To have that map be an isomorphism, its more like you need the inclusions to K to have certain properties
#groups-rings-fields
fair obsidian
The idea is that if K also has the universal property, then you get unique maps F(A) *_H F(B) to K and K to F(A) *_H F(B)
Whose composition, in either order, by uniqueness must be the identity maps
Ok essentially there are 3 things we are using. 1. The free product with amalgam has the pushout universal property, 2. the diagram with F(AuB) instead has the universal property, 3. any two objects with the universal property are naturally isomorphic.
lament dawn
with the same universal property? and so the free product with amalgam and the diagram with F(AUB) have the same universal property? I'll obviously need to read up on universal property from cat theory before moving forward with that, but that's the conclusion youre drawing?
Using this from wiki, I suppose
Universal properties define objects uniquely up to a unique isomorphism.[1] Therefore, one strategy to prove that two objects are isomorphic is to show that they satisfy the same universal property.
fair obsidian
Yeah exactly that
Sorry, maybe I misunderstood what you said. Are you familiar with universal properties, maybe the pushout one specifically?
I can go over a simple universal property, the pushout universal property, and/or show how two objects with the same universal property are naturally isomorphic
lament dawn
I'm only really familiar with a couple universal properties, not with the overlying general theory
I also cannot find that paper wiki sites lol, Jacobson (2009)
fair obsidian
Oh lol
Well if you are interested, I can give you a pdf that has a good amount of examples and goes through the theory
lament dawn
that would be great!
fair obsidian
But the idea of showing that two objects with the same universal property are isomorphic isn't too complicated, the general idea is always the same
delicate bloom
does $\lim_{n \to \infty} \frac{ \text{ Number of nonabelian groups of order n}}{ \text{ Number of groups of order n}}$ exist and if it does, is it 1?
cloud walrusBOT
Merosity:
delicate bloom
I guess there are probably like some choices of subsequences that converge to specific different numbers now that I think of it
hot lake
final gulch
Yeah there can't be a limit
The subsequence of primes is constant and equal to 0
The classification of groups of order p^3 (I just googled it) gives you another subsequence that is constant and equal to 2/5
manic trail
In the same spirit, how about $\lim_{n \to \infty} \frac{\text{Number of nonabelian groups of order} \leq n}{\text{Number of groups of order } \leq n}$
cloud walrusBOT
Ridder:
manic trail
It seems like this limit does tend to 1
carmine fossil
Proof?
manic trail
I don't have a full proof but there's some evidence
carmine fossil
All prime order groups will be abelian
manic trail
So I have a source that $\text{Number of nonabelian groups of order } \leq N$ is $CN + O(\sqrt{N})$ with $C = \zeta(2)\zeta(3)\zeta(4)\dots$
cloud walrusBOT
Ridder:
manic trail
abelian groups*
fair obsidian
There should be a not terrible way of counting the abelian groups, so you really just need an estimate on the total number of groups
manic trail
Yep exactly
You just need some lower bound on the number of nonisomorphic groups
And have it being bigger than linear
$C \approx 2.29\ldots$
cloud walrusBOT
Ridder:
smoky cypress
It’s non abelian groups of order < n, it’s non abelian groups exactly of order n
carmine fossil
Probably related to a power of 2
knotty mason
wha the h*ck
olive mirage
by computer. There is no nice formula for it.
but already by order 16 there are isomorphism types of groups you could not write down in terms of the "classical" groups
neat ginkgo
when proving the general linear group is a group, i can use the fact that the determinant is an isomorphism to show the set closed under matrix multiplication, and i can use the fact that matrix multilication is associative to show the group is associative, right?
like in this whole field of math im very confused as to what facts i already know i can use
olive mirage
Determinanant is a homomorphism, not an isomorphism.
carmine fossil
yea,You can use det(AB)=det(A)det(B) to show closure
neat ginkgo
yeah homomorphism* my bad
olive mirage
If I assigned this to students I would want them to assume that matrix multiplication was associative, I wouldn't want them to assume det(AB)=det(A)det(B), but your mileage may vary.
carmine fossil
How would you do without that assumption?
olive mirage
it's two by two matrices, you're going to need to check it anyway
because you need to prove the inverse of these matrices have integer entries
although I guess that is veyr easy if you know that shortcut formula fro 2x2 inverses
if you knwo that, then I wouldn't mind using the multiplicativity
but you're right it is totally ambiguous what facts are fair game, and you just have to do your best to decide and ask the prof when you're not sur.e
neat ginkgo
one time i was supposed to show that the set of complex numbers with modulus 1 form a group, and i had this whole ass proof about how the inverse of every element will have modulus 1 too, but my TA just looked at me oddly and wrote |1/z| = 1/|z| = 1 on the board
which makes sense but like
im still confused am i allowed to do that?
olive mirage
I would definitely not use that fact. But I think TAs will have a different perspective than PhDs.
neat ginkgo
i guess i still dont understand abstract algebra well enough
carmine fossil
Can't you use polar form to show that?
olive mirage
Like, what you say about absolute values is true, but generally I wouldn't expect students to use it.
It's certainly true that the more things you assume the shorter your proofs can be. But I think in general, the less technology you use, the better understanding you're demonstrating/gaining.
neat ginkgo
right? because i feel like |1/z| = 1/|z| is something that follows from the very thing i have to prove
olive mirage
but this is why the first year of grad school in math is just all of undergrad math again. You get to do the content again knowing all the other content and forming these connections.
carmine fossil
one time i was supposed to show that the set of complex numbers with modulus 1 form a group, and i had this whole ass proof about how the inverse of every element will have modulus 1 too, but my TA just looked at me oddly and wrote |1/z| = 1/|z| = 1 on the board
@neat ginkgo So how are you supposed to do it?
