i have help channels muted i feel like they're more used by people who just want help to study for their classes or highschoolers than people genuinely interested in math (i might be wrong tho sorry if so)
#groups-rings-fields
chilly ocean
viscid pewter
no, they are
coral spindle
Help channels are also for high school questions; the advanced section is more appropriate for questions about this kind of topic.
coral shale
i dont think we should be encouraging questions that users can find the answer to with a bit of research
if they do their research and have questions about it still, sure.
That doesnt look like whats happening here.
thorn delta
I think some people learn by first obtaining some superficial understanding of what it is they're learning first. My main reason for pushing back is because I've seen people ask questions like this, but with more advanced topics, and not get the kind of response y'all gave L3.
For example, i somehow doubt that if i went into #algebraic-geometry and asked a random basic question like "is a locally ringed space kinda like X", I would be just immediately directed to the definition in Hartshorne.
left estuary
So this is a closed group?
1,0 are both contained in the set and applying that function to 1 gives you 0 and 0 gives you 1
left estuary
So what’s the problem?
coral shale
youve written an interval and an expression
also 'closed group' isnt a term
not the way you mean it probably
left estuary
A group can be finite or infinite
coral shale
That interval qualifies as a set yes
but that expression, i dont see how to interpret as a binary operator
coral shale
???
no, the definition of a group is precise
it must be exactly as stated
or... its not really a group
left estuary
So must be exactly a binary operator
coral shale
yes it must be.
left estuary
Alright
coral shale
ie. something that takes in 2 things from the set and outputs something from the set
noting any of these somethings may be the same
left estuary
So what about [1,0] operation is multiply
coral shale
ok that qualifies as a binary operator
then proceed to check the axioms hold
or not
south patrol
my brain hurts
left estuary
Next would be the number identify?
elder wave
you should read up about groups
south patrol
oh okay i see what you are doing now lol
elder wave
at least the definition
south patrol
lol
left estuary
at least the definition
I have
chilly ocean
i think they should read about basic set theory first
south patrol
the only interpretation of this i can tell is that you are asking whether $x \mapsto \frac{1-\cos(\pi x + \pi)}{2}$ is a well-defined function ${0,1} \to {0,1}$
coral shale
im assuming
cloud walrusBOT
potato
coral shale
but u need to check for everything in that interval
not just 0 and 1
An introductory book would hopefully explain with examples going through the axioms to make it clear
thorn delta
So what about [1,0] operation is multiply
not sure if L3 is talking about [0, 1] or {0, 1}
coral shale
{0, 1} indicates the set with 2 elements
chilly ocean
odd
coral shale
u mean that?
left estuary
Ye
coral shale
[0, 1] = reals from 0 to 1 inclusive
left estuary
Oh right cause it’s like a range
coral shale
yh ok, then try checking if {0, 1} under multiplication is a group
coral shale
1x0 1x1 0x0
thats the first step
left estuary
Next is number identity right?
south patrol
what is the multiplication you have
south patrol
you needn't show it's commutative to show it's a group anyway
left estuary
But it’s a requirement right?
south patrol
no
left estuary
Order in a set under the operation shouldn’t matter
chilly ocean
No.
left estuary
But I thought that would mean it’s not a group
coral shale
The axioms are what they are - nothing else should be assumed
none mention commutativity
to be clear thats when x * y = y * x for all x, y
coral shale
i use * to mean a general binary operator
left estuary
I’m a programmer so all that’s going to do is confuse me
coral shale
eg. you can write +(1, 2) to mean one plus two if you're clear about it
left estuary
Like when statistics uses | instead of the symbol is an element of
coral shale
you will find after the initial introduction the group operation is written as if it were multiplication usually
ab means f(a, b)
a^2 means f(a, a)
thats the usual standard adopted in texts
rustic crown
It doesn't make sense to exponentiate that generally. In a group we can only talk about a^n where n is an integer. This is just given by repeated multiplication of the element, or its inverse if n was negative
left estuary
Oh ok
chilly ocean
It doesn't make sense to exponentiate that generally. In a group we can only tal...
This makes me wonder whether there is some "natural" group-like structure that allows exponentiation by elements of some other ring R or group or whatever. So that when R=Z you recover the case of groups with the usual integer exponentiation
left estuary
Also is [a,b] the same as a->b
coral shale
Also is [a,b] the same as a->b
not sure what this is.
left estuary
-> arrow sign
coral shale
yh but i dont know what this notation means in this context
whats [a, b] and whats a -> b
left estuary
[a,b] where a and b are both integers and I think maybe i should use the arrow as a limit so as a approaches b where a and b are both integers
left estuary
uh there is no context just thinking of ideas
cunning dust
Convergence in integers just means that at some point they're equal?
left estuary
Yeah
coral shale
cunning dust
So what are you trying to do with [a,b]?
left estuary
[0, 1] = reals from 0 to 1 inclusive
^
coral shale
yh but thats not actually group theory related, was pointing out that didnt mean what u thought it did
also, you should check out #proofs-and-logic message
this would be a lot easier to take in with the foundations to go with it
and then some mathematical notation will be clearer
left estuary
Fair but right now I’m going to keep asking questions because it’s working and working well
cunning dust
There's no way to know if it's working unless you know the material or are being tested on it by someone who does.
And you're almost better off reading a book/watching videos on the subject than just asking questions. There's just way too much material to learn for you to learn it through asking questions, it's just impossible for anyone. For example there will be things that you wouldn't have even thought to ask questions about since you can't possibly think of everything on a subject you're unfamiliar with.
left estuary
Maybe i should clarify I do read and watch videos but then I create examples for myself and ask questions on where I don’t fully understand something
Because similarly it’s almost impossible for a book or video to answer every question on a subject
cunning dust
Watching individual videos won't help much either. You can't learn analysis by watching 3b1b for example. You'd need to follow some syllabus which textbooks and recorded courses inherently provide. And you should watch/read actual textbooks and lectures, for example watching 3b1b won't be enough, even his dedicated playlists on specific topics, like his linear algebra and calculus playlists, aren't enough to teach you their respective subjects. You need something structured which also ideally gives you problems to solve
cunning dust
Because similarly it’s almost impossible for a book or video to answer every que...
