delicate orchid]

This makes me think of a subgroup

[open sluice]

i thought that was for groups

yeah

[steel light]

The ideal generated by x^2 + 1 is all polynomial multiples of x^2 + 1 right? So any polynomial that has x^2 + 1 as a factor

polynomial in R[x] tbc

[toxic zephyr]

yea (x^2+1)g(x)

[steel light]

Okay

[toxic zephyr]

what's the difference :^)

[delicate orchid]

Subgroups only have one operator :3

[crystal turtle]

I have seen this for ideals as well

[noble saddle]

I've seen ideals use both, but parens are much more standard

[crystal turtle]

Interesting. Noted

[steel light]

I feel like I’m on the cusp of it

[toxic zephyr]

trying to remember. iirc aren't ideals like the ring version of normal subgroups

[cloud walrus]

tutor named finger (nix)

[noble saddle]

kind of, but I don't think that's a good way to think of them. Originally ideals were created to capture notions of divisiblity

[delicate orchid]

They’re like normal subgroups in that they’re the structure you need to make quotients make sense

[steel light]

It’s like all you’re left with after you quotient is x^2 + 1

[toxic zephyr]

yeah that's what i mean

[steel light]

Since it’s not factorable in the reals

[toxic zephyr]

i like using the same notation for similar concepts

[steel light]

But how does that imply that R[x]/(x^2 + 1) is C

[noble saddle]

you shouldn't think of normal subgroups and ideals as the same thing/concept

[steel light]

Or like help us define C or whatever

[long nebula]

Try writing down a basis for R[x]/(x^2 + 1) as an R-vector space

[toxic zephyr]

what is x^2

[long nebula]

It might help you to think about this

[steel light]

x^2 + 1

[delicate orchid]

Specifically, x+(x^2+1) squared

[steel light]

Oh yes

Yes

[toxic zephyr]

it's been a while for me but i always thought of them like how wew said. defined to make quotient stuff make sense. what perspective do you prefer

[steel light]

What about it?

[delicate orchid]

It’s -1+(x^2+1)

[steel light]

Um

Say what

[delicate orchid]

Bros going to actually make me write this out on a phone keyboard at 3am

[steel light]

Is this because of what you said about identifying r with s iff r - s is in (x^2 + 1)

[toxic zephyr]

x^2+(x^2+1)
subtract off a multiple of the ideal so you get something of degree 0 or 1

every element of R[x]/(x^2+1) is equivalent to degree 0 or 1 polynomial

(prove that)

[delicate orchid]

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

[noble saddle]

Ideals were defined in order to make sense of divisibility in rings more complicated than Z, like Z[i]. They can be quotiented by, but lots of things in math can be quotiented by. Do you think of equivalence relations, group actions, or topological subspaces as normal subgroups?

[crystal turtle]

i mean for at least some things normal subgroups and ideals behave similarly. I'm not sure it's a bad way to think of them

[noble saddle]

but that is not a fruitful metaphor beyond "yes I can quotient". For one thing, normal subgroup is esentially meaningless in an abelian group

[toxic zephyr]

can you be more specific about what "make sense of divisibility" means? seems just as vague as "so quotient stuff makes sense"

[crystal turtle]

of course, they're not direct analogs sure. But for basic intuition it helps. Especially for basic stuff where you're still learning, say, iso theorems

[topaz solar]

[steel light]

I don’t understand either of your explanations tbh, but I do understand it if I think of it like “x^2 is identified with -1 because x^2 - (-1) is in (x^2 + 1)”

[crystal turtle]

you can quotient in a manner that is consistent with the group/ring operations

That is much more and much stronger than "you can just quotient"

[steel light]

Does anyone actually use the coset definition

[topaz solar]

Yes, they’re exactly the things which give an equivalence relation which is a subalgebra of X x X

[noble saddle]

Try to think about how divisibility and containment in an ideal are related in Z. Try to extend that to non PIDs

[steel light]

This is infinitely more intuitive and easy to work with

[topaz solar]

But the inclusion divisibility stuff is novel

[toxic zephyr]

just like how a=a-n mod n, x^2+(x^2+1)=x^2-(x^2+1)+(x^2+1)=-1+(x^2+1). that is, we can shave off a multiple of the ideal generator

[steel light]

So is it safe to just always think of quotient rings like that in practice?

[noble saddle]

it's literally how ideals were first defined

[topaz solar]

Rarely

[topaz solar]

Yes I’m saying it’s novel to rings, not normals

Not that it’s a new idea now

[noble saddle]

oh gotcha, agreed

[steel light]

Okay

I think I understand it better

How about another quotient ring

[toxic zephyr]

Q[x]/(x^2-2)

[noble saddle]

in practice when we talk about ideals, it's either because we're factoring things in integer rings or we're trying to understand zeros of functions (because factoring functions is what tells us about zeros)

[steel light]

Hahah

This is similar to the last one

[toxic zephyr]

my prof said in practice ideals are used to tell us things about rings by quotienting them and examining the behavior of the quotient ring 🙃
it seems arbitrary to say "to make quotient rings make sense" is a "bad" way to think about them

[noble saddle]

it's not arbitrary when that's not what they were invented for, not what people originally thought of them as, and not the thing people most seek to do with them contemporaneously

like yes, you can quotient and people do, but that's not why we come across ideals/what they are there for

[toxic zephyr]

yeah but it's still a good example.

