cardinality
#groups-rings-fields
vital quail
prime cloak
We don't know
vital quail
i see
prime cloak
They tell us to use bezout equality
For m and a power of p well chosen
I have tried but I can't figure it out
untold sapphire
this makes $G$ finite
prime cloak
How ?
untold sapphire
if every element has a finite order then G is finite
vital quail
no
untold sapphire
(if its finitley generated)
prime cloak
(if its finitley generated)
How do you know that ?
vital quail
uhh
No. This is the General Burnside Problem -- must a finitely generated periodic (this is what you call "torsion": every element has finite order) group be finite? -- to which a negative answer was given by Golod and Shafarevich in 1964. Please see the linked article for more information.
in any case this is besides the point
prime cloak
Yep
vital quail
@prime cloak so to show that x |-> x^m is an isomorphism
prime cloak
I'm sorry to have sent you a french subject tho
prime cloak
Ok but
untold sapphire
```10
No. This is the General Burnside Problem -- must a finitely generated per...
TIL, whoops
vital quail
lol yeah
prime cloak
They make us show bellow that
vital quail
@prime cloak let's just say that x |-> x^r is your inverse to x |-> x^m
can you find such an r?
prime cloak
<@!184381299371409408> let's just say that x |-> x^r is your inverse to x |-> x^...
Ok
vital quail
i mean
try doing what i said earlier don't worry about it
that part is easy enough
prime cloak
How do I show x^r is in G
vital quail
x is in G
prime cloak
Ok
prime cloak
Yes I got it
But there is an issue
Why would they add m doesn't divide p if we don't use it
prime cloak
Ok nice
vital quail
you are trying to see if there's an r for which x |-> x^r is the inverse of x |-> x^m
what does that mean?
prime cloak
Composed left and right
vital quail
if we assume x |-> x^r is indeed an inverse
vital quail
well yes but tell me specifically what that means wrt an element x in the group
let the homomorphisms be M, R respectively
M(g) = g^m and R(g) = g^r
okay i mean you said it already so i will just notate it
it just means that for all x in G, R(M(x)) = R(x^m) = x^mr = x
(the same as composing the other way of course)
prime cloak
Y
vital quail
why?
because in the exponent you have integers and your normal integer multiplication
which is commutative
@prime cloak so then if x^mr = x what does that mean
prime cloak
Mr = 1 ?
vital quail
not necessarily
vital quail
could be (1 + ord x)
prime cloak
Ok
vital quail
x^(1+ord x) = x*x^(ordx) = x * 1 = x
prime cloak
Ok ok
vital quail
and similarly x^(1 + k*(ord x)) = 1
since it's x*(x^ordx)^k
so all that means is that mr = 1 (mod ordx)
prime cloak
Ok
vital quail
now what do you know about p-primaire groups
prime cloak
But wait I don't understand
When you compose both reciprocal morphisme you get neutral element
Not x ?
vital quail
when you compose inverse homomorphisms you get the identity homomorphism
prime cloak
Ok
prime cloak
I get it
vital quail
now what do you know about p-primaire groups```so now this
you know mr must be 1 (mod ord x)
recall the definition of p-primaire @prime cloak
prime cloak
There exist a k such that x^p^k=e
Ok and how do we use this
From our equality mod (order of x)
vital quail
okay so in general i want you to prove that x^n = 1 means that ord x | n
for any group and any x in that group
prime cloak
So x|p^k ?
prime cloak
Ok
vital quail
i already kind of gave the idea above
but just do it here
it's self-contained
x^n = 1 implies n is a multiple of (ord x), why?
hint: write n as a quotient plus a remainder
prime cloak
Isn't this in the definition of a cyclic group
vital quail
i dont know what you mean
this holds for any group
not just a cyclic one
okay, what that hint means is: write n = q(ord x) + r where 0 <= r < ord x
now try expanding x^n
@prime cloak
prime cloak
Ok
vital quail
now can you simplify x^(q*ordx)
prime cloak
Into x ?
vital quail
i.e. does writing it as (x^something)^something
prime cloak
^q
vital quail
in one of the two ways possible
help simplify it
you can either write it as (x^q)^ordx or (x^ordx)^q
which way simplifies it?
prime cloak
2nd ?
prime cloak
x
vital quail
no, 1
prime cloak
Lol yes
vital quail
and then 1^q is what
prime cloak
Sorry didn't think
vital quail
its ok
prime cloak
1 ?
prime cloak
Ok
vital quail
now remember that r < ord x
and ord x is defined to be the minimal positive number such that x^(ord x) = 1 when x has finite order
so what does that force r to be
@prime cloak so remember
0 <= r < ord x
prime cloak
n
vital quail
can r be positive?
can r be positive
r < ord x
ord x is the smallest positive integer such that x raised to it is 1
prime cloak
but it cant be
vital quail
r can't be positive
prime cloak
0
vital quail
yes
correct
so that means
n = q*ordx + 0
= q*ordx
so
we have showed that n is a multiple of ordx
that is all then
prime cloak
yes
vital quail
Okay so now we can use this
So we have mr = 1 (mod ord x) in our problem
And we know that there's a k >= 1 such that x^p^k = 1
What does that say about ord x?
prime cloak
it is less than or equal to p**k
vital quail
something more
what did we just prove?
we proved that x^n = 1 means n = q*ordx
(for some q)
prime cloak
ord x divides n
prime cloak
p**k= q* ordx
vital quail
yes, and what can q be?
on the left we have only factors of p
so on the right we only have factors of p
prime cloak
oh ok
prime cloak
where are you bringing me ... too much information
vital quail
right so
we have x |-> x^m
we are hoping that we can find an r such that x |-> x^r is our inverse homomorphism
that is equivalent to mr = 1 (mod ord x)
now, ord x is some power of p, and recall that m is coprime to p
prime cloak
ok
prime cloak
yes
vital quail
indeed
as p^t only has powers of p as factors
but m doesn't have any p factors
so they share no factors
prime cloak
yeah
vital quail
indeed, m is coprime to p, so there's an r for which mr = 1 (mod p)
so is mr = 1 (mod p^t)?
uh yeah not necessarily
hm
prime cloak
I have a question
How do you know where to go and how to go there ?