A good book gives you the tools to answer these questions yourself.
left estuary
But that’s how I learned calculus
barren sierra
You said you're a programmer right?
left estuary
Ye
left estuary
No I learned the other way around
barren sierra
Learning math is similar in a sense
You should be doing problems
You cannot learn without actually doing math. Reading is not enough
left estuary
I’m doing problems that I’m making as examples
barren sierra
Active learning, not passive
left estuary
You cannot learn without actually doing math. Reading is not enough
That’s what I’m doing
barren sierra
Doing problems you create won't test your knowledge
cunning dust
But that’s how I learned calculus
Calculus is much much different from more advanced math. For one, you can learn it through pure memorization. And I doubt that if you learned it through watching individual videos without doing a lot of practice problems that you learned it well.
left estuary
Making problems gives me problems doing problems helps me learn the material and asking questions about the problems is what allows me to fully understand them
barren sierra
How are you going to test your knowledge since writing good problems at a given level necessarily means you understand all the material at that level
left estuary
I write problems that confuse me or that I don’t quite understand
coral shale
If i tried making my own calc problems when learning i would likely end up with an impossible integral
cunning dust
Making problems gives me problems doing problems helps me learn the material and...
If you're the one making the problems you have no idea if what you're doing is correct. You have no objective metric to judge yourself on. You're not an expert at the subject that you're currently learning, you don't know what kinds of problems you should be doing/know how to do
left estuary
Which is why I’m asking questions
cunning dust
That doesn't solve the issue of you not solving problems written by experts. You don't know what kinds of problems you need to solve, there's no way for you to know: you're just learning the subject.
cunning dust
Plotting graphs means nothing
coral shale
thats great, but i feel like u still would miss out. as slurp says
if u tried a set of (harder) calc problems would u be able to do them I wonder
left estuary
Probably but when that problem comes up I go back and learn it
Like differential equations?
burnt flower
If i tried making my own calc problems when learning i would likely end up with ...
Sin(sin(x))
coral shale
like idk, but im thinking like some university entrance level Qs
or 1st yr uni Qs
problem sets
left estuary
Qs as in Quaternions?
burnt flower
Qualifying exams
cunning dust
Lmfao
left estuary
Oh lol
cunning dust
Specifically, how did you learn calc if not through a dedicated series by a professional?
left estuary
I’ve done this type of stuff for years though
This is how^
left estuary
By typing in integration equations
left estuary
If you cant compute integrals by hand
There just anti derivatives
I prefer the term Inverse derivative sounds more accurate
coral shale
So do you know of the fundamental theorem of calculus?
Integration by parts?
Those are important calc topics
rotund aurora
whats the point of this conversation?
coral shale
to try convince them to learn the right way
left estuary
I tried to learn how my school was teaching me and it was slow
thorn delta
bring l3 to the dark side of math bookwormery
left estuary
Those are important calc topics
I hope so
cunning dust
Can you integrate x*e^x?
left estuary
Isn’t that a lambert W function
coral shale
rotund aurora
no
left estuary
yes it is
coral shale
thats how you would go about solving it if that expression was equated to something
left estuary
Ye
coral shale
but that is not the wambert func
left estuary
I’ve done those before
cunning dust
It's like the classic example of integration by parts
rotund aurora
I dont think Lambert solves anything lmao, thats why I said "no", unless Im wrong
left estuary
I dont think Lambert solves anything lmao, thats why I said "no", unless Im wron...
Same concept as imaginary numbers just pretend that it exists
But there is a way to approximate it
coral shale
this is tangenting off either way. No - integrating that expression has nothing to do with wambert
left estuary
wambert LOL
cunning dust
The lambert W function is related to xe^x (it's the converse relation), it is not related to this integral
lavish spoke
You're thinking of inverting xe^x, integrating it is just ibp
Unless you're meming lmao
cunning dust
Yeah that's what I said
Though I understand that ibp is not part of every calc curriculum
lavish spoke
You could also just differentiate xe^x and then integrate the result or etc
coral shale
who calls it lambert w, its clearly wambert
left estuary
I never took a calculus curriculum
lavish spoke
wambert's l function
left estuary
Anyways next thingy
delicate bloom
Walmart L functions??
left estuary
You're thinking of inverting xe^x, integrating it is just ibp
Yes wambert is the inverse of xe^x
left estuary
Walmart L functions??
Whataburger L food
left estuary
Can you integrate `x*e^x`?
I found the graph btw
cunning dust
You shouldn't need to graph this to solve it
left estuary
I use the graph to organize my work
thorn delta
desmosTeX
left estuary
I don’t necessarily use the actual graph
coral shale
if u try to continue this route
left estuary
I struggle on that
coral shale
at some point one needs to learn how to structure their work properly
in a way the reader can understand
and well, books do give a good indicator on that
left estuary
I sometimes use programming logic to explain shit lol
For example I was using logical stuff to explain how Spanish works
Didn’t finish it but was working pretty well
left estuary
For example I was using logical stuff to explain how Spanish works
Found it
cunning dust
People are trying to use this channel for its intended use now, we should probably migrate to another channel
left estuary
Like a flock of birds
chilly ocean
Suppose G is a group of order 2n where n is an odd integer. Show that there is an injective homomorphism from G to a symmetric group S_2n on 2n letters such that an element h of order 2 in G goes to the product of n transpositions.
So if you look at the action of G on itself by left multiplication, this is faithful and you get that |f(h)|=2, but I'm unsure where to go next.
A product of n transpositions has order 2, but why can't there be other elements with order 2?
coral shale
A product of n transpositions has order 2, but why can't there be other elements...
(12)(23) = (123)?
disjoint i assume
well literally if your decomposition into disjoint cycles contains something other than a transposition it cant have order 2
the order of the product is the gcd of the disjoint cycle lengths
chilly ocean
disjoint i assume
Oh I see, yes, probably disjoint
chilly ocean
the order of the product is the gcd of the disjoint cycle lengths
OK I understand, but now why must it be n transpositions?
coral shale
i didnt think of the rest
unique valve
Consider the permutation group of left translations on G, and what the element of order 2 looks like
toxic zephyr
in general for a function $\func{f}{X}{Y}$ which is not surjective, is it okay to redefine it as a function $\func{f}{X}{im(X)}$? or is it always better to just redefine a new function $\vfunc{g}{X}{im(X)}{x}{f(x)}$? i would like to do this for a certain module homomorphism to show that the domain is isomorphic to the image.
cloud walrusBOT
nilpotent nix
toxic zephyr
i did the defining a new function for my original proof, but i'd like to know if being lazy and just redefining it is acceptable
thorn delta
if f : X -> Y, ill usually just write f : X -> im(f) when i need a surjection onto the image.
