groups-rings-fields

[south patrol]

There's also like: it's enough to prove it in the case of R -> R/(x), by first iso etc, and then it becomes the statement that if r is a nonzero zerodivisor then rx is either zero or a non-zero zerodivisor

[tardy hedge]

i dont get what u mean here

[south patrol]

Well so suppose f:R -> S sends all nonzero zero divisors to non-zero things and R has a non-zero zero divisor. If f(x) = 0 then f factors as R -> R/(x) -> S, and clearly R -> R/(x) has the same properties as f, so it's enough to assume f = (R -> R/(x))

But then like in this case we are saying: suppose R has non-zero zero divisors but (x) doesn't. Then x=0

[tardy hedge]

we can conclude x = 0 from here?

[south patrol]

yeah. suppose r is a non-zero zero divisor. then xr is a zero divisor, hence zero by assumption. But this implies x is a zero divisor, so x = 0 by assumption again

[white oxide]

Here, is B an A-module by restriction of scalars (namely we define ab = f(a)b where f: A \to B)

[tribal moss]

Yes.

[woeful cairn]

Trying to solve the equation x^3=6 in the finite field F7, does x^3=6 mean the same as x^3 mod 7 =6? so would 24 be a solution to this equation (it satisfies it) but it's not an element of F7?

[velvet hull]

if 24 is indeed a solution, then you'll have to look for what 24 is mod 7

[woeful cairn]

3

so...? I mean what I'm trying to ask is 24 satisfies x^3 mod 7 =6 but it's not an element of F7, so does it count as a solution

[velvet hull]

well, is 3 an element of F7?

[woeful cairn]

yes

[velvet hull]

so you tell me if you think it's a solution

[woeful cairn]

yes? so it doesn't matter 24 is not in F7 since 24 mod 7 is in F7

[velvet hull]

24 is in F7, we just say that it is "equal to" 3 mod 7

but that's beside the point

[woeful cairn]

hm ok

[delicate orchid]

No it isn’t? F_7 is a 7 element set

[velvet hull]

If you choose to interpret F_7 as a collection of equivalence classes of Z

Not saying you’re wrong

[delicate orchid]

Then it is a set of seven equivalence classes and contains no numbers at all

[velvet hull]

I just disagree with the statement that 24 isn’t in F7

[delicate orchid]

That’s like saying aquamarine is in the set {blue} because the vibes are the same

[delicate orchid]

This a much better interpretation

[velvet hull]

This is not something I care to argue about, feel free to disagree with what I said

[limber tapir]

is it? I would imagine that thinking of the elements as equivalence classes is probably the best interpretation

[delicate orchid]

That’s an even better one

[limber tapir]

and then 24 \in F7 becomes very intuitive abuse of notation

of course one means [24] \in F7

or however you want to denote equivalence classes

but I guess you're not wrong, there is a lot of room for confusion with abuse of notation

[delicate orchid]

My point is that implicitly using this abuse of notation while guiding someone through a simple conceptual task is bad pedagogy

[limber tapir]

yeah I agree

[delicate orchid]

You should be precise about what you say, and then when they’re comfortable with the concept you can be more handwavey

Although even at the post grad level people will look at you funny if you say that 24 is in F_7 I cannot lie 😭

But then we’re all hypocrites cause everyone does finite field (at least, of prime order) computations in Z then reduces mod p at each step

[limber tapir]

i will say that I think it will be enlightening to look at how F_p or Z/pZ (if you haven't already) are defined using equivalence relations, it will make such questions much more intuitive

[limber tapir]

i guess the word class should really fall somewhere in that sentence, you're right

[languid trellis]

Is 1/2 in Z_3? Because on one hand 3 doesn't divide 2, but 1/2 = 3/6 and 3 divides 6. How should I think about this?

[south patrol]

I should say like this is extremely bad notation lol since Z_p usually refers to the p-adics or sometimes Z/pZ and both are different to this one lol

[south patrol]

But yes it should be all rational numbers which have a representative of the form they say

[languid trellis]

I see

Idk what to think of the pedagogy. On one hand I appreciate giving an example of a ring with one prime (up to associates). On the other hand. ew

[south patrol]

This would usually be written Z_(p) but yeah it's an important example

A cute example imo would be F[[x]] for F a field

[languid trellis]

I don't have enough experience with formal power series to see it

Lemme think

[south patrol]

Yeah I mean there is a little work needed

[velvet hull]

Any local ring that’s a PID would be what you’re looking for

But I don’t have any examples at hand other than Z_<p>

I’ll think about it

[south patrol]

The actual Z_p works too ig

[languid trellis]

For the time being I think I'll treat this as a curiousity and move onto the meat of the chapter

Thank you both 🙂

[rocky cloak]

This is the rational p-adic integers, so not that bad notation I guess

[south patrol]

Idk I think it's pretty bad lol

The fact it is close to Z_p kinda makes it worse imo

[unique valve]

Z localized at p

[velvet hull]

Yeah I was gonna say, Z_<2> is standard notation for localised rings

I guess you would just have to add the brackets to distinguish it from the p adics

But it’s not good

[south patrol]

There's also the issue that A_f is often used to mean A[1/f] lol

[dull ginkgo]

This is Ore Localization in respect to a multiplicative set

The complement of a prime ideal (I.e the maximal (p) ) is multiplicatively closed

So in a sense this ring is ZS^-1 where S = Z\(p)

[dull ginkgo]

To be more specific it’s the equivalence relation and it’s down to the representative

[tardy hedge]

What chapter are u learning rn? Also what text was that

[elfin wraith]

Looks like D&F

[tardy hedge]

Hmm

[dull ginkgo]

I need to get back into algebra at some point

[delicate orchid]

📉

[dull ginkgo]

Fuck off britoid

[languid trellis]

ch2 of kenneth and rosen number theory

[languid trellis]

Gotcha. Ty

[dull ginkgo]

RS^-1 works basically for any comm ring provided S has some property in relation to zerodivisors I forget

[south patrol]

No need

Just it might be zero lol

[dull ginkgo]

@languid trellis adding to this, the case where S is the complement of a prime is like… why it’s called Localization. It localizes the ring

In a way it’s like the OPPOSITE of a quotient

It “kills” off the ideals intersecting the multiplicative set, because under localization they contain the identity so become the whole structure. Thus the only ideals left over are strictly in the complement

Consider when the complement of the multiplicative set is a prime ideal. Then that’s made maximal :3

[swift tundra]

Does anyone have an example of a ring where there are non-trivial nested prime ideals? I am trying to get more intuition for why we want the set of prime ideals to have a minimal element.

Also on the topic of localization, I know R_(P) gives you a local ring with unique maximal ideal P (well technically P/(R-P) I guess). Is there another way to localize a ring to get a semi-local ring (only finitely many maximal ideals)?

[velvet hull]

Let k be a field. in k[x1,...,xn], there is the tower of prime ideals <x1> subset <x1,x2> subset... <x1,...,xn>
this ascending chain construction works for any collection of algebraically independent irreducible polynomials

[dull ginkgo]

Finitely many maximal ideals hmm. We’d want the multiplicative set being localized to have a complement probably containing only finitely many maximal ideals

which you might have to find one

[dull ginkgo]

By the logic of this

[swift tundra]

Ahhh ok that makes sense, thanks!

[swift tundra]

Hmmm ok I’ll think about this more thank you. Maybe Atiyah Macdonald will have some examples when I get to localization

[dull ginkgo]

Let me specific, maximal within the set

[swift tundra]

Yes ofc

[dull ginkgo]

It doesn’t seem like that would naturally occur much?

Like you can define a prime set where it’s complement is multiplicatively closed, and you can maybe look for those?

[swift tundra]

Hmm so a prime ideals is a prime set and then look more generally?

[dull ginkgo]

Maybe think about prime sets and find one where there is only finitely many ideals maximal within that set

Idk when that happens, probably tricky

Is this for an exercise or a general quandary

[swift tundra]

Is the union of prime ideals a prime set?

[dull ginkgo]

Ooooh

Yes!

[swift tundra]

Just general quandary I guess lol

[dull ginkgo]

Complements are multiplicatively closed

And the complement of the union would be the intersection of multiplicatively closed sets

I.e multiplicatively closed

Good call

Arbitrary union in fact

[swift tundra]

Mmmmm oh yes that makes sense!

[dull ginkgo]

I haven’t done any AA for an extended period of time so I apologize if I’ve been rusty

I’m in the engineering and functional analysis mines rn

[south patrol]

AA batteries

[swift tundra]

So maybe we could even use HChans example for this?

[thorn jay]

The only correct way to pronounce this is with a small scream

AAA is a bigger scream

[swift tundra]

Union of (x_1) (x_2)…(x_n)

[dull ginkgo]

@swift tundra Maybe. Take a semilocal domain. Take a maximal ideal, chop out 0 from it (multiplicatively closed), then localize by that

Then it would still be semilocal

You’ve basically just murdered one of the maximal ideals

[karmic moat]

does this mean you're sober or relapsed?