@vital quail what about tryingto show that x**m=e is the same as x=e
so it is injectiv
vital quail
right, you can try to show bijectivity
just a sec
there's kind of an issue with my original idea
let me see
prime cloak
and find x such that x**m=y
vital quail
oh okay nevermind
so
what we did is good
but now we should use this instead to show bijectivity
@prime cloak
i see now how to do it
prime cloak
use my thing ?
vital quail
well first let's show surjectivity
prime cloak
call it y please
prime cloak
ok
vital quail
now what might we take x to be
prime cloak
y**-m ?
vital quail
that would give 1
prime cloak
+1
vital quail
or rather not 1 lmao
prime cloak
y**(-m+1)
prime cloak
+1/m
vital quail
look, let's think about it like this
prime cloak
y**1/m) would work
vital quail
not necessarily
prime cloak
ok
vital quail
look
can you just find an r such that y^r = x^mr
or uh rather what i wanted was
x^mr to be x
in which case by raising both sides to r we would get x = y^r
so is there an r such that x^mr = x?
remember that this is equivalent to mr = 1 (mod ord x)
right but does m have an inverse mod ord x
that's the question
i.e. is m coprime to ord x
is m coprime to p^t?
since ord x is just a power of p
remember that m is coprime to p
you already answered this above actually
remember?
ord x has only p factors, m has no p factors, so they have no common factors
prime cloak
yeah
vital quail
okay then that means that indeed, there exists an r such that mr = 1 (mod ord x)
so then y^r = x^mr
= x
right
since y = x^m
so y^r = x^mr
and mr = q_ord(x) + 1 so that x^mr = (x^ordx)^q*x = 1*x = x
okay?
prime cloak
same thing as b4
vital quail
yes
so then x = y^r is our element that gets mapped to y under the map
hence the map is surjective
right
prime cloak
what is r
prime cloak
ok i get it
sturdy marsh
What's the problem?
prime cloak
injectiv now ?
sturdy marsh
If you show injective you dont even need to show that it is surjective
the map sends each element to some other element in the cyclic group generated by that element
which is finite
and injective from finite set to itself implies surjective
vital quail
that's true
but it's probably more helpful for chakou to prove things like this
just for practice at least
woven obsidian
I'm trying to show that k[X,Y]/(Y^2-f) with f in k[X] is normal
With f squarefree
In the case char(k) neq 2 it is easy using either minimal polynomials or involutions. But I can't seem to handle the case char(k)=2
A necessary and sufficient condition for a+bsqrt(f) with a,b in k(X) being integral is that a^2-b^2f is in k[X], but I'm not sure where to go from here
sturdy marsh
idk if it works in char 2
woven obsidian
Hmm really?
It's an exercise in Reid's commutative algebra, 4.6, but he doesn't mention the characteristic
I think it works for "simple f" e.g f=X^3
woven obsidian
Well that is not squarefree, but the point is that its normalization is just k[X]+Y/X k[X]
I.e we just divide away the square
Because we have a nice parametrization of the curve right?
By letting t=y/x. We get t^2=x. T^3=y And k[t] is a ufd etc so the normalization is just k[t]
I don't think I use the characteristic neq 2 in this case
woven obsidian
Yeah I know, but we can reduce to the case f squarefree by doing as I did above
So the point is just knowing that we're "done"
woven obsidian
Hmm that's true
sturdy marsh
yeah so the thing is 'smooth'
so it should be true
actually idk
char 2 always makes me uneasy
chilly ocean
Suppose that an Abelian $p$-group $G$ can be written as a direct sum of countably many finite cyclic groups. Let $a_1,...,a_k$ be generators of $k$ distinct summands of this direct sum, and let these elements have orders $p^{n_1},...,p^{n_k}$ respectively. Then for what values of $m$ are $a_1,...,a_k$ linearly independent over $\mathbb{Z}_{p^m}$?
cloud walrusBOT
lugita15
compact needle
I'm trying to show that k[X,Y]/(Y^2-f) with f in k[X] is normal
It's not true in characteristic 2. Consider the example f=X^3+1, then Y^2-f = (Y+1)^2-X^3, so it's isomorphic to k[x,y]/y^2-x^3. Over an algebraically closed field, you can always shift f -> f+a to make it not squarefree, and then Y^2-f=(Y-sqrt(a))^2-(f+a) will have a singular point at Y=a,X=sqrt(a).
*X=doubled root of f+a
thorn delta
yo does anyone have a hint on this?
sturdy marsh
f has a multiple root iff f and f' have a common factor
thorn delta
hmm but that doesn't immediately contradict irreducibility does it? Because the common factor might not come from F[x]?
sturdy marsh
gcd doesnt care about field
thorn delta
yeah, exactly hmm
sturdy marsh
so if they are not coprime over C, then they are not coprime over F
as f is irred, this implies f' is an element of F
(as the degree is strictly lower)
thorn delta
so if they are not coprime over C, then they are not coprime over F
this isn't obvious to me...
as f is irred, this implies f' is an element of F
as in f' has degree 0? This isn't clear to me either
sturdy marsh
For the second thing, if f' wasnt 0 then that contradicts irreducibility
try proving the first thing
oblique river
you don't even have to work over F here. F is a subfield of C, so we can just think of f(x) as an element in C[x] anyway
especially since the problem is asking about roots in C, not roots in F
brofibration is correct that it's all equivalent, but based on how the problem is stated, it's probably easiest to just forget about F all together
sturdy marsh
the irreducibility condition is over F
if it were over C it is trivial
as linear ones are the only irreds
oblique river
oh -- you're right
sturdy marsh
the gcd thing is useful to know
oblique river
agreed
I think it's easier to prove "coprime in F implies coprime in C" than the converse though
(as a hint for kxrider)
thorn delta
oh okay i see. if gcd(f, g) = (unit) in F[x] then fq + gr = 1 for some q, r in F[x]. It follows that any factor coming from a larger field that divides f and g would have to divide 1
oblique river
that's it!