You would have to use |1/z| at some point of time
Which would be same as using that formula
olive mirage
It would not be the same. You can effectively reprove that without knowing it in advance.
carmine fossil
So,you can use z=e^ix?
neat ginkgo
but how do you know |(x + yi)^-1| = 1/|z|
that's the assumption i didnt know i was allowed to make
olive mirage
When I assign that exercise, I definitely don't expect students to assume that, but if they happen to have had complex analysis and so they do it without thinking, that's fine.
carmine fossil
So,Work out the inverse in terms of the original? That just seems like extra work for no reason
olive mirage
it's not for no reason if you don't know how inverses of complex numbers work off the top of your head. And it is not especially messy. I'm always crystal clear with what things students can assume are associative, because I don't want people checking associativity very often.
carmine fossil
Ok,It's not that much work
olive mirage
there's on a miniscule handful of examples where associativity doesn't come from the fact that addition and multiplication of complex numbers as well as compositions of functions is assosciative.
(multiplication of matrices is composition of functions, and hence is associative, for example. And I'm plenty happy for students to just take that as a given instead of thinking about that particular thing in depth.)
carmine fossil
Ofc,There are some people who want to prove associativity of matrices by hand
olive mirage
I'm pr;etty sure I did that once. But if I asked students to do that:
A) they wouldn't learn anything from it
B) I would have to look at it
I'd make grad students do it 😛
but every minute you spend looking at the proof that addition on an elliptic curve is assosciative is a minute you never get back.
chilly ocean
but how can you find groups using computer
like guessing the operations on the elements and checkin if it works?
olive mirage
well, you use the fact that p-groups are all solvable. So your group of order 1024 has a subgroup of order 512
and so you can recursively do that.
Have you read the classification of, e.g., groups of order 8?
carmine fossil
So,It is pretty trivial?
olive mirage
haha I wouldn't dare to use the word trivial in this context 😛
but it is purely algorithmic. You have a huge tree of cases you end up chasing.
(not unlike the proof of the classification theorem of finite simple groups)
I'm not sure if I could do the classification of groups of order p^3 off the top of my head, and I'm an algebraist 😛
order 4 there's a trick, but order p^2 is already a tiny bit tricky
neat ginkgo
Have you read the classification of, e.g., groups of order 8?
@olive mirage is that the "every group of order 8 is iso to exactly one of these 5"
Z8, Z2xZ4, Z2xZ2xZ2, D4 and Q8
olive mirage
yes, though I meant the proof of it
neat ginkgo
oh ye
i didnt know that was a famous proof
i thought it was just some problem we were given lol
olive mirage
a lot of finite group theory exercises are basically teaching you the tools for proving classification theorems, because back in the day thta is what group theory was.
devout crow
maybe this is the wrong channel for this question, but can you have an uncountable direct sum of groups?
fair obsidian
There's one non-abelian one right?
I think this is the right channel
When you say direct sum, are you talking about abelian groups? I'm familiar with the terms product and free product
devout crow
yeah, direct sum of, in my case, Z/2Z
for a countable sum any element of this group is represented as a finite binary string
idk how an uncountable sum would work
fair obsidian
Yeah same pretty much the same for uncountable sum
It's just that you index the entries with an uncountable set
Instead of a sequence of entries, which is like indexing with the natural numbers
fair obsidian
But you are still only allowed to have at most finitely many 1s
Yup it should pretty much be the same
devout crow
ok thanks :)
fair obsidian
No problem
devout crow
i guess the difference is in the number of elements in the group then right
fair obsidian
Yeah there should be more
And if you are interested, one way to think of any size sum of Z/2Z is as a set of functions
For example (Z/2Z)^5 is like the set of functions from the set {1, 2, 3, 4, 5} to Z/2Z
And then you add the functions component wise, so the group operation is (f + g)(x) = f(x) + g(x)
The only restriction you have to place is that each function is non-zero for finitely many inputs (which doesn't change anything if you only sum finitely many groups).
But then a countable sum is like functions N to Z/2Z and (one type of) uncountable like functions R to Z/2Z (with the same restriction that only finitely many inputs give a non-zero output)
devout crow
oooh nice
smoky cypress
I think that how “S copies of a set X” defined, where S, X are sets. It’s just the set of functions S->X
smoky cypress
Is there a nice description of the algebraic closure of Z/(p)?
wind steeple
U_n, F_p^n ?
olive mirage
Yeah, and you can formalize that as a direct limit
bleak abyss
The idea is that you can think of F_p as a subset of F_{p^2}
And so on
Then take the union
olive mirage
yeah, every element is in F_q for some q. Then the tricky part is if you have two elements, how do you add them
and the information about the inclusions Sloth is giving are the answer.
bleak abyss
(Also what've you been up to zeta?)
olive mirage
If you want to formalize this it is a giant confusing mess. But if you don't think about it it all makes sense
smoky cypress
Wait so you do something like $\bigcup_{n=0}^\infty\bF_{p^{2^n}}$?
cloud walrusBOT
Whoever:
olive mirage
I've been in and out, classes start soon and I have a new boyfriend. I'm teaching a class on open problems in math.
bleak abyss
Not just 2^n, everything
A class on open problems damn
An idea how you'll spin it?
smoky cypress
But like, is F_{p^2} necessarily a subset of F_{p^3}?
