like on two sets you can have all kinds of functions, but there only needs to be one bijection for them to have the same cardinality
#groups-rings-fields
chilly ocean
I understand
so my aim would be to construct a function
that is bijective and homomorphic
and since G, G' is arbitrary thats the key
stone fulcrum
In case it wasn't clear, if two groups are isomorphic they're "essentially the same group but the elements may have different labels"
chilly ocean
Yea ive seen that definition
stone fulcrum
Cool cool lol this convo would be weird if you haven't seen that
chilly ocean
in the case of groups with order 4, the reason why there are 2 groups is because there does not have to be a generator right?
namely in the second case that it has 2 subgroups of order 2
sharp sonnet
in general not all groups are cyclic
chilly ocean
alright
stone fulcrum
@chilly ocean
It is a good observation that V4 can't be cycled with one element, which differentiates it from Z4
Also note that Z4 has elements of order 4, where V4 has none
chilly ocean
ic
chilly ocean
Is by any means this isomorphism right? $Z[x]/(x^{2})\otimes Z[y]/(y^{2})\simeq \mathbb{Z}[x,y]/(x^{2},y^{2})$
cloud walrusBOT
SMC:
chilly ocean
I am a newbie in homological and cohomological algebra and Hatcher does not really make it easy for me 😦
sour plume
Seems good, right? You could for example write down a morphism $\mathbb{Z}[x,y] \to \mathbb{Z}[x]/(x^{2})\otimes \mathbb{Z}[y]/(y^{2})$ mapping generators in the (fairly) obvious way, prove that this is really a morphism and finally that the kernel consists of things which are spanned by $x^2$ and $y^2$
The important thing is that, using that $x^2 = y^2 = 0$, all elements in both spaces can basically be written as just finite sums, namely $a + b \cdot x + c \cdot y + d \cdot xy$, where $a,b,c,d \in \mathbb{Z}$
chilly ocean
should we not treat xy and yx differently?
xyx it's not clearly zero while yxx has to be, is this not true?
sour plume
I'm assuming you're working with commutative polynomials
I think that's the context in which hatcher works too?
chilly ocean
that's what im not getting
it should not be commutative
i mean i am almost sure that my isomorphism is not right
the problem is that i don't understand why nor how it should be isomorphic to
sour plume
Hm, I see
chilly ocean
Let x and y be the generator in H^{1} i can tell that
sour plume
I think the problem is a difference in multiplicative structure or so?
chilly ocean
$x\smile y= - y\smile x$ via degree
cloud walrusBOT
SMC:
chilly ocean
but i do not understand how to relate this to a ring structure
yes, the problem has to be related to some cup product property, I am almost 100% positive about htis
this
and via kunneth theorem i am almost sure that the structure has to be the tensor product on the left, the tricky thing is that you cannot "smash" polynomial ring in that way when considering tensor product
sour plume
Isn't that exactly the problem though? In $\mathbb{Z}[x,y]/(x^{2},y^{2})$, all multiplications are commutative, where in your ring, you want $xy = -yx$
cloud walrusBOT
Lartomato:
sour plume
And if you understand the tensor product as a tensor product of graded rings, this antisymmetry is baked into the definition of the tensor product
or, in the definition of the multiplication on the tensor product of graded rings
So the isomorphism that you started with holds if you view the tensor product as over $\mathbb{Z}$-modules, but if you consider it as a tensor product of graded rings, it's not true anymore
cloud walrusBOT
Lartomato:
sour plume
Something like that?
chilly ocean
yes ok now I understant that the problem is that i am mixing the notion
So the point it's that I should be happy in seeing it as the tensor product of the two ring with the relation stated aboved about degree in cup product? it does not traslate in any easier algebraic structure that i should be familiar with, is this right?
Because i read reference about exterior algebra and stuff like this but they were never touched in my algebra courses
sour plume
Yeah, I think you got it. This is about the tensor product of graded rings, so you'll need to think about the multiplication a little differently.
chilly ocean
Appreciated
woeful flint
Is it the case that $\mathbb{Q}(\sqrt[3]{4}) \subset \mathbb{Q}(\sqrt{2} + \sqrt[3]{4})$
cloud walrusBOT
Kraft Macaroni:
cloud walrusBOT
bertwit:
steady axle
bertwit:
@cloud walrus Thus was dumb sorry
steady axle
@woeful flint I claim $\mathbb{Q}(\sqrt{2} + \sqrt[3]{4}) =\mathbb{Q}(\sqrt{2} , \sqrt[3]{4})$. \ Suppose not. Degree of $\mathbb{Q}(\sqrt{2} , \sqrt[3]{4})$ is $6$. So the minimal polynomial of $ (\sqrt{2} + \sqrt[3]{4})$ must have degree $2$ or $3$. Show that this is not possible over $\mathbb Q$
cloud walrusBOT
bertwit:
obsidian zealot
How can I do this problem
sharp sonnet
by using the definition of kernel
sharp sonnet
blissful ice
Kernel is the set of values that get mapped to 0 (in this case 0+12Z)
astral aspen
stone fulcrum
Start forming the cosets.
You've got the subgroup <(4,4)> which is one of them.
