vast quiver
oh yeah you right, you just find a 3 generator presentation of Z/4Z or K_4
#groups-rings-fields
chilly ocean
overkill is fun
upper pivot
as a rule of thumb always do these problems with topology
its more elegent and intuitive 🙂
(by ggt bais coming in lol)
prisma shuttle
can someone help me understand this proof
for context R = Z_2[x]/(x^3+x^4)
i understand how they get the first equation
but then after that i don't understand how they get from that equation to the other equations
toxic zephyr
i'm having some difficulty with working with the dihedral group (trying to investigate a specific equation in D4 and then generalizing into Dn). is there an isomorphic group that might be more straightforward?
next obsidian
Well it’s hard to know because dihedral group means anything isomorphic to that group
Here’s the two ways to describe it I know
1: group of symmetries of a regular n-gon
2: group described by the presentation <r,f| f^2 = e, r^n = e, frfr = e>
barren sierra
For multiplicative cyclic groups
say I have some number n = pqr for primes p, q, r
how can I quickly find a generator for (Z/Zn)*
there is no good way right?
latent night
In aluffi the hilberts nullstellensatz is presented as "If k is an algebraically closed field, then all the maximal ideals of the polynomial ring k[x1, x2 .... xn] in n variables are of the form (x1 - c1, x2 -c2, x3 -c3 ... xn - cn). However, when they prove the theorem they prove "If k is a field and F is a field extensions of k, and F is a finite-type k-algebra,then F is an algebraic extension of k".
I don't see exactly how these are the same.
chilly ocean
if m is a maximal ideal in your polynomial ring k[x_i], you should look at the field k[x_i] / m and apply your theorem as well as algebraic closure
you will find that the natural map k -> k[x_i] / m is an isomorphism. take c_i to be the inverse image of x_i mod m. then x_i - c_i is in m for each i, by which m = (x_1 - c_1, ..., x_n - c_n) since this ideal is maximal
delicate orchid
how can I quickly find a generator for (Z/Zn)*
If you just want a generator take n-1. If you want to find all of them then idk lol!!
wooden ember
If you just want *a* generator take n-1. If you want to find all of them then id...
But unless I’m having a massive brain fart (Z/nZ)^x is not cyclic in general
delicate orchid
But unless I’m having a massive brain fart (Z/nZ)^x is not cyclic in general
it is not
but they wanted "a generator" so I'm assuming they're talking about one of the cyclic cases
or they are simply unaware
the more I think about it the odder the question becomes actually
wooden ember
the more I think about it the odder the question becomes actually
Yeah exactly
delicate orchid
we basically require all of the primes to be identical or one of them to be 2 and the other pair to be equal in order for it to be cyclic
so that's a good check lol
chilly radish
How important are all the explicit cone and cylinder constructions in homalg really
Am I going to need this later
I know that there's a definition of chain homotopy using the cylinder
dapper nebula
They are important
arctic storm
@inland otter @tribal moss Thanks mates!
agile burrow
How important are all the explicit cone and cylinder constructions in homalg rea...
I've used the cone a few times for some of the proofs about lifting to projective resolutions but I haven't used the cylinder except for like one exercise
tough raven
IDT that's true if it's maximal two-sided since non-division rings with no nontrivial two-sided ideals exist.
But if it's maximal as a left ideal or as a right ideal (and still a two-sided ideal), I think this is true.
chilly radish
I've used the cone a few times for some of the proofs about lifting to projectiv...
Ah cool cuz I don't wanna spend too long on this stuff
Just a bunch of LES chasing
lethal dune
so Tor functors are what makes M⊗N -> M' ⊗N -> M''⊗N ->0 into LES but how do I actually calculate Tor_i of M, N. I saw some constructions using free resolution but I didn't understand it
oblique river
I think you are confusing Ext and Tor. Ext is for Hom and Tor is for tensor products
Where are you reading about this? Typically one would calculate using resolutions. Understanding resolutions is a prerequisite to understanding ext/tor
agile burrow
Ah cool cuz I don't wanna spend too long on this stuff
yeah, the only two things I really remember are mapping cylinders let us replace morphisms with monomorphisms while keeping the induced map in cohomology, and mapping cones of quasi-isomorphisms are exact
lethal dune
I think you are confusing Ext and Tor. Ext is for Hom and Tor is for tensor prod...
I meant Tor* sry
lethal dune
Where are you reading about this? Typically one would calculate using resolution...
Aluffi
woeful flint
Is it true that maximal totally real subfields of cyclotomic fields are galois extensions of Q?
I genuinely have no idea and I can't find stuff online
oblique river
Is it true that maximal totally real subfields of cyclotomic fields are galois e...
Cyclotomic fields are abelian, and all subfields of abelian fields are galois
oblique river
Yes this is the standard computation. You need to understand resolutions before you can understand higher tor/ext
lethal dune
I have understood it but what does it mean to "throw M away and tensor the free by by N obtaining complex M*xN?"
like how's the second complex is even exact?
like if we throw away M
oblique river
Discard the M on the right
lethal dune
like how's the second complex is even exact?
ignore
oblique river
And tensor the rest of the complex with N
rustic crown
like how's the second complex is even exact?
it's not!
oblique river
This gives you a new complex
Which, as you realized, is no longer exact
So we can take homology of it and get something interestinf
If it were exact, then homology would be zero
oblique river
Discard the M on the right
Sry i said left here but i meanr right
lethal dune
okay
rustic crown
maybe think it like this. you have the category of modules. after that you saw how you could construct the category of chain complexes. so one thing you can do is try to understand a module using slightly easier modules (eg. free modules).
there is a way to embed Mod_R into the Ch_R, just send a module M to the complex ... 0 -> 0 -> M -> 0 -> 0 ...
but there are other complexes which mimic this complex. one such is the complex you get when you take a free resolution and throw away M. it's a complex and at each index it's homology matches with that of ... 0 -> 0 -> M -> 0 -> 0 ...
okay maybe this made things a little more complicated 😓
lol i realize idk how to explain this stuff
.<
cloud walrusBOT
Kraft Macaroni
oblique river
yes, you're correct about the degree
you could either: find its galois conjugate, which you know must be the other root of the min poly, and then multiply the linear factors together
like (x - a)(x - b) where a = your element and b = its galois conjugate
or, since you know it's quadratic, just look for a quadratic relation
square it
and then look for a linear relationship between a^2, a, and 1
hallow pecan
Vector space is closed under addition and multiplication. But isn't this trivial if we already "defined" as $+: V \times V \to V$ and $\cdot: \mathbf{F} \times V \to V$?
rustic crown
yea you would usually use the word closed when talking about subspaces
a subspace of a vectorspace V is a subset W which is closed under addition and scalar multiplication
since you already have a meaning of + when talking about addition of two elements of W, it makes sense to ask if the result again lies in W
else like you said, if you define + : V x V --> V, then closed-ness is already built-in
hallow pecan
Ah I see. Thank you!
long geyser
the first sentence is an error, right?
set of polynomials in ring R shouldn't be equivalent to functions from N to R, as polynomials have finite support.
south patrol
Yeah seems so
next obsidian
It’s only about the size
Well, that’s how I would interpret it to give them the benefit of the doubt
Wait is that even true?
