pq
#groups-rings-fields
formal ermine
median pawn
it's in Z[i]
terse crystal
I am not sure. Then three groups are finite abelian groups the statement seems to be true
formal ermine
but then a,b are the same
terse crystal
Because you need the real part in pZ
agile burrow
If T is finite then it's true, but it's my understanding that this is hard to prove
formal ermine
I was thinking the wrong direction lol
oblique river
Youre not going to get any mileage thinking avout abelianf examples
It is true (and easy to prove) if your groups are abelian
terse crystal
Yeah. Abelian case it’s true
lethal dune
If T is finite then it's true, but it's my understanding that this is hard to pr...
maybe fundamental thm of f.g. abelian groups will do
since it's ⊕, groups are abelian
terse crystal
Structure of finitely generated modules over pid things can prove it
oblique river
agile burrow
But for non abelian groups this is much harder
lethal dune
what even is ⊕ for non abelians? Free product?
probably the intended question is for ×
agile burrow
Yeah, I took it to mean direct product
terse crystal
https://math.stackexchange.com/questions/1122839/when-are-two-direct-products-of...
Gee, that’s so brilliant. Thanks a lot
terse crystal
what even is ⊕ for non abelians? Free product?
Oh… my bad, yeah I meant product, should have used Π
median pawn
how would u see that f is surjective?
terse crystal
how would u see that f is surjective?
x+yi+I=f(x+ya)
That’s an interesting proof, which book is it from?
Yes
median pawn
That’s an interesting proof, which book is it from?
it's from stackexchange
even i'm quite fascinated by the proof
how does the last cardinality argument work out? i.e., why is the cardinality of Z[i]/(a+ib) = a^2 + b^2?
median pawn
idk what that is yet
terse crystal
idk what that is yet
Z-Linear f: Z^n->Z^n, f represented by a matrix A, A=PDQ for a diagonal matrix D=diag{d_1,…, d_n}, and P,Q invertible in M_n(Z), so f is “conjugate” (not exactly conjugate, you can get what I mean, commutative square with both columns isomorphic) to the linear map represented by D, which is of the form e_i->d_i e_i. Clearly the cokernel of this map, has cardinality Π|d_i|=|det(D)|=|det(A)|
terse crystal
I see
south patrol
Well like apparently if G,H abelian and G x Z cong H x Z then G cong H
Easy for fg. but general case 
terse crystal
Yeah, strange that I didn’t think about it before, only now someone asked me about that question and I realized I never thought about finite groups case
the proof in that link is beautiful, they didn’t even use much group theory
coral shale
Ok let me make another stab - other than the fact no one (book) uses it, what is 'wrong' with this notation? Is there some underlying notion I'm missing that makes it not make sense.
I understand why using unconventional notation is bad in general, yes.
cloud walrusBOT
prisma ibex
What is this meant to notate?
coral shale
R/I
coral shale
or G/H for abelian G
prisma ibex
That's insane notation
coral shale
why tho 
prisma ibex
Nobody will know what you mean when you write it
coral shale
Is it inconsistent conceptually
prisma ibex
Yes
coral shale
How?
prisma ibex
A quotient is not like R-I
coral shale
im missing why
chilly ocean
I think its because R/I is supposed to denote equivalence classes so resembles a quotient by an equivalence relation
prisma ibex
Yup
Also if you do this for finite groups/rings and think of what the order of the quotient is
Then you're dividing cardinality
Not subtracting
coral shale
Can the same argument not be made for direct sums
agile burrow
direct sum is established notation and \otimes has other uses
prisma ibex
direct sum doesn't make sense for groups anyways
you have a direct product, and then this is the notation you expect
coral shale
hmm ok
chilly ocean
I think this notation is used for Minkowski difference
prisma ibex
the other worry is that \ominus does have another meaning, although it's not used often, it basically lets you remove summands from a direct sum
coral shale
im more trying to figure why multiplicative notation is used
over additive notation
and also vice versa in certain places
prisma ibex
for quotients it's because of the equivalence relation thing
there's no way around that
for stuff like Abelian groups you'll usually see additive notation, so like \oplus instead of \times for the direct product, but this is because the direct sum and direct product are the same thing in this case
for groups and rings there is no notion of direct sum, you only have the direct product and you use the multiplicative notation accordingly
prisma ibex
right this is another situation where you use direct sum usually
prisma ibex
but again you also use the "multiplicative" notation for quotients in any case
again because of equivalence relations
coral shale
makes sense ty 👌
pliant forge
Ok let me make another stab - other than the fact no one (book) uses it, what is...
i think this one is used in Lie Theory iirc
$Y \ominus X := \text{Log}(X^{-1}*Y)$
cloud walrusBOT
Eso
solar glacier
Question that may sound trivial and I think I know the answer but wanna confirm
If V(S) is the zero set of a set of polynomials
And a is a constant
Does V(aS) = V(S)
quiet pelican
Does V(aS) = V(S)
Yes
solar glacier
Oh what I thought
quiet pelican
Assuming a is non-zero
quiet pelican
(Or maybe not a zero divisor over a general ring)
pastel cliff
can someone verify this is right rq pls
chilly ocean
can someone verify this is right rq pls
this is wrong, phi(r) doesn't make sense
cloud walrusBOT
Blitz
chilly ocean
is what you want
as for im(phi), you have i and n
i and n are supposed to be the same thing
pastel cliff
completely unrelated (jumping around) but is there an intuitive way to see what coproducts are
perhaps im over complicating 
rustic crown
coproducts in which category 
long geyser
product but you flip the arrows
chilly ocean
completely unrelated (jumping around) but is there an intuitive way to see what ...
in most equational classes (which are categories of (universal) algebras given by some set of equations), they represent a free construction
for instance, the coproduct of groups is their free product
rustic crown
you can think of it like a spicy disjoint union 
tribal moss
Start with a disjoint union, but if the objects in your category must come with binary (or higher) operations, you'll need to invent new elements that these operations can produce when they're used on inputs from different sides of the disjoint union. Keep inventing new elements as long as you need more of them, and only make them the same as something that already exists when the axioms of your algebraic thing forces you to. In most cases you'll end up with an infinity of new elements. Rarely you get forced to stop after your new thing has become only slightly larger than the disjoint union, such as for abelian groups.
young fiber
Start with a disjoint union, but if the objects in your category must come with ...
Is it always well defined
tribal moss
No, some categories don't have coproducts in the first place.
