translations rather ?
#groups-rings-fields
hot lake
hot lake
oh no
spiral wolf
Lol
upbeat juniper
but it was about the action of G on 3-tuples, where each entry is from A
hot lake
is that guy that wanted to do music category theory still here
I don't remember their name
upbeat juniper
what interesting things can you do with that
upbeat juniper
lol
golden pasture
compositions that are too structural sounds like a homework for a class
upbeat juniper
reminds me of 12-tone harmony
chilly ocean
@carmine fossil I don't quite understand that short argument, but there's a group morphism N_G(P) -> Aut(P) by the conjugation action with the kernel clearly C_G(P), so you have |N_G(P)/C_G(P)| dividing |Aut(P)| = p^r (p - 1) (where you use cyclicity of P) for that smallest prime p dividing |G|, and P <= C_G(P) <= N_G(P) so |N_G(P)/C_G(P)| also doesn't have prime factor p and divides |G|. |N_G(P)/C_G(P)| divides p-1 while dividing |G|, so the smallest prime p here matters (so that the index = 1).
carmine fossil
Let's say a is the generator of P,then the group gPg^-1 is uniquely determined by gag^-1,so the number of groups in the orbit of conjugacy action of P is simply the number of elements in the orbit of conjugacy action of a,i.e.,
G/C_{G}(a) . C_ {G}(a) comes out to be the same as C_{G}(P)
@chilly ocean Is there a mistake in this argument?
chilly ocean
Ok I see. It's the same argument as the one that I just gave. Just reorganized in a somewhat different way.
carmine fossil
Have I used that p is smallest prime implicitly?
hot lake
are you assuming that two different gag^-1 will give two different groups ?
carmine fossil
if g1 (g2)^-1 is not in C_(G)(a) yes
hot lake
carmine fossil
I am just counting the number of elements in orbit under conjugation
chilly ocean
I think you should still restrict your conjugation action to N_G(P) instead of to all of G so you can compare the index N_G(P) : C_G(P) to the size of Aut P (which is easily computed using the fact that P is cyclic), then using p being the smallest prime factor of |G|
hot lake
yes but you are also counting the number of groups of the form gPg-1
if g1ag1-1 = g2ag2-1 then yes, the groups generated by them are the same
chilly ocean
but I gotta go out now. Have fun guys
carmine fossil
So,something like "a gets mapped to something different, but the two groups come out to be same"?
hot lake
but I don't see why you shrug off having to prove that if the groups generated by g1ag1-1 and g2ag2-1 are the same then g1ag1-1 = g2ag2-1
carmine fossil
My bad
hot lake
basically you have to show that if gag-1 is in P then g is the identity gag-1 = a
I think
and that's where p being the smallest prime of |G| comes in
carmine fossil
Isn't it more like if gag^-1 is in P,gag^-1=a?
hot lake
uh yes
carmine fossil
Thanks
Consider the quotient group HN/N,since H is a subgroup of G, HN/N is a subgroup of G/N implying |H|/(|H inter N|) divides |G|/|N|,which implies a factor of |H| divides |G|/|N|,but since |H| and index are coprime,we get that factor to be 1, i.e,H belongs to N
hot lake
where did |H| / |H inter N| come from
carmine fossil
|HN|=|H||N|/|H inter N|
hot lake
oh right N is normal in G so good stuff happen
I was thinking HN could be stupidly large
broken crag
would anybody care to discuss the possible number of sylow subgroups of any group of order 96?
@river nebula this is for you
river nebula
thanks
broken crag
oh you can type here now
olive mirage
I will do the hard part: 96 = 2^5 * 3
river nebula
I've figured most of the part
olive mirage
96 is the smallest order where some basic conjecture breaks... like there are two nonabelian groups with the same conjugacy class sizes that arent' isomorphic or something?
river nebula
I was actually mislead by someone who had told me that they intersect trivially and thus subgroup of order 3 is unique so normal
olive mirage
yeah that 32 is going to make for an annoying flower petal diagram
river nebula
yeah, i wanted to know how to approach such problems in general
one possible way is to use the result that |HK|=|H||K|/(H int K)
olive mirage
it depends a lot on what your goal is. Typing SmallGroups(96) into gap is a great start.
I don't think you get around that doing something with this big a power of two is going to be very tedious by hand
but there are some of these where some sort of neat trick pops up
river nebula
yeah, so just for the sake of clarity, this isn't too basic right?
olive mirage
Too basic for what? I don't think it is doable in half a page, if that is what you want to know
river nebula
Yeah, I mean I just wanted to clear the basic concepts of all the undergrad courses in order to prepare for interviews of doctoral programmes, and not dive too deep in them.
olive mirage
I would not ask grad algebra students to do this
of course, ymmv
but usually no one is going to ask anything too intense in a doctoral program interview
you should be able to do, e.g., order 24
river nebula
Yeah
olive mirage
and count the number of sylow subgroups of all your favorite groups of order 24
and understand the actions etc.
latent anvil
@olive mirage want to hear a neat group theory problem I did with a friend a while back?
olive mirage
Sure!
latent anvil
(up to isomorphism) how many semidirect products of An and C2 are there for each n?
I did use GAP for parts of this one
olive mirage
This is just asking how many elements of order 2 there are in S_n right? So the nth telephone number?
latent anvil
No, I don't think so?
olive mirage
except when n=6 😛
latent anvil
Are you thinking of taking the internal semidirect product with the subgroup generated by a transposition in Sn? Because those are all isomorphic to Sn
olive mirage
aren't we counting maps from C2 to Aut(An) = Sn (most of the time, offer not available in all areas)
latent anvil
And yeah this is ""secretly"" about n = 6, the other cases are boring
We're not counting the number of maps, because two different maps can give isomorphic semidirect products
olive mirage
oh I see, you want "up to isomorphism"
latent anvil
Yup
olive mirage
and so I'm going to guess that there is the direct product, and probably all of those maps I described give you Sn?
though it would be equally plausible for it to be 50-50
latent anvil
Yeah, because you can view them as an internal semidirect product
olive mirage
but then S6 is annoying
oblique river
hi i'm jumping in because this is interesting
latent anvil
Since the map Sn -> Aut(An) is an iso for n >= 3, n ≠ 6, all maps are just conjugation internal to Sn
So you found all of them
yaaay!
