m: A x A --> A ("multiplication") and i: A --> A ("inversion") and a map from the terminal object of C into A which is the "identity" element
#groups-rings-fields
oblique river
and they satisfy all the relations you would expect from a group
like associativity, and product of element and its inverse is the idnetity
etc
The map A x A --> A induces a map Mor(X,A) x Mor(X,A) --> Mor(X, AxA) --> Mor(X,A)
the first map being the universal property of products
and the second one induced by the map A x A --> A
alternatively, if you take the definition of a group object to be "an object A in C for which Mor(X, A) is a group for all X, in a natural way, compatibly with maps X --> Y"
then you don't really "know" what the operation on Mor(X,A) is immediately
raven fern
consider the following axioms for a bounded, complemented lattice:
cloud walrusBOT
katia
raven fern
is it possible to prove these theorems from these axioms? and if so, how?
cloud walrusBOT
katia
latent anvil
Take b = 0 and a = ¬0 and apply 8
That plus 5 and 2 gives ¬0 = 1
¬1 = 0 can by proved in a similar way
chilly ocean
To keep track of things in my head, I’d like to call a property a local property if
M has the property iff M_p has the property for all prime ideals p
Often we get something better
M has the property iff M_p has the property for all prime ideals p iff M_m has the property for all maximal ideals m
I’m trying to think of a good term form this (just for the purposes of keeping track of these in my head)
Since maximal ideals correspond to points in the spectrum we could call this pointwise-local
Is this an okay name or is this likely to fuck me up
Should I just call them both local and be careful
next obsidian
Well actually
Points in the spectrum are just primes. The maximal ideals are the closed points, so geometrically this is a closed-pointwise property
chilly ocean
¬1 = 0 can by proved in a similar way
This is a better image of what I thought your picture was
chilly ocean
Points in the spectrum are just primes. The maximal ideals are the closed points...
Ty
chilly ocean
Okay obviously solutions online are not always perfect
But what the fuck
Surely we can just say if phi:R to S is an isomorphism of rings, if u is a unit in R then phi(u) is a unit in S??
golden pasture
overly explicit moments
raven fern
¬1 = 0 can by proved in a similar way
thanks, that leaves theorems 3-5 remaining
latent anvil
I love it
thorn delta
Where is finite dimensionality used here? Its definitely not needed to make V into an End_F(V) module, but I didn't seem to need it to prove that V is a simple module either.
Basically, if A is a submodule != V, != 0, all I did was define a linear map which sends every element in a basis for V to an element outside of A. Since T(A) \subset A (A is closed under the action of end_F(V)), i came to a contradiction
next obsidian
I can’t immediately think of any issue / non-issue but authors will throw in assumptions unnecessarily if they make people feel more comfortable sometimes.
Sometimes this is finite generation, noetherian, etc.
next obsidian
Basically, if A is a submodule != V, != 0, all I did was define a linear map whi...
It’s not clear to me that this works actually
I’m worried about sums shoving something back into A
If it’s finite dimensional you can remedy this by taking a basis for A and then extending this to V
And then you know for certain you’ve sent sometbinf in A outside of V
For infinite dimensional it might still work but it’s not as clear to me
¯_(ツ)_/¯
sturdy marsh
it is true for infinite dim
thorn delta
If v \in V\A, I defined T to be the map sending a basis for V to v.
Then, T(A) \subset span_F(v)
and therefore v \in span_F(v)\subset A
and since v \notin A, contradiction
sturdy marsh
and it's probably easier to prove it directly
instead of doing a contradiction thing
show that there is a linear map taking a non-zero vector to any vector
(by completing the non-zero vector into a basis)
thorn delta
I mean, that sounds a bit stronger than what I need. I basically just sent everything in a basis outside of A, and it seems like that was enough
sturdy marsh
and here's another cool thing, if you assume F is algebraically closed (and V finite dim), then if V is a simple A-module for some sub-algebra A \subset End(V), then A = End(V)
(this is not easy to prove)
thorn delta
huh, interesting. Also, I see now that framing my proof for contradiction was unnecessary lol
oh, and i don't think my proof works
sturdy marsh
?
thorn delta
I said that T(A) \subset A always, and since T(A) \subset span_F(v), we must have v \in span_F(v) \subset A.