I think the fiasco of defining a new function g : X -> im(f) such that g(x) = f(x) blah blah is just extra unneeded text
everyone will know what you are talking about
shell agate
you can just restrict the image, it's usually written $f\vert^A$
cloud walrusBOT
Blaxapate
shell agate
If $f\colon X\rightarrow Y$ is a function and $\Im(f)\subset A\subset Y$, it is possible to define $f\vert^A\colon X\rightarrow A$
cloud walrusBOT
Blaxapate
shell agate
sometimes called the corestriction
white grotto
why do we have to impose for a ring homomorphism that f(1)=1
f(x)=f(x)*f(1) then if f(1) isnt equal to 1 then there's a problem
so isnt it obvious from the condition f(xy)=f(x)f(y)
just like f(0)=2f(0) implied f(0)=0
for groups
white grotto
even in a ring with one?
woven obsidian
Why do you think it would?
woven obsidian
What's the problem?
woven obsidian
How do you know f(x) and f(x)*f(1) are different
white grotto
if they're not... for all y in Imf, then ofc f(1) is the unit dont you see?
if f(x)*f(1) =f(x) for all x
like saying if yx=xy=x for all x
then y ofc is 1
it's literally the definition
woven obsidian
You don't know it's true for all x, you know it's true for all f(x)
woven obsidian
Anyway, an easy counterexample is f(x) = 0 for all x
white grotto
Anyway, an easy counterexample is f(x) = 0 for all x
yes
woven obsidian
It's a ring homomorphism if you drop the condition f(1) = 1
A more interesting example is $f: \mathbb{Z} \to \mathbb{Z}$ with $x\mapsto 2x$
cloud walrusBOT
AoiKunie
white grotto
ah
woven obsidian
Hmm nvm that's not a ring homomorphism lol
woven obsidian
It's definitely defined, I just defined it. But it's not multiplicative
Here are some examples https://math.stackexchange.com/questions/270883/in-a-ring-homomorphism-we-always-have-f1-1
white grotto
could you please screenshot
woven obsidian
An example from the link: $f: \mathbb{Z}_6 \to \mathbb{Z}_6$ given by $x\mapsto 3x$
cloud walrusBOT
AoiKunie
white grotto
ah
woven obsidian
It is multiplicative because $f(x) * f(y) = 3x * 3y = 9 xy = 3xy = f(xy)$
cloud walrusBOT
AoiKunie
woven obsidian
The point is that the coniditon f(x) = f(x) * f(1) does not imply f(1)=1 in general
if your ring is an integral domain it does though
Actually, it suffices that there is at least one element in the image of f that is not a zero-divisor
Because then you can cancel this f(x)
woven obsidian
np
hollow mica
If M_1, ..., M_n are pairwise commuting nilpotent operators on a vector space of dimension n, how do I show that M_1 ∘ ... ∘ M_n = 0
wooden ember
hint: ||commuting operators share an eigenvector||
lethal dune
That doesn’t say anything
hollow mica
well I'm not sure how to make sense of the eigenvectors when the eigenvalues are all 0
wooden ember
have you tried using induction
note that if M_n commutes with the other operators
it commutes with M_1 o .... o M_n-1
south patrol
If `M_1, ..., M_n` are pairwise commuting nilpotent operators on a vector space ...
Interesting
Rather than caring about eigenvalues etc
wooden ember
oh you have a simpler way?
lethal dune
Not seeing with induction either
wooden ember
ill write it out because maybe im begin stupid
south patrol
Ye you wanna do what you said I think
wooden ember
ah wait
i had a block diagonal instead of block triangular matrix that's where i was going wrong
hmm
south patrol
Okay so like M_1,....,M_(n-1) are commuting nilpotent operators on im Mn, so we get the result by induction
I think
hot lake
it's not obvious how to do induction since the n in number of nilpotent operators has to be equal to the n in the dimension of the vector space
lethal dune
then you have to assume all of them have a block of size 1
south patrol
Okay so like M_1,....,M_(n-1) are commuting nilpotent operators on im Mn, so we...
(This is fine right?)
south patrol
But we have induction
lethal dune
one of them could be jordan block of size n?
wooden ember
We can’t argue by Jordan here anyways since it’s not necessarily over an alg closed field
hot lake
yeah that should work
wooden ember
Okay so like M_1,....,M_(n-1) are commuting nilpotent operators on im Mn, so we...
Im not sure about this step
lethal dune
Im Mn is dimension < n and the Mi map it to itself
okay
wooden ember
Why is that an endomorphism on imM_n
wooden ember
Right ofc
south patrol
Owo
wooden ember
Yeah sounds good then
hot lake
you still need to do the case n=2 then
lethal dune
n=2 is trivial no?
hot lake
how do you prove n=1 => n=2
lethal dune
just assume the jordan form, only 2 possible nilpotents then
wooden ember
how do you prove n=1 => n=2
The same argument no?
south patrol
I have two operators M1, M2 on a space of dim 2, then M2 maps V onto im M2 and by induction M1 vanishes on that
So like exactly the same argument works
So ye just need n=1
hot lake
ah yeah Im slow lol
wooden ember
It’s a neat result
south patrol
So ye
I was gonna come up w a meme solution but need to think a lil
okay I mean I have a funny solution that works over an infinite field I think
lethal dune
sure
hollow mica
why does the fact that im M_i is an invariant subspace give the result
wooden ember
why does the fact that `im M_i` is an invariant subspace give the result
Induction by considering M1…M_n-1 as an operator on imM_n
south patrol
Consider M = a1 T1 + ... + an Tn, this is clearly nilpotent and so M^n = 0. Now M^n is the sum kf terms (a1 T1)^k1 ... (an T^n)^kn where k1 + ... + kn = n. Now we should be able to show all the terms vanish by picking appropriate a_i and killing them off 1 by 1 lol
But that is more of a meme
hollow mica
Induction by considering M1…M_n-1 as an operator on imM_n
I don't see how to apply the inductive hypothesis here because it deals with a vector space of a different dimension
wooden ember
You’re inducting on n
Which is both the dimension and the number of operators
Oh wait o see your point
rotund aurora
Is $S^1\times \bR^+\approx \bC^\times$? Where $S^1$ is the multiplicative group of the complex circle ?
wooden ember
@south patrol imM_n could have dimension less than n-1
cloud walrusBOT
Croqueta
wooden ember
Is $S^1\times \bR^+\approx \bC^\times$? Where $S^1$ is the multiplicative group ...