[dull ginkgo]

Provided the maximal ideals are still Lin indep, otherwise you should remove the Jacobson radical too prob

[swift tundra]

Maybe algebra isn’t the place for me, these poor ideals 😢. Just constantly being killed

[dull ginkgo]

I laughed out loud at this

[karmic moat]

,w annihilator

[dull ginkgo]

All your ideals in life are killed by algebra yes

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by aanzhuu

[karmic moat]

okay well that wasn't helpful

https://en.wikipedia.org/wiki/Annihilator_(ring_theory)

[south patrol]

Ann(M) = { r | r.M =0}

[karmic moat]

yeah just they were talking about killing off and stuff so i thought of the word

[south patrol]

Ann(amono)

[karmic moat]

nooo vakil is already ann'ing me

[dull ginkgo]

ANOTHER ring HAS mapped into End(M)

[south patrol]

Lol what

[dull ginkgo]

M is left R-Module iff ring maps into End(M)

[south patrol]

Sure lol

[dull ginkgo]

Back to the functional analysis mines I go bye chat

[karmic moat]

yeah but no one will take that seriously because first and foremost it's "Ass"

[dull ginkgo]

Tits group

[karmic moat]

haha

true

[dull ginkgo]

I rest my case

[rocky cloak]

Take some finite set of maximal ideals and invert those elements not contained in either of them.

Then the resulting ring should have those as its maximal ideals by the prime avoidence lemma

So for example of you invert all primes in Z except 2 and 3 you get a ring with two maximal ideals (2) and (3)

[dull ginkgo]

Do you agree with the interpretation of localization behaving the opposite of wuotienting

[rocky cloak]

Kinda yeah.

The primes of R/p are the primes containing p, while the primes of R_p are those contained in p. So in that sense their opposites

Localizations are also epimorphisms though, so in many ways their similar.

[dull ginkgo]

In Quotienting, ideals in the output are ideals containing the quotient ideal, but in localization, the ideals are the ideals outside of the localizing set

[rocky cloak]

Yes that's right

[dull ginkgo]

Epic thank you for the insight

That’s how I remember a lot of localization stuff :p

[elfin wraith]

Huh, I didn’t know that, that’s a nice fact

Honestly I don’t know enough about localisation, my comalg class didn’t spend much time on it because the lecturer “doesn’t like geometry” and in my noncom class

[dull ginkgo]

Noncom localization is a nightmare

[swift tundra]

Oh wait that makes so much sense! Thank you. Also a good use of the prime avoidance lemma; my professor had us prove a weakened version of it but I was wondering what exactly it would be useful for.

[tribal edge]

\begin{tikzcd}[row sep=1.5em, column sep=1.5em]
& \text{Euclidean Domain} \arrow[d] \
& \text{PID} \arrow[dl] \arrow[dr] \
\text{Dedekind} \arrow[d] & & \text{UFD} \
\text{Noetherian} & &
\end{tikzcd}

[cloud walrus]

Slomenist

[tribal edge]

hey
we know that euclidean => pid => ufd => integral
i'm trying to expand this diagram to include artinian rings, local rings, integrally closed rings and other interesting types of rings

[grim bridge]

dedekind

[tribal edge]

can somebody help me ? with as much implications he knows as he can

[tribal edge]

¿

[karmic moat]

artinian iff noetherian and dimension 0

okay i should say the disclaimer: i'm assuming rings to be commutative and unital

[tribal edge]

yes

[karmic moat]

regular local implies ufd

[tribal edge]

do you know somewhere where there are similar iffs
and NSCs for the inverse to be true ?

[karmic moat]

NSC?

[tribal edge]

necessary and sufficient conditions

[karmic moat]

also idk too many iff's off the top of my head

[tribal edge]

i'm looking for them, summarising all this mess up would be very helpful to remember things pretty smoothly

[karmic moat]

serre's normality conditions give you an iff for a commutative noetherian ring to be normal

[south patrol]

nerd

jk nice

[karmic moat]

🤬 first you bully me for vakil now you bully me for this

a module is reflexive iff it satisfies S_2 (this requires noetherian, and everything henceforth will also assume noetherian)

[tribal edge]

nice
i was reading bourbaki books on commutative algebra
there isn't much interesting things like these

[karmic moat]

i learned these through a class i took

[tribal edge]

did you rely on a particular book in class ?

[karmic moat]

R is a UFD if and only if the following statement holds: if an ideal I is reflexive, then I is a free R-module

no

he kinda just did this all off the dome

[karmic moat]

this is due to auslander iirc

[tribal edge]

yeah
nice

[karmic moat]

a ring is normal iff its poincare series is in Z[t]

okay that's everything i found for "if and only if" in the lecture notes

[dull ginkgo]

It has to do with ideal structure

[karmic moat]

there's a million of these everywhere bc commutative algebraists love this kinda stuff

look up like idk

"cohen-macaulay" or "gorenstein" and you'll find a million things

from wikipedia

equivalent characterizations for gorenstein

[tribal edge]

oh wow that's interesting
now i'd like to read your profs full course/notes ;-;

[karmic moat]

i would send but unfortunately:

1. i typed the notes and there's a bunch missing
2. there's some stuff about my classmates' questions and stuff in there

so i can't send for the sake of privacy

[tribal edge]

thanks i'll look them up

[tribal edge]

alright it's fine ;)

[karmic moat]

https://www.math.utah.edu/~chau/files/Cohen-MacaulayRings.pdf

if you look through bruns-herzog (and subject yourself to looking into that boring ass book) you probably find a lot of "if and only if" statements for cm rings, gorenstein, etc etc

[dull ginkgo]

Ideal hell

[karmic moat]

Multiplicities and Chern Classes in Local Algebra by Roberts has a bunch of open questions and homological conjectures and stuff somewhere in there, you can prob find stuff there

for integral closure/integrally closed rings, see Integral Closure of Ideals, Rings, and Modules by Swanson Huneke

[tribal edge]

too much stuff to learn but kinda fun in the end
it's even more fun and interesting to solve problems like these, mysterious beautiful problems

[tribal edge]

nice, i would
thanks

[karmic moat]

https://stacks.math.columbia.edu/tag/00AO

you could look here too

[tribal edge]

yeah i used to read a lot in this website
it has a lot interesting things

[south patrol]

Cute

[karmic moat]

https://arxiv.org/pdf/2506.21764
from the abstract: As an
application, we prove that generic Gorenstein local rings, non-trivial connected sum of generalized
Golod-Gorenstein rings satisfy the Tor-vanishing property and consequently the Auslander-Reiten
conjecture.

Idk what golod gorenstein nor auslander reiten are but yeah commutative algebraists love this stuff

[tribal edge]

yeah

[tribal edge]

these are even more advanced

[karmic moat]

Yeah

[tribal edge]

crazy research papers

[karmic moat]

Every paper and its paper-mother has “betti number” and “koszul complex” in it

[spark veldt]

I've been reading about Zorn's lemma but never actually attempted it. Is my proof correct?

Let $R$ be a commutative ring with $1$, then $R$ has a maximal proper ideal.

We seek a \textit{maximal} proper ideal, so why don't we create a set of all proper ideals and see if it is partially ordered. Set
$$P={I\subseteq R:I\text{ a proper ideal of $R$}}$$
We show that $P$ is partially ordered. As before, with respect to set inclusion. Any $I\in P$ is a subset of itself (reflexive), if $I_1\subseteq I_2$ and $I_2\subseteq I_1$, then $I_1=I_2$ (antisymmetric), and if $I_1\subseteq I_2$, $I_2\subseteq I_3$, then $I_1\subseteq I_3$ (transitive). Let ${I_\alpha}{\alpha\in J}$ be a chain in $P$. We recall the definition of an ideal of some commutative ring $R$: $U\leq R$ is an ideal if $U$ is a subgroup of $R$ under addition and for every $u\in U$ and $r\in R$, both $ur$ and $ru$ are in $U$. Now let's actually consider the union $I'=\bigcup I\alpha$. Clearly this union is an upper bound of ${I_\alpha}{\alpha\in J}$ with respect to set inclusion. Now we verify that it's a proper ideal of $R$. Let $v_1,v_2\in I'$, then because of total ordering, there must be some $I\beta $ such that $v_1,v_2\in I_\beta$, hence $v_1+v_2\in I_\beta\subseteq I'$. Now if we take $v\in I'$, then $v\in I_\beta$ and $-v\in I_\beta$ for some $\beta\in J$, so $-v\in I'$. We show that $I'$ is not trivial: $I'$ must not contain $1$. Had $1\in I'$, then there is some $\beta\in J$ such that $1\in I_\beta$, contradicting the fact that $P$ is the set of all proper ideals. We verified that every chain has an upper bound in $P$, so by Zorn's lemma there is a maximal proper ideal.

[cloud walrus]

bluepianist

[spark veldt]

oh kinda missed proving the ui and iu are both in U for any i in I'. But other than that, is my attempt ok?

[karmic moat]

The overall idea looks okay

you should show that I’ absorbs elements under multiplication, and thus is an ideal (so far, you’ve only shown it’s a subgroup under addition). Idrk what you mean by U here since U only appears as a definition and not actually in the proof; you should instead be showing that for any i in I’ and r in R, ir is in I’

Some minor nitpicking about the proof writing itself; this is more of my opinions so you can ignore what you want:

• for your set P, you should write I \subset R and not I \subseteq. I know that you said “proper ideal” which means I is never R, but again this is just some nitpicking
• recalling a definition in a proof often leads to more visual clutter within a proof

Aside from that looks fine

[spark veldt]

ah yeah sorry slight typo there should be ri and ir. thank you!!

[karmic moat]

If you liked this Zorn’s lemma argument you can also prove similar results, like “every vector space has a basis”

An aside but worth noting: you’ve proved that every commutative ring with unit has a (ie, at least one) maximal ideal. A ring with a unique maximal ideal is called a “local ring”

[spark veldt]

ah i see okayy

[spark veldt]

yeah this is what i dipped my toe into. after getting a feel of what the argument looks like roughly i decided to try the commutative ring one by myself and super happy i got the general idea right :)

used to be super scared of this lemma because the wording itself is so abstract but now i finally get the gist

[karmic moat]

Yeah but in practice it’s very intuitive

Glad you liked it

[next obsidian]

Simple and intuitive

[twilit vapor]

Is there a term for the relationship between the ring $\mathbb{Z}q$ and $\mathbb{Z}{kq}$ where k and q are integers? Is it a homomorphism?

