Okay, yes
#groups-rings-fields
meager flint
woven delta
So there is an Algebraic invariant, the number of nth roots of unity
bleak abyss
Alright I guess it's time for me then
I finna git learnt
(Serre Rep Theory commentary time specifically)
meager flint
So if I have a $3\times 3$ matrix with determinant -1, we know that $\lambda_1 \lambda_2 \lambda_3 =-1$, so that either $\lambda_1 =\lambda_2 =\lambda_3 =-1$ or $\lambda_1=\lambda_2=1$ and $\lambda_3=-1$. But apparently it also works out if $\lambda_1=c$, $\lambda_2=\bar{c},$ and $\lambda_3=-1$, where $\bar{c}$ is the complex conjugate of $c$. This is probably dumb but I don't understand this last case is true
Where the $\lambda_i$ are the eigenvalues of the matrix obviosly
bleak abyss
Alright now I'm actually working on Serre
bleak abyss
Alright back, continuing
upbeat burrow
I'm going through a proof of a corollary following Nakayama's lemma. I'm not understanding how some of it is done.
$M$ is a finitely generated $A$-module, $N$ a submodule of $M$, $\mathfrak{a} \subseteq \mathfrak{R}$ an ideal. Then $M = \mathfrak{a}M + N \Rightarrow M = N$
cloud walrusBOT
𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫:
upbeat burrow
In the proof, they look at $M/N$. $M/N = (\mathfrak{a}M + N)/N = \mathfrak{a}((M+N)/N)$.
bleak abyss
Okay so let's say $\rho:G\to GL(V)$ is a $G$-rep, $W$ is a stable subspace, and $W^0$ is a stable complement. For $x\in V$, we write $x = w + w^0$ for $w\in W$, $w^0\in W^0$. Then $\rho_g(x) = \rho_g(w) + \rho_g(w^0)$, so as long as you know how $G$ acts on $W$ and $W^0$, you know how it acts on everybody. So think of the abstract direct sum $W\oplus W^0$, and have $G$ act on it by $g\cdot (w,w^0)\mapsto (\rho_g(w),\rho_g(w^0))$. Then consider the map $\tau:W\oplus W^0 \to V$ given by $(w,w^0)\mapsto w+w^0$. This is clearly an isomorphism of $G$-reps
upbeat burrow
How does the ideal factor out like that?
bleak abyss
Write some elements out
upbeat burrow
And why do we get to say by NAK $(M+N)/N = 0$
cloud walrusBOT
𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫:
upbeat burrow
Shouldn't it be $(\mathfrak{a}M + N)/N = 0$?
cloud walrusBOT
𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫:
bleak abyss
Okay if you're gonna be writing in TeX I seriously am disinclined to see that ridiculous nickname over and over
upbeat burrow
Sorry
bleak abyss
You'll need to change your global discord username
cloud walrusBOT
Dami:
upbeat burrow
Right
bleak abyss
Well, we're looking at $(\mathfrak{a}M + N)/N$. That contains $(\mathfrak{a}M + \mathfrak{a}N)/N = \mathfrak{a}(M+N)/N$. But the point here is we're modding out by $N$ anyway.
cloud walrusBOT
Dami:
upbeat burrow
So that's enough to say that $\mathfrak{a}(M + N) = \mathfrak{a}M + N?
bleak abyss
Really the way I prefer to see it is this
upbeat burrow
OH!
bleak abyss
Also if you changed your global username I can get rid of the "new nick" nickname now that your main one is acceptable
upbeat burrow
this proof makes sense now
Nah, that's fine
Although I'll bug someone to change it if I can come up with a nick worth having.
bleak abyss
I mean "new nick" is one of our generic "request a new nickname" things. But sure
upbeat burrow
Thanks so much for the help!
smoky cypress
oh my goodness wtf is this
bleak abyss
Time for some more Serre
So he's doing the tensor product directly for vector spaces
cloud walrusBOT
Dami:
smoky cypress
What is the center of $GL_n(\bR)$?
cloud walrusBOT
Whoever:
fading wagon
what's your guess?
smoky cypress
Im not sure, but I think matrices with nonzero diagonals
fading wagon
consider a matrix with mostly zeros
solemn hollow
do you think the elements in the diagonal can be different?
smoky cypress
Yeah
fading wagon
if AC=CA and BC=CB, then (A+B)C=C(A+B)
latent anvil
Another way to say that A commutes with all B is B^(-1) A B = A for all B
But that's the same as the change of basis by B
So A has to look the same under any basis
fading wagon
Note that A is upper triangular
you can find the jordan normal form https://en.wikipedia.org/wiki/Jordan_normal_form
latent anvil
Oh here's a cute one: think about what happens when B is a permutation matrix
No matter how you swap the basis vectors, you need to get the same matrix
I think that gives it to you
smoky cypress
Ok so the center is kI, where k is any nonzero scalar and I is identity
What I found by google
Lol
fading wagon
can you prove it?
smoky cypress
Not really because I don’t know much linear alg
latent anvil
The matrix A has to have some complex eigenvalue λ, with eigenvector v. I'll show Aw = λw for all w. Let B be any invertible matrix such that Bw = v. Then since A is in the center, B^(-1) A B = A, and B^(-1) A B w = B^(-1) A v = B^(-1) λv = λ(B^(-1) v) = λw. Taking w to be the basis vectors, we see A has to be λI
@smoky cypress does that make sense?
thorn delta
very cool
latent anvil
Suppose G and H act on X, and that their actions commute with one another. Further suppose G acts transitively. Then if for some g in Z(G), there is an h in H and an x in X such that gx = hx, then for every y in X, we must have gy = hy. To see this, pick some f such that fy = x. Then gy = f^(-1)gfy = f^(-1)gx = f^(-1)hx = hf^(-1)x = hy
Taking G to be GL and H to be the nonzero reals, we recover the above
fading wagon
To see this, pick some f such that fy = x.
What if such an f does not exist?