Prove it
But no
wooden ember
You need R^*
chilly ocean
you need (R^*)^+
hollow mica
<@356844891722219520> imM_n could have dimension less than n-1
that's what I mean!
rotund aurora
They asked to show that SO(2,C) is isomorphic to S^1\times R^+, but SO(2,C) is trivially isomorphic to C^\times
I suppose they meant R^*, not R^+
rotund aurora
tru
south patrol
<@356844891722219520> imM_n could have dimension less than n-1
So?
hollow mica
oh wait strong induction?
wooden ember
Oh no just embed imM_n into a space of dimension n-1 lol
hollow mica
strong induction works too right
wooden ember
I don’t see how strong induction applies here
Since you’d have to also reduce the number of operators considered
hollow mica
yeah
wooden ember
First few don’t preserve imM_n, only the first n-1
south patrol
I mean
Just take some subset of the M_i if im M_n has smaller dim than n-1
No need to embed in a bigger space ig
wooden ember
First few don’t preserve imM_n, only the first n-1
Yeah this was wrong
hollow mica
oh so it works
wooden ember
Since each operator preserves imM_n
south patrol
Yes
wooden ember
I need to stop trying to help people when I can’t even help myself lol
hollow mica
If S,T commute then S(Tv) = T(Sv)
wait also why does this extend to more than two operators
induction again?
hollow mica
each operator preserves the image of another operator
south patrol
Well of T_m in particular
hollow mica
so why should the composition preserve it too
hollow mica
oh I'm dumb
if you apply one operator, the result stays in im Mn
apply another, still stays
I see
cloud walrusBOT
Joe 1
quiet pelican
Well, they must all conjugate it to something in <(12345)>, and you know what conjugation in A5 (well, S5 but it mostly transfers) looks like
chilly ocean
Consider the permutation group of left translations on G, and what the element o...
You mean the action of G as a subgroup of the automorphism group?
quiet pelican
I mean, since you know what conjugation looks like, it doesn’t take that long
But yes I would think
chilly ocean
Yea, but also G has size 16?
chilly ocean
Suppose G is a group of order 2n where n is an odd integer. Show that there is a...
bumping this
white grotto
Yea, but also G has size 16?
yes 🤦♂️
white grotto
why K[X] is principal if K is a field
chilly ocean
pick a non-zero element of your ideal of minimal degree and use division
white grotto
so P0 is minimal degree, any element can be written P=QP0+R, R=0, so P0 divides everything
right?
where did i use K being a field?
white grotto
pick a non-zero element of your ideal of minimal degree and use division
r u still here?
chilly ocean
you tell me, you're the one who wrote the proof
white grotto
Lol that was my question tho
white grotto
degree of R is smaller than P0
but it wouldnt have had to be strictly smaller if i wasnt in a field?
chilly ocean
just making sure :^)
the polynomial ring over a field is a euclidean domain
that's what lets you do this division in the first place
white grotto
wdym
white grotto
what do you mean i cant do this division
chilly ocean
where did i say that?
white grotto
its not..
chilly ocean
you must be in a euclidean domain to do this
white grotto
why?
chilly ocean
... by the definition of a euclidean domain?
lol
you can't do this in Z[x] for example
white grotto
why are we talking about euclidean domain, i didnt study that i am asking about fields
chilly ocean
ffs
chilly ocean
the polynomial ring over a field is a euclidean domain
THE POLYNOMIAL RING
if k is a field then k[x] is a euclidean domain
chilly ocean
then how do you even know you can perform that euclidean division????
chilly ocean
so P0 is minimal degree, any element can be written P=QP0+R, R=0, so P0 divides...
how did you do this???
chilly ocean
i can do multiplication and addition, but i can't divide x by 2 in Z[x] like this
white grotto
thats division tho...
chilly ocean
okay, let me clarify
white grotto
so P0 is minimal degree, any element can be written P=QP0+R, R=0, so P0 divides...
this is decomposition
chilly ocean
you cannot do this "decomposition" in Z[x]
are you just criticizing me for abusing terminology? are we talking about math anymore?
white grotto
okay got it
white grotto
are you just criticizing me for abusing terminology? are we talking about math a...
im not sure what you mean, and no
ruby sundial
how did you do this???
dam dude chile
white grotto
so the ideal generated by x and 2... what would it be generated by in R[X]
chilly ocean
Z[x] is not a principal ideal domain
ruby sundial
ye take 1,x
chilly ocean
(2, x) is the classic example
solar glacier
^^
chilly ocean
dam dude chile
it's really fucking irritating to be pinged three minutes after posting something being asked "are you still here?"
ruby sundial
oh 😂
chilly ocean
don't do that, dazzlefishe
white grotto
okay
solar glacier
I have a question
how come when I divide an integer by a gaussian integer and round up or down I end up with distinct GCDs
are they all the same up to plus/minus
white grotto
well i think i got it thanks tteppa
solar glacier
or do they all have same norms which is ok for them to be distinct
for example when dividing 85 by 1+13i i get either 7+6i
or 6-7i depending on rounding
chilly ocean
Let G be a non-abelian simple group. Let H be a subgroup of index 7 in G. Determine the number of conjugates of H in G.
