I am beginning to think the problem isn't trivial, because in the quadratic case we used the fact that, if D is always a quadratic residue, then it is a square.
#groups-rings-fields
cyan marten
oblique river
latent anvil
That makes sense. There's probably no high level way to see it
But I'm still curious about the cubic case, even if there isn't a nice proof...
Ie if a cubic polynomial has a root mod each prime it has a root
cyan marten
It probably has to do with Hasse's principle.
golden pasture
isnt the goal literally to prove that here
actly if a polynomial has a linear factor mod all p
does it necessarily have a linear factor over Z
delicate bloom
good question, that's what I was trying to work out, how can we associate all the separate linear factors as really being just one?
golden pasture
yea somehow there needs to exist a way of figuring like the linear factors mod all p are the same linear factor hmm
cyan marten
does it necessarily have a linear factor over Z
Yeah, that's what I proved. If it always has a root then its discriminant is always a quadratic residue, which is impossible unless the number itself is a square.
vestal snow
Since Affine varieties are the closed sets in their topology, wouldn't it make sense to define quasi-affine varieties as intersection of closed sets with X?
sturdy marsh
they arent the only closed sets
vestal snow
My bad, I meant that affine varieties form a closed basis in their topology, so shouldn't quasi-affine varieties be defined to be the closed basis in the subspace topology
instead, they are the open basis in the subspace topology
sturdy marsh
I am not sure I understand your question, that is the definition of a quasi-affine variety (at least the one I know)
vestal snow
I am asking why that definition
as opposed to defining it as the intersection with a closed set
sturdy marsh
idk, because you can prove a bunch of theorems for open subsets of varieties?
vestal snow
Hmm I'll take your word for it
sturdy marsh
as opposed to defining it as the intersection with a closed set
also, the intersection of two closed subsets in A^n is pretty complicated
the intersection of varieties might not look like a variety at all
the standard example is the intersection of y=x^2 with the x-axis
you get nilpotents in the structure sheaf
vestal snow
isn't the intersection of any number of varieties a variety?
sturdy marsh
no
vestal snow
Wait
sturdy marsh
there's the counterexample
sturdy marsh
integral, separated, locally of finite type
but as radiateur-man pointed out the other day, some books don't require varieties to be irreducible
vestal snow
I haven't gotten that far yet, so I was guessing what a variety was based on the definition of an affine variety
I know affine varieties need to be irreducible
sturdy marsh
oh wait, you're doing it in the classical sense
yeah that notion of quasi-affine is just useful
🤷
vestal snow
oh wait, you're doing it in the classical sense
Me too dumb for sheaves
At least right now
sturdy marsh
nah sheaves make stuff easier
it's a matter of getting used to the scheme formalism
vestal snow
I went through the first few sections of Qing Liu, but I lost motivation
So I'm doing a little bit of classical algebraic geometry from some guy's notes
and then hopefully that motivates sheaves, schemes, bundles etc.
sturdy marsh
maybe
bundles will not be too hard to motivate ig
and sheaves (or some of them) will play the same role that bundles play
vestal snow
What book did you use for AG?
vestal snow
I've been told Vakil is pretty good for self-learning
Though its like 800 pages long IIRC
sturdy marsh
I didnt read all of it lol
he even marks stuff that you should skip while reading it for the first time
vestal snow
Oh that's nice
I'm mainly learning it for ANT, so I would probably end up reading Qing Liu once I'm done with the basics from Vakil
sturdy marsh
if you're into arithmetic stuff then schemes will not be hard to motivate
vestal snow
How long did it take you to do the essential parts from Vakil?
sturdy marsh
I didn't self-study
took a 2 quarter course
which covers pretty much all the essential parts ig
vestal snow
Considering AG is usually a year long course
sturdy marsh
depends on what you mean by a first course in AG
vestal snow
Chapters 2, 3, and 4 from Hartshorne are typically taught over two semester right?
But I guess it depends on your school
sturdy marsh
dunno
we didnt do all of 4
or all of 3
we skipped the stuff on formal schemes as well
vestal snow
I'm probably gonna try and do as much as I can over the summer
sturdy marsh
and also, it took me way longer than 2 quarters to digest all of what we did
a lot of the stuff in the course is starting to make sense now
vestal snow
Did you take it as a grad or and undergrad?
sturdy marsh
undergrad
vestal snow
Ah okay. My school only offers it as a grad course, so maybe that's why its longer
sturdy marsh
oh you meant the course
the course was a grad course
we dont have an undergrad AG course either
sometimes it's offered as a variable topic course
vestal snow
@sturdy marsh Do you have a list of the important sections in Vakil?
I couldn't find it in the preface
sturdy marsh
nope, but vakil tells you if something should/may be skipped
vestal snow
like in text?
sturdy marsh
yup
vestal snow
Alright
sturdy marsh
right before the section
vestal snow
Thanks
sturdy marsh
also, you should skip if there's a star iirc
or two stars
he mentions it in the intro
burnt pond
For an abelian group T, why isnt it possible to have
a nontrivial extension of two different one-dimensional representation, i.e
if 0->V->U->W->0 is an exact sequence of smooth T-representations and V,W are one dimensional, i.e. characters of T, why does V=/W imply that U is the direct sum of V and W?
context: they use this as an argument in case 1 to apply schurs lemma on each component to get the result
vestal snow
Can someone explain why the notation for a coordinate ring over an affine variety is the same as the ring of regular functions over a quasi-affine variety?
next obsidian
idk haha
vestal snow
if the field is algebraically closed, then its obvious because any polynomial (function over X) in n variables will have at least one solution
Do you remember the main idea of the proof?
next obsidian
It's in Hartshorne
I don't think it's that simple
It's not even immediately obvious that global regular functions are simply rational functions
vestal snow
Isn't that how they're defined?
vestal snow
next obsidian
yes
but this is local
you can have a function which has a description on two opens
which agree on the intersection
vestal snow
Oh so f might not be g/h at points outside of U?
next obsidian
but you cannot define it on the entire set as a single rational function
yes
I can't think of an example off the top of my head
vestal snow
It's fine. I just wanted to know if it was true or not
How long did it take you to do chapter 1 from Harthshorne?
vestal snow
I might switch to that because these notes are kinda bad in defining things
next obsidian
then switched to something else, Gathmann's notes
fading wagon
Looks very algebraic topology?
next obsidian
since we hated Hartshorne's
This is algebraic geometry
anyway, we did that briefly
carmine fossil
Is hartshorne horrible?
