oh
#groups-rings-fields
next obsidian
a is not coprime to rs
since it shares a factor with r
so a isn't a unit mod rs
but that equation says it is!
sterile garden
Oh wait, I missed these.
k doesn't need to be coprime to r. If we choose k < r, then ks < rs
so a^ks != e
sterile garden
Oh right
next obsidian
but the proof above shows it is coprime to r
if it satisfies the equation
ks + jr = 1 mod rs
(for some j)
sterile garden
the a you're using above is a different a than my group generator?
next obsidian
yeah
that's just the gcd
(assuming the gcd is not 1)
I'm just running out of letters haha
all we need is a nontrivial common factor of k and r
next obsidian
Yup
sterile garden
I had the idea of this, but couldn't show it.
next obsidian
Then by adding multiples of s and r
sterile garden
This is quite clear
next obsidian
you can make k,j be in the ranges we want
then you just show there's only one solution there
and boom
next obsidian
Yup via contrapositive
it does
Suppose we had two b's which worked
(or two c's)
then b = a^{ks} and b' = a^{k's}
then we do the shenanigans to make k,k' be in the range we want
they (along with whatever j we have)
form two solutions (k,j) and (k',j)
so we get k = k'
but then b = b'
sterile garden
This was bothering me all of yesterday. I don't have a good group theory foundation and am trying to build it up a bit.
next obsidian
this is mostly elementary number theory I thin
once you translate some of the gruop theory it turned into just number theory, which is common to a lot of group theory
TBH at first I was just saying stuff, I wasn't even sure why what I was saying would imply it
I just have quite a bit of experience with this stuff, and usually if we have "prove if (a,b) = 1 then..."
the coprimeness just does soooo much work
so I did what felt right, and that's just cuz I like groups a lot and have a lot of experience with them, point simply being that you'll get there
next obsidian
I wasn't sure either
it just felt right, and I just went with contradiction
since if something feels true and you're not sure why, just assume it's false and see what breaks
woven obsidian
I'm a bit unsure about the step where we assume that the number of T:s is also d
For reference, this is thm 4.33
light tusk
I think the number of T's should be at most d because they are assumed to be algebraically independent in k[Y1,..., Yd], and the last inequality would be $r \leq h(r) \leq e \leq d$
cloud walrusBOT
Apopheniac:
woven obsidian
Yeah, I thought something like that. But I'm not sure how to prove d+1 elts can't be independent
next obsidian
I’m not quite sure what it’s doing, but are you asking why you couldn’t have d + 1 algebraically independent elements in k[Y_1,...,Y_d]?
That’s isomorphic to a d-variable polynomial ring and the d indeterminates form a transcendence basis, so you can only have d algebraically independent elements
light tusk
Also $k[T_1,...,T_e] \subset k[Y_1,...,Y_d]$ is a finite extension
cloud walrusBOT
Apopheniac:
dim escarp
Does any finite field extension of rational numbers subextension of cyclotomic extension? Just curious.
oblique river
no, only abelian extensions
that's a pretty important theorem btw
the fact that all subextensions of cyclotomic extensions are abelian just follows from basic galois theory but the converse (that every abelian extension is a subextension of a cyclotomic field) is difficult but also really important
it's called the kronecker-weber theorem
dim escarp
Oh right. Subextensions should be abelian
I guess it is hard to prove
Btw it's relatively easier to prove that there is field extension of Q with galois group Sn right?
oblique river
it's not hard to prove that subextensions of abelian extensions are abelian
the S_n fact is easier to prove than "every abelian extension is contained in a cyclotomic field" but harder than "subextensions of abelian are abelian"
dim escarp
I mean, subextension of abelian is trivially abelian
oblique river
just checking, but yes
dim escarp
I guess it's feasible to find function f with Gal(f) = Sn over Q
Tho how feasiblee is it?
How hard to find it
oblique river
kind of hard
steep hull
I believe it’s a result of Hilbert
dim escarp
Hilbert?
dim escarp
What result you mean?
steep hull
For every n, there exists an f in Q[x] such that Gal(f)=S_n over Q
sturdy marsh
dim escarp
Interesting that Hilbert did this kind of work
I thought all their work would be supreme advanced that undergrad cannot even conceive
golden pasture
hilbert was kinda old tho
dim escarp
Why thonk :/
dim escarp
Was Hilbert that old?
golden pasture
he was 1900s ish
dim escarp
I mean that is fairly recent
dim escarp
Compared to ppl like Gauss, Pascal, Fermat..
golden pasture
it is like the start of what we usually call abstract algebra
dim escarp
I see, so Abstract Algebra is a fairly recent field
golden pasture
like for groups galois had the idea in 1830 but it really started off at like 1880s
golden pasture
yup
dim escarp
What even
golden pasture
die at 20 from simping someone? yes
dim escarp
Ideal before ring is fine tho
..maybe group is defined after ring and field
Now that would be.. orderly hell
golden pasture
nowadays you define ideal as a subset of a ring with properties
field is defined way earlier i think
golden pasture
seems like 1880s as well
dim escarp
Or is it me who don't get groups
golden pasture
more like
dim escarp
What is Galois Group then
golden pasture
in the past people didnt have the idea of like needing to concretely define such algebraic structures
you can see the ideas of these in like super early on in math
prob pre-1800s
jus no formal definition yet
carmine fossil
"just Intuit math"
dim escarp
How did Galois do things like S5
carmine fossil
That was prob how people did math in the olden days
dim escarp
Or is Galois theorem not by Galois
golden pasture
oh ye and limits and stuff were only really formalized in 1800s/1900s
dim escarp
I mean limits are indeed hard
When I first learned it, I struggled at getting wth it is
Too wiggly
Actually
How is advancedment between 1900s and 2000s so huge
Compared to like 1800s and 1900s
dim escarp
- imho intuit math worked better than formalism which struck hard by limitations
golden pasture
prob like start of 1875s things got fyn
dim escarp
Now it's nearly impossible to do new stuff in math
golden pasture
uh
carmine fossil
"femat's last theorem is trivial by Intuition"
dim escarp
I mean
carmine fossil
Ask fermat
golden pasture
there is
dim escarp
Intuition does not give you Fermat last theorem
dim escarp
Certainly
dim escarp
Nevertheless it's going to be simply too hard to learn I think
golden pasture
carmine fossil
What is too hard to learn?
