groups-rings-fields

[dull ginkgo]

It’s a pain to prove

[glad osprey]

Ah, the Jacobson commutativity theorem? I've heard about that

[dull ginkgo]

I spent forever trying to prove it myself

Unfun

[rocky cloak]

They're called roots of unity at least

[glad osprey]

Right, I guess that term works in any ring, not just in C

[dull ginkgo]

There’s also principal roots of unity, which allow you to generalize the Discrete Fourier transform

[naive lance]

Thank u

Yes its clear if we localize and look at the nilradical of A_p

[wide brook]

Is there a finitely generated torsion R-module, where R is a PID, which is infinite? I think if R=Z all such modules(where M=T(M))are finite. But is this true for all PIDs?

[dull ginkgo]

F[X] is a PID for any field F. Let F be infinite (or F(t)), then F is a simple F[X] module (F[X]/(X)) that’s infinite and torsion as Xx = 0 and generated by 1

[dull ginkgo]

Any simple module over a commutative ring is either torsion or torsion-free (as T(M) is a submodule) so you can pretty much always try to find one lol

[wide brook]

Simple is if it’s generated by one element right?

[dull ginkgo]

That’s cyclic

Simple can be equivalently stated as

• no proper/nontrivial submodule
• every nonzero element generates R
• is isomorphic to R quotiented by a left (right) maximal ideal [it’s annihilator] as a module

[next obsidian]

No

[wide brook]

Ok thanks! Haven’t done this yet. So I’m gonna have to think about it and come back at you

[next obsidian]

Structure theorem says that any torsion module is a finite direct sum of things of the form R/x for x in R nonzero

So it suffices to show each R/x is finite

But now factor x = x_1^n1•…•x_k^nk where x_i are irreducible

[dull ginkgo]

@next obsidian

[next obsidian]

Then the Chinese Remainder Theorem says that R/x = (R/xi)^ni and each R/xi is a field

Oh Hurb

I was replacing size with length

I was going too woke and thinking about how Artinian algebras are finite LENGTH

[dull ginkgo]

Any chain of prime ideals of my thang stabilizes after 6

[next obsidian]

That measures height

Not length

[dull ginkgo]

[lean sail]

i did this proof the other day with a large amount of help from @dull ginkgo and @glad osprey . i still feel that i am missing one thing. i dont think ive shown that there are exactly p - 1 elements in each equivalence class.

[charred iris]

Indeed you haven't

(I also think the wording in the symmetry part can be improved)

[lean sail]

any hints as to how i can show that each equivalence class has p - 1 elements?

[charred iris]

If you have x in S, what other elements are guaranteed to be in the same equivalence class?

[lean sail]

so if we let $[x]$ be an equivalence class of $S$ containing $x$...

[cloud walrus]

proofman

[charred iris]

In fact the definition of ~ explicitly tells you what they are

[lean sail]

then $[x]$ contains: $x, x^2, \dots, x^{p - 1}$ since the elements would start repeating at $x^p$?

[cloud walrus]

proofman

[charred iris]

Yes

[lean sail]

but i know that i am not proving anything by saying that

[charred iris]

Well you have to also say why these p-1 expressions are distinct elements of G

[lean sail]

yeah @dull ginkgo tried explaining this and i didn't quite understand... so we have to show why $x$ and $x^2$ (for example) are distinct (not equal)?

[cloud walrus]

proofman

[charred iris]

Yes

Because it might be the case that some of x, ..., x^p-1 are the same and so you have fewer than p-1 elements in the equivalence class

[lean sail]

ok, here is where i get really confused, because, for two things to be in an equivalence class together, doesnt that automatically mean that they are not distinct? since they are "equivalent"?

i think that i dont have a clear definition on what distinctness really is

[charred iris]

Why would it be wrong to say that in [x] we have infinitely many elements x, x^2, x^3, etc.

[lean sail]

consider $x^{p + 1}$, it is just $x x^p = x$

[cloud walrus]

proofman

[lean sail]

so we get a repeat

[charred iris]

Yes, now how do you know that we have no repeats in x, ..., x^p-1

[lean sail]

i think i can show this

but is that enough?

[charred iris]

In the same way it's wrong to say we have infinitely many elements x, x^2, ..., it might be wrong to say we have p-1 elements x, ..., x^p-1

(We do have p-1 and not fewer, but you need to prove that)

[lean sail]

After I show that the elements in each equivalence class are distinct. Am I finished or is there another step?

[charred iris]

Well by now you've shown that there are p-1 distinct elements in [x], and that there aren't any others

So you've indeed shown that there are exactly p-1 of them

[lean sail]

Before we show distinctness, do we already know there are less than or equal to p - 1 elements?

[charred iris]

You've said that the list x, ..., x^p-1 will have all the elements of [x], because every element is some x^n and any n > p-1 is extraneous because of the repetition (or isn't even in S). So you have at most p-1 elements x, ..., x^p-1 in the equivalence class

[lean sail]

I said that, that’s true. Is that enough for a proof of that part?

(Plus showing distinctness)

[charred iris]

Yes

Because you've explicitly listed the p-1 elements of [x]

And explained why there aren't any more

[lean sail]

I could say something like consider that every element after p is a identity multiplied by a element of S sorry for bad English

[charred iris]

Yes (and x^p and multiples of it aren't even in S)

[lean sail]

Well, $x^p = e$ so I’m not sure what you mean

[cloud walrus]

proofman

[charred iris]

You explicitly don't have e in S

[lean sail]

Yeah makes sense, we don’t have $e$ in $S$, but $e x = x \in S$

[cloud walrus]

proofman

[lean sail]

Anyway, @charred iris thanks for the help!

[dull ginkgo]

So this is the last time I’m going to go nuts trying to solve this problem

Let’s say we have a PID R, and some collection of vectors in R^N

We can write these as a matrix, with the columns being the vectors. This matrix, call it M, represents the map from D^|S| (S the collection) to D^N

We can find invertible matrices: U (|S| x |S|) and L (N x N) such that:
UML = K where K is a diagonal matrix

How can we use this diagonal matrix to find basises for the kernel and image of M

[rocky cloak]

I guess you mean K = LMU?

Anyway, you can find the kernel of K quite easily as just those basis elements with 0 on the diagonal. Then the kernel of M is U^-1 times the kernel of K.

Similarly a basis for the image of K is just given by the nonzero columns, and the image of M is just L^-1 times the image of K

[dull ginkgo]

Thank you man

[lean sail]

Had one more question about this… do I need to mention anything about $n < 0$?

[cloud walrus]

proofman

[charred iris]

Since x^-1 is x^p-1, it falls under the same repetition thing

You can mention it if you want

[lean sail]

I actually think it doesn’t make sense because when we talk about n number of elements a negative doesn’t make sense

[lean sail]

But yeah your statement is definitely true

[wide brook]

Hello! I’m trying to find the invariant factors of pM. Where R is a PID, in which p is irreducible and M is a torsion R-module with invariant factors p^a1 | p^a2 | ……p^as. What I’ve thought so far is that pM=pR/(p^a1) oplus …….. pR/(p^as)= (p)/(p^a1) oplus …… (p)/(p^as). Now, my intuition tells that pM must have invariant factors of the form p^b1|…..|p^bt where t=<s and bi=<ai but I don’t know how to prove this or how to move forward

[next obsidian]

You just have to analyze what pR/p^nR is isomorphic to

And it’s very suggestive what it should be

[wide brook]

Is it not (p)/(p^n)?

[next obsidian]

I mean, what is this as a module

The way it’s written suggests an answer

And it turns out that answer is true and it’s not difficult to show it

[wide brook]

It’s emmmm a cyclic module(?) sorry I’m pretty stuck

[next obsidian]

I mean this is division

Just turn your brain off and pretend you’re 13 years old

You’d say this is?

[wide brook]

(1)/(p^(n-1)) ?

[next obsidian]

Right, so this would correspond to R/(p^n-1)

So find a surjective map R -> pR/p^nR with kernel (p^n-1)

And you’ve now rewritten pR/p^nR as a cyclic module so you know its invariant factors

[next obsidian]

Do this to each summand here and you’re done

[wide brook]

Ohhhh alright, so I just prove it with first isomorphism theorem then?

[next obsidian]

Yes

[wide brook]

Thanks!

[next obsidian]

[wide brook]

Sorry last question, we know that R-> pR/p^nR is surjective because R->M where r->rm is surjective and pR/p^nR is a submodule of M, correct?

[next obsidian]

Wut

You have to make the map

And then it’s surjective by obviousness of the construction

Idk what M is here, but if you’re taking M = pR or something then this is basically true but not because pR/p^nR is a submodule of M, but because it’s a quotient of it

Addition and multiplication are just what you’ve called your two operations so the question doesn’t really make sense

I think you’re thinking of addition and multiplication as being inherited from something else like taking examples of R,C or F_p, but these are just symbols

[wide brook]

M is the initial R-torsion module that I had which had p^ai as its invariant factors

[next obsidian]

Then no

M won’t be surjected on by R unless M is itself cyclic

Which is to say M has only a single summand

[wide brook]

Oh it had to be R^as for this to work?

[next obsidian]

Yeah

But this is entirely irrelavsnt

There’s a clear surjective map here to take with the correct kernel

[wide brook]

Something like r->rmod(p^(n-1))?

[next obsidian]

That doesn’t land in the correct module

The image contains the class of 1 (and also you’re taking the wrong ideal to quotient by i guess)

You’re mapping onto pR/p^nR

[wide brook]

Hmmm is it then r-> (p+p^n)r?