@latent anvil
latent anvil
I said that G acts transitively
fading wagon
oh yeah
latent anvil
I don't think this is a very general result
latent anvil
But it's nice to know that the only actual linear algebra is in the eigenvalues
The rest of the proof is really group theory
chilly ocean
If (1,2)(3,4) is an element of S4, what does it mean to multiply it by itself? Meaning ((1,2)(3,4))^2
I'm not sure if I understand what (1,2)(3,4) means...
Is it f(1)=2, f(2)=1, f(3)=4, f(4)=3?
for some f in S4
ok nvm I got it I think
wind steeple
multiplying is composing permutations
paper birch
"The conjugacy class of a permutation is determined completely by the decomposition into disjoint cycles" can someone explain to me why this is true?
bleak abyss
Show that two k-cycles are conjugate
And that anything conjugate to a k-cycle is a k-cycle
And then generalize to a product of disjoint cycles of prescribed lengths
paper birch
mmmhhh, thanks!
meager flint
Let $T$ be a linear operator defined on the space $\mathbb{C}^{2\times2}$ of all complex matrices by the rule $T(M)=AM-MA$. I'm having trouble determining the eigenvalues of $M$ in terms of the eigenvalues of $\lambda_1,...,\lambda_n$ of $A$.
cloud walrusBOT
yeahbitchphysics:
meager flint
help D:?
stone fulcrum
What's A?
latent anvil
Probably an arbitrary 2×2 complex matrix?
@meager flint the eigenvalues of M or of T?
The eigenvalues of an arbitrary matrix will be unrelated to those of A
meager flint
T sorry
latent anvil
Suppose λ is an eigenvalue of T. Then for some M, λM = AM - MA, or MA = (A-λI)M. Let v be any eigenvector of A, with eigenvalue μ. Then μ(Mv) = M(μv) = MAv = (A-λI)Mv, so Mv is an eigenvector of A-λI with eigenvalue μ
I'm not sure if that's helpful
If λ is an eigenvalue of T, then A-λI and A have the same eigenvalues
Is what that proves
I'm not 100% sure but I think so
meager flint
At this point anything helps lol thank you
chilly ocean
It depends on the matrix A
What can be said for sure is that, if $A$ is diagonalizable, then so is $T$ with eigenvalues $\lambda_i - \lambda_j$ for $1 \le i,j \le n$
cloud walrusBOT
trex:
chilly ocean
The argument being that, if $A$ is diagonalizable and $(e_1, \dots, e_n)$ is an eigenbasis of $\mathbb C^n$, then each elementary matrix $e_{ij}$ defined by $e_{ij}(e_k) = \delta_{jk}e_i$ is an eigenvector of $T$ for the eigenvalue $\lambda_i - \lambda_j$, hence you get an eigenbasis for $Mat_{n}(\mathbb C)$
cloud walrusBOT
trex:
humble bramble
Is the set of zeroes of x^(2/3)+y^(2/3)-1 algebraic? 
somber bramble
there appear to be uncountably many of them
hot lake
yes
humble bramble
How should this be of any use? For clarification: def.: $X\subset\mathbb{A}^n_\mathbb{R}$ is algebraic iff there exits $S\subset\mathbb{R}[x_1,...,x_n]$ s.t. $X=V(S):={x\in\mathbb{A}^n_\mathbb{R}\mid f(x)=0\forall f\in S}$
wind steeple
$$X\subset A^n_R$$ is algebraic iff there exits $$S\subsetR[x_1,...,x_n]$$ s.t. $$X=V(S):={x\in A^n_R: f(x)=0\forall f\in S}$$
cloud walrusBOT
Zak:
Compile Error! Click the 
reaction for details. (You may edit your message)
Axion:
latent anvil
Is it true that (k[X]/I)^× = k for a radical ideal I?
Where X is several variables being adjoined
bleak abyss
Group of units? @latent anvil
latent anvil
Yeah. Jacobian answered
I was being dumb
I was trying to think about the units of k[x, y]/(yx(x+1)(x-1))
How to show x - 1 isn't a unit
bleak abyss
olamf
bleak abyss
Tru dat
pale elbow
hey - can someone get me started on understanding the vocabulary and grammar of abstract algebra?
It's been pretty hard for me to come to terms with the holo-s and the homeo-s, the iso-s, the -ives, the -ism's and what have you
stone fulcrum
There's a homomorphism. That's a special mapping between two groups such that their operation is respected.
An isomorphism is a homomorphism that is invertible. The inverse is also a homomorphism.
Holo- and homeo- belong to other subjects, ignore those for now.
rich patrol
What does invertible mean?
Is it that after the first mapping, it cannot be reverted?
?
Yeah idk abstract alg
Nvm I can just Google it
pale elbow
@stone fulcrum would an example of a homomorphism be to just toss two variables next to a binary operator like a plus sign?
i might be misunderstanding the concept of mapping
prisma ibex
homomorphism means different things depending on what structure you are preserving
pale elbow
@prisma ibex I want to know about ALL the structures!
pale elbow
@stone fulcrum A group is a specific algebraic object with X properties
I am not familiar with those properties
pale elbow
I donate to wikipedia 😎
bleak abyss
That'll help you learn algebra faster tru
pale elbow
are given by regular maps on the variety.
there are many things that I do not understand in regards to the wikipedia article on algebraic groups, but this is one that feels like it would be a very good choice to figure out first
given by sounds like it has a really really specific meaning
thanks
equipped with a binary operation this sounds incredibly specific. Why is this wording the one used?
a set equipped with a binary operation
prisma ibex
it's an operation that takes in two things and spits out one thing
stone fulcrum
A set, like real numbers, matrices, ect.
A binary operation is like +, or ×, ÷. Anything that takes an element on the left and right, and gives another
prisma ibex
yea?
pale elbow
i have yet to have a teacher or professor describe a mathematical object using that word
stone fulcrum
Just a set that has a binary operator
pale elbow
and it scares me
prisma ibex
??
it just means it's part of the structure
if you like you could write
"a set along with a binary operation..."
means the same thing
pale elbow
is there a formal description of a set with any operator?
prisma ibex
??