Now, take an element not in <(4,4)> and add it to every element in <(4,4)>, this will give the coset that contains that element
solemn rain
i dont know what 'compute' means
but just keep doing (a,b)+(factors of 4, factors of 4)
where a in z6 b in z8
stone fulcrum
'Compute' meaning 'give a simpler representation for this group'
stone fulcrum
Oh yeah @solemn rain sorry if I confused you. I didn't mean "representation" as an actual "representation"
Compute means "express as a simpler group"
woven delta
Well in this case it's not hard to show that the quotient is abelian
So a good representation might be as a product of cyclic groups
woven delta
What's the problem with that?
stone fulcrum
Lol yeah it's a bit of a circular definition isn't it
woven delta
It means that there is a natural map of Z to any ring
solemn rain
what
oblique river
how is it circular
stone fulcrum
I assume they mean nCk as a regular integer, when they say "taken nCk times"
solemn rain
on rings
oblique river
n choose k is a perfectly well-defined integer
chilly ocean
It just means add 1 nCk times
woven delta
So you define nCk in Z
solemn rain
they define n choose k in another way buncho
in the remark
n choose k is 1+1+1... nchoose k times
oblique river
lmao
solemn rain
wtf does this mean
woven delta
And then take the image of that under the map from Z to R
oblique river
just listen
stone fulcrum
For example
3C1 = 3 in N
So in R it's 1 + 1 + 1
oblique river
n choose k is an integer
solemn rain
yea goti t
oblique river
then in R
solemn rain
yea yea
chilly ocean
1 may not be 1
oblique river
add up 1 that many times
solemn rain
yea got it
oblique river
that's "n choose k in R"
solemn rain
why cant n choose k be n choose k
oblique river
n choose k is an integer
chilly ocean
because 1 in R may not be an integer 1
oblique river
what does "n choose k in R" mean
solemn rain
ty
oblique river
not helpful
solemn rain
u talking to me?
chilly ocean
lmfao
woven delta
Ik lol
chilly ocean
u can always induct
oblique river
yes that is one way
solemn rain
yea cool
man rings are awesome as fuck what seperates them
from like the things we did in algebra in middle school
or high schol
whats missing?
oh nvm
woven delta
Lol
stone fulcrum
Inverses haha
stone fulcrum
They're "like matricies" that's how I think of them
chilly ocean
lol what
solemn rain
yea matricies are big part but like
stone fulcrum
Except even rings get to be commutative sometimes
solemn rain
polynomials are like BIG shit here
woven delta
Most high school students don't know what a real number is
chilly ocean
why
solemn rain
i dont know what they are
okay guys for problem 26
the part that i am not sure about is 3 which is not shown
proving that any integral domain has char p
or 0
stone fulcrum
solemn rain
yea 3 isnt there
proving that any integral domain has char p or 0 where p is prime
so what i tried was this
i was just wanna check solutions
woven delta
Let n be the minimum number st n*1 = 0
solemn rain
so now i use this map defined above
and call it phi
wait
damn ur fast boys
thats not what i didi
can u check for me solution?
woven delta
Lol
latent anvil
:^)
solemn rain
now let this map defined above be called phi
latent anvil
it was not an entirely serious suggestion
solemn rain
suppose for the sake of contradiction R is an integral domain that has composite char not = 0
let char(R) = mn
for integers m,n
phi(mn) = 0
phi(m)phi(n) = 0
m and n are not 0
contradiction
?
chilly ocean
I mean jsut using def of id should be neough?
solemn rain
im bad
woven delta
You don't
woven delta
Just say "suppose $nm1=0$, then $(n1)(m*1)=0$"
cloud walrusBOT
Liquid:
solemn rain
whats nm*1
woven delta
Gross
woven delta
So I'm using that to signify adding up 1 nm times
solemn rain
yea
yea damn
wow thats cool okay got it
okay 1 more
R and S are rings with homomoprhism f : R--->S , if x is a nilpotent element in R then f(x) is nilpotent in S
proof: let x be nilpotent in R
dawn kiln
wrong
chilly ocean
qed
dawn kiln
glad i could help
solemn rain
XDD
wtf
let x^n=0 for some n in N ( positv intg
phi(x^n)=phi(0)=0=phi(x)^n
since phi is a group homomoprhism
qed?
woven delta
Yeah this is trivial
solemn rain
yea thats subjective no?
chilly ocean
no
solemn rain
do i prove
that (a+b)^p = a^p + b^p in ring of char p
using binomial theorem?
and removing the terms that have p since they are 0?
chilly ocean
ye
solemn rain
col
do i hve to read all this shit?
ppl who did dummit and footer
for examples?
or jsut one or two will suffice?
chilly ocean
those examples are cool read all
solemn rain
they look so boring XD
bleak abyss
I mean I will say you can literally write that example in like
2-3 lines lol
Namely if (2,x) = (a(x)), then a divides 2 and x. Uh oh.