[nimble folio]

In generel, if i want to show $V \cap W = {0}$, should I show that $0 \in V, 0 \in W$ and $v \in V,v\neq 0 \implies v \not\in W$ and $w \in W, w\neq 0 \implies w \not\in V$?

[cloud walrus]

clubsoda14

[dull ginkgo]

Alternatively:

if it's in the intersection then it's 0

Would you like an example of this type of idea?

[nimble folio]

Yes please

Oh

I think I know how to show that

But go on anyways

[dull ginkgo]

Lets say we have two vector spaces V and W, and a map f from A to B, and g from B to A, such that f \circ g is the identity on B (i.e f(g(b)) = b).

Assume a is in the intersection of the Ker(f) and Im(g) in A.
Then a = g(b) for some b, and f(a) = 0, then 0 = f(a) = f(g(b)) = b, so a = g(b) = g(0) = 0, thus a = 0

we have shown that any element in the intersection of Ker(f) and Im(g) (subspaces of A) must be 0

[nimble folio]

Bruh

Ok not your fault

but the example you gave me is literally the HW i was trying to solve LOL

[dull ginkgo]

ARE YOU FUCKING SERIOUS LMAO

[nimble folio]

Ya lol

Thats hilarious

I'm not reading it

But I think I can figure it out

[dull ginkgo]

i'm sorry, that's funny tho

[nimble folio]

No its ok

[dull ginkgo]

It's like the classic example of that idea that isn't like, just free modules where it's by def lol

[nimble folio]

The problem I'm doing is a little different actually

I'll write it up and the proof ( i think) i just finished

[dull ginkgo]

Typically this route is via contradiction

assume it's in one, show it's not in the other

but the problem is that you have to ALSO do the other direction

so sometimes it's easier to just show it's 0 if it's in the intersection sorta deal

[nimble folio]

Right

Because its shorter to do that

[dull ginkgo]

yeah, sometimes the other route is easier

i had an example but it slips my mind or is more general (involves modules)

[nimble folio]

Let $\text{dim}_F (V) \leq \infty$ and $T \in \text{Hom}_F (V,V)$ such that $T^2 = T$. Prove that $\text{Im}(T) \cap \text{ker}(T) = {0}$

\begin{proof}
Let $v \in \text{Im}(T) \cap \text{ker}(T) $. Then $v \in \text{Im}(T)$ and $v \in \text{ker}(T)$. So there exists a $w \in V$ such that $T(w) = v$ AND $T(v) = 0$. Since $T(w) = T(T(w)) = v$, it must be the case that $T(v) = v$, establishing that $v = 0$.
\end{proof}

[cloud walrus]

clubsoda14

[dull ginkgo]

yep

[nimble folio]

So T(v) = 0 and T(v) = v implies v=0?

[dull ginkgo]

v = T(v) = 0

transitivity

[nimble folio]

oh LOL

[dull ginkgo]

Btw

you really don't need the dim here

it holds for like, abelian groups too lol

[dull ginkgo]

Here's a follow up question: show Im(T) (+) ker(T) = V REGARDLESS of the dim

[nimble folio]

Yep

thats part b

It does explicitly mention in the problem that V is finite dimensional

[dull ginkgo]

that's there to throw you off lol

it might be easier to just view it as an abelian group :3

[nimble folio]

Hm ok

I'll try and figure it out on my own

[dull ginkgo]

i say this because I got stuck on the one I mentioned (a bit different) when the answer was literally right in front of my face the whole time and I didn't realize it

for the second part

[nimble folio]

We haven't actually covered direct sums yet

And when I was self-studying modules over the summer I was always confused on how to show a module M is equal to, for example, the direct sum of submodules

I'll probably finish the rest of this problem later this week when we cover it

[dull ginkgo]

Here's another funny bit

We can have "internal" direct sums, which is when we add submodules together who have trivial intersection, and "categorical" direct sums, also referred to as coproducts sometimes

In actuality, the "internal" direct sum is isomorphic to the "categorical" direct sum of the two submodules "on their own", as their own thing

so we have elements in V + W, of the vorm v + w, in the vector space M. But as elements of V (+) W, they are of the form (v,w)

but we can map V (+) W into M by sending (v, w) to v + w in M. this is injective (monomorphic)

this happens a lot to the point we say they are the same thing

[nimble folio]

Yes I remember that

[dull ginkgo]

this makes me get fucky about the notation sometimes

[nimble folio]

From Dummit and Foote

[dull ginkgo]

in this case, we just want to show any element of V is of the form T(b) + a, where T(a) = 0

[nimble folio]

where a,b are in V?

[dull ginkgo]

yep

[nimble folio]

Ok

I'll see if I can work it out on my own

[dull ginkgo]

It's 2 small steps but you can overlook it easily from experience

@nimble folio how goes it

[nimble folio]

Let $x + y \in \text{Im}(T) \oplus \text{ker}(T)$. Clearly $x+y \in V$.

Let $v \in V$. Note that $T(v) \in V$. So $T(v) = T(v + k - k)$ for some $k \in \text{ker}(T)$. Then $T(v) = T(v+k) - T(k) = T(v+k) + 0$.

[cloud walrus]

clubsoda14

[nimble folio]

Very weird

Don't really know what to do

[surreal dagger]

If I understand the problem right,
you can "notice" that you can "add zero" to an arbitrary v and this gives you the shape you want

[nimble folio]

yes it gives the right shape

but v is not necessarily in the image of T (atleast I think)

My guess then is, since v = v + 0, and T(v) is in V

then T(v) = T(v+0) = T(v) + T(0)

But this just looks weird

[exotic verge]

getting back into my algebra book after a break

did i get the forward direction here right

for this

ignore the set minus isntead of / lol

[hidden wind]

what is an example of an integral domain that is not euclidean

[quiet pelican]

Take any Z[sqrt(-p)] that isn’t a UFD
So like Z[sqrt(-5)] works

[hidden wind]

whew

i have yet to come to Z[sqrt(-5)], am excited to see how unique factorization breaks down

[quiet pelican]

It’s the standard example of a non-UFD

[chilly ocean]

Another example: Z[x] since it is not a PID (But it is an UFD)

[quiet pelican]

2*3 = (1 + sqrt(-5))(1 - sqrt(-5))

And 2 is irreducible (by computing norms) but isn’t prime (look at that factorisation of 6)

[brisk oak]

What is []_n ? 😭

[rocky cloak]

Equivalence classes modulo n, presumably

[rocky cloak]

polynomial ring in more than one variable.

[brisk oak]

Hmm then [am]_n is not equal to [a]_n I think, which nchoosek is assuming

[rocky cloak]

If [m]_n = [1]_n it is. So I think that's where the mistake lies

[exotic verge]

yes

[brisk oak]

Oh well I guess it's easy to fix.

am + bn ≠ 1

am ≠ 1 (mod n)

But m generates Z/nZ so there exists an integer a such that am = 1 (mod n) - contradiction

Cool proof, never seen it before

[surreal dagger]

By adding zero I meant v=v+a-a=(v-a)+a
where (v-a) is in the kernel and a is in the image, then you have it
So now you just have to find that a

[mighty kiln]

Follow-up exercise

[hidden wind]

[lean sail]

@charred iris @glad osprey @dull ginkgo after a couple of weeks of working on this. here is the full result. thought you might be interested. thank you for all the help. of course, still looking for criticism and feedback. this is kind of a long one to read through

[glad osprey]

I'm not sure how you get that a-b must divide p (in the distinctness part)

[lean sail]

since $x^p = e = x^{a - b}$, it follows that $a - b$ must divide $p$, since $a - b$ and $p$ are coprime, the only way for this to happen is if $a - b = 1$

[cloud walrus]

proofman

[glad osprey]

Why does x^p = e = x^(a-b) imply that a-b must divide p? This is true if a-b is the order of x, but not in general

[lean sail]

im not really sure how to answer this question

[glad osprey]

I think it's better to use bezout's identity again for that part

[lean sail]

you say that this is true if a - b is the order of x, but not in general, can you show that this is the case?

let me put it another way, if $x^{a - b}$ is equal to $e$, then $x^{a - b}$ is some power of $x^p$

[glad osprey]

if x^2 = e for some x, then x^6 = e and x^8 = e, but 6 does not divide 8

[cloud walrus]

proofman

[glad osprey]

Yes, if p is the order of x, then p divides a-b, not the other way around. And if x != e and x^p = e, then p is the order of x since p is prime

but I'm not sure how many of these results you can use without proof, since you haven't gotten to the part of the book about order and subgroups and stuff

[lean sail]

i was going to say something like that

[glad osprey]

so if you use bezout you don't have to talk about order

[lean sail]

i think in this example, 2 is actually p, 2 divides 6 and 2 divides 8

[glad osprey]

well, if x^2 = e then the order of x is either 2 or 1

[lean sail]

in $S$ defined in the solution, $x^p = e$

[cloud walrus]

proofman

[lean sail]

so in your counter example, what is $p$?

[cloud walrus]

proofman

[lean sail]

i guess im just trying to understand your counterexample

what is a, what is b, what is p?

[glad osprey]

I'm saying that if x^a = e and x^b = e, you can make no conclusion about whether a divides b or vice versa. However, when x != e and x^p = e when p is prime, then p is the order of x, and the result follows. You can prove this easily, either specifically when p is prime (in which case you could use bezout), or in general when p is the order of x (in which case the proof would be slightly different but still pretty easy)

[lean sail]

im stuck at the first part of this message, why is it that we can make no conclusion about whether $a | b$ or $b | a$

[cloud walrus]

proofman

[glad osprey]

if you want a concrete example: 1³ = 1 and 1⁴ = 1, but 3 does not divide 4 or vice versa

[lean sail]

i laid out a couple steps in my attempted proof, where i tried to show that a - b divides p, do you know which step is incorrect?