golden pasture
i mean it is impossible to learn every field probably
but idt anyone expects anyone else to be such a chad
dim escarp
I don't think I'd be able to like read modern papers even after decades of learning
carmine fossil
Erdos
golden pasture
uh
really you only need like
depends on field
actly
like there are some very accessible paper and others that are super specific to a field
most ppl get phd before 30
so
dim escarp
I think whatever paper would require lots
Too much for smone like me to even conceive
golden pasture
eh not rlly you're seriously overestimating how deep fields are
dim escarp
I don't think I'd ever get a chance in pure math tbh
Which is why I'm trying to go for buzz nonsense field like applied math
golden pasture
i think everyone can do pure math lol it is just interest and resources
dim escarp
Interest and resources?
How would that help
Okay I guess I'm so dumb for anything
I wonder how would I get any job or sth then
dim escarp
Wouldn't it be "that and high IQ"
dim escarp
That's contrary to what I'm hearing
carmine fossil
Why would a bunch of random puzzles determine how good you are at math
dim escarp
Colleagues? Idk
golden pasture
dim escarp
Oh and some graduates
dim escarp
Also by IQ I mean how actually smart a person is
golden pasture
i wouldnt continue XD but basically anyone can do math
just choose a field you like
and learn
dim escarp
Hmm, I see
I don't think really anyone could do math, especially after getting crushed during my abstract algebra homework
golden pasture
if it isnt hard are you learning?
dim escarp
Though
chilly ocean
did you solve that problem about irreducible degree p polynomial btw?
chilly ocean
what was the solution?
dim escarp
I just calculated the function
Afaik x^3(x-2)(x-4)...(x-2p+6)+2 becomes irreducible for p>4
golden pasture
sounds right
chilly ocean
ah
dim escarp
The problem "Degree of Q(a + b + c)/Q where a, b, c are prime roots of primes" was particularly hard, I could not solve it myself until asking abt it here
golden pasture
yea it will be difficult i doubt im able to solve it like 3 months ago
you jus kinda like gain experience as you solve these type of problems
golden pasture
I don't feel like I'm gaining experience.., but I'll push through
it takes time to realize
just look back like one year later
obsidian path
I'm trying to determine the units in R=Z6[x]/<x^2+2x>
I know that
1 = (a + bx)(c + dx) => ac + (ad+bc -2bd)x => ac =1 and ad + bc - 2bd = 0.
what do I do next?
vital quail
little algebra exercise: compute order of sylow p-subgroup of GL(n, p) 
golden pasture
I'm trying to determine the units in R=Z6[x]/<x^2+2x>
I know that
1 = (a + bx)(...
next just brute force all the possibilities
maybe a bit more efficient you know that like x and (x+2) arents units
and product of nonunits are also not units
which saves you from checking a bunch of stuff
next obsidian
Is that ring local?
dim escarp
Interesting one, it isn't even domain
next obsidian
If there’s only a single maximal ideal containing (x^2 + 2x) you get the units for free
But I don’t feel like computing ideals of the polynomial ring over Z/6Z lol
woven obsidian
That’s isomorphic to a d-variable polynomial ring and the d indeterminates form ...
I guess this has something to do with Galois theory
obsidian path
would determining the zero divisors be easier to do, and then the units are the rest of the elements?
golden pasture
Is that ring local?
no cuz (2)+(3)=(1)
golden pasture
would determining the zero divisors be easier to do, and then the units are the ...
that is iff local ring
obsidian path
ah damn
woven obsidian
It's a ring with only one maximal ideal
woven obsidian
I hadn't heard about it until I took a course in commutative algebra
The nice thing about them is that a ring being local is equivalent to the non-units forming an ideal, which is sometimes easy to prove/disprove
obsidian path
ah, sounds fun actually
obsidian path
next just brute force all the possibilities <:KEK:586240877358350341>
to be honest I don't even know how to brute force this like.. how does it make sense to solve this even
dim escarp
Oh, only one maximal ideal. ..too highly specific tho
uncut girder
you can take any ring, and take a prime in that ring and localize at that prime
golden pasture
Oh, only one maximal ideal. ..too highly specific tho
local rings are actually quite common in the sense we see it very often because it is super nice
obsidian path
oh nevermind
wait no
to bruteforce, you multiply <x^2+2x> over all the polynomials in Z6 one by one, find which ones are =1?