chilly ocean
Suppose that $G$ is an Abelian group, and $X$ is a subset of $G$ with the property that whenever $x_1,...,x_n\in X$, $k_1,...,k_n\in\mathbb{N}$, and $k_1x_1+...+k_nx_n=0$, we must have $k_1=k_2=...=k_n=0$. Then is $X$ linearly independent over $\mathbb{Z}$?
oblique river
isn't that the definition of a linearly independent set?
there is no linear dependence
chilly ocean
there is no linear dependence
Read it carefully, you’ll see a distinction between $\mathbb{N}$ and $\mathbb{Z}$.
oblique river
oh, then no it's false and there's an extremely easy counterexample
G = Z, X = {1,2}
chilly ocean
oh, then no it's false and there's an extremely easy counterexample
Oh right, thanks.
chilly ocean
Does the Prufer p-group contain any subgroups which are isomorphic to a direct sum of finitely or countably many finite cyclic groups? https://en.wikipedia.org/wiki/Prüfer_group
woven obsidian
It's not true in characteristic 2. Consider the example f=X^3+1, then Y^2-f = (Y...
I see, thanks. But to find the normalization for "simple curves" is it still possible to parametrise?
E.g is normalization of k[X,Y]/(Y^2-X^3) still k[t] with t=y/x?
cloud walrusBOT
inhyak
oblique river
not always, no
if there is such a map, we say that the map G --> G/H "splits"
(well, more precisely, if the composition G/H --> G --> G/H is the identity, then we say the map G --> G/H splits)
no
Just because G/H --> G --> G/H is the identity doesn't mean the map is anisomorphism
becaus we're not requiring that the other composition G --> G/H --> G is the identity
np!
woven obsidian
Say that I have an ideal in k[X,Y,Z] like (X^2+YZ,Z^2X,Y^3)
Now I want to intersect this with k[X,Z], my first instinct is to just put Y=0 in the generators, but that's not correct right?
Now that I think about it, it seems like it might be correct. Please correct me if I'm wrong
Usually when you just do things with generators things doesn't work out
golden pasture
Now I want to intersect this with k[X,Z], my first instinct is to just put Y=0 i...
this is essentially like quotienting with (y) so yes it works
if you're super paranoid can do things very explicitly
every element is
a(X^2+YZ)+bZ^2Y+cY^3
taking mod Y we get
etc.
latent anvil
it's not quotienting out by Y though?
latent anvil
Consider (x^2-y, x+y). The intersection is (x+x^2) but the quotient is (x), I think
golden pasture
hmm true
latent anvil
Usually when you just do things with generators things doesn't work out
this is the right intuition imo
but maybe that's just because my brain is small
woven obsidian
Yeah but it felt like it might work in this case
Ah I see why it doesn't work now
latent anvil
I don't see why it would contain x^2
woven obsidian
Is there a strategy for projecting that doesn't involve gröbner bases?
cloud walrusBOT
lugita15
chilly ocean
Never mind, I figured it out.
Suppose you have two Abelian groups $G$ and $H$ and you know that their torsion parts are isomorphic. What additional information do you need to conclude that $G$ and $H$ are isomorphic? What else needs to be the same about them?
cloud walrusBOT
lugita15
woven obsidian
If I is a polynomial ideal and {g1,..,gn} a subset of I whose leading terms generate the ideal of leading terms in I, is it then true that the gi:s generate I also?
hot lake
yes I think so
next obsidian
this looks like an exam
white nymph
@next obsidian it's an old prelim exam I'm studying from
next obsidian
Ah
IIRC I just did this explicitly
Just like conjugate it and then sneak in an e via something like gg^-1
viscid pewter
i think it is?
next obsidian
Or it might be, but it isn't waht you reallly want to be looking at
you want to look at the stuff generated by it
since this is just supposed to be the commutator subgroup I think?
next obsidian
exactly
next obsidian
I mean maybe if you assume it's closed
viscid pewter
meh
next obsidian
then this is the commutator subgroup??
viscid pewter
yee
white nymph
yeah i proved its a subgroup assuming it was closed
next obsidian
I mean if it's closed then
white nymph
as instructed
next obsidian
isn't it immediate
viscid pewter
it is
next obsidian
I guess you need to know inverses
next obsidian
I'm pretty sure normalcy is like
white nymph
(xyx^-1y^-1)^-1 = yxy^-1x^-1
white nymph
yeah that was my first attempt and I could not show it to be of the form of the members of H
viscid pewter
yeah, when i was first going through commutator subgroups i had to google the proof i think
next obsidian
It's not like this is clearly a kernel of anything IIRC
white nymph
i also tried sticking in identity in various places but could not figure out a clever way to do so
as one of you suggested
next obsidian
I mean you could try showing zH = Hz
viscid pewter
okay so the notation is just [x, y] = xyx-1y-1
imma just spitball a little here
actually no
hmm
next obsidian
Okay so
H is the intersection of all normal subgroups N for which G/N is abelian
like
This isn't obvious
But it is a true fact
white nymph
lol
next obsidian
And if you can show this abstractly you can show H is normal like that
viscid pewter
mathe
next obsidian
normally you'd show H is nomral
viscid pewter
okay there was an explicit proof tho
next obsidian
I mean okay so if G/N is abelian then
next obsidian
(a + N)(b + N) = ab + N = ba + N = (b + N)(a + N)
then you get that aba^-1b^-1 is in N
so H is a subset of all N for which G/N is abelian
Showing the other direction is where it might be harder
viscid pewter
okay so
next obsidian
Like normally the other direction is just "H is norml and G/H is abelian"
viscid pewter
not assuming anything now
g[x, y]g-1 = [x, y][[x, y]-1, g]
okay that is not as clear as it is when in handwriting
if u is in H
gug-1 = u[g, u-1]
so
[g, u-1] = v is in H
uv is in H so we're done?