No
Lmfao
long geyser
I wouldn't care if it didn't change the size also
but I believe the distinction makes it countable
that said, still not finite so the proof stands
warm wyvern
does artin cover dual vector spaces?
and where do you first get introduced to this topic?
fluid portal
and where do you first get introduced to this topic?
Linear Algebra section of Abstract Algebra
fluid portal
Chapter 11 of Dummit and Foote covers them.
Artin doesn’t cover a lot, but it’s really short, so you gotta expect that
chilly radish
First place I really needed dual spaces was functional analysis
And ig some duality stuff for (co)homology with field coefficients
And lefschetz fpt
void cosmos
its defined to you
warm wyvern
it's not defined as such
it's stated to be as such
also, what does "each block with lambda=0" mean?
void cosmos
yea
warm wyvern
and k_i^block only concerns blocks with lambda=0?
void cosmos
yes
warm wyvern
how about when lambda=/=0?
how do we get the size of those blocks?
oh
ok, for that you just set T to T-\lambda I
void cosmos
yes
warm wyvern
makes sense
ok, hmm
but how would we add the numbers k_i^blocks if we don't know how man blocks there are and their size beforehand?
obsidian sleet
jcf 
( 
)
rustic crown
but how would we add the numbers k_i^blocks if we don't know how man blocks ther...
we're going the other way.
you know abstractly you can decompose the matrix into blocks. now since we're interested in blocks with eigenvalue lambda, we look at T - lambda * I. This converts all those blocks into nilpotent matrices.
so from the information of k1, k2, ... could we figure out the sizes of those blocks?
since the blocks whose eigenvalue was not lambda stay invertible, they're not gonna contribute to the value of k_i. with the explicit description of each block you can see that k_d - k_{d-1} is the number of blocks with size at least d. this means that number of blocks with size d is exactly 2k_d - k_{d+1} - k_{d-1}
torn niche
if $\alpha$ is transcendental then why the evaluation map $$ k(x) \to k(\alpha)$$ is not onto
cloud walrusBOT
Lizard
torn niche
k is a field
rustic crown
it is surjective
of course k[x] --> k(alpha) isn't. it's surjective if and only if alpha is algebraic.
rustic crown
oh i don't think its a typo
if alpha is algebraic, then you can't even define the map k(t) --> k(alpha)
because if alpha satisfies the polynomial f(t) then where does 1/f(t) go?
rustic crown
yep
rustic crown
fluid portal
Lmao look at page count 
lethal dune
wait my copy only has ~1000 pages
fluid portal
wait my copy only has ~1000 pages
Lmao dw I fused DF, Einsbud, and Hartshorne
stuck cosmos
What exactly is a fundamental representation?
stuck cosmos
delicate orchid
never fear, I am here
delicate orchid
What exactly is a fundamental representation?
do you know about weights of representations
stuck cosmos
do you know about weights of representations
Nope
I did see mentions of it in a lot of places, but, no clue.
delicate orchid
ok this might get a bit dicey cause I'm not too comfortable with them 
stuck cosmos
Go ahead anyway, my guy.
delicate orchid
basically a weight is a homomorphism from an algebra to a field, kinda like a linear character of a group if you know about them
delicate orchid
yeah it's just a 1 dim rep of the algebra
ahhh I can't remember how it goes
basically
we can formulate a "weight of a representation" using a "weight space" by some means
stuck cosmos
So, fundamental would mean lowest weight?
delicate orchid
and a fundamental representation is one such that it's "highest weight" (which is what you think it is, it's the weight such that all other weights are lower than it) is it's fundemental weight, which I think of as kinda being a "unit vector" of this weight space
stuck cosmos
Oh, lol
delicate orchid
this explanation is bad but it's the best I can do
now I retreat back into the comfy world of finite groups 😌
stuck cosmos
Wait, what's a fundamental weight?
delicate orchid
they're some weights that form a basis of a subalgebra of the algebra we're taking a representation of, I think
why are you asking about this btw
just curious
stuck cosmos
The professor just threw the term in the notes and never really explained.
chilly ocean
stuck cosmos
I presume it's like, post-grad stuff
Yeah, I'm doing a master's in mathematical physics.
severe cedar
is left multiplication by $[a] \in \mathbb{Z}_n^{\times}$ an automorphism? i think it's not even a homomorphism, right?
stuck cosmos
Left multiplications are not a homomorphism
delicate orchid
I cannot imagine why you'd need to know about weights for representations of groups
cloud walrusBOT
capitalsigma
stuck cosmos
I cannot imagine why you'd need to know about weights for representations of *gr...
The professor never really explained what fundamental representations are; he just gave examples.
delicate orchid
I cannot imagine why you'd need to know about weights for representations of *gr...
the whole point of them is they generalise eigenvalues and eigenspaces for algebras - tbh if it's just thrown in there without explanation and doesn't appear elsewhere in the notes I'd just ignore it and learn from past papers lol
delicate orchid
The professor never really explained what fundamental representations are; he ju...
yeah that sounds like physics 
()
severe cedar
ok, i don't know wtf this question is asking then: prove that for any $[a] \in \mathbb{Z}n^\times$, multiplication by $[a]$ defines an automorphism $\alpha{[a]} \colon \mathbb{Z}_n \to \mathbb{Z}n$, given by $\alpha{[a]}([k]) = [ak]$
stuck cosmos
There are no past papers; it's the first time the course has been offered at the institute.
cloud walrusBOT
capitalsigma
severe cedar
is this just an incorrect question?
stuck cosmos
What's $\mathbb{Z}^{\times}_n$?
cloud walrusBOT
SK2099
delicate orchid
the group of units of Z/nZ
severe cedar
the multiplicative group mod n
delicate orchid
good, cause it's bad
stuck cosmos
LOL
severe cedar
lol
stuck cosmos
But, yeah, left and right multiplications shift the identity
severe cedar
oh, i see, the groups are different, it's asking if multiplication by a unit is a homomorphism in the additive group
severe cedar
yeah, cool, thanks!
delicate orchid
np
rustic crown
nope >.<
stuck cosmos
Okay, let's do one more.
rustic crown
it can probably mean many things depending on the context
delicate orchid
reps of lie algebras
rustic crown
oh nope, me no know lie stuff
stuck cosmos
Where does equation (22) come from?
delicate orchid
good lord
stuck cosmos
delicate orchid
I have to google what the spin groups are everytime bro 
stuck cosmos
That fancy I is a unit quaternion.
stuck cosmos
I have to google what the spin groups are everytime bro <:monkey:651816857153699...
Unit quaternions.