[cloud walrus]

request new nickname in modmail

[south patrol]

What is Z_q?

I don't know what you mean by "it a homomorphism" - i think you have got terms confused

[twilit vapor]

its a ring of integers

modular ring

Z/qZ basically

[knotty badger]

So you should be able to view Z_q as a quotient of Z_kq

I’d have to check the details but this at least works on the level of groups

Yes this follows from Z being the initial ring essentially

Z_q is not a subring of Z_kq though

It does work as a subgroup, however

[thorn jay]

Yep, you can make a lattice out of the rings Z/nZ, its dual being the lattice of ideals, i.e. the integers ordered by divisibility

[tardy hedge]

For f: R->S surjective, prime ideals of R containing ker are prime ideals in S under f right

[karmic moat]

Yeah

[tardy hedge]

If S is an integral domain, is every prime ideal of R prime in S under f?

[karmic moat]

If f is surjective it doesnt matter what your ring is (aside from commutative with unit)

If you drop the condition that f is surjective and you want S to be an integral domain, it’s false

[tardy hedge]

Is there a typo here?

Idk what u mean

[karmic moat]

Fixed

[rocky cloak]

For example
Z -> Z/2
then the prime (3) maps to the improper ideal, which isn't prime

[karmic moat]

What I mean is there’s no assumption on R and S (eg integral domains)

Z -> Z[i] is another one

[rocky cloak]

Perhaps a more enlightening example
k[x, y] -> k[x]
x |-> x
y |-> x^2

Then the prime ideal (y) maps to (x^2)

[tardy hedge]

Ok yea because its not like ideals in R are either contained in ker or contains ker

[south patrol]

Note also that this can be viewed in terms of the correspondence theorem
like for f: R -> R/I, the assignments p |-> p(R/I) and q -> f^{-1}(q) give a mutually inverse correspondence between primes of R containing I and primes of R/I

[tardy hedge]

Yea i was verifying that if in the case where S is a domain then prime ideals (or i guess any ideal) contained in ker will be prime in S which i mean yea cuz S a domain

[south patrol]

ye note that anamono's example is very rich, like if F is a number field (i.e. finite extension of Q) then studying how pO_F factors into primes is a big thing

[karmic moat]

Shoutout atiyah macdonald

[south patrol]

lol for what

[tardy hedge]

I only learn math i learned a year ago a year later

[south patrol]

oh lol nice

Atiyah Macdonald my beloved

I think it was a very formative book for me

[velvet hull]

what's your opinion on it?

going through it right now, and it feels very brief

[tardy hedge]

But ideals under extension is kind of a different topic no?

[karmic moat]

PREAAAAAAAAACH

[velvet hull]

the book feels like it can say a lot more

like 150 pages is on the low end for textbooks

[tardy hedge]

How does that relate to what i was asking before?

[karmic moat]

Yeah but this example illustrates that the image of a prime isnt necessarily prime

You need to use extension

[tardy hedge]

Oh ok yea

Cuz image of ideal isn’t necessarily an ideal anyway too

[karmic moat]

The beauty is in the exercises

[south patrol]

Excellent book

[wraith cargo]

atiyah and macdonald is kinda ass ngl

[south patrol]

I think yeah like a lot of magic is in doing the exercises and i think this is what helped me a lot lol

and also feels very slick

[tardy hedge]

Lol

[wraith cargo]

I am forever a matsumura commutative algebra enjoyer

[south patrol]

I was weird thoguh cause I had learnt very little about rings and modules before the book

[tardy hedge]

Yea i like reading matsumura

[south patrol]

but was still kinda readable once i looked at the correspondence theorem

[tardy hedge]

Atiyah macdonald is nice to have tho

[south patrol]

lol

i think like atiyah macdonald is the sort of book one can work through relatively quickly (obviously can still take quite a while! esp if doing the exercises lol) and that's nice about it

[tardy hedge]

I bought cohen macaulay rings by bruns and herzog recently lol idk when im gonna be at the level to actually learn from it properly

[karmic moat]

Hartshorne if hartshorne was good

[tardy hedge]

Still kinda above me

[karmic moat]

RETURN IT

[south patrol]

One amusing thing is uh this is based off an old version of the commutative algebra university course I took, and it's interesting how much stuff is covered in the book that wasn't in our book

[tardy hedge]

Lmao

[karmic moat]

THAT BOOK IS BORING AS FUUUUUUCK

[tardy hedge]

😩

[south patrol]

maybe kiand just wants to fall asleep at night quicker

[tardy hedge]

Precisely

[south patrol]

Maybe the best way to learn commutative algebra is to start with the subject you are learning comm alg for and work backwards

[tardy hedge]

Did u guys like dummit and foote

That used to be my go to

[karmic moat]

Never read it

I learned group theory through a course

[south patrol]

not rly a fan personally

[rocky cloak]

People have motivation to learn stuff? Smh

[tardy hedge]

I used it for the ring and module chapters mostly

[south patrol]

well if you don't have motivation to learn stuff don't read a textbook and touch grass

Found an error in a proposition I wrote down months ago... Game over

[karmic moat]

Didnt even know it had those tbh

[karmic moat]

Errata on arxiv soon I hope

[tardy hedge]

Dude it goes up to homological algebra and even some alg geometry lol

[karmic moat]

Oh nice

[tardy hedge]

I dont think those sections are good thi

Skips proofs

[karmic moat]

Use atiyah macdonald

[wraith cargo]

the only proper way to learn commutative algebra is to look at hartshorne, understand how little you know, and then read matsumura because that's what he cites

[tardy hedge]

The authors got tired

[karmic moat]

Hartshorne part lowkey true when I started I didnt realize how bad my CA was

So I went and read atiyah macdonald bc it has a bunch of exercises on ag

[tardy hedge]

Maybe thats when one starts to actually get good at ca?

At least i am hoping ? Lol

[rocky cloak]

Grass? I'm telling you this alggeo agriculture terminology is getting out of hand

[wraith cargo]

imho if you're learning AG from an algebra text you're doing something wrong lol

hartshorne

my belved

taught me everything I ever need to know about life

I remember trying to read eisenbud and I fucking hated that book

[karmic moat]

I wouldnt use it as a primary source but it introduces the basics of ag

In this sense I think it’s good

Because imo ca on it’s own is really dry

[south patrol]

not on arxiv rip

i have been hoping to put it out soon for the last 25 years

[karmic moat]

But when you have exercises relating to stuff like NT and AG it lets you learn with applications in mind

[wraith cargo]

ig that's true

tho maybe I'm insane cuz when I was reading about CA just the story of the algebra was interesting enough to me lol

I didn't feel like I needed to know the applications a priori

but everybody is different

and if you're different from me you're doing some wrong smh

[karmic moat]

Clausen core

Yeah I guess when I started learning CA I found it interesting as well

Then I started doing AG and, yk, comparison is the thief of joy

And then CA on its own became very dry to me

[rocky cloak]

I feel this too hard rn

[tardy hedge]

Basic question but for a ring R we know all maps out of R are controlled by the ideals of R, but what about maps going into R? Is it just that the possible maps into R are given by the subrings of R?

I mean the subrings dont give the map like ideals do but

[south patrol]

Hug

[next obsidian]

Maps into R are maps out of Spec R

#factsandlogic

[tardy hedge]

S -> R gives Spec R -> Spec S thing?

[tough raven]

What do you mean by the first statement?

[dull ginkgo]

Yes

[mighty kiln]

Every ring morphism factors as surjection followed by injection

[tardy hedge]

So meaning what in context to my question?

[mighty kiln]

I assume that's what you meant by "determined by ideal / image"

That is, every morphism R → S is given by R → R/I → S

Where the first map is quotient by an ideal and the second is inclusion of subring

[covert cliff]

I am very rusty on algebra but I seem to be stuck here, I want to show that if $V$ is a real vector space of positive dimension then for linearly independent vectors $v, w \in V$ the wedge product $v \wedge w$ is nonzero. The definition I have for the wedge product is that it is the algebra defined by the relation $v \wedge w = - w \wedge v$ for all $v, w \in V$

[cloud walrus]

Khush

[covert cliff]

Not sure how this leads to $v \wedge w$ being nonzero

[cloud walrus]

Khush

[covert cliff]

I know there is a way with the alternating map but I want to avoid that since the text I am going over does not cover this

[covert cliff]

my other issue with this definition is that couldn't v \wedge w = 0 for all v and w?

[elfin wraith]

Yeah I’m also slightly lost, the we all know has me worried I’m being dumb though

[kind temple]

the exterior square A^2V is universal with respect to bilinearity and antisymmetry.
so if V x V -> A^2V is the map taking (v,w) to v \wedge w, and if f : V x V -> W is any alternating bilinear map, then there is a unique linear map A^2V -> W which f factors through.
any vector space W for which there is an alternating bilinear map V^2 -> W will therefore be a quotient of A^2V via the universal property of A^2V.

if v and w are linearly independent, extend them to a basis of V and let Proj(v), Proj(w) : V --> F be the linear functionals which pick off the coefficient of a vector written in the extended basis corresponding to the v and w components, resp.
formally, Proj(v) is the dual vector corresponding to v, likewise for w.

consider the alternating bilinear map f : V^2 -> F defined by (x,y) |-> Proj(v)(x) Proj(w)(y) - Proj(v)(y)Proj(w)(x), which is the alternation of the bilinear map (x,y) |-> Proj(v)(x) Proj(w)(y).
by the universal property, there is a unique linear map A^2V -> F such that x ^ y |-> f(x,y). but f(v,w) = 1, so v ^ w can't be zero, otherwise, it would be in the kernel of this map.