next obsidian
I got it
Either what you said
or do the following
zghg^-1h^-1z^-1 = zghg^-1z^-1h^-1hzh^-1z^-1 = [zg,h][h,z]
viscid pewter
pain
white nymph
brutal
next obsidian
lol
viscid pewter
i couldn't have come up with my own one
this is just what i wrote down after googling it last time
next obsidian
I just wrote shit out
and kept adding until it worked
lol
I think I've proven this 3 times
white nymph
so for gug-1 = u[g, u-1], u = [x,y], right? for x,y in G?
next obsidian
each time doing this exact calculation
white nymph
if so, i dont follow how gug-1 = u[g, u-1]
next obsidian
Just do my calculation :^)
viscid pewter
no
white nymph
can we start fresh somewhere, its been hard to follow two different ideas
viscid pewter
okay
shhh
watch this
gug-1 = gv-1g-1 let's say
v in H
gv-1g-1 = v-1(vgv-1g-1) = v-1[v, g] = u[u-1, g]
next obsidian
I think that overcomplicated it a bit, what you originally wrote had g and u^-1 backwards
you can just see that
u[u^-1,g] = uu^-1gug^-1 = gug^-1
viscid pewter
fine
white nymph
i see
thanks much guys
what did you google to find the stackexhcnage on this? commutator subgroup?
viscid pewter
something like commutator subgroup normal
white nymph
right on - thanks
chilly ocean
Hey, does anyone know any theorems on 3 deminsional dihedral groups
viscid pewter
they aren't dihedral any more
they're called coxeter groups, i think
well not quite
the symmetries of the three dimensional regular polyhedra are a type of coxeter group
as are the symmetries of the two dimensional regular polygons
chilly ocean
oh
viscid pewter
they're pretty neat, reading about them on wikipedia
haven't actually followed the heavy maths tho
chilly ocean
were trying to do a project on "dihedral groups" and we thought maybe we can try and explore more than 2 dimensions
but thanks, yeah i wasnt finding much on "3 dimensional dihedral group" maybe because... THEY DONT EXIST
viscid pewter
yeah lol
vital quail
@next obsidian @viscid pewter so you know the 5/8 problem that's like "show that if the probability two randomly selected elements in a finite group commute is greater than 5/8, the group is abelian"
vital quail
there's a proof that relies on [G:Z] < 4 forcing G abelian, and then with the constraint that [G:Z] >= 4 you can bound a certain sum that gives that probability and get that it's <= 5/8
which basically relies on using Z(G) as a very coarse measure of the abelianness of G
however you can get a much more refined measure using commutator subgroup
as you can see here
vital quail
ah i see
woven verge
Just a nobby question, are ideals consists only integers? And not rational?
thorn delta
no, ideals are special subsets of rings
woven verge
Oh, so if a ring is rational then it subset the ideals must be rational also?
thorn delta
what does it mean for a ring to be rational?
woven verge
Rational is constructed from the integers? Idk about rings
thorn delta
do u literally mean the rational numbers like 1, 1/2, 34/453,...? The rational numbers are a field, so its only ideals are 0 and the whole thing
woven verge
do u literally mean the rational numbers like 1, 1/2, 34/453,...? The rational n...
So the rational has also a zero ideal?
Now I get it. Thanks
thorn delta
hmm im not sure we're on teh same page here. Every ring has a 0 ideal 
vital quail
and they have ideals
whose elements are of course also matrices
so you can see, it's a more general concept
@woven verge
woven verge
Yeah, I’m reading Wikipedia about the bezout identity and Euclid algorithm. I am starting on learning abstract algebra
Ok. Thanks
white nymph
I'm looking for help/hints on part b.). It says to recall that $L_\mathbb{Q} (K)$ is isomorphic to $M_2(\mathbb{Q})$, but that is not something that I know off hand/is not obvious to me. I figure if I were to know such an isomorphism, I could inject $K$ into $L_\mathbb{Q} (K)$ via $\varphi$ from part a.) and then I could restrict that unknown isomorphism to $\varphi(K)$ and get something in $M_2(\mathbb{Q})$ isomorphic to $K$, but I'm sorta at a loss at the moment.
If there's a better way to go about this/think about this as well, I'm all ears.
cloud walrusBOT
EricW
chilly ocean
I'm guessing L_Q(K) is isomorphic to M_2(Q) because it is linear transformations on a 2 dimensional space
white nymph
yeah that makes sense to me - i was just wondering if there was an explicit isomorphism i could use because I was unable to figure anything out by just fiddling around aimlessly. i think i lack some conceptual understanding for this problem
like 1 and sqrt2?
im still at a loss here - not even sure what question to ask
right so if we called such a map f
f(1) = sqrt2 so corresponds to column vector or whatever for matrix representaiton [0,1]
and f(sqrt(2)) = 2 corresponds to 2,0
so we get matrix
0 2
1 0
?
am i following at all?
cool
i dont understand what to do from there
i see we matrix represented one linear map
yes
hmmm i follow the notion - i get worried because it says to "explicitly describe a subring" and i was imagining i would have a set of matricies that i would then show is a subring
astral galleon
is the zero ring not a subring of \mathbb{Z} since there's no identity element present ?
white nymph
perhaps i can do that with this [f(e1), f(e2)] idea tho?
alright - ill give it a whirl
white nymph
I'm uncertain about the closure of the subring. does it just follow from $\varphi$ being a ring hom as shown in part a.)?
or do i need to do something to show that $M(\varphi(\alpha)) * M(\varphi(\beta))$ is still in the subring (for $\alpha$ and $\beta$ in $K$?
cloud walrusBOT
EricW
white nymph
as well as + being closed ofc
cloud walrusBOT
wstayman
next obsidian
Classification of finite abelian groups
cloud walrusBOT
wstayman
next obsidian
but yes
I think there's only 4 possible things it can be by the classification
so you can just enumerate them all
white nymph
ahhh of course, thanks
the idea of playing around with a set of matrices was stuck in my head
fading wagon
how can we visualise symplectic groups? Any geometric intuition we can rely on?