But then there's n = 6
olive mirage
the external automorphism group of S6 is C2
oblique river
well, more just viewing from the sidelines for now
olive mirage
but I'm unusure if that means there is just one more
latent anvil
Oh wait umm
There is actually one more case
So if you take an order 2 element in Sn\An and act on An, you get Sn back
But the same logic doesn't apply if you take an element of order 2 in An and let that act by conjugation
It's no longer an internal semidirect product
olive mirage
ahh yeah,
latent anvil
This is to finish n ≠ 6 but it's also relevant for the n = 6 case
olive mirage
I seem to recall there being some doublecover of An that was denoted by 2An or something
latent anvil
Group props says this is a thing but it has normal subgroup C2 and quotient An
Instead of normal subgroup An and quotient C2, which the semidirect product should have
latent anvil
I had never heard of it ¯\_(ツ)_/¯
olive mirage
that's right, Sn is not a double cover of An
latent anvil
Is it somehow a topological double cover? Like if we build K(An, 1) or something?
err
that goes the wrong way doesn't it, covers of K(An, 1) have to have fundamental group a subgroup of An
olive mirage
anyway, what happens with A6?
latent anvil
It turns out that there is a non-inner automorphism of A6 of order 2 and that all such automorphisms are conjugate in Aut(A6), so they give isomorphic semidirect products
the set of fixed elements by the automorphism (the one I looked at) is this weird subgroup of order 10 in S5 <= S6 and you can use this to argue that you don't get S6 or A6 x C2
like it's this copy of D10, you can determine that it's index 2 in its normalizer in S6 pretty easily and determine that that normalizer has no elements of order 2 outside the copy of D10
but in the semidirect product you have this new element in the normalizer, the order 2 automorphism
For A6 x C2 you can say that if there were a central order 2 element its product with the automorphism you've added in would be in the centralizer of that subgroup and in An, and then from that you can argue that the product is trivial and so the automorphism is central in the semidirect product, which isn't (we started by saying its set of fixed points has order 10)
anyways it's a neat little fact to accompany Out(Sn) = 1 for n != 6, Out(S6) = 2
you get exactly one more semidirect product
olive mirage
that's ver ycool!
latent anvil
getting the data bout conjugacy classes in Aut(A6)/the specific automorphism/its set of fixed points was left to GAP 😛
sturdy marsh
What do you mean by Sn is not a double cover of An?
What is a covering map of groups?
The map induced on the geometric realizations of the nerves?
olive mirage
just a two to one homomorphism
sturdy marsh
What's the map?
olive mirage
to say a group is a double cover of another group just means there exists such a thing.
sturdy marsh
oh okay
balmy dagger
If I have a question about abstract algebra do I ask it here or in the questions channels ?
olive mirage
Here is good
balmy dagger
I’m struggling with proving this here.
K is a finite field.
sturdy marsh
x to x+1 is a bijection from K to itself
if you make the sub substitution t = x +1, the second sum can be rewritten as $\sum_{t-1 \in K} t$
but by summing over t-1 you are summing over precisely each element of K
as t to t-1 is a bijection from K to itself
cloud walrusBOT
Brofibration:
balmy dagger
Bare with me a little bit as I’ve only learned what fields are today, I know what bijection means for a function but I can’t link the dots with fields 😅
sturdy marsh
K is a set
you are summing over elements of the set
the bijection is just changing the "order" in which you're summing the terms
Foe example, if we're working with F3, then the first sum "is" 0 + 1 + 2
the second sum is 1 + 2 + 0
balmy dagger
ooooooh it just clicked I think
Because we are summing elements all the +1 does here is shift where we start summing from not the overall value.
Is that correct ? 😅
sturdy marsh
yeah pretty much
in general if you're summing over a finite set you can sum over it in any order you want
and these bijections from the set to itself are doing the reordering
the reason the value doesnt change is you're summing over the same set
balmy dagger
Aha and here since we are summing all the elements of a finite field the answer will be 0 per the definition correct ?
sturdy marsh
no
take F2
the sum is 1
the sum over any finite abelian group will end up being the sum of all the order 2 elements
balmy dagger
Oh okay interesting.
Well thanks a lot.
I hope with more practice I can get used to thinking about these things.
chilly ocean
Does anyone have some literature about metacyclical groups?
i.e. G is metacyclical if there is some normal subgroup N of G so that N and G/N are cyclic
I've searched the internet but it seems there's nothing concrete
glossy yoke
Hempel classifies metacyclic groups here: https://www.tandfonline.com/doi/abs/10.1080/00927870008827063
That's C.E. Hempel. I wonder if he is related to John Hempel. That guy wrote a book on 3-manifolds.
@chilly ocean
chilly ocean
Much appreciated! I'll look into it
viscid pewter
how many
cinder bone
anyone know anything about ultraproducts and Łoś's theorem?
wind steeple
that's more a #foundations topic imo
you can ask you're question here if you're unsure
chilly ocean
quick (i hope) question: If something is Z mod n, isnt it always a group?