I mean the operators are functions on sets
so yes, sets equipped with functions
pale elbow
binary operation that combines any two elements to form a third element in such a way that four conditions called group axioms are satisfied, namely closure, associativity, identity and invertibility the wiki article describes a group as a mathematical object whose smaller objects include a set and the binary operator described above
stone fulcrum
It's a structure that consists of a set, and a binary operator
pale elbow
I like thinking in terms of neat little boxes whose inner workings i dont quite understand
yeah, that makes a ton of sense
i like that description
One of the most familiar examples of a group is the set of integers together with the addition operation
this makes SO much sense now
is there a name for groups equipped with N-ary operators?
with a generalized set of axioms
prisma ibex
yea that's a topic of universal algebra
don't get too far ahead of yourself
don't get too far ahead of yourself
pale elbow
universal algebra is THE BEST
sorry @prisma ibex i live to get too far ahead of myself - as recompense, I found out today that groupoids with respect to abstract algebra are called magmas
in category theory they're also known as virtual groups
prisma ibex
what
prisma ibex
uhhh
pale elbow
i thought it would be relevant since your nickname on the server is ngroupoid
pale elbow
from what I understand virtual groups and magmas are different, yes
prisma ibex
okay but you're using the term groupoid and I don't think it means what you think it means
pale elbow
Category in which every morphism is invertible.
magmas are algebraic structures with a single binary operation
prisma ibex
yea those have nothing to do with magmas
groupoids with respect to abstract algebra are called magmas
what do you mean by this then?
pale elbow
huh
I've never heard this terminology before, and it should be avoided
groupoids mean something very very different
pale elbow
@prisma ibex im a wikipedia warrior
magic owl
nG I actually had this exact same argument before
With someone who had this defn of groupoid
Deja vu
urban acorn
rant
i hate it when people specify closure under operations as axioms of algebraic structure types
if the set you have in mind isn't closed under the operation you have in mind, then that operation isn't a well-defined operation on the set
golden pasture
its like the most useless axiom
urban acorn
That's like specifying a complex number as a number of the form a+bi with a,b real numbers and with a not being a rubic's cube
mild laurel
Eh
delicate bloom
having sets that aren't closed gives rise to interesting ways to close them though
mild laurel
Especially when you look at subgroups
You want this operation to be closed under the subgroup
E.g. if you take a group G with a subgroup H, then the operation restricted to H gives you a function from H x H to G but you want this to be H x H to H in order for it to be a subgroup
You can define the binary operation of a group G to have to be G x G to G, but this really amounts to the same thing as closure
Saying the operation isn't "well-defined" isn't really correct
magic owl
I’d actually disagree
And say that closure only makes sense as a part of the defn of subgroup
Of course an operation $GxG->G$ is gonna be closed under $G$
cloud walrusBOT
MaxJ:
magic owl
You get that immediately from the defn of group
mild laurel
I mean that's what I'm saying
Closure is essentially the fact that the codomain of your operation is G
urban acorn
closure is essentially the fact that the codomain of your operation is G
The closure of the codomain of an operation under the operation is trivial, but that's not a good definition of closure because it doesn't generalise to other cases in a way that agrees with what we mean when using the word
the positive reals are closed under real number addition, but the codomain of real number addition includes non-positibe reals.
Define a binary operation on a set G as a function GxG -> G.
then closure is implied by saying that it's equipped with "a binary operation on the set"
what you said if I understand you correctly is that this is essentially the same thing as explicitly declaring closure in the operation but with extra steps
but if you think about the way in which it has to be formalised
every function we talk about needs to have a codomain
so if you want the closure not to be implied by the codomain of the operation but be an independent group axiom
then you'd essentially be saying something like * : FxF -> F with F a superset of G such that for all g,h in G, g*h in G
that's the unnecessarily complicated one
and a specification of an operation on a set by this definition with a lack of closure on the set is not well defined because a well defined specification of a function needs to associate each element in the domain to exactly one element in the codomain, and a specification which associates values outside the codomain sometimes thus isn't well-defined
as merosity said, we do care about sets that aren't close under some operations
fair and square
but the operations aren't operations on the set
delicate bloom
duh
urban acorn
i don't deny that this is trivial
bleak abyss
I really don't think there's much to it lmao, most of the time people don't even say the word closure for the whole group but just a subgroup
magic owl
^
bleak abyss
I mean you'll find one or two people who do it and then just like, roll your eyes at them or something
I don't think it's common
magic owl
I’ve never seen it in a book on group theory
delicate bloom
if I ever saw it, I just skimmed over it and moved on with my life
urban acorn
"In mathematics, a group is a set equipped with a binary operation that combines any two elements to form a third element in such a way that four conditions called group axioms are satisfied, namely closure, associativity, identity and invertibility." - Wikipedia
most people dont be like that
but wikipedia do
😦
paper birch
mmhhh, very odd
I've never heard of it being used as an axiom
They do add a footnote
delicate bloom
rational numbers are closed under multiplication, so if we multiply rational numbers together infinitely many times we get a rational number
they could possibly be saying it to emphasize that they're outside the realm of analysis
urban acorn
rational numbers are closed under multiplication, so if we multiply rational numbers together infinitely many times we get a rational number
do you mean finitely many times?
chilly ocean
Hi, I'm having trouble proving that if g is an element of group G then there exists only one homomophism f of cyclic group $\left(\mathbb{Z}, + \right)$ such that $f\left(1\right)=g$
cloud walrusBOT
Godel:
chilly ocean
Not sure if my proof shows anything lol
Assume h is another homoomprhism with such properties, then there exists a in Z and b,c in G such that f(a)=b and h(a)=c, but f(a) = g^a = b and h(a)=g^a=c. Does it imply b=c ?
final gulch
if f : Z -> G and f(1) = g then what is f(n)?
chilly ocean
g^n
final gulch
good so you have determined f completely
chilly ocean
wat
chilly ocean
But is it because of what I wrote above? I mean it seems kinda obvious
if there exists a different one then it differs in some point
final gulch
I guess you can write it out in any level of formal detail you desire
chilly ocean
Ok so its just obvious then
final gulch
but from f(1) = g you have already determined what f(n) is for every n
chilly ocean
I mean that's a definiotn of a homomorhpism right?