• is this a field?
• what ring/field is this isomorphic to?

the purpose for why something was invented isn't always the best way to think about them

[topaz solar]

$\bigg[\prod_p \overline{\bQ_p}\bigg]/\mathfrak m$ where $\mathfrak m$ is the maximal ideal generated by the indicators of the form $1_{P-A}$ for $A\in U$ a Ramsey ultrafilter on the set $P$ of primes

[cloud walrus]

Dragonslayer Sharp

[crystal turtle]

sharp what the fuck

[noble saddle]

fair, but it is when it's still how they are thought about today

[crystal turtle]

i think "how they are used today" and "how they make sense to a beginner in a specific context" are two different things

Not disagreeing with you, but for some contexts it matters

[toxic zephyr]

but just one more thing, we can write the quotient ring in the following way

$\bR[x]/(x^2+1)=\bdef{a+bx:x^2=-1}$

[cloud walrus]

tutor named finger (nix)

[noble saddle]

but the question is "how should I think about ideals?" and while a beginner might think "thing I can quotient by" that doesn't make it a good intuition

[toxic zephyr]

you can neglect to write the +(x^2+1) as long as you're clear you're still in a quotient ring

[topaz solar]

I joke, but ya know if you have arbitrarily many Galois extensions L/K all of the same degree n, we can say Gal(prod L/prod K) = prod Gal(L/K) modulo the same ultrafilter, and prod L/prod K is again Galois of degree n

[toxic zephyr]

but this is why it's clearly isomorphic to C

[topaz solar]

Which is

[crystal turtle]

fair enough

[noble saddle]

I say this as someone who in my younger years thought of ideals as like normal subgroups

[toxic zephyr]

bro #advanced-algebra pls you're scaring me 😭

I don't wanna see Galois in this channel ;-;

[steel light]

Yeah nope sorry Nix LOL I don’t know how this works out

I figured out how to see that x^2 gets identified with 2

[toxic zephyr]

yeah

[steel light]

But I don’t know what the other elements are

[toxic zephyr]

a+bx+(x^2-2)

[steel light]

Why though?

[toxic zephyr]

you can always subtract off a multiple of the ideal

[steel light]

That’s what I tried doing

[toxic zephyr]

so if you had an cx^2, you could yeet off a c(x^2-2)

dx^3? yeet a dx(x^2-2)

etc

in general, for a polynomial quotient ring, R[x]/(g(x)) you only have to concern yourself with terms of degree strictly less than the degree of g

just like how in Z/nZ you only really concern yourself with elements between 0 and n

it's the same idea

[toxic zephyr]

but given this and the fact that x^2=2, what ring/field is this isomorphic to? @steel light

[steel light]

Q[sqrt(2)]

[rotund aurora]

do ultrafilters show up in nt

[topaz solar]

Not that I know of

[crystal turtle]

they can if you try hard enough

[topaz solar]

Pester delta idk

[crystal turtle]

find some use for them

[topaz solar]

Some stuff is probably equivalent to em

[steel light]

I think I see why

[rotund aurora]

what does this mean

[toxic zephyr]

yes exactly

[noble saddle]

someone in grad school made me sit through a talk on ultrafilters. let me see if I can track down what that was about

[rotund aurora]

I have only seen them in the context of hyperreal numbers

but

[topaz solar]

Ooh pass it

[rotund aurora]

actually

[noble saddle]

they definitely come up in trying to figure out the maximal ideals of C[x_1, x_2, ....] if that counts as NT adjacent

[rotund aurora]

I think they are a thing in combinatorial nt

[fervent dagger]

I'm 9 and i don't understand these

[topaz solar]

<@&268886789983436800> classic

[crystal turtle]

huh

[fervent dagger]

Just kidding

💀

[rotund aurora]

https://youtu.be/IS9fsr3yGLE?si=ODSuRhMf4ZdFyfeu

here Tao talks about ultraproducts

i think they are related to ultrafilters lol

now I realize is not quite the same xd

[long nebula]

Stop posting in this channel if you have nothing to add

[topaz solar]

The title sounds like hyperreal/NSA stuff

[long nebula]

Indeed!

[toxic zephyr]

if you want another example, you can show that

$\bQ[x]/(x^3-2)\cong\bQ(2^{1/3})$

[topaz solar]

Ultraproducts are products mod an ultrafilter so

[cloud walrus]

tutor named finger (nix)

[rotund aurora]

right right

did you guys know

that telling whether two hyperreal number fields are isomorphic depends on CH or something

[toxic zephyr]

but maybe you've had more than enough examples @steel light d:

[topaz solar]

Tossing that in the archive though

[rotund aurora]

ok not the actual statement, but something like that

[topaz solar]

Yeah, if CH then they’re all saturated therefore iso

[noble saddle]

I don't remember exactly what happened, but here's a paper in my inbox with ultrafilters involved: https://arxiv.org/pdf/2008.11905.pdf

I don't understand anything about it anymore

[crystal turtle]

very fascinating that ultraproducts seem to be showing up in a few places

[topaz solar]

Tossing that in the bin too

[crystal turtle]

I would but i don't understand enough just glancing at it

[noble saddle]

I definitely remember there being something simpler, but can't find it now

[rotund aurora]

I want to read more model theory

in fact, I will

in a few minutes

[topaz solar]

Yeah I’m definitely not up to par for that paper L

[noble saddle]

for context, weight-monodromy is basically the weil conjectures

[rotund aurora]

mmh ig if you do things over non-archimedean fields then they will show up

[rotund aurora]

how do you show that the fraciton field is l(t)

and that its a field iff l/k is algebraic

[rocky cloak]

That's a good question. I think it's also true in the nonunital case, but I'm not totally sure.

Anyway, there is another way to prove that J(R) is nilpotent that more directly uses artinianess.

[cosmic nexus]

Hi sorry if this isn't the right place to ask. And sorry for the potentially stupid questions. But I'm just learning up on Galois theory and admittedly I do have a lot of basics to catch up on. So I have a couple of questions to have a better theoretical understanding.

1. We usually talk about adjoining the field $\bQ$ some root of a polynomial. But what if we had just start with $\bR$ instead and then if there are imaginary roots, just adjoin $i$? Because then it's a very simple extension, but then the automorphism group acting on the field extension $bR(i)$ would just be the complex conjugation, which is a solvable group, so the polynomial then would be solvable?

2. What do we really mean by permutation (automorphisms) of the roots? Because if we write any polynomial as linear factors $f(x)=(x-x_1)\cdots (x-x_n)$, switching the roots around in any way won't change the polynomial since commutativity lets us do that. Why do some polynomials have Galois groups $A_4$ for example where only certain kinds of permutations are allowed when if we just did any transposition between two roots, it doesn't change what the polynomial is? How does permuting the roots really have any relation to how the general formula for the solution looks like?