Because the identity is 0,
any 2 elements (* or +) always get another element inside the mod n
and the inverse always is whatever gets you to mod n
viscid pewter
sometimes
golden pasture
specially primes
viscid pewter
(z6, x) is not a group: 2x3
chilly ocean
How does * work? idk if ive seen it tbh
chilly ocean
So like if it's (z6, *) how does that work? can it be cyclical?
viscid pewter
okay so it's like 2x3 = 4x3 mod 6
so there aren't inverses
it's not a group
i think
golden pasture
2*3 literally 0
chilly ocean
yeah 0
golden pasture
we call these elements zero divisors or nonunits(they are generally not the same)
viscid pewter
multiplicative groups like that are only groups if n is a prime
golden pasture
the units do form the unit group tho
viscid pewter
i'm bad at maths
golden pasture
so the numbers coprime to n
viscid pewter
i do not know these fancy words
golden pasture
i.e. if n=9, {1,2,4,5,7,8} is a group with *mod9
viscid pewter
yeah so
and yeah if n is prime then everything is coprime to it so all the numbers work
right
chilly ocean
How does * work? With + you start from 0+whatever number youre doing
So (z6, +)
2: (0,2,4)
3: (0,3)
4: (0,2)
5: (0,5,4,3,2,1)
But if it's *, 0 x anything is always 0
golden pasture
say we take mod 5, our group elements are {1,2,3,4}
say mod 12, group elements are {1,5,7,11}
chilly ocean
isnt it 0,1,2,3,4?
chilly ocean
ok ok so in * there is no 0
golden pasture
yes
chilly ocean
ok makes sense
chilly ocean
now let me scroll up to see what you guys were talking about with inverses.
So let me see, is this the right way to find inverse?
a*a^(-1) = e
So like it would be: a^(-1) = (e/a)?
nova plank
What is /?
chilly ocean
divided by
nova plank
How is division defined?
chilly ocean
😢
nova plank
Hint: It's not
golden pasture
you can define division
as long a is coprime
usually via extended euclidean algorithm
chilly ocean
So like for the example cat gave, z5,*
How would you find the inverse? because that's what can determine if its a group or not right?
golden pasture
if you want to be fancy x^(φ(5)-1) is a inverse
viscid pewter
a group doesn't have a single inverse (unless it's trivial), each element has an inverse
golden pasture
ok so prove that
chilly ocean
yeah i got that
chilly ocean
And for it to be a group each element has to have an inverse?
viscid pewter
yes
chilly ocean
ok ok starting to click
golden pasture
the numbers coprime to n forms a group under multiplication mod n
viscid pewter
2x3 = 1 mod 5
golden pasture
elementary reason is extended gcd
viscid pewter
so 2 is the inverse of 3, 3 is the inverse of 2
4x4 = 1 mod 5, so 4 is its own inverse
golden pasture
fancier reason is you can define inverse with fermat little theorem
viscid pewter
and 1 is the identity and the identity is always self-inverse
chilly ocean
ok ok got it thanks
glossy yoke
this is hurting my feelings. Z/nZ with multiplication is never a group because 0 has no inverse. What is a group is the group of integers mod n coprime to n.
viscid pewter
yeah sorry
glossy yoke
oh, well, if n = 1, then it's a group.
chilly ocean
ooooh so like someone said earlier, if its prime
viscid pewter
(z/nz - 0, *)?
chilly ocean
if n = prime
vital quail
lol
viscid pewter
if it's prime, everything apart from 0 works
vital quail
you can just specify U(Z_n) if you want
vital quail
viscid pewter
i'm bad at maths
vital quail
Z_(n) ari
golden pasture
mm
chilly ocean
do you not write it as Z_(n)?
vital quail
anyways the superior notation is (1)/(n)
golden pasture
the ring Z localized at the ideal (n)
vital quail
@chilly ocean different object
but you could see Z_n for Z/nZ
don't write Z_(n) though
chilly ocean
that's like dihedral groups right?
vital quail
that's a different thing
a dihedral group is one made of a Z/nZ and also reflections
golden pasture
honestly im late night prob drunk posting now
chilly ocean
@golden pasture lmao
viscid pewter
rsrs
vital quail
so the r^n is your Z/nZ
golden pasture
dihedral is <a,b|a^2,b^n,ab=b^-1a>
chilly ocean
ohhhh D_n
viscid pewter
u wot
chilly ocean
yeah yeah my bad
viscid pewter
C_n? why have you done this
vital quail
why not
golden pasture
disagree
wait no
agree
so i have a excuse to use chinese in every half sentence
vital quail
yeah
golden pasture
and be as dumb as possible
vital quail
hanzi for math
chilly ocean
well we probably will have to anyway
glossy yoke
different words have different connotations. even in math. it's important.
viscid pewter
anyway the mathematic
vital quail
viscid pewter
mwbmd are you happy
vital quail
@golden pasture
chilly ocean
Ok well i got my misunderstandings, understood. thnx guys, ily ❤️
glossy yoke
examine, for example, the "concept with attitude" entry on n-lab.
viscid pewter
cool
vital quail
but it's just hanzi anyways
golden pasture
literally fake
vital quail
nah it's good ari
imagine using greek stuff
anyways people run out
there's only like at most 50 greek things
viscid pewter
the real flex is using morse
golden pasture
chinese prob have sufficient characters
viscid pewter
oh my gods yes
golden pasture
imagine if scholze knew chinese
vital quail
lol
golden pasture
水-module
vital quail
running out of notation? just godel number
golden pasture
i shall now call jigsaw puzzles 凸 凹 = 口
viscid pewter
anyway i have a thing to say
consider the monoid of all groups with the operation as the direct product
no question btw
it's just really cool
chilly ocean
I guess Chinese characters as math symbols seems weird to me because it is not an alphabet
golden pasture
weird for non chinese cuz isnt alphabet
weird for chinese cuz it has a actual meaning
astral galleon
Let $G={ 2^{\psi} }$ and $K= {2^{\psi } }$ where $\psi \in \mathbb{Z} / {0,-1,1 } $ let $f: K \rightarrow G /K$ where we have that $f(\psi) = 2^{\psi}$ $\forall \psi \in K$ does $f$ from a homeomorphism ?
Note that $G/K$ is the quotient group
chilly ocean
what even is a conjugate?
astral galleon
@chilly ocean did I leave something out in my question
chilly ocean
Oh sorry that was just random. I didnt know you guys were in the middle of something
astral galleon
nahhh I think whatever dicussion was finished I decided to ask my question will you look at it
chilly ocean
im the worst person here lmao.
