It also doesn't assume functorial factorizations and the author insists that that is good
#groups-rings-fields
hidden haven
I don't fully understand why
Well from what I understand it is because not all model cats are cofibrantly generated or have an obvious functorial factorization
People also sometimes recommend Quillen's manuscript for model cats
I think it is called Homotopical algebra
and sometimes I recommend my masters thesis 🙃
slim kayak
Yeah, sharing is caring
hidden haven
The slice spectral sequence and KR theory
It has a chapter on model cats and one on stable homotopy theory
slim kayak
spectral sequences 
hidden haven
My beloved 
lethal dune
agile burrow
moldi 
slim kayak
I'll go through that chapter at some point, just a lot to take in
lethal dune
oh you love ss, prove it works 🔫
slim kayak
Waltuh
hidden haven
oh you love ss, prove it works 🔫
Sure 
hidden haven
→
slim kayak
Sometimes I believe spectral sequences main motivation was scaring away mathematicians from different fields
lethal dune
that's why it's called spectral sequence
slim kayak
Spectral as in spectre
lethal dune
👻
hidden haven
I was scared of them too but after properly working out why they work (to write it down in my thesis) I really like them
slim kayak
No one can say that it is wrong, Ravi made that up on the spot
hidden haven
I was more of a cat theory person and any explicit calculation left me shook and it still kinda does but the proof that they work is so good
lethal dune
I was scared of them too but after properly working out why they work (to write ...
I never looked at the proof of why they work... never will lol
though the article you once shared was very helpful
slim kayak
Spectral sequences make my mental RAM overflow so it's hard to push through
hidden haven
I just sent a proof 
just work out what the diagrams mean
hidden haven
though the article you once shared was very helpful
Was that Ravil's article?
slim kayak
I'll just plan on doing the exercise and scramble back at the definitions 10 times a minute until I have a proof down
hidden haven
yeah, that's what you're doing in the diagram no?
Yes, but this is carried out in greater generality and actually works as a formal proof and not just intuition
hidden haven
Spectral sequences make my mental RAM overflow so it's hard to push through
Same, that's why I prefer diagram calculus proofs of it
delicate orchid
true!
lethal dune
cool
slim kayak
The what?
it is in my thesis too
slim kayak
"The diagram obviously sequences"
hidden haven
slim kayak
The mathematical community hasn't dared not feeding it yet
hidden haven
put a topology on it and now it's a topology problem
Actually a thing you can do with string diagrams in cat theory 
hidden haven
They are very cool
delicate orchid
true
slim kayak
Actually a thing you can do with string diagrams in cat theory <:KEK:58624087735...
Like, topology on the category or diagrams?
hidden haven
The diagrams
lethal dune
it is metrisible 
slim kayak
Groethendieck and his consequences
hidden haven
You can associate a topological space that you can draw to a natural transformation
It is 2-dimensional
slim kayak
Uh huh
hidden haven
And you can draw some obvious deformations of it
slim kayak
I mean, exactly
lethal dune
cool!
hidden haven
Sounds too fancy if I describe it this way 
But you just have diagrams on which you can see manipulations and you don't have to do those algebraically
And it makes some hard proofs visually obvious
lethal dune
I kinds sorta stopped caring about very abstract math so idk if I'll ever look into these
lethal dune
And it makes some hard proofs visually obvious
example?
delicate orchid
horseshoe lemma
lethal dune
😐
hidden haven
One that I have used a lot in diagram chases in topology: Suppose you have a functor of 2 variables F, and F(a, -) is left adjoint to some G_a for each a. Then there is a unique way to make G functorial in a such that the natural isomorphism of homs for this adjunction becomes natural in a.
An example is that if you define tensoring with a as the left adjoint of Hom(a, -)
If this left adjoint exists for each a, then it is functorial in a as well
lethal dune
interesting!
slim kayak
What field does that fall under at that point?
hidden haven
Cat theory lmao
slim kayak
It sounds really cool but also really disconnected from anything other than pure cat
hidden haven
It sounds really cool but also really disconnected from anything other than pure...
In homotopy theory, you often want to construct closed monoidal categories, which require this adjunction of 2 variables between some formal tensor product and some formal hom. This theorem gives you a unique bifunctor structure on the other as soon as you have it on one
And you can actually use a version of this in diagram chases which is weird
To transform diagrams under adjunctions
The more standard examples you would find have to do with monads
Working with monads is a lot easier if you use string diagrams
The reason I don't mention that is that I haven't yet used monads for much outside of cat theory, but for that you can ask diligent clerk he is a big monad fan
slim kayak
I mean, I know what the words mean or have at least heard of them. I guess I'll have to see down the line how useful and interesting it turns out to be
hidden haven
And you can actually use a version of this in diagram chases which is weird
An example of this: these 2 commutative diagrams are equivalent. The existence of a diagonal map in the top diagram from the bottom left object to the top right one making it commute is equivalent to the existence of the same for the bottom diagram. I found this very difficult to prove without string diagrams
Maybe I just missed something silly but I spent a lot of time stuck on this part in a paper and once I realized string diagrams resolved it it was easy
hidden haven
An example of this: these 2 commutative diagrams are equivalent. The existence o...
If you wanna try this on your own, the symbols are
I think T left adjoint to U, and M_n left adjoint to (A maps to A_n)
And A and B are objects that can be evaluated at n in this way
Nothing else is needed
slim kayak
This brings back memories of proving the magic/diagonal base change cartesian square. Not sure if that is a good thing
hidden haven
None of these are cartesian squares though
This is a common lifting problem you would see when working with model categories
A commutative square in which you want to find a diagonal map
lethal dune
what's wrong with just compose?
hidden haven
The other diagonal lol
hidden haven
hidden haven
This is a part of a proof that characterizes cofibrations of sequential spectrum objects under their projective model structure
slim kayak
idk much model cats
hidden haven
Like we know that the fibrations and weak equivalences for these objects are the levelwise ones
And this characterizes the cofibrations
By spectrum objects I mean this works in any pointed model cat
But you can take these to be the usual spectra too as a special case
hidden haven
hidden haven
so to prove they are fibrations of some sort?
I mean it is pretty close to what you said here 
slim kayak
cool!
Be positive, you can cosplay as your non-math friends listening to you talk about math
lethal dune
I'm always doing that
grave pebble
someday I will torment you back
slim kayak
Oh wow, that's how I got into math too lmao
hidden haven
Yo nice
slim kayak
Bonus points you can pretend a little to understand whenever the mathematical physicist start talking about their stuff
grave pebble
well, I learnt a bit of abstract algebra so I know what Haskell programmers are talking about
lethal dune
i wanted a year drop so joined math and got lazy
teal vessel
I got into math because there were answers. English teachers had a nasty habit of contradicting themselves.
slim kayak
English classes in high school consisted out of us gaslighting our teacher into convenient interpretations of the text
coral spindle
well, I learnt a bit of abstract algebra so I know what Haskell programmers are ...