I feel like it isn't
Oh wait
It can be
delicate bloom
like a natural injective homomorphism?
smoky cypress
Is there a natural injection?
olive mirage
F_{p^2} and F_{p^3} both inject into F_{p^6}
it's not natural, which is what makes the formality here messy.
it also depends on things like "do you assume the algebraic closure of F_p exists?" before you construct it. Because if you assume it exists, then F_q is just the splitting field of x^q-x in that algebraic closure
and that neatly gives you all the inclusions
smoky cypress
I just want like a nice description of the algebraic closure
Like how you can easily do computation in F[x]/(f(x)) by doing polynomial division
Probably hoping for too much
olive mirage
yeah, the answer is "do it in a finite field"
(which I know is not very satisfying)
computer algebra systems have to bend over backwards to handle this
delicate bloom
if it helps, the multiplicative group of the algebraic closure of F_p is the roots of unity of order relatively prime to p
olive mirage
yeah, you get to choose one to be easy, multiplication or addition
and the other will suck
delicate bloom
if you want a field containing p power roots of unity you have to go to a larger field containing it 😛
well sorta, just talking about that multiplicative group structure
olive mirage
yeah, you described the view that makes multiplication super easy
smoky cypress
Let F be a field and E be a splitting field of an irreducible polynomial f(x) in F[x], then is it possible that f(x) is not square free in E[x], and is it possible that f(x) is a square of some polynomial in E[x]?
fair obsidian
Yeah it is, I think the usual example for getting an inseperable extension in characteristic 2 works
Are you familiar with it?
smoky cypress
I don’t know what inseparable extension is yet
fair obsidian
Ok so take the field F2(t) of rational functions in some variable t with coefficients in the field of 2 elements
Btw the thing you are asking about it basically an inseparable extension
Anyways, consider the polynomial x^2 - t
Being degree 2 it is irreducible as it has no roots
smoky cypress
Ah, but if you find a splitting field of that polynomial, you see that one root is the negative of the other, and since in a field with characteristic 2, we have -1=1, so the roots are actually the same
So it’s a square
fair obsidian
Yeah exactly
You can do the same thing in characteristic p
And you get that the polynomial is a perfect pth power
fair obsidian
The idea is that the map x to x^p is a field homomorphism in characteristic p
And so (x - s)^p = x^p - s^p
Now usually this isn't a problem
As the map, being a field homomorphism, is injective
So if the field is finite, it is also surjective
But it might not be if the field is infinite. Hence why you need to use something like F2(t) to get an element without a pth root
Does that make sense?
smoky cypress
Not quite
fair obsidian
Is there a part in particular I could explain better?
Actually I'm kinda starving, I'll be back in a bit
neat ginkgo
i dont get algebra i swear
or maybe its just ugly
so cosets are a thing, and G/H where H norm G is a quotient group
G/{e} would then be like
a set of |G| cosets each containing a single element of G
and G/G would be a single coset
is this right
and [x, y] is notation for x^-1y^-1xy
and then it just says a commutator subgroup is [G, G]
?
what the hell does [G, G] look like
and for some reason G/[G, G] is an abelization of G
god damn it
i dont understand anything, i didn't have classes because of corona
our prof just gave us some stupid notes
and we have to email him for every question its such a waste of time
fair obsidian
a set of |G| cosets each containing a single element of G
and G/G would be a single coset
Yeah that is right
solemn rain
65
neat ginkgo
what does [G, G] look like
neat ginkgo
i mean i have the definition for [x, y] when they're elements
but G is a group
ok
oh
so its like
solemn rain
let x be in [G,G]
neat ginkgo
so its like
yeah ok
god why wouldn't he just explain that
or give a little example
even if only 5% of people weren't sure about it is it really not worth it
solemn rain
haha idk
i have this weird kink where i like struggling
with this
with anythign in math
just like bdsm
uh oh anyways
any more questions
solemn rain
what do u think
solemn rain
[x,y] = x^-1y^-1xy
neat ginkgo
neat ginkgo
i think so
since x^-1y^-1xy is in there
as well as x and y
then y^-1xy is too and x^-1yx
solemn rain
everything is there
groups are closed under the operatrion
and inverses
those are parts of the definition
neat ginkgo
when im doing math im always trying to think of like
edge cases and shit
and trying to have a visual intuition
it just doesn't work here
neat ginkgo
im trying to think of an example
solemn rain
here are tips
neat ginkgo
of an element
neat ginkgo
that would not be in the [G, G]
solemn rain
but listen
at first
try to just turn everything into definitions
just work algorithmly literally
after sometime
these definitions will turn into ideas
thats later
once u get the sense the mathematician who defined this shit had
u got it
but that comes osm uch later so justt
reduce everything to its definitions
and u will get it
forget inutition
neat ginkgo
fym ur bad at math
solemn rain
thank you
solemn rain
neat ginkgo
wait no i was wrong
holy shit
not every element from G is in [G, G]
ofc
if it was then it would just be G god damn it
solemn rain
umm i dont understand u but u got it so thats cool
neat ginkgo
here
solemn rain
i thought this was about closure
if x and y are in the set is xy
in
but anyways u got it so no worries
any other question?
neat ginkgo
im just bruteforcing some examples and trying to figure out some stuff
so if G is abelian, [G, G] = {e}
neat ginkgo
hence G/[G, G] would just be a single coset of G that is abelian
no its not hard for me to see i just figured that out, didnt read it yet lol
solemn rain
okay
solemn rain
wow
neat ginkgo
is that useful
solemn rain
groups that have this property are called perfect groups
yes
want a spoiler for such group?