So actually less than one line
bleak abyss
DF takes far too long yeah
solemn rain
cool
shiould i switch?>
i am still in beginning rings
so it shouldnt be that bad right?
or hsould i just hang on?
bleak abyss
Eh if you haven't gotten far you could probably stand to switch yeah
solemn rain
all i know are rings , homomorphisms , quotient rings, and ideals
and some examples like group rings polynomial rings and matrices
oh and integral domains
are math textbooks supposed to be just def them proof literally
like lang?
or this like df?
or in between/
woven delta
Lang has good problems
woven delta
Dummit and foote has some bad problems
woven delta
No
woven delta
Yes
woven delta
I have a copy of DF in my bag rn actually
solemn rain
no need
just pick a section in lang ( that is in df obv )
and watch
i really hope we are on the same page and talking about grad lang
not undergrad alg lang
woven delta
If there is a section in Lang that is not in DF
solemn rain
but id saay both are the same
woven delta
Then I win automatically
solemn rain
umm
well df is big so i can have a chance
i mean what would lang have df doesnt?
df has everything no?
woven delta
No
woven delta
Okay the problem in Lang that tells you to do all the problems in a homological algebra book
woven delta
It does
sharp sonnet
not in 3rd edition i think anymore
sharp sonnet
like it was in first edition of lang
solemn rain
okay new rule
sharp sonnet
and maybe second
woven delta
He never specified the edition
solemn rain
yea yea now i do u player
woven delta
So I'm okay with it
woven delta
There are a ton of sections in Lang not in DF
sharp sonnet
but like im pretty sure that lang's section on homological algebra still is very different from DF
solemn rain
tbh iw ould doubt but idk
woven delta
He's right
upper pivot
i have 3rd edition on me rn
solemn rain
dont u have a pdf of it?
steep hull
This is so annoying to watch
solemn rain
ik tree3 but he started it 🤷♂️
woven delta
No I didn't
woven delta
🐶
I know that Lang talks about ses's a lot earlier than DF
So there are probably many ses questions early on not in DF
solemn rain
well that cancels out
you know why?
i have my ace in the hole
groebner bases
he doesnt talk about that does he
lang
woven delta
But I doubt that there is much intersection
Why does that matter
You made a statement that DF has every problem in Lang
solemn rain
damn i agree tbh
1 sc
@solemn rain
wait the other account is gone hwo the fuick\
okay i agree
upper pivot
i mean DF is still garbage tho
solemn rain
yea i tried to read atiyah mc for rings but turns out too hard for me
im really getting sick of the long texts
woven delta
No shit
woven delta
You don't know linear algebra
upper pivot
half of it modules lmao
solemn rain
i dont like artin
sharp sonnet
then read linear algebra book first
woven delta
Read Hoffman kunze
upper pivot
how do u do AM w/o LA
solemn rain
i dont like hoffman kunze
woven delta
Or even axler
solemn rain
i wanted to try serge lang ug algebra
but turns out it has like
3 sectios on rings only
woven delta
I honestly like axler
solemn rain
so i thought it wouldnt be that much of good
it doesnt talk about these domains
which have their own chap in df
ufd pid ..
upper pivot
it almost definately does?
woven delta
I like the way DF does linear algebra tbh
upper pivot
what else would u teach in intro ring lmao
upper pivot
even then why would it not talk about that hmm
solemn rain
and he does like
solemn rain
modules in a very weak way compared to what is ee in df
like he doesnt tlak about sequences
which are alot of shit in df
thats how i compare atlesat
exact sequences to be exact hahahaa
sharp sonnet
artin seems to be the way to do algebra and linear algebra in one go
also artin covers some nice things that other linear algebra books do not
solemn rain
i dont like artin for some reasona
upper pivot
i personally recommend it, and it really picks up in later chapters
woven delta
Lol
upper pivot
read one of his ring theory chapters lol
woven delta
That's dumb
solemn rain
i jsut want to learn vector spaces just as i woudl learn groups
sharp sonnet
read finite dimensional vector spaces by halmos
woven delta
Or axler
sharp sonnet
its the underdog LA book
sharp sonnet
yeah or axler
woven delta
Axler doesn't like matrices so much lol
upper pivot
lmao..... too old school
solemn rain
XD its like the first text on VS
woven delta
But matrices are nice
bleak abyss
Yeah you just need to cure yourself of this anti-matrix syndrome lol
It's one thing to be like
solemn rain
i dont like them from HS
bleak abyss
I don't wanna put 25 matrices into rref
upper pivot
go do some physics ull start appreciating matrix more
solemn rain
i felt so bad at myself for not being good at gaussian stuff and turnign a matrice into fuck -fuckery echelon form
so i just quit it
woven delta
Go calculate some homology
solemn rain
idk what that is
woven delta
You'll appreciate matrices more
solemn rain
df doesnt assume LA from reader
upper pivot
sharp sonnet
df is also not a LA book
bleak abyss
DF doesn't assume non-trivial linear algebra background
solemn rain
thats when i knew it i can learn LA from it
@bleak abyss i knwo how to add matrices , and i google multiplicationevery time
upper pivot
artin does a lot of LA. its a bit matrix heavy but most of it is "abstract". i think you just havent seen matrices enough lmao
bleak abyss
But the question is whether just reading it gives you enough proficiency as if you went through like
Hoffman-Kunze
solemn rain
i started hoffman kunze cuz jacobian recommended me it ( he was the nicest guy ever )
i enjoyed the later chapters but the first chapters were so fucking hard for me
and it was the first time to read an actual math textbook
and mind you im very bad at math
so it didnt work out
bleak abyss
The answer is to improve lol, if that's too hard you need more basic experience with proofs
solemn rain
i meant like 2 months earlier
now idk about it but i just have this bad feeling so i dont try it
upper pivot
rich patrol
Convo moved to #math-discussion if you want to catch up
stone fulcrum
@solemn rain
Let x,y be generators, and
xy² = y³x
yx² = x³y
What group is this?