[kind temple]

maybe there is a way to not extend to a basis of V...

[tribal edge]

Let $K$ be a field and $L$ an algebraic extension over $K$. Then:
[
\mathrm{Card}(L) \leq \mathrm{Card}(K[X] \times \mathbb{N}).
]
\textit{Proof attempt:} Inject $L$ into $K[X] \times \mathbb{N}$ via
[
a \mapsto (P_a, k_a),
]
where $P_a$ is the minimal polynomial of $a$ over $K$ and $k_a$ is the position of $a$ among the roots of $P_a$.
but this proof seems to work only when $P_a$ is separable, so only for few types of fields, and not for any arbitrary field $K$

[cloud walrus]

Slomenist

[tribal edge]

i'm looking for different ideas to solve this problem

forgot to mention that one of the hints given is to consider the set : \
$${(P, x) \in (K[X] \setminus {0}) \times L \mid P(x) = 0}$$

[cloud walrus]

Slomenist

[kind temple]

Corollary 5.9 (discussion)

[cloud walrus]

c squared

[kind temple]

i think this gets around the choice issue

[glad osprey]

How does it fail when L is inseparable? The map will still be injective AFAICT, depending on how you define "position among the roots". You'll just have fewer roots than the degree of the minimal polynomial

[rocky cloak]

You get some trouble if k=1 in your first case. Better to pick d=0

Well, you would be picking
a=b=c=d

Which isn't possible

Having one being 0 and the others non-zero is always fine.

Otherwise one needs to be careful

[covert cliff]

I see, so I would have to show that this universal property holds?

[cursive spindle]

I feel like this was mentioned here previously

[kind temple]

yea. you use the universal property of the tensor product and the universal property of the quotient vector space

[muted heron]

Yes I did lol

[cursive spindle]

oh lol

[muted heron]

Btw deltoid can u help me in #advanced-algebra

[tardy hedge]

hmm i was about to ask a question but i wasnt sure if what i was asking was really dumb or nt

not*

[crystal vale]

[white oxide]

If $G$ is free abelian, then does it have no nontrivial elements of finite order? For let $a \in G$ be an element of finite order. Then there exists $b \in \langle a \rangle$ where $\langle a \rangle$ is not infinite cyclic. Furthermore, such a representation is unique. This contradicts the fact that $G$ is free abelian.

[cloud walrus]

okeyokay

[velvet hull]

you can use the universal property

[tardy hedge]

Like G a free Z-module?

[tardy hedge]

The universal property of free modules as in if A is basis then any set map A->M factors through A->F(A)->M?

[velvet hull]

there is a (natrual) bijection between set maps and homomorphisms
that's one way to state the property, yes

[rocky cloak]

I'm not sure what exactly you mean by your penultimate sentence there.

What representation is unique in what way and what is it contradicting?

[tardy hedge]

Ok, it seems like a lot of the time people use universal properties in this kind of way

Potato said smth similar when i asked abt localization

[south patrol]

Real

[velvet hull]

oh yeah localization has a universal property (and adjoint pair) as well but I don't think that one is as nice

[south patrol]

It is nice

[tardy hedge]

Bijection between morphisms R—>S sending all stuff in M to units, and maps M-1R -> S?

[south patrol]

Using M for a set of elements is melting my brain

[tardy hedge]

Ik that was silly

[velvet hull]

If a ring homomorphism sends S to units

then it extends to a homomorphism in the localized ring

[tardy hedge]

Ya that is like the first way when i said the free module one right

[velvet hull]

Saying that it factors is not strong enough, although probably sufficient for proving most things

it's a bijection

[tardy hedge]

Saying it factors uniquely would be incorporating the fact that theres a bijection there right

[velvet hull]

that would give you an injection

[tardy hedge]

Right

Then surjection part is just if you have F(A)->M you know the A->M that would fill in the diagram?

[south patrol]

ye

[tardy hedge]

This fact of it being a bijection where is this helpful / used?

[south patrol]

but the diagram is sort of abusive cause the important thing is that F(A) -> M is a map of modules

[tardy hedge]

Yeah

Similar thing with tensor products tho right

One arrow is map of abelian group and other is like bilinear map

[velvet hull]

these are instances of adjoint functors

[south patrol]

Sure i mean this is an abuse for ye

adjoint functors sometimes

Thoguh idk i wouldn't recommend it idk

[tardy hedge]

Wouldnt recommend what

Adjoint functors were briefly introduced once but i did not learn it whatsoever

[south patrol]

writing diagrams which aren't in a category rly lol

[tardy hedge]

Im not sure what u mean

[tardy hedge]

Kind of confusing myself over the uniqueness part. Isnt it obvious that given F(A)->M there is a unique A->M by just precomposing with A->F(A)?

[tardy hedge]

We were talking about atiyah macdonald earlier : i do like how a/m is small and compact enough to more easily bring around with you and read

Some texts u just cant do that with cuz too big and annoying

[south patrol]

Yes.

[south patrol]

Oh I mean like if u draw a commutative diagram to understand the free module construction then it will involve a bit of abuse cause ur mixing sets and modules

[thorn jay]

Then use the forgetful functor

[south patrol]

Ye you can say like unique F(M) -> N such that the composite F -> U(F(M) -> U(N) is what U started w lol

[thorn jay]

Mhm

[solemn wadi]

Need help with a problem

A set G with operation * is associative and has left identity and right inverse .....prove it is a group

Tricky one seems

[knotty badger]

oh yeah these asymmetric defs are weird

[tough raven]

I think this is false, actually.

At least I feel like I've heard that this is false.

[charred iris]

it is not false

Let e be the left identity, g in G, and h the right inverse of g. then hg = h(eg) = ...

unless I messed up, eventually you show h is also the left inverse (not through just a chain of equalities though), and then showing e is a right identity is easy

[tough raven]

Are you sure the theorem you know isn't left identity + left inverses ⇒ group?

That is indeed a folklore theorem.

[charred iris]

Given that I've proved it for myself just now, reasonably sure

[tough raven]

That's neat! I would like to see that proof if you're willing to share it.

[charred iris]

I made a mistake 😔

[elfin wraith]

Many such cases

[thorn jay]

Don't we all

[coral spindle]

I will write the left identity as e. For each element a choose an inverse a^-1 such that aa^-1 = e.

Lemma 1: (a^-1)^-1 = ae.
Proof: Consider a . a^-1 . (a^-1)^-1. Associating one way gives us e . (a^-1)^-1 = (a^-1)^-1. Associating the other way gives us ae.

Lemma 2: ab = ac = e implies b = c, so right inverses are unique.
Proof. ab = aeb = (a^-1)^-1b and same for b using lemma 1. Now left multiplying by a^-1 gives us eb = ec so b = c.

Lemma 3: (ab)^-1 = b^-1 a^-1
Proof: (ab).(b^-1 a^-1) = a(b b^-1)a^-1 and you can see the rest, I hope, by applying lemma 2.

Lemma 4: If ae = e then a = e. The proof of this was unfortunately wrong.
Proof: left multiply by a^-1 and apply lemma 1 to get e = a^-1 e. Then right multiply by a to get a = a^-1a = e.

Lemma 5. a^-1a = e. The proof of this relied on the incorrect proof above.
Proof: by lemmas 1 and 3, e = (a^-1 a) (a^-1 a e) = a^-1 (a a^-1) a e = a^-1 a e. Apply lemma 4.

I think you should be able to finish things from here.

Please do let me know if there are any mistakes here.

And yes, I had coffee at 6pm, how could you tell?

Darn, I just noticed an important mistake. Nonetheless I'll leave up some of these here in the hopes they help.

[kind temple]

this is not true. Consider the set $G = {e,a}$ with binary operation $\ast : G \times G \to G$ given by $x \ast y = y$. $\newline$

\textbf{Associativity: } $(x \ast y) \ast z = y \ast z = z = y \ast z = x \ast (y \ast z)$ $\newline$

\textbf{Left Identity: } $e \ast x = x$ for all $x \in G$ $\newline$

\textbf{Right Inverses: } $x \ast e = e$ for all $x \in G$ $\newline$

$G$ is not a group: $x \ast a = a$ for all $x \in G$, so $a$ does not have a left inverse, and therefore can't be a group. Alternatively, if $x \ast a = e$ for some $x \in G$, then $a = e$, but we know that $a \neq e$, so $a$ can't have a left inverse in $G$. $\newline$

[cloud walrus]

c squared

c squared
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[kind temple]

column is left component, row is right component

[kind temple]

if G is a semi-group with a left (resp., right) identity and every element has a left (resp., right) inverse with respect to the left (resp., right) identity, then G is a group, as shown here

[tribal moss]

It's even easier to say: G is not a group because there's no right identity.

[kind temple]

no right identity, i think you mean, right?

[tribal moss]

Whoops yes, fixed.