[glad osprey]

where? I just see that you write x^(a-b) = e, so a-b divides p

[lean sail]

really? let me go back and check

[lean sail]

yes, you are right

[lean sail]

ok, i am thinking about this example... thank you for your patience

[glad osprey]

honestly, I think your time would be better spent reading the next couple of chapters instead of continuing this proof

[lean sail]

i have seriously considered this, but at the same time i don't want to give up

[glad osprey]

Read about orders, and the theorem that if x^n = e then ord(x) divides n. Then try to prove it

I just don't think it's very productive fighting with a proof when you don't have a good overview of what is going on. With more theory under your belt this proof could be 2-3 lines

[lean sail]

how would you do this with bezout? i am going to attempt this and then move on.

start with (a - b)m + pn = 1?

[glad osprey]

Yep! To be clear, you can use bezout to derive the contradiction directly, you don't need to deduce that p divides a - b (though you can if you want)

[lean sail]

does that look ok?

[glad osprey]

what do you mean by (a-b)u = 1 and then a - b = 1?

[lean sail]

theres a missing step there

a - b must divide 1

the only way for that to hapen is if a - b = 1

[glad osprey]

why?

[lean sail]

are you saying that this is wrong? a - b is a positive integer, such that 0 < a - b < p

thus in order to divide 1, a - b must be 1

[glad osprey]

Yes, it's wrong, I can't see why it's true atleast

[lean sail]

ok, how would you approach this with bezout?

[glad osprey]

like, why would a-b divide 1? Do you have a reason, or are you just guessing?

[lean sail]

no im not guessing

how to explain this...

(a - b) or u must divide 1

[glad osprey]

it would be useful to hear your argument, so we can find the mistake

[lean sail]

like if xy = z, x or y must divide z

[lean sail]

wait, are you not sure if there is a mistake?

[glad osprey]

This is true, but how is it relevant? We have x^(a-b)u = x, but why do we get (a-b)u = 1? We can't just equate the exponents like that

for real numbers we can take the log base x on both sides, but that doesn't work here obviously

[lean sail]

i am getting really confused about this conversation. my assumption is that you know what the mistake is, but are trying to sort of guide me to finding it. is this true or false?

[glad osprey]

You've made an unjustified leap in your proof. That's the mistake. And I'm trying to find out why you think you could make that leap

[lean sail]

oh, the answer to this is simple. it's because i thought we could equate exponents (as you stated)

[glad osprey]

Ok good. It's important to identify when you've made these kinds of mistakes, because it means there's a misunderstanding that needs to be corrected

[lean sail]

when you say "obviously" here, it is not obvious to me, i really dont understand why we can't equate exponents in group theory

[glad osprey]

it's because exponentiation in a group is not injective: for example x^1 = x^(p+1), but p + 1 != 1

think about functions in general: f(x) = f(y) implies x = y is only true if f is injective

(it's pretty much the definition)

Anyways, the proof is really simple: you need to use the fact that x^(a-b) = e. Just start the same way you did and try again

[lean sail]

i honestly have no idea why i started doing that... it does not say in the book that we can do this.

this is like relearning algebra all over again

[lean sail]

im going to give this a try, thanks for the help and patience

then im going to move on to subgroups, order, cycles, whether or not i get this right!

[lean sail]

@glad osprey honestly... what did i do wrong this time? 😩

[glad osprey]

the step from line 3 to 4 is wrong

[lean sail]

oh thats ridiculous... i need to step away from this probably...

[glad osprey]

yeah, might be worth taking a break if you keep making mistakes

[lean sail]

think i got it, if not, i am shutting down discord, turning off my computer, done until tomorrow

[glad osprey]

Nice, good job!

[lean sail]

for probably the 100th time, thank you for your infinite level of patience

[wild jasper]

In the proof of theorem 13.6 in Dummit and Foote, why is there any need to extend $\varphi$ given the definition?

[cloud walrus]

mh_le

[crystal vale]

I think it is true for a,n > 1, n | phi(a^n -1).

Because since a in U(a^n -1) and a^n = 1 in U(a^n -1), so n| phi(a^n -1).

Is it correct?

[chilly ocean]

My copy of Dummit and Foote seems to be numbered differently. Can you post a screenshot?

[chilly ocean]

The definition only defines phi on F and on x. But there are other elements in F[x] like x²+1, and you need to define phi for those as well

And turns out there is an unique way to extend the map from F u {x} to all of F[x] such that the resulting map is a homomorphism

[coral spindle]

They have indeed described it twice

I think they just wanted to describe it in symbols and then in words.

[wild jasper]

ah I see

[barren sierra]

I've argued existence

but my argument is nonconstructive

is there a constructive way to do this?

if you're curious about the non-constructive solution

I mean ok this gives me a constructive solution if I have a particular {p_1, ..., p_m}

[wild jasper]

thanks 🙂

[barren sierra]

but I'm thinking something more along the lines of Lagrange interpolation polynomials

where that's an explicit general construction

if that makes anysense

[south patrol]

Probably works. I think you can do something like dictionary-ordering the p_1,...,p_m [viewing as elements of 2n-dim space] and then weighting the coordinates by very different amounts to separate the points

That definitely works over R and should over C w minor adjustments

[barren sierra]

Yea I tried this weighting thing and couldn't prove anything formal

Probably the weights have to be in terms of the points themselves but then it gets tricky in the multivariate case

[nimble folio]

Silly question

if $W \xrightarrow{f} X \xrightarrow{g} Y \xrightarrow{h} Z$ are maps where $f$ is surjective and $h$ is injective, what does $g$ have to be for $h \circ g \circ f$ to be injective?

[cloud walrus]

clubsoda14

[quiet pelican]

Without conditions on f, you can’t guarantee it

(Further conditions)

[nimble folio]

What conditions?

[quiet pelican]

f needs to be injective

And in that case, g has to be injective on the image of f, and h has to be injective on the image of gf

[nimble folio]

So f being surjective does not help

[quiet pelican]

f being surjective doesn’t help

[nimble folio]

If f were surjective, then it would have to be an isomorphism

ok

[quiet pelican]

Yes

[nimble folio]

Thanks, I don't this I can use this idea to solve my problem

[lean sail]

@glad osprey i found an interesting problem in the next chapter 🙂

[glad osprey]

Interesting, I don't see immediately how to prove that. I see why n <= |g| though

[lean sail]

it seems very close to the one i was doing before, it seems like it will use a lot of the same concepts, but not about primes. im guessing this time i will have more tools under my belt

[glad osprey]

I think I found a proof, it is fairly similar to the proof of how if g^n = e then ord(g) divides n

[south patrol]

There is also a nice proof if you know a bit about cyclic groups

[south patrol]

Sounds good

I assume you did like ||let N = |g|, put N = nm + r for 0 <= r < n, then 1 = g^r * stuff in H so g^r in H and r = 0 or we get a contradiction||

A conceptual way I like is ||note you can assume G = <g> and H = <g^n>. Then this is clear by Lagrange, say.||

[lean sail]

Thank you for hiding, I am not interested in a solution yet

But still like to see what people are talking about here even if I don’t understand it yet 🙂

[glad osprey]

Ah, I was thinking about something like that, but wasn't sure how to prove that ||H = <g^n>. Actually <g^n> is always a subgroup of <g>, so why doesn't this imply that every integer divides ord(g)?||

[glad osprey]

yep, did it that way

[south patrol]

||If n doesn't divide ord(g), then you can find m < n with <g^n> = <g^m> essentially by the same argument lol. For example if G = Z/3Z and we consider the subgroup generated by 2 then it just ends up being all of G||

[crystal vale]

Yes

[glad osprey]

I see, you can even say ||<g^n> = <g^(gcd(n, |g|)> if you want to find the smallest exponent, right? But I don't quite understand how the proof works out. I see why you can say G = <g>, and I think for the same reason H = <g^n>. But <g^n> has order |H| = |g| / gcd(n, |g|), so how can we use |H| divides |g|?||

[south patrol]

Well ||n was assumed minimal!||

[glad osprey]

Ahh, that helps 🙂

[south patrol]

||n = |G|/|H| and |g| = |G| ||

||My point was if you know standard facts then this 'trick' of replacing G and H makes it fall out. I guess it's a helpful reduction||

[glad osprey]

Yeah, it's a nice proof 🙂 I've done one semester of algebra, but I feel like I should be even more familiar with groups. I guess it's just a matter of doing lots of exercises then you get used to seeing things that way

[celest cairn]

Hi 🙂

[crystal vale]

I am stuck at the second part, yes the commutator subgroup contained in H but it doesn't help

[crystal vale]

Ah got it ( MSE )

[rotund aurora]

||there is an element x of order 2 not in H such that G=H cup xH=H cup Hx, and xhxh=hxhx=1 for all h in H. Then for h_1,h_2 in H you have xh_1h_2xh_1h_2=1 --> h_2xh_1h_2=xh_1-->xh_2xh_1h_2=h_1-->h_1h_2=h_2h_1||

[daring burrow]

Is C true because R/(x) is integral domain my answer key saying it is not true

[grave sedge]

(2)(x/2) is in <x> but neither (2) nor (x/2) is

[wild jasper]

Where can I find a reference to the universal property of polynomial rings?