Again, sorry if the questions are dumb.

[cloud walrus]

Electrical_Data

[maiden ocean]

The inclusions of k(t) and l into l(t) induce an inclusion from the tensor product (check that it is injective using that everything is fields). Thus the field of fractions of the tensor product is contained in l(t). Now writing an element of l(t) as f/g with f,g in l[t], you can ||write f as sum a_i t^i with the a_i in l and hence as the image of sum t^i (x) a_i and likewise for g. Hence the field of fractions contains l(t) and so they are equal.||

[rocky cloak]

Let J^k = J^k+1 . Consider the set of ideals I contained in J^k for which J^k I is nonzero. By Artinianess this has a minimal element S. This is necessarily principal.

If J^k S is a smaller ideal, then J^k S = J^k (J^k S) = 0, which is a contradiction. So J^k S = S. In particular there is a j and s such that js = s. And then you can use nakayama.

Alternatively, you can pass to the universal unital ring and show that the radical is preserved.

[rocky cloak]

1. A polynomial being solvable (by radicals), means that we can write the roots as a combination of the four basic operations and nth-roots of rational numbers. You might also say solvable by radicals over F, where F is some field other than Q.

Over the real numbers every polynomial is solvable yes, so that case is not very interesting.

2. So the Galois group consists of field automorphisms. Such automorphism will necessarily permute the roots of the polynomial, but not every permutation of the roots can be extended to a field automorphism.

For example consider the polynomial

x^4 + x^3 + x^2 + x + 1 = (x - w)(x - w^2)(x - w^3)(x - w^4)

Where w is exp(2pi i/5), there are 4!=24 ways to permute these roots arbitrarily. But a field automorphism s, must satisfy s(w^k) = s(w)^k, so it would be impossible to have a field automorphism that for example swaps w and w^2 , because

s(w^2) = s(w)^2 = (w^2)^2 = w^4 =/= w

[cosmic nexus]

Thank you for the reply. I get no. 2 now. For no. 1, does that mean there does exist some general formula using radicals using just the coefficients (which can have irrational values) of a degree 5 polynomial?

[rocky cloak]

No, there are polynomials with rational coefficients which are not solvable. So no general method can exist

[cosmic nexus]

I see that makes sense. So I suppose we study from Q a lot because most of the computation is nontrivial and if we can prove that for a smaller field there isn't a general formula for polynomials degrees five and higher, then for any larger field containing Q, those polynomials in that field in general won't have a solution in radicals using the coefficients. Thank you!

[somber sleet]

guys if I0m asked to construct a galois extension of order 28, how would you proceed?

I thought about creating a radical extension with the root of unity, but I'm confused about the order

[rocky cloak]

Here's a hint: 29 is prime

[somber sleet]

Right??? So it is isomorphic to (Z/29Z)*

Which means that I need a 29th root of unity

Or am I completely off?

[rocky cloak]

Exactly

The splitting field of x^29 - 1 is a Galois extension with Galois group (Z/29)*, hence has degree 28

In fact with a little bit of divisibility trickery you can use this to construct a Galois extension of any degree

[steel light]

Huh

Uh

Let me see why

Hm

So if we have a prime ideal P subset A, we're guaranteed that at least one factor of an element in the ideal is from P

[coral shale]

yay feather learning stuff idk about

[steel light]

So consider an element x in P, we write x = x1 cdot x2 cdot x3 cdot ... x_n

One of these x_i must be in P

Now if we intersect over all P

[crystal turtle]

what do you mean by "factor" of x?

[delicate orchid]

write x as yz

y or z has to be in P

[steel light]

y and z are factors of x in Wew's example

[crystal turtle]

but where are these x_i from i mean?

[steel light]

Oh

Our ring I suppose

[crystal turtle]

this is correct by induction btw that at least one is in P. Keep thinking outloud I guess maybe I shouldn't interrupt

[delicate orchid]

isn't this proof pretty non-trivial lol

[crystal turtle]

yeah this is definitely a non-trivial one, for at least one the directions

[wraith cargo]

one direction is easy
The other one requires Zorn's lemma

[crystal turtle]

but anyways keep thinking sorry idk why I interrupted

[steel light]

I'm thinking

Not sure how to write the next part or if it's right

[steel light]

An x in the intersection can only be in the intersection if at least one of its factors is common to all of the ideals

I'm not sure how to justify this next part

[rocky cloak]

Which direction are you working on?

[steel light]

I think I am doing intsction of all prime ideals => nilradical

[delicate orchid]

I think this is intersection => in nilradical

(the hard direction lol)

[steel light]

o

[crystal turtle]

(this needs not be true: consider 6 in 2Z \cap 3Z, it's has no factors common to both ideals)

[steel light]

But the idea is that all of the factors of x must be common to all of the elements of the ideals

[steel light]

Fuck you

😭

[void cosmos]

how do i know? i just want it to be true

[steel light]

Anyways

[crystal turtle]

sorry 😭

[void cosmos]

so i just put it on the test paper and run

its much easier and makes more sense form e

for me*

[steel light]

zamn I cannot go forwards

[summer path]

I think there's usually a statement you want to prove before hand

[steel light]

I am wrong

Thank you Ryx

LOL

This is the hard direction?