[velvet hull]

Groups don’t have right identities what do you mean

[kind temple]

the identity of a group is a left and right identity

[velvet hull]

Trolled a little too hard xd

[kind temple]

oof my bad lol

[tribal moss]

I spent entirely too long time trying to work out the free semigroup-with-left-identity-and-right-inverses on one element to check whether it's a group. After much paper, I think it turns out to be isomorphic to Z×G where G is C Squared's example:

left identity (0,e)
right inverse of (n,p) is (-n, e)
the generator is (1,a)

[kind temple]

thats cool

[tribal moss]

With n > 1 generators, I think the free thing is the product of

• the free group on the generators, and
• a set of n+1 elements (one for each generator plus a designated identity) with the x*y=y operation

[kind temple]

im going to work through this later actually

[tribal moss]

The crucial calculations for me were

• aa'x = ex = x
• a'ax = a'aex = a'aa'a''x = a'ea''x = a'a''x = ex = x
  where the prime denotes the right inverse. So inverses are effectively two-sided anywhere in a word that's not the rightmost symbol. And if the rightmost symbol is an inverted generator, we can replace it with any other inverted generator:
• a' = a'aa' = a'e = a'bb'
  so all the inverted generators map to a single symbol in the x*y=y factor.

[rocky cloak]

I think interpreting right inverse to mean
x x^-1 = e
for a left identity e is kinda weird anyway.

If right inverse instead means
y x x^-1 = y
then it is true. And you can replace the existence of a left identity by the set being non-empty.

[kind temple]

how are inverse elements defined for semigroups?

like b is a left inverse of a if aba = a? something like that?

bab = b for b some right inverse of a?

[mighty kiln]

Left/right cancellative?

[kind temple]

so if b is a right inverse to a, then ba is right cancellative?

that just feels like saying ba is a right identity

[tough raven]

Gah. I was sure there was a simple example but I tried toset with min and didn't think of "second operand".

[topaz solar]

It’s similar, but not quite the same

Since cba need not be c, but cba=dba implies c=d

[kind temple]

oh true

thanks for that

[tough raven]

I think @ jagr2808's proposed definition is that b is a right inverse to a iff ab is a right identity.

[topaz solar]

I mean, multiplication in naturals bigger than 2

That’s cancellative with no identity

And no absorptive elements

(And commutative)

[topaz solar]

This I think is better than being cancellative, but a pseudo-inverse for cancellation is still nice to have exist etc

There’s the uhh what’s it called inverse semigroups where a weird sort of inverse occurs

like xyx = x?

[kind temple]

i guess this is what i was thinking it should be

[topaz solar]

I mean it’s kinda screwy since you can get inverse-like stuff even without identities

But if you have identities, things kinda should work with them

[tribal moss]

That's true, but requiring yxx'=y for all y just implies that xx' is a right identity, so that would also make the problem statement "kinda weird".

[rocky cloak]

It's weird either way, but at least this way the statement you're asked to prove is true

[tribal moss]

That is an advantage, I suppose.

[coral spindle]

Lovely

[kind temple]

im sure the caffeine was worth it tho lol

[coral spindle]

Hmmmm

Well I feel great this morning!

[kind temple]

so in the van kampen theorem, say we decompose our space X into two open path connected sets A and B with path connected intersection.
then pi1(X) is the amalgamated free product of pi1(A) and pi1(B) over p1(A n B)

how do you form the amalgamated free product if you decompose your space into more than 2 sets?

[tough raven]

The details are in Hatcher

[mighty kiln]

If all your subsets are path-connected with a common basepoint, you have a bunch of π1 maps

Just take the colimit

[tough raven]

but it still works as long as all three-way intersections are path-connected

[kind temple]

yes

[south patrol]

I love how the free product is a coproduct. Great terminology

[kind temple]

hmm... maybe i am misunderstanding something. do i amalgamate over the free product of all of the pi1(Ai n Aj)?

i guess i don't know what diagram im taking the colimit over

[mighty kiln]

Consider the category of all intersections of your subsets

So A, B, C, A∩B, A∩C, B∩C A∩B∩C, if you have 3 sets

[kind temple]

okay

[mighty kiln]

Each inclusion, say A∩B → A, induces a map of π1

[kind temple]

yes

[mighty kiln]

This colimit gives the π1 of the whole space

[kind temple]

so we only consider the inclusions from the intersection of two sets?

[mighty kiln]

A∩B∩C → A∩C too

All the inclusions

[kind temple]

okay

at least i have the right diagram. will think through it later. is there standard notation for this?

[tribal moss]

#alg-top-geo-top though.

[kind temple]

yea, i guess i was hoping to get a description for groups in general, so i thought to ask here

[tribal moss]

And a direct limit is a colimit.

[south patrol]

Yeah this came to mind too oof

[tribal edge]

Let $E$ be a finite-dimensional vector space and $|\cdot|$ a norm on $E$. Prove that the subgroup of isometries of $\mathrm{GL}(E)$ acts transitively on the unit sphere $S$ if and only if $|\cdot|$ is a Euclidean norm.

[cloud walrus]

Slomenist

[tribal edge]

hmm and that's equivalent to show that "... iff the norm |.| satisfies the parallelogram law"
by Fréchet-von Neumann-Jordan theorem

[rocky cloak]

Here's a fun argument:
Calling your group G, note that G is topologically closed in GL(E), and that it's bounded (wrt the operator norm for example), so G is a compact topological group, and E is a continuous finite dimensional representation.

Then there exists a G-invariant inner product on E.

Note that the ratio of the given norm and the one induced by the inner product is G-invariant, so if the action is transitive it is constant on the entire sphere, so the two must be the same (up to a scalar).

[coral spindle]

Yes

That’s how we define cosets

[velvet hull]

yes

[glad osprey]

A coset gH is by definition for a subgroup H, but you can define gH = {gx for x in H} for any subset H of a group. But as I said, the nice properties of cosets come from H being a subgroup

Yeah pretty much, but I don't think I've come across gH without H being a subgroup

[coral spindle]

It’s a bit gauche to pit people against each other like this

[glad osprey]

Lol, that ping looks weird, but it's in reference to what I wrote in #advanced-lounge

Another reason to keep discussion in a single channel

[coral spindle]

The endless benefits to asking in a single channel

That’s not really the issue. The problem is that I now don’t know if there’s some context I’ve missed, if the user you pinged was simply wrong about something, or if you just forgot to say something. It puts me on the spot in an unpleasant way

But we’re lingering on this too long. Please in the future ask in just one channel

[covert parcel]

i have a conjecture

A graph is supertough if for all |S|>1 where S is a subset of G, the number of components of G-S<|S|.
Conjecture: If a graph is Eulerian and supertough then it is Hamiltonian.

.idk if this is the right channel; this is graph theory

[next obsidian]

#combinatorial-structures

[covert parcel]

ty

[alpine plank]

Is there a characterization of rings R such that GL_2(R) is generated by elementary matrices?

[dull ginkgo]

I know this is true for R that are Euclidean domains but not necessarily true for PIDs so that’s pretty strong

[solar dock]

I guess this question may go here. Can any group be obtained by quotients, sums and products of only ordered groups? I assume such operations are not order preserving at all.

[south patrol]

Free groups are ordered groups, so yes

[solar dock]

So basically I am asking if any group can be decomposed into orderable bits.

Ah

Nice I guess

So for like a G module M we can express it as sums, quotients and products of modules on ordered groups? @south patrol

Where I take module as a functor from G to Ab.

[south patrol]

That seems like quite a different question and i'm not sure

[white oxide]

Is $\implies$ true since $h\bigl(f(a)\bigl) = h\bigl(f(a) \cdot 1\bigl) = h(a \cdot 1) = h\bigl(g(a) \cdot 1\bigl) = g(a) \cdot h(1) = g(a)$?

[cloud walrus]

okeyokay

[white oxide]

And the converse is true for [h(a \cdot b) = h(f(a) \cdot b) = h(f(a)) \cdot h(b) = g(a) \cdot h(b) = a \cdot h(b)]

[cloud walrus]

okeyokay

[velvet hull]

just to be clear, the statement is assuming that f and g are A algebra homomorphisms themselves

if f and g are just ring homomorphisms then their composition doesn not necessarily preserve the A algebra structure

[south patrol]

you say h(g(a).1) and that doesn't really make sense

i think it's a bit clearer if you decorate the dots or something

$h(f(a)) = h( f(a).1) = h( a \star 1_B) = a \star h(1_B) = g(a).h(1) = g(a).1 = g(a)$ where the $\star$ denotes using the $A$-module structures

[cloud walrus]

Prismatic Potato

[south patrol]

(and . is multiplication)

[south patrol]

but this is good

[solar dock]

Think this may reduce to a question about how the functor from Grp to the category of G-module categories behaves? Where the morphisms are taken naturally as functors?

If that functor preserves limits and colimits I think it should be fine?

If this is the case this is pretty no?

[south patrol]

Idk they seem like very different questions because you're now fixing the group right

[south patrol]

Oh

[solar dock]

Not even the full strength is needed just those quotients, sums and products that implicate ordered groups

Tho that may be sufficient for the full thing

[solar dock]

Ye still haven't checked tho will in a minute

[gusty thistle]

If L/K is a field extension do does there always exists a (non-unique) map between their algebraic closures that extends the inclusion? Is there an easy direct way to see this or is it a theorem?

[south patrol]

Fix an algebraic closure M/L. Now let M' = { elements of M algebraic over K}. This is an algebraic closure of K

But also more generally: if you have any field extension L/K and an algebraic closure L' of L, then for any algebraic extension of F/K you get a map F -> L'. You do a Zorn's lemma thing, but the important step is that whenever you add an algebraic element a to K, you can pick an element of L' which satisfies the minimal polynomial of a and hence get a map K(a) -> L'

You can then apply this, taking F your chosen algebraic closure of K

[south patrol]

This is basically the same as how you show that algebraic closures are unique, of course - and if you know that algebraic closures are unique then you can just use the line above (and compose with an isomorphism between your algebraic closure and the one i gave you)

[gusty thistle]

Oh lol

Thank you that makes a lot of sense

[tribal moss]

Hmm, would it be an answer to this that the free group with G's elements as generators maps into G, and that makes M itself a module over an ordered group?