The category of fields, for example, doesn't.
young fiber
I meant like is it possible that there can be more than one way to combine two objects
Or do you have to add the minimal number of elements
Even then, couldn't there be different structures in the resulting objects while still satisfying the axioms of the category
rustic crown
if you wanna combine the objects a and b, then you're looking for an object c with morphisms a --> c and b --> c. if you do this "in the most efficient way" then you get c as the coproduct.
rustic crown
Or do you have to add the minimal number of elements
so yea this idea would make sense
tribal moss
Or do you have to add the minimal number of elements
In general it's the other way around: you'll need to add the maximal number of elements and only reuse outputs when the algebraic laws you need to satisfy imply you must.
young fiber
Hmm
tribal moss
(Except only ones that can be made from you original disjoint union in finitely many applications of the operations).
Thus assumes we're looking at a category of algebraic structures. It might not work at all in categories defined in different ways.
languid tusk
Can someone explain the Euclidean algorithm? Where do you get q from?
I don’t think my textbook explains this
long geyser
your textbook probably does explain this, look for the division algorithm
tribal moss
The q_i's are the quotients of the division in each step.
languid tusk
The q_i's are the _quotients_ of the division in each step.
I mean…. Yeah, but where does the number for q come from?
Where does 5 come from?
tribal moss
Divide 57970 by 10353. What do you get?
languid tusk
I'm a dumbass 🤦♂️
I'm sorry
I just took a cursory glance at it and it seemed to be arbitrary
But...
Fucking derp
pastel cliff
i and n are supposed to be the same thing
this was a typo 
fixed those two things ty
pastel cliff
is it ok to think of annihilators and torsion like orbits and stabilizer?
lethal dune
pastel cliff
is that a no 
i dont get this shit lol
ok i'll rephrase - what does it mean for a module to "have torsion"
from what i understand, the torsion of a module is the subset of the module M that can be annihilated
elements of the torsion be like
and if that's right then this proof should be valid 
pastel cliff
alternate attempt but stuck on second claim there now
lapis plume
Hi I am solving a problem
Idk how to list out all the left cosets
Can anyone help?
lethal dune
alternate attempt but stuck on second claim there now
Try M=R
quiet pelican
Idk how to list out all the left cosets
Take an element of G, list out its coset
Take another element of G not in that coset, list out its coset
Take another element of G not in either of those cosets, list out its coset
Etc until you’ve listed every element
cloud walrusBOT
blanket
solar shore
ahh
tender wharf
it would be
tender wharf
ababab...
tender wharf
if its abelian its nicer
solar shore
right yeah
tender wharf
np
solar shore
i had to make sure because i was thinking about it in the context of the inverse
tender wharf
ah no
solar shore
like where the a b would switch
yeah i forgot i should write them out before tryna assume anything lol
ababa...
woops
solar shore
right
although i do have one more question
say |ab| has order n
then does |abab| has order n/2?
formal ermine
then does |abab| has order n/2?
only if 2 divides the order of |ab|
formal ermine
n
coral shale
use an example
integers mod 5
3
has order 5
3*2 = 1 has order 5
(let a = 1, b = 2 so a+b = 3)
solar shore
gotcha
that helps
can i bother y'all for one more question
it goes back to the abababa... thing
formal ermine
you aren't bothering us
solar shore
im having a bit of trouble proving that $|ab| = |ba|$
cloud walrusBOT
blanket
formal ermine
just go ahead and ask
formal ermine
im having a bit of trouble proving that $|ab| = |ba|$
is the group assumed to be abelian
solar shore
errr
ive been told thats called an automorphism?
oh
wait actually i think i just got it
formal ermine
yes, conjugation is an automorphism
solar shore
so this is what i wrote down
formal ermine
ba and ab are conjugate to one another, so what can you conclude
solar shore
they have the same order?
formal ermine
yeah
solar shore
ah
formal ermine
automorphisms preserve orders
solar shore
hrm i dont think we're allowed to use that idea quite yet because automorphism doesnt come up until later in the book lol
i got a bit curious and got ahead of myself
formal ermine
do you know what a homomorphism is
coral shale
breh illu
solar shore
it preserves smt afaik
coral spindle
Observe $(ab)^n = a(ba)^{n-1}b$
cloud walrusBOT
Boytjie
coral shale
smh
formal ermine
im stupid
solar shore
idk if my logic is right, but i ended up writing abab... = e then a(babababa,,,) = a then a^-1a(babababa) = e thus |ba| = n
solar shore
Observe $(ab)^n = a(ba)^{n-1}b$
ah
solar shore
idk if my logic is right, but i ended up writing abab... = e then a(babababa,,,)...
is this correct tho?
coral spindle
It is wrong
lapis plume
Take an element of G, list out its coset
Take another element of G not in that c...
Okay thx
south patrol
Is it clear that if R^n admits a normed real division algebra structure (so n=1,2,4,8 ofc) then it is unique up to isomorphism? It's trivial for n=1 and n=2 is standard, but n=4,8 seem messier - maybe I can just adapt the argument for n=2, though
lapis plume
I had proven that HK is a subgroup of G if H,K are subgroups and G is abelian. But I have some further question, is there any example that G is NONabelian and H,K are subgroups, but HK is NOT a subgroup?
quiet pelican
I had proven that HK is a subgroup of G if H,K are subgroups and G is abelian. B...
Yes there is
My natural example would be the group of formal strings of a, b, a^-1, b^-1, and the subgroups generated by a and b
I suspect there’s a nice example in some small symmetric group
But that’s marginally harder to find
coral shale
A good way to come up with examples, is to look at your proof
and figure out which part needed the property
and base your example on this point
blissful moon
how do I show that there exists a unique irreducible representation M_n -> GL(C^n)?
lethal dune
Number of irred representation = number of conjugacy classes
sweet echo
im looking for a notion of an "infinite grassmannian", im not looking for Gr(2,infty) but something like Gr(infty,infty), does anything like this exist in the literature?
perhaps the algebraic geometers might also know I'll ask them too after a while
coral shale
@ grass
rotund aurora
What is an example of a prime ideal of infinite height?
rustic crown
(x1, x2, ...) in k[x1, x2, ...]?
pastel cliff
Try M=R
huh? doesnt this make it not general enough?
super late sorry
repost + my incorrect attempt thus far
i was trying to use the fact that R/I is a field if I is maximal
and ideal generated by a unit is necessarily maximal iirc
delicate orchid
and ideal generated by a unit is necessarily maximal iirc
this is not true, a maximal ideal cannot be the whole ring
oblique river
repost + my incorrect attempt thus far
I would try to use specific modules.