Doesn't follow
sturdy marsh
you have v \in T(A) \subset A
thorn delta
Nah, i don't think so. I might be able to remedy this with a special choice of basis, but T doesn't necessarily map anything in A to v
at least i don't see how
sturdy marsh
sure, a problem could come up if everything in A goes to 0
but pick a nonzero thing in A
complete nonzero thing to a basis for V
send it to v
this gives you a linear map
thorn delta
yeaaa, i see now. I said you're idea sounded stronger than what i needed before, but its of course exactly what i needed haha
latent anvil
@raven fern i think the right way to prove the remaining things is to show a <= ¬b iff a ^ b = 0 and ¬a <= b iff a v b = 1
Here's a proof sketch, you'd need to justify some steps I'm skipping:
If a <= ¬b then a ^ b <= ¬b ^ b = 0, so a ^ b = 0.
If a ^ b = 0 then a = a^1 = a^(b v ¬b) = (a^b) v (a^¬b) = 0 v (a^¬b) = a ^ ¬b <= ¬b
The equivalence of ¬a <= b and a v b = 1 is similar
vestal snow
Can someone help me understand the construction of a Boseck basis for holomorphic differentials?
I needed help understanding why the set in theorem 1 is indeed a basis of holomorphic differentials
The paper that contains the proof is in German so I couldn't understand what was going on
compact needle
The paper that contains the proof is in German so I couldn't understand what was...
A way to think about it: dx is a meromorphic differential on $\mathbb{P}^1$ which is holomorphic except at the point at infinity. It will pull back to a meromorphic differential on the smooth curve $C$ (determined by $F$) which is regular away from the points over infinity. Since these are curves, all meromorphic forms differ by multiplication by some rational function. If you multiply $dx$ by some rational $h(x)$, as long as the degree of the denominator is at least 2 more than the numerator, $h(x)dx$ is regular at infinity. The $x^\nu g_\mu(x) dx$ have this property, with the added bonus that you know exactly where the poles are. The question then becomes what rational functions $u(x,y)$ will cancel the poles (of the pullback to $C$) while not introducing new ones
cloud walrusBOT
Turgul
vestal snow
@compact needle Thanks for the answer. I don't really know much AG, but how would one compute the order of $x^{\rho}y^{\mu}g_{\mu}(x)dx$ at some point P?
cloud walrusBOT
Have a Banana, Bitch
compact needle
<@!525505750093791233> Thanks for the answer. I don't really know much AG, but h...
Having thought about it a bit more, this is more complicated than I had assumed. It seems if any of the m_i>1, then the curve given explicitly by the equation is not smooth. So I don't know if your reference really wants to work with non-smooth curves or if you have to study the normalization. I don't have the energy to try to figure out the normalization of the curve. If they mean use the singular curve, then I don't even know what it means to be holomorphic at a non-smooth point. Sorry for not being more helpful
vestal snow
It's totally fine! Though I don't think the smoothness should be a problem. Since k(x,y) is a function field in one variable, it corresponds naturally to a smooth projective curve
compact needle
There definitely is a unique smooth curve associated, but it sure is a lot easier (to me at least) to compute orders of poles/vanishing when you have affine charts that are smooth
I already didn't think that the projectivization of the equation gave something smooth at infinity, but I thought you could just use the x-line and be careful you didn't add any poles at points over infinity. Turns out, still singular even on the remaining points, if the m_i>1, so in principle, you might need a separate chart for each corresponding point
vestal snow
Oh okay
One more question
This was related to a research problem that I'm working on
And I think it would be worthwhile to examine the case when all m_i=1
Do you know any special properties we'll get in this case?
compact needle
Not special properties per se, only that I actually know how to compute orders of vanishing of functions, hence of differential forms. I have only really thought a lot about a single curve produced as an AS-cover of something, that was an AS-cover of an elliptic curve, and it's been a while
vestal snow
Oh okay. Thanks for the help. Would you be okay with sharing your notes/knowledge with me in the future?
compact needle
I'm happy enough to chat. I don't really have much in the way of useful things written down
vestal snow
Thanks!
compact needle
Good luck! I find some of that stuff interesting, but I really don't know that much about curves in char p
simple valley
is there a slick proof of $GL_2(\bZ_2) \cong S_3$
cloud walrusBOT
mniip
simple valley
one that doesn't come down to exhaustive enumeration
I mean I can pick a presentation of S3 I guess
carmine fossil
(r^2=1,s^2=1,(rs)^3=1) is probably the most convenient one
wind parrot
GL_2(Z/2) acts transitively & faithfully on the nonzero vectors in (Z/2)^2
simple valley
@wind parrot so?