final gulch
this uniquely specifies f
final gulch
don't overcomplicate things
chilly ocean
Oh also, kinda stuck in other one: Let $f : G \to H$ be a monomorphism. Let $G' \leq G$. Show that $f : G' \to f\left(G'\right)$ is an isomorphism.
cloud walrusBOT
Godel:
final gulch
any function maps onto its image
chilly ocean
Wait, if its mono then it obviously has to have the same caridnality?
final gulch
f(G') is just by definition the set of h in H with h = f(g) for some g in G'
chilly ocean
ok yeah
chilly ocean
I have to show that if $f : G \to H$ is a monomorphism then $o\left(f\left(g\right)\right) = o \left(g\right)$. $o\left(f\left(g\right)\right)$ dividing $o \left(g\right)$ follows from definition of a homomorphism, dont now how to show the other way. Tried this way but not sure if its correct: $\ \ $ Assume $o\left(g\right) $ doesnt divide $o\left(f\left(g\right)\right)$, so if order of g is n, then let order of f(g) be an+r. We have: $$e= f\left(g\right)^{an+r} =f\left(g^{an+r}\right) = f\left(g^{an}\right) \circ f\left(g^r\right) = f\left(e\right) \circ f\left(g\right)^r = e \Rightarrow r=0$$
Doest that show that o(g) divides o(f(g)) or not at all?
mild laurel
what does g(g) mean
cloud walrusBOT
Godel:
chilly ocean
forall g in G
chilly ocean
Well at one point, you do have to use injectivity..
ah okay my bad, your proof can be made to work
but you have to explain the very last $\Rightarrow$
cloud walrusBOT
trex:
chilly ocean
hmmm
I think it will be ok to explain it by saying that f(g)^ (an+r) = f(g)^r because kernel of f has to be trivial?
wait no, that doesnt explain it quite well, needs more.. let me think
Well, ker f is {e} so g^(an+r) has to be equal to g^(r)?
and r was chosen to be < n
so implies r=0?
g^{an+r} is equal to g^r, but that is just by definition of the order of g
What does use ker(f)={e} is the fact that $f(g)^r = e \Rightarrow g^r = e$. This in turn implies $r = 0$ because we assumed that $r < n$ and because it follows from the above that $n$ divides $r$.
cloud walrusBOT
trex:
fringe nexus
given a min poly x^3+x^2-1 how can i construct two non similar real 3x3 matrices?
we have 1 real root of degree 1, so im kinda stuck
I don't think complexifying works either because the other roots have real parts
i think i have to use jcf but idk how
idk where to ask this :(
woven delta
So given a minimal polynomial, you can construct a jcf matrix
And then just take the identity matrix or something
@fringe nexus
fringe nexus
yes but the issue is
i want a real matrix
and the factorization gives me one real root
woven delta
Oh lol
fringe nexus
and two complex roots
the real root has degree 1
im like 99% sure theres some trick to this but i cant figure it out
fringe nexus
ok
woven delta
Now think about what that tells you
fringe nexus
hmmmmm
does it really tell me anything though? are they the real and complex eigenvalues
woven delta
So the thing is that you are only looking at similar wrt real matrices, right
woven delta
So you have M^-1(A+iB)M= M^-1AM +I M^-1BM
fringe nexus
yup
fringe nexus
can i like sum the two complex jcf together
no right
:(
since complex eigenvalues come in conjugate pairs
woven delta
Honestly this sucks cause there are some obvious real examples
Using shit like determinant and trace
fringe nexus
wdym?
woven delta
Similar matrices have the same determinant and trace
Oh probably the answer is if you take the jcf as A+iB, the matrices are A and B
Have you tried those matrices?
fringe nexus
well these jcf's are easy to find since the blocks all have size 1 so we just split it up into the real and complex parts on the diagonal?
i don't think these jcf's work they have different eigenvalues don't they
woven delta
Oh you want them to have the same polynomial
fringe nexus
(x - 0.754878) (x + (0.877439 - 0.744862 i)) (x + (0.877439 + 0.744862 i))
woven delta
Lmao I didn't understand the question
fringe nexus
no i want them to satisfy this polynomial
woven delta
Oh okay
woven delta
That's a bit easier
fringe nexus
woven delta
No, modulo invertible your polynomial has the same information
As that statement
Okay, I think I understand
woven delta
So x^3 +x^2 -1=0 has one real root, call it r
Consider the polynomial rI
And the other polynomial is the one whose char polynomial is the whole x^3+x^2-1
@fringe nexus
fringe nexus
wait hold on
woven delta
The question didn't say anything about char polynomial lol
fringe nexus
no but the minimal polynomial has to divide that polynomial
woven delta
Yeah
fringe nexus
so I was solving for minimal polynomials
fringe nexus
ok
woven delta
So that's all good
woven delta
I think you can do it with jcf
fading sparrow
mild laurel
It's not the subspace topology
mild laurel
It's the hyperbolic topology
fading sparrow
Any suggestion how to tackle compactness?
mild laurel
There are a ton of theorems you can use here, let me find some of them
fading sparrow
I was suspecting that there's something wrong with my approach
But the domain is correct, right?
fading sparrow
ermm
yes
it should be SL_2(R)
a typo i guess
now i'm embarassed
still, any suggestions on compactness?
mild laurel
Hm
the hyperbolic thing doesn't really help here
I think your fundamental domain is correct
fading sparrow
I can't find any notes on the topology of hyperbolic space
and the task is to show that it's no co-compact
my mistake there
south coral
Do any of you have opinions about Judson's Abstract Algebra: Theory and Applications ( http://abstract.ups.edu/ )? Specifically, how does it compare to more traditional (?) textbooks? I've heard about Lang, and Dummit & Foote but to be honest I'm not familiar with these either.
(My university's abstract algebra course (which I'm taking in spring) uses Judson)
upbeat burrow
@tribal pasture Almost. You have your definition of order wrong.
Order of an element a is n such that a^n = e.
Not a^n = a.
Otherwise it looks good.
tribal pasture
Oh indeed. Smh. So I should rewrite (0,1); (0,-1) as (0,0) right?
upbeat burrow
Your second and third lines
tribal pasture
Yes yes.