neat ginkgo
hold up
neat ginkgo
so i just need to find a way to get X or Y from x'y'xy
do you know about solvable groups
@solemn rain nope
solemn rain
okay
so G/[G,G] can be thought as
how much can you make this group abelian
so for [G,G] = G
the 'abelianization' would be trivial
so
first
neat ginkgo
i think i still don't get abelization as well as i should
when we say that a quotient group is an abelization of G
do we mean like
solemn rain
define G^n = [G^(n-1),G^(n-1)]
solemn rain
a solvable gruop is a group such that G^n = {e} for some n
neat ginkgo
oh
solemn rain
a perfect group would be a group such that G^1 = G
neat ginkgo
what algebra course is this taught in usually
neat ginkgo
oh well im a computer science major we have only 1 algebra course
maybe it's even dumbed down for us a bit
lol
neat ginkgo
wym
solemn rain
okay thats cool
sad theres no galois theory but cool
so anyways we left the problem
u were looking for perfect groups
neat ginkgo
i can take algebra 2 next year
hold up i think i need to understand something before that
solemn rain
what
neat ginkgo
is G = G/{e}
solemn rain
yes
neat ginkgo
so when we're talking about quotient groups
we're not talking about their elements as cosets
we're talking about
the union of all the elements in all cosets
solemn rain
that would be the original group
the cosets partition
the group
you know about partitiosn right
partitions*
neat ginkgo
i know what partitioning is yea
solemn rain
yea
the cosets are the equivalence classes of the equivalence relation a= mod b H iff a=bh for some h in H
they partition the group H
the quotient group is the set of all cosets
so G/H = {gH | g in G}
neat ginkgo
the elements of all cosets G/anything will always contain all of G
neat ginkgo
so when you say G = G/{e} what do you mean exactly
neat ginkgo
but G/G isnt equal to G right
solemn rain
what
whats ==
G/G = {e}
if thats what ur asking
do you know the first iso theorem
neat ginkgo
an example of a group which is equal to its own commutator subgroup
no lol thats what ive been trying to figure out
neat ginkgo
simple groups have only {e} and themselves as normal subgroups right
solemn rain
yes
neat ginkgo
and direct products i know
solemn rain
okaay
solemn rain
check what
neat ginkgo
no i mean i had to check if i know what simple groups are
i had to ask you
because im not sure if we were talkign abou tthe same thing
neat ginkgo
why
solemn rain
uh fuck i gtg
neat ginkgo
LOL
solemn rain
you know about alternating groups rihgt
neat ginkgo
yeah
neat ginkgo
jesus
solemn rain
better
try to prove that any non-abelian simple group is perfect
should be cool
tbh thats like the only thing i know about perfect groups but should be cool
solemn rain
cool
neat ginkgo
if H is a normal subroup of G such that both H and G/H are cyclic, prove that G is generated by two elements
before proving this i would usually try to think of an example and see why its even true in the first place
but i dont think its true at all (???)
if G is cyclic, let H = {e}
all the conditions are satisfied
H is a normal subgroup since its trivial
H is cyclic
G/H = G/{e} = G which is cyclic
but G is not generated by two elements
manic trail
G is not generated by two elements if it is cyclic?
If it's generated by one element then it's also generated by two elements, right?
neat ginkgo
🤦♂️
carmine fossil
I don't think a generator can generate another generator
neat ginkgo
yeah i thought the set of generators must generate disjunct sets
carmine fossil
Do you want the answer?
neat ginkgo
yes sir
carmine fossil
Anyway,the question is pretty simple
gH=(a^r)H(because G/H is cyclic
g=(a^r)h for some h
g=(a^r)(b^m) because H is cyclic
g is an arbitrary element
neat ginkgo
that was easy
so technically
the generators don't have to generate disjunct subsets
carmine fossil
They have,I think
neat ginkgo
(a^r) is an element of G right
carmine fossil
Yes
neat ginkgo
any element
carmine fossil
But not an element of H
neat ginkgo
oh yea
carmine fossil
So,a is not in h b is in h
neat ginkgo
so there is no p such that a^p = b
carmine fossil
Yes
carmine fossil
if H is a normal subroup of G such that both H and G/H are cyclic, prove that G is generated by two elements
@neat ginkgo I think The fact that H is non trivial and proper should be part of the question
neat ginkgo
yeah it messes with me too
carmine fossil
Because in both cases,it will be 1
neat ginkgo
the fact that it works for {e} if you assume that the generators can intersect
carmine fossil
What about H=G?
woven obsidian
I have a question about quotient rings
If k is field, then I know that if p is irreducible in k[x], then the quotient ring is an integral domain(even field). How would I show that this is also true in k[x1,x2,..,xn]?
carmine fossil
Can a subgroup generated by a generator intersect another subgroup generated by a bunch of other generators non trivially?
if H is a normal subroup of G such that both H and G/H are cyclic, prove that G is generated by two elements
@neat ginkgo
Should H be proper and non trivial?
elder valley
doesn't need to be
carmine fossil
Because if H =G, G is generated by one generator
neat ginkgo
so the solution to this problem for H = {e} is "G is generated by some x and e"
i mean if this is true it kinda sucks
because it means every group which is generated by n elements is also generated by n+1 elements
unless n+1 > |G|
but whatever
elder valley
i mean every group G is generated by |G| elements
carmine fossil
Then, What would be the point in counting the number of generators?