wind parrot
Well an action of a group G on a set of n elements determines a homomorphism G-> S_n
The fact that the action is faithful implies this map is injective
hot lake
any 2 nonzero elements form a basis so you can send them to any of the 3, and the last one has to be sent to the last one
simple valley
what does transitivity give us in this context?
wind parrot
Oh yeah I guess not necessarily transitivity, but rather what zef said, gives surjectivity
simple valley
that is slick indeed
rigid cave
Is there any way of understanding why the eigenvalues of a rotation are complex using algebra, i.e isomorphism's or different tools? I have thought about this for awhile but I am not yet that comfortable with this kind of math since I'm still in high school. I don't even know if this questions makes sense, it just came to me a few weeks ago.
rustic crown
if you look at the rotation matrix in R3, then the axis is preserved which corresponds to an eigenvalue of 1. Product of eigenvalues is the determinant... which for a rotation matrix should be 1 since its doesn't scale or change orientation...
rigid cave
But in R2 the eigenvalues of a rotation are i and -i, right?
oblique river
real eigenvalues would imply there's some sort of scaling going on
but there's no scaling in a rotation
rustic crown
But in R2 the eigenvalues of a rotation are i and -i, right?
well a rotation in R2 can be imagined like a rotation in R3 with z-axis as the axis... and i would suspect the eigenvalues are e^i*theta and e^-i *theta
hot lake
if you try to rotate a line of slope i it stays a line of slope i
rigid cave
Yes okay, I understand. How about this: is there any way of bringing the concept of eigenvalues and eigenvectors into algebra?
I have no idea if this makes sense
rustic crown
the notion of eigenvalues is completely algebraic... so i don't really understand the question
rigid cave
A linear transformation can be seen as a homomorphism. Can we expand this concept even further?
carmine fossil
rigid cave
Well okay I might be out of my mind
carmine fossil
See module homomorphisms
rustic crown
a linear transformation is a homomorphism of vector-space (or modules if you want the scalars to be from a ring)
rigid cave
yeah okay. I don't even know what I am asking anymore. I need to take nap asap. Thank you thought!
carmine fossil
A ring(R,+,•) is a abelian group wrt +,and a monoid wrt •,along with an axiom which defines how • and + interact with each other(Talking about distributivity). Are there any interesting structures (by that,I mean things people study) which can be created by changing the "interaction" axiom?
rustic crown
Is there any way of understanding why the eigenvalues of a rotation are complex ...
maybe look up Jordan canonical form?
rigid cave
Yup I will definitely look that up later
cyan marten
A ring(R,+,•) is a abelian group wrt +,and a monoid wrt •,along with an axiom wh...
A lattice?
But that's not accurate (here we have two monoids).
carmine fossil
cyan marten
A ring(R,+,•) is a abelian group wrt +,and a monoid wrt •,along with an axiom wh...
I've just looked up nearrings and semirings, neither of which is an example of this. You can just see a list of all things that people and study and check.
carmine fossil
Ok
latent anvil
@chilly ocean check
twilit pawn
Is there an infinite perfect field, such that it's algebraic closure is an infinite extension?
mild laurel
Is Q not an example for this?
sturdy marsh
it is
twilit pawn
thanks
im taking this algebraic geometry course and a good amount of field theory is being assumed that i dont know too well 😦
plus it's been like 2 years since i did anything algebraic lol
so forgive me if i post more dumb questions in the future
latent anvil
oh zoph is back
golden pasture
twilit pawn
ok so i solved the problem that prompted my previous question, but i didn't use one of the assumptions. I feel pretty good about this solution and i think my professor was just playing it safe. but lemme know if i messed anything up
didn't use that k is infinite
mild laurel
L isn't an extension of k, its an extension of k(y). If k were a finite field F_q, then F_q(y) is no longer perfect and so the extension might not be separable
scarlet estuary
group theory spoiler: everything is just matrices
celest brook
everything is matrices
scarlet estuary
group theory spoiler 2: also everything is just subgroups of S_n
twilit pawn
wack
celest brook
S_n is matrices
granite horizon
just wanna die ngl
celest brook
f
twilit pawn
L isn't an extension of k, its an extension of k(y). If k were a finite field F_...
ok so i can say f is separable in k(y) since k(y) is infinite and f is irreducible. at that point, is there anything wrong with modding out by y^n for some n big enough to not effect anything in our polynomial, so the extension becomes finite?
yeah can't figure out how to make this become a finite extension :/
twilit pawn
think i got it 🙂
thorn delta
hint: linear algebra.