stone fulcrum
It is haha.
woven delta
Design an algorithm that will tell if a finitely generated group is the trivial group
@solemn rain
warped bay
really?
woven delta
Yeah
warped bay
oh, seems about right
chilly ocean
why is it impossible
solemn rain
yea thats what i hate about these shitty stuff
i i would have never knew that
if u hadnt told me
dont do this again
solemn rain
wdym no
woven delta
I will do this again
solemn rain
no u wont
chilly ocean
bare minimum if its hard then no
woven delta
Also have you decided if you are going to leave the server
Or do every problem in axler
warped bay
this is why i love abstract alg
woven delta
Do every problem in axler then
solemn rain
okay im realy down to learning lin algebra since u guys know better
but can u choose another text
?
i hear axler is p bad
and not hoffman too
woven delta
Nah it's alright
solemn rain
are u sure? im of low math matuirty 🤷♂️
woven delta
Except for the fact it doesn't do determinants
But that's not too bad
You'll encounter those elsewhere
It's more important you do the rest of the stuff
chilly ocean
i have a question
solemn rain
jsut ask ok
woven delta
upper pivot
i think you should honestly do some intro to proof mo2men
solemn rain
from me tbh
upper pivot
oh wait u already did intro to proof?
solemn rain
yea
smoky cypress
Bruh
solemn rain
i really dont know whats missing aside from that countable stuff
smoky cypress
That’s not abstract algebra @chilly ocean
solemn rain
but okay u know btter
rich patrol
@chilly ocean wrong channel; see #prealg-and-algebra
upper pivot
where did you do intro to proof
upper pivot
yeah i know it
solemn rain
^ first book ever
solemn rain
yea why not
chilly ocean
bruh if u read it once ure good
solemn rain
what am i missing
upper pivot
like try doing some exercises rn w/o reading ch
if u can solve them then ur good
if not, yeah you need to review
(of vallemen)
solemn rain
axler looks nice
as in like book
not content
like its beautiful
yellowish
blue
woven delta
The content is good
Do you know the proofs of the theorems in algebra
Or do you just cite them
solemn rain
what theorems in algebra
woven delta
Like let's say first isomorphism theorem
solemn rain
the isomoprhism is basically defined as
phi(n) = f(Pi(n))
where phi is the homomorph
and pi(n) is natural
f is the iso
this goers to f(nH) = phi(n)
woven delta
Well obviously
solemn rain
i just gave out the isomoprhism
woven delta
Like at the stage you are you can't just say "the rest is details"
solemn rain
okay
okay ur right
let G and H be groups
f:G--->H be homomorphisms
define phi:G/ker(f) ---> img(H)
phi(gK) = f(g)
where K is ker(f)
first the function is well defined
suppose gK = hK
--> h^-1g is in K
so f(h^-1g) = 1 ---> f(g) = f(h)
now phi is also a homomprihsm
phi(gK)phi(hK) = f(g)f(h) = f(gh) = phi(ghK)
thats it?
phi is naturally surjective ig
@woven delta
inejcitvyt
XD
okay
yea mb
okay so suppose phi(gK) = phi(hK)
chilly ocean
proof of 1st iso thm: define the mappings, rest are details
woven delta
Details are hard
solemn rain
yea sorry
about that
what am i doing now
am i done
no
suppose phi(gK) = phi(hK) ---> f(g) = f(h) ---> f(g)f(h)^-1 = 1 ---> f(gh^-1) = 1 ---> gh^-1 is in K
---> gK = hK
am i done ?
woven delta
Looks fine
solemn rain
cool
woven delta
But anyway the whole "rest is details" attitude is bad
solemn rain
yea i thought those were easy tbh
woven delta
Because a lot of people just memorize theorems
And cite them instead of actually doing math
solemn rain
yea i know about those
and i just didnt want to be one of them
but
there are theorems that i cant prove
that ik about
like fundamental theorem of abelian groups
i cant do this 1
or
woven delta
You shouldn't cite them then
woven delta
Fundamental theorem of finitely generated abelian groups isn't so bad tbh
You should know how to do cauchy
solemn rain
yea just boring induction xd
woven delta
It's the only nontrivial step in the proof of sylow
chilly ocean
tbh I feel like I cant prove any multivar calc theorem
woven delta
No you don't
raw moth
proof of cauchy is like
solemn rain
jk
raw moth
not that bad
woven delta
Because you don't know how to do cauchy
raw moth
or you could understand it
chilly ocean
cauchy proof was like very tricky right?
chilly ocean
my prof said its his favourite proof
raw moth
given one step it's p. easy
upper pivot
hmm the proof i saw for first sylow doesnt use cauchy, but proofs cauchy as colloraly
chilly ocean
and its been unproven for a long time iirc
solemn rain
i really wanted ap roof of the fundamental theorem tho
but the text just asid
'yea you take those in modules u prove more generlized stuf bye'
it didnt prove it for me
raw moth
then prove it yourself
woven delta
^
solemn rain
nah
woven delta
The proof isn't so bad
solemn rain
i cant do it
upper pivot
yeah the proof i know uses matrix
raw moth
what is the theorem
chilly ocean
wahts the fundamental theorem of algebra?