[knotty badger]

It suffices to just investigate ring homomorphisms out of them

[wild jasper]

I just realized how simple it was. Thanks 😊

[knotty badger]

Yeah np, I think often you don’t have to memorise these things but can rederive them quickly

[wraith nexus]

|| isn't this false for G=S3, no element x order 2 with xhxh=hxhx=1 all h in H||

[rotund aurora]

hx is not in H therefore has order 2, this is by hypothesis

the same with xh

[wraith nexus]

o

[rotund aurora]

I think the case of S_3 is fine, (123)(12)=(13) which has order 2, etc.

like H is the subgroup generated by (123) and the rest of the elements are the transpositions (12), (23), (31)

[wraith nexus]

yeah it is I'm dumb

[glad osprey]

Is an F-algebra the same as an algebra over a field?

[chilly ocean]

It probably means an algebra over F. So if F is a field, probably

[glad osprey]

Here's how our lecture notes define F-algebra: Let F be a field. A ring R is an F-algebra if

• (R, +) is a vector space over F
• for all a in F and x, y in R: a(xy) = (ax)y = x(ay)

Wikipedia defines an algebra over a field as a vector space with a bilinear product. So the bilinear product in this case is multiplication in R?

[south patrol]

I mean an algebra over a field F is an F-algebra like the two are synonyms. The issue is more that there are distinct definitions of an algebra

They are similar, except Wikipedia's definition is more general in that rings are usually assumed to have identity whilst Wikipedia's definition doesn't imply there is an identity, and algebras in Wikipedia's sense needn't be associative either

[south patrol]

Everthing in the first sense is a algebra in the second sense yes by using the multiplication

Condition 2 in definition 1 implies multiplication in R is F-bilinear

[glad osprey]

I see. And this wiki page describes a generalization of algebras over a field or something completely different? https://en.wikipedia.org/wiki/F-algebra

[south patrol]

That is different yeah

[glad osprey]

okay

Aluffi seems to call it K-algebras, but I can't seem to find where he defines it 🙁

[south patrol]

I mean like

K or F is just the name of the field lol

[chilly ocean]

Replace the K by whatever field you are using

[south patrol]

C is an R-algebra

Q(i) is a Q-algebra

Etc

[glad osprey]

Right, makes sense. Do you have a favorite book or source you like to look up definitions in, if wikipedia isn't good enough? I've used Fraleigh a lot, but it's often not comprehensive enough

[dull ginkgo]

Let’s say we have a diagonal n x m matrix D of R^N over integral domain R, would the kernel be the basis vectors where it is 0 along the diagonal or 0 entirely, and the image basis be just the scaled basis vectors corresponding to the diagonal?

[chilly ocean]

Good thing you changed it to integral domain

Isn't it just: they have to be 0 in the coordinates where the diagonal isn't 0? (and for the coordinates where the diagonal is 0 or doesn't exist, they can be anything)

[south patrol]

Well it's more that conventions differ here tbh

[south patrol]

Yes if [for convenience] you assume that De_1 = ... = De_k = 0 and De_{k+1},...,De_n !=0 then it dies on the subspace <e_1,...,e_k> and is an isomorphism when restricted to a map <e_{k+1},...,e_{n}> -> <e_{k+1},...,e_{n}>

[glad osprey]

Yep. What's interesting is that our textbook doesn't require integral domains to be commutative, which I think is very unconventional

[south patrol]

That is, have never heard of anyone doing that

[lapis latch]

that's what the term domain is for...

[dull ginkgo]

Now let’s say our ring wasn’t integral, would the torsion submodule of the image be generated by the basis vectors with a zero divisor on the diagonal

[swift prawn]

this is from Ty Lams book on noncommutative rings

I was just wondering what it meant here by additive index of [R_3: R_2] and why it is 2

additional context: R_1 is the rational quaternions, R_2 is the integer quaternions (so of the form a + bi + cj + dk; a,b,c,d \in Z)

I assume here additive index means their index as abelian groups. I'm just wondering why it is 2

nevermind

I think it's just first iso theorem

[rocky cloak]

That is what it means.

Like they said R3 = Z^4 with basis as given, hence R2 embeds as 2ZxZ^3

And so the quotient is Z/2Z

[swift prawn]

yep

that makes sense thank you

[lethal cipher]

So I know if R is a commutative ring with char F=p (p prime) that for any x,y in R, (x+y)^p=x^p+y^p

Is this also true if R is not commutative?

Hmm, no. This already fails when charF=2

[chilly ocean]

Yeah, it can fail

I wonder if it can ever be true for a noncommutative ring, though

It can't be true in char 2, because (x+y)² = x²+y²+xy+yx so you'd need xy=yx for all x,y

[lethal cipher]

Yea. Okay, weird question: is there some name for elements that commute with everything in the ring?

Because if you have (x+y)^p where y commutes with everything, then it holds

[chilly ocean]

The set of all elements that commute with everything is called the center of the ring

[lethal cipher]

Okay, so it's also called center like it is for groups

[chilly ocean]

Yeah

[lethal cipher]

I wasn't sure. Thanks

[chilly ocean]

More generally you could use any x,y such that xy=yx

[grim hawk]

If I am trying to form an isomorphism between a group of functions and a product of two cyclic group can I simply say an isomorphism maps one of the generating functions to the generator of the first cyclic group and then map the second generating function to the generator of the second cyclic group?

[rocky cloak]

I think if you take F the field with 3 elements and consider the alternating algebra of 2d space, then it's not commutative, but has (x+y)^3 = x^3 + y^3

[fleet mortar]

if i have a non prime element in a ring (say j), does that mean j=ab where neither a nor b are units?

[chilly ocean]

No, that is the definition of reducible, not non-prime

Your claim is true for Unique Factorization Domains

[fleet mortar]

ok thank you

is there a clean definition for non-prime

[chilly ocean]

An element is prime if whenever it divides a product then it divides one of the terms of the product (and also if it isn't an unit. this is automatic if you allow for empty products)

So an element x is nonprime if it is either an unit or if there are some a,b such that x divides ab but divides neither a nor b

[fleet mortar]

thanks

[chilly ocean]

For a counterexample, consider the ring Q+xR[x]. Here x is irreducible but it divides (sqrt(2)x)^2 despite not dividing sqrt(2)x

[arctic trail]

Man, my mind is fried from studying too much

Can anyone help me with this:

A finite group G is said to be primitive if there exists a maximal subgroup H, such that the core of H (that is the maximal normal subgroup of G contained in H) is 1.
I know that a primitive solvable group has a normal minimal subgroup (which is abelian).
Does anyone know how to prove that this minimal subgroup is in fact unique (i.e. the minimum), and that it's the entire centralizer of N?

It's easy to see that if N,N' are two distinct minimal subgroups of a primitive solvable group, then N intersection N' = 1 and therefore given n in N and m in N' we have: m^(-1)nmn^(-1) in the intersection and thus mn = nm and so N' is in the centralizer of N.

[glad osprey]

Let D be a division ring and $R = M_2(D) = \begin{pmatrix}D & D \ D & D \end{pmatrix}$. Then $I = \begin{pmatrix}D & 0 \ D & 0\end{pmatrix}$ and $J = \begin{pmatrix}0 & D \ 0 & D\end{pmatrix}$ are both left ideals of R. Isn't this true for any ring D?

[cloud walrus]

sheddow

[rocky cloak]

It is yes

[glad osprey]

thanks 👍 I don't know why our lecture notes have these arbitrary restrictions in every statement... maybe to trick me into generalizing them?

[chilly radish]

It might be that the general case is not relevant to whatever you're learning, but I agree it is annoying to impose conditions that are unnecessary

On the other hand, this might be in the context of a simple ring (no 2-sided ideals) that still has one-sided ideals, in which case this is a good example

[icy totem]

H is a subgroup of Dn not only made of rotations iff it is generated by a rotation (the identity count as a rotation) and a simmetry.

Is this true?

[chilly ocean]

By symmetry you mean a reflection?

[icy totem]

Yes sorry

[dull ginkgo]

D_n is generated by a "rotation" element (of order n) and a symmetry involute element

So consider what generates your subgroups

[icy totem]

I tried to prove it showing that <r^d, sr^h>=<r^d> *<sr^h> for every d, h

Where r is a rotation of order n

[dull ginkgo]

Very confused what your use of symbols is here

Oh direct prod

Not semidirect i take it

[icy totem]

No, just product

The set of elements that is the product of an element of the former subgroup and one of the latter

[dull ginkgo]

Consider the intersection of your cyclic subgroup r^d and generated group sr^b

Try to describe the generated subgroup of sr^b

[icy totem]

Its just {id, sr^b}

[dull ginkgo]

… no

[icy totem]

🤨

[dull ginkgo]

What is b

Miscommunication maybe

Any b?

what is (sr^b)^2

[icy totem]

Any b different from multiples of n

sr^b is a reflection, no?

[dull ginkgo]

You said for any b

[icy totem]

For b=0 still I have s that is a reflection

So it actually works for any b id say

[dull ginkgo]

How do you define D_n

So i can work in that context

Presentation? Geometric?

[icy totem]

Presetation is better

As {id, r, ... r^(n-1), s, sr,...}

[dull ginkgo]

So we have the presentation:
r^n = s^2 = (sr)^2 = e

[icy totem]

Yes

[dull ginkgo]

Actually easier route: conjugation!

(sr)^2 = srsr = e, thus srs = srs^-1 = r^(n-1)

so consider the cyclic group generated by sr^d

(sr^d)^2 = (sr^d s )r^d = (srs)^d r^d = r^d(n-1) r^d = r^dn = e

Nvm you are right, I messed up the first time

[icy totem]

Ah okay lol

[dull ginkgo]

Sorry man

[icy totem]

I was rereading my lecture notes to see where i fuckedup

Np

[dull ginkgo]

First time i did it i made an error :c

[arctic trail]

if D_n geometric

if D_2n presentation

[dull ginkgo]

So let me get this straight real quick:

When I have a n x m matrix M, let M* be the induced map from R^m to R^n

Lets say we have such an M where we have an invertible n x n matrix A and m x m matrix B such that AMB is diagonal n x m matrix D.