[gusty thistle]

Ah that’s good

[alpine plank]

preimage of a point

you talk about fibers of functions at a point

would help if you shared the context you've seen the word used

what? fibres are subsets of the domain

[knotty badger]

ah right

so the fibers of $\varphi$ refer to the collection ${\varphi^{-1}(h) : h \in H}$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

of the domain

usually this is phrased in the language of equivalence classes

but i suppose fibers work too

[alpine plank]

they are equivalent, surjections, partitions and equivalence relations

[knotty badger]

well i don't think $\varphi$ is assumed to be surjective here

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[alpine plank]

Yeah i suppose they have to specify they aren't including the empty set

I think you are fine, usually these concepts are introduced using the language of cosets/equivalence relations as pseudo was saying.

[knotty badger]

ah so i guess it's like

usually fibers come up when you want to talk about "indexed families" of things

like an indexed family of sets $A_i$ for $i \in I$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

or an indexed family of vector spaces $T_p M$ for $p \in M$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

hm, not quite

for example, in topology, you often want to consider an indexed family of open subsets

well the idea is that you have some index set $I$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

for example you could have a sequence of sets $A_1, A_2, A_3, \dots$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

in this case the indexing set would be $\mathbb{N}$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[alpine plank]

Another example is the projection on to x axis R^2 \to R , the fibers are vertical lines (I guess this sort of thing is why they are called fibers)

[knotty badger]

mhm

[tardy hedge]

For A->B ring hom, q an ideal of B, why is A/q^c isomorphic to a subring of B/q? (This was in context of q is a primary ideal but idk if thats relevant here)

What I thought is that for A->B we can post compose with B->B/q to have a map A->B/q , and q^c is in the kernel of that, so we have a map A/q^c -> B/q, but I dont see why this would be injective

[knotty badger]

there's a nice story with how dependent type theory lets you talk about these

because dependent type theory is built to handle precisely these kinds of "indexed families", which they call "dependent families"

e.g. in maths "an indexed family with indexing set I" is something that might worry you a litte

[south patrol]

In the USA they don't teach about fibres. But fibers do come up

[knotty badger]

is this some kind of function from $I$ to, like, the set of all sets?

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

that would be worrying

luckily there's a different way to think about these which is reasonable

[velvet hull]

Suppose that f(a + q^c) = 0.
Then f(a) + q = 0

[south patrol]

First iso theorem

q^c is essentially by definition the kernel of A -> B/q

Well tbf you know that aha but yeah first iso

[tardy hedge]

I didnt for some reason i thought i could just say q^c was in the kernel

So i thought the kernel could be bigger

Oh q^c must contain the kernel obviously cause 0 is in q lol

[south patrol]

well also like

Preimage of 0 under projection B -> B/q is q

Preimage of q is q^c by definition

[tardy hedge]

Oh so if we write the map A->B/q like A->B->B/q, we can track the pre image of 0 over each step? I guess this is probably a basic property of how preimages work right

[south patrol]

Yes

To both

[tardy hedge]

Ty

[solar dock]

Can you give the explicit map? I don't think that quite works for what I'm trying to do but I would be happy to see it. Cause I'm looking to actually generalize the norm here to G modules on ordered groups et al. By inspiring myself from vector spaces and the natural way to extend the norm to complex values.

[tribal moss]

Uh ... the elements of G map to themselves, and by the universal property of the free group, saying where the generators map to is enough to define a group homomorphism uniquely.

[solar dock]

ha

hmmm

Haaaa I see

Hmmm but it doesn't necessarily preserve ordering on the free group no?

Not quite what I had in mind, its more like

uhhh if G is an ordered group I want to define a norm on it

And if G is not I want to define a norm on it such that it breaks down into pieces of ordered groups each a norm as the above

A norm on a G module for G an ordered group here being a definite positive map that satisfies homogeniety and triangle inequality

[solar dock]

So to be exact your idea is to use [\phi:<G>\rightarrow G][g\mapsto g] ?

[cloud walrus]

Bushy (Diff geo #1 Fan)

[tribal moss]

Basically I'm not sure what it is you're asking, so my proposal was less "here's a solution" than "perhaps I can understand you better if you describe why this is not a solution".

I presume that the G you start with is not an ordered group (because otherwise M would by your specification already be "a module over an ordered group").
So I'm trying to understand what exactly what it means to "express M as sums, quotients and products of modules on ordered groups" in this case -- that is, what does it mean to express a G-module as a sum/quotient/product of some modules over one or more different groups?

[white oxide]

What is meant by rational integer coefficients? Aren't by definition, the coefficients in f(Z), which are not necessarily integers

[velvet hull]

coefficients, in terms of Z-scalar multiplication

[white oxide]

Is this problem missing the assumption that M is flat? (Since the hint says to tensor with the exact sequence)

[velvet hull]

no, it holds in general

[white oxide]

Oh okay

[velvet hull]

well, the isomoprhism holds in general

let me try and figure out what the proof is trying to do

[white oxide]

yeah bc I don't know what information we can gain from tensoring if the resulting sequence isn't necessarily exact lol

I was just going to rawdog it with the first iso theorem

[rocky cloak]

Tensoring is always right exact, which is what you use here

[white oxide]

I guess we get $\mathfrak{a} \otimes M \to A \otimes M \to A/ \mathfrak{a} \otimes M \to 0$ is exact

[cloud walrus]

okeyokay

[white oxide]

Yeah I figured

[south patrol]

You know what is annoying

Right adjoints are left exact

[white oxide]

idgi

[knotty badger]

right cause rapl or something

type shit

[solar dock]

So what I mean when I say we can express a G module M as a sum/quotient/product of some modules over one or more different groups lets assume the family ((G_i){i\in I}) is the following idea that M is isomorphic to an object of (this is gonna be a little ugly) [<(G_i){i\in I}:Modules>] which I will take to be comprized uniquely of A-Modules where A is obtained as direct sums, products or (covariant) quotients of groups in the family ((G_i)_{i\in I})

[cloud walrus]

Bushy (Diff geo #1 Fan)

[topaz solar]

I mean, it’s not like every group has nice orderings definable

[tribal moss]

How should the G_i relate to your original G, if at all?

[topaz solar]

Like, what about an infinite p-group, that should be pretty screwed for having any orderable subgroups

[topaz solar]

And, further, how do you combine such modules into a G module?

I guess a G module should be a G_i module for subgroups, but idk how to really glue? I might just be stupid though

[solar dock]

That part is the mystery part that I don't know and will have to check

But there is some notion of isomorphism of modules and some notion of direct sum of modules with scalars from different groups

[rocky cloak]

Like if G' is an ordered group that surjects onto G, then M will also be a G'-module

Is that what you want or something else?

[solar dock]

Ah hmmm that does help

But that implies G can be ordered?

[rocky cloak]

Like a G-module is just a homomorphism G -> Aut(M), so you can compose with any map G' -> G

[rocky cloak]

It does not

[solar dock]

Hmmm waiy

wait

Oh

Yes yes okay thats also a nice tool

[rocky cloak]

It's still unclear to me what you actually want though...

[solar dock]

Okay so what I actually want is to be able to do something like:

[n:M\rightarrow G][1. n(0)=e][2.|a|n(v)=n(a*v)][3.n(v+w)\leq n(v)+n(w)] Where |a| works the same way as with the additive ordered group (\mathbb{R})

[cloud walrus]

Bushy (Diff geo #1 Fan)

[rocky cloak]

So should |a| be a scalar or an element of G?

[solar dock]

Its an element of G

And M is a G module here

Basically a very fucked up norm

[rocky cloak]

Right so |a| just gives the inverse of a if it's negative and a otherwise I'm assuming

[solar dock]

Yes

[rocky cloak]

But then you definitely need G to be ordered to begin with for this to make sense

[solar dock]

Yes

But I wish for the module decomposition so I can do something like

[rocky cloak]

2. Seems impossible though.

n(0) = n(a*0) =/= |a| n(0)

[solar dock]

[n:V\rightarrow \mathbb{C}][n(v)=n_{Re}(v)+in_{Im}(v)][|z|n(v)=n(z*v)] where V is a (\mathbb{C}) vector space

[cloud walrus]

Bushy (Diff geo #1 Fan)

[solar dock]

Ah shit

[rocky cloak]

Maybe n should be valued in the group algebra instead

[solar dock]

hmmm

[solar dock]

This is weird, it holds in usual vector spaces no?

Ah wait

thats multiplication

I see your poiny

point*

Hmm but that risks losing order no? @rocky cloak

[rocky cloak]

It would yeah

[solar dock]

Hmmm

Maybe there is some way to extend it?

[rocky cloak]

So I guess you would need to ditch / modify 3

[solar dock]

I'd hate to touch 3 since it allows for analysis

But rings are useful in their own right

Perhaps I need to modify 2.

Ah hmmm

Okay so I think I have a really shitty way of bypassing it by simply doing the following

[n:M\rightarrow {g\in G: 0\leq g}] definite positive and satisfies trig identity wrt ordering.