[solar vessel]

2Z and 3Z

[crystal turtle]

You might want to start with the other direction lmao

[summer path]

Yes

[crystal turtle]

fuck

[steel light]

My thinking was like (ignore that my logic was flawed)

[summer path]

The other direction is pretty easy

[crystal turtle]

it was understood what I meant

[summer path]

Like 2 lines

[delicate orchid]

so lets pick a MF r in N(R) so like we have r^n = 0 in this hizzouse

[steel light]

All of the factors have to be common to all of the ideals so we're basically taking the product of the same factor some number of times

[delicate orchid]

don't shush me that's literally the definition of the nilradical

[steel light]

I just don't know how I'm supposed to get that this is nilpotent

But that's probably because my logic is wrong

[crystal turtle]

btw, this factor stuff makes no sense in general rings: not all (even commutative and unital) rings have a notion of "unique factorization"

such rings are called UFDs, or "unique factorization domains" (even then, there are a few technicalities compared to how we think of factorization of positive integers, but they are mostly moot)

[steel light]

Maybe I should just read the proof

[crystal turtle]

try to come up with nilpotent ==> in all primes on your own

[steel light]

Okay

[crystal turtle]

other direction yeah lmao read the proof it's wacko bozo stuff

[steel light]

Let R communital (haha) ring

Let r in R

By defn it is nilpotent, so there exists some n > 0 s.t. r^n = 0

Uh

I don't know

I'll guess though

But I don't think it's right

[formal ermine]

0 is contained in an ideal right?

[steel light]

If rrrr... = 0 then it is in (0)

Yes wait let me finish

[summer path]

I think usually you start with the statement of S some multiplicative set and if I subset R, then there is some maximal ideal P wrt P cap S = empty, and that all such P are prime

[formal ermine]

Real

just localize

[summer path]

Then the proof is much more doable from there

[rocky cloak]

After having a think, nonunital rings can be Artinian without being Noetherian. Just take any group that is Artinian but not Noetherian and make it into a ring by having everything multiply to 0

[crystal turtle]

this is presented well before multiplicatively closed subsets and localization

this is literally 8 propositions in

[steel light]

I want to say that (0) is a subset of every other prime ideal, but I have zero clue if that's actually right or not, and I don't want to think about it too hard rn lol

[crystal turtle]

yes

[delicate orchid]

of course it is

[steel light]

So r^n is in every prime ideal

[summer path]

You don't need localization for this though

[formal ermine]

I just ate oreos but it had some funny rainbow particles in the white stuff. was it safe to eat that?

[steel light]

QED?

[delicate orchid]

you've shown that 0 is in every prime ideal

not r

[formal ermine]

yeah

now use primeness

[steel light]

But r^n is in 0

[crystal turtle]

r^n = 0

[steel light]

So one of the factors r belongs to 0...? Since 0 is prime

[formal ermine]

0 isn't necessarily a prime ideal

[crystal turtle]

(0) = {0}

[steel light]

Yes

[delicate orchid]

yeah what if 0 isn't prime lol

[steel light]

[formal ermine]

0 being prime implies your ring being an integral domain

[steel light]

I need to make a freaking dictionary

[crystal turtle]

So you have r^n = 0 is in your prime ideal P

[steel light]

Yes

[delicate orchid]

and when you're an integral domain you don't have any non-zero nilradical elements

[crystal turtle]

How do you use primeness to get r in P

[steel light]

Okay

Let me think

[formal ermine]

also feather becoming algebruh pilled??? no way

[steel light]

Since it's a prime ideal P at least one of the factors belongs to P

But the factors are all r

So r is in P

[formal ermine]

technically correct, yeah

but it's weird to say it like that

[delicate orchid]

stop factoring things

this is like the one case where it works

[formal ermine]

if xy in P then either x in P or y in P

[steel light]

What do you call y and z for r = yz

[formal ermine]

then you proceed by induction

[steel light]

Why can I not call those factors

[summer path]

What even is there to factor

[steel light]

Is that not what a factor of a number is

[formal ermine]

who says these are numbers

[steel light]

Sorry not number

Element

Whatever

But isn't that what a factor means

[formal ermine]

it's weird to talk about "factors" unless you're in a ufd

[crystal turtle]

I have not seen this proof before hmm i'll look into it. AM just zorn's it up to show every non-nilpotent element has a prime ideal it's not contained in. Which I guess is probably similar, taking I = {x, x^2, ...}?

[delicate orchid]

I'd say a factor y of x is an element such that (y) contains (x) but meh

[formal ermine]

because they're not uniquely determined

[steel light]

Then what do I call the things that multiply to make another thing

[formal ermine]

things that multiply to make another thing

[steel light]

[delicate orchid]

this kind of skips over the actual cool part of this proof and it makes me sad

so I'm going to just do it myself

[steel light]

Okay

Can you send it in one message

Easier for me to follow since so many ppl are typing lol

In class rn

[summer path]

I guess the proof is pretty much the same I think, it just generalizes a bit? I think?

[delicate orchid]

r^n = 0 in every prime ideaL
r^n = r*r^{n-1}, one of these is in every prime ideal, if it's r we're done
if it's not r then r^{n-1} is in every prime ideal
but r^{n-1} = r*r^{n-2}
proceed inductively until we reach r^2 = r*r, now we don't have a choice. So r must be in every prime ideal

[steel light]

Okay

I see my mistake

But I don't see why it's a mistake

[crystal turtle]

which if it's not nilpotent, it does not contain 0 (any multiplicatively closed subset containing 0, of course, has no such disjoint prime ideal)

[steel light]

So the defn of prime ideal involves a binary multiplication

I thought you could just extend that to multiplication of n things

[delicate orchid]

what if we "factored" r^n as r^2r^{n-2}

[steel light]

And it means at least one of the n comes from your prime ideal

[delicate orchid]

we cannot conclude shit about r from that

[summer path]

This is like the least necessary explicit induction proof

[crystal turtle]

yeah I think ik what you're referencing. AM has some stuff along these lines in the exercises

[delicate orchid]

not a single explicit induction proof is necessary

[steel light]

OHHHHHHHHHH

[delicate orchid]

not one

[steel light]

This is what I was doing but I did it "all at once" which is wrong

[summer path]

[delicate orchid]

and believe it or not when teaching a beginner sometimes putting more detail in that you deem neccesary is a good thing

[steel light]

But I don't understand why this is wrong

[delicate orchid]

so perhaps back off slightly

[steel light]

Is it because that just isn't the definition of prime ideal?

The definition involves a binary multiplication

So we HAVE to do it with two terms?