The image of D* has the nonzero diagonal column vectors of D as a basis, and the kernel of D* has a basis of the respective standard basis vectors corresponding to the zero columns of D.

Lets say I have already computed U = AM and V = MB (in the process of row reduction).

Then is there a way that I can just pull the basis vectors of the kernel and image by considering the corresponding column vectors from these two matricies U and V?

It's frying my brain because it's so much going on at once lol

[old hollow]

Show that if every element in a group $G$ has order 1 or 2, then $G$ is a vector space over $\mathbb{F}_2$.

[cloud walrus]

a.b.s._.0.

[old hollow]

I can see why G is abelian, so that makes sense

but I don't know how to get the field action, or why it relates to the order being 1 or 2

[lapis latch]

In the same way that an abelian group is a Z-module, your field action is nx(n in F_2, x in G, the group written in additive notation)

You’ll see that’s well defined since they’re all order 1 or 2 and abelian ensures it follows the vector space axioms

[old hollow]

ok interesting

[old hollow]

if every element in G has order 1, 2, 3 then is it a vector space over F3

I proved your statement but I went case-by-case which was a bit ugly

i wonder if this generalizes

[lapis latch]

It would need to be abelian and no element can have order 2, because if you have 2x=0 then if the vector space axioms hold that should imply x=0 since 2 is invertible in F_3

Basically it’s only the groups which are some direct sum of F_p

[old hollow]

oh you mean 2x=0 implies 2(2x) = 0 implies x = 0

hmm

somehow this is not easy to think about for me

[lapis latch]

The big generalization is that if you have a ring R and an R-module M with annihilator I then M is an R/I module, so when R=Z you get a vector space when I is some prime ideal I.e Z/pZ

[dull ginkgo]

To be specific, here’s the chain of ideas:

• Every Abelian group is a Z-module
• If M is an R-module, and Ann(x) is the annihilating ideal of M (elements r of R such that rx = 0 for all x in M), then for any subideal J of Ann(x), M is naturally a R/J-module. I am suppressing the sidedness here, but it's uniform
• For commutative ring R and maximal ideal M, R/M is a field ( if x is not in M, then M + xR = R, so there exists a y in R: xy + M = 1. Quotienting by M gives an inverse)
• The maximal ideals of Z take the form (p) for some prime p
• A vector space is a module over a field

[long obsidian]

If S is a multiplicative set in a ring A what are the ideals of the localization S^-1 A? Are they of the form S^{-1}J for ideals J containing S ?

[dull ginkgo]

I never thought about this in the context when A isn't integral

horror

[dull ginkgo]

I think so? the localization forces the elements of S to become units, so if an ideal intersects S, the localization i'd think would make it become S^-1 A

otherwise yeah, I'd assume S^-1 J since otherwise we wouldn't have closure

this is just my first guess, i haven't worked it out

trying to reformulate like, all of linear alg in the module theoretic framework I have mentally

has been proving to be the biggest struggle I've faced thus far with AA

like viewing them as Linear Maps between powers of a ring, basises being bijective maps from R^|B|

Matrix multiplication needing to be reversed to compose the actual maps

brain hurty

You also have to deal with opposite rings

[old hollow]

so in this case, since G has all elements of order 1 or 2, then we can say the action of Z on G is "degenerate" in the sense that the ring homomorphism Z --> End(G) is not injective, because ng = 0 for any n in 2Z. Therefore quotienting by the degenerate part 2Z gives a faithful action Z/2Z --> End(G), and this makes G a module over the field Z/2Z, and hence a vector space over Z/2Z

[dull ginkgo]

it's not quite degenerate, just not faithful

[old hollow]

oh is degenerate an actual term

I just meant not faithful

like has collisions in some sense

[dull ginkgo]

That quotienting by the annihilator is basically quotienting out the redundancy to make it more faithful

[old hollow]

yes exactly

[dull ginkgo]

Then yes :3

[old hollow]

ok lit

thank you sm

[dull ginkgo]

It's just a lot of random simple arguments being smashed together lol

[old hollow]

yeah

the field part at the end is the most irrelevant

I'm interested in the concept of quotienting by the redundant part

to get a faithful action

its cool

[dull ginkgo]

i mean it's like, parallel to group actions lol

[old hollow]

yeah

Interesting question:

[dull ginkgo]

G acting on S is a group map from G to Sym(S) (technically Aut_set(S))
R giving M a left R-mod structure is a map from R to End(M)

So like

You could say we have a "monoid action" of M on End_set(S) if you wanted i suppose

[dull ginkgo]

though note right modules have a map from R^op to End(M)

faithful means the respective map is injective, so you can quotient out the kernels respectively to make it faithful

[old hollow]

Let $G$ be all the roots of unity in $\mathbb{C}$, that is, all complex numbers $x$ such that $x^n = 1$ for some $n \in \mathbb{N}$. Prove that $$G = G_2 \times G_3 \times G_5 \times \cdots,$$ where $G_p$ are the roots of order $p^n$ for some $n \in \mathbb{N}$.

[cloud walrus]

a.b.s._.0.

[dull ginkgo]

chinese remainder theorem

i think

[old hollow]

wtf

[dull ginkgo]

It'd be a subgroup of the circular group

consisting of torsion elements

pretty sure you can use CRT on that lol

[old hollow]

circular group?

hmm

yes thats true

ok

i'm thinking more along the lines of "the data of an element of G consists of (n, k) where n is the exponent and k iterates over the nth roots of unity"

[dull ginkgo]

It'd be the torsion subgroup of it actually

[old hollow]

but this is probably complicated

[dull ginkgo]

fuck I think there is a related group and I can't find the name for the life of me

brb

[old hollow]

I want to decompose the exponent n = p1^r1 ... pk^rk

and then obviously there should be some correspondence between the p_i's and the G_{p_i}'s

[dull ginkgo]

https://en.wikipedia.org/wiki/Prüfer_group FOUND IT

:3

[old hollow]

wtf

yeah p-groups

so each G_p is a p-group

oh shit this is literally the group

[dull ginkgo]

roots of unity lie as a subgroup of the unit circle under multiplication obviously, and being a root of unity is literally being torsion verbatim. SOooooOOoOoOO

Consider the inverse of the isomorphism e^(2pi x i) of R/Z onto the unit circle right

Imagine some element x + Z in R/Z is torsion, i.e n(x + Z) = nx + Z = Z, then nx must be in Z, i.e x is a rational

then use prime decomposition

:3

[dull ginkgo]

not quite

[crystal vale]

I can count the left zero divisor as a zero divisor, right?

[onyx pecan]

Can someone help with part b of this problem? I've been banging my head against a chalk board for the better park of an hour playing with relationships between (gh)^n, (gh)^{n+1}, and (gh)^{n+2} and have gotten no where

[crystal vale]

This is tricky

Can you prove that hg^n = g^(n)h ?

By using that you can also prove that hg^(n+1) = g^(n+1)h

[onyx pecan]

Wait how do you jump from the above to this conclusion?

[vagrant zinc]

This kind of exercise is very frustrating because it is so trivial and difficult to write, I think you know the answer :D

[onyx pecan]

wait i dont is the issue 😢

i think i must be overlooking something extremely trivial in all of my scratch work

but idk

[crystal vale]

Use (gh)^(n+2) = g^(n+2)h^(n+2) = ghghg^nh^n

[vagrant zinc]

Ese ejercicio lo he mirado un monton de veces y siempre se me olvida como es la demostracion, por que literal sale en 3 paso al ser tan trivial
Suponga que (gh)^{n}=(hg)^{n} esto puede ayudar

[long obsidian]

If I have some nested ideals I \subset J in a ring R. How can I show there is a surj ring hom R/I\to R/J?

[mighty kiln]

It should be easy once you can write down the ring hom

More generally, if you have ideals I ⊂ R, J ⊂ S, then a ring map f: R → S such that f(I) ⊂ J induces a ring map R/I → S/J

[long obsidian]

Oh I see thank you and in particular this can so that if J an ideal of R and R is a subring of S then there is a ring hom h:R/J \to S/JS where JS denotes the ideal generated by J in S. Since the inclusion map j:R\to S has that property you mentioned j(J) \subset JS. Thanks!

[surreal dagger]

Atleast for cute examples, If I want to determine the prime spectrum of a Ring is the way to go:

1. Know Spec of the basic building block Rings and Polynomial Rings.
2. Then when im inspecting some quotient the quotient is either isomorphic to one of the basic Rings or im arguing by the correspondence theorem

Is this the general strategy?

[arctic trail]

love the Prufer group

[winged void]

hello i have small question

\textbf{Definition 1.9.} Let ( R ) be a ring. An element ( a \in R ) is called a \textit{unit} (or invertible) if there exists a ( b \in R ) such that ( ab = ba = 1 ). (Note the peculiar language usage: the identity element is indeed a unit, but not vice versa.) The set of units of ( R ) is denoted by ( R^* ) and is called the \textit{unit group} of ( R ) (since it is indeed a group, see Theorem 1.12). One also encounters the notation ( U(R) ) (unit (English) = unit).

An element ( a \in R ) is called a \textit{left unit} if there exists a ( b \in R ) such that ( ab = 1 ), and a \textit{right unit} if there exists a ( c \in R ) such that ( ca = 1 ).