So the number of conjugates has to be greater than 1 or else H is normal
But what do you do from there?
lament grove
I am interested in finding the unitary irreps of the free group
has this been done before?
solar glacier
damn three questions lol
question if in the gaussian integers the norm of a doesnt divide the norm of b then a doesnt dividee b in Z[i] i.e., their GCD is 1
white grotto
question if in the gaussian integers the norm of a doesnt divide the norm of b t...
yes
white grotto
its easy to see if a divides b then norm of a divides norm of b
white grotto
yes. and its how you prove it
solar glacier
question
white grotto
a divides b. b=ac so lbl=lal*lcl
solar glacier
how come when i divide 85 by 1+13i
and round up or down i get either 7+6i or 6-7i
is this ok as both have same norm
white grotto
division isnt unique in gaussian integers
white grotto
they're unique up to multiplying by 1,-1,i,-i
solar glacier
ahhhhhhhhh
makes total sense now
then ive been doing it correctly all along
thanks so much
cause i used the online calculator to cheeck my answer
and it was sometimes off by one of the factors u gave
white grotto
but can you tell why?
solar glacier
oh this is true
white grotto
and units in the gaussian integers are exactly the elements with norm 1
and that happens to be the roots of unity in gaussian integers
solar glacier
true
white grotto
so 1,-1,i,-i
solar glacier
makes sense
white grotto
i dont remember doing so
rotund aurora
Google exists
white grotto
just need a quick answer, like are they cyclic
white grotto
no
rotund aurora
neither Wikipedia?
rotund aurora
so go to Wikipedia
idk why you cant use google, but you are using internet, just use some different search engine
will work fine
wooden ember
what structure do finite abelian groups
short answer: it's complicated
i dont really get why the fourier transform is defined as it is for non-abelian finite groups
like for abelian ones it makes sense
rotund aurora
wooden ember
you're decomposing on an orthonormal basis of functions from G to C
white grotto
<:this:489575168314966026>
i have the memory of a goldfish
is it normal to have read that, and said that, and have no memory of it?
wooden ember
and for non-abelian groups apparently the implicit basis you're doing this on is the matrix entries of irreducible reps
but i dont see where that is in the definition of the fourier transform
i tried to recover that interpretation with the fourier inversion formula but im not really getting anywhere
rotund aurora
is it normal to have read that, and said that, and have no memory of it?
Do you want me to diagnose you with premature alzheimer or amnesia?
white grotto
Do you want me to diagnose you with premature alzheimer or amnesia?
first one
or the other i ve had a lot of concussions
rotund aurora
concussions is not algebra
white grotto
concussions is not algebra
wdym
south patrol
(v quick - is there any nice notation for v1 \otimes ... \otimes v_n?)
stuff like otimes_{i=1}^{n} v_i looks a bit weird to me but maybe it is standard?
wooden ember
like small v_i's?
cause it's standard if you're actually writing the tensor porudct of many modules but i have no idea about the standard for writing simple tensors
i agree with you that would look weird
rotund aurora
perhaps the \prod notation? No idea tho
white grotto
why if a=1 mod 4 then a can be written as a=b^2+c^2
is there some name to this law?
white grotto
nice thanks a lot
solar glacier
what structure do finite abelian groups
fund theorem for finite abeelian groups
tells you what theyre iso to
version 2 (of the fundemental theorem) in dummit and foote
question
cloud walrusBOT
MyMathYourMath
solar glacier
can i start by claiming theres no integer with norm equal to 2
then suppose 2=ab where a b are non units
its just that i cant really induct on n and how do I use that n >3 is square free
coral shale
The approach is likely to write a and b in an explicit form and go from there
cloud walrusBOT
MyMathYourMath
solar glacier
where do I go from here
cloud walrusBOT
MyMathYourMath
solar glacier
cani just claim theres no element of R with norm 2
coral shale
2
solar glacier
but i wanna know this method too!
coral shale
oh that kindof norm
solar glacier
yeah
coral shale
well i cant remember it
solar glacier
a^2+nb^2
solar glacier
**MyMathYourMath**
what do I do with all the x_i and y_i
coral shale
equate imaginary coefficients?
chilly ocean
Is there some generalization of the concept of R-module analogous to how groups are generalizations of abelian groups?
rotund aurora
Do you mean something like an R-module but where the modulants need not commute with addition? Like, some action of a ring on a non-abelian group
chilly ocean
Does every compliment of a normal subgroup come from a split extension?
chilly ocean
Do you mean something like an R-module but where the modulants need not commute ...
yes
chilly ocean
Does every compliment of a normal subgroup come from a split extension?
That's a good exercise, what have you tried?
chilly ocean
That's a good exercise, what have you tried?
One direction I understand. For the other, if I am trying to construct a s.e.s sequence when I know that H is complement to N. I'm not sure which surective map you would put for G-> H.
next obsidian
yes
I’m pretty sure this is forced by the axioms to have addition commutative
This is true for rings at least using like, distributivity or something
(1+1)(a + b) = a + b + a + b
but also 2a + 2b = a + a + b + b
So now subtract a from the left and b from the right
Now b + a = a + b
coral shale
is non associative algebra like done/used anywhere
next obsidian
Yes, Lie algebras are an example that pop up a lot
If you ask rep theorists they’ll say “associative, unital algebra” oftentimes
Because for them an algebra by default has neither associativity nor a unit
coral shale
ic
next obsidian
It’s basically just the data of a symmetric bilinear form
If you also don’t care about commutativitiy, just a bilinear form kekw
chilly ocean
I’m pretty sure this is forced by the axioms to have addition commutative
right, that yes was stupid
next obsidian
It’s a valid question, just one that ends up not panning out
A stupid question is asking for associativity to no longer hold 
muted epoch
can anyone help me with these questions?
delicate bloom
can anyone help me with these questions?
I'd probably start looking at the orbit-stabilizer theorem first
oh and of course Burnside's lemma
chilly ocean
Why isn't 0 always required to not be in a multiplicatively closed subset?
next obsidian
Because we care about these to localize at them
If you localize at a set with 0 in it the result is 0
fading wagon
so it depends whether you are fine with including the zero ring
chilly ocean
If you localize at a set with 0 in it the result is 0
I understand. Every element in the localization will be equivalent to 0/0.
narrow marsh
This is about a polynomial time reduction of the compressed word problem on a free product to something else, what does the something else mean?