upper pivot
we are talking fundemental theorem of finitely generated abelian groups
chilly ocean
its not if I think thats the polynomial thing
raw moth
should I just google the statement
woven delta
That's another theorem you should know
woven delta
Yeah
solemn rain
yea lmao Xdx
upper pivot
its just play with matrix to get nice form
solemn rain
@woven delta i mean why didnt the text put it at first then
upper pivot
basically
raw moth
like even without matrix
woven delta
What?
woven delta
Because it's usually proven in an analysis class
solemn rain
analysis a prereq of algebra?
woven delta
Cause the algebra proofs are hard
But the analysis proofs are easy
It's a theorem about the complex plane
Which is an analytic object
upper pivot
theres a nice proof that uses the fundemental group of the circle
which is somewhat algebraic
upper pivot
one step at a time
woven delta
You should learn what countable means first
woven delta
Okay mo2men, prove that a countable union of countable sets is countable
Not here though
solemn rain
okay where
woven delta
In general
hearty oyster
Does anyone know the name of the algebra you get when you take the Dynkin diagram for B_n or C_n and remove the orientation? Wikipedia calls it BC_n, but I can't find any other references to that name
sinful zephyr
Hi, I need help showing that given a ring homomorphism, the preimage of a prime ideal is a prime ideal
chilly ocean
what did you try
sinful zephyr
I showed that it's an ideal first, and we're using the definition that if p is a prime ideal, then if AB subset p, either A subset p or B subset p
so if f(p) = p' where p' is prime, I let AB subset p
but I wasn't able to construct a something like A'B' = p' that let me use the fact that p' is prime
sour plume
Is this for when A and B are ideals themselves?
sinful zephyr
yes
sour plume
I'm a bit confused with what you're doing: "so if p = f(p') where p' is prime, I let AB subset p" wouldn't you rather want the preimage?
Or is this a typo
sinful zephyr
that was a typo
sour plume
Ah, alright, so you assume AB is a subset of f^{-1}(p'). So you can conclude that f(AB) is a subset of p'
And then it seems to go pretty straightforwardly if you use f(AB) = f(A)f(B)
(which should be a consequence of f being a ring morphism)
sinful zephyr
but then aren't f(A) and f(B) not necessarily ideals?
sour plume
Ah, I guess that's true. I thought images of ideals still might be ideals
chilly ocean
wait arent they?
sinful zephyr
if you let f(x) = x from Z to Q be a homomorphism
sour plume
Hmm, maybe something can be done with the ideals generated by f(A) and f(B). I'll think a bit
chilly ocean
Need assumption its surjective as well I think
sinful zephyr
Our instructor noted that f does not need to be surjective, since someone asked
sour plume
Hm, okay, so first one can show that if f(AB) is in p, then the ideal generated by f(AB) is in there as well. Is this by any chance equal to the product of the ideals generated by f(A) and f(B)?
sour plume
Yeah, I'm also not 100% sure about that, but I've got a semi-good feeling about it
because otherwise I'm out of ideas lol
The inclusion from left to right is true at least 
Yeaaah okay now I'm also not so sure about that anymore though
A small silly question: Are your rings commutative?
sinful zephyr
yes
sour plume
oh
sinful zephyr
well I assume we can just take that for granted
sour plume
Yeah that seems important, 'cus in the noncommutative case this is false
But in the commutative case I think (f(AB)) = (f(A))(f(B)) is pretty easy
Given an element in the right-hand side, by distributivity it's a linear combination of elements $r f(a) f(b) = r f(ab)$, and those are also in the left-hand side
cloud walrusBOT
Lartomato:
sour plume
(I reordered a bunch of things which needs commutativity)
I think this should lead to the right idea, yeah
(using that now (f(A))(f(B)) is in p, hence by primeness (f(A)) or (f(B)) is in p)
sinful zephyr
how would A relate to (f(A)) then?
sour plume
f(A) is in (f(A)), so if (f(A)) is in p, then f(A) is in p, and then preimaging should give you what you want
(of course note that A =/= f^-1(f(A)), but A lies in that preimage)
sinful zephyr
so if (f(A)) in p', then A subset f^{-1}((f(A))) subset f^{-1}(p') = p
with messy parentheses but I think I get it
sour plume
Ye if you write it down slowly and nicely it'll clear up
That was fun, even though I thought things were noncommutative half the time, lol
and: https://math.stackexchange.com/questions/399579/inverse-image-of-prime-ideal-in-noncommutative-ring for noncommutative rings this is not true
sinful zephyr
tyty
golden pasture
If every prime ideal of $A$ is maximal is equivalent to $A/\mathfrak{N}$ is absolutely flat(AM chapter 3 exercise 11), then doesn't this imply $\mathbb Z$ is absolutely flat(which it isnt)
cloud walrusBOT
ariana:
golden pasture
oh right XD
shrewd halo
For an isomorphism between 2 representations of a group, must it be the case that the same vector space isomorphism also gets applied to the images of the group elements? Or is it possible to transform the elements and the space in different ways that still yield an isomorphism?
oblique river
usually you talk about isomorphisms of representations of a fixed group G
you just require a phi: V --> V' such that phi(gv) = gphi(v)
oblique river
np
shrewd halo
Actually, follow up: I ask this because I feel like there should be an isomorphism between two representations of Dn that choose conjugate roots of unity as their "rotation generator". The representations I had been working with were over C^2, where rotations multiplied the two coefficients by opposite nth roots of unity and the flip would swap the two coefficients. Would the isomorphism I'm looking for in this case just be a swap between the two coefficients?