Unless it is the minimum number of generators required
elder valley
smaller generating sets are often better to work with
carmine fossil
But,You don't say cyclic groups are generated by 2 elements
elder valley
no, but that's not a false statement
unless G={e}
there's no harm in adding more stuff to a generating set. but you can't always take things away and still have a generating set. so talking about the smallest generating set can be helpful
neat ginkgo
so
if H is a normal subroup of G such that both H and G/H are cyclic, prove that G is generated by two elements
means "prove G is generated by a minimum of 2 elements"
ok i got a new question 🤡
i a permutation from S9 p is odd, then what is <p> ∩ A9
i have a suspicion that it's just <p^2>
woven obsidian
@dapper nebula How would I show the ideal not being prime contradicts the irreducibility? The problem is I'm not sure if we have a factorization algorithm here as we do in k[x]
elder valley
i think there's some unique factorization involved there
woven obsidian
Hmm
Perhaps I would need to show that k[x1,..,xn] is a UFD
That would imply irreducible --> prime
There are a few exercises in my book outlining the proof of that so I guess I'll try doing them
Thanks for the answers, it helped me pin down what I had to prove
Yeah
Yeah it is outlined in my book's exercises so I'll check them out. Uses something galled Gauss' lemma
neat ginkgo
how deep into abstract algebra is galois theory
there's "splitting field of a polynomial" as part of the course content of my algebra 2 class, guess that's that?
can you explain galois theory to me like im 5 years old
in 1 sentence
got it
stone fulcrum
@neat ginkgo
Field extensions
Elements of those fields
And polynomials over those fields
All have properties in common. Knowing a lot about one gets you info on the others. Galois theory is about looking into these properties.
neat ginkgo
ill take algebra 2 if all my other classes aren't that hard 
gonna have to take one of these fellas, algebra 2 looks like one of the hardest
scarlet estuary
what in the world is analysis 4
sharp sonnet
at my uni it is functional analysis
neat ginkgo
idk one thing i know is it goes over "Riemann–Stieltjes integrals"
wait when im multiplying two matrices from GL_2(Z_2)
when this applies
what does ai1*b1j + ai2*b2j mean
we have 2 operations here
only + is defined in Z_2
is this just ai1+b1j + ai2+b2j
or am i actually multiplying two elements from z2
suppose i have GL_2(Z_3)
and for some cij i get
2*2 + 2*2
if its really multiplication i get 4 + 4
even if + is the group operator of Z_3
+ : Z_3 -> Z_3
4 is not in the domain
bleak abyss
There is multiplication in Z/2 though
Z/n always has multiplication, you just multiply in Z and take mod n
So 4 is 1 mod 3, hence 2*2 = 1
neat ginkgo
when you say Z/n you mean Zn right
bleak abyss
Yeah
neat ginkgo
so what you're saying is, the elements of a matrix can only be from a ring
bleak abyss
Yes, you need the multiplication of elements to define multiplication of matrices
smoky cypress
Is it true that by definition if a polynomial (with field coefficients) is separable then it is square free
I don't know why this definition here makes it seem like what I just said was not true
fair obsidian
Ok I kinda get what you mean actually
So separable is defined differently by different authors
smoky cypress
Are the different definitions at least equivalent?
fair obsidian
No
smoky cypress
Damnit
fair obsidian
I believe that usually you define a polynomial to be separable if it itself has no repeated roots
smoky cypress
So that means that if F has characteristic 0, p is a prime polynomial, then p^2 is separable
fair obsidian
But some people like this author defines a polynomial to be separable if its irreducible factors are separable in the previous definition
smoky cypress
Oh ok
fair obsidian
So that means that if F has characteristic 0, p is a prime polynomial, then p^2 is separable
Yeah in the above definition
smoky cypress
But the thing is, he assumed the polynomial to be square free like 2 pages ago
fair obsidian
Above in the passage you quoted lol not the first one I said
smoky cypress
So idk if that still applies in this page
fair obsidian
I don't think so
Probably he'll start assuming things about the irreducible factors
For Galois theory this definition actually might be slightly nicer as you get to say all of the polynomials are separable in a perfect field, not just the irreducible ones
fair obsidian
Which I guess you only really care about when you talk about splitting fields of not irreducible polynomials cause otherwise you only care about irreducible polynomials lol
smoky cypress
Because of E is the splitting field of f iff E is the splitting field of the product of all irreducible parts of f
Yeah
fair obsidian
Yeah
It's just a different definition, you push back the not having repeated roots one level to the irreducible factors
smoky cypress
I guess it doesn’t quite matter in galois theory
This definition might actually slightly nicer I suppose
neat ginkgo
is every quotient group G/H isomorphic to some subgroup of G
looks like this is the case for only finate abelian groups
fair obsidian
Yeah I'm pretty sure you are right, in general the quotient group G/H need not be isomorphic to some subgroup of G
I do think that it is always true for finite abelian groups, but I'm not completely certain, I would have to check. But there are definitely counterexamples when G is finite non-abelian, or when G is infinite
Someone can correct me if I'm wrong by I think that since it holds for cyclic groups of prime order, by the classification of finite abelian groups you can show it holds for an arbitrary finite abelian group.
Do you have any questions? I can provide some examples if you want
woven delta
Why is prime order enough?
I would think you need cyclic groups of prime power order at least
fair obsidian
You are right, you need prime power order
Although it's not hard to prove it for arbitrary cyclic groups
elder valley
i think the quaternion group is a small counterexample. taking H={1,-1}
neat ginkgo
oh my god
i can't find the english word
for a concept
Θ: G × X → X
Θ(e, x) = x
Θ(g, Θ(h, x)) = Θ(gh, x)
please someone tell me what this is called
neat ginkgo
if i were to literally translate it from my language it would be "effect"
a group action
@oblique river thank you
jesus
LOL
oblique river
what is your language?
neat ginkgo
serbian
oblique river
ah okay, yeah sorry I can't help too much more then
but I'm glad we got this one figured out
neat ginkgo
what is an example of a subgroup H of a group G, which has a different operation than G
carmine fossil
Actually, They always have the same operation
neat ginkgo
thank god
i was just explaining this to a classmate and he was convinced they don't always need to have the same operation
and then i actually checked m profs notes
carmine fossil
You used the inclusion being a homomorphism definition ?