thorn delta
idk why ur apologizing. did u figure it out?
obsidian path
cause perhaps this should go in the lin alg chat not abstract alg
is what I thought you meant haha
my geometrical intuition says this has to be true as dim(R^2)<3
thorn delta
i think you have the right idea. dim(C) = 2 as a real-vector space
obsidian path
ah yeah that makes sense although I'm not too confident about my complex knowledge
uncut girder
<:AYAYAsmug:587222984553529348>
I love this image
upbeat bloom
Take ring R[x,y] ( R is for reals ). Is there a maximal ideal containing ( x^2 + y^2 -1 ) which isn't of the form ( x-a, y-b ) ?
golden pasture
consider like
carmine fossil
I have a feeling that's the only way how you get the x and y terms to cancel
golden pasture
what are the maximal ideals in C[x,y]
C is algebraically closed
now, take that ideal, and intersect with R[x,y]
rustic crown
Another way to think would be... quotienting by a maximal ideal would give a field extension, and since C is algebraically closed all you can hope is either the extension is C or its R. you don't want it to be R.... so it makes sense to include x^2+1 in the ideal.... and see what happens
upbeat bloom
Thanks for the help, will think along these lines
twilit pawn
can someone explain the dual of an affine plane curve? the notes i'm using don't really do that...
twilit pawn
ok update, i've made sense of what that is, but i need some help with other stuff. looking for an explanation of intersection multiplicity in the projective plane...
will take a reference if anyone's got one
bleak abyss
@twilit pawn So I'm reading [REDACTED]'s notes
bleak abyss
And it seems like multiplicity is defined for line \cap curve in affine space right?
bleak abyss
So I think you'll probably just say take an affine chart
twilit pawn
i like sort of thought that, but also it makes me sad
i'm not really confident doing any computation that way basically
but i will try, i'm working on other stuff in that pset that i think i can work out
i'm actually like 70% there so it's actually going ok
bleak abyss
I mean the other possibility is that a line in P^2 is just given by ax + by + cz = 0
twilit pawn
that is what a line in P^2 is
bleak abyss
And you can change coordinates linearly such that it's just x = 0 anyway
bleak abyss
Or maybe y=0 or z=0 whatever you need
twilit pawn
right
ok so like for problem 4 in this pset
you can basically say for (a) and (b)
"by definition in projective space, it's what happens on an affine chart, and we know this for affine space"
(c) i feel should be straightforward but i haven't had a good idea to do it
this is the problem
bleak abyss
"has an asymptote along the line l"
Goddammit this is too much geometry for me
I mean I guess if you just word it out it kinda makes sense in some weird way but... What's the def of an asymptote again?
Like super formally
I just think of it in words as "You never really touch it"
twilit pawn
"the distance between the curve and the line approaches 0 as x or y -> infinity"
according to wikipedia
which is pretty reasonable as a def
bleak abyss
I mean so basically you literally touch the line at infinity
But since P is a smooth point
The intersection multiplicity of \ell and C at P should also be 1 right?
twilit pawn
think of y-x vs y-x^2
bleak abyss
This sounds gimmicky but I think in the smooth case it should be kind of the extreme possibility right?
twilit pawn
ok i have an idea maybe?
P = [X:Y:0]. wlog X \neq 0
so just take P = [1:Y:0]
ok, now in the affine chart (Y,Z) |-> [1:Y:Z]
being smooth at P, literally means some partial derivative of the homogenization is not vanishing
our affine space is R^2, so the partial in Y and Z ought to give the tangent at P
so C has this tangent in these affine coordinates that is not zero. for Z very small we're approximating C well along the tangent
but when Z is very small, Y is very big 🙂
if we normalize back to Z = 1 to think about our original curve
bleak abyss
Wait aren't we dehomogenizing to get to the affine curve? And rehomogenizing to get to the OG curve
twilit pawn
you plug in Z=1 to get the original affine curve
bleak abyss
Oh I thought of the "original curve" as the projective one
twilit pawn
yeah we are being asked about the affine curve in (c)
I think this roughly checks out, gonna write it out more carefully
this is a similar idea to 3 actually
3 was the most analysis flavored so i got it first lol
bleak abyss
Yeah I guess. I wish he was more specific when he said asymptote here lol
bleak abyss
Does he have a book he's referencing at all?
twilit pawn
no
bleak abyss
Figures
bleak abyss
Wait here's something
Looked at his website and in one of the older times he taught the class he gave a syllabus
twilit pawn
ok so yeah he talked about all these books, but he also said he's not really following them
bleak abyss
It's possible one of these mentioned defines asymptote in the AG way
twilit pawn
"this class will prepare you to read chapter 1 of hartshorne" 🙄
bleak abyss
Lol
Are there any people who wanna do AG and didn't test out of that class?