oblique river
it's not clear to me that those are necessarily isomorphic
for example, if your group if Z/4Z, then having a generator act on C by multiplication by i
is not isomorphic to having it act by multiplication by -i
shrewd halo
Well that's cyclic, no? I agree in that case
granite compass
Hi! Is this correct deduction: Since $\frac{1}{\sqrt{2}}=\frac{1}{2}\sqrt{2}$ we have that $Q [\sqrt{2}] = Q (\sqrt{2})$ and thus ${1,\sqrt{2}}$ is a basis for $Q(\sqrt{2})$ and now we can form an isomorphism between the basis of $Q(\sqrt{2})$ and $Q (i)$ which is analogously ${1, i}$. Thus they are isomorphic as Q-vector spaces
oblique river
but like, I don't see how making your group more complicated could fix that
oblique river
oh hmm
cloud walrusBOT
PAHUS:
granite compass
Nice! Thanks!
shrewd halo
Alright, I think I formulated it better. Say w is an nth root of unity. One such representation would have its rotation generator be
(w 0
(0 w * )
and the other's would be
(w* 0)
(0 w)
These two should be isomorphic if I choose my vector space isomorphism to be a swap between the two elements, and my group isomorphism to send those generators to each other. For some vector (a,b), applying the first group element and then applying the isomorphism will yield (bw*,aw). But applying the isomorphism first and then applying the second group element (aka the image of the first group element) will also yield (bw *, aw). Is that not an isomorphism between the two?
chilly ocean
can you have infinitely many generators in the group?
snow flint
https://crypto.stanford.edu/pbc/notes/group/ anyone know if this is an ok intro to group theory
chilly ocean
@chilly ocean Q is not finitely generated
chilly ocean
It is as a Q vector space
smoky cypress
Suppose $G$ is a non abelian group. Is it possible for $f:G\to G$ defined by $f(x)=x^n$ for some integer $n$ to be a homomorphism
cloud walrusBOT
Whoever:
sharp sonnet
I was about to say take n=1
upper pivot
lol
sharp sonnet
or -1
upper pivot
-1 is homo iff abelian tho
mild laurel
If some integer n has the property that n(n-1) is relatively prime to |G|, then this being a homomorphism implies that G is abelian
iirc this is almost an if and only if condition, except for a couple counterexamples
oblique river
here's a fun fact
if the function f(x) = x^n is a homomorphism for 3 consecutive integers n
then the group is abelian
upper pivot
wrong
upper pivot
ok
snow flint
can someone lmk if the above link is a decent intro to group theory
granite compass
So I've been stuck on this for a long time; If I have a field extension L/K and A is a collection of algebraic elements in L w.r.t. K. How do I prove K(A)/K is algebraic?
I think I can represent all x in K(A) as f(a_1,...,a_n) for some algebraic {a_i}, but thats about it.
mild laurel
Showing that the sum of two algebraic elements is algebraic and the product of two algebraic elements is also algebraic would be enough
mild laurel
@granite compass
golden pasture
@snow flint well it's ok if you just want a online but like better to use a book with proofs and exercises
snow flint
yeah, thats waht i imagined. Its just that like i want something to fill in the gap now that classes are over, without fully investing in like an actual textbook or smth
thx
golden pasture
l i b g e n
chilly ocean
@snow flint Lang’s “Undergraduate Algebra” maybe
golden pasture
jacobson basic algebra is uwu if you can lin alg
chilly ocean
what I have for now it's the following: An element $\varphi \in H^{n}(S.(X)\otimes S.(X))$ correspond to an homomorphism:
$$\varphi: S{i}(X)\otimes S{j}(X)\to \mathbb{Z}$$ where we let $i,j$ be all the integers that sum to $n$.
For every element in $S{n}(X\times X)$ we have a correspondent element in $(S.(X)\otimes S.(X)){n}$ via EZ. The composition $\varphi\circ EZ$ is then an homomorphism from $S{n}(X\times X)\to \mathbb{Z}$ and we define $\alpha{n}(\varphi)=EZ*\varphi$
cloud walrusBOT
Stephen:
chilly ocean
Any hint? Does this seems a viable way?
hot lake
what's the S thing and what cohomology groups are the H^i(X,Z) again ?
chilly ocean
By S I mean the chain complex it might be C.(X) in the literature? I think Hatcher refers to it as C.(X) the set of all continuos map from the i-th simplicial ∆->X
The cohomology groups are the coker/ker of the cochain complex Hom(X,Z)
I don't know why the latex erased all my _ in the indexing of the groups, sorry, @hot lake
wind steeple
it's bc you copy pasted the text on discord which parses _ with text styles, to keep all the symbol you need to modify your message to get de "source code" (not parsed message) and copy/paste it
chilly ocean
Thanks I will try next time
For the math related question any hint on how to proceed? Or is this just a dead end?
chilly ocean
I even tried a more "theoretical" approach but it seems too obscure to me
We know that $EZ$ is chain homotopic to the identity, this implies that $H_{i}(S(X)\otimes S(X))\simeq H_{i}(S(X\times X))$.