[cloud walrus]

Mootje

[winged void]

R* is not a ring or a subring

right

but it is a group

right

?

[chilly ocean]

Yes

Unless R is trivial, 0 won't belong to it so it is not a subring

[winged void]

i see thanks a lot

and one more question a ring can be seen as two groups this is how i looked at i t

like a group of addition

and other group of multiplication

at the same time

[winged void]

thx

[chilly ocean]

From a ring you can obtain its additive group and its group of units. You lose information by doing this. In particular, the group of units will always have less elements than the original ring (unless it is the trivial ring)

[winged void]

i see but i mean in general a ring

is two groups right

like one of addition

and one of multiplication

liek to images

[winged void]

this is what i mean

[chilly ocean]

It's more than that. You also need information about how the nonunits multiply

This does make me wonder if given a ring (R,+,*) if you can recover it from (R,+) and from (R*,*)

[winged void]

ah i see makes sense

[rocky cloak]

R* meaning units, or just the monoid of R under multiplication?

[chilly ocean]

Units

[rocky cloak]

Then no, something like Z[x] and Z[x1, x2] have the same units

[chilly ocean]

Yeah, that works

[glad osprey]

You can think a ring as an abelian group (addition) and a monoid (multiplication), where the multiplication distributes over addition (a monoid is a group without inverses)

If an element in a ring has two or more right inverses, then it has no left inverse. What about if it has exactly one right inverse? Is it guaranteed to have a left inverse then?

[rocky cloak]

They actually have the same multiplicative monoid as well

[delicate orchid]

holy moly!

[rocky cloak]

Say xy = 1, then x(yx - 1) = xyx - x = x-x = 0,
so x( y + yx - 1) = 1
if the inverse is unique that means y = y + yx - 1 which means yx = 1

[tardy hedge]

OH HELLL NAHH!!

[chilly ocean]

Is it the free monoid on irreducible elements + 0?

[glad osprey]

Nice, thanks 🙏

[chilly ocean]

I'm never thought about it but I guess the mult monoid of an UFD should be free monoid on irreds + 0

[rocky cloak]

Times the group of units, but yes

[chilly ocean]

Right of course

[long obsidian]

I think k[x,y]/<x,y>=k is this true?

[quiet pelican]

Yes

Use the first isomorphism theorem on f: a + sum_{i, j} a_ij x^i y^j -> a

[long obsidian]

Thanks I see how that follows from first iso

[wooden fulcrum]

Hello

How much linear algebra is needed/best to know before starting to learn about modules?

I've a class coming up that will spend probably like 2/3rds of the semester on modules, my uni's information system is down rn but from what I remember it's supposed to be covering flat, projective, injective modules, resolutions and idk what else

I've had kind of an all-over-the-place kinda introduction to math, I've had abstract algebra but linear I've only kinda seen the basics, no spectral theorems or stuff pertaining to operators, multilinear, tensor products, affine spaces, inner product spaces etc. none of that, will I be missing that in this course?

I'm going through LADR but I don't think I can feasibly cover all of it including the exercises in the time left before the module part of the course starts

[void cosmos]

ig for the intuition but like

u can do exercises and like get the definitions of these kinds of modules

without needing like deep linear algebra

in a sense modules are slightly "lower-level" than vector spaces

by lower level i mean it in a computer-sciency sense

[dull ginkgo]

hell

[rocky cloak]

The reason computers where invented

[south patrol]

Ye i don't see much value in doing exercises like these tbh

[south patrol]

Wdym

[delicate bloom]

I'm guessing he means cause all fields are rings

[south patrol]

Well that was why I was asking aha

[delicate bloom]

lol

[void cosmos]

Like how like doing a high lvl programming language is like

Pushing unto registers into assembly in a way

High level jordan canonical forms are like

PIDS and moduels

The fundmental theorem i mean

[south patrol]

I guess I don't really know what you mean though I do know about high vs low level programming

[void cosmos]

like how linear algebra hides some of the technical lower lvl details

U dont see the PID structure theorem when doing jordan form in a wah

Wah

Way

[arctic trail]

is it related?

[dull ginkgo]

I thought that’s smith

[arctic trail]

I don't see any resemblance a.a

[gusty moon]

hi @barren sierra

[barren sierra]

what are u still confused on

[gusty moon]

why f_i is in some I_j

[barren sierra]

Suppose not

[gusty moon]

like I totally understand that it's in I

[barren sierra]

but I is the union of all I_j

[gusty moon]

but what if we chose specific ideals in the chain that doesn't include it

[barren sierra]

I mean how could something be in I but not be in any I_{j_i}

[gusty moon]

oh

that is the reason then

[barren sierra]

the point is that there is some I_{j_i} such that f_i is in I_{j_i}

[gusty moon]

yes

[barren sierra]

we can't choose at random

[gusty moon]

this helped

I understand now

[barren sierra]

but we know some j_i exists

cool

[gusty moon]

thank you so much @barren sierra

[barren sierra]

ofc

[gusty moon]

@barren sierra hi

I do have another question

I don't understand how f = f + 0 satisfies condition (ii)

[barren sierra]

Ok so the remainder of division of f by G is 0

wait no

[gusty moon]

yes but we don't know that from the start

[barren sierra]

other direction

[gusty moon]

yea

[barren sierra]

ok so f in in I

[gusty moon]

the condition is there is a g in I such that f = g + r

yes

[barren sierra]

so we want to show that r = 0 is the r such that f = f + r = f + 0 satisfies conditions (i) and (ii)

you say you understand how it satisfies (i) so lets look at (ii)

[gusty moon]

yes

[barren sierra]

I mean like

take g = f (that's what they're doing by writing f = f + 0)

f is in I

so r = 0 is our desired remainder no?

[long swan]

thats what i thought too

[barren sierra]

kinda just writes itself

[gusty moon]

oh I was still thinking f was in k[x_1,...,x_n]

my bad

[barren sierra]

all g

[gusty moon]

😐

thanks!

[barren sierra]

we're assuming f is in I of course lol

yea np

[gusty moon]

yes

[long swan]

what book is this btw @gusty moon

[barren sierra]

Cox, Little, O'Shea - Ideal, Varieties, and Algorithms

[long swan]

ah

[gusty moon]

@barren sierra actually recommended it when I asked for books on alg geom

it's pretty good

but I think im only able to read 1 section per day

[barren sierra]

it's a nice elementary introduction to algebraic geometry

[barren sierra]

that's a much faster pace than me lol

especially if you're doing the exercises

[gusty moon]

oh im not haha

[barren sierra]

bruh

then you aren't really learning the math

you need to be doing the exercises

[gusty moon]

hmmm

[barren sierra]

math is not a spectator sport

[gusty moon]

they look difficult

[barren sierra]

the only way to learn is to do

[gusty moon]

hahaha

[barren sierra]

yea that's the point

they poke and prod at your understanding

slow down

[long swan]

spamakin recommended me sagan's book on the symmetric group a long time ago for rep theory and algebraic combinatorics

[barren sierra]

go back

do exercises

[gusty moon]

hmmm

[long swan]

idk if you remember lmao

[barren sierra]

yea it's a good book

I don't remember lol

[long swan]

yeah mightve been my only interaction with you

[gusty moon]

lool

[gusty moon]

ok I will

[barren sierra]

did you end up looking at any of it?

[long swan]

so far not much past the first chapter i dont think, but i'm gonna go back to it for sure

[barren sierra]

nice nice

[long swan]

i wish i was better at deciding what to go through and splitting my time

maybe it just comes with experience

[barren sierra]

I feel that

I'm splitting my time between too much math right now

[gusty moon]

I wish I was better at understanding literally any math paper

I only understand textbooks

but papers are like greek

[gusty moon]

which areas?

[long swan]

what ive settled on is like roughly following what im taking in school and then out of the topics not explicitly listed finding a way to relate them to the subject of the courses im taking and its worked out pretty well

[long swan]

yea its pretty tough for most areas i think

[gusty moon]

ok

[long swan]

well obviously its tough in general but i meant from a specialization standpoint as well

[barren sierra]

various things in algebraic geometry

[dull ginkgo]

In a way though sometimes it’s nice to do a separation of church and state thing with math vs your schooling

College gives you that luxury more

I tend to use math as my way to kind of decompress and keep away from tech and in a sense, more mindlessly work on stuff closer to logic than trying to retain constant rules or constantly trying to adhere to guidelines

It’s hard to explain to others without sounding pretentious, but it is my way of coping i guess

[long swan]

ah i guess its pretty different if youre not majoring in math

[dull ginkgo]

Yeah, i intentionally chose a route that uses some math but isn’t directly pure (engineering), so often times what i work on is more of like fun puzzle side projects that aren’t constant reading

[long swan]

yeah i see wym

[dull ginkgo]

Which is why I enjoy exercises and tasking myself with difficult problems or trying stuff on my own especially in algebra

Tldr I HAVE FREE WILL

[long swan]

tbh i have a pretty PDEs/measure theory heavy courseload this semester so ig i'm doing a fair bit of applied math as well

cause naturally a lot of it is physics

i like it a lot tho

[dull ginkgo]

That’s how i can justify my two week long excursion into Jacobson’s simple module and finite dimensional division algebra shit to flush out Galois theory

It has the same prereqs as Galois theory but is just less relevant so that’s why it’s held off til later lol, it’s just kinda neat

[long swan]

https://en.wikipedia.org/wiki/Brauer_group this kind of stuff?

i havent heard of the book actually ill check it out at some point

[dull ginkgo]

Jacobson density theorem, Jacobson Bourbaki endomorphism-commutator correspondence for finite-dimensional division rings extensions, crossed products of rings, and some Morita shenanigans

[long swan]

ah

yeah idrk shit about it yet tbh although some is familiar

[dull ginkgo]

None of it is particularly difficult frankly, just a bit interesting

[gusty moon]

ah ok

[long swan]

yea i see what you mean about same prereqs to galois theory

[dull ginkgo]

It’s like an intermediate correspondence between the standard one of (Galois) Field extensions and the “automorphism-stabilizer” subgroups

[long swan]

Yeah

nice

[dull ginkgo]

Tldr division rings of endomorphisms instead take center stage which THEN correspond to the groups in the case of fields

[crystal vale]

If A is left ideal and B is a non-empty subset of R, then AB is left ideal right?