next obsidian
then just jumped to chapter II
fading wagon
This is algebraic geometry
Oh I forgot what the field is called.
next obsidian
Hartshorne is terribood
Idk
I hate it but also spent like fucking hundreds of hours doing the exercises
which I'm now 50% done with! (at least from chapters II and III)
vestal snow
I don't think I've ever 100% any book
next obsidian
III.5 and III.7 because it needed II.7
and I think II.8
and III.9-12 because of similar stuff / just didn't feel like it
vestal snow
Is it worth doing all of the exercises?
next obsidian
You need to
or at least most of them
there's some like computational examples that don't serve any theoretical purpose
but are probably still worthwhile doing so you can actually practice computing sutff
Hartshorne honeslty is only worth it for the exercises like
at this point there's other sources to learn definitions and see proofs
(altho I couldn't find some of the thms in other sources but idk maybe I'm bad)
Also FWIW
I did III.1-4, III.6, III.8 in like...
legit 2 weeks?
The things are very short
and few exercises
it's just cohomology so if you know some homological algebra it's really quick to get through
There's still some exercises from those sections but
once you hit II.5
if you ever get disheartened just jump to chapter III
Really all you need are like II.1-3 and II.5 to do most of chapter III
you just have to know definitions, some basic results about schemes
and then be chummy with sheaves of modules
vestal snow
I'm so confused on which book I should use
next obsidian
lol
Just like
use them all
tbh
if you want to learn scheme theory
I recommend checking out Algebraic Geometry I by Goertz and Wedhorn
vestal snow
Vakil seems to be the most accessible, but he does everything over algebraically closed fields
next obsidian
it's really big, and doesn't touch cohomology
but for actual scheme theory proper it's very thorough
and spells out a lot of the stuff Harthshorne won't, and was made in 2010
so there's a lot of benefit to it being modern
Like, at the end there's really handy charts for what properties imply what
vestal snow
Then there's Qing Liu which is apparently the best for ANT stuff
vestal snow
Wait Vakil doesn't have cohomology at all?
next obsidian
and just roll with that and just have access to other stuff for when that doesn't make sense
Vakil does
He pushes it really far back tho
I was talking about Algebraic Geometry I
next obsidian
uh
Technically I started summer of 2019 but calling that a start would be so trash
I barely knew the definition
I started in like winter last year but didn't really do much, I just sat in class
I really started to do exercises like in the spring of this year making it like... 8 months now?
but I took hella breaks
vestal snow
That...
next obsidian
Idk, I've put in hundreds of hours now I think
vestal snow
doesn't sound so bad actually
next obsidian
Like I made for a big push to hit 111/222 by Jan 1st
and like I took logs
on 12/17 I had 85 and on 12/26 I hit 111
so in that 10 day span I did 26
but I was going at it like 12 hours a day
Also I jumped forward into cohomology which had a lot of like low hanging fruit in terms of exercises
idk, it really just varies
I think if you just say I'm gonna do 1 exercise a day or some shit
or like...
4 a week
just set some goal
and don't stress about actually hitting it
vestal snow
I still can't believe my advisor originally wanted me to learn 75% of AG in the last two week of summer
next obsidian
haha yeah idk what that was about
Don't do what I did
like
don't linearly solve exercises
it's fucking stupid
vestal snow
wdym linearly?
next obsidian
I just
in Spring
I said fuck this I'm gonna start fresh
just wnet back to II.1
and legit went down the list and solved every exercise in order
only skipping ones which required me to know varieties
and did that through II.1, II.2, and II.3
and for most of II.4
it fucking sucks, and is inefficient tbh
just do like 1/2 of each section
vestal snow
That's what I was planning to do
next obsidian
and if you get stuck on some stuff later maybe go back and see if any exercise seems useful
if you need an opinion on what's important you cn shoot me a dm
This summer I want to make a website
vestal snow
IMO 50% is where the sweet spot is
next obsidian
About a lot of AG stuff
and include like, my opinions on what exercises you need to do in order to move through
like here's the stuff you'll use a lot, and you'll need
here's some things not commonly talked about or hard to learn which are really helpful
just like, shit you'd need to learn from someone normally becasue a textbook won't tell you
Anyway, I would not recommend doing what I had done
when you come back to it later you'll be way more comfortable with the stuff
and it'll be easier
Also if you go linearly you'll get stuck on monster problems
which end up not fucking mattering at all really
or which do in the grand scheme of things matter but not right now
II.3.15 is an example of that
that problem is so fucked up, it's an entire section of Vakil
Or Chow's lemma?
Like wtf
vestal snow
Lmao
I think someone on here said something like
Hartshorne is a good way to test your mental fortitude
AG, not so much
vestal snow
you're in an abusive relationship with it
next obsidian
you need a hard goal to shoot for I think
haha
Like my goal is by the end of this school year to have done all the exercises in II and III
I really don't think it will happen, but I think I can hit like 90%+
vestal snow
what are your plans after you're done with it?
next obsidian
Drink
vestal snow
Very well
vestal snow
algebraic topology
next obsidian
Anyway, point is after I set an actual goal I've kept way more motivation for it
and I think my approach has improved
I'm okay just skipping a problem
I used to spend like 3 days stuck on something spending like 8 hours just trying shit
and then I would say fuck it and quit for a month
vestal snow
and then I would say fuck it and quit for a month
Wait is that not how you're supposed to study math?
next obsidian
I heard that it's not the best for long-term results haha
next obsidian
lmao I forgot that's a thing
fierce perch
I used to spend like 3 days stuck on something spending like 8 hours just trying...
This is me
How do you know when to back off
next obsidian
When I get frustrated
or if I seem to have just straight up no idea
sometimes just doing other stuff and coming back I have a better idea
I tend to stick with it as long as I have ideas that seem like they could work or I'm still making progress
when I just sit around for a while just writing down the same thing over and over again because I have all these weird ideas and it ends up always being me just saying the same thing in a different way I just try to move on since I think I'm not really being productive
bronze trench
help
I've reduced a problem to describing all semidirect products of P by Q where |P|=p^2 and |Q|=q
I'm missing a bunch of cases lol
but one that's bothering me rn is this
carmine fossil
Is p<q?
bronze trench
no
I saw online that that's a bit easier
this time p>q lol
in the case P=cyclic of order p^2 and q divides p-1 I'm basically done but I need to count isomorphism classes
and I'm convinced there's only 2 actually, despite there being many different homomporphisms to realise the semidirect product
so I have this but I'm really convinced of this (if I can't solve it I'm leaving it for the professor to read lol)
but I'd prefer to actually prove this or realise it's false and do it right 😂
oops typo there somewhere. I mean q=n(p-1)
carmine fossil
Take 2 cases:Case 1:The group with order q is the normal subgroup
Case 2:The group with order p^2 is the normal subgroup
For case 1: you will end up with the trivial semidirect product/direct product as the only possibility
bronze trench
The first part is to show case 2 always holds
so case 1 is a particular case of 2 when we choose to make it a regular direct product I guess
I have this written down so I in fact reduced it to the case where the semidirect products of P by Q, in that order
this is the actual question btw
carmine fossil
Yea,There are exactly 2 semidirect products
cyan marten
in the case P=cyclic of order p^2 and q divides p-1 I'm basically done but I nee...