And we simply define a new operation

[cloud walrus]

Bushy (Diff geo #1 Fan)

[solar dock]

Upon that above via

[|.|*n:G\times M\rightarrow P\times P\rightarrow P] [(a,v)\mapsto(|a|,n(v))\mapsto|a|n(v)=n(av)]

[cloud walrus]

Bushy (Diff geo #1 Fan)

[solar dock]

Well, partial operation

And we will ask [4. |a|*n(v+w)\leq |a|*n(v)+|a|*n(w)]

[cloud walrus]

Bushy (Diff geo #1 Fan)

[solar dock]

@rocky cloak does this seem sane to you? Think that actually does what I want

Analysis preserved, group structure and order preserved and it works well for like the other examples I guess?

[rocky cloak]

I guess so. Don't really know what it's for, but I guess so

[solar dock]

It should be a good way to extend classical analysis onto G-modules?

[rocky cloak]

Well I don't know that much analysis, but stuff like unitary representations of Hilbert spaces doesn't accomplish that for you?

I don't really know that the classical results one would want to extend are

[solar dock]

I have some crappy examples like reals mod 2π

With like a metric given by angle

Or rather norm

on a sphere taken as a module of R mod 2π where the addition is going around the sphere

as you would think

[solar dock]

That feels like a weird opposite direction to me, I am not that well read on that

Well rather than opposite like a dual of sorts in a weird way

[rocky cloak]

So what is the module for this example?

[solar dock]

This

I guess should just be R?

[rocky cloak]

So R under addition is the group?

[solar dock]

Yes

Its trivial but like normally not permissible

no?

Since that isn't really a metric

I actually got here from diff geo

[rocky cloak]

I mean, it is a metric. It's just the standard metric on the circle

[solar dock]

Not quite

Its like a metric and an equivalence relation

No? Am I thinking wrong

Its angle mod 2π

[rocky cloak]

Well
d(x, y) = min{ n(x-y), n(y-x) }
would be the metric I guess

[solar dock]

Whats n here?

[rocky cloak]

The norm thingy you defined

[solar dock]

Huh ye that seems to work I guess?

But bleh I did it the easy way kinda?

In my head at least

I'll find non trivial examples

Imma sleep now

[rocky cloak]

Well, I think the example is non-trivial enough. Probably better to figure out what you would want to do with such examples

[solar dock]

I mean one idea I had was to do differentiation with this

As the diff geo fan that I am

Cause I think little o calculus carries

so I can do some weird variant of inf dim diff geo here

And also lipschitz stuff

They do be

existing

Geodesics and stuff I do be wındering if the usual Banach space results extends

[solar dock]

On this front actually its enough (and sufficient) to have a positive cone [P\subset G] Such that [1.\forall x,y\in P\implies xy, yx\in P][2. e\in P][3. P\cap P^{-1}={e}] Since you can prove [a\leq b \iff a^{-1}b\in P] is a total preorder if succh a set exists. And the positive cone [P:={g\in G: 0\leq g}] serves the exact purpose so they are equivalent. The thingy I wish to study is how to firstly ruin this, and how to basically obtain any group like this cause I got some stupid ways to glue Modules, all I really need is the ordering on like the components of some stupid decomposition as I've given above to extend the bellow norm like function of interest to G modules.

[cloud walrus]

Bushy (Diff geo #1 Fan)

[topaz solar]

Have you considered treating G as the value group in a valuation

[solar dock]

Hmmm I have to google that

[topaz solar]

That unit circle is U(1), unitary group of dim 1

[solar dock]

Oho I see

Yes

Wait

[topaz solar]

Anyhow, my original work is on analysis on groups and bad functions that are almost actions on them, so they’re not too far from reps or such

[topaz solar]

You can do a good bit of analysis on groups or modules w/ metrics as well

[solar dock]

Hmmm very interesting

[topaz solar]

As in, genuine metrics

[solar dock]

Can I yoink a paper?

[topaz solar]

Well, keywords are: harmonic analysis, geometric analysis, functional analysis, unitary groups, unitary representations

This isn’t quite as general as what you want, but it works well enough for locally compact or metric groups

Now, I do have some stuff I can show wrt partially ordered groups

Gimme a sec

[solar dock]

This would be a life saver actually, much appreciated

[topaz solar]

Here

Check references, other papers by this author, etc

For genuine metric-y groups or at least locally compact topological groups, look at (abstract) harmonic analysis

For merely metric without compactness conditions, I can possibly help a bit with specific questions if you need, though I’m no expert

[solar dock]

I've heard quite a bit about (abstract) harmonic analysis but I'm still kinda green(?) behind the ears as someone in between ug and masters so

Its like a hard area for me to break into

[solar dock]

I'll be taking advantage then

[topaz solar]

I’ve had a few schizo sketches about looking at some infinite dimensional unitary groups in a diff geo-esque sense, Banach-Lie kinda stuff

[solar dock]

I probably have just the tool for this since diffeology is made for this kinda shit

In some sense

Or its cousins

[topaz solar]

And I have some (unfortunately lackluster) scribbles on analysis on groups under “pseudo-compact” kinds of conditions, low regularity, and relations between these

[topaz solar]

Well I was specifically looking to make like a very rigid differential structure to invoke DE type results, since some rep theory results look a lot like ODE ones

[solar dock]

I do know Banach Manifolds and Frechet Manifolds as well as Lie Grps embed into its category maybe that helps? Since its real easy to manipulate for that kinda?

[solar dock]

Oh

[topaz solar]

Hey maybe that is the move, obviously my attacks hadn’t worked yet

So don’t just trust me when I say what it should be, if I actually knew I’d prove it

[solar dock]

https://tenor.com/view/conniptions-dark-souls-mob-psycho-rage-gif-27403059

Fr fr

[topaz solar]

Now, with regards to your positive cone thing

Consider the valued fields

There’s a valuation group, but that’s not the norm

[solar dock]

I'll try my stuff if I get far enough ;-;

[topaz solar]

So, perhaps you should look at valuations rather than norms

[solar dock]

Do you have like a book for this stuff?

[topaz solar]

(Which are a logarithm away, morally)

[topaz solar]

I dunno dawg, Atiyah-Macdonald? It has some valuation ring stuff

[solar dock]

Thats a statement and a (amorally)

[solar dock]

I'll dig about probs

[topaz solar]

Ershov has some fancy multi-valued field things, but it’s a touch too logic-y for your purposes

[solar dock]

I actually ripped that off from convex analyais

[topaz solar]

Anyway, valuations are a thing in commutative algebra, valuation rings are kinda a move

[solar dock]

I'll look into valuation rings, had the wiki open tho I really do need to sleep

,time

[cloud walrus]

The current time for godofbushes is 03:08 AM (+03) on Sat, 12/07/2025.

[topaz solar]

Check the paper I sent an image of for extra ordered group data, but linear orders on groups are tricky

Go to sleep, consider what exactly you’re trying to do

[solar dock]

I'll be reading it if my smooth brain can comprehend

[topaz solar]

I or jagr or someone can help once you have more direction on how you’re trying to proceed

More algebraically or analytically, or maybe it really fits better in a weird condensed setting idk

[solar dock]

I'll be burdening you guys with my weird questions here then.

[solar dock]

I'll see where it grows once I have some more examples

hopefully

[topaz solar]

May all be analytic

[solar dock]

I at least want to look at geodesic stuff

Possibly using my favorite reference of "metric spaces of non positive curvature"

I'll cleanly write what I learned today first tho probs.

And possibly get a decomposition theorem that won't make me feel shame when trying to present it to @tribal moss

[tardy hedge]

For R noetherian M an R-module, and ideals I and J such that rad(I) = rad(J), why if I^pm= 0 for some p means J^qm = 0 for some q? Can i just have a hint for how to show it?

[velvet hull]

is m an element of M?

[velvet hull]

anyways, in Noetherian rings, for every ideal I in R there exists an n such that sqrt(I)^n \subseteq I

[tardy hedge]

Is this a basic fact that would be found in atiyah macdonald or smth?

Probably need to look at generators right

[velvet hull]

it might be in there, I haven't gotten to the noetherian section yet

but it's a pretty simple lemma

[south patrol]

I would say this is a good exercise

[tardy hedge]

Ya im gonna do it

[south patrol]

I guess you want "is contained in I"

[velvet hull]

oh, right

[south patrol]

(e.g. consider (x,y^2) contained in k[x,y] - obviously no power of (x,y) is (x,y^2))

[tardy hedge]

Bc x is not in any power of (x,y) right lol

[barren sierra]

Well x is in the first power

(and only the first power)

But y^2 is in second power only

[cobalt heath]

Maybe I post here, this suits more here since ring Chinese Remainder theorem

Let's think of DFT on power of 2 dimension.
So DFT matrix is factorized under bitreversal permutation, which is the basis of Cooley-Tukey method
DFT also corresponds to the Chinese Remainder theorem of some ring, so the factorization should correspond to some intermediate steps of the CRT.
The question is, how exactly?

It feels grating to follow through how permutation goes around

[velvet hull]

in terms of representation theory, there are exactly n irreducible representations of the cyclic group on n elements in C (namely you just pick an nth root of unity as your generator).
Hence by the Schur orthogonality relations, this implies that every representation of the cyclic group on n elements in C^k can be decomposed into a direct sum of the irreducible representations, which corresponds to the DFT

not particularily sure how useful this version of the DFT is for you, but there's some keywords for you that may help out

see https://en.wikipedia.org/wiki/Discrete_Fourier_transform#Representation_theory

[cobalt heath]

Yeah sadly, I don't think this would help me track where each element goes via the DFT

To put some detail, the factorization is of form
DFT = V W P where (IIRC) W is a block matrix, and P is a permutation.