Like “because this is true for all modules, it’s also true for this particular module”
And maybe just show directly something like “if I is an ideal of R then I = 0 or I = R”
Or “every nonzero element of R is a unit”
Or something
pastel cliff
just to make sure im actually getting the statement right too: it's saying that for a ring R that "acts" on any abelian group in such a way that only 0 can be annihilated, R must be a field
perhaps that's a roundabout way of saying it
oblique river
Not quite…
Well sorry
I didnt know what you meant by annihilated here
But yes I see
pastel cliff
vaguely that i can be taken to zero by some r \in R
oblique river
If M is an R-module and if rm = 0 for some r in R and m in M then either r = 0 or m = 0
(Is what you are given)
pastel cliff
yeah ic
oblique river
Actually wait im wrong
Lmao
The definition of torsion submodule is onky for non-zero divisors
pastel cliff
oop
oblique river
So what you can conclude is that if rm = 0 then m = 0 or r is a zero divisor
You should double check your definiton of T(M) though
Or aee if you are assuming that R is a domain here
Cuz that will make this easier lol
pastel cliff
"T(M) is the subset of M consisting of elements m for which there is some r \in R so that rm = 0"
oblique river
Hmmm ok that doesnt make sense because for any m in M you could take r = 0
And you would conclude that T(M) = M for all M
pastel cliff
*R is specified as commutative and with id.
but that doesnt change much i dont think
oblique river
So the definition of T(M) on wikipedia requires r to not be a zero divisor
But if you adopt that definition here then the claim is false i think
And R = Z/6Z would be a counterexample
It’s not a field but satisfies the property that “for any R-module M and for any m in M, if rm = 0 then m = 0 or r is a zero divisor”
blissful moon
Number of irred representation = number of conjugacy classes
how do you find the number of conjugacy classes of the representation then?
rotund aurora
(x1, x2, ...) in k[x1, x2, ...]?
Could it be finitely generated?
pastel cliff
But if you adopt that definition here then the claim is false i think
i thought about it some more - wouldnt it be a different zero?
rustic crown
Could it be finitely generated? <:pandaHmm:858834883760750602>
idk but aren't there some weird noetherian rings whcih have infinite dimension?
pastel cliff
like if we let r = 0_R, then rm = 0_R and therefore is not our torsion
pastel cliff
it only goes to 0_M if m = 0_M
rotund aurora
Im just curious. Just learning the definitions so far
rustic crown
but infintie dimension is not the same as infintie height tho
true
pastel cliff
det wanna take a look at something super quick
pastel cliff
repost + my incorrect attempt thus far
it's this
ignore the start of that proof it's wrong
rotund aurora
You have things like these right, but I think you require the whole ring to be noetherian, and not just the ideal to be finitely generated (?)
(I have not examined the proofs yet)
pastel cliff
well
buncho said the question might be wrong since we're not assuming r to be non zero-divisor
but i think that's wrong bc it's different zeros
im also not entirely sure how to prove it, quotienting feels a bit silly
rustic crown
yea i'm not sure about the definition of torsion lol
m in M is torsion element if there exists a non-zero r or non-zero divisor r such that r * m = 0
if it's non-zero-divisor ig R = Z/4Z is weird
because all non-zero divisors are units here
pastel cliff
the defn i have technically doesnt require those things
rustic crown
what's your def?
pastel cliff
"T(M) is the subset of M consisting of elements m for which there is some r \in ...
.
pastel cliff
but if r = 0_R
wont that be in the ring as opposed to M
or does the module structure mean it's still in M
pastel cliff
ah 
pastel cliff
not necessarily in prof's notes but wikipedia does
pastel cliff
ok then suppose it's just non-zero
rustic crown
yee then it true
pastel cliff
then i shall simply assume non-zero
rustic crown
okie :3
pastel cliff
prof's notes
rustic crown
M is a submodule of M 
okie you probably want non-zero divisor now that i think about it
else you dont' have closure under addition?
because r*s can be 0
yea that maeks snese
R=M=Z/6Z has 3 and 2 as torsion elements if you don't put the "non-zero-divisor"
but their difference is 1
which is not torsion
@pastel cliff
(sowwy for the ping, didn't want you to waste time latexing)
pastel cliff
you can always ping
pastel cliff
It’s not a field but satisfies the property that “for any R-module M and for any...
but also then this
rustic crown
hmm right
rustic crown
yee
pastel cliff
i asked before about direct sums - based this defn' is it just a product with mostly 0's in each index?
and how does this equate to the categorical defn of flipping the arrows in the product diagram
thorn delta
intuitively the elements of the direct sum are just finite formal sums of elements of each term in the product
maps from each term induce a map on the direct sum by "summing together" all the maps
pastel cliff
maps from each term induce a map on the direct sum by "summing together" all the...
this sounds like the flipping arrows bit
do you know an example by any chance
like what is the direct sum of Z/4 and Z/6
or Z/5 and Z/3 if that might be more enlightening
thorn delta
so, for finite direct sums it just coincides with the direct product
pastel cliff
this is the problem im working on
why is it specifically a direct sum in this case then
thorn delta
you could use direct product. it doesn't matter. i think direct sums are just more common notation for finite sums/proudcts
thorn delta
do you know an example by any chance
you could think about direct sums vs. products of vector spaces. For example, what would be a basis for the countable infinite direct sum of copies of k (k a field)? What about the countable direct product?
pastel cliff
you could think about direct sums vs. products of vector spaces. For example, wh...
if i understand correctly, direct sum of copies of a field mean that it's mostly 0 except in a few indices right? if so then take 1_{K_i} for each nonzero index and you have a basis no?
that sounds wrong
thorn delta
yea, that's correct
pastel cliff
thorn delta
and the product is a bit of a meme question. afaik constructing a basis for that vector space is as difficult as finding a well ordering of the reals
however, you can show that its basis is not countable
even though the product is countable
pastel cliff
you get uncountability from a countable thing?
that sounds like the opposite of useful
thorn delta
well, its just a different beast i guess
pastel cliff
well in any case
for the question above then, is it enough that img and ker makee up the whole module?