Spin(3) at least.
delicate orchid
wait that actually makes sense
how terrifying
what's the sigma with the squiggle
stuck cosmos
The sigma with the squiggle takes you from the quaternions to 2 by 2 complex matrices.
rustic crown
so if you're rotating on the x-y plane the unit quaternion for that rotation is k and its corresponding matrix is
[i 0]
[0 -i]```
multiply that by alpha/2 and exponentiate to get
[exp(alpha/2 i) 0]
[0 exp(-alpha/2 i)]
matrix exp isn't painful with diagonal matrices :3
.<
stuck cosmos
so if you're rotating on the x-y plane the unit quaternion for that rotation is ...
The signs don't match
rustic crown
they don't?
cloud walrusBOT
SK2099
arctic storm
This is confusing me. Does it really matter whether you write the j as a superscipt and the i as a subscript?
I am guessing not
In that case it only implies that a tranposition induces a horizontal mismatch on the indices right?
lethal dune
the indexing scheme is related to tensor notation. You can ignore for now
arctic storm
ooh okay I see thanks
warm wyvern
we're going the other way.
you know abstractly you can decompose the matrix into...
oooh
that makes so much sense
so k_i^block was just to demonstrate how k_i changes as i increases
(also, please don't hesitate to ping me in case you wanted to reply to me :))
north sand
prisma shuttle
without any mod necessary, it's just 0
wait i have a quesiotn then
cuz its for the proof of this top equation
can't u just say its equal to 0 because when u plug in 0 and 1 in the first equation x^{m+1}+x^{2m}+x^{2m+1}+x^{4m} evaluate to 0 in Z_2
why do u have to plug in these monomials and stuff
unless ima interpreting wut they mean by mod (x^3+x^4) wrong
chilly ocean
can't u just say its equal to 0 because when u plug in 0 and 1 in the first equa...
You mean, why can't we say the polynomial is zero because it evaluates to 0 for everything we plug in?
prisma shuttle
You mean, why can't we say the polynomial is zero because it evaluates to 0 for ...
yea
like \equvi 0\pmod{x^3+x^4}
chilly ocean
because there's a difference between a polynomial and a polynomial function
prisma shuttle
ohhhh
prisma shuttle
wait then like
do u get the m\geq 2 case
i don't get how they simplify eveything to x^3
chilly ocean
polynomial functions are a quotient of all polynomials
oblique river
i don't get how they simplify eveything to x^3
they are using the relation x^3 + x^4 = 0
chilly ocean
To be precise, they are quotient by ideal (x^p-x) where p is the characteristic of the field
oblique river
rewrite this as x^4 = -x^3, or just x^4 = x^3 since 1 = -1 in Z_2
so you can replace all x^4 with x^3
but now multiply both sides by x
x^5 = x^4
and x^4 = x^3,
chilly ocean
basically, a^p = a for all a in characteristic p is why it doesn't work
oblique river
so x^5 = x^3
now multiply both sides by x again
basically, in this quotient, you have x^3 = x^4 = x^5 = x^6 = ...
so all higher powers of x can just be replaced by x^3
chilly radish
can someone explain why this equation is true in Z_2[x]
Can I ask what you're reading that cares so much about this specific ring
delicate orchid
I love char 2......
chilly radish
Literally the worst characreristic
delicate orchid
agreed
say (0) is a prime ideal in integral domains and no one bats and eye
say 0 is a prime in integral domains and every1 loses there minds...
chilly radish
🃏
prisma shuttle
Can I ask what you're reading that cares so much about this specific ring
pages 8 and 9
toxic zephyr
so i know how to prove that a linear map is injective, but what is the general process i could for a map which is not? for example, something like T(X)=XAX^(-1)AX, where X and A are square matrices (and A is given). let's say 2x2s for simplicity.
well... for matrices maybe i could just look at the determinant. but what about for various other groups like the dihedral or symmetric group
next obsidian
Well, if by linear you mean a group homomorpjism, you just check to see if the kernel is nontrivial
Sometimes you can try to solve some equations or look at what’s going on and get a good idea as to what might be sent to 0 which is nonzero to begin with
toxic zephyr
Well, if by linear you mean a group homomorpjism, you just check to see if the k...
i can use a nontrivial kernel for a map that isn't a homomorphism?
toxic zephyr
But why are you considering a map that isn’t a homomorpjism?
it's for a project that i'm working on. i need to know how to show that a map like T(X)=XAX^(-1)AX is or is not injective on a group.
toxic zephyr
fair enough
next obsidian
If the map isn’t playing nice with the algebraic structure, then you can’t bring the power of algebra to bear on the problem
So this basically just turns into “hope you figure something clever out”
toxic zephyr
ill do my best lol
dull ginkgo
So, stupid thing I noticed
If we have a ring, K, which has a characteristic k > 0, then every element in the additive group of K has order k if K is an integral domain
is this useful in any way
delicate orchid
it's one half of showing that every field with char = p > 0 has a copy of F_p in it I guess
and cyclic groups are nice in general
dull ginkgo
I wanted to prove that every ring with a cancellative property (aka, 0 singleton is a prime ideal) is a commutative ring (just a step), and came across that
I wanted to avoid class equations
chilly radish
Maybe a dumb question, suppose I have two CSAs $A,B$ and an element $a\otimes 1$ which is known to be a unit. I wanna show that its inverse must be $a^{-1}\otimes 1$, but a priori $a$ is not a unit itself
cloud walrusBOT
ShiN
chilly radish
what do
dull ginkgo
wtf is a csa
chilly radish
what
no
central simple
i.e. its center is the base field and it has no nontrivial two-sided ideals
chilly radish
C* ALGEBRAS ARE ALSO NOT COMMUTATIVE
dull ginkgo
systemic discrimination of noncommutative algebras
prisma shuttle
for a ring R, define s(R) be the minimal degree for which a vanishing polynomial exists (aka a polynomial that evaluate to 0 for all inputs from R)
i asked a question earlier about Lemma 8 in this paper here
and i understand how lemma 8 shows that s(R) <= 4 (it shows a vanishing polynomial of degree 4 for R)
but idk how to show that s(R) = 4 (as in the remark that is left to verify by the reader)
does anyone understand how to show s(R) = 4
and note that brackets in the def of R denote the generated ideal, so its Z_2[x]/(x^3+x^4)
sly crescent
Is it possible for a Lie algebra to be simple over one field but not another?
agile burrow
Is it possible for a Lie algebra to be simple over one field but not another?