AB is set of all finite sum of a_ib_i, a_i in A and b_i in B.

[gusty moon]

right

[crystal vale]

Okay thank you

Let R be a ring (without unity) and P be a prime ideal then R/P has no zero divisor, right?

In general we know that if R is commutative and R has unity then R/P is integral domain but if I don't take this condition, then ?

[crystal vale]

Converse is not true, right? Because P can be R so R/P has no non-zero divisor

[next obsidian]

No this isn’t true

Wait nvm sorry

This is weird I’ve only ever seen this where B is also an ideal

But it isn’t necessary wtf

[surreal dagger]

If I want to calculate A := R[X,Y]/(Y-X^2,X+Y) Y=-X holds in A so theyre basically polynomials in X such that -X-X^2 holds i.e R[X]/(-X-X^2)? Is this correct? And the easiest formal way to show this is isomorphic would be to give an isomorphism from R[X,Y]->R[X]/(-X-X^2) with the kernel (Y-X^2,X+Y)?
R is just Real numbers

[crystal vale]

So it is true, right?

[rotund aurora]

Why don't you try proving it?

[crystal vale]

Since A is left ideal so for any r, ra_i in A so ra_ib_i in A so also finite sum r(a_1b_1 +..+ra_nb_n ) in A

[tulip glacier]

Given a ring K, we may define the polynomial ring (in one variable) over K to be the coproduct between K and the free ring with one generator.

How can we define, categorically, the root of a polynomial ?

[rotund aurora]

The map taking x-->x+1 is an automorphism of R[x] which doesn't preserve roots

[tulip glacier]

Uhh, I'm not sure I understand. 😦

It indeed doesn't preserve the roots of the polynomial. However, how does this prevent us from giving it a "category-theoretic" definition ?

[rotund aurora]

what is the coproduct of rings?

[tulip glacier]

Uhh, given a category C, we may take the coproduct A + B (if it exists) of two objects A and B of it. It is given by the universal property that any pair of maps from A to X and B to X correspond to a unique map from A+B to X (and vice versa). Eg. the cooroduct in Set is the disjoint union.

A coproduct of rings is simply a coproduct in the category Ring of rings and ring homomorphisms between them. Basically, it is the ring you obtain if you "freely adjoin" the two rings together (that is, when you don't establish any identities between the elements of the two rings other than those required by the algebraic structure of the ring).

[rotund aurora]

Yes I know what a coproduct is, I was asking about the particular construction in Ring

[tulip glacier]

Right. It is the ring that you obtain when you freely adjoin the two rings without establishing any fruther identities (other than those which already existed in the two separate rings). This is why the coproduct between a ring and the free ring on one element is its polynomial ring.

[rotund aurora]

the free ring on one element is just Z, right?

uhm

well yeah should be

[tulip glacier]

Uhh

Okay I see my mistake now

[chilly ocean]

Z[x]

Z is the free ring on 0 elements

[rotund aurora]

ok

but isn't R[x] the coproduct with R and Z?

[tulip glacier]

That would mean that the variable of the polynomial can also have negative powers...

[rotund aurora]

oh

yeah so the coproduct is just tensor product over Z?

[tulip glacier]

Okay I am actually not so sure if the polynomial ring is the coproduct with the free ring on one generator. Let me check again.

[tulip glacier]

Haven't thought about it. Hmm

[rotund aurora]

If it's the tensor product over Z it surely is

If you think that Z[x] is categorical and taking tensor products is categorical, why don't you think the usual definition of "root" is categorical? Evaluating at x is the same as giving an R-linear homomorphism R[x]-->R

[knotty badger]

Depends if you’re working in crings or not

[tulip glacier]

I am not sure I understand that evaluating at x part. Shouldn't it correspond to morphisms from the free ring on one element to our ring, somehow ?

[tulip glacier]

Hm, how so ?

[rotund aurora]

I guess for the R linear part you could lift a homomorphism Z[x]-->R and the identity R-->R to R otimes_Z Z[x] by the universal property of coproduct

[knotty badger]

If you’re working in non-commutative rings, evaluation isn’t necessarily a homomorphism

You can still define the polynomial ring of course

What ends up happening is that ring homs R[X] -> S naturally correspond to a ring hom R -> S, and a choice of element s of S, which commutes with everything in the image of the ring hom

[rotund aurora]

yeah that's a good point. I guess this adds more difficulty for a "categorical" definition

[knotty badger]

I mean what I gave is still a universal property

Since it’s an alternative description for Hom(R[X], -)

[rotund aurora]

I mean this is pretty uninteresting imo

[knotty badger]

That’s kinda subjective

I haven’t worked much with non-commutative rings but I think their polynomial rings are still sometimes useful

E.g. you can use them to give a very quick proof of Cayley-Hamilton

[rotund aurora]

In group cohomology you consider Z[G] where G is any group, so a non-commutative ring

[tulip glacier]

I think I see.

But it's just the evaluation of a polynomial that I'm not sure I 100% understood.

[knotty badger]

You can totally define polynomial evaluation

It just won’t be a ring homomorphism in general

[rotund aurora]

If you allow this kind of stuff, isn't all of algebra (ignore foundations to avoid subtler issues maybe) reformulable in terms of category theory?

[tulip glacier]

It is

[knotty badger]

I guess it depends what you mean by that

It’s not so much whether you can, it’s whether you should

You have the option to

You ain’t forced to ofc

[rotund aurora]

well to me the adjective categorical has more content than just writing something with words of category theory

[knotty badger]

Well idk universal properties seem pretty categorical

[prisma ibex]

water is wet also

[knotty badger]

I don’t exactly see how that’s relevant

[rotund aurora]

But is Ring categorical?

[tulip glacier]

I mean, Ring is a category, so yes.

[knotty badger]

I’m not especially interested in arguing semantics here

It’s more just whether it ends up being useful for what you want to do

[rotund aurora]

mmh doesn't follow. This is like saying that since 123456789 is an element of a group, it is group theory to ask whether it's prime or not

[tulip glacier]

And rings themselves can be defined categorically too - either as monoids in Ab (with the tensor product) either as their specific Lawvere theories.

[prisma ibex]

the main issue is that in the category of rings the polynomial rings aren't free objects anymore, they are only free objects in the category of commutative rings

[knotty badger]

Depends what you mean by free object

[prisma ibex]

there is only one sensible thing that this means here

[knotty badger]

I mean, Riehl talks about free objects in CTiC, essentially as a synonym for objects satisfying a universal property for maps out of them

[rotund aurora]

Oh I'm not trying to argue. I'm just saying that if the definition of categorical is too loose, then pretty much everything can reformulated in a categorical way, but this ends up being trivial imo

[knotty badger]

I don’t think it’s as settled as you claim

[prisma ibex]

so this screws up the universal property you are expecting for this evaluation morphism for polynomial rings; you still get the right universal property in Ring for the evaluation R[x]->R at an element x of R iff x is central

[knotty badger]

And/or objects initial in some appropriate category

[knotty badger]

Yeah I get this

I think just because you can reformulate something categorically, doesn’t mean you should

It just about whether it ends up being useful for what you want

[tulip glacier]

Yeah, I was interested in this "roots of a polynomial" thing hoping it would help me understand this at the most fundamental level etc.

[prisma ibex]

in the commutative case it's still worth thinking about, since this construction has nice properties. It breaks in the non-commutative case, but so do lots of other things that work in the commutative case

[tulip glacier]

Because a lot of very abstract mathematics (eg. algebraic geometry) is concerned with this, but so far I don't see any use for this "root of a polynomial" thing other than high-school-level polynomial functions

Like I try to get an intuition in the most abstract sense for this concept

[prisma ibex]

well in the commutative case this construction of evaluation and localization is absolutely central

[rotund aurora]

Maybe my example above wasn't accurate. But can't sets be defined categorically too, and therefore set theory is categorical (maybe there are some subtle issues of which I'm ignorant of).

[prisma ibex]

one of the main features of schemes is that you can play this game with evaluations and the ring that the evaluation lands in might vary along with the points you're evaluating at

it's still coming from the same universal construction

[tulip glacier]

I mean, yeah. By categorical I meant mostly things that cam be expressed solely in terms of morphisms and equality between them (commutative diagrams), and also things like universal propertirs. But yes, it was indeed an ambiguous term.

[knotty badger]

I guess for me I don’t concern myself with whether something is or is not categorical as a binary choice - it’s more about in what way, or to what extent, it’s categorical

[prisma ibex]

there are different levels of abstract nonsense you can bring to a construction and it's not always useful to be maximally abstract and categorical with these things

sometimes it does help, but usually you have some good idea of what you're trying to abstract to solve a particular problem, rather than just screwing around trying to discover construction (at least this is my experience)

[rotund aurora]

If you can define rings categorically you can define sets (at least with >=2 elements). But how would you define a set with 2/1/0 elements?