What if P is elementary abelian? Or are you specifically interested in the cyclic case?
cyan marten
and I'm convinced there's only 2 actually, despite there being many different ho...
Two homomorphisms give the same semidirect product iff their images are conjugate iirc. I know for sure it's true in the case where the image is a cyclic group.
bronze trench
What if P is elementary abelian? Or are you specifically interested in the cycli...
I haven't tacked the other case, I need to look at GL(2,p) and that scares me lol
bronze trench
Two homomorphisms give the same semidirect product iff their images are conjugat...
oh I might have a theorem or lemma somewhere saying that in the class notes, if so I'll try to use it to see they're all isomoprphic
cyan marten
I haven't tacked the other case, I need to look at GL(2,p) and that scares me lo...
That's what we discussed the other day, and I said it wasn't easy. You need to classify all the matrices of order q up to conjugacy.
In Dummit and Foote this is given as an exercise for q = 2 and there are eight matrices (up to conjugacy), namely
[*1 0 ; 0 *1]
where * can be + or - 1, and ";" separates rows.
bronze trench
oh ok, I'll look into that, thanks!
frank haven
If i am trying to decompose these product cycles
can someone tell me... is the leftmost one just (15432)?
or am i misinterpreting this whole question
sturdy marsh
yup, you got the leftmost one right
fading wagon
yeah just figure where 1, 2, 3, 4, 5 goes to
frank haven
ok sick, this stuff is weird to wrap my head around. but its pretty interesting stuff
fading wagon
also depends on your notation
frank haven
yeah, this is in artin. I figure that he just omitted the circle for product
sturdy marsh
im assuming the convention is right to left
fading wagon
yeah, but some books use left to right (shock!)
frank haven
wait thats so weird, i always see right to left for compositions
chilly ocean
can someone recommend a book/note of exercises for abstract algebra with answer ? the goal is to ensure achieving full mark in exams
chilly ocean
except http://homepage.math.uiowa.edu/~goodman/algebrabook.dir/book.2.6.pdf
and A First Course in Abstract Algebra, 7th Edition John Fraleigh
quaint ivy
maybe Dummit & Foote and just dig for solution manuals online? (some people have made these for the first few chapters, up to field theory or so.) You can also ask here or on Math Stack Exchange
old lava
(if you pay for slader, then all of d&f solutions are on there, I believe)
stoic rose
Does anyone has an idea how to do this? This is from Milne's algebraic groups
chilly ocean
maybe Dummit & Foote and just dig for solution manuals online? (some people have...
cool thanks, but exercises are quite basic.
proofs in this note are a bit interesting : http://people.math.harvard.edu/~ctm/home/text/class/harvard/55a/08/html/home/course/course.pdf
sturdy marsh
Does anyone has an idea how to do this? This is from Milne's algebraic groups
what's mu_p?
golden pasture
pth roots of unity im guessing?
lavish gale
whats lie alzerbra in laymans term
latent anvil
I could give a simplified explanation but tbh it would be a lie
chilly ocean
thing that goes [x, x] = 0 and sum_cyclic [x,[y,z]] = 0 
(maybe you need [x,y] + [y, x] = 0 instead of just [x, x] = 0)
stoic rose
what's mu_p?
p^th roots of unity
chilly ocean
whats lie alzerbra in laymans term
A Lie algebra is a linear approximation of a Lie group. If a Lie group was a smooth curve, a Lie algebra is line tangent to the curve
stoic rose
This is one motivation but Lie algebras are still interesting independently of Lie groups.
wind steeple
I saw on a paper that extensions 0->F->E->B->0 are classified by 2-morphisms B->AUT(F). But here F is abelian so 2-morphisms are juste 1-morphisms lol
Here : https://arxiv.org/pdf/math/0608420.pdf in the cohomology section
sour plume
The setting of this question is a very classical problem about group cohomology, actually! The way you can generally begin is by writing your group $G$, as a set, as $\mu_p \times \mathbb{Z}/p \mathbb{Z}$ with the maps $\mu_p \to G$ and $G \to \mathbb{Z}/p \mathbb{Z}$ being the "canonical" maps $x \mapsto (x,1)$ and $(x,z) \mapsto z$
cloud walrusBOT
Lartomato
wind steeple
I guess the article isnt that technical, it gives the idea
You can read the proof in the article you dont need any background
sour plume
You can do this because, since you have an extension of groups it holds that $G / \mu_p \cong \mathbb{Z}/p \mathbb{Z}$, so at least in terms of cardinality, any group $G$ which fulfils this must be just as big as the cartesian product of the two groups
cloud walrusBOT
Lartomato
sour plume
And then the question sort of becomes: What multiplications can you put on $G$ that respect the maps in your extension
cloud walrusBOT
Lartomato
sour plume
Looking at Zak's article, it goes a little bit overboard, but for a more general framwork for this question, the right kinda term is looking up "group cohomology"
stoic rose
I know a bit of group cohomology but I don't really know how it extends to group schemes 
Also for abelian groups I don't think any group cohomology is needed to classify extensions, they are just classified by Ext^1
But I don't know if we can define stuff like Ext^1 for commutative group schemes
sour plume
Hm, I know nothing about schemes either, I was kinda hoping that we can just work with \mu_p as some explicit group
i.e. roots of unity or so
But group cohomology is actually the same as applying the Ext functor!
stoic rose
Hum I just found out that commutative algebraic groups over k form an abelian category, so I guess we could indeed compute Ext groups
In general you can't describe group schemes with a single explicit group, they are really functors from algebras to groups
sour plume
oh no
Hm, so do I understand this extension of group schemes as "for every algebra that I insert into my functor, I have an extension of groups"?