We recall now that if our chain complexes its free we have: $H^{i}(Hom(Y,\mathbb{Z}))\simeq Hom(H_{i}(Y),\mathbb{Z})$ in our case, using the fact that all the $H_{i}(X)$ are finitely generated free abelian groups, we have then:
$$\bigoplus_{i+j=n} H^{i}(X)\otimes H^{j}(X)\simeq H^{n}(Hom(S(X)\otimes S(X)),\mathbb{Z}))\simeq Hom(H_{n}(S(X)\otimes S(X),\mathbb{Z}) $$ $$ Hom(H_{n}(S(X\times X)), \mathbb{Z})\simeq H^{n}(Hom(S(X\times X), \mathbb{Z}))\simeq H^{n}(X\times X)$$
the requested isomorphism.
cloud walrusBOT
Stephen:
sullen island
hi, im an undergrad year 2 learning about hopf algebras so im sorry if this question is silly as my understanding of hopf algebras is only rudimentary: why does the coproduct on an element $x^n$ result in a finite linear combination indexed by i, where i runs from 0 to n? shouldnt an element in the tensor product $\mathbb{C}[x] \otimes \mathbb{C}[x]$ be an infinite linear combination ? (since $\mathbb{C}[x]$ is itself infinite-dimensional)
cloud walrusBOT
xy:
sullen island
so why is $\Delta (x^n) = \sum\limits_{i=0}^{n} x^i \otimes x^{n-i}$ ?
cloud walrusBOT
xy:
sullen island
<@&286206848099549185>
halcyon siren
Is C[x] infinite dimensional?
sullen island
yeah
halcyon siren
Are you sure?
sullen island
its just polynomials with coefficients in C
halcyon siren
hmm
halcyon siren
yeye
sullen island
yeah so im not really seeing why the result of a coproduct acting on an element x^n results in a finite linear combination 
halcyon siren
could be a typo
I notice that there are other typos in that pdf
(I found it online)
upper pivot
hmm im not familiar with the comultiplication stuff itself. but it seems they are saying each basis element is that. theres still infinite basis of the form Delta(x^n)
also given a basis, an element is defined as a finite linear combo of elements of the basis, as there is usually no concept of covergence of infinite sequences in a general ring(i might be wrong as i said im not familiar with this specific thing)
sullen island
well comultiplication is like the "dual" of a multiplication defined on an algebra
so $\Delta : A \to A \otimes A$
cloud walrusBOT
xy:
sullen island
and it has to satisfy $(id \otimes \Delta) \circ \Delta = (\Delta \otimes id) \circ \Delta$
cloud walrusBOT
xy:
upper pivot
hmm i see
anyhow the C[x] tensor C[x] is still infinite dimentional, as theres infinitely many Delta(x^n) which are the basis of this
atleast from what im understanding
sullen island
yeah, so im thinking that in the most general form , the coproduct should result in an infinite linear combination of basis elements in the tensorproduct, which doesnt seem to be the case
upper pivot
its still gonna be finite combo of these basis
sullen island
😮
upper pivot
its like, whenever you see generated for general rings or something, its gonna usually refer to finite since there is no concept of converge of infinite sums for example
(theres exceptions ofcourse, like ring of formal series over some ring)
chilly ocean
what I have for now it's the following: An element $\varphi \in H^{n}(S.(X)\otimes S.(X))$ correspond to an homomorphism:
$$\varphi: S{i}(X)\otimes S{j}(X)\to \mathbb{Z}$$ where we let $i,j$ be all the integers that sum to $n$.
For every element in $S{n}(X\times X)$ we have a correspondent element in $(S.(X)\otimes S.(X)){n}$ via EZ. The composition $\varphi\circ EZ$ is then an homomorphism from $S{n}(X\times X)\to \mathbb{Z}$ and we define $\alpha{n}(\varphi)=EZ*\varphi$
@chilly ocean any help with this?
cloud walrusBOT
Stephen:
upper pivot
yeah, also just think polynomial rings
like R[x]
its generated by 1,x,x^2 .... as an R module right, but a polynomial can only have finitely many of these
cloud walrusBOT
xy:
sullen island
like, they chose these in particular to define the coproduct
upper pivot
its prolly cause it satisfies the commutative diagram when taken in the summation
sullen island
it certainly does, but is there a way to "derive" it? because from the looks of it, this is a natural, intuitive way to obtain a coproduct 🤔
(though im not seeing the intuition)
upper pivot
i mean yeah i suppose if you played around with it a bit you could force it out. like from what im seeing if we said $$\Delta(x^n) = \sum c_i (x^{a_i}\otimes x^{b_i})$$ then using the diagram we can figure out some stuff, and then get in some good guesses ig
cloud walrusBOT
JohnDoeSmith:
sullen island
hmm i see
upper pivot
also this is likely just 1 way to define the coproduct on this
upper pivot
apperantly another coproduct would be using binomial coefficients instead of 1 as coeffs of each x^i tensor x^(n-i)
chilly ocean
I have a subset of complex numbers, that consists of all complex numbers, whose modulus is equal to one, under multiplication. Which group is it isomorphic to?
Should it be isomorphic to the group of rotation of the circle with radius one?
oh, and also - would such group be a topological group?
solemn rain
I have a subset of complex numbers, that consists of all complex numbers, whose modulus is equal to one, under multiplication. Which group is it isomorphic to?