thorn delta
hm, im not exactly sure what you mean by that img and ker belong to different modules. If you know about exact sequences i'd recommend trying that
coral spindle
that sounds like the opposite of useful
Why should its use have anything to do with its countability? This is silly. As it turns out, the direct product is very important.
pastel cliff
yeah i didnt mean it that way, i figure if it's defined it has important uses
just wasnt what i was expecting i guess
maybe this might be a more enlightening question
what is that direct sum here?
coral spindle
@next obsidian it's ur twitter qn
It's... it's a direct sum yes
not a direct product
pastel cliff
i know im asking basic things but just to be sure then - since this is a direct sum this means that elements of that big sum are mostly zero except in a few indices
pastel cliff
yeah im speaking colloquially on purpose but ik that
"mostly" is nicer than "all but finitely many"
coral spindle
There's a word for that, it's cofinite
pastel cliff
i knew that, i just forgot i knew it lol
potentially silly question - the "all but finitely many condition" feels kinda random - what stops us from having zero in the first index that is only Z?
i guess nothing is stopping us but
thorn delta
if im understanding ur question correctly, then nothing. "all but finitely many are 0" can be equivalently just be thought of as "eventually 0" if there is a way to make sense of ordering of terms
like countable direct sums are eventually 0 sequences of elements from each term of the sum
pastel cliff
in the context of showing that it's isomorphic to that quotient tho, what does it mean if we do have 0 in the first index?
actually
it says "as a Z module" which is obv a ring and an abelian group
is that first index the ring then?
or is the whole direct sum our abelian group and the modul is Z x (Z \oplus Z/2 \oplus ....)
thorn delta
in the context of showing that it's isomorphic to that quotient tho, what does i...
well, if you have something nonzero in the Z index, then that element can't be torsion
pastel cliff
ok i can see that
opal osprey
Do analytical results about the assymptotic behaviour of the partition function and its generating function have any interesting uses in the representation theory of S_n or GL_n ?
I was curious if stuff like Hardy-Ramanujan expansion for p(n) and the Rogers-Ramanujan expansion actually have some uses in representation theory.
Also, I never really understood why conjugacy classes in S_n correspond to partitions of n
could someone give me some insight on that?
quiet pelican
Also, I never really understood why conjugacy classes in S_n correspond to parti...
This follows from the characterisation of conjugation in S_n
It’s essentially a “change of basis” so two elements are conjugate iff they have the same cycle type, and cycle types correspond to partitions of n
opal osprey
ok so, if I have some permutation σ in S_n, I can always find disjoint cycles τ_i such that:
σ = τ_1 ... τ_k
and the cycle type of σ corresponds to (l1,...,l_k) where l_i is the lenght of the ith cycle right?
quiet pelican
and the cycle type of σ corresponds to (l1,...,l_k) where l_i is the lenght of t...
Yes
opal osprey
but is this decomposition in disjoint cycles unique? I don't remember
quiet pelican
but is this decomposition in disjoint cycles unique? I don't remember
Yes
pastel cliff
up to reordering
opal osprey
oh, ok then
chilly ocean
There's a word for that, it's cofinite
Category theory smh
next obsidian
<@146419489083228160> it's ur twitter qn
Lmfao it is
kind jacinth
they take {0, 2} as a nontrivial subgroup, but why not {0, 1, 3} as well?
quiet pelican
they take {0, 2} as a nontrivial subgroup, but why not {0, 1, 3} as well?
1 + 1 = 2 is not in the set
kind jacinth
ohhh i see
thanks
and {0, 2} works because
for Z4 we dont care about 4 anyway
right?
since 4 isnt in Z4
quiet pelican
since 4 isnt in Z4
Well, it kinda is, but it’s equal to 0
kind jacinth
Well, it kinda is, but it’s equal to 0
u mean cz 4 mod 4 is 0?
quiet pelican
u mean cz 4 mod 4 is 0?
Yes
pastel cliff
no it's the trivial subgroup
kind jacinth
im trying to answer in the context of generators
is it still trivial subgroup?
because the nit would always be the case
but with {0, 2} we cant get all elements of Z4 so then its not that one right?
is it just the group itself?
what im trying to understand is:
Suppose that G is a group and let {gi : i ∈ I} be a set of elements in G,
where i ranges over some (not necessarily finite) index set I (that is a set whose
elements are used to label elements of another set). The smallest subgroup of
G containing all of the gi’s is the subgroup of G generated by the gi’s
formal ermine
what im trying to understand is:
Suppose that G is a group and let {gi : i ∈ I} ...
what part are you having trouble with?
kind jacinth
in the context of Z4 which one would it be?
possible subgroups are {0} {0, 1, 2, 3} and {0, 2}
formal ermine
yeah
<1> = the entire group
<0> = trivial group
<3> = the entire group because 3 is coprime to 4
<2> = {0, 2}
(<x> means the (sub)group generated by x)
if that's what you were asking
kind jacinth
hmm. I dont think im quite getting what they mean by generators
could u provide an example?
what do u mean by <1> the entire group?
formal ermine
all the possible ways you can add 1 and its inverse together
1, 1 + 1, 1 + 1 + 1, 1 + 1 + 1 + 1, etc
which just ends up being the entire group of Z4
kind jacinth
so then <1> and <3> the best generators of group Z4?
formal ermine
"best"?
kind jacinth
The smallest subgroup of
G containing all of the gi’s is the subgroup of G generated by the gi’s (this part)
formal ermine
okay so you have a set of elements
{ a, b, c, d, e }
you wanna find the smallest group that contains them
that's the group generated by those elements
kind jacinth
right
so if we have subgroup {a, b, c} and by doing a binary operation we can get the set elements {a, b, c, d, e} then we say that the subgroup {a, b, c} genrates the group?
formal ermine
yeah
kind jacinth
ok and we want the smallest one
formal ermine
the formal definition of it is this: (idk if it helps you though, when I learnt about this back then it was like really unclear to me)
,, \gen{M} \coloneqq \bigcap_{\substack{M \subseteq H \subseteq G \ H \text{ subgroup}}} H
for a group $G$ and a subset $M \subseteq G$
formal ermine
we look at all subgroups where our set is contained in
then take the intersection
(this works cuz the intersection of groups is a group)
kind jacinth
and the intersection has to give back the group? (then it generates the group)
kind jacinth
first part makes sense
its the converse where i am lost
why are we using gg^-1? we dont know if g^-1 is an element of H right?
south patrol
g and g are in H, so gg^-1 is in H (by the assumption)
kind jacinth
g and g are in H, so gg^-1 is in H (by the assumption)
ohhh h can be anything including g...
south patrol
Yes, the whenever refers to just any two elements
south patrol
Np!