And proceed inductively

[summer path]

Sorry, let's do an induction warmup....

[delicate orchid]

if r is in a prime ideal and r = xyz then we can only conclude that one of {x, y, z, xy, yz} is in the prime ideal

I think

[crystal turtle]

Perhaps show separately by induction that if x = x_1 * x_2 * ... * x_n is in a prime ideal P, then one of the x_i is in P

[steel light]

What about xz

Noncommutativity?

[delicate orchid]

oh yeah they're commutative rings

xz also

[steel light]

Get fucked old man

[delicate orchid]

I will be crying myself to sleep tonight

[steel light]

Sigh

Yeah

Okay

I will no longer make that mistake

-.-

Anytime I see a product I will NOT extend it to higher terms unless I prove it inductively

[delicate orchid]

it was just for the cool factor

[crystal turtle]

like you're right that it works, it might just be helpful to work through it

[steel light]

In the future

Tu/Th are my busy days

[crystal turtle]

:swag:

[steel light]

I have back to back to back classes

from 11:30 to 4 PM

[crystal turtle]

i have like 6 hours straight on tuesday this next semester

eww

[steel light]

Ew indeed

I wish I wasn't so needy

If I can get the proper environment set up I can get some fkn WORK done like I'll build the pyramids of Giza in a day on god

but it's unlikely I ever enter the proper environment

Anyways

Moving on

Time to read the other direction

Zorn's Lemma seems to show up a lot here

AM referenced it earlier too

[crystal turtle]

yeah

[steel light]

Is it about time I start caring about it and choice and stuff?

I've never actually used it before so I never bothered to memorize what it is lol

[crystal turtle]

(just assume choice and use Zorn freely)

[steel light]

Fuck yeah

[crystal turtle]

in fact it's pretty much just Zorn, not explicitly choice

[steel light]

what is a chain

[steel light]

Yeah Zorn equiv choice

[summer path]

No one actually cares about using choice I think

Just use it if you need it

[steel light]

Ryx and Sharp circlejerk about it frequently

-.-

The rest of us don't lol

[crystal turtle]

set theorists need a job /s

[summer path]

Lol

[steel light]

What is a chain

[crystal turtle]

i circlejerk it but I think rejecting choice is dumb lmao

[steel light]

That's like

[crystal turtle]

(altho I see the use for stuff like "statements internal to a topos" where you might not have choice but whatever(

[steel light]

the biggest green flag from u so far

poggers!

[crystal turtle]

yea and then I said this which is probably the biggest red flag I've said so far so uhhh

[steel light]

How is Zorn related to well ordering principle

They're like opposite sides of the same thing kinda

One least element one maximal element

[crystal turtle]

Zorn ~ choice I think is okay? don't remember actually
well ordering theorem ==> choice is easy. Other direction iirc is

[steel light]

Oh

They're equivalent?

Well ordering and Zorn?

[crystal turtle]

yup

[steel light]

Wow

Nice

I love how my professor is just like

[rocky cloak]

So for a set you can impose a well order on a subset. For example for a finite subset this is easy to do.

You can then also do it in a bigger set, and you can gather all possible well orderings of some subset into a big poset.

Zorn's lemma implies that this poset has a maximal element, and that would exactly be a well ordering of the whole set

[steel light]

"don't worry about the convergence of any of these laplace transforms or how we get them, just table it all"

[delicate orchid]

engineer detected

[summer path]

You can think of a chain like a bunch of subsets sorted by inclusion, and then you want to show that there is an upper bound of the chain (usually the union of all the subsets in the chain), then by Zorn you have that there's a maximal element

[steel light]

Indeed I am Wew

[crystal turtle]

interesting that doesn't seem too bad actually

[delicate orchid]

been thinkin lately

[crystal turtle]

me too

[topaz solar]

Choice -> WO is ez

[delicate orchid]

been thinkin bout HIGHER centres

[crystal turtle]

oh oop

[rocky cloak]

Yeah, Zorn is the one that is easy to prove implies all the other ones. Going the other way is worse

[delicate orchid]

hmmm today I will simply choose a well ordering :clueless:

[crystal turtle]

that must be what I was thinking of then. I imagine choice ==> Zorn is the hard one then?

[summer path]

A bunch of Zorn proofs kinda go the same way in that you just define a set, have a chain, show union over stuff in chain is still in the set, hit it with corn, and you're done

[topaz solar]

Apply choice to P(X), do ordinal function w by 0 -> f(X), n+1 -> f(X-U {w(i<=n)})

[crystal turtle]

okay for some reason I thought that was choice --> WOT

[summer path]

I'm not fixing the typo, it's better this way anyway

[crystal turtle]

btw, eventually AM will just be like "a standard application of Zorn give..." and not give details

but it spells out the first few at least

[summer path]

If you want another exercise which is pretty much the same thing, you can show every vector space has a basis

[steel light]

I don't understand what Sigma is

The set of ideals that don't have nilpotent elements?

[crystal turtle]

Sigma is the set of ideals which contain no powers of f

[steel light]

Oh wait no

Yes

Yes

Lol

I realized right before you said it

[crystal turtle]

[steel light]

Unrelated to this channel but

New technique learned for evaluating limits (at least at infinity)

Final Value Theorem

[delicate orchid]

got an actual question for chat

[steel light]

Ah but that requires the function be Laplaceable...just assume it :^)

[summer path]

Isn't that just like twice differentiable?

[delicate orchid]

I could probably show this myself but I cannot be bothered, let G(n) be the nth derived subgroup of G, is G/G(n) isomorphic to the nth centre of G?

or like what's the relation ykwim

[topaz solar]

Now show a maximal-in-\Sigma ideal is prime and gg

[summer path]

[topaz solar]

I do not believe so

If by nth center you mean the uhh

Upper central series?

[steel light]

This comes from the fact that if (x) and (y) were in p then p + (x) and p + (y) would just be p and hence it would not strictly contain p Yes?