stoic rose
And theses choices of extensions have to be functorial
And the resulting functor have to be representable by a scheme
sour plume
Ah, I see. Then maybe trying a "pointwise" approach and understanding the extensions of groups first would be a good first step nonetheless
stoic rose
Unfortunately in char p $\mu_p(R)$ is trivial when R is a field extension of k (or more generally has no nilpotents) and is quite complicated in other cases
cloud walrusBOT
radiateur-man
sour plume
Perhaps you don't need an explicit description of \mu_p(R) in the end, the main property that you need to work with seems to just be the "cyclicity" of all the involved groups. But eh, maybe there's something special and clever about group schemes that I'm not aware of which you could use instead
stoic rose
What do you mean by "cyclicity"? mu_p(R) is not cyclic
sour plume
Yeaah I was being a bit imprecise there, more like the "periodicity" that r^p = 1 for all the involved things
But yeah, not sure
stoic rose
Hum maybe
I think computing Ext directly in the category of commutative algebraic group could work actually, I'll try this
sour plume
Good luck!
frank haven
In a proof, say i am showing that two groups are actually the same group. Is it enough to show that each group is a subgroup of the other like with traditional sets?
carmine fossil
Yes
frank haven
ok cool
stoic rose
Hum, turns out the constant group scheme Z is not even projective in the category of affine group schemes over k
I'm not even sure if the category has enough projectives
So it'll be probably quite hard to compute the Ext groups
Also @sour plume I realized that the functor G -> G(R) is only left exact, so G(R) need not be an extension of mu_p(R) by Z/pZ(R)
sour plume
oh nooo
I've actually been looking for some math thing to study, maybe I should read about group schemes, sounds fun
stoic rose
They are quite cool indeed
sour plume
Are you studying them for any particular reason or is it just part of a course?
stoic rose
I'm just studying them for fun 
I'm currently learning a lot of AG stuff and groups schemes are quite cool
This exercise is giving me big headaches though 😅
next obsidian
It wouldn't surprise me if there weren't enough projectives
next obsidian
So, if I have modules $M,N$ and linear topologies on them given by ${M_\lambda}$ and ${N_\gamma}$, and a linear, continuous map $f\colon M\to N$, we get an induced map on completions making the diagram
$$\begin{tikzcd}
M\arrow{r}{f} \arrow{d} & N\arrow{d}\
\hat{M}\arrow{r}{\hat{f}} & \hat{N}
\end{tikzcd}$$
cloud walrusBOT
Chmonkey 2.0
next obsidian
Matsumura claims that the bottom map is actually uniquely determined by commutativity of this diagram and continuity, but I don't see it. Does anyone know why this is the case?
If you take an I-adic topology instead, or really any topology given by M_1 > M_2 >... then I see how it's uniquely determined, but not otherwise
sturdy marsh
is the thing in the bottom left not the completion?
next obsidian
That is the completion
Like, you can define f-hat using f explicitly
via blah blah inverse limit continuity yadda yadda
but the claim is that a linear, continuous map from M-hat -> N-hat making the diagram commute is unique
sturdy marsh
continuous maps are determined by their value on a dense subset right?
next obsidian
no
only into Hausdorff spaces :(
well... not "only into" but it is true in that case
sturdy marsh
right
next obsidian
and N-hat won't be Hausdorff in general
latent anvil
I think linearity saves you
Well, maybe not
If the 0 of N hat is closed you're 👌
Since then if g is another such map, ker (f^ - g) = (f^ - g)^-1(0) will be closed and contains M
@next obsidian
Not sure what the topology on N^ is, you tried explaining it to me but it was complicated
next obsidian
hmmmmm
Okay so uhhhh
Let's suppose we have just topological spaces X_alpha
each one is discrete yeah?
then under the product topology the set which is a singleton in the alpha-slot
and just X_beta for all beta not equal to alpha
that's open and closed right?
like... any open set in the product topology on all the X_alpha is also closed?
err... not quite
but any basis element is right?
I think this gives it to us then because then the set which is 0 in any single slot and the whole thing everywhere else is closed
so like, it won't be countable, but just pretend it is for now, then the sets below are all closed
{0} x M_1 x M_2 x...
M_0 x {0} x M_2 x...
so you can take the intersection of all of them and you just get the singleton (0,0,0,...)
aka the zero
next obsidian
like...
latent anvil
The complements should be open yeah?
next obsidian
I guess that's mega not true
latent anvil
But infinitely many terms aren't the whole space
next obsidian
uhhh
latent anvil
Hmmm this annoying
next obsidian
I mean I just need
to be able to get some closed set where the only element in one index
is 0
So the way the topology is defined is
latent anvil
I don't think there's such a closed set
latent anvil
oh I mean maybe there's one on your weird subspace?
next obsidian
oh right
latent anvil
I doubt it though
next obsidian
hrmmmmmmmmmmmmmmmmmmmmmm
latent anvil
Like, what are the opens in the product topology?
next obsidian
yeah...
latent anvil
Well yeah
next obsidian
or finite
latent anvil
So the complement should be open
next obsidian
for any point
latent anvil
points closed
next obsidian
err
no like
the other one
for any x,y not equal
there is some U nbd of x not containing y
and likewise for a ndb of y not containing x
latent anvil
Right
next obsidian
I think this is easier to try and show T1 on the product
latent anvil
Oh this seems easy
next obsidian
right
latent anvil
If they're in different factors
next obsidian
Yeah so like
latent anvil
just exclude y
next obsidian
suppose x and y differ in the alpha slot
take a U for x not containing y in the alpha slot
latent anvil
oh right
next obsidian
and everything else everywhere else
latent anvil
Right
latent anvil
Yup
latent anvil
Anyways yeah linearity lets you replace the diagonal by a single point
Via subtraction
next obsidian
yeah that is cool like
I was thinking maybe I could try something similar
but I couldn't figure out how to get that to work
so like knowing that f^ - g is conts
latent anvil
WAIT DOESN'T THIS SHOW THE DIAGONAL IS CLOSED????
next obsidian
is because of subtraction being linear right?
I don't think so...