sour plume
The only way I ever got intuition for symplectic groups is by accepting their importance in physics, see here: https://math.stackexchange.com/questions/3614247/what-is-symplectic-geometry
Algebraically they kinda feel sensible, because they're somewhat of a "skewsymmetric counterpart" to the orthogonal groups, and past SL_n and SO_n's, their Lie algebra is the last remaining one in the classification of simple Lie algebras, so they're undoubtedly important
but visualizing them without physics, i don't know; especially since the first "interesting" symplectic group shows up in dimension 4, since $Sl(2) = Sp(2)$
cloud walrusBOT
Lartomato
fading wagon
hmm I only seen PGL and PSL groups so far
idk how the finite analogues would work though
cloud walrusBOT
lugita15
chilly ocean
Never mind I figured it out.
astral galleon
$\textbf{Problem (ii)}$
Generalize Corollary $2.131$ by provin that if the prime factorization of an integer $m$ is $m= \Pi_n p_n^{e_n}$, then
$$\mathbb{I_m} \cong \mathbb{I_1} \times \cdot \cdot \cdot \times \mathbb{I_m}$$
Before getting to desert make things a little more clear for those who don't have the book on hand. Futhermore observe that,
$$U(\mathbb{I_m}) \cong \prod_{m}U(\mathbb{I_m}) \iff U(\mathbb{Z}/ m \mathbb{Z}) \cong \prod_m U(\mathbb{Z}/m \mathbb{Z})$$
$$ , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , \ , , , , , , , , , , , , , , , , , , , , , , , , , , , , , \iff { [a] \in \mathbb{Z}/m \mathbb{Z} } \cong \prod_m U(\mathbb{Z}/m \mathbb{Z}).$$
Now we consider the mapping $f:U(\mathbb{Z}/m\mathbb{Z}) \rightarrow U(\mathbb{Z}/p_{1}^{e_1}\mathbb{Z}) \times \cdot \cdot \cdot \cdot U(\mathbb{Z}/p_{n}^{e_n}\mathbb{Z})$ where our mapping is defined by $f(a) = [a]_m$ Now we conjecture that;
$$f(U(\mathbb{Z}/m\mathbb{Z}) \cdot U(\mathbb{Z}/m\mathbb{Z})) = f(U(\mathbb{Z}/m \mathbb{Z})) \cdot f(U(\mathbb{Z}/m\mathbb{Z})) , , , , (*)$$
cloud walrusBOT
Zophike1
astral galleon
Is this the right way to start $\textbf{Problem (ii)}$ ?
cloud walrusBOT
Zophike1
sturdy marsh
wait what are you trying to do?
astral galleon
I'm trying to show that it's an isomorphism by showing it's an homomorphism since homo implies iso @sturdy marsh
sturdy marsh
are you trying to prove the chinese remainder theorem?
and what's U(Z/mZ)
or U(I)
astral galleon
Units of the Multiplicative group of integers modulo n
sturdy marsh
ah okay
are you trying to show that the units of Z/mZ is iso to product of units of Z/p_i^e_i?
astral galleon
Yes !
sturdy marsh
okay
astral galleon
@sturdy marsh is it not clear ?
sturdy marsh
the notation is not what im used to
astral galleon
okay okay yeah that's why I sort of rewrote the problem
sturdy marsh
anyway, yeah define the obvious map and check that it is an iso
or if you're feeling fancy, the functor U(_) is a right adjoint
so it commutes with products
astral galleon
yeah i'm trying to do it by showing it's homomorphism therefore it's an isomorphism
took me hours to come up with the right path to proof it 🙂
sturdy marsh
just proving that it is a homomorphism doest imply that it is an iso
astral galleon
oof fair point
sturdy marsh
but it is not hard in this case
astral galleon
Wait but I thought Homomorphism implies Isomorphism ? @sturdy marsh
Or is it the other way around
sturdy marsh
No, an iso is a homomorphism with an inverse that is also a homomorphism
isos are homos
not the other way around
next obsidian
If you have a finite module M can you always take a projective resolution by finite projective modules?
I know you can do so if A is Noetherian, by starting with a finite free thing, then each kernel is finite, so we can just take a finite free resolution
cinder bone
can someone check this for me 🥺
I know the solution is right, but I tried a different way than my professor
thorn delta
it seems okay to me
viscid pewter
the size?
thorn delta
as a hint, a matrix is invertible if its columns form a basis for the vector space in question.
thorn delta
hm yea that could work here i guess
leaden finch
hmm
thorn delta
going along with my first hint, if you wanted to build a basis for Z2 x Z2, how many choices do you have for the first basis vector? (this is going to be a combinatorial argument)
viscid pewter
paaain
cinder bone
@thorn delta thanks king
thorn delta
bro exactly, either argument is ez
viscid pewter
oh fine if it's easy
thorn delta
i shouldn't say easy. I really mean most methods are going to be about an equal amount of work
sturdy marsh
If you have a finite module M can you always take a projective resolution by fin...
no
next obsidian
Yeah I talked to Shamrock and we figured it out aka he did
turns out if this is true then A is Noetherian lol
sturdy marsh
turns out if this is true then A is Noetherian lol
right, you just look at R/I for I not finitely generated
sturdy marsh
there's a pretty standard way to count the number of invertible matrices over finite fields
Look at how many options you have for the first column, second column, ...
the nth column shouldnt be in the span of the previous ones
for all n
stone fulcrum
@leaden finch
What are you asking?
woven verge
Let $p$ be a prime number, and $m$, $n$ be a non-zero integers, such that $p$ divides $m$, $n$ or $p|nm$ => $p|n$ or $p|m$
Proof: Assume that $p$ does not divide $m$ then the gcd of $p$ and $m$ is 1, and there exists integers $a$, $b$ such that: $1=ap+bm$, multiplying by $n$ we get: $n=apn+bmn$ But $mn=pc$ for some integer $c$, $n=(na+bc)p$ and $p$ divides $n$. I think this is obvious for you guys, but simple question is $(na+bc)=m$?
cloud walrusBOT
QQWWWWWEEE
vital quail
@leaden finch it would help to think about it in full generality
i.e. compute |GL(n, p)|
@chilly ocean #❓how-to-get-help
@chilly ocean 🇨🇦 🖕
carmine fossil
Why should (na+bc) be m?
carmine fossil
Let $p$ be a prime number, and $m$, $n$ be a non-zero integers, such that $p$ di...