[topaz solar]

If they terminate they have the same length at least

[delicate orchid]

yur

[topaz solar]

But if they just stabilize at a nontrivial point then uhh the lengths to reach that stabilization can differ

[delicate orchid]

ok I was just curious

[topaz solar]

Which kinda kills that

[delicate orchid]

oh ok that's interesting

lemme think about why

[topaz solar]

Ye it’s in D&F

[delicate orchid]

upper central terminates at a centralless groups right

[topaz solar]

Try S_n I think

[delicate orchid]

and uhhhh

the nth dudes terminate at a perfect group...? yeahhhh

obviously perfect => centralless but not the other way around

cool

[topaz solar]

If the upper & lower series go all the way, they have the same length though

Something like that

There’s a section in D&F which gives an example where they’re very clearly not the same iirc

[rocky cloak]

I mean for G = S3, G(1) = C3, and Z(G) = C3, but then G(2) = 0, and Z(G/Z(G)) = C2

[topaz solar]

Ye

[delicate orchid]

very sad!

there also doesn't seem to be a nice iterative formula for higher centres which is cringe

have to consider the kernel of some dog water map

[topaz solar]

Lower series > upper

[delicate orchid]

I don't know what either of those words mean

[steel light]

why does AM use the same symbol for nilradical and Jacobson radical

[crystal turtle]

nonono

one of them has a very slight curve

[formal ermine]

who cares about non pids anyway

[crystal turtle]

[steel light]

how do we know almost all of the x_i are zero

[white oxide]

what does lang mean by the definition of an operation?

[steel light]

Or is it just a part of the definition to guarantee convergence

[white oxide]

or rather how does an operation being well-defined imply that 1x = x here

[formal ermine]

probably the operation

[white oxide]

whoaaa! thank you that makes sense!

[delicate orchid]

yeah this awfully worded but it means a ring action

[formal ermine]

it's a requirement for those sums

like we only consider finite sums

[delicate orchid]

viewing A as a multiplcative monoid
and there we go, it's a monoid action

[white oxide]

ah ok that makes sense

[formal ermine]

"convergence"

[steel light]

Ok so if almost all of the x_i are not zero then we cannot define the sum?

[formal ermine]

Yes

[steel light]

STOPPPPPPP 😭 😭

[crystal turtle]

definitionally this holds (we don't want to consider infinite sums in arbitrary rings)

[steel light]

Gotchya

[delicate orchid]

it's literally by defintion yur

[south patrol]

vercongence

[delicate orchid]

WJHAT

[formal ermine]

like we could define it (see e.g. something like tate algebra) but it just doesn't make sense in this context

[steel light]

What does it mean for xy to generate an ideal? I understand what it means in the context of principal ideals ubt not in general

[south patrol]

So it means like

[formal ermine]

the second sentence defines it

[south patrol]

Well, at least the morally correct definition, is that given a subset S of a ring A, the ideal generated by S is the intersection of all ideals containing S

i.e. the smallest such ideal

A concrete description / alternative definition is given in the 2nd/3rd line there

[steel light]

How do we get the sum part from this? Why are the elements not just all of the products x_i y_i

[south patrol]

That needn't be an ideal.

[delicate orchid]

because that isn't closed under addition

[steel light]

Is it because if we add two then they're going to be closed under addition too

Oh

HAIWUHEQ

LOL

[delicate orchid]

so we take the group they generate

[south patrol]

I mean even if we take the trivial case of like one non-zero element

[south patrol]

why is this open cry lol

[crystal turtle]

no you're right

that's the correct way

[south patrol]

this is the typical notion of what it means for smth to generate smth ye

[delicate orchid]

the nerd way

[crystal turtle]

🤓

[south patrol]

lol

i mean it is well-motivated

[crystal turtle]

just obtuse for actually working with it lmao

[south patrol]

sure

but then it is hopefully clear once you know this definition what it looks like

[steel light]

This is like

[south patrol]

anyway

[crystal turtle]

but I agree for broad notions of "generation" this is a universal way to construct something

[steel light]

1000x more difficult than manifolds

at least manifolds I could intuit my way through stuff lmao

[delicate orchid]

I really really disagree with this statement

oh my god manifolds make no fucking sense what the fuck is an atlas??!?!? mfs think I can reaD?!?!?!

[south patrol]

hm i found it took me a while to get intuition for playing with algebra

[steel light]

be so fr wew

[south patrol]

but then eventually you should get the hang of it more

[steel light]

yeah

[delicate orchid]

the only way manifolds make sense is if they're a lie group

[south patrol]

just it's a different intuition to drawing and stuff lol

[chilly ocean]

wew has never seen a sphere in his life

[steel light]

I'm sure you'll be unsurprised to hear I don't have the proper algebra background potato

[south patrol]

oh i mean that's why i said it like

[delicate orchid]

I only see things up to homotopy so

[steel light]

LMFAO

[south patrol]

no i mean like

it's normal lol

[steel light]

wtf is an hcf

[south patrol]

highest common factor

= gcd

[steel light]

Oh

weirdchamp

[topaz solar]

Mfw im reading model theory and I’m seeing homotopies

Not a joke

[delicate orchid]

I wouldn't know why that would be a joke

[steel light]

OH?!?!?!?

[topaz solar]

Because agony

[steel light]

any algebraic number theory fans what the FUCK?!