Btw if theyre equal to themselves for n>1 then for n-1, theyre equal to identity. So I guess just including that might also be enough
upbeat burrow
If we have an ideal I = (x,y) in the ring A = k[x,y], k a field, then what is I^2? Is it (x^2,y^2)?
wind steeple
no
upbeat burrow
(x^2,xy,y^2)?
wind steeple
yes I think so
upbeat burrow
Okay, wasn't sure.
So if we have a second ideal, J = (x,y^2), why is I^2 strictly included in J?
wind steeple
you don't know the definition ?
I² is the minimal ideal containing {ab, a in I, b in I}
(minimal for the inclusion)
explicitely, I² = {sum xiyi, xi in I, yi in I}
upbeat burrow
So it would be (x^2,y^2) then wouldn't it?
wind steeple
I*I isn't contained in (x²,y²)
bc xy isn't in (x²,y²)
that seems weird that I² is included in J wtf
maybe I'm wrong
wind steeple
that's false
in Z
if I take I = (2,4)
I = (2) bc gcd(2,4) = 2
hum no I said nothing
oh
I'm dumb
lol
I² is in J
in I you have x²
x² is in J
y² is in J
xy in in J
upbeat burrow
But xy isn't in J
upbeat burrow
Oh derp.
wind steeple
so that's seems legit
upbeat burrow
and it's strict inclusion since x is in J, but x not in I^2
wind steeple
yes
latent anvil
I^2 = II = ((x) + (y))((x) + (y)) = (x)(x) + (x)(y) + (y)(x) + (y)(y) = (x^2) + (xy) + (yx) + (y^2) = (x^2, xy, y^2)
It's good to know that (a1,...,an) = (a1) + ... + (an) (which is definitional really) and that ideal multiplication distributes over ideal addition
The ideals of a ring form a "semiring", with additive unit (0) and multiplicative unit (1)
merry pollen
https://math.stackexchange.com/questions/1044124/conjugacy-class-of-a-group-of-order-12
Could someone explain the last 2 steps in the guy's answer
"But clearly Z(G)⊆ZG(x), so the center Z(G) either has order 1 or 3. If |Z(G)|=3, then x∈Z(G), but then it would have trivial conjugacy class as a central element, contrary to |CG(x)|=4. So Z(G) is trivial."
If x is in Z(G), why would it have trivial conjugacy class as a central element, and why is that contrary to CG(x) size = 4
mild laurel
What's the conjugacy class of elements in the center of the group
merry pollen
isnt it just every element in the group?
mild laurel
why is that?
merry pollen
well if its in the center it commutes with everything
merry pollen
er well by definition the conjugacy class C(x) = x' in G s.t. x' = gxg^-1 for some g in G, but x commutes with all the g's so x' = x
so Im wrong
for each element in the center
would it just be itself
mild laurel
Correct
merry pollen
I still don't understand why what i asked about is true though
mild laurel
"If x is in Z(G), why would it have trivial conjugacy class as a central element"
you understand why this is true now right
merry pollen
no
merry pollen
itself
mild laurel
So it's trivial
merry pollen
wait is the trivial conjugacy class the identity element
mild laurel
no
merry pollen
oh ok
mild laurel
Think about what the conjugacy class of the identity always has to be
merry pollen
So if Z(G) size is 3, why does that imply x is in Z(G)
mild laurel
Think
merry pollen
sorry what exactly is the trivial conjugacy class
mild laurel
Just the element
For an element x, x must be in the conjugacy class of x
The conjugacy class is trivial if it's just x and nothing else
merry pollen
Hmm I still dont see why x being in the center is contrary to |CG(x)| = 4
tribal pasture
Is this correct? I feel very doubtful of the last claim
steep hull
Those two quantities aren’t usually equal. Your solution is pretty close to being correct though. Just look at |<x>| instead.
tribal pasture
Why is it not though? I thought if gcd(m,n) =1 then <x^m> = <x^n>? Is it not true if m =1?
steep hull
Well the issue is that x has order that is a power of p by Lagrange. But if the order of x is p^k with k>0, then the order of x^p is p^(k-1).
steep hull
The general principle is that |<x^m>|=|<x^n>| if v_p(m)=v_p(n) (assuming the group has prime power order). The converse does hold if v_p(m),v_p(n)<=v_p(|<x>|).
tribal pasture
What is v_p?
tribal pasture
Perf!
steep hull
Does that make sense
tribal pasture
I am trying to see why is that true
steep hull
It only works for groups of prime power order
The general spirit is similar, however.
Try using Lagrange again.
tribal pasture
"group has prime power order" By this you mean that <x^m> and <x^n> are subgroup in G which has p^e order?
steep hull
Yeah, they are naturally subgroups and |G|=p^e
tribal pasture
Yep then the forward makes sense
steep hull
The hardest part of the proof follows from this general fact: if x^k=e for some k, then |<x>||k.
tribal pasture
v_p(m),v_p(n)<=3 By this do you mean that (v_p(m),v_p(n)) can be (2,3)? and not necessarily (k,k) where k<=3?
steep hull
Whoops I wrote something but meant something else. It’s correct now.
tribal pasture
Same question but with v_p(|x|) now instead of 3
steep hull
The v_ps can be anything up to v_p(|<x>|).
tribal pasture
I mean do they both have to be same?
steep hull
No
If v_p(|<x>|)=3, then we could have v_p(m)=0,1,2,3 and v_p(n)=0,1,2,3 for a total of 16 possibilities.
tribal pasture
ye ):
steep hull
That’s why the result’s useful; it applies in a lot of cases.
tribal pasture
I cant seem to find out why the converse holds
steep hull
Let me type out a full proof
tribal pasture
Is it because, if <x^m> = <x^n> then x^m =x^nq => m = nq But since m,n are prime powers thus q has to be a prime and hence m=n
tribal pasture
Okay
tame bear
just break into cases what the size of <x> could be
tribal pasture
What cases?
tame bear
the size of <x> needs to divide p^3, with p a prime
what are the only numbers that divide that
tribal pasture
Yeah I am aware of that technique and that was what I was suggested. Just wanted to see whether my methodology worked
Is this what works @steep hull
steep hull
We know that G has order p^e and that v_p(m)=v_p(n). We may write m=ap^k with (a,p)=1. Since <x^m> is a subgroup, we may suppose that |<x^m>|=p^r. But then we have that |<x^m>| divides |<x^(p^k)>| (by subgroups), so |<x^m>|<=|<x^(p^k)>|. We have that x^(ap^(k+r))=e, but then this would imply that x^(p^(k+r))=e. Therefore, |<x^(p^k)|<=|<x^m>|, implying they’re equal. Hence |<x^m>|=|<x^(p^k)>|=|<x^n>|.