Does it mean its reducible to both? or at least one?
idk if i should post here or #combinatorial-structures or #foundations LOL
wooden ember
This is about a polynomial time reduction of the compressed word problem on a fr...
maybe try #numerical-analysis too?
just to be sure im not doing something illegal here, if i have a handful of elements a_1,..., a_r of a finitely generated k-algebra i can always "complete" this set so that the algebra is just equal to k[a_1,....,a_n] =k[x_1,...,x_n]/I (with a_i = the image of x_i) right?
just by universal property of polynomial algebras?
south patrol
Not sure what you mean by complete the set
Can't you just take the map k[x1,...,xr] -> k[a1,...,ar]
wooden ember
I just mean complete because the algebra might not be generated by a_1,…,a_r
So I just want to add in a few more elements so that it is
And then yeah I take the evaluation
rotund aurora
If p_1,..., p_n are primes then sqrt p_1, ..., sqrt p_n are all linearly independent over Q right?
rustic crown
yee
you can also show that degree is that extension is 2^n, which gives you something more than just linear independence
which means sqrt p_i can't be written as any polynomial involving other sqrt(primes) with Q-coefficients
wooden ember
why can we assume that rho is unitary wlog in the proof of plancherel?
simple mulch
Hello
simple mulch
why can we assume that rho is unitary wlog in the proof of plancherel?
uhhh diaconis!!!
wooden ember
Yah
simple mulch
I knew that in the previous year but I can't recall 😐
Question: Consider $C_n$ and $C_m$. What would the operation be on $C_n \times C_m$?
cloud walrusBOT
mns
chilly ocean
do you know how direct products are defined as groups
G, H groups, the set G x H becomes a group with the operation (g, h)(g', h') = ...
simple mulch
iirc, it's pointwise, right?
simple mulch
So $(g,h)(g',h') = (gg',hh')
simple mulch
thanks
tired horizon
Can anyone explain to me whats an algebraic closure
If anyone can explain it in french it will be better
chilly ocean
Can anyone explain to me whats an algebraic closure
in what sense
in the french sense
parlez-vous francais?
tired horizon
Oui
chilly ocean
closures are usually constructed as "smallest thing with such and such property"
which is achieved by taking intersections of the objects you're constructing
for example, in universal algebra the algebra generated by some set is the intersections of all the subalgebras containing this set
oh wait. Now I remember
tired horizon
What i want to understand is
An algebraic closure in finite fields
What does a set thats algebraically close mean
Algebriquement clos in french
An algebraic closure of a field
chilly ocean
sorry I never heard of the concept, what I remembered is algebraic closure operators in universal algebra, idk anything about algebraic closure in field theory
which is a bold name btw
An algebraic closure of a field K is a field K' containing K such that every polynomial in K' of degree >=1 has a root
chilly ocean
an algebraic closure is an algebraic field extension containing your field which is algebraically closed. you can call it the algebraic closure because they're all isomorphic (fixing the base), so blitz is pretty much right (edited to bolden "algebraic")
tired horizon
Cause some polynomials don't have roots in R?
chilly ocean
you can even see C as adjoining the roots of x^2 + 1 artificially to R as C = R[x]/(x^2 + 1)
taking the polynomial ring R[x] and then quotienting
then x plays the role of i
coral shale
the smallest field containing them
chilly ocean
adding on
coral shale
that contains the original field
chilly ocean
*an* algebraic closure is an **algebraic** field extension containing your field...
Are they? What if you take $\mathbb R$ as subset of an algebraic closure of $\mathbb R(t_\alpha : \alpha \in I)$ where $I$ has cardinal larger than the continuum? This should result in an algebraic closure of $\mathbb R$ that is not $\mathbb C$ but bigger
i don't know the french equivalent
cloud walrusBOT
Just A Dishwasher
tired horizon
Yeah I get it now
coral shale
wut. i thought alg closures are isomorphic
chilly ocean
like attaching the roots of x^2 + 1 to R
coral shale
yes
chilly ocean
wut. i thought alg closures are isomorphic
i might be wrong but unless you require the algebraic closure to be "minimal" in some sense instead of just being an extension that is algebraically closed i don't think that's true
coral shale
alg closure extends original field
coral shale
i might be wrong but unless you require the algebraic closure to be "minimal" in...
yh it has to be minimal i thought
otherwise u can keep extending
tired horizon
If we take a polynomials ring and quotient it by a polynomials that doesn't have a root in the field K
chilly ocean
algebraic extension
tired horizon
That quotient extends K?
coral shale
An algebraic closure is not just an extension that is algebraically closed. Thats how it was defined for me
chilly ocean
i didn't say just "algebraically closed extension"
coral shale
otherwise C would also be an algebraic closure of Q
chilly ocean
oh TTeppa said algebraic extension, nvm then, sorry
i used to think the term was used for any extension that is algebraically closed because i my eyes skipped the word "algebraic", it makes much more sense this way ;-,
coral shale
And here I will advocate symbolising everything because words are dumb, but get told why thats sht again 
chilly ocean
symbols are words too
coral shale
english grammar articles and conjunctions are fluff
untold basin
Hello
If I have that |Gal(L/K)| = 1 for a finite extension L/K, can I conclude that L = K ?
chilly ocean
what do you think?
untold basin
I want to say yes but I can't convince myself formally xd
because G(K/K) = {Id_K}
and if |Gal(L/K)| = 1 it means that Gal(L/K) = {Id_L} but Id_L fixes K
tribal moss
So what you need to do is assume you have an a in L\K and somehow use that assumption to show there must be a nontrivial K-automorphism.
rotund aurora
Are they? What if you take $\mathbb R$ as subset of an algebraic closure of $\ma...
That wouldnt be an algebraic extension, so thats not an algebraic closure
coral shale
we resolved this dw
rotund aurora
uhh lmao Sorry didnt bother to read down and I thought the discussion was over haha. My bad
like you can take X to be a set with k many variables, and form the algebraic closure of Q(X). If k is continuum, thats C right, but if its larger it will be well idk.
but something containing C
tribal moss
Is N a notation for "continuum many" here?
rotund aurora
uhm I just realized I made many assumptions
okay, the transcendence degree of C/Q is c
warped viper
is this correct ?
coral shale
okay, the transcendence degree of C/Q is c
is that number of things u need to adjoin to Q
rotund aurora
is that number of things u need to adjoin to Q
its the cardinality of a maximal algebraically indepenent subset (over Q) of C
coral shale
ooh interesting
tribal moss
If I have that |Gal(L/K)| = 1 for a finite extension L/K, can I conclude that L ...