I can probably count on one hand how many times i've used this test to check if something is a subgroup lol
chilly ocean
i feel like so many people reference this result when it's basically just as easy to check the definition of a subgroup
people on this server LOVE to say "just do the subgroup test!" and it always just feels like a way to get around remembering the definition of a subgroup
chilly ocean
non-empty.
pastel cliff
people on this server LOVE to say "just do the subgroup test!" and it always jus...
do you feel the same way about subring test
chilly ocean
yes
pastel cliff
kek
coral shale
non empty; closed under xy'
the proof method is the same though, ur not saving yourself thinking, yes
kind jacinth
I can probably count on one hand how many times i've used this test to check if ...
oh thats good to know. So instead of doing the subgroup test i can use this as a shortcut right?
chilly ocean
this is the subgroup test
kind jacinth
i meant instead of showing its closed, associative, has same identity etc
chilly ocean
that's the definition of a subgroup, i would not call it a test
kind jacinth
yeah true
coral shale
i meant instead of showing its closed, associative, has same identity etc
identity (if its in the subset) and associative are automatic
check the definition of a group
its for all elements that the identity acts as an identity
so for a subset of elements, itll also be one
same story for assoc - that condition needs to hold for all triples
Youre just checking closure and inverses. (and checking the identity is in your subset)
kind jacinth
yeah makes sense. its the same binary operation so its clearly also associative
smoky cypress
I don't see why is every element of G_2 is determined by its restriction on G
Here the definitions are this:
Where we identify G_a in G_{a+1} using inner conjugation
chilly radish
I don't see why is every element of G_2 is determined by its restriction on G
If you have g,h that identify on G then gh^-1 is in the centraliser of G_0, but it's trivial, so g=h. I think that's the idea?
chilly radish
No problem
livid willow
Hi, I am currently working with Ideals and quotient rings in my rings course. I have found different sources for this but are all ideals subrings?
If so, ideals like 2Z, 3Z.... do not have the multiplicative identity.
Also is it necessary to impose that the multiplicative identity should be present?
chilly ocean
any ideal which contains a multiplicative identity is the whole ring
because if 1 is in your ideal and r is in your ring, then r = r1 is in the ideal
so if you required ideals to contain the multiplicative identity, they would all be trivial
livid willow
How would it be the entire ring if 1 belongs to ideal?
chilly ocean
i explained this in my second message
livid willow
Ah okay okay
So then most ideals are never subrings (except the trivial ones), correct?
sage lodge
are all ideals subrings
yes, with the additional property that they are closed under multiplication from the outer ring
chilly ocean
if "ring" for you means "ring with unity" then the only ideal which is also a subring is the entire ring. if "ring" does not necessarily mean "ring with unity" then refer to the above message
sage lodge
oh right, rings not rngs
livid willow
Oh I see. My course so far only covers commutative rings with unity so I forgot to explicitly mention that.
sage lodge
then you should listen to TTeppa
livid willow
Yes, I got it now, thanks
chilly ocean
silk bramble
sage lodge
is there a polynomial in k[x1,...,xn] that is irreducible and doesn't have any zeros regardless of the field k?
trying to solve part b
the previous exercise, for reference
V(I) is the set of points in k^n such that all f in I vanish
chilly ocean
what happens when you skip affine varieties and go straight to schemes
sage lodge
sorry i haven't learned about them yet
chilly ocean
not talking to you
sage lodge
oh lol
i see, the spectrum of a ring
i don't think i can use the same ideal they give in the hint for 8c because x^2 + 1 might have zeros over another field
e.g. Z/2
chilly ocean
but you can still find some polynomial with no roots
you have assumed that k is not algebraically closed
chilly ocean
by assumption (and one that is not a constant, of course)
kind temple
i think the ideal generated by that polynomial should be maximal
sage lodge
by assumption (and one that is not a constant, of course)
should this be in a single variable?
i thought the definition of algebraically closed was that every nonconstant polynomial in k[x] has a root in k
kind temple
just replace x with x1
sage lodge
so we have some nonconstant polynomial in k[x1] that doesnt have a root in k
chilly ocean
i think the ideal generated by that polynomial should be maximal
it doesn't need to be, since you can just take any maximal ideal containing it
chilly ocean
should this be in a single variable?
strictly speaking, yes. but look at 8(c)
chilly ocean
you could try to do the same thing, but instead of x^2 + 1, whatever polynomial you had over k with no roots
chilly ocean
i guess i just have prove that it's maximal
give it a shot. i also think you can do what i suggested in an earlier message
but i do not think there is anything so special about the real numbers that would prevent you from generalizing your proof to this case
sage lodge
i actually haven't done the proof for the real numbers
but i think it would be something using the division algorithm
sage lodge
give it a shot. i also think you can do what i suggested in an earlier message
got it, thank you
csquared as well
chilly ocean
i don't know about the easiest way, but what are the approaches you've seen so far?
i would just try to do it directly. assume x is a product and go from there
perhaps i'd also try to square that product and set it equal to (yz)^2
next obsidian
There’s a far easier way
In an integral domain prime => irreducible
Note that the element you’re quotienting by is irreducible as it’s linear in y
So the quotient is an integral domain
Oh hmm
Maybe I lied hehe (I wanted to say x is prime cuz correspondence but it isn’t hehe)
Well my updated way to do it is that the grading by degree descends to the quotient cuz the ideal you quotient by is homogeneous
Since x lives in degree 1 it can only factor as the product of a degree 0 and a degree 1 thing, but degree 0 things are invertible
chilly ocean
hot
lapis plume
Yes there is
My natural example would be the group of formal strings of a, b, a^...