This was also how I realised I wanted to do math however many years ago
lethal dune
originally I wanted to be an engineer or physicist lol
it's a very convoluted character arc
coral spindle
But now all I want to be is someone who chats in a discussion channel
lethal dune
longer than onc piece
slim kayak
I just started self-studying a real analysis book in HS, I don't even like analysis these days so a wonder I stuck around
grave pebble
But now all I want to be is someone who chats in a discussion channel
truly like the taverns of the enlightenment
lethal dune
analysis
teal vessel
I mean, the questions to which there are not (yet) answers still are pretty self-explanatory. it follows a set of rules to show if an answer is satisfactory or not. English was a whole bunch of "you have to do it this way because I said so" and then me failing because I did it that way instead of how they actually wanted it, which was a different way they didn't tell us.
slim kayak
Immediately abandoned it after getting to the topology section and started doing munkres
coral spindle
Seriously can yall go to #math-discussion or something this is getting too much
slim kayak
Abelian rings? I sure hope they are
lethal dune
Seriously can yall go to <#359052581022203914> or something this is getting too ...
hiring mobs lol
coral spindle
We don't speak of rings being Abelian; that word is reserved for groups and Lie algebras
If you're a pedant at least. And trust me.
hidden haven
Varieties?
coral spindle
I am 🤓
grave pebble
haven't gotten to Lie algebra
teal vessel
don't lie to me
grave pebble
just the one
slim kayak
We don't speak of rings being Abelian; that word is reserved for groups and Lie ...
Okay, abelian is reserved to those two in normal people math
teal vessel
petition to stop using "commutative" and simply say "Abels"
"addition abels over the reals"
"the abelian property of multiplication"
coral spindle
keep saying it maybe we'll start laughing
slim kayak
Localize over a multiplicatively closed set
slim kayak
Yeah so? You scared of 0?
coral spindle
At risk of being simple and intuitive I'm going to ask again if anyone knows if this construction I'm coming across a lot has a well-used name.
grave pebble
0 ate 9
ashen heron
<@&268886789983436800> check past messages, seems like a spammer
grave pebble
noo dont snitch
grave pebble
you guys cant be feeding the troll
coral spindle
Can y'all stop shitposting for a second? I'm literally trying to use this channel for math.
slim kayak
<@&268886789983436800> check past messages, seems like a spammer
We almost had our algechill mascot
grave pebble
Can y'all stop shitposting for a second? I'm literally trying to use this channe...
don't worry I won't ignore you
rocky cloak
At risk of being **simple** and **intuitive** I'm going to ask again if anyone k...
Which construction is that?
coral spindle
I'm typing it out.
lethal dune
imagine using math channels to ask math
coral spindle
I'm dealing with the situation where we have a group $G$ with a subgroup $N \unlhd G$ and a near complement $H \leq G$, where $NH = G$. We have a natural surjection $N \rtimes H \to G$ given by $(n,h) \mapsto nh$ and the kernel of this map is then going to live in $N \cap H$.
cloud walrusBOT
Boytjie (never-to-be-glomed)
coral spindle
This is if you like the 'inner' definition
There is a way to make an outer definition, where one specifies the action of H upon N and the kernel, and I have worked out all the details of this. In fact I've posted this before so if you really want to you can search 'nearly semidirect' in this channel to see what I did.
I would like to know if this has been done in a book somewhere, because while it would be fine for me to do some exposition on this in the work that I'm doing, it would be more convenient not to have to reinvent notation and terminology that already exists.
upper pivot
Yes we call groups abelian as to not confuse with commutative (I.e not quantum) ofc
rocky cloak
I'm dealing with the situation where we have a group $G$ with a subgroup $N \unl...
Isn't it enough to just think of it as the semidirect product modulo N\cap H, or what sort of exposition about it would you want?
coral spindle
The outer definition
coral spindle
I would really just like a name and notation so I can talk about this without doing some annoying exposition
If it doesn't exist, as you say it's fine I will just write it like that
frigid lark
Is there a way to get from there are infinitely many a,b, non zero, rational, such that -4a^3 - 27b^2 is the square of a rational, to the statement here?
Also I swear I'm reading an algebra textbook
It will also be sufficient to show that given a field K/Q, there exists an element a in K, with minimal polynomial over Q, f(X) = X^3 + aX + b for a,b integer
glass thorn
If two matrices commute, what would be the easiest way to tell if they generate a cyclic group?
Specifically I'm dealing with SL_2(R)
Or GL_2(R)
glass thorn
I'm guessing just Jordan normal form?
round hull
one surefire way is to notice a = -3 b = 2 works
and then you can distribute powers of 6
wooden ember
this has come up in past topology exams and my teacher hasnt really given my a satisfying answer... does anyone know of reliable ways to show that an element has finite order given a presentation of a group?
in general i know dealing with presentations is difficult and im assuming this is probably a difficult problem in general, but for small examples?
only ways ive figured out are either checking in an abelianization, but that rarely works, or looking for a representation of the group but that's usually difficult and off topic from the topo course anyways
coral spindle
this has come up in past topology exams and my teacher hasnt really given my a s...
This doesn’t fall into the framework of adian-rabin, but even so I would put money on it being undecidable.
wooden ember
Yeah exactly
But it still comes up as an implicit question in exams 😭
And I never know how to deal with it
(We often want to show a fundamental group given by a presentation has a subgroup iso to Z to identify its isomorphism type, say a normal subgroup to give it a semidirect product structure for example)
And when I asked the teacher he said « you’re right, I’d be fine with you just mentioning the generating element » but that’s way too unsatisfying
slim kayak
(We often want to show a fundamental group given by a presentation has a subgrou...
Isn't this just a question about whether the group contains a free subgroup
There are some ways to go about it. What kinda spaces are you taking the fundamental group of?
wooden ember
depends, sometimes it's nice surfaces, other times it can be abit more gross
we had a p-divisible cylinder once
and sometimes it's not based on a space at all and he's just testing us on working with group pushouts
so it can be anything really, just you wont really have more than 3 or 4 relators
hidden haven
this has come up in past topology exams and my teacher hasnt really given my a s...
From the context it sounds like you instead want to show that the element has infinite order. Maybe you can come up with a map G → ℤ/nℤ such that the element maps to 1 for arbitrarily large n
wooden ember
Oh yeah oop that’s what I meant typo
wooden ember
From the context it sounds like you instead want to show that the element has in...
Yeah I’ve tried stuff like this as well it never really worked out because it would factor to the abelianization so you’d need the element to have infinite order in an abelianization
Which is easy to check but in general isn’t true
hidden haven
Fair
How about replacing ℤ/n with other small groups which have elements of order n
wooden ember
Like Dn
wooden ember
Yeah that’s a good point I could try stuff like that
I guess it’s in the same spirit as finding a representation
hidden haven
True
wooden ember
Anyhow I can tell it’s generally non trivial from people’s reactions so I’ll only give time to such a question if have time remaining
They’re worth very few points
slim kayak
so it can be anything really, just you wont really have more than 3 or 4 relator...
Hm, you are given a presentation?
wooden ember
Yeah, or we have to find one
slim kayak
Well, in that case this is more or less a combinatorial group theory problem. You can usually throw together generators to get something of infinite order
wooden ember
Okay so let me take a simple example
<a,b| aba^-1b^-p>
It seems obvious to me b has infinite order
slim kayak
If you have something neat like just a free product you have that the commutators actually have infinite order always. There were some analogues way of finding such elements for free products with amalgamations
wooden ember
If you have something neat like just a free product you have that the commutator...
Oh really
Interesting
slim kayak
Yeah, the kernel is actually a free group with index equal to the abelianization
b to the power minus p?
slim kayak
Yeah okay. You mean reduce b^n actually by replacing b^p with aba^-1, which means that if n mod p = m you have ab^(n-m/p)a^-1 b^m.