Yes, sl(2, C) is simple but sl(2, F_2) is not
sly crescent
Yes, sl(2, C) is simple but sl(2, F_2) is not
Those are two different Lie algebras
sly crescent
sl(2,C) is simple over both R and C, and sl(2,F_2) only exists over F_2.
rustic crown
does anyone understand how to show s(R) = 4
find an element in R which doesn't satisfy a polynomial of degree 3 or lesser. ||what about x in R?||
prisma shuttle
find an element in R which doesn't satisfy a polynomial of degree 3 or lesser. |...
could u elaborate a bit more
ima kinda confused
wdym "doesn't satisfy a polynomial"?
rustic crown
we would like to show that s(R) = 4, so there should be at least one element in R which doesn't have a vanishing polynomial of degree 3 or smaller
prisma shuttle
we would like to show that s(R) = 4, so there should be at least one element in ...
oh so u mean like find an element which doesn't vanish for any polynomial of deg less than 4?
rustic crown
yep!
prisma shuttle
why would x satisfy that?
rustic crown
because going modulo x^4+x^3 won't do anything to a polynomial with degree < 4
smoky ivy
How does one show
$$\mathbb{F}{p^n} \subset \mathbb{F}{p^m} \iff n \mid m?$$
cloud walrusBOT
lewis
smoky ivy
I actually only need to prove =>
bc I already have <=
I just don’t see a way to prove “=>”, I’ve tried pretty much everything I had on mind
nvm, I think I can see it:
$\mathbb{F}{p^m}$ is just some field extension of $\mathbb{F}{p^n}$, so consider the degree $[\mathbb{F}{p^m} : \mathbb{F}{p^n} = k$ for some $k \geq 1$. This means that $\ mathbb{F}{p^m}$ is a $k$-dimensional vector space over $\ mathbb{F}{p^n}$ and therefore isomorphic to $(mathbb{F}_{p^n})^k$, which shows that $p^m = p^{kn} \implies m = kn \implies n \mid m$.
wooden ember
F_(p^m) is a vector space over F_(p^n)
cloud walrusBOT
lewis
smoky ivy
yeah
smoky ivy
welp, latex on phone is not the best
latent night
What is the initial object in R-alg?
chilly radish
Maybe a dumb question, suppose I have two CSAs $A,B$ and an element $a\otimes 1$...
I solved it it was stupid
C(1\otimes B)= A \otimes 1 and centralisers are closed under inverses
chilly radish
Weibel proves that a projective R-module is a direct summand of a free module, but he says that this is iff but doesn't prove the other direction. Is it immediate?
agile burrow
It shouldn't be more than a few lines, you need to use the fact that free modules are projective along with the composition P -> F -> P
fluid portal
Can someone give an intuitive definition of Cayley’s Theorem
chilly radish
It shouldn't be more than a few lines, you need to use the fact that free module...
Btw he says that there being an incluson i from P to F and a projection p:F to P such that pi=1_P implies that P is a direct summand. Is there a way to see this without using e.g. the splitting lemma?
And yea that was easy walter
The converse
agile burrow
hmm, i'm not sure about an easier way to see that a retract implies being a summand besides the splitting lemma since the proof of that essentially does everything by hand
chilly radish
True ig
agile burrow
Can someone give an intuitive definition of Cayley’s Theorem
I think Cayley's theorem is best understood in terms of group actions, do you have experience with actions?
chilly radish
Eh actually it's easy.
Clearly ker p and Im i intersect trivially, and you get the direct sum decomposition by considering ip of an element
fluid portal
I think Cayley's theorem is best understood in terms of group actions, do you ha...
Yes
agile burrow
Ok great, so every group acts on its underlying set via left-multiplication (just the group operation). Recall that the data of an action on a set is nothing more than a group homomorphism from G to the symmetric group on that set; in this case, we have a homomorphism G -> Sym(G) (the symmetric group on the underlying set of G).
We want to show that this map is injective to see that G is a subgroup of Sym(G), so we need to compute the kernel of this homomorphism. An element g is in the kernel if it acts trivially on the group, so lets suppose g acts trivially on the group. In particular, g * e = e, which implies that g = e. Thus, the kernel of the homomorphism is trivial, hence we can identify G with a subgroup of Sym(G).
woeful flint
is there a nice way to write $\gamma_5$, the 5th root of unity, using radicals?
cloud walrusBOT
Kraft Macaroni
fluid portal
We want to show that this map is injective to see that G is a subgroup of Sym(G)...
So, Sym(G) acts on Obj(G’ (copy of G)) which produces an action g(s). Alternatively subgroup (G) only correlates part of the permutations on an obj (G’). So with the former we set G an action on the left side (left action) with an object G’ (which is really G) and a right action which is the byproduct of g(s) that can be denoted as g x s such that G correlates ALL permutations on G’..? Alternatively, having all permutations means that we have all symmetries.
agile burrow
I'm not sure what you mean by the right action. We won't be able to recover all of the bijections from G' to itself (or even all of the symmetries of G') because the homomorphism from G to Sym(G') isn't usually surjective, only injective. This is because the action of G on itself determines some of the bijections from G to itself, but not all of them.
We can only show that the homomorphism is injective
fluid portal
I see thanks
prisma shuttle
can someone explain how to go from the second line to the third line in this series of steps
i don't understand why x^3Q^3(x) = x^3Q(1)
oblique river
For the same reason as I gave for the other thing. All powers of x higher than x^3 are just equal to x^3 in this ring. Maybe jt would help to write out Q(x) = a + bx + cx^2 + …
Then x^3 Q(x) = x^3(a + bx + cx^2 + …) = ax^3 + bx^4 + cx^5 + …
Now replace the higher powers of x with x^3 and factor out x^3 again
@prisma shuttle
Also why do you care about this ring so much? Youve been asking about it for days lol
prisma shuttle
Also why do you care about this ring so much? Youve been asking about it for day...
oh i get it now
its for this paper ima reading
i don't really care about the ring lollll
its just an example ima trying to understand 🙂
thx for all the help
oblique river
I think you need to just write tbinfs out more
Instead of staring at “x^3 Q(x) = x^3 Q(1)” for days wondering what it is
Just expand both sides
And it becomes clear
prisma shuttle
And it becomes clear
k thx for the advice
yea idk wut i was doing ahhh
inability to expand ig
wooden ember
Not sure where to ask this but is there an easy to obtain the dual curve of an algebraic curve in P^2(C)?
I’m asking because I’m trying to exercise applying Kippenhahn’s theorem to compute the numerical range of a matrix but I’m having some trouble getting the dual curve of det(uH_A+vK_A+wI)=0
And this is in a simple example with a Jordan block (the homogeneous polynomial above become 8w^3 -4wu^2-4wv^2=0)
Hmm there’s a way around that calculation but I’d still like to know
devout crow
is there a nice way to write $\gamma_5$, the 5th root of unity, using radicals?
this should help
long obsidian
I'm trying to decide whether the set of continuous functions with compact support C(X)={f:X->R} on a manifold X is a ring with or without unity through pointwise addition/multiplication. I think there couldn't be a unit I_R because it would have to have compact support and to be a multiplicative identity suppI_R should contain the support of every other function C(X) which seems difficult for a compactly supported function
chilly ocean
what is the support of the constant 1 function?
long obsidian
Hmmm maybe this depends on whether the manifold itself is compact
chilly ocean
that's true. how so?
long obsidian
Well if the manifold is compact then maybe being a compactly support function is a cheap property that is easily satisfied.