I sure hope not dude
latent anvil
It's the preimage of 0 under subtraction N×N->N
latent anvil
I mean sure
next obsidian
ummmm
next obsidian
oh shit...
dude
I'm dumb so like
Ummm
wait okay this is weird so
I explained the linear topology right?
latent anvil
Oh sorry not a topological group in general
For a topological group, T1 iff hausdorff iff identity closed
next obsidian
oh okay that makes more sense
so we know that umm
M under the linear topology by M_lambda is hausdorff iff bigcap M_lambda = 0
we know that M^ is T1 because it's built up as an inverse limit of M/M_lambda under discrete topologies
but M isn't T1
So I guess M is T1 iff bigcamp M_lambda = 0
which makes sense!
suppose it weren't
say we have x in there as well
latent anvil
I have a new proof
latent anvil
Topological groups are just lie groups that are scared to come out of the closet, and lie groups are manifolds so they're hausdorff
next obsidian
-_-
latent anvil
topological groups btfo
latent anvil
How was that horny???
next obsidian
M^ is Hausdorff
latent anvil
Am I horny for lie group???
next obsidian
anyway cool 
next obsidian
yeah
Okay so if uh
X and Y are Hausdorff
nvm
I hate having to work with these inverse limits explicitly
hahaha
especially when I need to start thinking about the topology as well lol
oh lmfao
Matsumura writes the Jacobson radical as rad(A)
so I saw for a in I, 1 + a is a unit so I < rad(A)
and I was so confused
latent anvil
oh hey this is nifty
My original proof only showed uniqueness of f^ as a linear continuous map
But this new one shows it as any continuous map
oh wait nvm, even if g isn't linear f^ - g is still continuous and the proof shows it's zero on a closed dense subset
i retract my oh hey
next obsidian
@latent anvil bruh you know the like, if M/IM is generated over A/I by some finite set you can pullback the generators to generators of M over A?
You get to do this with arbitrary rings and ideals if A is complete wrt I and M is separated for the I-adic topology
burnt pond
So, im still trying to figure this out. If $A$ is an abelian group und $\chi_1,\chi_2$ are characters $\chi_1\neq \chi_2$ of $A$ and $\pi$ is some complex representation of $A$ why does the sequence $0->\chi_1->\pi->\chi_2->0$ split, i.e. $\pi\cong \chi_1\oplus \chi_2$?. For finite $A$ one can use maschke's theorem i guess, but what if $A$ isn't finite?
cloud walrusBOT
Ceana
lavish gale
I appreciate your input but i don't know Lie group yet 😔 @chilly ocean
sly nexus
does anybody know why if I have a unitary representation $(\pi,V)$ that admits a proper non-zero subrepresentation $(\pi|W,W)$, then $(\pi|{W^\perp},W^\perp)$ is a subrepresentation(so that I can write it as a direct sum)?
cloud walrusBOT
Fractal
sly nexus
tag me if you do 
sour plume
The important property to check is that $\pi |_{W^\perp}$ maps $S^\perp$ into itself, right? And I believe unitarity of the representation gives you a very direct way to check that
cloud walrusBOT
Lartomato
sour plume
o wait hmmm
If $v \in W^\perp, w \in W$ and $g \in G$ where $G$ is the represented group (or whatever you're representing), then $\langle \pi(g) v, w \rangle = \langle v , \pi(g)^{-1} w \rangle = \langle v, \pi(g^{-1}) w \rangle$, and this is zero since $\pi(g^{-1})w \in W$ and $v$ is perpendicular to $W$
cloud walrusBOT
Lartomato
sour plume
Hence $\pi(g) v \in W^\perp$ for any $v \in W^\perp$ and $g \in G$, thus we're happy. @sly nexus does this make sense?
cloud walrusBOT
Lartomato
sour plume
o and that S^\perp above should be W^\perp
sly nexus
oh
didnt think of left multiplying by the inverse 
thank you @sour plume
it does make sense
sour plume
Naisu! I did have to stop and pause for a moment as well, lol
sly nexus
Im still trying to conceptualize subrepresentations better
I had seen that in AA before, but now it appeared on qm outta nowhere 
sour plume
Gotta be real careful with how symmetries of classical systems transfer to their quantized counterparts 
but representation theory of unitary groups gives you a lot of very cool insight about important quantum numbers like spin 
mint gulch
Do you know if the proposition "Every Principal Ideal Domain is a Unique Factorization Domain" is equivalent to AC?
All the proofs I saw use AC
sturdy marsh
PIDs are Noetherian
you normally don't need the full AC to deal with noetherian stuff
mint gulch
Noetherian is that every ascending chain of ideal converges to an ideal?
Or something like that
sturdy marsh
every ascending chain is eventually constant
mint gulch
Ok, ok
And to prove that PID are Noetherian don't you need AC?
I don't know the proof, it just sounds me
sturdy marsh
don't think so
if you have a chain
you just take the union
and by the PID condition, it is generated by an element
carmine fossil
And that element is in some ideal in the union
sturdy marsh
yup
mint gulch
Lol, I used that without knowing in my proof that PID => UFD
sturdy marsh
and for PID implies UFD, you just need to prove that a noetherian domain s.t. height one primes are principal is a UFD
mint gulch
But I used Zorn Lemma
sturdy marsh
I think you might need choice if you're switching between the various definitions of Noetherian
but probably not the full version
latent anvil
I think UFD iff "ascending chain condition on principal ideals" and "irreducible => prime" should be true without choice? Not sure though
mint gulch
Which version?
latent anvil
I don't see where the proof involves choice, at least
enigsis, it might be easier if you post your proof
We can try to think about whether the application of AC is redundant together
mint gulch
I mean, I have seen some other proofs that use Axiom of Choice
That's why I ask
But let me write it
golden pasture
idt you need to prove noetherian for PID->UFD?
latent anvil
the question is whether you need choice
golden pasture
i dont think so either?
sturdy marsh
idt you need to prove noetherian for PID->UFD?
you probably don't, it's just that I was using: A noetherian domain is a UFD iff height one primes are principal
golden pasture
ah lol
mint gulch
There you go
I did it a little fast I used a theorem without saying nothing
In the last part I used that, if $(a)$ is a maximal ideal, then $a$ is irreducible
cloud walrusBOT
Enigsis
mint gulch
I think is $\iff$ but I only use one direction
cloud walrusBOT
Enigsis
latent anvil
I think I see why AoC is (probably not necessary)
Since PIDs are noetherian, any ascending chains actually stabilize after a finite number of steps
So you shouldn't need the full strength of zorn's to prove Σ has a maximal element
probably
I would need to think harder about it and it's NYE what do you people want from me
Hmm it's possible you need some kind of countable choice... I want to say that if I isn't maximal, you can find a bigger J, and by iterating this process you get an infinite chain of proper containment. But this needs some form of choice, I think?