But you can try to check what the kernel of Z into your group is and see whether any power of b is equal to the trivial word
Or identity
wooden ember
The kernel of Z into your group?
hidden haven
But you can try to check what the kernel of Z into your group is and see whether...
The problem is in proving that a given word is not equal to the trivial word
wooden ember
Not sure what you mean by that
hidden haven
That is an undecidable problem
hidden haven
<a,b| aba^-1b^-p>
In this case you can make this act on a line with a being scaling by p and b being translation by 1
slim kayak
Good old word problem
wooden ember
Aight I think I’ll stick to looking for representations or actions they seem the most effective
slim kayak
It is in undecideable in general, but possible when your group is finitely related
It often is fine, you just go through your routine. But proofs about word problems in general groups make me miserable
hidden haven
I hate words
slim kayak
Me too
hidden haven
Garrrr
hidden haven
slim kayak
At least we have Cayley graphs now, no more of that nerdy Nielsen transformation stuff
?
Why am I getting pepe worried 
slim kayak
The kernel of Z into your group?
Send n to b^n
See if for n not equal 0 b^n is trivial
The right notion here is that of reduced word length. A bit informal, but if n > p you again get the reduced form ab^ka^-1b^n-k. You can convince yourself that no relation could decrease the length of this word any further, so since the word has non-zero reduced length it isn't representing the identity
wooden ember
The right notion here is that of reduced word length. A bit informal, but if n >...
sure i can convince myself but not rigorously
hidden haven
Why am I getting pepe worried <:pepeworry:1093036439735894056>
Idk sounds scary
hidden haven
> cool stuff
> group theory
south patrol
Basedd
coral spindle
blocked and reported
chilly ocean
Just read Artin
Also this is wrong. A group is not commutative by nature. What you are trying to say is that if there is an inverse property such that for all x there exists y such that xy = yx = e; that is, an element that is both a right inverse and a left inverse.
long nebula
Also this is wrong. A group is not commutative by nature. What you are trying to...
They didn't claim it to be commutative?
teal vessel
is that not what was written there? "for all a there exists b such that ab=ba=e"
long nebula
xD
coral spindle
😭 don't summon them
rotund aurora
can you have vector spaces over a monoid, instead of over an abelian group
long nebula
Wdym by over an abelian group
coral spindle
You can certainly have a group acting on a monoid.
But you cannot have a ring acting on a monoid in the same way; we need negation in that case.
rotund aurora
it should be a field action
long nebula
What's (-1)*v then?
coral spindle
^
rotund aurora
mmh right
slim kayak
You can do some cool stuff with them
Yeah like skipping the chapter when you see them mentioned
teal vessel
if I have to "prove that the matrix A is the linear transformation that rotates the x,y plane [etc.]" I just have to show that for some generic vector (x y) the matrix produces a rotated version of x and y, right? i.e. throw x and y into polar form and then do some angle sum and differences
coral spindle
Yes
teal vessel
that seems... underwhelming for a "prove" statement lol
coral spindle
OK
delicate orchid
or just take the determinant
teal vessel
also, what is "extending to" a homomorphism? we love when books use new language in the exercises
slim kayak
or just take the determinant
If you know A is orthogonal, ig?
teal vessel
if my assumptions are accurate, I can pretty easily do it, it's just funny words
coral spindle
also, what is "extending to" a homomorphism? we love when books use new languag...
If I define a homomorphism only on generators for a group, then it is defined for the whole group. For instance if I know f(x) and f(y) then I know f(xy), f(x^2) , f(y^2), f(xy^-1) etc.
teal vessel
ah, that property, yeah.
coral spindle
You of course need to check that this satisfies f(1) = 1 as usual
But this is the idea in any case
The catch is that a priori there is no reason for two words in the generators to produce the same result. For example if your group is Abelian then xy = yx but it may be that f(x)f(y) =/= f(y)f(x) the way you chose it, so this needs to be checked.
slim kayak
Well, and like check well-definedness
coral spindle
Woah really :O
teal vessel
poe's law really being strong on that one lol
delicate orchid
If you know A is orthogonal, ig?
oh yeah you can do (x, 0)(0, 1/x)
I've been working over Z far too much
sly rain
What is the "correct" way to think about field extensions and the field theory stuff you usually do (in a intro abstract algebra class) leading up to galois theory?
Or like useful mental models/perspectives to look at these concepts etc.
Maybe this is pretty vague, but I guess I take anything you deem useful (e.g. for Ideals, I read to think of them as kernels and that was helpful for me, so maybe there is something similar for field theory)
delicate orchid
I like to think about them as algebras over a field - so like a vector space but you can multiply stuff
and how that stuff multiplies is deterimed by what you're quotienting out by
rotund aurora
What is the "correct" way to think about field extensions and the field theory s...
a lattice (in the sense of order theory)
and you can also draw the diagram
delicate orchid
example, Q[swagggg] = Q[x]/(x^2+x+1), is a 2 dimensional vector space over Q, so elements look like a+bx - such that x^2+x+1 = 0
slim kayak
Very important edit
sly rain
and you can also draw the diagram
Can you expand on that, im not sure what you mean.
coral spindle
Croq is referring to galois theory stuff
I don't think it's helpful to see it that way yet.
slim kayak
example, Q[swagggg] = Q[x]/(x^2+x+1), is a 2 dimensional vector space over Q, so...
Oh right, you can visualize algebraic field extensions by a single element as coordinate rings
delicate orchid
can you not do it with multiple elements as well
just by quotienting like Q[x_1, ..., x_n] instead?
slim kayak
Maybe? The interesting bit is anything not falling under primitive element theorem umbrella no?
rotund aurora
Can you expand on that, im not sure what you mean.
i was saying that if you consider all (or some) field extensions of Q (or some other field), they are partially ordered by set inclusion, and furthermore, given two field extensions Q<K and Q<L there is a maximal field extension of Q contained in both K and L (think about this as an analog of the gcd of two natural numbers), namely K intersection L, and there is a minimal extension of Q that contains both K and L, namely the composite KL.
delicate orchid
I don't know what that is
is that the one where all field extenstions of Q are primitive
rotund aurora
this does not involve Galois theory strictly, because Galois only enters the picture when you relate that to symmetry groups
slim kayak
I don't know what that is
Finitely many intermediary extension iff there exists an element generating the field extension
sly rain
Thanks, thats more clear. Also ty wew lads
I will be incorporating these
slim kayak
So if your field extension by multiple elements has only finitely many intermediary extensions it is actually generated by adjoining a single element
delicate orchid
right I see
rotund aurora
Can you expand on that, im not sure what you mean.
so if you are given a field extension, it is often helpful to try to decompose it into smaller extensions. So thinking about that is certainly useful
delicate orchid
so you can have a big fack off extenstion that isn't of the form F[X] for some (possibly infinite set) of elements X
is that what you're saying
slim kayak
Sometimes even finite but yeah
Idk how you get the usual coordinate ring thingy from that, but I am welcome for enlightenment
delicate orchid
I'm struggling to see how this doesn't contradict with the universal property of the polynomial ring
there's probably some nuance I'm just not aware of
slim kayak
Universal property of polynomial rings?
delicate orchid
given two rings R, S and some arbitrarily sized set X of elements of S, then any ring homomorphism between R and S extends into a unique evaluation map from R[X] -> S
in my head I was taking R = F, S some field extenstion of F, and the map being the inclusion
slim kayak
Oh, that characterises polynomial rings uniquely?