Hi, I don't see how to prove that, could someone tell me please ?
latent anvil
I think we need more context? Is this for elements of like a matrix group?
That equality can't be strictly true, you'll need some kind of exponential map in a general lie group
chilly ocean
yes for a matrix group
the point is that we assume g to be possibly written as exp(iX) with X the generator, so the equality is for the first order approximation ?
latent anvil
The issue is that "g_ε = … up to O(ε^2)" doesn't make sense unless g_ε is varying somehow, or we know something about it being close to the identity
Which is why I asked if it's a curve
astral galleon
Not to interrupt anyone but can someone help me with my question
astral galleon
@viscid pewter what was unclear about what I posted
viscid pewter
i don't understand the mapping
you've reused, what is it
psi
is it just f(x) = 2^x?
astral galleon
pretty much but I excluded 0,-1, and 1
viscid pewter
ohhhhh
right so
those aren't groups right
where's the identity
without 2^0
i'm bad at maths btw
astral galleon
shoot good point 😦
viscid pewter
ok phew this isn't secretly mega-advanced and completely logical
latent anvil
Zophike, forward slash means something very different than backlash
viscid pewter
it might be a special character, might need to escape it by doing \/
latent anvil
Yeah I'm saying Z/{-1,0,1} is not the integers except 0,-1,1
viscid pewter
cloud walrusBOT
shamrock:
chilly ocean
Lads, is it true that a product HK of a solvable, normal subgroup H and a nilpotent, normal subgroup K is nilpotent in G?
I figured you'd look at the chains of H and K and consider another chain {H_iK_i} and that this should be terminating in at most m+n steps where m is the length of the chain for H and n for the chain of K. But I can't seem to get this to work.
latent anvil
Taking G = H and K = 1 this says solvable => nilpotent, right?
@chilly ocean
and that's not true, eg S3 is solvable but not nilpotent
chilly ocean
Should've said that K is nontrivial
It seems like the proof depends on the structure of K which if nilpotent would give you that K is isomorphic to a direct product of p-subgroups
And if G is finite you'd have that abelian/cyclic quotients for the chain of H
There's a way to combine all of this but eh
marsh fractal
Hey guys, so when we have an embedding map, why can we say that it is then contained within that field?
is it just because the structure remains the same
but the objects of the set is not exactly identical
oblique river
@marsh fractal whenever you have an embedding from one object X into another object Y, you can consider X as being contained in Y by identifying X with the image of the embedding
marsh fractal
the image of the embedding is that exactly the object X?
oblique river
yes
oblique river
yes
marsh fractal
cool thanks
oblique river
np
marsh fractal
im trying to understand why Zp is an embedding map to any finite field
oblique river
I mean, be careful how you phrase it
marsh fractal
yes
oblique river
for every finite field, there is a prime number p and an embedding Z/pZ into that field
oblique river
Z/pZ is more standard than Zp
because Zp is often used to refer to something else entirely
marsh fractal
Zp is the integers mod (a prime) right?
the equivalence classes of those integers congruency
soz thats broken english but is Z/pZ something different?
marsh fractal
ok cool
oblique river
some people also use Zp to mean that
but I don't like that notation because Zp is more commonly used to denote the p-adic numbers
marsh fractal
ah ok
my lecturer said she preferred the Z/pZ notation
I just didnt know why
if you have time, can chat in the maths voice if youd like?
oblique river
nope, sorry, I have to go in like 3 minutes haha
oblique river
np and gl
marsh fractal
tyty
light tusk
Mapping the multiplicative unit 1 to 1, there can only be one choice for the homomorphism. It's only a homomorphism if the field you're mapping into has characteristic p.
languid meteor
speaking of homomorphisms, could I just let this be f(a) = a bar (residue of a mod 15)
carmine fossil
Yes,but why do you think you can?
languid meteor
because I can define a bar + b bar = (a+b) bar and a bar * b bar = (ab) bar so f(a+b) = f(a) + f(b) and f(ab) + f(a)f(b)
carmine fossil
There is one more thing
f has to be well defined
In this case f(a) should be f(a+525k) for all k(your choice of representatives shouldn't change the output)
languid meteor
could I do this by showing that if a bar = c bar and b bar = d bar then (a+b) bar = (c+d) bar and the same for multiplication
because then my f doesnt depend on specific choices of a
carmine fossil
Show if a bar = c bar in Z(525)
a bar=c bar in Z(15)
languid meteor
ah ok that makes sense
it's immediately clear that this ring hom is surjective but is there a way I can phrase exactly why?
I cant quite articulate it
carmine fossil
Take x bar in Z_15,we note that there is an y bar in Z_525,(y bar is the equivalence class of x in Z_525) such that the output is x bar
languid meteor
gotcha, thanks so much!
languid meteor
ive actually also been trying to find these two ring homs but been finding some difficulty. I wanted f to map to maybe the residue mod 15 and g to have something to do with inverses in modular arithmetic but I cant quite link everything
carmine fossil
Do you want the answer?
Define f: f($\overline 1$)=$\overline{10}$ and g:g($\overline 1$)=$\overline 1$
cloud walrusBOT
DrunkenDrake:
languid meteor
how are we defining the equivalence classes? Because aren't they sets of all numbers that give a certain remainder when divided by n
carmine fossil
aRb if n divides a-b
If so,a and b are in the same equivalence class
For Z_3 n is 3
For Z_15 n is 15
So 1 bar in Z_3 means the set {...-2,1,4,7,10...}
1 bar in Z_15 means the set {...-14,1,16,31...}
languid meteor
idk why this is confusing me, I guess its not clicking what im actually doing here. So im mapping the equivalence class of 1 in Z_3 into the equivalence class of 10 in Z_15?
carmine fossil
Yes
languid meteor
but if im mapping from Z_3 into Z_15 for f, doesn't 1 bar in Z_3 contain elements not in Z_3 = {0,1,2}?