Oh nice thanks
lapis plume
A good way to come up with examples, is to look at your proof
Yes but then to prove the xy^{-1} is not in HK, it may be problematic since I cannot show that it is in fact due to nonabelian. It may just imply that I cannot express the term exactly the same as the selection criterion of HK.
zinc loom
hi does isomorphism follow between two groups if they have the same presentation?
zinc loom
i believe so. i can map the two generators from one group to the generators of the other and say that they have the same presentation
sage lodge
which is correct: "ideal of R," or "ideal in R"?
summer path
i'd say of, since an ideal is really a special subset of a ring
although if someone were to say "ideal in a ring", i probably wouldn't think that much of it
sage lodge
interesting, i'm reading a book which mixes both
next obsidian
I think of makes more grammatical sense, but no confusion arises using either so it’s whatever
white oxide
why is the identity permutation considered a cycle? my book says that it can be represented by an orbit containing the integer 1, (we are dealing with Sn here for context) but the definition they give requires that any permutation must have at most one orbit containing more than one element. however, there is only one element in the orbit containing 1.
summer path
im not sure if this helps, but the identity permutation can also be written as e = (1)(2)(3)...(n)
(im not really sure where the confusion is)
it fixes every element and hence is a 1-cycle
summer path
i went and googled it again because i thought i was being gaslit lol
hot lake
why is the identity permutation considered a cycle? my book says that it can be ...
At most one means one or less
Zero is less than one so that's acceptable
(the identity permutation has zero orbits with more than one element)
coral shale
im not sure if this helps, but the identity permutation can also be written as e...
This is also a good way when needing to count perms
proper moth
https://mathworld.wolfram.com/BinaryOperation.html
for the definition of an operation, what exactly does "f uniquely associates each pair of elements in A to some element of A." mean?
does it just mean that it's a valid function? i.e. that f(a,b) maps to at most one value?
lusty marlin
does it just mean that it's a valid function? i.e. that f(a,b) maps to at most o...
Yes, it just means that it is a function from A×A to A.
coral shale
does it just mean that it's a valid function? i.e. that f(a,b) maps to at most o...
Yes, a binary operation is just a specific kind of function f: S x S -> S
And S non-empty too apparently (that might just depend on convention)
proper moth
also quick question to clarify. is it valid to do the operation on the same element (i.e. (a,a))? e.g. a set of {0,1} with addition, is a valid operation 1+1=2, which is not in the set, so it's not a group?
or is a /= b required?
lament dawn
or is a /= b required?
It is not
robust pollen
Is a split monomorphism from a projective module to itself an isomorphism?
rustic crown
do you have finite generation?
else it's false
take F to be a free module of infinite rank, then F --> F⊕F = F is a split mono
tender wharf
also quick question to clarify. is it valid to do the operation on the same elem...
there's no 2 so that's already disallowed
that would not be valid
elder wave
This is a useful way to think about it
just a common illum L
pliant forge
Let f: A \to A' morphism between k-algebras, where all simple left A'-modules are of dim 1 over k. I'm not sure how to show that a simple A'-module is a simple A-module, where module structure is inherited from f.
rustic crown
if you have a non-trivial A-submodule then what would be its k-dimension?
pliant forge
non-zero
rustic crown
(your simple A'-module had k-dimension 1)
pliant forge
that necessarily means the A-modules k-dim its inherited from has to be 1?
rustic crown
yep, it's a k-algebra homomorphism after all. you didn't change the underlying k-vector space structure of that simple A'-module, just defined a new A-action on it
coral spindle
Does anyone know an example of an Abelian group which has two non-isomorphic structures as a vector space? I think there should be an example over the field R, but can't think of one
delicate bloom
like as a vector space over Q vs over R or am I making a mistake
white oxide
At most one means one or less
ohh right thanks
rustic crown
ig you can do it using some inseparability stuff. k = F_p(x) then V = k can be given two k-vector space structures, a * v = av and a.v = a^p v
can't be iso as one has dim 1 and other has dim p
coral spindle
like as a vector space over Q vs over R or am I making a mistake
Over the same field. I'm aware that there is at most one structure of an Abelian group as a Q-vector space
coral spindle
ig you can do it using some inseparability stuff. k = F_p(x) then V = k can be g...
Very nice example. Is it possible in char 0?
Not with the same trick ofc
rustic crown
oh ig a similar trick does work, hehe
earlier it relied on a non-surjective field map Frob on F_p(x). we can do the same for the field C.
pick a trancendence basis of C/Q say B. let f : B --> B be an injective but non-surjective set function. and define the map Q(B) --> C extending f. now since C/Q(B) is algebraic and C is alg-closed you can further extend this to a field map C --> C which is not surjective. the image would be algebraic over the image of Q(B), but C is trancendental.
anyway, the construction doesn't matter, in the end i have a non-surjective field map C --> C
use this to define two non-iso C-vector sspace structures on V = C
coral spindle
Damn that's cool
Would this also work for R over Q?
I'm not familiar with some of the arguments here
rustic crown
the same trick won't work though 😦
coral spindle
Very sad
rustic crown
if you have a field map R --> R over Q, then it's going to be continuous. which would force it to be identity
rustic crown
i'll have to recall the details, but the main idea is that squares are positive and squares go to squares
rustic crown
rustic crown
hehe :3
coral spindle
or maye my key is roken
rustic crown
yee that works
so if a > 0, then a = b^2 so f(a) = f(b)^2 > 0
so it sends positive stuff to positive stuff
coral spindle
Uh-huh
rustic crown
which means if a < b, then (b-a) > 0 so f(b) -f(a) > 0
so f is an increasing function
f restricted to Q would be identity
coral spindle
Now I wonder
rustic crown
if r is any real, pick rational sequences a_n increasing to r and b_n decreasing to r
coral spindle
Does every such finite-dimensional example induce a field automorphism?
So wlog we can restrict to a map of vector spaces k -> k^n
Sorry not of vector spaces
an isomorphism of abelian groups k -> k^n
If we then project we get a surjection k -> k that is not an injection
I guess there's no guarantee that this respects anything
There's probably some really disgusting field which provides an example but I don't have the energy for it
white yoke
I have a question about representation theory, I don't think it's very difficult but idk
How does this follow? (The part highlighted in yellow)
No one could help me in the help channels
coral spindle
Consider $\dim V_i^{\oplus \dim(V_i)}$.
cloud walrusBOT
Boytjie
white yoke
Consider $\dim V_i^{\oplus \dim(V_i)}$.