This
vital quail
the phrasing of that is so weird
but i assume it's just p | ab => p | a or p | b in Z
this should go in #elementary-number-theory @woven verge
woven verge
Ok
carmine fossil
How do I show commutative local rings have a multiplicative identity?(defining commutative local ring as a ring with a unique maximal ideal)
Or is existence of multiplicative identity part of the definition?
latent anvil
I think 4Z would be a maximal ideal of the ring 2Z if you don't require them to have units
carmine fossil
Is 2Z considered a local ring?
vital quail
2Z is not considered a ring 🤔
carmine fossil
Technically rng,But in some texts,it's consider to be a ring
latent anvil
Oh yeah arch good point
I should've clarified by something something about not requiring them to have 1
next time I guess
Oh also I think I meant to say 2Z/8Z has a unique maximal ideal 4Z/8Z, sorry
chilly ocean
you think you can discriminate against rings like that?
what if i say a human without a unit is not a human
that's pretty discriminatory
in fact, degrading
leaden finch
can someone help me with Latex
uncut girder
Definitions are arbitrary
cloud walrusBOT
viscid pewter
what even is this
sharp sonnet
what
delicate bloom
probably they're asked to find all abelian groups of order 5^4
sharp sonnet
probably of order 675
chilly ocean
Yes
sharp sonnet
those are not all of them
viscid pewter
same
sharp sonnet
also this claim of isomorphism
chilly ocean
Use the fundamental theorem of finite abelian groups to describe all abelian groups of order $675 = 5^{2} \cdot 3^{3}$
cloud walrusBOT
inshallah
chilly ocean
Z_675 contains element of order 225
Is there an easier way to recognize these I’m so bad at just knowing the numbers, like ik it’s x 3
Am just terrible at like looking for these things
carmine fossil
Just use the fundamental theorem of abelian groups?
viscid pewter
i mean
okay i'm going to look really really dumb if this is wrong
but z5 is like 0, 1, 2, 3, 4 right
so the proposition is clearly wrong?
steep hull
Multiple of 5 should be multiple of 4
paper flint
Oh, okay. Does my proof seem fine then?
viscid pewter
uhhh i think you might also want to consider an element y with order 5 and see how they interact? not sure
paper flint
Unless the group is Abelian, it doesn't help much, since $$(xy)^5=(xy)(xy)(xy)(xy)(xy)$$in the general case.
cloud walrusBOT
Ted
paper flint
I don't think I can extract anything sensible from this unless the group is Abelian.
paper flint
If I do have to consider two elements, I'll have to reframe my proof I guess. I'll try that, thanks!
viscid pewter
np
carmine fossil
You can't say anything about xy in the general case
viscid pewter
if x^a*y^b = e we don't need to bother
if it doesn't and the order of x^a*y^b = 5, what does that imply
paper flint
You can't say anything about xy in the general case
Is it justified to leave the general xy case, then?
steep hull
You don’t need to worry about interactions. Just take the set of all cyclic subgroups of order 5. They have trivial intersection and each one contributes four elements.
You proof is basically right Ted
paper flint
Oh, okay. Thanks both of you. 😄
chilly ocean
Well if Ted is all done, here's a quick one. Does this answer look good?
paper flint
Looks good to me, although I'd use some parentheses when or/and are being used in a single line.
chilly ocean
ok cool. thanks ❤️
paper flint
So the first implication may look better as $$x\in A\text{ or }(x\in B\text{ and }x\in C)$$
cloud walrusBOT
Ted
chilly ocean
yeah gotcha 🙂
paper flint
Also, this may be better suited to #proofs-and-logic or #discrete-math , although I'm guessing you encountered this in prelims of an abstract algebra book.
sturdy marsh
@next obsidian Do you know which book your class on stacks is going to follow?
Is it the one by Olsson?
next obsidian
Yeah
sturdy marsh
👍
delicate bloom
you can check by thinking through the frobenius automorphism
next obsidian
Then Harris for moduli of curves
delicate bloom
you're just squaring so it's not too bad
@chilly ocean
there's only one finite field with p^f elements, and they are all isomorphic by relating the elements with a field automorphism
maybe don't worry about it, they'll cover it if it's important lol
delicate bloom
I didn't check because someone else already did
next obsidian
It was
delicate bloom
I thought archsys answered you
next obsidian
I was replying to brofibration
delicate bloom
looks like he deleted his message or I'm hallucinating
fierce perch
@next obsidian is it going to be online? I heard a rumor the lectures would be ”open”?
next obsidian
It's online, idk if it's open
viscid pewter
what is this
paper flint
To prove that the subgroup H of a cyclic group G is necessarily cyclic, is it sufficient to show that every element in G generates a subgroup? Or should I try to begin with an arbitrary subgroup and work with that?
chilly ocean
every element in G will generate a subgroup because that's how cyclic groups work; what you have to show is the converse, that every subgroup (in a cyclic group) is generated by some element
you should begin with an arbitrary subgroup
paper flint
Okay, thanks!
chilly ocean
you can try considering some simple examples which might help you get started with your proof
like the integers and their subgroups
paper flint
Yes, will try that. I was thinking about integers.
paper flint
I cannot. 
Working with an arbitrary subgroup, I picked an element x and showed that by closure all multiples of x(and their inverses) are in the subgroup, but how do I show if a given element in the subgroup actually generates the entire subgroup?