[delicate orchid]

ok time to compute the higher centres of D_1 million for a laugh

[crystal turtle]

huh

[delicate orchid]

I kinda forgot how boring Z(D_n) was

[crystal turtle]

L

[delicate orchid]

sharp give me an interesting one

[night onyx]

It means that ab is the smallest ideal of A containing all products xy for x in a and y in b

[delicate orchid]

yeah it's kinda cool how there's like a lattice isomorphism hahahahahahHAHAHAHHA

[crystal turtle]

lattice of integers

[steel light]

dude

ideals don't generally have distributivity...

my life is ruined

[crystal turtle]

🤯

[steel light]

agh that's gonna be a pita to get used to

[chilly ocean]

ideals aren't ideal, whodathunk

[topaz solar]

C_n \wr S_n

[delicate orchid]

that has trivial centre I think

ah no

[topaz solar]

Dang it

[delicate orchid]

diagonal subgroup

[formal ermine]

we should just go back to ideal numbers

[topaz solar]

Ooh

[delicate orchid]

it's centre is abelian

[formal ermine]

"centre"

[topaz solar]

woah

[formal ermine]

it's Zentrum

[delicate orchid]

so the 2nd centre is uhhhh uhhhh no, the whole group (the 2nd term of the upper series is trivial)

[topaz solar]

Dang it

[steel light]

why not just call them points

[delicate orchid]

what would we be calling points

[steel light]

(x1, ..., xn)

[chilly ocean]

elements of rings are functions, not points

[crystal turtle]

zip it german child

[delicate orchid]

call them functions out of [n] into A yeah hahahha

[185db klaxon] based alert!

[steel light]

what?

[delicate orchid]

everything is a function dw about it

[steel light]

what?

[crystal turtle]

[delicate orchid]

in fact elements of rings are TWO functions at once

[steel light]

what?

[chilly ocean]

algebraic geometry

[steel light]

oh

polynomial rings

you mean?

[south patrol]

This is pain

[steel light]

what?

😭

[south patrol]

Idk what wew is on about ignore him

[steel light]

feels like everyone is teasing me

:p

[delicate orchid]

for each a you can assocaite the function given by multiplying by a and adding a on

two functions - three if it's non-commutative!

[south patrol]

you can associate any number of functions wit a

[steel light]

[chilly ocean]

yeah feather. polynomials are the functions on affine space you care about

[steel light]

yeah

me who doesn't know what an affine space is (but I've seen it used in the defn of varieties)

[delicate orchid]

yeah watch me

[chilly ocean]

someone more versed in algebraic geometry than me can finish my train of thought

i have things to do

[topaz solar]

Ok so, the easy idea might be uhh

C(X, R)?

Im not well versed in the AG functions, but should look similar since uhh

Locally ringed space

[night onyx]

affine space is just k^n where k is a field, and you think of polynomials in k[x_1, ..., x_n] as functions k^n -> k. Algebraic sets are the just set of zeros of a bunch of polynomials

[topaz solar]

And uhh there’s rings of functions on varieties or such iirc

[delicate orchid]

coordinate rings :GangnamStyle:

[night onyx]

yeah, basically if you have an algebraic set V in k^n, you can look at k[x_1,...,x_n] / I(V) where I(V) is the set of polynomials which are zero on V, which if you think about it is essentially "polynomials functions restricted to V", where you identify polynomials which are equal on V (ignoring their values outside of it)

[hollow mica]

is this the the same as the definition of affine space on wikipedia? which says it’s a set with a vector space’s additive group acting on it transitively and freely

[steel light]

It is more than a little obnoxious that ideals do not behave ideally

Under operations

Lol

[hollow mica]

or ig what you wrote is a particular example? (k^n acting on itself by vector addition, but of course your guy has more structure)

[steel light]

what????

So the set of all x in A that "absorb" elements of b into a?

[delicate orchid]

slurrpppp

[steel light]

Is that a yes

LOL

[topaz solar]

What if a is prime

[steel light]

Um

[delicate orchid]

Id go with "move" rather than "absorb" but yeah

[steel light]

If a is prime thennn

[delicate orchid]

or perhaps where sharp is going

[solar vessel]

dawg icl I've never heard of this b4

[steel light]

Then x is in a or b is a subset of a

[delicate orchid]

I haven't either lol

[steel light]

Yes?

[delicate orchid]

yur

[steel light]

I'm the fkn goat

[hollow mica]

this looks like a painful definition to work with

[night onyx]

mmm I'd have to think about it, it seems like that's a more abstract definition, when I think of affine n-space it's usually in the context of algebraic geometry where it's just coordinate vectors (or points) over a field, but my guess is it's the same thing just from a different perspective

[steel light]

This is how AM define annihilators

I've seen you guys use that term

Isn't that important?

[delicate orchid]

wh

[topaz solar]

A weird way to reach annihilators but yeah I can see it

[delicate orchid]

oh my god are you defining Ann(I) as (I;0)

[steel light]

(0; I)

lol

speaking of

wtf is this notation

isn't this nonsense

[delicate orchid]

I might hit the griddy at this rate

[steel light]

oH wait

I keep forgetting

(x) = principal ideal generated by x

not just x in parentheses -.- or like x as an input of the Ann function (functor? whatever the technical term is)

[delicate orchid]

yeah although you can just define it on elements

in the obvious manner

[hollow mica]

ahh that’s good to know. I’ve never actually cracked open an AG book but I know it’s about affine spaces. So one day I decided to visit the wiki page for affine space and I was kinda disgusted by the definition lol

[steel light]

I'm so confused with the union condition

Zero divisors can be zero right?

[delicate orchid]

why would that matter

[steel light]

But this condition on the union implies that they can't be

[topaz solar]

0 is gonna be in there always

But consider

What’s Ann (0)

[steel light]

Uh

[topaz solar]

[steel light]

Don't spoil

oh

0

[topaz solar]

No

[delicate orchid]

no it's everything

[steel light]

Wew

-.-

[delicate orchid]

0x = 0 for all x so x in Ann(x) tada

[topaz solar]

Yeah so that’s why U(x\neq 0)

[steel light]

Oh yes

Okay yes

[topaz solar]

[night onyx]

lol yeah I get that, really you think of affine space as k^n together with the ring of polynomial functions on k^n (kindof like in topology you consider spaces and continuous functions over those spaces), AG is all about polynomial functions lol

or rational-polynomial functions

[steel light]

This stupid fucking (x) notation is going to give me cancer

[delicate orchid]

(x) is basically just x anyway

[steel light]

Yeah

I know

I just keep forgetting the parentheses meawn something else here

[delicate orchid]

Affine space is..,. it's k^n s-s-swemi dwiect GL_n ..... :tearsofthenile:

it kinda is doe...