What about it?
rustic crown
can you write that in terms of d_i?
white yoke
oh it's d_i+ .. + d_i , d_i times so is just d_i^2
rustic crown
white yoke
<:eeveeKawaii:556858499930259471>
Thanks det
toxic zephyr
can someone please explain the difference between |G : H| and |G/H|?
chilly ocean
there's a difference?
formal ermine
there's a difference?
rustic crown
there's a difference?
toxic zephyr
lmao okay
elder wave
toxic zephyr
idk what |G:H| is
this is what confused me lol
i guess from context its obvious theyre the same but idk im a dummy
chilly ocean
strictly speaking i wouldn't use |G/H| if G/H weren't a group (ie if H weren't normal)
but like. it's clear what it should mean anyways
rustic crown
what if you wanna talk about G/H as a pointed G-set?
chilly ocean
rustic crown
toxic zephyr
chilly radish
idk what |G:H| is
It's a less commonly used but still used notation for index
I've seen ggt people use it
rustic crown
formal ermine
let L/K be a field extension, when does the quotient L/K have a field structure?
coral spindle
The quotient doesn't even typically have the structure of an additive group ring; K is not an ideal
formal ermine
so it's just regular boring ring theory?
I thought there might be something more interesting considering both are fields
coral spindle
Well it's more like fields are nasty algebraic structures
to be more precise
Fields do not form a variety of algebras
(Now we wait for blitz to arrive)
rustic crown
do you mean L/K as abelian groups?
chilly radish
Category of fields sucks
elder wave
imagine not having products
chilly ocean
(Now we wait for blitz to arrive)
veriety of algebras, also called an equational class, is a class of so called (universal) algebras, that is structures consisting of a signature which is data that tells you about the n-ary operations on this etc.
Those equational classes are formed using universal quantifiers over variables and equality
the condition that fields have to have at least two elements, or that 0 =/= 1, prevents the class of fields to be expressible as an equational class
here you go
equational classes of universal algebras form nice categories
but in absence of being an equational class, we might not have this
wet zodiac
analog zephyr
Hi
oblique river
let L/K be a field extension, when does the quotient L/K have a field structure?
The strongest sense in which a quotient L/K would make sense is as K-vector spaces
In which case the quotient is another K-vector space of dimension one less than dim_K(L)
analog zephyr
It's been a while since I stop studying math (group theory to be more specific). I feel like I'm forgetting everything
oblique river
Thats what happens when you dont regularly use things youve learned
But thats ok, its nothing to feel bad about
analog zephyr
Last thing I've studied was Factor Group
oblique river
If you want to relearn and if you have specific questions, you can ask them here
analog zephyr
I have many things to worry about. Job, house, payment many things
The problem is the time
But i hope I'll get the time for math again
charred crescent
i still don't really understand factor groups for some reason
can someone help?
sonic coral
i don’t understand how to show H is normal to K and K is normal to G
upper pivot
i don’t understand how to show H is normal to K and K is normal to G
for H is normal to K, you should just manually verify that, its not a lot of work.
sonic coral
okay, i knew i could do that but i didn’t know if that was the intended approach
sonic coral
i feel like it’s K
upper pivot
and dont worry about "intended approaches", thats never useful to think about.
sonic coral
but like if i go with what i believe to be the intended approach i would’ve just manually verified K being normal to G
and then i don’t think about things like your hint
upper pivot
well there is no harm in that either
if you tried manually verifying K is normal in G i think you probably would have eventually stumbled on what i said
sonic coral
guess we’ll never know
upper pivot
hmm i guess let me put it like this
if you have an idea just go ahead and try it i guess
dont hold yourself back due to it not being "intended approach"
thats all i wanted to say
sonic coral
fair enough. i agree with you. my issue was just that I knew i could easily do it that way, and was wondering if there was any other way to do it, which is exactly what you gave me
upper pivot
fair
carmine fossil
fair enough. i agree with you. my issue was just that I knew i could easily do i...
Well there's actually a more elegant way. If H is a subgroup of G such that [G:H]=2, H is normal in G
Actually I think the whole point of this exercise is to see why that's nice since manually showing it is kinda annoying
sonic coral
Well there's actually a more elegant way. If H is a subgroup of G such that [G:H...
i’m working toward that
it’s the bottom lemma
i’m at the part where i show that normality can be transitive, what group is H for this? is it the same H from b)? if so, then it’s also K
sonic coral
for the lemma then. I know that there are two distinct cosets
i don’t know how that helps me though
carmine fossil
Think of right cosets
Well also left cosets
Can an element of gH be in H?(g not in H)
That's your hint
flat treeBOT
Please read #❓how-to-get-help
summer path
merry pond
wrong channel
kk
solar shore
currently struggling with some homework
cloud walrusBOT
blanket
next obsidian
Try a nonabelian group 👍
solar shore
OH
next obsidian
I say this because
next obsidian
You can find one in a nonabelian group
So you should think “hmm what nonabelian groups do I know” and it’s probably only a few
And that’s good because you should be able to come up with an example in that
solar shore
i only know GL and SL atm
solar shore
i feel like i should know more but none are coming to mind
next obsidian
That probably works too
solar shore
is that symmetries?
ah
i dont think we went over it too much
i remember more D_n
ill try something with D_n
it has to be greater than D_2 right?
iirc D1 and D2 are the only abelian
next obsidian
Yeah
solar shore
gotcha
quick question
are D_k and D_m subgroups of D_n only when k and m are divisors of D_n?
id figure because something like a triangle has half the reflections of a hexagon
but im not sure
cloud walrusBOT
blanket
next obsidian
Umm…
I don’t this this is true
Because it is impossible to come up with an example in an abelian group
summer path
^proving this might be a worthwhile exercise
chilly ocean
I have an intro Lie Theory / Rep theory question:
basically I'm trying to find su(3) subalgebras of g_2 by looking at root diagrams
so this is my root diagram for g_2
and I found the first one which is just the inner hexagon but I am wondering if this below image can also be justified as a diagram of su(3)
Because its rotated and the arrows are kinda off from the typical diagram for su(3)
sage lodge
Can we show that every proper ideal is contained in a maximal ideal?
next obsidian
Yes
sage lodge
What would be the strategy?
next obsidian
Zorn
sage lodge
oh
lethal dune
🌽
chilly ocean
why is it that a subgroup H of G is normal if the index [G : H] = 2
quiet pelican
why is it that a subgroup H of G is normal if the index [G : H] = 2
Because gHg^-1 = H is equivalent to gH = Hg, but then if the index is two, both of those sets are just the complement of H in G
chilly radish
(If g is not in H, otherwise they're both H, clearly)
novel parrot
if $(c) = (b_1, b_2)$ then $c = u_1b_1 + u_2b_2$
cloud walrusBOT
ActiveChapter
novel parrot
why are u_1 and u_2 coprime?