This proof technique basically gives the converse too
We proved that if m=ap^k with (a,p)=1, then |<x^m>|=|<x^(p^k)>|. Hence it’s enough to show that if k,l<=v_p(|<x>|), then we have that |<x^(p^k)>|=|<x^(p^l)>| iff k=l. One direction is trivial. For the other, if |<x>|=p^r, then p^r is the smallest positive integer n such that x^n=e, so the smallest positive integer multiple of p^s that has the property must be p^(r-s) for s<=r.
That’s the full proof.
Changing the approach slightly allows you to prove such results for any group.
merry pollen
@mild laurel its just x
steep hull
Yeah it’s its own conjugacy class
Do you know the orbit-stabilizer theorem?
@merry pollen
merry pollen
yes
steep hull
Wait so you get why the centralizer of the element x has order 3 then, right?
Meaning there are three elements in the group that commute with x.
steep hull
x is trivially not the identity. Hence we know that e,x are in this group of order 3, but we also know that all groups of order 3 are cyclic. Therefore, the group must be {e,x,x^2=x^(-1)}.
mild laurel
If x is in the center
merry pollen
but how is that "contrary to |CG(x)|=4"
mild laurel
so that x has trivial conjugacy class
What's C_G(x)
If x has trivial conjugacy class
merry pollen
conjugacy class of x
mild laurel
And what is this set if x has trivial conjugacy class
merry pollen
which set are you talking about
mild laurel
What is the conjugacy class of x if x has trivial conjugacy class
mild laurel
And what's the cardinality of that set
steep hull
@merry pollen after you deduce that the centralizer of x is {e,x,x^(-1)}, it’s enough to show x^(-1) isn’t in the center, which is easy.
mild laurel
@steep hull No I'm not sure where you're going
steep hull
@mild laurel I showed by orbit stabilizer that the subgroup that commutes with x has order 3, meaning that it is cyclic and generated by x itself (as x commutes with itself). Hence it’s in the form {e,x,x^2=x^(-1)}. Anything not in the subgroup isn’t in the center, so it’s enough to show that x^(-1) isn’t in the center, but the centralizers of x and x^(-1) have the same order.
mild laurel
Works as an alternative proof I guess
steep hull
What was your proof
mild laurel
I didn't have one
I was just explaining https://math.stackexchange.com/questions/1044124/conjugacy-class-of-a-group-of-order-12
from which he asked a question
steep hull
Oh okay. I think the solutions are quite similar actually. I just didn’t explicitly mention the center until the end.
Centralizer is still the standard term for one element subsets, right?
@mild laurel
merry pollen
sorry why is "If |Z(G)|=3, then x∈Z(G)" this true again
steep hull
Essentially the center is a subgroup of centralizer, and we know that the centralizer of x has order 3 and includes x (as it commutes with itself)
merry pollen
oh right thanks
mild laurel
@steep hull yeah
magic owl
could use another hint if anyone is around. have to show A_4 is generated by any 2-cycle and 3-cycle combination. I know that i just have to show I can make all 3-cycles, and WLOG we (1,e) and (a,b,c) as generators with the 3-cycle having (1,c,d)
latent anvil
What do you mean by "2-cycle and 3-cycle combination"?
@magic owl
I'm confused because 2-cycles aren't in A4
inner acorn
Just a notation question
When they write g^(phi*) do they mean phi*(g)?
similarly for x_i^phi meaning phi(x_i)
Okay, from the context of the proof, this seems to be what they mean - have been exposed to g^phi as being the elements g in G fixed by phi xD so didn't want the ambiguity .... but reading a lil further gives gave me context
inner acorn
I figured out they meant homomorphisms, this is just a proof of the universal property of free groups
x_i are letters, and ε_i denotes if they're in X or X^-1 etc....
worthy kindle
So I have a group $(A, +)$, $B$ a subgroup of $A$ and $C$ a subgroup of $B$, and I was wondering whether or not
$$\fun\psi{A/C}{A/B}{\pi_C(x)}{\pi_B(x)}$$
was a well-defined groups morphism, and idk how to proceed
cloud walrusBOT
Tuong:
worthy kindle
algebra is happening
latent anvil
That looks right to me. If x and y are congruent mod C, they're also congruent mod B
xy^(-1) in C <= B
solemn hollow
wait A/C and A/B are groups?
latent anvil
Yes, this is called a quotient group
mild laurel
I'm assuming A is abelian?
worthy kindle
Oh yeah, sorry
solemn hollow
right so that makes C and B normal
latent anvil
If you pick an element of A/C, it looks like xC for some x in A. We want to send this to xB in A/B. How do we know that this is well defined (i.e. independent of representative x)? Choose any other y in A such that xC = yC. This means that y = c^(-1)x for some c in C. But since C is contained in B, we also have y = b^(-1)x for some b in B (namely b = c). Thus yB = xB
@worthy kindle does that make sense?
worthy kindle
Yup
latent anvil
It's a group homomorphism because multiplication in A/B and A/C works by picking representatives and multiplying in A
worthy kindle
ah so it doesn't necessarily work when C isn't supposed subgroup of B
worthy kindle
alright, thanks!
mild laurel
This is not the right channel
chilly ocean
eh ;p
what is the right channel then
.-.