For example, the set of even integers is generated by 2, but it also subsumes the subgroup generated by 4, 6, etc.
sharp sonnet
what do you know about the elements in a subgroup of a cyclic group? @paper flint
paper flint
what do you know about the elements in a subgroup of a cyclic group? <@!69784846...
All of them generate a subgroup?
The union of subgroups generated by all the elements of a (cyclic) group is the group itself?
I'm sorry if I'm not making sense, I'm kinda clueless :3
sharp sonnet
how does a cyclic group look like?
marsh fractal
So to show this
I get why its true, but not sure how to prove it with just simple algebra
like a-b = nq for some q in Z
sharp sonnet
do euclidean division by n on an arbitrary element of Z
marsh fractal
ok i kind of get the line of thinking ill go with that
so like a = b(n-k) + qn
with 0 >= k >=n
vestal snow
How do we get the last equality?
I think the book is using the strict triangle inequality, but to do that we need that v_P'(y_1) is not 0
And I don't see why that must be the case
next obsidian
To prove that the subgroup H of a cyclic group G is necessarily cyclic, is it su...
If you have Lagrange, any subgroup divides the order of the cyclic group. So let G by the cyclic group of order n, and H be a subgroup of order d | n. Using a generator of G, you can produce an element of order d by taking suitable powers.
Now any h in H has order dividing d, you can show there’s exactly d of these by looking at powers of the generator of G and figuring how which elements have order d. This is enough to tell you any subgroup of order d is unique since they necessarily have exactly the same elements as you only have d elements that could be in it.
Now take your element h of order d, consider that the cyclic subgroup <h> has order d, then <h> = H
I think there’s other ways to do this but this is one which uses minimal theory
marsh fractal
do euclidean division by n on an arbitrary element of Z
im still not so sure about this
sharp sonnet
I think there’s other ways to do this but this is one which uses minimal theory
its not, the minimal way is to notice that every element in the subgroup is of the form g^x for a generator g of G, then take the smallest positive integer of the x's and show that this generates the subgroup
sharp sonnet
im still not so sure about this
what's the issue?
i think your formulation of euclidean division was wrong
if you divide an arbitrary integer a by n, you get a = qn + r with 0<=r< n
now qn is in nZ
and hence [a] = [r]
marsh fractal
ok yeah didnt know the 0<=r<n is assumed
sharp sonnet
that's just how euclidean division works
if r were not in this range, you could modify q accordingly
marsh fractal
yeah is there a point in showing that it works that way
sharp sonnet
what do you mean? euclidean division is really useful
in this case it lets you construct a unique representative for every element of Z/nZ
(you also need a bit more but wtv)
marsh fractal
no agreed, i mean should one show that 0<=r<n
like it make sense obv with them just being a multiple of a remainder but should this be shown
sharp sonnet
one should absolutely show this at some point
but i think most texts on algebra do this
marsh fractal
ok
sharp sonnet
as you need euclidean division all the time
marsh fractal
thats maybe where im stuck
sharp sonnet
you essentially look at the set of numbers a - nq as q ranges over the integers
then there is a smallest positive element, which will be your r
there is a proof on wikipedia, that is a bit more algorithmic https://en.wikipedia.org/wiki/Euclidean_division#Proof
marsh fractal
ok ty that makes sense
next obsidian
If (R,m,k) is a DVR which contains its residue field [which I think you have an embedding k -> R -> R/m = k which is the identity on k], I think you get a height 1 maximal prime in R[x]
Does anyone have a single redacted idea on how to show this
Okay, so new idea, m = (r) for some r in R, can we show that inside of R[x], that (r,x) is a minimal prime containing (r)?
If so you get it by the Hauptidealsatz
okay I retract this statement, any height 1 prime is necessarily principal
Okay also (r) is just straight up a prime hurb
paper flint
its not, the minimal way is to notice that every element in the subgroup is of t...
This is cool!
delicate bloom
sure
the product is simpler one to do
so maybe start there
the sum argument will be a bit messy but it's elementary at least
or maybe it won't be so bad, maybe I'm thinking in particular if you're proving the integral closure, so it might actually be alright too I don't recall
I'd start by writing out a polynomial for each with summation notation personally
$f(x)=\sum_{i=0}^I a_ix^i$ and $g(x)=\sum_{j=0}^J b_jx^j$ with $f(\alpha)=0$ and $g(\beta)=0$
cloud walrusBOT
merowo (◡‿◡✿)
delicate bloom
then I'd start by writing down $h(\alpha \beta)=0$ is the polynomial $h(x) = \sum_{k=0}^K c_kx^k$
cloud walrusBOT
merowo (◡‿◡✿)
delicate bloom
and try to work out what I might have to do to this
ultimately I want the c_k in terms of the a_i and b_j coefficients
there's more than one way to try to think about it or try to solve it I'm sure, but this is sort of my general thing, at least putting a name to things clearly gives you a hand hold into reasoning with them further
cloud walrusBOT
needs protection
delicate bloom
your second f should be g, but I'd start backwards from this
cloud walrusBOT
needs protection
delicate bloom
f(a)g(b) won't get us really what we want
we want h(ab)
it's the roots that multiply, not the polynomials
try writing out h(ab) as a sum, that might help to look at
cloud walrusBOT
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delicate bloom
this is a bit too random, yeah
you don't get anything for multiplying them or plugging both in like that, that I can see at least
what we really want is a way to solve for the c_k or show they exist
cloud walrusBOT
needs protection
delicate bloom
nope
plugging in a*b into those doesn't really get us something cause f(ab) !=0 or anything
we need to create some h so that h(ab)=0
yeah exactly
cloud walrusBOT
needs protection
delicate bloom
yeah good
cloud walrusBOT
needs protection
delicate bloom
that's fine, they're in a commutative ring
or, specifically we're in Q right haha
cloud walrusBOT
needs protection
delicate bloom
oh no
I thought you meant the exponent rule lol
(ab)^k = a^k b^k
yeah what you put is bad haha
good thinking though
trying stuff out at least
lmao
putting the 'abstract' in abstract algebra
cloud walrusBOT
needs protection
delicate bloom
compare it to one of the other polynomials like f or g
and stare at them side by side, and pray maybe
cloud walrusBOT
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delicate bloom
hopefully something pops out at you
cloud walrusBOT
needs protection
delicate bloom
well K is undetermined
so it could be the same as I or J
well, here, notice in h(ab)
what if we just look at it as being a polynomial in b?