delicate orchid
yeah so the evaluation map would just be like
say X = {s_1, s_2, ..., s_k} or whatever
then the extension of the ring homomorphism R -> S would be R[x_1, ..., x_k] with x_i |-> s_i
I hope I'm making sense
slim kayak
Yeah, just a bit taken aback by this universal property
Polynomials are such a basic thing in my mind, that feels kinda like saying "universal property of matrices"
delicate orchid
this is a basic property, it just looks complicated!
slim kayak
Yeah, it's pretty useful result
delicate orchid
it's just saying "evaluating a polynomial in formal symbols and coefficients in R at some elements in S is a ring homomorphism"
this is the intuition anyway
slim kayak
I am not cat pilled enough yet 😔
delicate orchid
keep it that way
slim kayak
Sadly I need to use the universal property of colimits and adjoints fairly regularly the past few weeks, so it is too late to change course
Anyways, what exactly was your issue?
delicate orchid
so
is your thing saying that there are field extenstions that do not look like F[x_1, ..., x_n] where n can be infinite if I so please
slim kayak
Express extensions as R[a1,....,a_n], or as an algebra
And then divide by kernel to get coordinate ring representation?
delicate orchid
yur
I'm starting to see where there might be some problems for transcendental extenstions
south patrol
Oh lol universal property is how I think about polynomial rings lol
delicate orchid
but that's just iso to F[x] irregardless
delicate orchid
Oh lol universal property is how I think about polynomial rings lol
ofc YOU would
slim kayak
Yeah
south patrol
It's how you work with them right lol
delicate orchid
I don't work with them
but it is very useful
actually no I lie, I do use it
just not explicitly
south patrol
Im not even that cat pilled lol
OK nah I take that back like among friends I am probs most into them
delicate orchid
among
slim kayak
No
south patrol
Amongst
slim kayak
Among them?
devout dirge
Does anyone knows a bit magma (the algebra computation application) and knows how to calculate the number of Conjugacy Classes in GL_4(F_2[x]/x^3)?
slim kayak
This universal property has been occasionally useful so far, what r u doing where you need it constantly?
delicate orchid
Does anyone knows a bit magma (the algebra computation application) and knows ho...
do you have a presentation for this group already?
south patrol
Every time you define a map out of a poly ring
delicate orchid
cause if so it's just #ConjugacyClasses(group)
delicate orchid
This universal property has been occasionally useful so far, what r u doing wher...
working with algebras
being able to write R-algebras as quotients of R-polynomial rings is so handy
hidden haven
Polynomials are such a basic thing in my mind, that feels kinda like saying "uni...
I can try to say why this shouldn't be surprising. Suppose I give you a set X and ask you to generate the free commutative ring on it, ie the most general commutative ring generated by this set. Since you know how this works for groups, you would do something similar: take the set of formal sums and products of elements of this set. Using distributivity, it is enough to just take sums of products. If there are repeated terms, you can combine them and put an integer coefficient up front. What you end up with is the polynomial ring over the integers with variables from X, ℤ[X]. The universal property says that to describe a map from this ring, it suffices to describe it just on the free generators. More generally, R[X] is the free R-algebra generated by X.
delicate orchid
very well worded
hidden haven
slim kayak
Wait a sec, are polynomial rings the free objects of rings?
hidden haven
For commutative rings yes
slim kayak
I feel enlightened
hidden haven
Isn't it weird that no one tells you this
What else is big math hiding
Learn cat theory to find out
hidden haven
Ye
delicate orchid
sick
rotund aurora
category theory is such a conspiracy
hidden haven
Who is conspiring
delicate orchid
me and u
lethal dune
co-inspiring
hidden haven
rotund aurora
life is too short to learn category theory
lethal dune
cats are good and all, I prefer 🐶🐕
delicate orchid
that's a quip
hidden haven
Dogegory theory
rotund aurora
dog theory --> doge theory --> hodge theory
south patrol
Categories
delicate orchid
that's a quip
lethal dune
cat gore eyes
lethal dune
do it
devout dirge
do you have a presentation for this group already?
Isnt there a way to write GLn(finite ring)?
delicate orchid
Isnt there a way to write GLn(finite ring)?
I think just GL(n, ring) would work
and if that doesn't work maybe try MatrixGroup<n, R | >?
south patrol
Hm if $\mathfrak g$ is a Lie $k$-algebra and I have a representation $\rho: \mathfrak g \to \mathfrak{gl}_V$, is it normal notation to write things like $xyz(v)$ for $v \in V$ and $x,y,z \in \mathfrak g$? I feel like this is slightly weird since it's written like we have a product in $\mathfrak g$, but it's used a decent bit in some notes I was reading without mention lol
delicate orchid
I'd write (xyz)v
rocky cloak
It's normal write xv for x acting on v, so by extension writing something like xyzv shouldn't be a problem. Writing something like $\rho(xyz)v$ or $(xyz)v$ seems weird to me. As it seems to imply you can multiply x, y and z
cloud walrusBOT
jagr2808
delicate orchid
just keeps it clear what's from where 
rocky cloak
I would have written xyz(v), then
south patrol
Sorry
That was a mistake with what I said i'd err towards writing lmao
Sure makes sense jagr thanks
Yeah I guess I was worried because of the potential ambiguity like implying you can actually multiply stuff
But it should be clear from context
rocky cloak
If it's not clear from context, then I guess you would have to go with x(y(z(v)))
hot lake
you can also see x,y,z as elements of the universal enveloping algebra of g and then the product makes sense inside of that algebra (and you can then enlarge the reprentation to the algebra and everything is defined so that xyz(v) = x(y(z(v))) )
south patrol
Oh that's nice atually thank you
rotund aurora
in a ring, how many derivatives can you have satisfying Leibniz's rule, linearity and (a^{-1})'=-a' · a^{-2}
rocky cloak
in a ring, how many derivatives can you have satisfying Leibniz's rule, linearit...
Depends on the ring, for example if R = Z[x], then D(x) = p uniquely defines a derivation for any polynomial p.
white oxide
can i get a hint for this? i'm considering the principal ideal generated by some nonzero a in the integral domain D and am trying to show that it's a unit, a little bit stuck tho
trying to use the decreasing chain of ideals condition somehow
my first thought was that i could say ar = 1 for some r in D because <a> is a subring with unity
but that seems too powerful
idk
south patrol
That doesn't necessarily follow
white oxide
also it doesn't use the decreasing chain of ideals condition
how tho, i thought an ideal was a subring and since it's a subring of an integral domain doesn't it have to contain unity
like intuitively it seems wayyy too powerful so i know it's wrong
but idg why it's wrong
south patrol
Well by definition it should contain the unit lol
But it isn't just the set of powers of a necessarily
Like think about Z
rocky cloak
An ideal is a subring only if you don't require rings to have 1
south patrol
Pick a non zero which u wanna show is invettible
The only ideals you should contemplate are ones defined in terms of a and stuff
Because we have no more info at our disposal
Try to find a nice descending chain
south patrol
Also I should perhaps add for context that this condition is known as being Artinian
So this says Artinian + integral domain => field
So like perhaps slightly harsh giving this as an exercise when it is commonly just given in books but should be doable
rocky cloak
So like perhaps slightly harsh giving this as an exercise when it is commonly ju...