carmine fossil
Ah yes,That one
neat ginkgo
this is called the "inclusion being a homomorphism definition" ?
didn't know there were other ones
carmine fossil
(H,*) is a subgroup of (G,.) If inclusion(seeing an element of H as an element of G) is a group homomorphism. H is a subset of G
Which is same as above
neat ginkgo
ah yes
neat ginkgo
is every group G of order m such that m = p1p2p3..pn where pi is a prime factor of m and for every i =/= j => pi =/= pj
G isomorphic to <p1> x <p2> x .. x <pn>
i know that Z10 is isomorphic to Z2 x Z5
and that Z4 is not iso to Z2 x Z2
trying to generalize that
solemn rain
first of all
for this notation Z_n means cyclic group of order n
so
first of all : Z_n x Z_m = Z _(nm) iff (n,m) = 1
solemn rain
crt for group theory :thonnk
smoky cypress
I don't think every group of square free order is isomorphic to <p1> x <p2> x .. x <pn>, if by <pk> you meant the cyclic group of order pk
For example S_3
But if you assume the group is abelian then it is true
fair obsidian
non abelian groups don't exist
woven delta
Lol
vestal snow
When defining a the stalk of a sheaf, is the order relation on the directed set in the direct limit reverse inclusion?
U =< V iff V \subseteq U?
pastel cliff
hi again @fair obsidian my question is prompted by a numberphile video on why 1 isn't prime. a comment mentioned primes and irreducible numbers as separate in some instances but i dont understand why. any help? this is just curiosity
fair obsidian
Sure, so first we can probably talk about the definition of irreducible and prime in math
So when you talk about a number being prime or irreducible you need to give some context. We can start by working in the integers
An integer p is called prime if it is not 0, 1, or -1, and if whenever p divides an integer ab, so a product, the prime p divides either a or b
So by definition we exclude 1 or -1 from being prime
as you mentioned
So you might be able to see that a number like 3 has this property, while a number like 8 doesn't
Do you see why 8 doesn't have this property?
pastel cliff
yeah that much i understand but how much of a technical definition is needed for what a prime number is
i know it goes slightly beyond just "a number divisible only by 1 and itself" but is it really necessary
fair obsidian
Yes, if you are working with other objects it is necessary
pastel cliff
pardon my ignorance im still just a freshman math major but what do u mean by objects, that's the second time u say that i think
fair obsidian
Ok so do you know polynomials?
pastel cliff
yes lol
fair obsidian
Cool so we can talk about prime and irreducible polynomials. We can also talk about the prime or irreducible Gaussian integers (which are numbers a + b*i for integers a and b and the complex number i), or even weirder collections of rational numbers or irrational numbers.
And so sometimes we want to generalize what it means to be prime or irreducible to these contexts, and a more technical definition is required to talk about what we really mean.
And by different objects I meant things like these, the polynomials, collections of rationals/irrationals, Gaussian integers, etc.
Maybe I can give some examples
oblique river
hi -- sorry to interrupt, but I just wanted to add some additional context here
vman has given some good examples of other "objects" in this context. I think the gaussian integers (numbers of the form a + bi) are a great example. we can "do arithmetic" on them in the sense that we can add and multiply them
but now we need to be very careful about what we mean by "prime"/"irreducible". here's just a basic observation: in the realm of gaussian integers, whatever our definition of "prime" will end up being, 5 is certainly not going to satisfy that definition, because we can factor it in a nontrivial way! 5 = (2+i)(2-i)
"nontrivial" here just means "not silly", which I am being purposefully vague about now but am happy to elaborate more on (and vman can as well), but for an example, 5 = 5*1 is a "silly" factorization, as is 5 = (-5)*(-1) and 5 = (5i)*(-i), since that just "reduces" to 5*i*(-i) = 5 * 1.
this example is just to motivate that we need to be careful about what constitutes "prime" -- you can't just make a statement like "5 is prime" unless you've specified where that statement is happening! 5 is prime in the integers, but not in the gaussian integers
pastel cliff
sorry to interrupt does that mean a gaussian integer is just a complex number?
oblique river
one of the subtleties in this discussion which I think Vman was going to go into in a little more detail is that there are multiple ways to formulate "prime" in the integers. We can say "p is prime if whenever p = a*b, either a or b is equal to 1 or -1" and we can also say "p is prime if whenever p divides a*b, then either p divides a or p divides b". For regular integers, these are equivalent, but there is no reason for these to be equivalent in more general "number systems"!
a complex number with integer "coefficients"
2 + i is a gaussian integer
but pi + i is not
because pi isn't an integer
pastel cliff
understood
once again sorry to interrupt but would it be possible to continue this discussion later? im supremely interested, i love math, but i have a preexisting conflict in the form of sleep
need to wake up in like 5 hours lol
fair obsidian
Lol very fair, I'd be happy to keep discussing sometime tomorrow whenever I am awake, but you can also post a question that came up from this discussion anytime and someone will probably answer it
oblique river
yeah :) I don't wanna speak for vman (and it seems like you two have a history of talking about math) but you can always ping me if you see me online, and if I'm around my computer then I'll be happy to talk about this
this is the area that I do research in
so I love talking about it haha
pastel cliff
oh hell yeah
thank you both very much :) im home for quite some time and im following a free mit course in discrete maths but i keep getting sent on tangents which lead to rabbit holes and i love it
this has been one of the more fun ones so i'll definitely bother you again
fair obsidian
That's awesome, I'd love to talk again sometime too
chilly ocean
Hello, I have a problem that I am trying to solve. I'm supposed to find all finite subgroups of $(\bQ\setminus{0},\cdot)$, where $\cdot$ is the usual multiplication for $\bR$
cloud walrusBOT
Jeffrey:
knotty mason
what did you try?