I think you need dependent choice for the argument you're employing
But that's a lot less choice than your argument suggests
This is very cool!
now this doesn't rule it out for pids
mint gulch
@latent anvil Could you say how can I use the axiom of Dependent choice in my argument?
latent anvil
Well you use zorn's yeah?
mint gulch
Yes
latent anvil
To conclude that any set of ideals in a ring in which every ideal is finitely generated has a maximal element
I claim you can do this with dependent choice
Suppose this isn't true
Choose a sequence x1, x2,... such that xi is properly contained in xi+1 for each i
You can choose this by dependent choice and the fact that every ideal in your collection is properly contained in a larger ideal
The existence of such a sequence is impossible by your argument
The bit about the chain
mint gulch
Hmm, If $\Sigma$ is finite, certainly has a maximal element. If is infinite, take a chain, you can make a sequence of that chain, right?
cloud walrusBOT
Enigsis
mint gulch
I mean, the sequence you construct
A sequence of properly contained ideals
Or Am I using AC without knowing there?
How do you define the DC axiom? I don't understand Wikipedia page on it
latent anvil
I think you're using DC here
mint gulch
Hmm, the bad part of my argument is that is with contradiction, then, it's hard to see why I used Zorn Lemma
I think maybe I need it because I relaid in Maximal Ideal and Maximal Ideal <=> b is irreducible in PID
mint gulch
https://math.stackexchange.com/q/59003/408287
The same here
Maybe if suffices to use DC, my question would be now
$PID \implies UFD \iff DC$
cloud walrusBOT
Enigsis
latent anvil
I think you'd have more luck asking this in #foundations
mint gulch
Ok, thanks
golden pasture
:o
latent anvil
lmfao first response is gindi gindying
Yeah I am very surprised
I would've expecteded this to not require choice
But it does! (although a relatively weak form)
bleak abyss
Oh yeah I remember I was sitting in commalg a bit in undergrad since it was just ring theory but at a bearable pace
And the way we phrased the proof it basically was, either Koenig's lemma or weak Koenig's lemma
mint gulch
I would've expecteded this to not require choice
Saying that isn't provable in ZF alone is the same that saying that needs AC?
mint gulch
I mean, if the theorem isn't provable in ZF alone, we need to add an axiom to prove it
latent anvil
Huh I figured it required DC dami
Well I think I was able to show that "ACC => existence of maximal element" for posets is equivalent to DC, and that's basically how you're using choice here. But because this is only for PIDs you might only need something weaker
bleak abyss
Yeah I mean probably not full blown AC but (weak?) Koenig coming up in the proof in a non-trivial way vaguely suggests to me that if you can do things just in ZF you'd need to put some legit effort
latent anvil
You can't do it in just ZF
bleak abyss
That's about as much as I'm willing to say about logic given that I don't do logic lol
latent anvil
See the MO thread
I also don't do logic lol
And should probably not be talking about this
mint gulch
Me neither hahaha
I just asked because I have seen some weird stuff like Every Vector Space Has a Basis equivalent to AC
latent anvil
I was looking at this paper when googling for your question
"six impossible rings"
The technical details went over my head but it was interesting
mint gulch
Can you post it?
mint gulch
Thank you very much
old hollow
I don't really understand what "the" finite field of p elements is
rather, what F_p means
does it just mean Z/pZ as a field?
broken vale
all finite fields with p elements are isomorphic
old hollow
Oh I see
broken vale
and when i say isomorphic, i mean as fields (which is really just the same thing as saying isomorphic as rings)
old hollow
What is an example of a finite field with p^k elements (where k > 1)
broken vale
lol yeah uh
old hollow
wot
broken vale
so let R = Z/pZ and let f(x) in R[x] be an irreducible polynomial of degree k (which always exists apparently)
then R[x] / (f(x)) is a field with p^k elements
i stole this from wikipedia btw https://en.m.wikipedia.org/wiki/Finite_field#Properties
old hollow
oh thank you
broken vale
i don’t know why such f(x) always exists but i can explain why that’s a field with p^k elements if you want
old hollow
Nah I’ll spare you, I need to refresh my memory of quotient rings
God that is unintuitive, R[x]/f(x)
broken vale
it why i hate algebra like why can’t there just be simple examples smh
and it’s / (f(x)), not / f(x)
it’s the ideal generated by f(x) which is a subring of R[x]
old hollow
Wait the () notation means ideal right
mint gulch
it why i hate algebra like why can’t there just be simple examples smh
Well, there are simple examples. In Z/3Z you have that x² + 1 is irreducible there, since there's no root, then you attach i to that field, then, you get the field (Z/3Z)(i), the elements in the field can be written as
a + bi, with a,b in Z/3Z, then, you have 9 possibilities, this is |Z/3Z(i)| = 3 ^2
It's the same result as (Z/3Z)/(x² + 1)
In this field actually, -1 = 2, then i = √2, if I am not wrong
crisp badge
If you are happy with the idea of splitting fields of polynomials, then the field with p^k elements is the splitting field of x^(p^k) - x over F_p @old hollow
cloud walrusBOT
Genie66
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
crisp badge
This gets around having to show existence of irreducibles of degree k (which will in fact turn out to be factors of x^(p^k)-x)
old hollow
@mint gulch oh actually that’s a nice example, thanks
sterile garden
I'm supposed to show that in Z[i], 2 = u(1-i)^2 where u is a unit. Now I did this just by solving for u, but is there a way to show this using the fact that N(a + bi) = a^2 + b^2? I've also showed that 1-i is irreducible in Z[i].
I guess could I say that N(2) = 4 = 2^2 = N((1-i)^2)?