I am thinking of how W corresponds to an intermediate CRT isomorphism,
when you consider DFT as
$\bC[X] / (X^M - 1) \simeq \bC^M$

[cloud walrus]

Absta

[tribal moss]

I mostly lost track of the discussion afterwards, but this is actually impossible.
Since a(bv)=(ab)v for a group action, this implies |a|·|b| = |ab|. (That looks innocuous; it is true for R after all. But the absolute value for R negates with respect to a different operation than the multiplication it preserves).
If a is any positive group element, presumably a^-1 should be negative, so |a^-1|=a, and we have a·a = |a|·|a^-1| = |a·a^-1| = |e| = e.
But that means a = a^-1, which would have to be both positive and negative, which is absurd.

[solar dock]

I was trying to define a new operation there actually and not use the group operation but that has its own problems it seems

Actually its a constraint on the function n

[tribal moss]

Oh, I'll just shut up then. :-)

[solar dock]

Sharp also pointed your exact point out and I wad like ;-;

And then I messed with homogeneity and axiome 3

So basically its become something like

[n:M\rightarrow G][1.n(v)=0\iff v=0] [2. \forall a\in G,~|a|*n(v+w)\leq |a|*n(v)+|a|*n(w)] Where (|a|n(v)=n(av))

[cloud walrus]

Bushy (Diff geo #1 Fan)

[solar dock]

With a*v the action of G on M

And |a|*n(v) is made into an operation on the positive cone (P:={g\in G: 0\leq g}) and the absolute value is defined via the operation and order on G

[cloud walrus]

Bushy (Diff geo #1 Fan)

[solar dock]

@tribal moss So ye that was my way of "solving" that problem but I'll have to come up with more examples than just U(1) with norm in R mod 2π via angle on the 1-sphere from its action on (0,1)

[tribal moss]

What does |a| mean now?

Or is |a| = n(a*v) n(v)^-1 the definition of |a|? In that case, don't you need additional axioms to ensure that RHS doesn't depend on a choice of v?

Also in (1), is 0 an element of G now? That would suggest G is written additively, so what is * between group elements in (2) and |a|*n(v)?

Ah, I think I perhaps I get it -- the whole |a|*n(v)| is a single symbol with two inputs a and v?
If so, it would be less confusing just to unfold it in the statement of (2).

But then 2 just says

n(a*(v+w)) <= n(a*v) + n(a*w)
which follows from n(v+w) <= n(v) + n(w) and also implies it by setting a=e.

Do you have a main example in mind that you're trying to model?

[white oxide]

If G is a free abelian group, does it have no elements of finite order? For if a was a nontrivial element of finite order, we would have a nontrivial element of Z x ... x Z of finite order

[weary frost]

Yes

[white oxide]

What is the purpose of the hint in c? Is it that we have $n_1 a_1 + \dots + n_m a_m - \frac{1}{2} a_{\alpha} = 0$, where $\frac{1}{2} \neq 0$, contradicting the ${a_{\alpha}}$ being a basis for $\mathbb{Q}$?

[cloud walrus]

okeyokay

[weary frost]

If you express 1/2 a_alpha in the basis you add it to itself and have a_alpha in the basis and go from there...

[white oxide]

What's wrong with my proof tho

[weary frost]

1/2 is not in Z lol

If you multiply by 2 it's fine

[white oxide]

Oh ya

Thanks

[tardy hedge]

wheres that from?

[white oxide]

Munkres lol

[tardy hedge]

Bruh

Their alg top section i guess?

[white oxide]

yeah I couldn't understand Rotman's proof of Van Kampen so I'm supplementing with Munkres lol

[tardy hedge]

nice

i dont know much any alg top yet

topology is pretty difficult for me

[regal flume]

how do I find the direct limit of the groups Z^2 with the homomorphism ((1,1),(1,0))?

[rocky cloak]

You just listed one group, do you mean two copies of the same group or a system with a loop?

In either case the system wouldn't be directed, so maybe you just mean colimit...?

[regal flume]

all of the groups are Z2, the indexing set is N, the homomorphism is the matrix, to get from one group to another who's indexes differ by one.

[regal flume]

hopefully that makes more sense?

directed colimit, colimit direct limit all mean the same thing to me (I think it shouldn't lol)

[rocky cloak]

Yeah, the homomorphism is an isomorphism so the direct limit is just Z^2 again

[rocky cloak]

"direct limit" usually means "colimit of directed system"

[regal flume]

Is there a reason why one would write it as $\mathbb{Z} \times \tau \mathbb{Z}$ instead?

[cloud walrus]

Xetrov

[rocky cloak]

Maybe for ease of notation, writing elements as
a + tau b...

idk, is tau something specific?

[regal flume]

eigenvalue of matrix?

your pfp is so cute

[rocky cloak]

The eigenvalues of the matrix should be the golden ratio and its conjugate.

I wouldn't say it's that natural for it to appear there, or for it to be notated by tau

[regal flume]

tau happens to be the golden ratio lol

[rocky cloak]

Okay, but what about the other eigenvalue? Why wouldn't that appear?

[regal flume]

That's what I'm thinking

i also have no justification

because I'm just reading

[rocky cloak]

You can try to post what you're reading if you want, maybe there's some context...

Either way it's just ZxZ anyway, so just notation I guess

[regal flume]

when do things like Z[1/2] appear in colimits?

can't seem to find an example

[regal flume]

I'll show later

[wraith cargo]

wdym "when they appear"

[regal flume]

direct limits of groups

i recall seeing that times X was the direct limit of some groups Z2

I'll need to look at the book again anyhow

[rocky cloak]

It is the direct limit of
Z -2-> Z -2-> ...

[regal flume]

x2?

[rocky cloak]

Yes -2-> is multiplication by 2

[regal flume]

R/Z is S1 right?

why am i not just identifying all of the integers with each other, so it looks like a dyadic solenoid with one side squished?

[tribal moss]

It R/Z means S1, it's when the / means a group quotient.

As a topological quotient space, it becomes a different space (called something like "flower with infinite petals"; I don't recall the exact phrasing).

Two different operations, notated with the same symbol. You're supposed to be able to guess from context which of them is meant.

[regal flume]

something like a wedge sum

[dull ginkgo]

What’s the relation here

[regal flume]

i can't remember the name either

[dull ginkgo]

Or is it the pushforward (or pullback I forget) of the congruence relation?

Wouldn’t it be fine as a topological group

[regal flume]

this makes sense cheers

[tribal moss]

If X is a topological space and A some subset of it, the notation X/A often stands for quotienting by the equivalence relation "a~b iff a=b or a and b are both in A".

[dull ginkgo]

I see

[regal flume]

i did topology before I did group theory (and only doing it because my research requires me to learn it)

not a massive fan of group theory at all

actually that's a lie

I find it reasonably elegant

but the nitty gritty details get infuriating to work with

[tardy hedge]

In the case where sqrt(I) = (f1,f2) can I say sqrt(I)^2 = (f1^2, f1f2, f2^2), sqrt(I)^3 = (f1^3, f1^2f2, f1f2^2, f2^3), etc? Then sqrt(I)^i = (f1^i, f1^(i-1)f2, f1f2^(i-1), f2^i), so choose i to be one more than max of n1,n2 that makes f1^n1 in I and f2^n2 in I?

I guess the part im unsure of is generators on the product of ideals

[south patrol]

Ye exactly

Wait uhh

[tardy hedge]

did I do some generators wrong

[south patrol]

Nah it's good

[tardy hedge]

in the case with sqrt(I) = (f1, f2, f3) would you choose i to be like, two more than the max of n1,n2,n3 that make fj^nj in I?

[dull ginkgo]

Group presentation NIGHTMARE NIGHTMARE NIGHTMARE

[regal flume]

do not even get me started

WHAT DO YOU MEAN THE GENERATORS OF THE FUNDAMENTAL GROUP FORM A FREE GROUP

WHAT DO YOU MEAN THERE'S LIKE TEN DIFFERENT WAYS TO COMPUTE A HOMOLOGY

THERE'S A BASIS FOR THE CW COMPLEX? WHERE?!

why did I even choose this as my project

that's just me tweaking out dw

[long geyser]

can someone steelman the case for writing Mod_R to mean left R modules and R_Mod to mean right R modules

because it looks crazy to me

[dull ginkgo]

Easier than R^op mod

After all a left R module is a right R^op module

[long geyser]

huh? I'm asking why not use R_Mod to mean left R modules

[dull ginkgo]

Sometimes we have cases where we are working with left and right R modules simultaneously

[long geyser]

and?

[rocky cloak]

I guess default module being left module and Mod_R being the default notation.

The you the the throw in {}_RMod as an afterthought

I usually write $\operatorname{Mod}R$ for the category of left modules.

[cloud walrus]

jagr2808

[rocky cloak]

But then I just write $\operatorname{Mod}R^{\rm{op}}$ for right modules

[karmic moat]

fwiw rotman uses {}_RMod for left R-modules lol

[cloud walrus]

jagr2808

[karmic moat]

and Mod_R for right R-modules

[long geyser]

Yeah that makes sense, I first thought it would make right R modules awkward, but yeah then I realized once you establish this isomorphism Mod_R^op is perfectly fine notation for right R modules

[long geyser]

I've seen this notation every other text, which is why this surprised me

[rocky cloak]

I'm also a fan of R-Mod vs Mod-R, but this subscript on the left is weird to me

[karmic moat]

yeah typesetting subscript on the left is annoying

[rocky cloak]

Well, you would make a macro for it hopefully

[karmic moat]

oh yeah that's a good idea

[wraith cargo]

huh

what's K_1 and K_2?

lol

er yes

are you trying to show that K/H is a normal subgroup of G/H?

yeah this is true

[tribal moss]

How about Mod^R for right modules?