I think so
yes
well, originally I had thought that if R is a finite ring, then every element of R is either a zero divisor or a unit
is that not true?
obsidian path
I see..
woven obsidian
It should be pretty easy to find zerodivisors here, try to factorize x^2+x
dim escarp
I mean, that's specific in my eye
golden pasture
well it occurs very often cuz youll localize at primes quite often in the future
theres also a lot of properties (local properties) that if you prove for when it is localized at primes/maximal ideals, it implies it is true
so essentially proving for local rings is sufficient for quite a lot of things
dim escarp
Oh, is this the local vs global thing
glossy yoke
what are you having trouble with?
glossy yoke
so typically you can show this by first defining a bijection, then showing that it respects addition, then showing it respects multiplication. Are any of these steps giving you trouble?
scarlet estuary
you never specify what the function f actually is
it seems that it maps $\begin{pmatrix}a&0\0&a\end{pmatrix} \mapsto a$
cloud walrusBOT
Namington:
scarlet estuary
which is indeed the correct isomorphism
but you need to actually say this, rather than just have people assume it
anyway, you havent proven that its injective or surjective in that image
leaden finch
oh okay
hmm so where do i put that the matrix maps to a?
for the one you were referring to
scarlet estuary
...when you introduce the function f?
you cant just randomly start talking about a function f without telling the reader what it actually is
is the point
generally these sorts of proofs start with "Consider the function f: S -> R defined by [blahblahblah]. I will show that this function is an isomorphism:"
and then you proceed to do the stuff you did
vital quail
so a proof structure would look something like
"Define f : S -> R by (a 0, 0 a) |-> a. This is injective [etc] and surjective [dfasfdsaf] and structure-preserving (show that f(ab) = f(a)f(b) etc) hence an isomorphism, qed"
@leaden finch in the format of your notes as in the picture you posted, you could just put "define f: S->R by (a 0, 0 a) |-> a" right below "isomorphism"
wispy glen
the case m neq 0 is easy. pl give a hint how to do the case m=0
from dummit and foote
What is your proof in the case m \neq 0?
@open torrent
let u have norm m. 1=uv+w for some v and w. N(w)<m
so w=0
when m=0, N(w)=0.
this doesn't imply w=0
oh! got it
wispy glen
Let F be a finite field. Prove that F[x] contains infinitely many primes
first I'm trying for F=Z_2.
since F[x] is pid, irreducibles and primes coincide.
I'm trying directly from definition. ie to find f(x) s.t. f(x)=g(x)h(x) implies either g or h is constant.
I'm struggling partly because polynomial multiplication is complicated
please give a hint
sharp sonnet
copy Euclid's proof
dim escarp
Wow, Euclid works here as well?
spiral wolf
\tex I need some clarification for the following notation for elements of the Dihedral groups
Let's say we have an element of $D_{30}$, written as $(12,-1)$. Is this the same as the element $r^{12}s$ given presentation $\langle r,s, |, r^{15}=s^2=(sr)^2=1 \rangle$?
cloud walrusBOT
Espio:
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reaction for details. (You may edit your message)
chilly ocean
We build rings like ((R, +, 0, i), *, 1) where i is the unary inversion and * is the multiplication and 1 in the multiplicative identity element
But this is sometimes not commutative, and how come?
(R,+,0,i) is an abelian group or sometimes called a commutative group
sharp sonnet
- is commutative, but * might be not
chilly ocean
But how is that possible? Rings are made of abelian groups
sharp sonnet
only (R, +) is an abelian group
chilly ocean
So I guess I am confused of the structure
In my mind, if we have something
Like I mentioned above
A ring consists of an abelian group, the multiplicative identity and the multiplication
Since it consists of the abelian group, it must be commutative
Since an abelian group is commutative
How can it not?
sharp sonnet
yes, but the abelian group is additive
you have an additive abelian group
think of the set of matrices
with matrix addition and matrix multiplication
matrix addition is commutative, matrix multiplication is not
chilly ocean
Multiplication in abelian groups is commutative???
sharp sonnet
a group has just one binary operation
a ring is a set R with two operations + and *
such that (R, +) is a commutative group
chilly ocean
By the way silly question but is there any reason we use the letter R? Just because for Ring or what?
Is it connected to the reals?
twilit pawn
R for ring yeah
the reals happen to be a ring tho
and also just to give a succinct answer to what you asked above: a ring is an abelian group (R,+), where you have another binary operation * on R. That second operation doesn’t have to be commutative (but it is associative, distributes over addition)
twilit pawn
yes
chilly ocean
And only addition is necessarily commutative?
oh my god
Thanks
That was my confusion
twilit pawn
you got it
twilit pawn
R = space of 2 x 2 matrices with matrix addition and matrix multiplication is a nice example of a non-commutative ring btw
chilly ocean
Thanks now it makes so much more sense now
twilit pawn
algebra pro tip: when a binary operation is commutative, we usually use “+” to denote it. if it’s not necessarily commutative, we use “*”
just a convention
chilly ocean
Okay cool
spiral wolf
a lot of time people use K for rings
i usually use K to avoid confusion with the real numbers R
opal osprey
Where can I find a paper summarizing each one of the fundamental steps and corollaries used to give a proof of the classification theorem of finite simple groups?
I intend to give it a read, the page on the subject in Dummit and Foote's book made me interested in its proof. I know it's really huge and difficult. But slowly and with some pace I may be able to read it all.
oblique river
I don't think such a thing exists right now
opal osprey
It will prolly take a year or so I'm pretty sure.
oblique river
the original proof is over 10,000 pages long
opal osprey
Oh
oblique river
combined over hundreds of separate papers
I believe there is a group right now working to condense that
opal osprey
That's good news to hear.
chilly ocean
Is Dummit and Foote's book considered undergraduate material?
oblique river
because within all of those pages there is some redundancy, and some arguments can eb simplified
but I don't know what their progress is like
opal osprey
I guess Feit-Thompson's proof may be readable.
opal osprey
the outline on wikipedia is probably the best you'll get haha
Yeah, I've checked this article right after finishing the book's chapter on the subject.
Ncat prolly has an article on the subject
Lemme see
Yup
chilly ocean
Is Dummit and Foote's book considered undergraduate material?
yes
opal osprey
I'll also read this one when I get some time
But hey, what about Feit-Thompson's theorem?
oblique river
also long and also complicated
opal osprey
Y'all know where I can find a pdf for the article.
That one has 255 pages according to Dummit and Foote's book.
opal osprey
also long and also complicated
Yeah, as I said. I'll give myself some time to read it.