tender wharf
what's this notation
chilly ocean
Ideals
novel parrot
$\langle c \rangle = \langle b_1, b_2 \rangle$ sorry
cloud walrusBOT
ActiveChapter
novel parrot
do you know why u_1 and u_2 are coprime ?
chilly ocean
I guess they mean something like how you can complete a space by using Cauchy sequences
But sometimes you can just embedd it into another space and take closure
The former being "intrinsic"
Analogously here, probably
novel parrot
@chilly ocean can you help me?
chilly ocean
Probably if I take an effort to
novel parrot
if you dont mind, im desperately stuck
chilly ocean
I'm guessing we are in a UFD?
chilly ocean
Even better ig
So coprime was defined as no prime divides both ig
So take a prime element p and suppose it divides both
Write b_1, b_2 in terms of c, divide by c
Conclusion, p divides 1
Contradiction
Yeah?
novel parrot
Write b_1, b_2 in terms of c, divide by c
wym here?
novel parrot
dont we need c to be unit
chilly ocean
Nah
novel parrot
can you write in latex?
chilly ocean
$cx = cy \implies x = y$
cloud walrusBOT
Blitz
pastel cliff
is there a nice way to see that the direct sum agrees with the categorical coproduct
well maybe agrees is the wrong word since it is the coproduct in the category of modules right?
chilly ocean
direct sum of modules then?
chilly ocean
I think you can just say that its the coproduct of abelian groups and all the morphisms are actually R-module morphisms
for a "nice" way to see it
pastel cliff
those kinds of arguments tend to fail me you're saying "it's like that in category of abelian group and that's the same as category of modules so it carries over" right
chilly ocean
yeah, well, and the fact that all those morphisms actually agree too in this case
other than that, you can just prove it directly
probably the best way to do it
pastel cliff
will probably do that then
elder wave
yeah i wasn't sure if you're asking for a way to verify the universal property or a conceptual approach to it
pastel cliff
yes 
elder wave
pastel cliff
chilly ocean
elder wave
try working out the case where you only take two Modules/Ab groups
pastel cliff
yeah i'll be back about that eventually
we just showed that dimension "makes" sense for vector spaces
and for R modules where R is comm. and has id/
and in the proof of it, which isnt too involved, prof used quotient of dir. sums = direct sum of quotients
which again, isnt too hard to see
elder wave
and for R modules where R is comm. and has id/
what do you mean here
for arbitrary R modules dimension is kinda
eh
if you direct sums of quotients were involved you were probably talking about length?
but you need some extra structure on your modules for that
pastel cliff
if R is a comm. ring w identity and R^n \cong R^m, => n = m
chilly ocean
hint: it's true in linear algebra
pastel cliff
i mean i just looked over it so doesnt count
i'll explain it later tonight before i review stuff again so it's not fresh in my memory
elder wave
Fun fact: Rings who satisfy this have their own name
Well not their own name
But the property is called IBN (Invariant basis number)
It doesn’t hold for all rings
pastel cliff
chilly ocean
but all rings are commutative and with identity
pastel cliff
surely
elder wave
True
Blunder on my side
Sebb I’m still wondering about the thing I tagged you earlier with
How did you show that dimension makes sense for general R-modules
pastel cliff
if R is a comm. ring w identity and R^n \cong R^m, => n = m
this was the statement
chilly ocean
with one exception. banach algebras dont need to be commutative. those guys are special
elder wave
Still this doesn’t seem related to the module
oblique river
with one exception. banach algebras dont need to be commutative. those guys are ...
No those are not rings, they are algebras
elder wave
That’s a property of the underlying ring
oblique river
Algebras need not be commutative :^)
chilly ocean
grrr
elder wave
(Im not trying to grill you im genuinely curious)
pastel cliff
i havent looked at this part yet but prof said the relevant isomorphism are all iso's of R/I modules
all good
here i can send his notes one sec
elder wave
The quotients of free modules might not be free
chilly ocean
when i asked for the proof i was honestly just hoping you'd say "quotient by a maximal ideal and apply the hint"
soooo long and technical
pastel cliff
yeah the hardest part of this class so far is digesting my profs notes only to realize something couldve been said much more concisely
and he does say it in class most times
but also goes so fast in lecture 
oblique river
Honestly the first part of the proof should just be its@own proposition
Homs M -> N descend to homs M/I -> N/I. An isomorphism is taken to an isomorphism.
vagrant zinc
Guys, a question I was reading about cyclic groups and then I get this example in which 5 is a generator of 12 but then, I know that the definition is given by
<a>={a^{n}/n in Z}
i tried but i can't figure it out
oblique river
It shouldnt really be included in this proof
pastel cliff
Guys, a question I was reading about cyclic groups and then I get this example i...
english speaking server
tender wharf
5 is coprime to 12 lol
vagrant zinc
english speaking server
look :'c my bad
pastel cliff
Homs M -> N descend to homs M/I -> N/I. An isomorphism is taken to an isomorphis...
i see
elder wave
okay but this is exclusive to free modules
vagrant zinc
5 is coprime to 12 lol
11mmmmmm
elder wave
now i see what's going on
novel parrot
$(c) = (b_1, b_2)$ then $c = u_1b_1 + u_2b_2$ in a PID
chilly ocean
this channel just cannot get a break
cloud walrusBOT
ActiveChapter
novel parrot
why is u_1 and u_2 coprime?
elder wave
i think this is something worth mentioning that dimension as a measure of "size" does not make sense for arbitrary modules
novel parrot
i need desperate help :2
pastel cliff
this channel just cannot get a break
#abstract-chillgebra
too many things fall under the scope of this channel no?
elder wave
not really
oblique river
i need desperate help :2
I thought someone already gave you a proof
vagrant zinc
You have the definition of the subgroup wrong
novel parrot
I thought someone already gave you a proof
i didnt understand them
vagrant zinc
This is what appears to me before they explain the Z12 generator to me
oblique river
The group operation on Z/12 is written additively
hot lake
the operation in Z/12 is +, not *
oblique river
Not multiplicatively
novel parrot
Bananas can you help
pastel cliff
**ActiveChapter**
what do you know about coprime elements of a PID
maybe that's not phrased correctly
start with the definition of prime elements probably
hot lake
you have conflicting notations for what the group is
oblique river
Yes but i dont want to just repeat what was already said
novel parrot
what do you know about coprime elements of a PID
if u1 and u2 are coprime then (1) = (u1, u2)
vagrant zinc
Not multiplicatively
aaah ok, I forgot about the operation, I'm very dumb 😦