Then conclude that 2 = u(1-i)^2 since N(u) = 1?
delicate bloom
I see it as $2 = 1^2+1^2 = (1+i)(1-i)$ and then it happens in this case we can factor out i from the first term to get $i(1-i)^2$
cloud walrusBOT
Merosity
delicate bloom
yeah, I guess here is suggestive to then ask/solve when can we factor out a unit from a+bi to get a+bi?
sterile garden
er, what?
delicate bloom
well 2 is sort of exceptional in that it ramifies in Z[i]
this was cause we were able to write $u(a+bi)=a-bi$
cloud walrusBOT
Merosity
delicate bloom
in general a prime would ramify if for some unit u we end up with $(a+bi)(a-bi) = u(a-bi)^2$
cloud walrusBOT
Merosity
delicate bloom
so it's worth solving for what cases this holds for in general
delicate bloom
oh, it just means 2 is a square times a unit
no other primes can be written as a square of an element times a unit
sterile garden
In Z[i]?
sterile garden
So if we work with Z[\omega], are the ramified primes different?
delicate bloom
5=(2+i)(2-i) for instance won't simplify down to u(a+bi)^2
yeah they are different
sterile garden
And are they still defined as being written as a unit times a square?
delicate bloom
for quadratic extensions yeah
but in general no, depends on the degree
that's a bit more complicated to think about, but in just the case I brought up it's simple enough to play around and solve by a bit of algebra that it's only possible for 2
if p=a^2+b^2 then showing a+bi = u(a-bi) for a unit u can only happen for p=2
delicate bloom
yeah let me know if you get stuck or anything, it should only take a minute or so
sterile garden
@delicate bloom I'm maybe a bit stuck. I know that p = 1,2,3 mod 4. Obviously this works with p = 2. It can't work with p = 3 since squares are congruent to either 0 or 1 mod 4, so a^2 + b^2 = 0,1,2 mod 4. I just need to show that this doesn't work with p = 1 mod 4.
(At least this is the strategy coming to mind for me)
delicate bloom
ah I'm thinking something more basic
delicate bloom
there are only 4 units, 1, -1, i, -i and just test them
1*(a+bi)=a-bi for starters
this means b=0 so it's entirely real, so p=a^2
well that's not prime obviously
sterile garden
Oh I see
delicate bloom
yeah good, now i(a+bi)=a-bi only slightly trickier but not much
frank haven
Yo is this channel free for me to post a question?
sterile garden
b = -a
delicate bloom
we're almost done @frank haven
frank haven
ok, ill wait
sterile garden
and again b = -a for multiplication by -i
delicate bloom
should be b=a for -i
delicate bloom
haha so now what can we do with this relation
delicate bloom
keep going almost there
sterile garden
Ah, I've got it.
So a^2 = b^2 = 0 or 1 mod 4.
So either p = 2 mod 4 or p = 0 mod 4. The latter isn't possible, so p = 2 mod 4 => p = 2.
delicate bloom
no no mod stuff
sterile garden
But that works!
sterile garden
Ah, so a = +/- 1
delicate bloom
unless a=1, p is not prime
delicate bloom
yeah +-1 good catch haha
frank haven
Alright, can i send my q now?
delicate bloom
yeah go for it
frank haven
I'm doing some introductory group theory and came across a question on artin relating to orders of elements
I think |ab|=mn
but i dont know how to prove that the order of the subgroup generated cannot be less than mn
sturdy marsh
what if b = a^-1
sterile garden
I don't think that's true
frank haven
what is the order then? my logic was, since its abelian $(ab)^{mn}=a^{mn} b^{mn}=(a^m)^n(b^n)^m=e^ne^m=e$
cloud walrusBOT
Oily Maccaroni
frank haven
maybe i am just being stupid
sturdy marsh
that only implies that the order divides mn
frank haven
so the order is less than mn?
frank haven
how can i go about finding the order, is it related to gcd(m,n)?
sturdy marsh
try looking at the case when m and n are coprime
frank haven
im stumped tbh
sterile garden
Sometimes I find it helps to work with a group like (Z/nZ)*
frank haven
like start with an example?
sterile garden
To maybe gain a bit of intuition
frank haven
then try to generalize?
frank haven
actually, now that im reading the question again it just says "what can you say about the order of ab" do you think it is enough to say |ab| divides mn, or is there more that can be said explicitly
like is it actually possible to find the order explicitly given arbitrary elements?
sterile garden
I think you can explicitly find it.
uneven jackal
there's an identity that could help prove that...
frank haven
so i think in the case that gcd(m,n)=1, |ab|=mn. Then in the arbitrary case im just trying to wrap my head around it
i suppose i could by assuming m>n WLOG
uneven jackal
have you studied arithmetic ?
frank haven
ive done a bit of discrete maths
so im familiar with divisibility and modular arithmetic
but im rusty
uneven jackal
arithmetic is a very important prerequisite to group stuff
frank haven
yeah i know
uneven jackal
you can use bezout here
frank haven
i just finsihed an analysis course so its been a little while
ok so gcd(m,n)=mx+ny
im familiar with that
i think
maybe that isnt bezout
uneven jackal
well what can you deduce quickly from that ?
delicate bloom
i suppose i could by assuming m>n WLOG
it's possible m=n
frank haven
oh thats true as well i suppose, but that doesnt matter cuz in that case the order of |a^2| is just |a|
or half |a|
maybe
uneven jackal
I might mistake for a similar exercise, but it should be useful
delicate bloom
not necessarily, you can have a != b but both have the same order
frank haven
oh right that makes sense
delicate bloom
klein 4 group for instance has elements with this property
in some sense this is a bit of a hint though solving the case when m=n should end up naturally being part of your solution
it's like the opposite case of what you already did of them being coprime
here they're the total opposite of coprime, they're equal haha, so you can think, how does this affect the order for things intermediate?
frank haven
im really trying to work this out, is a good approach to just trying raising ab to powers and see what happens?
or is that futile?
i also think cases is a good approach, start with m>n then m=n
old lava
you can also think of it by seeing that (ab)^s = a^s b^s, and if s is to be the order, then what's the smallest s so that both a^s and b^s are 1
old lava
it would be the lcm(m, n)
frank haven
oh right
old lava
or mn/gcd(m, n)
old lava
so ya, you can deduce that the order is at most lcm(m, n)
of course, if you get something like b = a^{-1}, then it doesn't matter what the order of a and b are, the order of ab is 1
so it's not sure that it will be lcm(m, n)
old lava
why do you need max
frank haven
lol i dont
old lava
I think the best you can say is that the order of ab divides lcm(m, n)
frank haven
ok, yeah that makes sense
in the case where they're arbitrary i guess its hard to pin down an explicit formula?
old lava
ya, it's hard to figure out an explicit formula
because you can have the inverse shenanigans going on
frank haven
so in the case where gcd(m,n)=1 then LCM(m,n)=mn?
old lava
yes
frank haven
then what i found above holds
ok, i think im happy with that answer. i wish artin had a solution manual to see what he was going for
thanks for all the help, and the patience
im pretty new to algebra
sterile garden
Huh, apparently you can show something else (sorry, I had to look up).