It's just that I want to do something to consolidate some of my knowledge on the subject.
oblique river
I'm not sure if that's really a good use of your time
the proof is highly technical and I'm not sure if it would make you a better group theorist
opal osprey
I see, thanks for the clarification anyway.
chilly ocean
wiki says the proof uses some heavy rep theory
and idk how accessible that is right after doing GT from d&f
i'd guess not very
oblique river
yeah, I would recommend learning some finite group representation theory first
or just reading about p-groups as the classification of p-groups is more accessible (although still really technical) but a little more concrete
opal osprey
I was intending to read the proof right after finishing the book actually.
opal osprey
I know...
oblique river
I don't think that whatever rep theory d&f does would really be enough anyway
does d&f do any at all?
chilly ocean
d&f has some i think
oblique river
in any case even after reading d&f it's not going to be enough to take you to the level of feit-thompson
opal osprey
I see it
But yeah, it's a really huge book
But I'll try to maintain you know
I actually have tried to read it twice and didn't finish it all
oblique river
you don't need to finish it all to get something out of it
opal osprey
I've always stopped reading after ring theory
oblique river
honestly I feel like you'd be better to just pick up a book on rep theory
instead of trying to read the rep theroy section here
similarly with geometry
uncut girder
^
chilly ocean
ty i will check it out
i say as my alg class hasn't even finished the GT part of d&f 
opal osprey
It's just that I don't want to be an algebrist at all I guess. And since D&F talks about a lot of subjects in a somewhat concise way, I think it's perfectly suitable for me.
At least for now
It will do the job
chilly ocean
lmao don't want to be an algebraist and read entire d&f?
oblique river
"somewhat concise" = "over 1000 pages"
opal osprey
Algebraist, damn. Didn't have any idea how to write that in English, thanks.
opal osprey
"somewhat concise" = "over 1000 pages"
I see each of the chapters as small presentations of the given topic.
I'm pretty sure he doesn't go full detail on representation theory of finite groups.
It's prolly more concise than an entire book about it.
oblique river
wait a minute also
why would you want to read the entire proof of classification???
or feit-thompson?
if you don't want to be an algebraist
lol
chilly ocean
oblique river
I think you should reevaluate your priorities
I am an algebraist and I have no desire or need to read either of those
opal osprey
Not the entire proof, just the main ideas. But I guess you are totally right.
chilly ocean
thats why hes not an algebraist
opal osprey
I thought that the proofs would have interesting insights on the subject, so that's why they would be important to read.
spiral wolf
\tex I need some clarification for the following notation for elements of the Dihedral groups
Let's say we have an element of $D_{30}$, written as $(12,-1)$. Is this the same as the element $r^{12}s$ given presentation $\langle r,s, |, r^{15}=s^2=(sr)^2=1 \rangle$?
cloud walrusBOT
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uncut girder
Group theory is more interesting when its applied to other subjects imo, like Algebraic topology and galois theory
opal osprey
But it seems that they are just really technical and don't present so many interesting new ideas from what you are describing.
So yeah
I wouldn't really need to read the proof if it really is like what you are saying.
oblique river
correct. they are extremely technical
opal osprey
I just didn't have a clue about how it was like.
And thought they'd have interesting new ideas.
oblique river
I mean, in some sense the ideas are interesting because they can be used to prove hard things
but "interesting" is relative
also -- are you a student somehwere?
or are you learning on your own?
opal osprey
I'm learning on my own
oblique river
oh, I was going to suggest asking a prof who knows you better or something for recommendations
opal osprey
Maybe next year I will be able to do that.
But currently that's not possible.
Anyway, thanks a lot for the clarification.
Also, I have already studied the theory of finite vector spaces over the reals/complex.
Would this together with group theory be enough to study representation theory of finite groups?
oblique river
yep
opal osprey
Niceee
So I'll read some rep theory right after finishing his chapter on the subject.
Thanks
oblique river
gl
next obsidian
There's a book specifically on Feit-Thompson
It doesn't require too too much background as far as I know, but it's very dense
I plan to read it eventually
opal osprey
Mind telling me the name of the book?
sturdy marsh
I plan to read it eventually
lmao nerd
latent anvil
This is the math version of having a really cool idea for a novel/startup
next obsidian
I don’t know the title off the top of my head. If you google Feit-Thompson book or something I’m sure you’ll find it
My old TA has a copy he hasn’t got around to reading from Berkeley’s library
opal osprey
I suppose you may be referring to "local analysis for the odd order theorem", right?
There's also "character theory for the odd order theory"
But it was written by a different author it seems
So not sure which one, if any, of these books you are referring to.
next obsidian
I've gone back and checked and my TA said that it's over two books, the first one being "local analysis for the odd order theorem"
If I had to guess I bet the other one is the second
opal osprey
hell, even the cover of the two books look the same lmao
i'd also bet it's this one
thanks man
next obsidian
NP
dim escarp
I heard about a (possibly troll) question, I wonder if it is true:
"Let G be a group of order factorized into powers of three prime integers. Is G cyclic if its all proper subgroups are cyclic?"