[prisma ibex]

0 and 1 is trivial lol

but yes 2 is already subtle to think about

[knotty badger]

Yes yes of course

[prisma ibex]

should being a set with 2 elements be a property or an additional structure on the set

[knotty badger]

At least for me, abstraction isn’t a goal unto itself, nor is it universally better

[rotund aurora]

how? I just said >=2 because rings have >=2 elements

[knotty badger]

In fact I often tend to prefer things when stated more concretely

[chilly ocean]

There's the trivial ring with 1 element

[tulip glacier]

You can define any set categorically.

Take the category with just one object and just one element. Then, Set is isomorphic (as a category) to the free cocompletion of this category - that is, every set is a colimit of this unique object.

The empty set is the colimit over the empty diagram. The singleton is the colimit over the identity functor on this category. The set with 2 elements is the coproduct of this object with itself etc..

I know this sounds strange, but when I studied material set theory (eg. ZFC), I've always found it unintuitive why sets are determined entirely by their "membership" relation. This categorical definition actually helped me understand this as it is just a special case of Yoneda's lemma.

[knotty badger]

There is a kinda circularity in this definition in that saying what it means to be “cocomplete” requires a notion of set-sized

So I view this more as a property that Set satisfies rather than a way to define Set itself

[tulip glacier]

Fair enough.

[chilly ocean]

You'd need to consider different morphisms than functions for sets to be determined by the membership relation rather than by cardinality

[rotund aurora]

"Take the category with one object and just one element" (I suppose by element you meant arrow, i.e., the identity). Why can you do that?

[chilly ocean]

Namely functions that preserve the unary relations R_X for every set X where R_X(Y) iff X\in Y, I think

[tulip glacier]

Yeah, I meant the identity morphism, my bad.

I mean, any mathematical system has some axioms/concepts which you have to take for granted. What I meant to show is that at least on a conceptual level, if you take categories for granted, you can define sets. But yes, it's nonetheless an informal argument because it can lead to circularity.

[chilly ocean]

There is ETCS

I think it's easier/more intuitive to start with set theory though, but I guess some people might disagree

[knotty badger]

I quite like set theory honestly

And I would agree that it’s easier/more intuitive to start with it

[rotund aurora]

Allowing to talk about "1 object" and "empty" seems like set theory (when you haven't even started with category theory yet). Therefore I'm skeptical about the claim that Ring is categorical

[tulip glacier]

Right. However, we don't have a formal definition of "categorical" so this whole distinction is unnecessary. When I asked the original question what I meant by "categorical" was something along the lines of commutative diagrams, unuversality etc.. I haven't thought this through this fundamentally as to where to draw a line. It was more like a vague orientation towards what type of answer I waz looking for.

[chilly ocean]

You can define the category with just 1 object and 1 identity morphism up to equivalence as the terminal category

And the other one as the initial one

You do need axioms implying they exist though

[knotty badger]

Yeah that’s generally the issue with defining things by universal property

You do actually have to check that they exist at all

Though of course, universal properties aren’t just useful for defining things

[rotund aurora]

Don't you define terminal categories with respect to some category?

[tulip glacier]

It is the terminal category in the category Cat.

[rotund aurora]

yeah, okay

[tulip glacier]

If you take the category Cat and all it does for granted (as axioms), you can obtain this terminal category.

[rotund aurora]

How much do you need to ensure that Cat is really Cat (or at least, that it approximates it to a decent degree)? Would assuming that the category Set is a thing enough?

[tulip glacier]

Yes. Knowing what Set does can be enough. Look for example the foundational system named ETCS. It is an alternative to traditional set theory which approaches sets from a category-theoretic perspective. It defines Set as being an elementary topos which satisfies three extra convenient properties:

• having a natutal numbers object (the equivalent of ZFC's axiom of infinity)
• satisfying the axiom of choice (that is, every epimorphism is split)
• being generated by one element (that is, being the cocompletion of the category with one object, what was discussed earlier).
  The elementary topos part means, broadly, that every set has a power set (a set of its subsets), and that you can form special cartesian products where you can establish any relation you want between the various components (you can take all limits).

But I am not too familiar with it at a rigurous level. All I know is only this conceptual presentation (you can find it on nLab), but the real thing has far more subtlety. However, if you want to learn more, look up ETCS.

[tulip glacier]

If you take these concepts (function, equality of functions etc.) for granted, you actually don't have to define a category in order to define a set. It's just that sets are defined, axiomatically, in a way such that you can transpose this whole definition to a very specific category (Set) if you happen to have defined categories already. It just defines sets in a "category-theoretic" style.

Ok, this whole discussion about abstraction now has actually made me really curious about the reason we care about polynomials over rings etc. at the most fundamental level (eg. for algebraic geometry). I've never actually seen a usecase in which the underlying ring/field was anything besides the complex numbers (or a subset such as the reals). And obviously this is not enough to motivate concepts as abstract as schemes or all the magic Grothendieck did.

Actually, if I think about it, polynomials aren't that motivated even for traditional applications such as complex numbers - why are they so important ? The only thing that sets them apart from other functions is that they are the only endofunctions of C you can form with its traditional ring operations - so we still need to motivate why, abstractly, we care about polynomials and roots in the context of an abstract ring.

What do the common roots of a set of polynomials actually mean, in the most abstract sense ? Why are they so important ?

(Would've asked this in #algebraic-geometry but I think the question is too basic)

[chilly ocean]

Polynomials over a commutative ring R represent every possible operation you can apply to an element of an R-algebra using the ring operations

Two polynomials are equal iff the operations they represent are the same over every R-algebra

[dull ginkgo]

Polynomial rings essentially are the "most general" extension of a ring by a single element

I.e the indeterminate, usually X, doesn't satisfy any relations relative to the ring that we are constructing the polynomial ring over

And we can extend maps from R to U by sending X to some element, for each element

[tulip glacier]

I see. And why do we care specifically about their roots ? And more generally, about roots of a set of more polynomials ? This seems like a very specific thing. Why the identity and not another element ?

[tulip glacier]

Also @dull ginkgo ^^^

Like, in the context of freely adjoining an element to a ring - why do we care about what elements of the ring we can replace this new element with, such that we get the identity ?

[dull ginkgo]

Well for the commutative case, especially when it's an integral domain, roots allow us to decompose the polynomials down to smaller factors due to a theorem called the remainder theorem (which also holds over general noncommutative rings btw)

This is especially useful when we have well behaved rings

Polynomials are also quite natural

anyway I need to get back to class, I can elaborate later to the best of my ability :3

[tulip glacier]

Okay, thank you!

[chilly ocean]

Another reason is that roots encode algebraic dependence

[tulip glacier]

What do you mean by this ?

Oh, the solution of any equation in a ring is the root of a specific polynomial, no ?

[chilly ocean]

If a polynomial in (x1,...,xn) has a root (y1,...,yn), this means the elements y1,...,yn are in a sense related

You might ask what is special about roots (=0) instead of more generally finding where the polynomial is equal to some other value

But it is equivalent

p(x)=a <=> p(x)-a = 0

And p(x)-a is a polynomial

If you were using semirings then it would no longer be equivalent

[tulip glacier]

I see. So looking after idetities subsumes looking after any other value too.

So then, looking for the set of common roots of a set of polynomials means that you look for all tuples of elements which are related in a specific way.

[arctic trail]

second most

the free algebra being the first

[dull ginkgo]

by a single element in a commutative way i meant

[south patrol]

Indeed a polynomial ring over a commutate ring is the free commutative algebra on the generators

[delicate orchid]

wait.... are you guys telling me that polynomial rings are free objects in cring??!?!

[dull ginkgo]

CRINGE

[delicate orchid]

I know you are but what am I

[dull ginkgo]

super mario galaxy 7

[rotund aurora]

the cringe category

[rotund aurora]

This is a bit like asking why do number theorists care about numbers.

Something that hints towards Grothendieck stuff is Hilbert's nullstellensatz, you should check that

[delicate orchid]

number theorists do NOT care about numbers

[prisma ibex]

I've never actually seen a usecase in which the underlying ring/field was anything besides the complex numbers

[prisma ibex]

the reason why algebraic geometry behaves so well abstractly/categorically is the entire theory is built from localizations

you do end up seeing polynomial rings showing up in this context because you're playing the game of building geometry from localizations with commutative rings, and polynomial rings satisfy a universal property there that makes them show up repeatedly

there are other contexts in which you can build geometry out of localizations like this which subsume the usual algebraic geometry (e.g. analytic rings)

those richer contexts are interesting, but since the usual setting of algebraic geometry embeds into them, a lot of the constructions and properties of algebraic geometry get inherited

[chilly ocean]

Analytic rings is condensed math stuff?

[prisma ibex]

yes

there are other ways to approach analytic geometry that existed long before this but they don't enjoy the sorts of localization properties I'm talking about here

[prisma ibex]

If you like, another justification is that categories of modules are somewhat universal or distinguished among certain kinds of categories (e.g. among Abelian categories by Freyd-Mitchell embedding) and algebraic geometry is fundamentally about localization and gluing of categories of modules

[chilly ocean]

Are analytic rings related to rigid analytic spaces?

[prisma ibex]

they subsume them yes

you can do all the usual rigid analytic geometry (actually better than we could before) in this setting

In the setting of analytic stacks again the fundamental operation is gluing and localization of categories of modules

in algebraic geometry you usually define things in terms of the structure sheaf, where you're gluing (sheaves of) commutative rings

but it's the gluing of categories of modules that is more fundamental here

[chilly ocean]

I'm guessing more generally you could glue arbitrary abelian categories?

[prisma ibex]

so long as you have a good notion of localization sure