Actually if (as the notation suggests) you know the extension is Galois, doesn't your conclusion follow directly from the fundamental theorem of Galois theory (instead of mucking around with elements)?
coral shale
its the cardinality of a maximal algebraically indepenent subset (over Q) of C
curious why u said C as opposed to R
(cardinality wise, thats whats usually used right)
rotund aurora
curious why u said C as opposed to R
because we are talking about the extension C/Q
formal ermine
is there any shortcut to showing that the subset of K-algebraic elements in L for a field extension L/K has a field a structure?
rotund aurora
Algebraic independence is beautiful, because many of the elementary results are analogous to those of linear algebra concerning linear dependence. Eg. there always exists a transcendence basis (in a transcendental extension at least)
coral shale
is there any shortcut to showing that the subset of K-algebraic elements in L fo...
so this subset is L intersect K bar right
intersection of 2 fields is a field... (right?)
formal ermine
so this subset is L intersect K bar right
huh
rotund aurora
Algebraic independence is beautiful, because many of the elementary results are ...
and the cardinality of any two transcendence basis for the same extensions is the same, so that lets you define the transcendence degree
untold basin
Actually if (as the notation suggests) you know the extension is Galois, doesn't...
The extension isn't Galois
tribal moss
is there any shortcut to showing that the subset of K-algebraic elements in L fo...
Shortcut, compared to what?
tribal moss
Most of the axioms you get for free by L being a field. You just need to check closure under addition, negation, multiplication, reciprocal.
formal ermine
every c in K(a, b) for a,b in \bar{K}_L is algebraic
coral shale
is my argument circular or something
formal ermine
so we can just choose a \pm b, a \cdot b, \frac{a}{b}. does that work?
rotund aurora
Now that Troposphere is here, I'll ask a question
: Given an extension F<K such that all elements of K not in F are transcendental over F, can you find a field E such that F<K<E and F<E is purely transcendental?
tribal moss
The heck are you waiting for me to show up before you ask field theory stuff?
rotund aurora
uh no, I had asked this already in here, but no one really provided an answer
tribal moss
I see.
rotund aurora
And I thought you might know something about it
Sorry if that offended you, I didnt intend it in that way
white grotto
example of a linear function on the space of sequences
tribal moss
No, sorry, it's just funny that I don't really feel knowledgeable about field theory. What little I know I have to reconstruct from first principles each time.
formal ermine
to show that a polynomial $f(x)$ is the minimal polynomial of $k$ over some field, I only have to show that $f(x)$ is normed, irreducible, and $f(k) = 0$?
formal ermine
and I'll do it again.
rotund aurora
That question probably touches on set theory (maybe Im wrong), algebraic closures or transcendence bases already do touch on infinity (but thats just a straightforward application of Zorn's lemma). Thats why I thought someone knowledgable about infinity could say something about it (I often see you talk about set theory and logic in the server, so Im assuming this is kind of your expertise, yes).
coral shale
Now that Troposphere is here, I'll ask a question<:catThin4K:811354320460578896>...
im assuming ur interested in 'can you always find' 
maybe im missing something, but i dont see why not pick E = F(x)
rotund aurora
because you have to include K as an intermediate field
coral shale
oh i got it the wrong way round sillynme
untold basin
Suppose x in L not in K
Wait but this is false lol
Gal(Q(cubic_root(2)) / Q) = {Id} and Q(cubic_root(2)) =/= Q
tribal moss
because you have to include K as an intermediate field
Why doesn't K(x) work?
coral shale
the top isnt true
why is that the galois group
isnt it C3 or something
oh wait dumb me
no that extension isnt galois
so can u clarify what is Gal in that case
untold basin
so can u clarify what is Gal in that case
In which case ?
coral shale
Gal(non galois extension)
untold basin
If I have that |Gal(L/K)| = 1 for a finite extension L/K, can I conclude that L ...
here?
oh
rotund aurora
so something like Q<Q(t, sqrt(t^2+1)) is totally transcendental but not purely transcendental
coral shale
i think we answered the Q assuming Gal operates on galois extensions only
untold basin
i think we answered the Q assuming Gal operates on galois extensions only
The definition in my course doesn't require a Galois extension to define the Galois group
coral shale
i think we answered the Q assuming Gal operates on galois extensions only
if u dont, then indeed uve found a counterexample to your original statement
untold basin
Only a finite extension
coral shale
so Gal is just all automorphisms of K in L I see
untold basin
if u dont, then indeed uve found a counterexample to your original statement
okay thank you
In my course we write G(./.) but I saw that here we write Gal xd
Actually it is all auto from L to L that fixes K
rotund aurora
because you have to include K as an intermediate field
So what you want to do is find a set of independent variables X such that F<K<F(X)
untold basin
mb sure
I'm trying to prove that L/K finite of degree 2 implies that L/K is a Galois extension
@tribal moss Thx btw, I guess you too have assumed that L/K is Galois
coral shale
I'm trying to prove that L/K finite of degree 2 implies that L/K is a Galois ext...
So where does considering |Gal(L/K)| come into this?
being slow
L/K not galois => |Gal(L/K)| = 1 is your thought processs?
untold basin
So where does considering |Gal(L/K)| come into this?
L/K is Gal <=> |G(L/K)| = [L : K]
untold basin
L/K not galois => |Gal(L/K)| = 1 is your thought processs?
Yes because we always have that |G(L/K)| <= [L : K]
coral shale
ok ok, im with u now
coral shale
Yes because we always have that |G(L/K)| <= [L : K]
stronger: I believe it actually has to divide right
not just leq
coral shale
so if you change your question from 2 to 3
you will see your approach indeed can't work
right
coral shale
L/K not galois => |Gal(L/K)| = 1 is your thought processs?
you can still think this but get stuck?
untold basin
And by contradiction, I have this : |G(L/K)| =/= [L : K] i.e |G(L/K)| > [L : K] or |G(L/K)| < [L : K]
And the fact that |G(L/K)| <= [L : K] is always true eliminates the first case
untold basin
you can still think this but get stuck?