Ya it was quite a while ago when I learned abt groups, so I had hard time constructing counterexample
My intuition was indeed saying that there was a counterexample, but I forgot how to make a group
Why thonk
thorn delta
@open torrent is the counterexample you have in mind abelian?
sturdy marsh
There's a super easy nonabelian example
dim escarp
What is it.. my memory ded
sturdy marsh
it's sorta troll tho
dim escarp
Lol wait why
golden pasture
whats the smallest nonabelian group you can think of
dim escarp
Nah
golden pasture
yea
dim escarp
@sturdy marsh well that was me sucking at description
thorn delta
no no no 5^0 doesn't count imo
golden pasture
oh needs 3 prime factors
golden pasture
then nonabelian groups of order 30
dim escarp
Why does it work
golden pasture
wait maybe not oops
dim escarp
He
thorn delta
pretty sure there aren't any abelian examples
dim escarp
Ya I thought about that but hassle to check that all its proper subgroups are cyclic
I'm sure that statement holds for abelian, ye
golden pasture
im thinking maybe some sort of semidirect product
dim escarp
Viburnum said that it was trivial hm
sturdy marsh
it should be easy enough for that group
assume you had a subgroup H of that semidirect product, if the Z/2 component of every element was 0, then we're done
so assume there is an element with of the form (g,1)
ord(g) is odd
so we have (0,1) in there
uhh wait
sorry
that was bs
dim escarp
Is there like online calculator for finite groups
sturdy marsh
this group isnt hard to work with
it's the dihedral group of order 30
there's a regular triangle inside a regular 15-gon right?
if there is then we have a noncyclic subgroup
thorn delta
<r^5, s> generates a subgroup iso to S3, right?
chilly ocean
You can probably even classify all groups of order pqr with some sylow bashing
thorn delta
im waiting for "Viburnum's last counterexample"
sturdy marsh
there's a triangle inside any n-gon
so dihedral groups dont work
😦
I think there cant be an abelian counterexample
finite abelian is not cyclic implies you have two summands corresponding to the same prime
and those two summands give you a non-cyclic proper subgroup
lmao apparently it's true
there are no counterexamples
thorn delta
r i p. it was probably S3 kek
dim escarp
Lmao ffs
golden pasture
:o
prime cloak
hello I need to find 4 morphism from R to U I already have the exponential and the neutral element of U
scarlet estuary
sorry, what's U?
scarlet estuary
well, if you have the exponential, consider the periodicity of sine/cosine
(im assuming R is the group of reals under + here)
in fact periodicity isnt necessary to consider, but its a good starting point
prime cloak
ok thx
prime cloak
I didnt find 😦
dim escarp
Sry what is complex number of modulus 1 again?
carmine fossil
Numbers on the unit circle
prime cloak
Maybe i could considere rotations
scarlet estuary
thats kinda the idea, yeah
if you can find an automorphism on U (intuitively a rotation of the unit circle), then you can compose it with the exponential map
you might notice something interesting
wispy glen
consider part b. x^2+2 \in R is irreducible in R but it does not belong to the set '±p where p is a prime in Z and the polynomials f(x) that are irreducible in Q[x] and have constant term ±1'.
where's the discrepancy?
south temple
If I have the field $F = \mathbb{Q}[x]/\langle x^3 - 5\rangle$ I know that I can express every element as a linear combination of $1, x, x^2$. Could someone give me a hint as to how I would find the multiplicative inverse of $x + 1$ in $F$ expressed as a linear combination of $1, x, x^2$? I tried solving for $a, b, c$ in
$$
(a + bx + cx^2)(x + 1) = 1 + (x^3 - 5)g, ; g\in\mathbb{Q}[x]
$$
but it's not really getting me anywhere...
cloud walrusBOT
vov&sons
south temple
doing polynomial division on $\frac{x^3 - 4}{x + 1}$ I get
$$
a + bx + cx^2 = x^2 - x + 1 - \frac{5}{x+1}.
$$
I think I'm just not seeing how to convert $5/(x+1)$ to an element of $F$.
cloud walrusBOT
vov&sons
next obsidian
Are you certain that’s a unit?
Anyway, my first thought is to take a general possible inverse of the form a + bx + c
Immediately a = 1
south temple
well F is a field
next obsidian
Is it?
south temple
Yes 100%
next obsidian
I haven’t thought about if x^3 - 5 is irreducible
It probably is
Lol
Yeah it totally is
Okay
So a = 1
So we deal with
1 + x + (x + 1)(bx + cx^2)
Oh hmm
Nevermind
south temple
You're trying to solve for x^3 - 4 right?
next obsidian
For some reason I thought x^3 = 0 not 5
I don’t think we can say that a = 1 then
Well... actually maybe we can
south temple
The thing is, when you try to just solve for a,b,c you end up having to set a=-b=c
From what I've tried
next obsidian
Well the constant ends up being a + 5c
south temple
And even if we could solve for a, b, c, then my long division shouldn't have a remainder?
Right?
next obsidian
So we can say that a + 5c = 1
I don’t think that’s necessarily true I’m unsure how long division works in a quotient like this
south temple
Where are you getting a + 5c btw?
next obsidian
(1 + x)(a + bx + cx^2)
The only terms that contribute to the constant are the normal constant which is 1*a
And the coefficient of the x^3 term since x^3 = 5
And the x^3 term has coefficient c
south temple
Ah right
next obsidian
So from a +5c = 1 I arrive at a = 1 - 5c
We have two other equations for the linear and quadratic term
The linear gives us a + b = 0?
I believe
south temple
Let me try and write this out one sec
next obsidian
And I believe the quadratic gives b + c = 0
I believe these let you solve for c
From which you can solve for a and b
I think c = 1/6?