chilly ocean
I am not sure where to start. I should ask if you or anyone has any hints to start.
chilly ocean
What do you mean?
next obsidian
So here's just one idea, any group contains the subgroup generated by any of its elements right?
like if you have g you have g^2, g^3,...
and g^{-1}, g^{-2},...
chilly ocean
Right
knotty mason
@chilly ocean do you know any finite order elements of the multiplicative group Q?
will tell you that 1/2 has infinite order: 1/2, 1/4, 1/8, 1/16, ...
chilly ocean
That's right. The only element that has finite order is 1.
knotty mason
yeah good! 1 has order 1, there is another element with finite order
chilly ocean
in the rationals
carmine fossil
There is one more
knotty mason
well done
carmine fossil
Any more?
knotty mason
now what subgroups can you make from those?
chilly ocean
${1,-1}$ and ${1}$
cloud walrusBOT
Jeffrey:
chilly ocean
Thank you very much @next obsidian @knotty mason @carmine fossil
latent anvil
very cool @next obsidian
Also @next obsidian I realized my proof earlier solves a stronger version of your problem
A -> B just needs to be integral
Actually wait I think it proves something a little stronger
latent anvil
Let R <= S with R integrally closed in S and suppose A -> B is an integral ring map
next obsidian
integral = affine + finite type
Or locally of finite type maybe
closed in S?
wut
latent anvil
Then if you have \begin{tikzcd} B \arrow[r] & S\ A \arrow[r] \arrow[u] & R \arrow[u] \end{tikzcd}
cloud walrusBOT
shamrock:
latent anvil
You get a factorization B -> R
wait so integral => finite type for scheme maps?
that's weird
because it's not true in ring maps, right?
take a polynomial ring in infinitely many nilpotents
latent anvil
ahh okay that's not surprising because now I don't understand either of the conditions on the right
latent anvil
i did not consent
next obsidian
universally closed = closed + any base change is closed
latent anvil
Gabe the problem from earlier was the square I posted but A -> B was assumed to be finite and S was the field of fractions of the valuation ring R
I think chmonkey got it by reducing some AG problem
next obsidian
Show finite => proper
latent anvil
oh you were using the valuative criterion of properness?
That was like the last thing I learned in Max's class lmao
next obsidian
sort of
I was using valuative criterion for universally closed
since I already had separated
since any affine morphism is separated
cuz separated is local
latent anvil
finite type + integral = finite for scheme maps
next obsidian
and maps of affines are separated
latent anvil
Not sure if that's what gabe is thinking of
vestal snow
How do we know that defining F(U) in this manner is well definied
It might be possible that U can be written as a product of basis elements in more than one way
How do we get that the groups obtained by taking the product over the union are the same?
latent anvil
I don't think it's obvious
I would define it using all basis elements contained in U, then there's no concern about well definedness
but I bet you can show it's independent of the cover chosen in some canonical way
Maybe show that if you refine a cover you end up with an isomorphic set of products and then show any two open covers by elements of B has a common refinement (take pairwise intersections maybe?)
neat ginkgo
is a field homomorphism defined exactly the same way as a ring homomorphism
except on fields
smoky cypress
Yes
Because a field is a ring
So it makes sense to define it that way
But actually field homomorphisms are really nice, since they're always injective if 1≠0
neat ginkgo
cool
we didn't do fields in this class since it was cut buy like 20% because of corona
but i have the exam tomorrow
should do good, been studying this for a while now
smoky cypress
good luck
neat ginkgo
Thanks!
vestal snow
When proving surjectivity, why do such U_i and s_i exist?
vestal snow
Nevermind I figured it out
neat ginkgo
i know that if H and K are normal subgroups and H intersect K is e, then HK=KH on an algebraic level
but god damn it i can't get it intuitively
Oh fuck nvm i get it
hkh'k' = e
so like
either they're all e
(Which they're not because h and k are generic elements)
so it must be that hkh'k' = hh'kk' =ee=e
is this the proof
oh it can be even more intuitive
hkh'k' = e
h*something = e only if something=h'
=> kh'k'= h'
=> kh' = h'k
Wow im proud lmao
latent anvil
Congrats!
vestal snow
Do direct limits commute with quotients?
next obsidian
But actually field homomorphisms are really nice, since they're always injective if 1≠0
@smoky cypress field with one element isn’t a field so 1≠ 0 always
latent anvil
@vestal snow I'm not sure exactly what you mean. Do you have a system of M_i's and N_i's and you're asking if colim (M_i/N_i) = (colim M_i)/(colim N_i)?
@next obsidian thought the same thing but by some conventions ring homomorphisms don't need to preserve 1
so you could have the zero map as well as injective maps
next obsidian
Cringe
Also direct limit = colimit and quotient = cokernel = colimit so they commute right?
vestal snow
@latent anvil yeah that's what I meant
next obsidian
Also what I said 😒
latent anvil
wait maybe not... I've confused myself
Yeah I didn't understand why what you said was true
vestal snow
colim (M_i/N_i) = (colim M_i)/(colim N_i)
We're still talking about this right?
next obsidian
Yes
latent anvil
Yup
next obsidian
I’m arguing by some abstract category theory nonsense
latent anvil
yeah I don't think that makes sense? The diagram you're taking a colimit of to compute the quotient changes
Right?
it's like colim_{i in I}colim_{j in J_i}...