[night onyx]

well it's important that if you have Ann(x) = {r | rx = 0} for an element and Ann(I) = {r | rb = 0 for all b in I} for an ideal, then Ann(x) = Ann((x)) so it's safe to write either/or

[solar vessel]

what the hell is tearsofthenile

[delicate orchid]

:uponthewitnessing:

[steel light]

So the radical of an ideal is all ring elements who have a power in the ideal

[topaz solar]

Yes

[steel light]

The nilradical of a ring is the set of all nilpotent elements is the intersection of all prime ideals

[night onyx]

yeah, it's basically "all roots of things in the ideal"

[solar vessel]

and that's why it has the root symbol

[steel light]

The Jacobson radical of a ring is the intersection of all maximal ideals

[solar vessel]

it's all coming together

[steel light]

Yes it is

[topaz solar]

What’s the nilradical of R/I

[steel light]

No spoil pls

um

[delicate orchid]

that's a bit mean to ask at this point lol

[steel light]

All the elements in R/I which are nilpotent (not an answer, just thinking out loud)

[solar vessel]

true

[steel light]

An element of R/I is a coset of r in R

[topaz solar]

Feather can handle it

[steel light]

I am not going to read this because it looks spoilery

Anyways

Thank you

[night onyx]

sorry bout that!

[steel light]

All good :)

So what cosets of r are nilpotent?

Hm

[night onyx]

all this ring theory just get's me all riled up!

[steel light]

Let's see

[night onyx]

real fired up you know

[hollow mica]

real juiced up

[topaz solar]

Well, what’s (r+I)^2

[white oxide]

here would we consider a morphism between two objects in the category of homomorphisms of M whose kernel contains N to be that of this example from hungerford? sorry i'm very new to category theory type stuff so it's not really trivial to me lol

[steel light]

The product of (r + I)(s + I) is the coset (rs + I)

Sharp I'm not going to read your hint either sorry

Not until I know for sure I messed up and I can't think of anything else

[topaz solar]

That’s literally what I suggested

[steel light]

I really really want to try and be better at doing this stuff more formally

Oh

LMFAO

Anyways :^)

[topaz solar]

Ye go from there

[formal ermine]

feather algebra arc???

[steel light]

So we are asking when (r+I)^n is nilpotent

This is just (r^n + I)

Which is nilpotent if it equals 0

When does it equal zero?

[night onyx]

You'd probably verify that if f : M -> M' is a module homormorphism whose kernel contains N, then there is a unique homomorphism f* : M/N -> M' such that F*(m + N) = f(m)

[steel light]

Well if r^n is nilpotent then you're left with the coset being I, but in the quotient ring this is equivalent to 0 right?

[white oxide]

okay, i'll think about how that's equivalent to the statement he posited

[steel light]

I still don't understand well what quotient rings look like but from everything we talked about yesterday this is what I understand so far lol

Anyways

[topaz solar]

When is x + I = y + I

[steel light]

When x - y is in I

[topaz solar]

what about nilradical of R/I

[steel light]

I don't understand

I'm just going to continue with where I was going

[night onyx]

A good way to think of quotient rings is by taking a ring R and imposing new relations on it, making it smaller. A good example is taking R[x] and then adding the relation that x = 0. This effectively replaces each occurence of x in a polynomial with 0, so you just have the constant values. Depending what relations you impose the new ring looks different, like if you take R[x] and impose the relation that x^2 + 1 = 0, you actually get the complex numbers

[topaz solar]

You gettin there

When is ||r^n + I = 0 + I||

[steel light]

I think r^n is nilpotent when r is nilpotent, since if we have (r^n)^m = 0 this means r^n = 0 which means r is nilpotent

[topaz solar]

Yes but uh

[steel light]

There are other cases to consider too

Let me remember where I am

So the nilradical of R/I is the set of all cosets (r + I) s.t. (r + I)^n = 0. This happens when r is nilpotent. So when is r^n + I = 0 + I?

When r^n - 0 = r^n is in I

[topaz solar]

Yes

The radical of I

[steel light]

But this is the radical of I

[mighty kiln]

[steel light]

Oh

[topaz solar]

[steel light]

So the

Wow

Nice

[topaz solar]

Ascended

[steel light]

So the nilradical of R/I is the radical of I

[mighty kiln]

Quotiented by I

[steel light]

I see where you're getting quotiented by I comes from but not why

Let me think

[steel light]

For it to be equivalent to 0 (which is what we require for the nilradical) we need to quotient by I

So of course the radical has to be quotiented by I

[steel light]

Loser (with love)

:3

I figured it out hehe

Oh

[topaz solar]

[steel light]

Nice

[topaz solar]

I saw it coming fr

[steel light]

So the radical of a subset E of A is the set of all e in A such that e^n in E?

[topaz solar]

Ye

I think

[white oxide]

in defining the A-algebra, do they mean the module E together with a bilinear map g: E x E --> E? or do they mean any module such that there exists some module E such that the map g is bilinear

[steel light]

Idon't understand the first step lmoa

the rest I do

but how do we know r(a + b) = r(r(a) + r(b))

[mighty kiln]

With the map

[topaz solar]

“module together with a”

[night onyx]

the fact that E is an A-module means you can scale elements of E by elements of A, and the map g allows you to multiply elements of E together (and bilinearity means addition and multiplication distribute)

[topaz solar]

There’s only 1 module involved

[maiden ocean]

a is contained in r(a) and likewise for b and r(b) so one inclusion should be fairly clear. conversely if x^n is in r(r(a) + r(b)) then it is y + z with y^m in a, z^l in b. then (x^n)^m+l is a sum of terms that are always divisible by either y^m or z^l, since term in (y + z)^m+l have the form c y^i z^j where i + j = m + l, so that if i is less than m j must be greater than l. hence x^n(m+l) is in a + b

[delicate orchid]

wow imagine doing it without just going "corrispondence theorem QED"

[steel light]

I don't understand the lsat step

"so that if i < m then j > l => (x^n)^(m + l) is in a + b"