But if the manifold is not compact. Maybe I could assume such an identify I_R exists. Then maybe in the complement of supp I_R there exists some compact subset such that a bump function exists with such support and that would be a contradiction to such a unity existing since the pointwise product would be zero since they have distinct supports
chilly ocean
that's a nice argument
sounds alright to me
if X is compact then every function is compactly supported, so you're in luck
chilly radish
I wanna show that if P is a projective chain complex each of its components is projective. I wanna do this without using the fact that a projective complex is split-exact. My thought was to lift to corresponding diagram to a diagram of chain complexes and use the lifting property there, but i'm having trouble with getting the proper diagrame to commute
So say we are looking at P_0
Then for some objects B and C I have an epi from B to C and a map from P_0 to C which I wanna lift to B
next obsidian
I feel like maybe you could do this by looking at complexes concentrated in a single degree
chilly radish
The natural thing would be to consider complexes concentrated at degree 0
next obsidian
And basically isolate a specific part of your complex
But idk, I forgot how I did this
chilly radish
But the problem is I think extending the maps into chain maps is impossible if you try and do that
Extending your.maps.to chain maps
next obsidian
What about this
Consider the map P• -> P_n which is 0 everywhere except at the n-th where it’s id
Consider a surjection F -> P_n of a free module, all this happening in degree n
Won’t this let you split off P_n into F and get that P_n is a summand of F?
By lifting this chain map
chilly radish
Consider the map P• -> P_n which is 0 everywhere except at the n-th where it’s i...
Is this a chain map tho?
chilly radish
This square doesn't commute
next obsidian
Or a cokernel
Like image of P_n-1 -> P_n
Before the P_n that’s the target
But that makes it hard to work with the free complex
chilly radish
Yea
next obsidian
Yeah I forgot how to do this lmao
chilly radish
That again gives the problem of extending to a chain map
next obsidian
Yeah I mean
When you have a complex of free modules it becomes easy thoigh
Because it’s easy to define maps
You just work with generators explicitly and stuff
chilly radish
Me and a friend solved this by using the fact it's split exact and showing that the boundaries are projective
But I was wondering if there was a nice way without that
next obsidian
Let me see how I did it
dull ginkgo
let $K$ be a ring, $K^+$ it's abelian additive group, $K^{\times}$ it's multiplicative monoid.
Well, there exists an injection
$\ f: Set(K) \hookrightarrow End(K^{+}) \ im(f) = K^{\times} \$
Besides this being just currying, does this have any use
I assume $(K^{+},f)$ completely characterizes K.
dull ginkgo
Currying can also be used to characterize a semigroup G by noting that there exists an injection $\k : Set(G) \hookrightarrow Set(End(Set(G)))\$
cloud walrusBOT
Mizalign
prisma ibex
idk if I'd call it a use but yes it can sometimes be clarifying to think of rings as Abelian groups with this action by endomorphisms
the same thing happens for modules where you think of these as Abelian groups with ring action like this
prisma ibex
it works fine for left/right modules
dull ginkgo
since left or right scaling is an endomorphism of the "vector" Abelian group
and right or left multiplication is an endomorphism of the scalar ring
prisma ibex
you just have to make a choice and stick with it consistently
like if you define R->End(M) as a left action then R^op->End(M) will be a right action
but you could also define it the other way around
as long as you keep the choice consistent it's fine
dull ginkgo
yeah
Well, actually
given the route I took in defining it, it doesn't matter
f is specified, which can be either left or right scaling, or both
Say for abelian groups
prisma ibex
for modules you have to make a choice unless R is commutative
prisma ibex
yeah then this is the distinction between left and right group action
dull ginkgo
then the function's L_x and R_x are not thesame
but still injective, and still characterize the group
they are seperate injections from the set of elements to the set of endomorphisms, with equal images
if they are the same injections, it's commutative
Thus, in the case of modules, yes left and right multiplication/scaling are different, but they are both injections which characterize the module i guess
Theoretically
For some set $S$, letting the set of injections from sets A to B be denoted Mono(A,B) by convention
$\$ Let $K = Mono(S,End(S)) \$
Can every element of $K$ can be considered as a semigroup when paired with $S$?
cloud walrusBOT
Mizalign
woeful flint
How many field embeddings are there $\text{Emb}_\mathbb{Q}(\mathbb{R},\mathbb{C})$. For some context I understand the notion of a totally real fields. I want to understand something a bit weaker i.e. if $\mathbb{Q}\subset \mathbb{K}$ is a finite field extension (it being Galois is not important for now) and $\mathbb{R}_K := \mathbb{R} \cap K$ is it always true that $\text{Emb}\mathbb{Q}(\mathbb{R}K, \mathbb{R}) = \text{Emb}\mathbb{Q}(\mathbb{R}_K,\mathbb{C})$?
cloud walrusBOT
Kraft Macaroni
cloud walrusBOT
det
woeful flint
Actually maybe I phrased this completely wrong
rustic crown
this is not too hard to show, by set theory number of such maps is at most |C^R| = 2^(|N| * |R|) = 2^|R| = 2^c
so you just need to find at least 2^c maps
woeful flint
What I want in the end really is that for any $\sigma\in \text{Gal}(K/\mathbb{Q})$ we have that $\sigma(\mathbb{R}_K) \subset \mathbb{R}$
cloud walrusBOT
Kraft Macaroni
woeful flint
But i think I was overcomplicating things in trying to find an answer
rustic crown
so you just need to find at least 2^c maps
pick a transcendence basis for R, say B and define the map f(b) = +-b, since C is algebraically closed, this map would extend to all of R. now we can choose the +- anyway we like and so if you show |B| = c, you're done.
rustic crown
But i think I was overcomplicating things in trying to find an answer
take K to be the splitting field of x^3-2. which is K = Q(cbrt(2), omega) and then K_R = Q(Cbrt(2)).
pick sigma in the galois group which sends cbrt(2) to cbrt(2) * omega so sigma(K_R) won't be entirely real.
rustic crown
woeful flint
How about if we restrict out view to $\mathcal{O}_K^\times \cap \mathbb{R}$?
cloud walrusBOT
Kraft Macaroni
rustic crown
i don't think that changes anything
cbrt(2) - 1 is a unit
one of its conjugates is cbrt(2) * omega - 1
which isn't real
prisma shuttle
because going modulo x^4+x^3 won't do anything to a polynomial with degree < 4
wait det idt this actually works
because if u look in the example the coefficients can actually be "polynomials" themselves
prisma shuttle
here's the example to refer to
so couldn't the coefficients sort of "increase" the deg
prisma shuttle
what's s(R) then? i probably guessed the definition last time
s(R) is the minimum degree of a monic vanishing polynomial where the coefs of the polynomial are in R
so for that example: xP + (1+x)P^2+P^4 that would be deg 4, and the coefs x, 1+x, and 1 are all in R
and so to prove s(R) = 4 we have to somehow show that there can't be such a polynomial with deg < 4
rustic crown
ah, i see. the single polynomial should kill every P in R
prisma shuttle
ah, i see. the single polynomial should kill every P in R
exactly
rustic crown
we should just work it out then. if a polynomial with degree at most 3 kill everything, then in particular it kills 0, 1, x and 1+x. if it kills 0, the constant term is 0
so assume the polynomial looks like aP + bP^2 + cP^3
this should be 0 when when plug P = 1, x, 1+x
so we have
a+b+c = 0
ax+bx^2+cx^3 = 0
a(1+x) + b(1+x^2) + c(1+x+x^2+x^3) = 0
this means c(x+x^2) = 0 and only way this happens is when c is divisible by x^2, but as the polynomial we want should be monic this forces c = 0
so the equations simplify to a+b = 0 and ax+bx^2 = 0
so a = b and b(x+x^2) = 0
by the same logic b = 0
so s(R) >= 4
woeful flint
Is it true that if $K \subset L \subset M$ then $\text{Gal}(L/K) \subset \text{Gal}(M/K)$?
cloud walrusBOT
Kraft Macaroni
coral spindle
I think you'll find it's the other way around
Give me a moment to think about this
coral spindle
No, sorry. My mistake.