like we have some family of Ms and Ns
next obsidian
I’m pretty sure you could do it more boots to the ground tho. I feel like the unique map both colimits and quotients have lets you do this
Idk
latent anvil
And M_i/N_i is the colimit of
->
N_i M_i
->
where the first map is 0 and the second one is the inclusion
oblique river
dumb question but how do you get a map from colim N_i into colim M_i
from maps N_i into M_i
vestal snow
Ooh I know this
latent anvil
can't you take the colimit of all the maps N_i -> M_i?
vestal snow
you take the map induced by the inclusion
using the universal property of the direct limit
oblique river
oh I see, you need the maps in the directed system to commute with the inclusions
latent anvil
I think the claim follows from exactness of the direct limit functor but I'm not 100%
oblique river
yeah in that case it's fine
latent anvil
like you have an SES 0 -> N_i -> M_i -> M_i/N_i -> 0 for each i
Take a direct limit and you preserve exactness
next obsidian
That sounds right to me
oblique river
direct limit functor is only exact in R-mod
oblique river
(what I mean is it's not exact in general but it is in R-mod)
oh hmm yeah that's reasonable
latent anvil
anyways chmonkey I don't think your argument is valid
latent anvil
it's like commuting series
When the bounds of the second depend on the index of the first
it works if they're independent but not in a case like that
next obsidian
Other banana, are these of Abelian groups / modules?
vestal snow
Kinda did
latent anvil
Do you know what a short exact sequence is?
latent anvil
Okay so here's another way to phrase your problem
It's basically equivalent
Yeah prove lim is exact
vestal snow
but I can figure that out on my own
next obsidian
You can google it. Since colim is well, a colimit it preserves the surject ivory
Surjectivity
latent anvil
sure, it should be nicer than trying to work with quotients
next obsidian
Since it’ll take an epi to an epi
latent anvil
a well?
vestal snow
I'll take you guys' word for it
next obsidian
So I think all you need to do is show it preserves injectivity / exactness in the middle
oblique river
I mean
latent anvil
I'm not sure what chmonkey is suggesting
next obsidian
I guess that’s true Buncho
latent anvil
oh is it obvious that it preserves surjectivity and injectivuty?
oblique river
because "preserves injectivity" is just "preserves the middle of 0 --> A --> B"
oblique river
yeah
next obsidian
oh is it obvious that it preserves surjectivity and injectivuty?
@latent anvil preserves surjectivity because it’s a colimit I’m pretty sure
latent anvil
ugh
next obsidian
Rip
latent anvil
Okay so Im α is a little gross
vestal snow
O no
next obsidian
This is from Liu not Hartshorne?
vestal snow
next obsidian
But either way yeah this is really where category theory kinda shined for me in the first place
vestal snow
Yeah this is from Liu
latent anvil
Once upon a time this problem caused chmonkey and several headaches lol
next obsidian
That equality doesn’t make sense in the strictest sense
latent anvil
So α_x : F_x -> G_x right?
Which means im α_x is contained in G_x
But (im α)_x isn't a subset of G_x
latent anvil
no it means something a little stronger
It's kind of weird
There's all this context floating around
vestal snow
Okay
latent anvil
We have a canonical map im α -> G which gives an injection (im α)_x -> G_x
Does that make sense?
next obsidian
You sit on it and it makes no sense
latent anvil
Not that deep in terms of like, extra category theory stuff
but very deep in terms of being a thinker
next obsidian
Then once you finally come to terms with what that = means
A lot of stuff makes more sense in AG
latent anvil
Understanding this problem made sheaves make more sense to me
okay so
We have two embeddings floating around
The image of α_x is literally a subset of G_x
So we have an inclusion
And (im α)_x admits a map into G_x
The really important thing is that the isomorphism is compatible with these embeddings
we want to think of the image sheaf as sitting inside G right?
vestal snow
Yes as a subsheaf
latent anvil
We make this identification of it and an actual subsheaf of G
So the = isn't just an isomorphism
It's an isomorphism compatible with that identification
vestal snow
wait
next obsidian
That means some diagram commutes
latent anvil
I should probably stop talking and let you think about this problem, sorry
next obsidian
Like if you have i: A -> B and j: C -> B being two inclusions to say A = C in this case means an iso k: A -> C such that i = j•k
The identifications here are i and j
vestal snow
Okay
next obsidian
We pretend A and C are inside B by pretending they’re equal to i(A) and j(C)
Is the general sort of idea here
next obsidian
It means isomorphic in a way that plays nice with the identifications we made
vestal snow
Got it
So the natural way of embedding Im (a)_x into G_x must commute with this isomorphism and...?
vestal snow
No
latent anvil
This kind of playing with identifications shows up there a lot
oh okay
also yes that's what the = here really means imo
Isomorphism compatible with the ways of embedding into G_x
in that you have a commutative triangle
next obsidian
The other identification of im(a_x) into G_x is literal inclusion?
vestal snow
Oh yeah lol
next obsidian
Right?
vestal snow
My bad
next obsidian
I think
latent anvil
yeah but I think it's easier if you forget that for now
lol
like just draw a triangle
vestal snow
okay gotcha
next obsidian
So one thing to note
vestal snow
So just so this doesn't become word vomit
Let's name these
i: Im(a_x) to G_x
j: Im(a)_x to G_x
next obsidian
I believe I solved this by showing one of the things satirized a universal property for the other
Then this gives an abstract isomorphism between the two, and one thing is that arguing by universal properties always comes with some sort of uniqueness on maps
So that might be enough to actually tell you the triangle commutes
Because you can argue both ways to go around the triangle fit into this “unique map such that...” business
vestal snow
If you have this latexed, can you DM me?