@mild laurel what is the channel? ;p
i can get help from
full raven
Guys guys quick question
Does every ideal of a ring have a generator set? Finite or infinite
Ok nvm i have now realised the entire ideal is a generating set
Another question
mild laurel
Rings are called noetherian if all ideals have finite generating sets
full raven
We have defined an integral element over a ring A in B as x belonging to B st A[b] is a finitely generated A-module
Is it true that if there is a finitely generated A-module that contains A[b] then b is integral
And I know that that was a dumb question lol
mild laurel
Yes that is true
full raven
How do you show it?
mild laurel
There's no super easy way iirc
The easiest is probably through the generalized cayley-hamilton theorem
brazen frost
when I'm modding out two groups, how do I know when I've found all the nontrivial relations? Ex. For the abelian group ${a+b+c, a+b-c, 2c}/{a+b+c,a+b-c} \simeq Z\oplus Z_2$. It's clear that I have the relation $2c \sim 0$, but how do I know that there aren't others? Like if I was trying to establish the isomorphism
cloud walrusBOT
s1dev:
brazen frost
I have the generators ${a+b+c, a+b-c, c+c}$ and then I mod out the elements (above). The group operations is abelian (+) and the group operation is flipping signs.
Does that clarify? I'm a bit of an algebra math noob
cloud walrusBOT
s1dev:
hot lake
a+b+c is a generator ??
brazen frost
lol yes
magic owl
Yeah haha
Well
If you have three variables and n equations
You know how much you can possibly determine them
So you can figure when you’ve sort of done all you can
hot lake
so you're basically looking at subgroups of Z^3 ?
the one generated by {a+b+c,a+b-c,2c}
and the one generated by {a+b+c, a+b-c} ?
brazen frost
I think so
hot lake
aren't they the same group ?
magic owl
Don’t forget you get three things
a+b+c =0
a+b-c = 0
And their equality to each other
You get 2c=0
and c=a+b
Hence we have c in <a,b>
Sorry wait maybe that’s bad I’m not at my computer
Basically you should get to a point where you can’t find anymore ways to simplify
brazen frost
So I should be able to manipulate this somehow to ${a+b, c}/{2c}$
cloud walrusBOT
s1dev:
brazen frost
although stupid question: It's not obvious to me that c is in my group
magic owl
Maybe it would be best
If you posted the full homology question
Bc in general we should be looking at groups generated by the abc
Not by sums of them
Then we quotient by sums
This makes everything a lot easier to compute
brazen frost
I'm trying to compute the simplicial homology of the klein bottle $H^\Delta_1(K)$
cloud walrusBOT
s1dev:
magic owl
Ok so yes I actually did this fairly recently
Ofc C1 should be <a,b,c>
So the cycles will be too
Then we mod out by boundaries
Still w same generators but new relations
Does this make sense
brazen frost
Mostly! I see that C_1 is <a,b,c> but aren't the cycles only a subgroup of that?
So when I compute ker∂_1 / im ∂_2 I have the subgroup generated by cycles mod the boundaries
actually, pg 107 of Hatcher explains this really nicely
I appreciate the help!
magic owl
Yeah sorry it depends on the map
For the Klein bottle assuming you’re using the delta complex structure
You get that the C_1 map is the 0 map
So you can do the trick I mentioned
In general you have to look more carefully yeah
potent lynx
Show that if G is a nonabelian finite group, then |Z(G)| <= 1/4|G|.
what do i need
to prove this
plush vector
@potent lynx
potent lynx
here
plush vector
wait
okay
try to think of G as a coset representative of Z(G)
use manipulation of the given then make conclusions
plush vector
yep
tribal pasture
Can I say something about x if Z(x) = Z(G) where Z is the centralizer?
errant pewter
tribal pasture
Why can the order of a be 2? I understand it can be 4 since there exists a conjugacy class of order 5 but why 2?
latent anvil
@chilly ocean suppose char F = nm for n, m≠1
Since n < char F, the element n*1 is nonzero
Same logic shows m*1 is nonzero
Where multiplication means repeated addition
But (n*1)(m*1) = 0
So the characteristic of an integral domain is not composite
latent anvil
It's a little involved and depends on what you already know
The first part is pretty easy
Any such field is a vector space over Fp
So it's isomorphic as an additive group to (Z/pZ)^n where n = dim Fq
The second bit is harder
@chilly ocean how much field theory do you know?
I mean Fp
The field with p elements
The easiest definition of Fq is that it's the splitting field of x^q - x over Fp
But you need to know what a splitting field is
How did you run into this stuff without knowing that?
No judgment, it's just that this isn't a great way to learn how finite fields work
If you know a little ring theory there's another easy construction
Well less easy
You should learn ring theory before trying to understand field theory
Before ring theory
My recommended order for how to learn things is (linear algebra ->) group theory -> ring theory -> field theory
And then after that it gets murkier
If you want to learn this stuff, dummit and foote is a good introduction imo
But Wikipedia and scattered pdfs are not
latent anvil
I think I've heard pinter's is an easier first course
latent anvil
Up to you
potent lynx
do groups then rings then fields
yea ig
i think some grad students can use it?
im not sure
latent anvil
It works for undergrad or grad courses imo
latent anvil
@chilly ocean you shouldn't just prioritize based on what has more content
If you want more content read Lang lol
latent anvil
Do not read Lang
potent lynx
i recommend gaillan tbh but df is better
latent anvil
Tbh I recommend Aluffi but it's a little fancier
And maybe not right for your first course
It doesn't matter
3th vs 4rd
potent lynx
allufi is grad algebra right?
latent anvil
I mean yeah technically
potent lynx
literally chapter 1 is cats
latent anvil
It was my introduction and I run an intro (honors) course which uses it
I think it can be good
The category theory sounds a lot worse than it is
latent anvil
It doesn't prove any theorems about categories or anything, and the first reference to functors is in like chapter 7
Spoiling?