carmine fossil
1 bar always meant the set {...-2,1,4...}
You just picked a representative 1 and called the set '1'
Z_3 is actually {
{...,-2,1,4...} , {...,-1,2,5...} , {...,-3,0,3...}
}
Which you refer to as { \bar 1, \bar 2 ,\bar 0}
languid meteor
ahhh I get you now, that makes sense
what I don't quite get now is why we're mapping 1 bar to 10 bar, I dont see the reason for that choice
carmine fossil
1 bar + 1 bar +1 bar=0 bar should map to 0 bar(identity maps to identity)
Which means 1 bar maps to 5 bar or 1 bar maps to 10 bar
(Everything in the corresponding equivalent classes)f:1 -> 5 and g:1 -> 2 also works
languid meteor
so what exactly is my identity element? It should be the element 1 in each set right?
languid meteor
ahh the additive identity, ok this makes a bit more sense now haha
carmine fossil
(Also,These may not be ring homomorphisms,I checked only for the group condition)
languid meteor
thanks so much for the help, im gonna have to look through this and brush up on my definitions
carmine fossil
How are you defining rings?with or without multiplicative identity?
languid meteor
without, ive just been including it because I spent all of last semester with vector spaces
not meaning to including it, its just in the back of my mind
carmine fossil
Ok,So Both are ring homomorphisms,in that case
olive mirage
I hate so much that what maps are ring homomorphisms depends on your definition of ring.
(although it makes a good lecture for graduate students I suppose)
carmine fossil
Why do some people define ring as ring without identity?
scarlet estuary
https://mathoverflow.net/questions/22579/what-are-the-reasons-for-considering-rings-without-identity
tl;dr there are some important rings-without-identity that come up in analysis
this is one example of a common divide between an algebraist's definitions and an analysist's
another being "does N include 0?"
a similar example, but with the divide being number theory vs everyone else, is "do 'multiplicative functions' only need to satisfy f(ab) = f(a)f(b) for coprime a, b, or for all elements in their domain?"
languid meteor
these ring hom exercises are kicking my ass today, I feel like the image for this ring hom is maybe R or maybe some weird subset of R. I can't really factor out the sqrt3's to make a nice expression because I have (sqrt3)^n etc. The kernel is all of the polynomials with a root at sqrt3 but idk if I can find some general expression for all such polynomials or if I can just say the kernel is that set
golden pasture
could you think of some nontrivial polynomials in the kernel
languid meteor
I have x-sqrt(3) if that counts as nontrivial
golden pasture
that isnt in Z[x]
languid meteor
ah, of course that might be my problem. I guess then I have x^2 + x^4 - 12
so they have to be polynomials of even powers and the constant term has to be minus the sum of sqrt(3) to the power of each of the x^2,x^4 etc terms
oh they dont even have to be even powers because I could have odd powers cancelling each other out in some telescoping sum
so im back to having a bunch of possibilities
golden pasture
not rlly you can have x^3-x^2-3x+1
can have odd powers
hint: ||look up gauss lemma||
languid meteor
ive tried looking for the gauss lemma and I see one but I dont think I know enough to be able to use it
for my image the even powers always give back integers, so I have all of the integers in my image and then (3^(2n-1))^1/2 times whatever integer I want
so is my image just like Z U [(3^(2n-1))^1/2] Z, where n = 0,1,...
golden pasture
lets suppose p(sqrt(3))=0
hm
the idea is that you can factor p unless p is minimal
languid meteor
right so I can get (x-sqrt(3)) times a bunch of other (x-a)(x+b) etc
golden pasture
factor p over Z[x]*
languid meteor
ah ok so the minimal polynomial here should be (x-sqrt(3))(x+sqrt(3)) because we need integer coefficients
which is x^2-3
and so then the kernel will be all integer multiplies of this because I can multiply by any integer and still be able to factor (x-sqrt(3))
chilly ocean
wdym? there are polynomials of degree higher than 2 in the kernel, you won't get those by multiplying by integers x^2 - 3
all polynomials of the form (x^2 -3) * p, for p in Z[x] are in kernel, the question is are there more?
languid meteor
ah ok, yeah that makes sense. I would be very surprised if there are more, 0 is in Z[x] right? so that's accounted for
chilly ocean
yes, I don't think there are any more since x^2 - 3 is irreducible
but I don't know how to justify it correctly
languid meteor
I also think that I know the image of this homomorphism is Z[sqrt(3)] but I have no idea how to show that
since I cant see how to factor out a sqrt(3) to have something of the form a + sqrt(3)b
carmine fossil
Take p(x)=x
chilly ocean
I mean bx + a for a,b in Z is already Z[swrt3]
carmine fossil
If you apply the evaluation to that,you get sqrt(3) is in the image
You also know 1 is in the image
languid meteor
right, so I know that sqrt(3) is in the image and 1 is in the image, but how can I generate the rest of the image from knowing this?
ah ok, I havent convinced myself that there arent other elements in the image though
chilly ocean
under this map all elements you can get are of the form sum of (integer*sqrt(3) + integer) I think, which is of the form c +d(sqrt3) for c d integers
languid meteor
ah I get it, I can obviously factor sqrt(3)^5 into sqrt(3)*sqrt(3)^4 for example and sqrt(3)^4 is an integer
chilly ocean
ye, sqrt(3) to any power is either integer or integer times sqrt3
carmine fossil
Don't forget to show that the image always lies in Z[sqrt(3)]
chilly ocean
thats what we just showed
carmine fossil
You could possibly have a larger field
olive mirage
@carmine fossil If you require that rings have identity, then ideals aren't subrings, and when you try to make the story of rings and ideals match the story of groups and normal subgroups it makes something that is supposed to be clean ugly.
chilly ocean
no, the observation that a* sqrt(3)^n is of the form integer times sqrt3 guarantees it lays there
carmine fossil
Ok,That's sufficient
chilly ocean
wait zeta wdym? Dont you always require ring to have identity?
golden pasture
you would end up proving some form of gauss lemma for the kernel
olive mirage
my definition of "ring" depends a great deal on context. If I'm writing as a number theorist, rings are commutative and have 1.