that's my hint, but I can give a better one if you don't like it lol
cloud walrusBOT
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cloud walrusBOT
needs protection
delicate bloom
well we want to show we can solve for the c_k
so we should equate $c_j \alpha^j = b_j$
cloud walrusBOT
merowo (◡‿◡✿)
delicate bloom
$c_j = \frac{b_j}{\alpha^j}$
cloud walrusBOT
merowo (◡‿◡✿)
delicate bloom
I think so, you could multiply by a^J to get it so that it's out of the denominator
$h(x) = \alpha^J g(x/\alpha)$ satisfies $h(\alpha \beta) = 0$
cloud walrusBOT
merowo (◡‿◡✿)
cloud walrusBOT
merowo (◡‿◡✿)
delicate bloom
although this doesn't work, something's off, cause alpha^k isn't rational in general, I guess I'm missing a step or misremembering something
like sqrt(2) is algebraic over Q simple example
so this argument doesn't work
maybe it can be fixed somehow lol I don't know, maybe you do just need to buckle down and do it the fancy way
ultimately, whatever you do in that way would correspond to something we could do to fix this proof, or side step it entirely
haha yw
urban acorn
So I'm relatively familiar with the theory of groups and of vector spaces, but not so much with other algebraic structures.
What can I expect from them? Obviously modules are much less nice to work with than vector spaces, but except that, how are fields and rings like? What kind of thing does the math around them look at?
Does the study of things like rings of polynomials tell you a lot about polynomials when you care about them in other contexts?
next obsidian
Can you be a bit more specific? It seems like you're trying to be specific by asking specific questions, like "how are fields and rings like?" which is good, but that as a question is still very vague.
The question about polynomials is a bit more tractable, over \R a polynomial is invertible iff it's just a constant
But in the general context of a ring R, an element of R[x] is invertible iff the constant term is a unit, and all other coefficients are nilpotent, i.e. a^n = 0 for some n
Right
urban acorn
as in you can multiply it by another polynomial to get 1?
next obsidian
Also modules aren't "much less nice to work with", at least if you come at it from the right angle :^)
Vector spaces have a lot of nice properties, but I find that makes them pretty boring haha
next obsidian
as in you can multiply it by another polynomial to get 1?
precisely
urban acorn
well, to some extent my original question was vague as a necessary result of its generality
because i'm not asking about specific properties of these algebraic structures
next obsidian
In another sense, the fact that you can extend linear algebra to things over rings and not just fields means modules are "nice" in the sense that they can be defined over more general thigns
urban acorn
more like "generally what to expect" from studying them in comparison to the algebraic structures i'm already familiar with
sort of, but not specific properties, it's difficult to explain
next obsidian
Sometimes you want to consider things as having an action by only integers, and since Z is a PID finitely generated modules over them have a complete classification
I for example used this on a Riemannian geometry homework
urban acorn
Sometimes you want to consider things as having an action by only integers, and ...
finitely generated
so modules don't have a dimension analegously to how vector spaces do?
next obsidian
Expect a lot less finiteness things. Group theory is almost always done in the context of finite groups for a really long time unless you have reasons to care about infinite groups
It's not very well-behaved
but even then you can still differ a bit over a PID
you can differ by torsion
A free module is analogous to a vector space as in you can take a basis but it still isn't as nice
and a general old module has nothing of the sort
urban acorn
What do you think of dummit and foote for self studying? It's what I used for group theory.
urban acorn
I haven't done math in a while since I studied group theory, and coming back to it I find myself disappointed by the lack of more interesting exercises.
It's not that the exercises are too easy per se, just that they only require rigorous application of what you learned without a place for much more thinking, and without illuminating any nice properties of the things you're looking at most of the time.
With some exceptions, maybe.
partially yes
how would you answer that?
lots of polynomials
makes sense to me seeing as polynomials to rings are like words to groups and linear combinations to vector spaces
take some elements in a group, a word would essentially be any formal way to combine them using the properties that define a group
and take some elements in a vector space, a linear combination would essentially be any way to combine them using the properties that define a vector space
but any formal way to combine elements by adding and multiplying becomes a polynomial
I see
so is that why we care about polynomials as functions, whereas given a word on a and b no one cares about the function that you get by plugging values into it?
I see.
is it still true in abstract rings that a polynomial as a function is in a 1-to-1 bijection with a polynomial as a formal expression?
but like, over a ring of order 2, wouldn't x^2 = x?
one moment
but like in GF(2), 0^2 = 0, 1^2 = 1
and those are the only elements so in general x^2 = x
the way in which I was using x has a hidden "for all x in GF(2)" in it
sure, so it's not true in general, just in that field
but how do you define polynomials as formal objects without already being given a field?
unless you force yourself to only use integer coefficients or somethng
cause like, in some abstract field you may have some element, and a polynomial in that field can have that element as a constant coefficient
but you can't predict the element unless you're already talking with respect to that field
maybe i should just go ahead and open an algebra book lmao
I see
chilly ocean
Hello ! I asked a question on the advanced-number-theory channel but it's maybe more suitable to ask it here because it is group theory : Can someone explain this theorem to me ? I'm not understanding the notion of linear invariants I guess
chilly ocean
does anyone have a good general definition / theorem (excuse my lack of jargon)
for subgroups of a dihedral group?
Like ive seen this. but its not very general, because of the a they use is defined differently depending on the cases that they use later on.
viscid pewter
ok
chilly ocean
sowwy 😢
viscid pewter
right so i don't see the problem with reusing a?
chilly ocean
i dont either, its just not well defined right? when we're talking about it being general.