@chilly ocean what really does matter is that you do exercises
latent anvil
Reading a textbook means going through the proofs on your own and doing as many problems as you can until you feel comfortable with it
latent anvil
Probably more than that if you haven't done this a lot
Part of learning math is learning how to judge correctness of a proof
You can post here if you want though, I think
@potent lynx yeah he does that at the Stagg
*start
latent anvil
The introduction says it's suitable for an upper lever undergrad course
In aloofi
Exefizes are weird
It's hard to know which ones to do or not
@chilly ocean I have a bunch of problem sets if you want
Which should have good exercises
Dm me if you're interested
Yeah they're for an intro course
It is
My problem sets expect a basic understanding of like set theory and modular arithmetic
I think that's all
smoky cypress
Can elementary symmetric polynomials e_i (0<=i<=n) in n variables be written as a polynomial of e_0, e_1, ..., e_(i-1)?
hot lake
what ?
tacit pebble
Can someone explain how lagrange's theorem works with infinite groups? Like if a group has order Aleph naught then subgroups have order ??? I know it's supposed to hold somehow I just don't know how
latent anvil
You can prove Lagrange's theorem by giving an explicit bijection between G and H×(G/H)
Where G/H denotes the set of cosets
tacit pebble
Hmm thank you affline, I think I get the proof where I'm getting tripped up is application of lagrange's thrm. Like say I had a group of order Aleph_1 call it G, what would she subgroups be? Would it be the set of H such that there is a bijection between Hx(G/H) and G?
I guess I should say the "possible" subgroups
latent anvil
The subgroups all have cardinality less than or equal to aleph1
I'm not sure you can say more
lyric falcon
Hey, does anyone have quick access to Dummit and Foote's abstract algebra, the second edition? My prof gave us some exersices for Monday (just numbers and pages) and I forgot to check them in the library
worthy kindle
smoky cypress
But I only have third edition
worthy kindle
depending on you geographical situation, you may or may not need a vpn
smoky cypress
You can probably find 2nd edition on libgen
@lyric falcon sorry for the ping but this might help you
lyric falcon
Yo thanks but on libgen there's only 3rd edition
Well, I have the numbers of problems and the section, so i guess I can just hope they didn't change the problems in that section
upbeat burrow
Don't we just need powers of one of the elements to be in the ideal for it to be primary?
upbeat burrow
Ah, nevermind. I got it 🙂
latent anvil
Can a proper subgroup of a p group contain all elements of order p?
mild laurel
You mean, contain all elements of order p that were in the p group?
mild laurel
Well at first I interpreted it as
Can a proper subgroup of a p group only have elements of order p, which is obviously true
latent anvil
No, |e| = 1
mild laurel
Oh yeah I'm dumb
latent anvil
I almost posted obviously true too
But caught myself
I guess the elements of your subgroup H act on the order p elements by conjugation
That's something?
But it doesn't feel right
mild laurel
So you're wondering if the smallest subgroup containing all elements of order p is always proper?
mild laurel
I mean it's proper in that case right
latent anvil
Nvm lol
mild laurel
Okay klein four group doesn't work
latent anvil
Oh yeah I think that's was why I asked in the first place
For some reason Z/pZ×Z/pZ is my prototypical p-group
Also Q8, but (and this is not a lie) I was thinking p = 4
Not my brightest moment
mild laurel
lmao 4 is prime right because 2 is not prime
mild laurel
Yeah I feel like it'd be hard to classify what p groups work and which ones don't
latent anvil
This whole line of questioning was for showing maximal subgroups of p groups are normal and of index p, but I was going about it entirely the wrong way
There's an easy proof by induction
bleak abyss
Time for some AG mothafuckas
cloud walrusBOT
Daminark:
bleak abyss
Okay there's something here which may have multiple meanings so let me be sure
cloud walrusBOT
Daminark:
bleak abyss
You can react to the texit message to have it delete the original
Also I should get back on it actually I got distracted
cloud walrusBOT
Daminark:
latent anvil
I want to do algebraic geometry right now!
This is a breakthrough
The first time I've felt anything but hate for it in a while
bleak abyss
God I cannot focus
cloud walrusBOT
Daminark:
bleak abyss
So now that's what it means to be a regular function on a subvariety of A^n. And we define regular maps in the obvious way
Okay so some examples
bleak abyss
Err, sorry on A^1\{i,-i}
cloud walrusBOT
Daminark:
bleak abyss
Oh fuck the first one is defined on the complement of (-1,0) whoops
bleak abyss
The other direction is painful
Fuck it brb I'm doing this on paper
Did it
Okay so these two guys are isomorphic boom
So now next example, cuspidal cubic
bleak abyss
So this actually starts to look like manifolds and charts
You can write any locally closed set as a union of sets of the form D(f), which are isomorphic to algebraic subsets of A^n
brazen frost
If G is a free abelian group, how should I think of Hom(G, Z)?
G being generated by a basis with coefficients in Z
bleak abyss
Should be isomorphic to G, no?
brazen frost
Really what I'm trying to wrap my head around is how to compute cohomology groups.
I have some sequence of boundary maps for a CW-Complex and now I'm trying to dualize everything
^ Boundary maps which are morphisms of some free abelian groups
So I want to take G -> Hom(G,Z)
If I have a basis such that I can write any member in G as $\sum_i \alpha_i \sigma_i$ for some generators $\sigma_i \in G$ and $\alpha_i \in \mathbb{Z}$ then would $Hom(G,Z)$ be the row vector $\alpha$ which acts on a column vector of coefficients?
cloud walrusBOT
s1dev:
latent anvil
I guess that makes sense? It's basically the dual space
Doing linear algebra over R=Z
Are you thinking about how you get an iso between them after choosing a basis?
I would think of α as the row vector a = [α(σ1) α(σ2) ... α(σn)]
And α(c1 σ1 + ... + cn σn) = a times the column vector [c1 ... cn]
brazen frost
that makes sense
latent anvil
I don't know how helpful this is
brazen frost
Yeah, I'm doing some homology calculation and I need to go from a complex of chains to a complex of cochains which is taking the dual
latent anvil
I don't have a ton of experience with this stuff
But thinking in terms of explicit row and column vectors seems too low level
And might not give good intuition since Z is not a field
Whatever works for you
brazen frost
is there another way to compute things? I have an obvious basis
latent anvil
Oh yeah you'll probably need to actually break out a basis for that
I take back what I said
Really it depends on the problem
brazen frost
Okay, well I'll try that and see if it gets me something that makes sense
latent anvil
Group theory thing I'm struggling with rn