Why is it harsh? Seems like a very reasonable exercise
south patrol
Okay true sure
slim kayak
Unless I am trippin the reverse direction seems easier
delicate orchid
field => integral domain + artinian? yeah that does seem easier
slim kayak
Integral but not field then infinite descending chain of ideals
slim kayak
field => integral domain + artinian? yeah that does seem easier <:opencry:586078...
Lol
south patrol
I see no reason to do it like that hm
Like if I did it like that I'd end up using contraposition again lol
delicate orchid
not knowing the proof to the statement the contrapositive seems easier
south patrol
Lol so true
delicate orchid
||succesive powers of a PI probably works maybe||
south patrol
It does
south patrol
Except rings satisfying the hypotheses of this theorem
rocky cloak
The proof should proceed about the same either way
slim kayak
Right, looks the same in the end mostly
The contrapositive feels a bit better at guiding towards it to me
indigo ridge
I have basic knowledge of Abstract Algebra I took a second year course, should I read Aluffis Algebra book or stick to something like dummit and foote?
slim kayak
The good old try proof by contradiction and then remove the first and last line method if you feel unsure why something should work
south patrol
Assume as many things as possible and wait for smth to break
long nebula
I have basic knowledge of Abstract Algebra I took a second year course, should I...
Both seem fine
slim kayak
Instead of avoiding obstructions try your best to crash into them
long nebula
I hear Aluffi doesn't have as many/as good exercises as Dummit Foote though
but Aluffi is a more "modern" approach
slim kayak
Does it have nearly as many? Dummit foote feels like it is 50% exercises at some points.
white oxide
what exactly do they mean by "underlying set"? does it just refer to the elements of some algebraic structure without the structure imposed on it if that makes any sense
long nebula
^
slim kayak
It's just a nickname
coral spindle
The mental model is that it should be the underlying set, as in the underlying set of a group or ring.
long nebula
But also that is the idea behind what the underlying set represents yeah
coral spindle
But again, this is just the name.
slim kayak
Like how sometimes injective homomorphism are called inclusions, even though the domain isn't an actual subgroup
south patrol
If only there was another term for these conditions lol
Okay nvm sure this hasn't used cats so far I guess
Wait no it has
coral spindle
math is not real
yes
south patrol
This isn't even the normal definition of a concrete category right
delicate orchid
it's the one I'd use
slim kayak
Some interpretation of the idea is to simulate the forgetful functor into Set by attaching sets to your objects in a way that has all the relevant properties
delicate orchid
if sigma is a functor anyway I didn't actually read it
south patrol
I've always seen them be faithful by definition
delicate orchid
oh yeah true
south patrol
As is this is just saying there's a functor to Set
slim kayak
It isn't representing anything particular, you are just useful extra structure
south patrol
Which like, would be every category
delicate orchid
yeah and then obviously you just map everything to the identity and a single object
south patrol
Lol yeah
Well this is technically a pair I guess sure
But still
Faithfulness is kind of the main point I thought
coral spindle
This isn't even the normal definition of a concrete category right
It is equivalent, but it's tricksy. (i) is really saying that it is a faithful functor.
Oh you said so
nvm
schlow boytjie
delicate orchid
what lol
south patrol
Dodgy potentially lol
coral spindle
But this does in particular mean it's faithful
slim kayak
Saying the morphism "Is" the function is a bit like
Can't one construct a counterexample or is there some yoneda magic going on
coral spindle
Like they are literally equal
south patrol
You could have the same function on the same sets but be different motphisms
Because of different objects
Like identity function on X when you have various different topologies
slim kayak
Yeah, although if your category is pretty large the morphism might force some restrictions on what your set functions can be
slim kayak
Possibly uniquely determining them, in which case this makes sense. But yeah
This feels like randomly slapping sets on objects, the connection seems weak
south patrol
Like why introduce stuff w categories if ur gonna talk like this lol
delicate orchid
^
introducing this concept before functors is moronic
and if functors have been introduced then why isn't this phrased in terms of functors
noble hedge
If the minimal polynomial of some $\alpha$ in $F$ is $m(x)$, then $\alpha$ should have multiplicity 1 in $m(x)$ right?
cloud walrusBOT
strobilanthes
noble hedge
Can't really think of a formal argument, but I'm guessing there's some contradiction that arises by divisibility of $m(x)$
cloud walrusBOT
strobilanthes
noble hedge
The problem I'm running into is that if $\alpha$ has multiplicity $k$ in $m(x)$, then there isn't really a way to construct some polynomial $g(x)$ such that $\alpha$ has a smaller multiplicity in $g(x)$ unless I potentially move up to an extension containing $\alpha$, but at that point how would I conclude that the divison results in a term where all coefficients come from $F$ and not the extension?
cloud walrusBOT
strobilanthes
coral spindle
Well unfortunately what you want to prove isn't true!
noble hedge
wonderful!
coral spindle
This is a property of fields called being perfect.
Here's an example of a non-pefect field, described in a mathoverflow post: https://math.stackexchange.com/questions/106632/examples-of-fields-which-are-not-perfect
TL;DR the example they give is the (family of) field(s) F_p(T) with the polynomial x^p - T being irreducible and having a factorisation in its splitting field into p equal linear factors.
noble hedge
Yeah that makes sense
just ran into perfect fields earlier today, not much experience with them
coral spindle
Well now you know
noble hedge
yeah!
the exercises I've been doing are slowly building up results for separability, this question popped into my head and I wasn't quite sure if it'd work or not
rocky cloak
I find it funny that perfect fields, perfect rings, and perfect groups have nothing to do with one another.
Wonder how many more uses of the word perfect there are
cloud walrusBOT
person2709505
delicate orchid
No that’s not correct
You can have groups where you can’t simplify down an expression like “aba”
coral steeple
Ah yes
Is it even possible to write an arbitrary element explicitly like that in that case?
delicate orchid
Free group on two elements comes to mind
delicate orchid
Is it even possible to write an arbitrary element explicitly like that in that c...
Wdym, aba is explicit
Oh
Nvm sorry yes
Then no, best you can do is a word in a, b
coral spindle
Is it even possible to write an arbitrary element explicitly like that in that c...
Sure, it's $a^{i_1}b^{i_2}\cdots a^{i_n}$ for $n \in \bN$ and $i_j \in \bZ$ for $0 \leq j \leq n$.
delicate orchid
Oh yeah duh
coral spindle
Typo!
cloud walrusBOT
Boytjie (never-to-be-glomed)
coral spindle
You can imagine this becoming unwieldy after we add more generators :)
coral steeple
Yeah that looks messy
coral spindle
So instead we just say that <a,b,c, ...> is the smallest subgroup of G that contains each element a, b, c etc.
That is a very useful definition :)
coral steeple
The wikipedia page for word seems informative though, thanks
delicate orchid
I can’t come up with any finite examples of this happening :(
thinkin bout SL(2,3)
clever beacon
What do we mean when we say that an inner product <-,-> is "non-degenerate"?
delicate orchid
Basically means that there isn’t an x such that <-,x> is the trivial map to 0 iirc
clever beacon
I've got an argument of the form, where v,w are in a v.s. V and T:V->V is linear,
"<v, Tw> = 0" and "<-,-> is non-deg." => "T == 0"
delicate orchid
Is this for all v, w in V?
clever beacon
ye
delicate orchid
Then yeah
clever beacon
oh
delicate orchid
Oh wait or w is just in the kernel of T
clever beacon
Because if it were degenerate to some extent, then there'd be some v that I could take where I couldn't make any further claim about Tw
delicate orchid
Oh wait or w is just in the kernel of T
But think wew lads, if every vector is in the kernel then T is zero
devout dirge
do you have a presentation for this group already?