if I'm teaching intro algebra, probably neither of those is true.
chilly ocean
Ive never seen a definition of ring without 1 or 0
olive mirage
yeah, whether you require 1 and whether your require commutative multiplication are the main disputes. But you will also see disputes about whether it has to be a subset of the complex numbers, or whether multiplication has to be associative.
it is not really fair to call them disputes, I suppose
the point is, you might want some of these and not others depending on context
And it is nothing compared to the hell that is people's definition of "curve"
chilly ocean
I think it shouldn't be called ring if it doesn't have two operations then, I've seen few books that defined rings and Ive never seen a definition that didn't require two operations, that's what I'm confused about. With 1 operation it looks more liek a group
wait I read that wrong
2 operations but no identity
but still
olive mirage
my takeaway is this: groups are intrinsically interesting object in a way that rings are not. Rings become interesting objects when layered with additional structure, but those extra structures are not compatible.
Rings are like topological spaces, and groups are like compact manifolds.
olive mirage
basically, point set topology is not intrinsically interesting, but is a fantastic tool to address interesting problems. It is an invaluable tool.
Compact manifolds, however, are intrinsically interesting in the sense that many people spend years studying them.
the Classification Theorem of Finite Simple Groups tells you something about how deep and important even finite groups are.
maiden ocean
wtf this is a very hot take zeta
rings are not intrinsically interesting
: (
i care about them
olive mirage
oh, I mean, they are plenty interesting in context.
and that context might be something as simple as an algebra structure.
or a topological space where the ring is the ring of reasonable functions
maiden ocean
i guess spec n stuff counts as additional structure
olive mirage
haha yes
maiden ocean
fair
the additional structure emerges super naturally so it feels weird to think about it as "separate" from the ring ig
olive mirage
and there you are almost always layering way way more structure on because not only is it an algebra, it is a finite dimensional algebra with conditions on ideal chains etc etc etc
I woudl argue the reason it looks like it arises naturally is because people present it well, and it makes sense in the examples of rings we are naturally inclined to think about.
olive mirage
(but it requires commutativity to even get started down that chain of thought)
extra conditions allow you to impose additional structure, so I think they're more or less the same.
maiden ocean
i guess
olive mirage
like Commutative Algebra is really the study of commutative rings, and is "interesting" in the sense that group theory is interesting that I outlined above.
maiden ocean
yeah thats fair
if you count even imposing like commutativity n stuff as additional structure then yea id say
rings without unity are not the most intriguing objects lol
olive mirage
yeah I would say when you take commutative rings with 1 that are finitely generated over Z or over a field then you get an interesting class of objects.
maiden ocean
isnt that true to some extent re: groups though?
olive mirage
but that theory is wholly different than the (also important and interesting) theory of group algebras.
I would say it is true with regards to infinite groups. I don't know a lot of examples of using infinite groups without imposing additional structure on them. But for finite groups, I think you're already "there"
chilly ocean
I guess point set topology is sort of a useful common language to me
olive mirage
yeah, that's how I think about it. I think point set topology is amazing. And I think it not being deep makes it more useful not less.
maiden ocean
yeah i was gonna say theres a lot of cool stuff you can say about finite or finitely generated groups but in full generality "a group" is pretty broad
point set topology is ok but courses on it kinda suck
golden pasture
point set literally set theory lol
maiden ocean
its like learning a language word by word without ever forming any sentences
if that makes sense
olive mirage
And this is why we make grad students learn it on their own 😛
maiden ocean
olive mirage
So basically, there is a very shallow theory of rings that is a foundation for a lot of very deep stuff, that was the analogy I was trying to make.
But the fact that there is no "the" definition of most math terms is upsetting to accept.
languid meteor
hey guys I was trying to find the kernel of this ring hom and im finding that some elements of 5Z[i] arent included in it. But im meant to have 5Z[i] as a subset of the kernel and I cant find my mistake
for a + ib to be in the kernel I need a - 5b = 0 (mod 13)
<=> a = 5b(mod 13)
<=> a = 5b + 13n, n is an integer
<=> a + ib = 5b + 13n + ib
so a = 5b + 13n and b = b
so now if I want to make sure that 5 + 5i is in the kernel, the condition that I just derived tells me that it isnt
because if b = 5, then a = 25 + 13n and I cant make a = 5 when n is an integer
so im just kind of stuck there
maiden ocean
does ignore you here mean "i figured it out" or "i still need help" lol
languid meteor
that was someone else, I still need help for this question
because I dont think there's any way that 5Z[i] can be a subset of the kernel of this ring hom
unless 5Z[i] is a different set than im thinking of, I think it should be all 5a + 5bi, where a and b are integers
languid meteor
yeah its probably a typo
I was thinking it could be nonstandard notation but the person who wrote this question uses that notation in the traditional way all the time
I guess I can find the kernel at least, it should just be the set of all a + ib such that a = 5b + 13n
burnt pond
You can show that the ideal (3-2i) is in the kernel and 3-2i ist irreducible
languid meteor
what can I do with 3-2i being irreducible in this context? I havent explored that concept much yet
burnt pond
Do you know that Z[i] is a pid? If 3-2i is irreducible then (3-2i) is a maximal ideal contained in the kernel. So either kernel =(3-2i) or kernel=Z[i]
Cant be the second one obviously
languid meteor
what is (3-2i) in this context, does it imply all multiples of this complex number?
I think I know that Z[i] is a pid and I have the concept of maximal ideals
burnt pond
Yes it's the ideal in Z[i] generated by 3-2i
languid meteor
perfect, thanks a million
do you know how I could split that up into cosets btw?