Giving us a = 1/6 and b = -1/6
This seems kind of suspicious tbh but this is what my mental math gave me lol
Actually I think this works lmao
south temple
@next obsidian wow, you're totally right. We can check this because this field is isomorphic to Q[cuberoot(5)] and the element x + 1 corresponds to 5^(1/3) + 1
next obsidian
Oh huh, lol
I just multiplied the two polynomials and saw it ended up working out lol
south temple
Taking the inverse of 5^(1/3) + 1 is exactly the value that we get by plugging in a, b, c
next obsidian
Cool
delicate bloom
I'd multiply out to get c*x^3 + (b + c)*x^2 + (a + b)*x + a and then plug in x^3=5 and then solve the system of 3 equations
delicate bloom
tl;dr
next obsidian
That’s what we did smh my head my head
south temple
Yea thats what Chmonkey suggested
delicate bloom
looked like a buncha foolery
delicate bloom
lol
wanna see a crazy solution?
$$\frac{1}{1+x} = 1-x+x^2-x^3+x^4-x^5+\cdots = (1-x+x^2)(1-5+5^2-5^3+\cdots)$$
$$ = (1-x+x^2)\frac{1}{1-(-5)} = \frac{1}{6}(1-x+x^2)$$
cloud walrusBOT
Merosity
delicate bloom
valid because we can see this being a subfield of a 5-adic field extension where this converges
and hey I got the same answer as you
next obsidian
🤬 you Mero
delicate bloom
🤣
south temple
Wow thats actually really cool
delicate bloom
p-adics having applications in the """real""" world
💪
this is why thinking of extension fields lying in C is a disability though, you can look in more places and you suddenly gain more freedom
south temple
How did you get the second equality in your image there?
south temple
I get that you're factoring 1 - x + x^2
But why is the stuff on the right 1 - 5 + ...
golden pasture
that is Z_5 stuff
wait how does it hold
oh are you like
doing something along the lines of
Z_5 = Z_{5^3}
kinda thing
could really just say (1-x+x^2)(1-x^3+x^6-...) tbh
delicate bloom
I was heating up tea what'd I miss
golden pasture
south temple
Wondering how are you factoring to get the second equality
delicate bloom
no no, x is actually a cube root of 5
delicate bloom
we're working in $\mathbb{Q}_5/(x^3-5)$ so we're in this field ext
cloud walrusBOT
Merosity
delicate bloom
it's a complete metric space
delicate bloom
so we can do infinite sums just fine, plus there's no "rearrangment issues" either for infinite sums
well wait did you get the simplification @south temple
I'll try to show you, but every power of 3 larger I simplify to 5
delicate bloom
yeah this was just for fun lol, they already had a solution @golden pasture
south temple
I'm wondering how that factors so nicely
delicate bloom
I was just flexing
golden pasture
thanks
south temple
To terms of multiples of x^3
golden pasture
could really just say (1-x+x^2)(1-x^3+x^6-...) tbh
this should be clearer
can see like
you group the terms
into chunks of 2
delicate bloom
basically like,
golden pasture
1-x+x^2
-x^3+x^4-x^5
x^6-x^7+x^8
etc.
delicate bloom
$(1-x+x^2)-x^3(1-x+x^2)+x^6(1-x+x^2) -x^9(1-x+x^2)+\cdots$
golden pasture
essentially yes
cloud walrusBOT
Merosity
south temple
Damn that's clever af
delicate bloom
now go learn Hensel's lemma and start your journey to learning p-adics
golden pasture
study completions
delicate bloom
all you need to know is that completions are taking all limits and chopping them up into equivalence classes by which ones vary by a limit that goes to 0, this forms a maximal ideal in the ring of limits and so when you mod out by it you get your comfy field
next obsidian
The next section of Matsumura is about completions
delicate bloom
is that an anime?
next obsidian
I ought to finish the exercises in my section, I think I’m mostly done anyway
I just need to type it up
...
R u srs?
It’s a commutative algebra book
delicate bloom
I didn't think they made others than atia McDonalds
next obsidian
This one is actually good
delicate bloom
good to hear, I might check it out eventually since my commutative algebra is nonexistant/weak af
chilly ocean
What is the complete classification of all countably infinite Abelian groups upto isomorphism? How complicated is the classification?
viscid pewter
not sure it exists
glossy yoke
You might be interested in this: https://simpletonsymposium.wordpress.com/2013/05/31/countably-generated-abelian-groups/
It features a gadget called the Ulm Invariant.
chilly ocean
@glossy yoke So as far as I can tell, the post doesn’t address the case when a group has some torsion elements and some torsion free elements, right?
viscid pewter
afaict they just take the torsion subgroup and the quotient of that is torsion-free, and go from there
golden pasture
<@155869697974206464> So as far as I can tell, the post doesn’t address the case...
ye it seems like it so we essentially have a extension of group with no torsion with a group with torsion, which unfortunately is also rather hard to handle as some extensions could have no torsion and others with torsion, i dont quite forsee any nice results here other than like "these are all possible groups, some of these have the correct torsion subgroup"
tho maybe with some fiddling around with the extension construction you could maybe get something
im thinking something possible would be like
this subgroup/subset? of Ext^1_Z(T(A),A/T(A)) that has torsion group T(A) have some xxx property
idek if it would form a subfroup lol
steep hull
So really we’re just taking “norms”: 1/(x+1)=(x^2-x+1)/(x^3+1)=(x^2-x+1)/6
delicate bloom
shhh delet this
chilly ocean
Under what circumstances is $\Pi_{i\in\mathbb{N}}\mathbb{Z}_{p^{n_i}}$ a $p$-group?
cloud walrusBOT
lugita15
green locust
when it has order p
latent anvil
Does p group mean any element has order dividing p? Do we exclude groups which have infinite order elements from our definition?
Sorry, I don't know what p group means for groups of infinite order
bronze trench
According to rotman, therefore my prof of my gt class, it's a group whose elements all have order a power of p
drowsy oasis
Are uncountable intersections of rings closed even though within a ring only finite unions and intersections are closed?
latent anvil
Lol sorry that's what I meant benny
I always try to talk math right after getting up and I always say the wrong thing
latent anvil
Are uncountable intersections of rings closed even though within a ring only fin...