I'm kinda stuck yes
shell agate
I'm trying to prove that L/K finite of degree 2 implies that L/K is a Galois ext...
this is not true, so there's no way you can prove it
coral shale
If I have that |Gal(L/K)| = 1 for a finite extension L/K, can I conclude that L ...
So it'll turn out if [L : K] = 2 this is indeed true, but this is precisely what you're trying to prove lol
coral shale
this is not true, so there's no way you can prove it
wait wut
my brain 😵💫
ok separability problem?
shell agate
ok separability problem?
yes
untold basin
We assumed this for the entire course
coral shale
ok, so i assume the question assumes char 0
shell agate
in this case yes
untold basin
Omg I hate this field
shell agate
and you can see that there's only one automorphism
shell agate
yes, you have to send sqrt(t) to sqrt(t)
formal ermine
to show that a polynomial $f(x)$ is the minimal polynomial of $k$ over some fiel...
?
elder wave
yes
coral shale
to show that a polynomial $f(x)$ is the minimal polynomial of $k$ over some fiel...
idk about normed
formal ermine
ok thanks
coral shale
oh ok i see the defn
elder wave
idk about normed
it's a requirement you put on it for it to be unique
coral shale
i think my course stated that some other way
untold basin
and you can see that there's only one automorphism
Wait i'm kinda slow for this
coral shale
monomial?
chilly ocean
monic
coral shale
ah yh monic
elder wave
the relevant ideal would also be generated by a*p
oh yeah that's the correct word for normed
it's "normiert" in german that's probably why illum called it that
coral shale
There's also a 'norm' definition so might be confusing if u call it 'normed'
shell agate
In characteristic 0 (or to be fair, in characteristic not 2), you can easily characterize extensions of degree 2, they are "Kummer", that is, [L:K]=2 implies L=K(a) for some a not in K such that a² is in K. That is, you add a square root of an element. From that, you can see that there are 2 automorphism of L fixing K: sending b to b (the identity) and sending b to -b (the conjugacy).
formal ermine
$X^3 - 2$ is irreducible by eisenstein with $2$ ye?
formal ermine
habit
coral shale
$X^3 - 2$ is irreducible by eisenstein with $2$ ye?
i dont see why not
chilly ocean
over Z, Q, yes
formal ermine
yeh
chilly ocean
it has a real root so not there though
shell agate
Wait i'm kinda slow for this
So, you have L=K(b) with b² in K. If you're in characteristic 2, b=-b. Take an automorphism of L fixing K, it must send b² to itself, so it must send b to b or -b; but b=-b, so really there's only one choice, it's the identity.
coral shale
so something like Q<Q(t, sqrt(t^2+1)) is totally transcendental but not purely t...
Given an extension F<K such that all elements of K not in F are transcendental over F, can you find a field E such that F<K<E and F<E is purely transcendental?
Doubt I can help, but to give it a go, K = F(x^2, rt(x) + x), you would pick E = F(rt(x)) right
and I suppose you're expecting the statement is true
rotund aurora
and I suppose you're expecting the statement is true
No idk
coral shale
Picking the right extension for E is a bit like finding the 'gcd'
untold basin
So, you have L=K(b) with b² in K. If you're in characteristic 2, b=-b. Take an a...
I understand thanks
rotund aurora
> Given an extension F<K such that all elements of K not in F are transcendental...
K is purely transcendental there I think
We have Luroths theorem, that says that if F<K<F(x) then K is purely transcendental (i.e. iso to F(x))
It is weird, idk. The question maybe doesnt make much sense
coral shale
Why not.
rotund aurora
Q(t,sqrt(t^2+1)) is totally transcendental but not purely transcendental
but it only has one "variable", and we know Q(t,sqrt(t^2+1)) not < Q(x) for some x
these things are very subtle, and I admit I should know more about it before talking about it
Because for example, Luroth's theorem doesnt hold in general for intermediate fields of an extension of the type F<F(x,y), but it does hold when F is algebraically closed
coral shale
I'm thinking to myself rn if your Q is equivalent to asking:
For S an algebraically dependent set over F, can you always find extension F(T) over F(S) with T algebraically independent
coral shale
these things are very subtle, and I admit I should know more about it before tal...
Well I know nothing and just going off defns
rotund aurora
You can take S=K right
xd
well I guess that doesnt apply, because then it would be trivially dependent
but yes
rotund aurora
I meant K=F(K) lol, but K is algebraically dependent over F even in the case when F<K is purely transcendental (simply because it contains F). What I said is nonsense, forget it
rotund aurora
I'm thinking to myself rn if your Q is equivalent to asking:
> For S an algebrai...
Yes
coral shale
I'm thinking to myself rn if your Q is equivalent to asking:
> For S an algebrai...
if it is, I think this is an easier Q to parse
for me at least
rotund aurora
The point of transcendence bases is the following: If F<K is transcendental, then there exists a set T such that F<F(T)<K with F<F(T) purely transcendental and F(T)<K algebraic
coral shale
yh I think I get that goal
white grotto
I diagonalizable iff A diagonalizable
i simply said, let B be the matrix of I, BM=AM. So, if M invertible then B=A . other wise take a sequence of matrices Mn that goes to M that is invertible, we have B=A for all n, and from continuity we deduce B=A
i feel ive done something terribly wrong
any clues?
chilly ocean
is it now...
rotund aurora
but it only has one "variable", and we know Q(t,sqrt(t^2+1)) not < Q(x) for some...
I guess we would embedd Q(t, sqrt(t^2+1)) in a field of the form Q(X) for some set X of variables. I guess it does make sense
white grotto
phi(M+1/n)=BMn=AMn
chilly ocean
and what does "M + 1/n" mean?
rotund aurora
I guess we would embedd Q(t, sqrt(t^2+1)) in a field of the form Q(X) for some s...
but no idea tho
chilly ocean
and why is this going to be invertible?
white grotto
long proof
chilly ocean
chilly ocean
i am speaking specifically about M + (1/n)I
chilly ocean
also, what is K?
white grotto
a field
chilly ocean
will 1/n always be defined here? what's the topology on M_n(K)?
what does 1/n mean if the field has characteristic n?
white grotto
i am not sure what you mean
yes itll always be defined
just nvm ill look at the sol
viscid pewter
i don't even understand the question