Do you have an idea how to get such a presentation?
Or maybe if in GAP it is easier somehow
tough raven
i was saying that if you consider all (or some) field extensions of Q (or some o...
Btw
This is not quite correct unless K, L are already embedded in a common larger field where you can take their intersection and compositum.
(The actual fields you get depend on the choice of embedding, so you can't just skip that part.)
frigid lark
just imagine that every field extension of Q is normal
summer path
new name idea
frigid lark
Every irred polynomial is separable is better
alpine island
What is the "correct" way to think about field extensions and the field theory s...
I think of them in the context of the ideals being a set of rules to follow when you do arithmetic in it. Like in R[x]/(x^2), just remember to simplify x^2=0 wherever you can. It's kind of a naive approach but it seems to work so far.
fervent gyro
I want to prove that for a noetherian ring A, a multiplicative subset S and M a finitely generated A-module we have this isomorphism. I have seen Rotman's proof of the equivalen isomorphism for Tor, without restrictions on M or A, on An introduction to homological algebra, proposition 7.17. I wonder if that could be replicated for this case of Ext, and how noetherianity and M being f.g. are important hypothesis
rocky cloak
I want to prove that for a noetherian ring A, a multiplicative subset S and M a ...
I can't quite find the proposition you mentioned. But for the argument I'm thinking of if we don't assume noetherian + finitely generated you might run into problems because S^-1 does not preserved arbitrary products.
lethal dune
don't see why you need Noetherian. S^1A ⊗ is exact and sends projectives to projectives
hidden haven
You need
S^(-1)(Hom(-, N)) and Hom(S^(-1) -, S^(-1) N)
to agree on a projective resolution of M. Do they agree on non finitely generated modules?
For finitely generated, this is fine by taking a finite presentation. The problem seems to be what jagr said.
You would need A to be Noetherian to get a finite presentation
next obsidian
You require the projective resolution to always consist of finitely presented things which can’t be guaranteed when the ring isn’t Noetherian, even if M is finitely presented
I think asking that M be coherent is enough though
As long as your ring is coherent
glossy crag
In the definition of the rational canonical form do we require that the blocks be companion matrices of polynomials f_1|...|f_m or can it be companion matrices of any polynomials?
wraith cargo
You need
S^(-1)(Hom(-, N)) and Hom(S^(-1) -, S^(-1) N)
to agree on a projective ...
You need finitely presented for that
hidden haven
Ye that's what I was saying
wraith cargo
Ah sorry missed that lol
slim kayak
In the definition of the rational canonical form do we require that the blocks b...
Another form just has the blocks equalling the companion matrices of the irreducible factors of your characteristic polynomial
But yeah, what you described also works and is a bit more general
delicate orchid
blocks 😋
slim kayak
Bro is craving these bad boys
hidden haven
Can't blame him
delicate orchid
minecraft
slim kayak
With shaders and the 4k texture packs
hidden haven
Kinda dull tbh
glossy crag
But yeah, what you described also works and is a bit more general
I was just thinking on how the uniqueness of the RCF should reduce to the uniqueness of the invariant factors, so this definition makes the most sense.
slim kayak
Idk what kind of uniqueness you ascribed to the RCF. When you derive the RCF via the structure theorem it is pretty clear why they have to be what they are
Plus they aren't really unique, they are unique up to associatedness
teal vessel
so the kernel of the action of some group G on a member of set A would be the set of all g in G such that g(a)=1, right?
(for some fixed member "a")
lusty marlin
so the kernel of the action of some group G on a member of set A would be the se...
The kernel of a group action of a group G on a set A is the set of elements of G for which g.a=a for all a in A.
teal vessel
... then why is this question asking me to take a fixed "a" in A?
only for part B then?
lusty marlin
The set of elements of G for which g.a=a for some fixed a is called the stabilizer of a in G.
As correctly written in part b
I don't see what the issue is
teal vessel
just some confusion regarding the conditions of the question. The fact that it asks to fix some a in A during the initial setup implies that (to an untrained student) the kernel of the action may depend on which a is being used.
lusty marlin
It doesn't
slim kayak
so the kernel of the action of some group G on a member of set A would be the se...
With kernel they mean the map from G into the permutations grip of A that assigns to each g the permutation on A
delicate orchid
that wouldn't be well defined
slim kayak
So the permutation is only then trivial if it fixes all elements. If you look at the action by a single element a you won't call that the kernel but instead the isotropy group
lusty marlin
So the permutation is only then trivial if it fixes all elements. If you look at...
Don't you mean 'look at the action on a single element a'?
slim kayak
My bad, isotropy group*
teal vessel
just looking for clarification because of poor instruction ordering.
I can work with that
delicate orchid
lots of words involved that all mean very similar things at a glance
teal vessel
yep, pretty much
slim kayak
Formally the kernel is the intersection of all stabilizers, but you can also think of it as the set of elements that act trivially on the set A
teal vessel
ideal ordering here would be something to the following effect:
Let G be a group acting on a set A. Show that the following are subgrps of G:
(a) the kernel of the action,
(b) {g in G | ga=a} for some fixed a in A
delicate orchid
yes
lusty marlin
ideal ordering here would be something to the following effect:
> Let G be a gro...
Yeah.
delicate orchid
Formally the kernel is the intersection of all stabilizers, but you can also thi...
thinking about the kernel like this I'd even swap (a) and (b) around lol
barren sierra
Formally the kernel is the intersection of all stabilizers, but you can also thi...
I'd argue the second statement is the definition and the first statement is an immediate property
teal vessel
one day I'll get fed up with this enough to be a textbook editor.
barren sierra
one day I'll get fed up with this enough to be a textbook editor.
I've debated doing this for the discrete math class that I'm an undergrad TA for
Absolutely hate the text
But alas I don't have 20 hours extra a week
Or even 10
But I relate to the sentiment
long nebula
Sometimes confusing textbooks stop and make you think, forcing you to understand better though
barren sierra
Ehhhhhhhhhhh sounds like copium
wraith cargo
Partially true imo lol
teal vessel
better order and clarity:
G.A, show the following subgroups:
(a) {g in G |ga=a} for some fixed a in A, called the stabilizer of a in G
(b) {g in G |ga=a} for all a in A, [insert reminder of the definition of kernel, or reference to the original usage earlier in the text]
slim kayak
Since this is mostly just running through the definition just saying "kernel of action" might be the better exercise since that asks you to remember what that actually means
teal vessel
maybe, I like showing the parallel
when two things have very similar structure, I like showing that similarity, but that's just preference
long nebula
I think it should be partially on the reader to find that similarity for themselves
fair quartz
i have these definitions given and i need to prove that all m_n are in the point spectrum of T
i figured that the easiest way to prove that would be by finding the nullspace of (m_n-T) and showing that its not equal to {0}
but im struggling to make any further progress from this
slim kayak
But you still figured out they are similar, which you might have forgotten soon after if you had done this exercise with ease and moved on
fair quartz
i tried writing it out as ((m_n-m_i)*a_i) = 0 (for all i of N) but the only solution i seem to get is that a has to be zero
long nebula
what do you mean by m_n - T? I'm confused
how are you interpreting m_n as a linear transformation
fair quartz
this definition for some lambda being in the point spectrum of an operator
(nicht injektiv = not injective)
slim kayak
So viele deutsche hier
long nebula
Yeah ich kann Deutsch lesen lol can you send a picture of the full problem
long nebula
ah okay so we want to show that for each n, there exists v such that (m_n I - T)v = 0
I see now
fair quartz
ah so we translate it to saying that Eigenvalue-equations exist?
long nebula
Yeah, lambda here is the eigenvalue
So let's first try this with m_1 I - T
Can you find a vector in the null space of m_1 I - T?
fair quartz
the only equivalent statement that i found in my script is that the nullspace of that (m_n I - T) is not just the zero-element
long nebula
Yes that's correct
So explicitly, m_1 I - T acts on a_n by sending
(a_1, a_2, a_3, ...) to ((m_1 - m_1)a_1, (m_1 - m_2)a_2, (m_1 - m_3)a_3, ...)
fair quartz
Can you find a vector in the null space of m_1 I - T?
i tried this before but i thought the only vector in this nullspace would be the zero vector
long nebula
Can you use that to find a_n not all zero such that (m_1 I - T) a_n is all zero?
long nebula
i tried this before but i thought the only vector in this nullspace would be the...