The idea is that if f is an automorphism of L:K then you can extend it to an automorphism of M:K, right?
coral spindle
If you can prove this, you should be done. Of course this is not a true subset but rather an isomorphic subgroup
So Let's say f is in Gal(L/K)
Since M has an L-basis (we know this since they are fields), we can certainly just define the lift of f as follows
So, let M have an L-basis V
We can assume that 1 is part of this basis
then we simply define f', the lift of f, as f'(x1) = f(x), and f'(v) = v otherwise, where v is some other basis element
Does that work as an automorphism?
It ought to, right?
woeful flint
I'd think so
one sec
the fact that it's an isomorphic subgroup actually complicates things quite a bit
coral spindle
I think we should be able to handle it
coral spindle
Yeah ok so the idea is that we send f to f' as described above
But first yeah, we just need to check that it works
woeful flint
if we want to gain any useful information about the actual field automorphisms
which I need to compute the field trace
rustic crown
Is it true that if $K \subset L \subset M$ then $\text{Gal}(L/K) \subset \text{G...
i don't think it's true.
consider the quaternion group Q8. you know that there is a galois extension M/K with galois group G isomorphic to Q8.
Now take H = Z(G) the center of G, so H has order 2. define L to be the fixed field M^H. since H is normal L/K is galois and its galois group is isomorphic to G/H = (Z/2)^2 but this isn't isomorphic to a subgroup of Q8.
(you could also force K = Q as the inverse galois problem is true for solvable groups)
in general all you can say is that there is a surjection from Gal(M/K) --> Gal(L/K) given by restriction
coral spindle
Great counterexample
oblique river
Is it true that if $K \subset L \subset M$ then $\text{Gal}(L/K) \subset \text{G...
as det correctly pointed out, this isn't true in general; all you can say is that there is a surjection Gal(M/K) --> Gal(L/K), and this is assuming that everything is Galois. M/K being galois does not imply that L/K is galois, so you need to make two separate assupmtions.
However, if M/K is galois, then M/L is automatically galois, and you do have a subgroup relationship there: Gal(M/L) is a subgroup of Gal(M/K). it's a normal subgroup if and only if L/K is galois, in which case the quotient Gal(M/K)/Gal(M/L) is Gal(L/K)
tall mountain
does this argument work
@north sand since they helped me yesterday with a similar problem
rustic crown
Yep looks good. I just feel one should justify why inverse image of maximal would be maximal as that really needs the map to be surjective. But if you've proved it before then all cool
dull ginkgo
can this be generalized
tall mountain
Yep looks good. I just feel one should justify why inverse image of maximal woul...
yeah that was one of the exercises I did before
dull ginkgo
For two rings A and B, If C is a maximal ideal of A, then C x B is a maximal ideal of A x B
actually no
rustic crown
Yea, that's true I think
A x B/C x B = A/C x B/B = a field
Not only that every maximal ideal of A x B looks either like C x B or A x D
(where C is maximal of A and D is maximal of B)
dull ginkgo
A x B/C x B = A/C x B/B = a field
bruh what
the first half
I get if A/B is a division ring then B must be a maximal ideal but
the first equality
rustic crown
oh should have parenthesised it properly. (A x B)/(C x B)
dull ginkgo
that's a quotient rule I haven't seen
I assume that holds for any algebraic quotient
rustic crown
Yea it does. It's a simple consequence of the usual iso theorems
If I, J are ideals of A and B respectively then we have the maps A x B --> A --> A/I and similarly a map A x B --> B --> B/J
dull ginkgo
oh, yeah
rustic crown
This induces the map A x B --> A/I x B/J
Easy to see surjective and that kernel is I x J
dull ginkgo
alright, cool
rustic crown
dull ginkgo
wait, the image of a ring homomorphism need not always be ideal so when can we actually apply first iso
rustic crown
First iso just says that a surjective map R --> S factors through the quotient and induces the iso R/ker = S
rustic crown
Oh yea :p
tall mountain
my book doesn't really say much about the iso theorems for rings
just states the first one
oblique river
they're basically the same as the isomorphism theorems for groups
just replace "group" with "ring" and "normal subgroup" with "ideal"
and things like that
chilly radish
In weibel prop 2.3.10, he wishes to prove the contravariant hom functor Hom(-,R(I)) is exact, but he just proves it turns monos into epis. Is this sufficient? Because I think I saw an example where a monofunctor is not exact. Is he just sweeping stuff under the rug and there is more left to check?
Also, he requires the functor to be additive, but I don't see where additivity is being used here
cloud walrusBOT
chilly radish
A,B are abelian ofc
Is it because we exact functors are additive? Is an adjoint to an additive functor necessarily additive?
(I know this kinda toes the line between #groups-rings-fields and #category-theory )
I'm not sure why he didn't just draw out the full diagram
next obsidian
I don’t see what you mean
@chilly radish
You already know Hom(-,A) is always left exact
All you need to do is show surjecticity on the right
chilly radish
Oh right oop
next obsidian
And that’s what he did
next obsidian
chilly radish
I keep forgetting that lmao
Anyways adjoints are additive right
Between additive cags
Cats
next obsidian
I don’t know
That might be extra data
Or maybe it comes for free if both functors are additive
chilly radish
I wanna show that they're additive
You mean both cats?
If one is additige i'k almost certain the other has to be
next obsidian
No if the functors are additive
I mean you’re asking if the equation in the adjoint is also linear right?
I’m saying I’m not sure, but maybe if F and G are additive then it comes for free maybe
chilly radish
Wdym by linear
next obsidian
An adjoint is expressing
chilly radish
I'm asking if L is additive, is its right adjoint also additive
next obsidian
Hurb
chilly radish
Or do you get that for free
next obsidian
I thought you’re asking if the equation
next obsidian
Hom(LA,B) = Hom(A,RB)
chilly radish
Yeye
next obsidian
Your question seems more likely to be true imo
chilly radish
Extra assumptions never hurt nobody
next obsidian
Lmfao
Apparently adjoint functors are automatically additive?