**something else in addition to what you've shown.
frank haven
really, would you mind sharing the link you found?
frank haven
thank you
sterile garden
cor 1
frank haven
huh, so o(ab) | LCM(m,n) and LCM(m,n)/GCD(m,n) | o(ab)
i guess that notation that i used i cringe
this stuff is so interesting, i can already tell that algebra is amazing and ive been studying it for less than a week
old lava
algebra is indeed amazing
frank haven
now i just need to actually learn it and stop diving into a wikipedia rabbit hole
lol
delicate bloom
I don't think about it in such a fancy way really
a raised to a multiple of m is the identity, b raised to a multiple of n is the identity
what's the smallest number that's a multiple of m and n?
well, that's of course the least common multiple of m and n
LCM(m,n)
uncut girder
intuition with N
sterile garden
Every non-zero, non-unit Gaussian integer is a product of irreducible elements. Show by induction on N(a).
So I know that N(a) = p => a is irreducible. I've shown this.
And I know everything in Z can be written as a product of primes
Does that mean I can say N(a) = p_1p_2...p_n and conclude that N(a) is a product of norms N(a_1)...N(a_n)?
So then my a_i's are all irreducible
I think that isn't true.
vital quail
@sterile garden you can write z = ab yes
now think about norms
if either of a, b is a unit you are done
sterile garden
@vital quail Okay, so I said that N(a) = 2. a is irreducible since N(a) is prime. Suppose the property holds for elements with norm at most n. Then if N(a) = n + 1. If a isn't irreducible, then we can write a = bc with b,c not units. So then N(a) = N(b)N(c) and 2 <= N(b), N(c) <= n so we're done.
vital quail
good job
carmine fossil
Every non-zero, non-unit Gaussian integer is a product of irreducible elements. ...
That's true for any Euclidean domain,btw
wind parrot
Its even true for any Noetherian domain
So being able to write things as a product of irreducibles is fairly weak, whats interesting is when this decomposition is unique
golden pasture
random late night thought can we classify all rings R such thar with spec R=X where X is some topological space of our choice
So say like X is discrete we have R is artinian with |X| maximal ideals (hence product of local artinian rings ...)
or say in the case of the topology on two elements {{},{x},{x,y}}:
clearly x must be the unique maximal ideal, and y is contained in x, so we need some local ring whose nilradical is prime but not sure if i can get any more info🤔 (obvious examples are jus any local ring+integral domain)
frank haven
Hey, is the condition in this question simply that we need n|m since otherwise the mapping of identity would not be identity
sturdy marsh
yup
frank haven
ok thanks
sterile garden
So now I'm working with $\omega = e^{2\pi i/3}$ and $N(\alpha) = N(a + b\omega) = a^2 -ab + b^2$. I'm asked to find all 6 units in $\mathbb{Z}[\omega]$.
cloud walrusBOT
IamDerek
sterile garden
I already know what the units are, but I'm not sure how to find them given that information.
delicate bloom
maybe helps to know it's basically a geometric series and factors as $a^2-ab+b^2 = \frac{a^3+b^3}{a+b}$
cloud walrusBOT
Merosity
delicate bloom
I don't know
sterile garden
Hmm, I'll see if this helps 🙂
delicate bloom
oh I think I see, I think we need to use the AGM inequality
sterile garden
AGM?
delicate bloom
at least in particular trying to reason out when a^2-ab+b^2 is positive
arithmetic-geometric mean
$\frac{x+y}{2} \ge \sqrt{xy}$
cloud walrusBOT
Merosity
delicate bloom
I don't know, I'm just sorta throwing ideas out, like ultimately we know a,b are integers so trying to force them into a corner essentially by trying to split the case when they're positive and negative apart
sterile garden
Yeah. I mean, I know by inspection that if a = 0, b = +/-1 then that works. Same vice versa.
delicate bloom
like in particular we know when a and b have opposite signs that a^2-ab+b^2 is positive so it must be 1
sterile garden
Wait, so could I say a = 0, b = +/-1, then a = +/-1, b = 0, and finally a = +/-1, b = -/+ 1?
errr, wait. That last pair is incorrect.
delicate bloom
yeah I get what you mean
I think maybe we could show it a kind of brute way by saying suppose there were larger and a=1+m and b=1+n
and show that it's >1
sterile garden
That would by my six solutions. The book tells me there's six of them. I don't know how I'd know I'm done there though.
Working with integers, aren't a^2 + b^2 >= ab?
delicate bloom
ah $(a-b)^2 \ge 0$ then $a^2-2ab+b^2 \ge 0$
cloud walrusBOT
Merosity
delicate bloom
if they're both the same sign $a^2+b^2 \ge 2ab \ge ab$
cloud walrusBOT
Merosity
delicate bloom
or does it not matter, haven't thought it through
opposite signs would mean $ab > 2ab$ because ab is negative
cloud walrusBOT
Merosity
delicate bloom
but that's ok because opposite signs is simple to see in $a^2-ab+b^2\ge 0$
cloud walrusBOT
Merosity
delicate bloom
so getting something larger than 1 is clearly broken by anything other than putting 0, 1, -1 in here
sterile garden
ah yes
and then that shows that the 6 I found are indeed all 6
Thanks! Math is frustrating. I feel I'm not cut out for this stuff haha
delicate bloom
haha just try to enjoy the puzzlement
sterile garden
Oh I do enjoy it. It just makes me feel dumb 🙃 Some day I hope to get better.
delicate bloom
oh I don't think that ever goes away haha
so I don't even think about it anymore, stuff just takes as long as it takes my brain to process, no control over that
sterile garden
A good way of looking at it
next obsidian
random late night thought can we classify all rings R such thar with spec R=X wh...
Go to the stacks project and look up "spectral topological space"
There's a proof in there I think that the topological spaces that are spectrum of rings are exactly the "spectral spaces"
actually here it is right here
latent anvil
random late night thought can we classify all rings R such thar with spec R=X wh...
Spec A discrete doesn't mean A is artinian
Unless you additionally assume A is noetherian iirc
The ring $k[x_1,\ldots]/(x_i^i)$ has a single prime but is not noetherian, and so not artinian
cloud walrusBOT
Schamrock
latent anvil
also tfw Alex and I both check #groups-rings-fields immediately after finishing jackbox
golden pasture
Unless you additionally assume A is noetherian iirc
ohh ye
golden pasture
Go to the stacks project and look up "spectral topological space"
hm doesnt this only tell us when it is possible for R to exist tho
latent anvil
So this is essentially order theoretic right?