This is not correct
fair quartz
i already thought there had to be something with the one element where the indices match up
fair quartz
can you do the same for any m_n?
should work the same, no?
long nebula
No problem!
teal vessel
so on both the stabilizer and the kernal, since the identity of G has to be a member of both, can't I simply show that the identity 1 can be decomposed as (g^-1)(g) where at least g is a member of said subset, and then instead of 1.a I now have (g^-1)(g).a, which can be evaluated as (g inverse).a because we know that g.a=a and because of the definition of action?
fair quartz
this might be a silly question but is this not the same as saying lambda is element of that set?
delicate orchid
so on both the stabilizer and the kernal, since the identity of G has to be a me...
I buy it :smoked:
fair quartz
ohh or does this specifically mean that lambda is not element of the set of m_n with noninteger indices?
teal vessel
I buy it :smoked:
can't tell if that's a 420 reference or a "successful demonstration" reference 😅
wraith cargo
this might be a silly question but is this not the same as saying lambda is elem...
Lambda is an element of the closure of that set
Or sorry
Not an element
Lol
That means that lambda is not an element of that set and is not the limit of any sequence of elements of that set
Tbh actually depends on the notation lol
slim kayak
so on both the stabilizer and the kernal, since the identity of G has to be a me...
Yeah, you just show that g^-1.a = g^-1.(g.a) = 1.a
fair quartz
so basically, lambda is any complex number that does not happen to be one of the m_n?
wraith cargo
Tho usually a bar means closure?
But your book could define it differently
slim kayak
Thinking it as a decomposition seems a bit more complicated that it needs to be
fair quartz
Tho usually a bar means closure?
But your book could define it differently
there is a chance that we dont have a definition given at all in our lecture
wraith cargo
so basically, lambda is any complex number that does not happen to be one of the...
It's even more stringent that that
It also can't be a limit of any sequence of elements of that set
fair quartz
i see 🤔
though in this case its inverted because its not an element of the closure right?
so this bar + not being an element basically just means that its either an element or the limit of a sequence of elements
slim kayak
Or alternatively, lambda is an element of an open set that doesn't intersect any of the m_n
wraith cargo
though in this case its inverted because its not an element of the closure right...
No
I was talking about the case where it isn't an element
fair quartz
oh boy 😂
but if the closure of that set is essentially just C without any of the m_n, wouldnt that mean, that not being element of that closure means that lambda has to be one of those m_n? (or a sequence limit)
fair quartz
It's even more stringent that that
It also can't be a limit of any sequence of ...
but that definiton makes a lot more sense with what i have to prove in that next part 😅
slim kayak
Take C, remove all m_n and anything a subsequence of the m_n can converge to. This is the space of your values of lambda
fair quartz
then i must be misunderstanding what a closure is
slim kayak
This is equivalent to any of m_n having a distance of at least epsilon from lambda
fair quartz
i went with this explanation from basic set theory
fair quartz
interesting 🤔
then my paper probably wants me to asusme that the bar means closure, because that definition immediately implies what i have to prove
slim kayak
Ngl I never seen the bar notation used in a context where one also considers the topology
It would also be a double negation if the bar meant complement
fair quartz
yeah thats where my original confusion came from 😄
slim kayak
Yeah no, you want to make the set of the m_n's into a closed set, and the closure is the smallest such set
fair quartz
gotcha!
wraith cargo
Ngl I never seen the bar notation used in a context where one also considers the...
Tbh set theory is the place where I see the most variation in notation for some reason
slim kayak
Probably cuz everyone knows the basic and every so often has to teach it
wraith cargo
Why do people still have different standards for which inclusion means what
slim kayak
So over time people just make up notation and throw it in there
wraith cargo
It's the thing that bugs me more than anything else
open sluice
formal ermine
maybe a stupid question but how does an automorphism R -> R induce an automorphism R/p -> R/p for a prime p? is it just [r] -> [phi(r)]?
coral spindle
Which is why you should post context 
south patrol
The context will probably be that the automorphism fixes p lmao
In which case yes this is just like first iso basically
next obsidian
It induces an iso R/p -> R/f(p)
Whether or not p = f(p) is a question which… is kinda like Galois theory kekw
bitter shard
hello, can someone help me to make a short exact sequence for M module and A,B submodules like that?
0 → M/(A ∩ B) → M/A × M/B → M/(A + B) → 0.
rocky cloak
hello, can someone help me to make a short exact sequence for M module and A,B s...
You pretty much have it already. Just map m -> (m,m) for the first map and (m, n) -> m-n for the second
bitter shard
You pretty much have it already. Just map m -> (m,m) for the first map and (m, n...
and because i want the image of first map equal to the kernel of second map, how i know that m-n cant be in the kernel of second map unless m=n ?
for example couldnt m-n be in A+B ?
rocky cloak
for example couldnt m-n be in A+B ?
So imagine m-n is in A+B, that means that m-n = a+b for some a and b. Hence m-a = n+b. Thus (m, n) is the image of m-a=n+b from the first map
Because m-a+A = m+A and n+b+B = n+B
white oxide
how can i find the other conjugates in C over R? i know that there are 3 other ones (since the irreducible polynomial for sqrt(2) + i over R is x^4 - 2x^2 + 9) and C is algebraically closed
another one is obviously sqrt(2) - i
don't know how to find the other two without resorting to the quartic formula or trial and error
formal ermine
how can i find the other conjugates in C over R? i know that there are 3 other o...
that's not the min poly tho?
or am I tripping
white oxide
that's not the min poly tho?
i just let alpha = sqrt(2) + i
then i squared both sides and did some arithmetic
and i ended up w alpha^4 - 2alpha^2 + 1 = -8
maybe i can't do arithmetic
formal ermine
yeah
x = sqrt(2) + i
(x - sqrt(2))^2 = -1
x^2 - 2sqrt(2)x + 3 = 0
rocky cloak
Over R any element has degree 1 or 2
formal ermine
^
white oxide
Over R any element has degree 1 or 2
oh right because c is a 2-dimensional vector space over R
thank u
wait im confused tho
say you do have like a quintic polynomial with coefficients in R
then there would have to be 5 zeros in C right
rocky cloak
That is the minimal polynomial over Q though, and the cunjugates are ±sqrt(2)±i
white oxide
since C is algebraically closed
formal ermine
then there would have to be 5 zeros in C right
accounting for multiplicities, yes
slim kayak
since C is algebraically closed
Yeah, all those roots can be expressed as products and sums of any one of the roots
white oxide
okay thanks ppl
barren sierra
if V is a vector space
delicate orchid
V would never...