The proof is simple too
next obsidian
Oooo and the equality of homs is also additive
chilly ocean
I think
chilly radish
Apparently adjoint functors are automatically additive?
That's crazy
Loke not really but still lol
chilly radish
Oooo and the equality of homs is also additive
Ye I was saying if they're additive (and they are), hom.preserves finite biproducts and everything is natural so u should get that it distributes
chilly radish
If I know a free module decomposes as a direct sum, can I necessarily take homogeneous generators?
like a basis
yea right? I can just take a basis and decompose each element into its homogeneous components and then those should be linearly indpeendent as well
just wanna make sure i'm not missing anything
once we remove all the zeroes
wooden ember
what's a lie algebra class like?
I know it's a pretty vague question but ive been given the option to take two third year classes, and i know for one ill be taking rings and modules, but im not sure what to take for the other since most of the classes have a good few prerequesites but lie algebras is one of the only ones that doesnt really have any and sounds interesting
chilly ocean
Lie algebras are vector spaces with a bracket
wooden ember
i guess my question is more in the direction of what "flavor" does the class have if you know what i mean
i know what i mean but im not sure how to communicate it lol
chilly ocean
So it'll be pretty standard for an algebra class I think. Matrices can sometimes be emphasized a lot I guess
wooden ember
the reason im thinking of taking it too is because it's only given once every 2 years
so i wont have the option to take it next year
chilly ocean
i guess my question is more in the direction of what "flavor" does the class hav...
The flavor is suffering
but you should take it neverhteless
There's kernels and stuff. Nilpotency. Sequences of ideals
wooden ember
these are the options just in case, specifically the ones that have 2,2 on the earlier columns
(other classes are second semester)
chilly ocean
Kind of like in group theory
wooden ember
so it's kind of like linear algebra with a ring theory flavor?
sharp sonnet
probably works towards the classification of semisimple lie algebras?
wooden ember
yeah that's the last part
chilly ocean
That's how the introductory courses typically go
sharp sonnet
pretty cool class tbh
mine was titled "lie theory" and we first did lie group, lie algebra connection which is kinda galois theory vibes
chilly ocean
I read this about Lie algebras
sharp sonnet
and the classification introduces lots of nice tools
chilly ocean
galois theory vibes sounds nice
You wouldn't find that in a Lie algebras course
sharp sonnet
you can always learn a bit more 🙂
the lie algebras is the fun part for an algebra brain
chilly ocean
you would find that in a course that deals with the relationships between Lie algebras and Lie groups, which would typically be titled Lie theory
or Lie groups
wooden ember
ah i think that's a grad course
they have a lie groups class im pretty sure
to be fair i originally wouldve wanted to take rational quadratic forms but it seems like they removed it from the list
sharp sonnet
i think its also a pretty useful class from a utalitarian pov
but i cant comment too much on that
wooden ember
useful is good
chilly ocean
utilitarian
You can see a bit how it looks like here
Just look at Humphreys' book if you want a quick introduction
wooden ember
in the other available classes is there one youd prioritize over lie algebras (making assumptions on my preferences?)
wooden ember
You can see a bit how it looks like here
oof chains of ideals
sharp sonnet
did you take rep theory yet?
wooden ember
nope but it's available on the second semester
and i plan to take it then if im allowed to
sharp sonnet
thats another class i would say is both very useful and also fun
wooden ember
yeah and it's also given once every two years so im definitely prioritizing it
chilly ocean
You should take the Diff manifolds course if you haven't already
wooden ember
You should take the Diff manifolds course if you haven't already
thing is in the prereqs they say analysis 3 and 4 + diff geometry
which are all classes im going to take this year
so idk if it's wise
sharp sonnet
quadratic forms wouldve been a nice class 
chilly ocean
Spaces of non positive curvatures and groups sounds like it would definitely be fun
chilly ocean
You should take the Diff manifolds course if you haven't already
that's for Lie groups, no
wooden ember
quadratic forms wouldve been a nice class <:blobcry:971797574916907028>
yeah
sharp sonnet
though probably not too useful
wooden ember
Spaces of non positive curvatures and groups sounds like it would definitely be ...
same kind of deal
chilly ocean
Lie algebras should be more algebraic
sharp sonnet
take alg top too
chilly ocean
that's for Lie groups, no
I mentioned it regardless of the Lie algebras course
they should take it because it's an important course to take
wooden ember
take alg top too
second semester and id be taking topology at the same time so probably not, ill keep it for the following year i think
my plan for second sem is algebraic number theory, algebraic curves and rep theory
sharp sonnet
algebraic nt brain moment
chilly ocean
they should take it because it's an important course to take
I agree then
chilly radish
they have a lie groups class im pretty sure
DON'T take this if you: A. have not taken manifolds (you will suffer) or B. It does not have a manifolds prereq (It will be a shitty class)
wooden ember
DON'T take this if you: A. have not taken manifolds (you will suffer) or B. It d...
lol it's a grad course anyways im not touching that shit
chilly radish
grad courses are not that scary
tight flare
Damn thats a lot of options you have there
wooden ember
maybe but ill stick to third year courses for now
chilly radish
alrighty then
tight flare
grad courses are nice from what i've seen so far
way more interactive because of the size and just more chill
chilly ocean
It depends on the courses
tight flare
fair enough
sharp sonnet
you should take more seminars, less classes
wooden ember
what's the difference?
tight flare
seminars are great as well
wooden ember
not sure we have those
tight flare
i had so much fun in my algtop one
chilly radish
idk if that's right for someone at narwhal's point
tight flare
probably gonna do another one about riemann surfaces
sharp sonnet
seminars are usually more open and the examination is a talk
ye well, at grad level
chilly ocean
I'm just wondering if Lie algebras are useful for something that doesn't relate to differential topology/geometry
wooden ember
i see, yeah i imagine that's more at grad level
tight flare
I'm just wondering if Lie algebras are useful for something that doesn't relate ...
they are i think
sharp sonnet
i feel like at some point you dont benefit too much from just classes (you can read a book yourself) and seminars are closer to what you do as a mathematician
chilly ocean
I'm just wondering if Lie algebras are useful for something that doesn't relate ...
I mean the original motivation was PDE's
tight flare
one of my profs does something with them that isn't really that
chilly ocean
but that didn't seem to go too well
tight flare
don't ask me what tho
wooden ember
i feel like at some point you dont benefit too much from just classes (you can r...
that's fair
chilly radish
once we remove all the zeroes
can someone give me a sanity check on this. I think the result is correct but I think my reasoning is sus
wooden ember
alright well thanks for the recommendations i think ill be taking lie algebras in the end yeah
wooden ember
can someone give me a sanity check on this. I think the result is correct but I ...
oh yeah sorry i drowned your question
chilly radish
lol it's all good
this is obviously true under mild assumptions on the ring (Since e.g. if R is a PID all submodules are free so you can take union of bases), but i'm not sure if it's always true
Wait I do think I don't need this
yea wait ok
yea ok
nvm
I think this is not even ture in general
woeful flint
I'm trying to solve exercise 46 chapter 5 in marcus' Number Fields
specifically part a
cloud walrusBOT
Kraft Macaroni
woeful flint
So far I have that $u = \overline{u}\times\zeta_p^k$ for some $k$ where $\overline{u}$ is the complex conjugate of $u$. But why does $\overline{u}$ need to be real. Surely there is something I'm missing.
cloud walrusBOT
Kraft Macaroni
oblique river
So far I have that $u = \overline{u}\times\zeta_p^k$ for some $k$ where $\overli...