By ring here do you mean a collection of subsets of X closed under complements and finite unions?
drowsy oasis
By ring here do you mean a collection of subsets of X closed under complements a...
my text uses closed under symmetric difference and intersection, but then it shows that it implies closed under union and set difference
latent anvil
Sure, that should be the same. I was trying to distinguish from the algebraic meaning of the word ring
drowsy oasis
oh sorry, i didn't know there was another meaning
latent anvil
Np
So no, a ring won't be closed under arbitrary intersections
Or even countable ones
latent anvil
oh is that still a ring?
I think it should be, yeah
Take A, B in the intersection
Then for ring, A, B are in the ring
So their symmetric difference and intersection are in the ring
so they're in the intersection
drowsy oasis
ye each of the intersection will be ok for the sym. diff and intersection
i found this result a bit surprising since individual rings are only closed for union and intersection for finite number
@latent anvil if S is a semiring is it true that if A is in S then A may not contain a finite expansion?
apparently this is so, but by the definition of a semiring couldnt we just take A = union A a single union?
latent anvil
Can you explain what a semiring is lol
Because I'm guessing it's not what I think of as a semiring
drowsy oasis
yes sure
latent anvil
also what's a finite expansion?
drowsy oasis
https://en.wikipedia.org/wiki/Semiring see semiring on sets section, although my text is a bit diff
the finite expansion is that finite union of pairwise disjoint sets
i guess this pages definition 3 is the same, just take F = empty set and then it is saying E = that finite expansion
but then i am saying cant you just take E = union E
latent anvil
Sorry, gtg. Ill help later if you're still confused
drowsy oasis
sure, thank you
this theorem says if S is a semiring and R(S) is the minimal ring containing S then it coincides with the elements of S having finite expansion.
Since not every semiring is a ring i am confused since it seems every element of S has a finite expansion
light tusk
do you mean "finitely generated"?
drowsy oasis
do you mean "finitely generated"?
I don't think so since that subset of S may be infinite
light tusk
what does expansion mean?
drowsy oasis
it means that if A is in S then A = union C_i where C_i are in S and pairwise disjoint
and union is finite
light tusk
If an element of R(S) didn't have a finite expansion, you could probably exclude it and still have a structure that contains the semiring S. Does that get at what you're asking? I think the assumption of minimality would allow you to exclude the case of some infinite sum/product that you couldn't get around.
drowsy oasis
One thing confusing me is how can any A not have a finite expansion by the semiring definition by taking A = union A
light tusk
are these elements, subsets, or what? what is A?
drowsy oasis
A is an element of the semiring
ok i think i got it now
the definition is saying if there is any subset of A also in S then you must have a finite expansion
and i think the theorem is saying take all the elements of A with a non trivial finite expansion then those generate R(S)
but that seems like an impotant thing to leave out, maybe i don't get it. i gtg , thanks for your help
dim escarp
Does that belong here (old reply)
valid scaffold
oh sorry i think i should ask in the linear algrebra channel
vital quail
@latent anvil @bleak abyss here's my writeup
vital quail
i was looking at the paper trail
latent anvil
This is also how I did it
vital quail
and i found that it apparently originated with erdos and turan
or something
oh cool
lol i thought my proof was questionable because i couldnt find it online
until i realized baez's long ass post was it
lol
latent anvil
oh yeah and one of my friends did further stuff on this
like with the weird interval bounds
latent anvil
I don't think it was original work
He was just giving a talk at a student seminar
vital quail
oh i see
apparently k.s. joseph showed in his 1969 ucla thesis (that was apparently not published or something, i only found like one link and it was behind a paywall i think) Commutativity in Non-Abelian Groups that the probability is at most $\frac{p^2 + p - 1}{p^3}$ where $p$ is the minimal prime factor of $|G|$
cloud walrusBOT
アークシス｜|Aut(ℤₚⁿ)| = 𝚷(pⁿ - pⁱ)
vital quail
and taking p=2 gives back 5/8
latent anvil
yeah, this is in the pdf I just sent
vital quail
oh i see
latent anvil
That bit isn't very hard
latent anvil
I mean
latent anvil
oh okay
vital quail
with C(g) taken to be |G|/p
vital quail
lmao frenchposting
latent anvil
lol
chilly ocean
vital quail
o thanks
chilly ocean
archsys putting together more interesting looking exercises than my algebra prof damn
vital quail
lol thanks
i just do this from time to time
if i find a problem i think is cool
i make a proof sketch
this was my first one
(lmao)
dim escarp
?
vital quail
@latent anvil wow this paper is really cool
imagine if like, i could forget the contents of this paper, and then this paper hadn't been written, and then i later went and wrote this paper in undergrad
that would be poggers
latent anvil
Yeah!
vital quail
lmao
vital quail
oh nice
@latent anvil https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.437.5938&rep=rep1&type=pdf also look at this stuff lmao
i am not sure what scares me more, the formatting or the way they're doing group theory
latent anvil
This paper looks like it fucks
latent anvil
Dummit and foote? I mean that's all I know of it
vital quail
oh I see lol
latent anvil
I should learn more rep theory
vital quail
Jacobson doesn't really do rep theory afaik
oblique river
@vital quail I think fulton and harris is a common rep theory book which does character theory pretty early iirc
vital quail
oh alright I'll check that out, thanks
also what are your thoughts on serre? that's the one I know about
chilly ocean
If $k_1,k_2\in\mathbb{N}$, $k_1\geq k_2$ and $X\subset\mathbb{Z}$, then does $X$ being linearly independent over $\mathbb{Z}{k_1}}$ imply $X$ being linearly independent over $\mathbb{Z}{k_2}}$, or vice versa?