I dont know where to draw the line so to speak
burnt pond
Hmm so you know that (3-2i) is maximal, thus Z[i]/(3-2i) isomorphic to Z_13. So representatives for the cosets are just 0,...,12?
maiden ocean
umm
i mean u can do that but you should be careful
0, 1, ... 12 are not representatives in Z[i]
actually they might be lol cuz phi(a + 0i) = a - 50 mod 13 = a mod 13
but like the representatives are in Z[i]/(3 - 2i) not in Z_13
languid meteor
how do I know what counts as a representative and what doesnt though?
like I wouldve just chose 0,1,...12
chilly ocean
@burnt pond why is it Z_13, I mean, in general how would you find what is z[i]/a iso to
maiden ocean
thats not the problem saint
the trick is that in general like
here the image of our cosets happen to correspond nicely to the representatives
bc 0 goes to 0, 1 to 1, 2 to 2, etc
up to 12
but thats not true in general
languid meteor
Z[i]/ker(phi) is iso to the im(phi) @chilly ocean so in this case Z[i]/(3-2i) is iso to Z_13 since (3-2i) is the kernel and Z_13 is the image
chilly ocean
whats phi here
maiden ocean
like saint u see more generally that if i had some random map phi: R - > Z_13 and you asked me for representatives of the cosets i couldnt just be like
"0, 1, ..., 12" right?
(where R is an arbitrary group/ring/whatever)
languid meteor
yeah that makes sense
maiden ocean
so the thing is like
just because here 0, 1, ..., 12 happen to be in our choice of R it doesnt mean that the structure corresponds
languid meteor
phi: Z[i] -> Z_13, phi(a +bi) = a - 5b(mod 13) @chilly ocean
maiden ocean
if we had some other choice of phi it could totally turn out that 0, 1, 2, ..., 12 all happen to be in the same coset
and they wouldnt represent all the cosets
chilly ocean
ok yeah thats what I was asking about, you can guess it here, but is there a general way (or just guess? was the map stated in the problem?)
maiden ocean
like if our map was phi(a + bi) = b mod 13
chilly ocean
oh the map was stated nvm
maiden ocean
0, 1, ..., 12 would all go to 0
it works out in this case but i wanna be clear bc its an easy trap to fall into lol
burnt pond
@chilly ocean For a number field K and any element x in the ring of integers O_K you know that the norm of x is #(O_k/(x)). So in this case O_k=Z[i] and the norm of (3-2i) is 13, so Z[i]/(3-2i) is a ring with 13 elements.
chilly ocean
like in general, is there a way to solve it for any x? seems quite hard
reminds of some CRT I guess
maiden ocean
i guess u have to prove that theres only one possible multiplicative structure on Z_13
viscid pewter
i don't know much, but i know that all rings are groups, and there's only one group with order 13
so would that do it?
maiden ocean
you need to further show that there is only one possible multiplicative structure
up to isomorphism at least
im sure it can be proved
oh there are two
bc uhhh
chilly ocean
its always a finite abelian group.
@burnt pond hmm yeah I guess that makes it more simple
viscid pewter
pain
chilly ocean
you dont need to prove anything sloth if you show theres only 1 group of ord 13
I think
maiden ocean
one group of order 13 =/= only one multiplicative structure on that group
that fulfills the ring axioms
viscid pewter
okay wait so ab = 0 for any a, b
viscid pewter
wouldn't that mean there's no multiplicative identity
viscid pewter
aaaa
chilly ocean
I mean, you know this ring needs to be iso to Z_13 wrt to addition, I think theres only few multiplicative systems that could possibly work, all probably iso to z_13
viscid pewter
aha, i've proved it
maiden ocean
yeah this is easy because if I is an ideal of R then |I| divides |R| = p
maiden ocean
dan
languid meteor
you know how earlier we were talking about the cosets of Z[i]/ker(phi), does this give me the cosets of the kernel or of Z[i]?
chilly ocean
ye
maiden ocean
cosets of the kernel doesnt really make sense
chilly ocean
he means the ideal
maiden ocean
you have cosets of Z[i] wrt to some kernel
chilly ocean
I suppose
languid meteor
im asked about the cosets of ker(phi) in Z[i]
languid meteor
is this asking for Z[i]/ker(phi) then? im kind of confused by what it's asking me to do
languid meteor
R a ring, r an element of R, I a subset of R. r + I := {r + i | i an element of I} is the coset of I
hopefully thats the right defn
languid meteor
it will just be r + ker which is just r right?
maiden ocean
(r is an element of ker(phi) rather than of R)
ah i see what you mean
what do you think "cosets of Z[i]" are vs "cosets of the kernel"
im kinda confused on ur terminology rn
languid meteor
cosets of Z[i] to me would be r + I where r is an element of Z[i] and I is a subset of Z[i]
maiden ocean
umm
ok the problem here is that you are saying "I is a subset of Z[i]"
which is very broad
the idea behind cosets is that they partition your group
like
if we fix I
ok let me use an example
consider the cosets Z + 4Z @languid meteor
what does this mean
languid meteor
so I take an element in Z and then add each element in 4Z to it to get new elements
is that how it works?
maiden ocean
kinda
languid meteor
and then the set of the new elements is the coset
languid meteor
I think they are
maiden ocean
yeah
so like the idea behind cosets is that we want to basically partition our entire group
with respect to some subgroup
(normal subgroup but whatever not the point)
like
1 + 4Z = 5 + 4Z = 9 + 4Z = ...
so our cosets are
{0, 4, 8, ...}, {1, 5, 9, ...}
and so on
so do u see how this just brakes down if instead of taking 4Z we say "take any subset of 4Z"
languid meteor
yeah that makes sense, we dont get a partition anymore
I think I saw somewhere that the quotient ring R/I is the collection of cosets of I in R
so then would that mean that Z[i]/ker(phi) be the collection of cosets of ker(phi) in Z[i]?