I mean, if u isnt real then its complex conjugate surely isnt real
Like, if u = a + bi then its conjugate is a - bi, which is only real if b = 0, i.e. if u is real to begin with
agile burrow
Given a double complex $L^{\bullet, \bullet}: \cdots \to L^{-2, \bullet} \to L^{-1, \bullet} \to L^{0, \bullet} \to 0$ is $H^{i}(L^{\bullet, \bullet})$ the complex $\cdots \to H^i(L^{-2, \bullet}) \to H^i(L^{-1, \bullet}) \to H^i(L^{0, \bullet}) \to 0$ with zero maps as differentials?
cloud walrusBOT
walter
agile burrow
sorry for weird indexing, I'm trying to show that additive functors preserve the cohomology of cartan-eilenberg resolutions but I'm not entirely sure what that actually means
oblique river
@woeful flint do you know dirichlet’s unit theorem?
Also sorrt walter I also dont know what that means
agile burrow
all good, i lowkey interrupted anyway
agile burrow
ah yeah, i'll crosspost if this doesn't get a response in a few hours
chilly ocean
I'm trying to solve exercise 46 chapter 5 in marcus' Number Fields
this belongs to #advanced-number-theory
dull ginkgo
Is there a way to prove that if every element of a ring has finite multiplicative order, then it's necessarily commutative
toxic zephyr
so for D4 (the dihedral group), is there no commutator of two elements that results in r1 or r3?
south patrol
Uh do you mean every element is nilpotent then
dull ginkgo
never heard of that term
yes
Maybe I can induce contradiction by assuming there exists elements xy such that
xy - yx ≠ 0
and assuming it's an element z
with some minimum n such that z^n = 1
and using infinite descent
oblique river
zero still has an order, specifically 1
how are you defining the order of an element?
dull ginkgo
$Ord^{\times}(x) = min({ n \in \mathbb{N}^* : x^n = 1}) \$
$Ord^{+}(x) = min({ n \in \mathbb{N}^* : \sum_{i=1}^{n}{x} = 1})$
dull ginkgo
oh
oblique river
in which case your ring is the trivial ring
oblique river
so then your question doesn't make any sense
cloud walrusBOT
Mizalign
dull ginkgo
so then your question doesn't make any sense
yeah, every nonzero
oblique river
a) so are you talking about additive order or multiplicative order?
b) i'm assuming you mean = 0 isntead of = 1 in the second definition
c) why not just write nx lol
dull ginkgo
mutliplicative
dull ginkgo
a) so are you talking about additive order or multiplicative order?
b) i'm assum...
nx pains me because I get tired of having to specify it's a z module algebra
oblique river
... every ring is a Z-module
dull ginkgo
someone literally made me use a different notation so "n isn't in the ring"
dull ginkgo
... every ring is a Z-module
e x a c t l y
oblique river
someone literally made me use a different notation so "n isn't in the ring"
point me to them and i'll call them dumb
dull ginkgo
alright
but yeah
Let $K$ be a ring such that
$\forall x \in K^{\times}(\exists n \in \mathbb{N}^* : x^n = 1) $
Then $xy = yx$ for all $x,y$ in $K$
i'm gonna try aproof by contradiction
cloud walrusBOT
Mizalign
dull ginkgo
it's a theorem by Jacobson i thought
wikipedia:
"Nathan Jacobson later discovered yet another condition which guarantees commutativity of a ring: if for every element r of R there exists an integer n > 1 such that r^n = r, then R is commutative.[1] More general conditions which guarantee commutativity of a ring are also known.[2]"
maybe I'll try using ring radicals
actually different radical
oblique river
there's the paper
where it was proven first
I don't think you're going to be able to prove this with a simple proof by contradiction
tall mountain
"algebraic algebras"
chilly ocean
What’s the impact of a group being solvable but not split solvable
For example let $\mathfrak{g}$ be the set of matrices of the form [X=\begin{pmatrix}0&\theta&x\-\theta&0&y\0&0&0 \end{pmatrix},\theta,x,y\in\bR.] Then $\mathfrak{g}$ is solvable, but not split solvable.
cloud walrusBOT
engineersanonymous
smoky cypress
@oblique river I remember you mentioning something along the line like if the nth power map on a group is a homomorphism for 3 consecutive n's then the group is abelian. Can you show me like the proof of this result or any results along this line?
next obsidian
@smoky cypress no proof, but here’s some stuff about this property
I think I was able to do this the first week I learned the group axioms, so it probably isn’t extremely hard
smoky cypress
Oh ok thx
woeful flint
What is the correspondence between the degree $[L:K]$ of a field extension $\mathbb{Q}\subset K \subset L$ where $K$ is a subfield and the index $[G:K^\dagger]$ where $K^\dagger$ is the subgroup of $G$ fixing $K$?
cloud walrusBOT
Kraft Macaroni
woeful flint
It feels like its something that should be part of the galois correspondence no?
coral spindle
$[L\colon K] = |\mathrm{Gal}(L \colon K)|$, provided that the extension $K : \mathbb Q$ is normal.
cloud walrusBOT
Boytjie
coral spindle
I forget the proof of this.
In my experience, this is stated as part of the Galois correspondence
Ah whoops, typo, 1 sec
No wait in fact I did get that right, had to double-check though
coral spindle
$[L\colon K] = |\mathrm{Gal}(L \colon K)|$, provided that the extension $K : \ma...
No I did make a typo, I meant that L : Q is normal
woeful flint
like say if the degree of L over K is n then the index of the galois group over the fixed field is n
woeful flint
No I did make a typo, I meant that L : Q is normal
o shit yeh
coral spindle
Oh right I see
woeful flint
Fortunately I don't need to prove this cuz im working in finite fields
sorry cyclotomic
not finite
coral spindle
Well if I'm not mistaken, $K^\dagger = \mathrm{Gal}(L \colon K)$ right
cloud walrusBOT
Boytjie
woeful flint
I acc don't know but that sounds right
coral spindle
Because every element of Gal(L:K) already fixes Q, so it must be a subgroup of G
So I think you can translate this into something specific
But like, in my head, the reason this doesn't talk about the index of K^\dagger in G is because the galois correspondence reverses order
woeful flint
yeah that's exactly my problem
I've actually found a way to side step computing this index
i.e. I dont need it to show what I needed to show
coral spindle
Ok sick
woeful flint
but it's still interesting for general culture