cloud walrusBOT
lugita15
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
sturdy marsh
Z is not a module over Z/nZ
chilly ocean
@sturdy marsh Hmm, it’s not? Is $\mathbb{Z}{lcm(k_1,k_2)}$ both a module over $\mathbb{Z}{k_1}$ and a module over $\mathbb{Z}_{k_2}$?
cloud walrusBOT
lugita15
chilly ocean
@sturdy marsh OK, so then if $k_1,k_2\in\mathbb{N}$, $k_1\geq k_2$ and $X\subset\mathbb{Z}{lcm(k_1,k_2)}}$, then does $X$ being linearly independent over $\mathbb{Z}{k_1}$ imply $X$ being linearly independent over $\mathbb{Z}_{k_2}$, or vice versa?
cloud walrusBOT
lugita15
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
sturdy marsh
these two notions may not have anything to do with each other I think
you probably want some condition involving something dividing something else
not something greater than something else
chilly ocean
@sturdy marsh OK yeah, suppose further that $k_2$ divides $k_1$.
cloud walrusBOT
lugita15
sturdy marsh
it is
oh wait i probably got it the wrong way around lol
if n divides m, then Z/nZ is a Z/mZ module
so no, Z/lcm is not a Z/k1 module
sorry
Yeah if M is an R-module, then the smallest quotient of R that the module structure descends to is the annihilator
anything you quotient out by needs to be in the annihilator
hot grotto
quick question for yall what is meant when it is said that a homomorphism is given by reduction modulo N?
thorn delta
id probably just interpret that as the canonical map G to G/N
hot grotto
ah ok
makes more sense now I've never seen the terminology before and its actually kind of hard to google
yeah looking at my book thats exactly what it is
thx
thorn delta
npnp
gleaming quarry
Heyo, can someone check If I got this down right?
https://cdn.discordapp.com/attachments/486844875632017408/782814697929310278/unknown.png
Given a polynomial
it would only be represented in this marked down area F_5[X] here?
As in
F_5[X] -> F_5[X], x --> some polynomial
and for the polynomial the coefficients must be in the field but the exponents can be any (natural) number?
In this context F_5[X] is not being divided by a irreducible polynomial
And for the case there are irreducible polynomial, I can use any such with a max degree of 5?
Still learning the foundations
gleaming quarry
It's just an empty coordinate system with the limits of F5
But thanks for the confirmation
thorn delta
C(f) denotes the content of f in D[x] and a \approx b means a is an associate of b. I am stuck on the underlined part. Why does p not divide C(f_1)?
dim escarp
Eh.. perhaps since it is a unit? Idk
latent anvil
if u is a unit and d | u then u = xd for some x, so (u^-1x)d = 1, which means d is a unit. But (I think lol) irreducible elements are by definition not units
dim escarp
Does it really saying that C(f1) is a unit?
That part is confusing to me
Then Ofc prime should not divide unit
thorn delta
if u is a unit and d | u then u = xd for some x, so (u^-1x)d = 1, which means d is a unit. But (I think lol) irreducible elements are by definition not units
ah yea okay, thanks
latent anvil
p isn't a prime here cascadar
Irreducibles still don't divide units, but I wanted to make the distinction
thorn delta
okay, I am pretty sure R is a typo for "D" which is a UFD. I'm not sure how p being prime makes it any more clear that p doesn't divide a unit.
south temple
Could some give me a hint as to how I would approach part (b)?
Since it's asking to use the FIT I'm thinking that $F[x]/\langle g \rangle$ is isomorphic to $\phi(F[x]) \subseteq E$?
cloud walrusBOT
vov&sons
south temple
Now because $\phi$ is a homomorphism we know that $\phi(F[x])$ is a subring. But how can I show that every element of $\phi(F[x])$ is a unit?
cloud walrusBOT
vov&sons
vital quail
@south temple you can show that F[x]/<g> is a field (due to g being irreducible) and then you can show that the preimage of an ideal is an ideal, so the codomain of a homomorphism from a field is also a field
ping me back if you need more detailed help with any of these parts
south temple
sorry, how do we go from preimage of an ideal is an ideal to codomain of a hom. of a field is a field?
@vital quail
south temple
hmm
vital quail
if you have a nontrivial 'a' in an ideal
south temple
uh well ker(phi) is an ideal?
vital quail
yes, since ideals are closed under multiplication by elements from the entire ring
so you just do *a^-1
anyways
lets say I is an ideal in your codomain
then the preimage of I is an ideal in a field
south temple
yeah I've proven this in a previous assignment
south temple
ok I'm following so far
vital quail
so now do you have any ideas for the other part
south temple
so the ker(f) is either 0 or F
vital quail
i meant for the other part as in showing F[x]/<g> is irreducible but uhh for this one
i mean yeah
south temple
oh I've shown that F[x]/<g> is a field by showing that g must be irreducible
vital quail
you can just show that (assume R nontrivial) any nontrivial ideal I in R is just R
south temple
I'm just not sure how to apply FIT
vital quail
okay cool
south temple
or rather I know how to apply FIT
but don't know how to confirm that phi(F[x]) is a field
so far we just have that phi(F[x]) is a ring
a subring of E
vital quail
so you have an isomorphism phi between F[x]/<g> and phi(F[x])
south temple
right
south temple
isomorphism always takes a field to a field?
vital quail
it was just a matter of showing that F[x]/<g> is a field and then that
south temple
oh I see, I have to show that in general isomorphisms map fields to fields
vital quail
well yes it's just due to our previous work
if you have f: F -> R an isomorphism with F a field and R some ring
and I a nontrivial ideal of R (assume R nontrivial) then the preimage of I is just F
of course, f is a surjective map since it's an isomorphism after all
so I is just R
so that proves R is a field
the only ideals are the trivial one and R itself
@south temple does this all make sense
south temple
I'm trying to understand it
vital quail
the preimage of an ideal is an ideal