barren sierra
😔 sadly it has
is the "group of translations of V" is the group of all maps T(v) = v + x or the group of all maps T(v) = cv + x, c a field constant and x an element of the vector space?
delicate orchid
I'd call the latter the Aff_1(V) so I'd go with the former
delicate orchid
kind of a weird way of phrasing it as this group is clearly just the underlying group of V
white oxide
how much linear algebra do i need to review before diving into representation theory of finite groups
i took two courses on it, one computational based
coral spindle
Do you know what an eigenvalue is?
coral spindle
Then you're probably good
white oxide
okay bet lol
coral spindle
Honestly I don't think it's very heavy on lin alg.
white oxide
i kinda forgot some shit though bcuz back then i would just read the chapter once then go straight to the exercises
i didnt rlly know how to study math
coral spindle
I guess you do need to know what a quotient space is
white oxide
really
delicate orchid
ermmm well errr well aktually there are some ermmm intricies which may be ermm slightly beyond the rather uhhh pedestrian nature of a uhh linear lagebra course
coral spindle
That's something I should say
white oxide
yea i know what a quotient space is
or at least
i learned about it
let's say that
delicate orchid
wait why tf do you need to know what a quotient space is
coral spindle
Well need is a strong word
delicate orchid
no but like
coral spindle
Yeah ig you can just use subspaces
summer path
You can probably just start and then review what you need as you go along
delicate orchid
yeah
delicate orchid
le epic semisimple ring
oh no no no no
white oxide
okay i will thank youu 😄
delicate orchid
oh no no no no
they do not know oh no no no their block algebras have trivial jacobsons oh no no no
barren sierra
wait why tf do you need to know what a quotient space is
because they're fun that's why
delicate orchid
They’re literally the most boring thing ever
Unless you’re, for some god foresaken reason, referring to quotient modules as quotient spaces
white oxide
why is Q(\sqrt(2), \sqrt(3)) a splitting field of $x^4 - 5x^2 + 6)$? it only contains $\sqrt(2)$ and $\sqrt(3)$ but the other solutions are not contained in it right or am i trippin
i give up on latex
oh wait
multiplicity
wait no
the zeros are +-sqrt(2) and +-sqrt(3)
oops
slim kayak
That's mad bruv
celest furnace
On top of today being chewsday (This was not meant to be offensive)
warm wyvern
@celest furnace you can safely skip those, they aren't strictly necessary to understanding the material
it's just that aluffi was like
written for grad student so he prolly thought they had experience with LA
summer path
I mean surely if you aren't an early undergrad, you have experience with lin alg
rustic crown
tubuwu
warm wyvern
On top of today being chewsday (This was not meant to be offensive)
how tf would that be offensive?
rocky cloak
how tf would that be offensive? <:opencry:586078614865444865>
Making fun of people's accents..?
warm wyvern
delicate orchid
I'll knok yer fookin block of
warm wyvern
that's silly
summer path
Tuwusday
celest furnace
MoondaY?
delicate orchid
yes way
delicate orchid
it's why full moons on a monday have a more powerful spirtual energy to them
warm wyvern
summer path
tubuwu <:eeveeKawaii:556858499930259471>
detuwu
celest furnace
But i thought sunday was the first day of the week
delicate orchid
yeah if you're jewish, sure
celest furnace
not joking
delicate orchid
cause then the sabbath is on saturday
warm wyvern
But i thought sunday was the first day of the week
are you a muslim?
delicate orchid
oh yeah it could be muslim actually
celest furnace
No
delicate orchid
apologies if I got them backwards
warm wyvern
rustic crown
do you not like 0 >.<
warm wyvern
subbath is jewish thing
delicate orchid
holy day of prayer for muslims is on a friday right
warm wyvern
yea
delicate orchid
I WAS RIGHT!
green: sane
blue: shitholes
dark blue: based
celest furnace
Dark blue is based
warm wyvern
dark blue are such chad ong
celest furnace
55% starts on sunday
long nebula
green: sane
blue: shitholes
dark blue: based
as a blue I agree
delicate orchid
smurf lookin mf
summer path
Never start week, only month
long nebula
time just goes
celest furnace
W history
warm wyvern
it'd be kinda cool if time didn't go either tbh
rustic crown
no week => no weekend 
delicate orchid
people worked six days a week
death
rustic crown
det can't think about working more than 4days/week
warm wyvern
>people worked six days a week
death
students be like:
delicate orchid
I think anyone would be like
god I'm so glad to have been born when I was it's unreal
summer path
Det works 4 days/week?
summer path
void cosmos
yo how is every unitary R-module a homomorphic image of some free R-module?
void cosmos
a module over a ring with identity :d
coral spindle
Hint: choose generators for your module M, let's say they form a set S.
Build a free R-module from S to get what you want.
void cosmos
i get to assume its finitely generated?
coral spindle
No
void cosmos
sorry wydm choose generators
coral spindle
If you want you can just choose S = R.
void cosmos
yea
coral spindle
Imagine you have some generators. Try building a free R-module using that information to get what you want.
void cosmos
and then get a homomoprhism from R to M cuz i have the inclusion map from the "supposed" basis ( R )
right?
is that what u had in mind
coral spindle
I think you should try working it out
delicate orchid
true by definition of free? lol
void cosmos
cool
delicate orchid
construct the homomorphism
night onyx
another hint: for any set there's a free module with that set as a basis, so there should be an obvious choice with what set to take, and how to construct the homomorphism
void cosmos
yea its the same construction as with the proof of having a basis <--> free in the category
like the same map i think
night onyx
yeah, the idea is that M has relations in it, and by taking the free R-module over M, you're essentially forgetting all those relations, and then by quotienting the free module by some submodule (the kernel of the homomorphism you construct), you're rebuilding the relations in the original module
lol in a way every structure is some free structure with some constraints (given by taking a quotient)
alpine island
Everything is just F[x]/(something)
night onyx
yup
and it's useful to keep that in mind because later when you're constructing various algebras, the general approach is to quotient the tensor algebra by the "right" submodule
alpine island
Well, unless you have infinite basis
delicate orchid
and it's useful to keep that in mind because later when you're constructing vari...
or the free algebra if you're working with some DISGUSTING specimens
night onyx
lol yeah, I like the perspective that taking a quotient is just imposing extra relations, and that quotienting free objects is like, the general way we construct algebraic structures, it made quotients make a lot of sense intuitively when I first started learning about them
delicate orchid
completely agree with this idea
night onyx
and it was kindof eye opening too, because suddenly all sorts of constructions just made sense, I think at the time I was learning about Clifford algebras
delicate orchid
oh god yeah
I don't know how you'd get by without thinking of clifford algebras as "tensor product but woahhhh funny quadratic relator"
alpine island
I don't know how you'd get by without thinking of clifford algebras as "tensor p...
I got to them with Geometric Algebra
I still don't know what the hell a tensor is
delicate orchid
grrrrrr
night onyx
lol well that actually fits in to what we're talking about, because tensor products are basically structures where you can "multiply" but the multiplication has no relations
delicate orchid
I will never respect geometric algebra as an actual field I apologise
it just smells like physics too much
night onyx
so in a weird way tensor products are like "free multiplicative structures"