S^1 ~ R/Z
consider the homomorphism phi:R--->S^1
phi(r) = e^(2pi*i *r)
look at the kernel
@chilly ocean
brisk granite
may I have a hint for this? I don't rly know where to start.
brisk granite
<@&286206848099549185>
chilly ocean
@solemn rain thanks
somber bramble
H isn’t free abelian so they can’t be isomorphic
and you shouldn’t be able to have an injection either
a free abelian group has infinite cardinality while H has cardinality 2
if H is Z/2Z and oyu have both an injection and a surjection then G must also have cardinality 2 and thus be Z/2Z
as that is the only group of that cardinality
I have the following situation:
I have a sequence $X_i$ of free groups on $2ⁱ$ generators with injections between them (can be more specific if necessary) and $X$ is what I believe is called the direct limit of them? (it’s the union of the $X_i$ modulo the relations that we consider things you can get to via the injection to be the same). Inclusions of each $X_i$ in $X$ induce a injective group homomorphisms. Then $X$ is free, supposedly. I don’t have to prove this but I’m supposed to provide a source and I’m bad at searching for them. Does anyone happen to know a theorem that implies that and a textbook that proves said theorem?
(I don’t know a whole lot of algebra and in particular have basically not done anything about free groups)
cloud walrusBOT
Sascha Baer:
sour plume
I think the proof wouldn't be too difficult if you already understand the direct limit of these groups; Construct in an analogous way the direct limit of the generators, then I think it should be not too difficult to show that the direct limit of groups you constructed is isomorphic to the free group generated by the direct limit of generators
So again disjoint union of all your generators, modulo the relations that you mentioned
Does that make sense? @somber bramble
somber bramble
it does, I’d have to see how long the proof ends up being
doesn’t sound like it’d be too long tho
sour plume
I think it shouldn't be terribly long, and it's probably less work than finding a very specific citation 😄
somber bramble
well it’s entirely possible there’s some well-known theorem that has this as a special case
I just odn’t know many theorems about free groups
sour plume
I don't think there is, this seems like a very general, almost category-theoretic statement actually
Because once you have some intuition for both the direct limit of the free groups, and the free group generated by the direct limit, it should become fairly clear how you can write down an isomorphism between them
just mapping generators to generators, checking that relations are fine
And direct limits are really useful objects to get some intuition for anyway, so the time spent understanding those is not wasted
upper pivot
hmm yeah this can definately easily be extended to free R modules in general if i did my math right
cloud walrusBOT
JohnDoeSmith:
spice bay
@upper pivot I'm not sure this holds when you work in R-modules, since every flat module is expressible as a filtered colimit of free R-modules, but there are flat modules which are not free
upper pivot
oh hmm
spice bay
I think over Z everything works out nicely because flat=free (if I recall correctly)
Maybe there's something nice about the diagram we're taking a colimit over being the natural numbers?
upper pivot
hmm i dont see a flaw in my reasoning for the general module case. i might be missing something. So my reasoning was as follows:
cloud walrusBOT
JohnDoeSmith:
upper pivot
hmm im not sure whats wrong here exactly @spice bay , but i do think i might have made a mistake (i only recently learnt about directed limits)
spice bay
@upper pivot that looks pretty much correct to me - I agree that a set of generators for M is the union over the generators for M_i, so if there is some relation among finitely many generators, then there is some j big enough so that this is a relation inside M_j, which can't exist
My concern is something like this: take $\mathbb{Z} \to \mathbb{Z}\times \mathbb{Z}$ as inclusion into the first factor. Then take $\mathbb{Z}\times \mathbb{Z} \to \mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}$ as inclusion into the first two factors, and continue like this. Taking the colimit of this (in the category of groups NOT the category of abelian groups, this is important) you get what's called the Baer-Specker group, which is not free.
cloud walrusBOT
brouwers_bane:
spice bay
Although @upper pivot your argument definitely seems correct, so I'm curious where the incongruency is here
upper pivot
hmm yeah there has to be an inconsistency if there are counterexamples
spice bay
It's an interesting problem - you can also prove it (for groups) with some topology and categorical yoga
upper pivot
i see, i dont know much cat theory presently though, but i'll keep this in mind for when i do learn it
last ether
Does anyone know an example of a nilpotent polynomial?
spice bay
@last ether depends in what ring - for example the polynomial $x$ is nilpotent in the ring $R[x]/(x^2)$ over any ring $R$, but it isn't nilpotent in $R[x]$
cloud walrusBOT
brouwers_bane:
upper pivot
try proving a general result actually
$a_0+a_1x+\dots a_nx^n \in R[x]$ is nilpotent $\iff a_0,a_1\dots a_n \in R$ are all nilpotent
cloud walrusBOT
JohnDoeSmith:
spice bay
^over commutative rings 🙂
upper pivot
yeah commutative right
thorn flint
Hi can someone help please?
I need to show that there is such an isomorphism
So I should start by showing that it’s a homomorphism?
spice bay
Consider the determinant map $\text{det} : \text{GL}_2(\mathbb{Q}) \to \mathbb{Q}$
cloud walrusBOT
brouwers_bane:
spice bay
Its kernel will be a normal subgroup of GL_2(Q)
thorn flint
Okay so I show that if for some x,y matrices in GL_2 and I apply a mapping, then it is the same as applying the map to each of those matrices and adding them together?
And the kernel is all matrices that map to 0?
spice bay
No you shouldn't be adding anything, G is a subgroup of GL_2(Q) under matrix multiplication not addition