which I guess is what im trying to find
maiden ocean
the situation is analogous so yea
so saintscratch what you are looking for is
the set of cosets {r + ker(phi)}
languid meteor
I always found the whole concept of cosets/quotients to be quite confusing I just could never understand why they were defined that way
viscid pewter
how would you prefer to define them
maiden ocean
the idea is that they basically are quotienting in the same sense that
they shrink our groups
languid meteor
I dont have a preferred definition, I just always found it jarring for whatever reason
maiden ocean
its not very intuitive when you first see it
viscid pewter
there's some intuition for it i find
the order of a subgroup divides the order of the group, and so on
and then you use the subgroup bc it contains the identity, so all the other shifts of it will be higher in some sense
idk
languid meteor
my uni teaches rings and fields before groups for whatever reason so I havent done groups yet
maiden ocean
i find it helps if you picture your group visually (as a number line or something)
???? thats really bizarre lmao
but it works the same mostly so whatever ig
languid meteor
yeah its so strange, I think its so they can give us rings and fields 2 next semester so in our final year we can do galois theory
languid meteor
we do groups alongside rings and fields 2 next semester
viscid pewter
build up
languid meteor
galois theory?
languid meteor
ah, yeah I agree. Especially since i've always been very interested in groups I was excited to do it asap
maiden ocean
thats wack but like
whatever ig
anyway the intuition works best i think if like
imagine the number line with the integers marked off
scarlet estuary
my undergrad introduced rings and fields first
just so it could define vector spaces though
didnt really go into much detail on rings/fields
uncut girder
Nice drawing g
maiden ocean
thanks pty 
languid meteor
our university jumped right into vector spaces in detail with only the field axioms as a starting point haha
then did rings and fields, then will do groups
scarlet estuary
being able to use jargon like "n by n matrices over V form a ring" is handy
viscid pewter
they might be following that one weird textbook
which is rings and then fields and then groups
maiden ocean
tbh the key idea here is that equivalence relations by identifying elements basically let us think of them as "the same"
this is why topology should be taught before groups 
viscid pewter
um
you pain me
anyway for your original question saint i might say you have cosets of the kernel in Z[i]?
it's just words
languid meteor
thanks so much for the help everyone I really appreciate it, im a bit behind in rings so im playing catchup this week 😶
dusk summit
how would one get to algebraic number theory or algebraic geometry with only introductory group theory? like what "path" of topics would I have to learn
solemn rain
learn ring theory
and field theory
galois theory
and basic NT
thats for alg NT
for algebraic geometry know those and topology and commutative algebra
dusk summit
topology at munkres level?
solemn rain
yes
next obsidian
You don’t even need Munkres level
You literally only need the first couple of sections of Munkres to get definitions and shit and then access to google
E.G. me
maiden ocean
chmonkey how much AT do u know
next obsidian
Literally none
next obsidian
Gross
maiden ocean
simply categorify
next obsidian
maiden ocean
ok i need to go be productive bye
dusk summit
mhm
latent anvil
Chmonkey was talking about infinity nonsense the other day
latent anvil
Hello
Niiiice
I found out today that the person will toledo dated is now a trans woman
maiden ocean
i remember i saw a post that was like "CANCEL will toledo for having dated cate wurtz in 2011 who is also CANCELED!!!"
and i cant figure out if it was ironic or not
maiden ocean
i googled what she did and i couldnt find anything
latent anvil
Or that it was at least a very toxic relationship
latent anvil
Which makes "cancel bc he was in the relationship" weird
Yeah sorry I don't mean to speculate
maiden ocean
shamrock is cancelled for speculating about will toledo who is cancelled for having dated cate wurtz in 2011 who is cancelled
maiden ocean
hi buncho : )
latent anvil
Noncommutative ring theory is interesting and I want to learn more
oblique river
cancelled for not talking about algebra in the algebra channel
latent anvil
Lol
oblique river
hi sloth! :)
latent anvil
Hi buncho
maiden ocean
chad yes
chilly ocean
chilly ocean
i am gay
oblique river
ok good
chilly ocean
i should make the rainbow bg on my pfp more obvious
oblique river
... I didnt even realize that was meant to be a rainbow haha
maiden ocean
shrink legoshi to make room for gay
oblique river
meowmst
carmine fossil
himst
oblique river
I manually made mine
chilly ocean
i manually cut out the background to mine
made it transparent
then slapped a rainbow
chilly ocean
carmine fossil
No it stands for t in ttera
oblique river
maiden ocean
oblique river
oblique river
titty?
maiden ocean
ok they have been successfully car seat headrestpilled
i mean i blackpilled htem at the same time but u know
whatever
chilly ocean
what does that even mean
maiden ocean
it was bound to happen eventually
idk if tterra does not understand what car seat headrest is or if he does not understand what it means to pill someone
oblique river
i dont know but it gets the people going
chilly ocean
the first one
maiden ocean
they are an indie band
for gay people who are depressed
or depressed people who are gay
oblique river
thats me :o
latent anvil
I love beach life in death
Prob my fave song from them
also I didn't realize tterra was gay. Hello fellow riemannian geometry homo
chilly ocean
topologay
carmine fossil
What is a man? A miserable little pile of secrets
latent anvil
latent anvil
What is a manifold? A miserable little pile of charts
maiden ocean
hey space cadet is good
latent anvil
That one is very good
maiden ocean
drunk with friends is also good
the best one of course is drunk on a work night
warning loud
like rly loud
latent anvil
I got into a fight with my brother and I want to get drunk but I also have a midterm tomorrow
maiden ocean
is it for francaise
latent anvil
oh lol drunk on a work night is a song
I thought you were meming
No that was today
maiden ocean
just listen to it sham
latent anvil
Analysis is tomorrow
maiden ocean
latent anvil
This is the best csh song
maiden ocean
YES
maiden ocean
disjecta membra kinda slaps
maiden ocean
latent anvil
maiden ocean
i use y outube instead of spotify bc of this
a lot of old stuff and smaller stuff is only on yt and not spotify
latent anvil
Yeah definitely
scarlet estuary
some great algebra discussion here
latent anvil
*mod voice* this should be in #point-set-topology and #gay
@latent anvil
maiden ocean
greek lesbian punk rock isnt on spotify either
latent anvil
Called it
scarlet estuary
the album in my name isnt on spotify 😦
maiden ocean
we need a gay channel in the advanced section