If so this is a total waste
Like they need to quickly get to the commalg and the scheme theory
twilit pawn
i mean that's how the first year program is
classes are meant for the people who tested out of the class
bleak abyss
Do you mean for those who didn't test out?
twilit pawn
no i meant what i said
bleak abyss
Wait so what I'm saying is more like
Idk compared to first year geometry/topology
If you're gonna go on in topology that's actually a good class
And if not, it's sorta burnt time but it's not overly specialized
twilit pawn
right but if you're going into topology you usually place out
and if you don't place out, it's not taught on a level that is helpful
in general
bleak abyss
and if you don't place out, it's not taught on a level that is helpful
Is that a true statement? Like it'll prob be true that people who end up doing topology likely placed but
Idk it feels like, say DCal at least, actually teaches a good class
bleak abyss
Yeah he's supposed to do it well and usually Benson too I think?
Andre does a very good job with the second and third quarters
twilit pawn
andre is good yeah
bleak abyss
And I heard Amie's overall good too
twilit pawn
there are some good profs and some bad
the general point is that usually the courses are less helpful for the students who have to take it
bleak abyss
But yeah AG just never seems to get it right and I don't think that's strictly a function of Emerton and Nori being oof
twilit pawn
anyway anyway, we can debate this another time i gotta finish lol
bleak abyss
It's more like, you can't simultaneously teach AG to people who need to know AG and those who don't need to
Yeah I'll see if maybe Reid defines asymptote or proves something with it so we have a working definition
latent anvil
Feels hard to formulate without some kind of limits
like, my thought would be something about a limit of tangent lines in P^2
hmm no that doesn't make sense, nvm
twilit pawn
i actually feel pretty good about arguing like this:
the tangent line isn't the line at infinity
so $\partial_z F(p) \neq 0$
cloud walrusBOT
doubledual
twilit pawn
here F is the homogenization of f
so following the tangent line from p keeps you close to the curve
and you will move in the Z direction
Z very small corresponds to the other coordinates going to infinity
i dont have a definition, but if i did that would be the argument
good enough for me tbh
twilit pawn
yeah this is very convincing imo
thorn delta
Let $R$ be a ring such that $R = M \oplus N$ impiles that $N=0$ or $M = 0$ (as $R$ modules). Then $R$ has the invariant dimension property.
Any hints? I think there is some connection to be drawn with the ideals of R. In particular, if I can show that R has a maximal ideal, then I'm done. Having a little trouble getting anywhere tho 
cloud walrusBOT
kxrider
sturdy marsh
If R^n = R, then what is n?
thorn delta
1?
sturdy marsh
yes, but why
oh before you do that I should probably ask what is your definition of invariant dimension prop
thorn delta
every free module over R has the same rank
sturdy marsh
oh nvm I misunderstood the problem then lmao
sorry, im a bit sleepy
(I thought it meant invariant basis number)
thorn delta
every free module over R has the same rank
oh i said the definition wrong. i meant to say something to the effect that rank is well defined oops
sturdy marsh
ah okay, it makes more sense now
alrighty, so prove that R^n = R implies n=1
and you should be done
thorn delta
hmm okay i think i get it. At first I thought maybe there was some subtlety with internal/external direct sums here, but this makes sense.
sturdy marsh
yeah it should be just R^n = R^n-1 \oplus R = R
thorn delta
how does it make any sense for R^n-1 = 0?
sturdy marsh
well, if n=1
thorn delta
oh yea, the idea is that if R^n = R then n = 1, duh
thorn delta
im probably being very dumb and tired rn, but I don't see how to reduce R^n = R^m to the case R^k = R. If m >= n, then R^n = R^m = R^m-n \oplus R^n, and uhh...
for example, if you have $R \oplus R^n \simeq R^{n+1} \simeq R^2 \simeq R \oplus R$, its not like you can just cancel out $R$'s on both sides lol.
cloud walrusBOT
kxrider
golden pasture
you can just spam a bit of linalg to show this is true for commutative R
thorn delta
i think i got it hold on
I think i way overthought this. R^n = R => n=1 shows that R has the invariant dimension property, and sums of free modules with the invariant dimension property have the invariant dimension property 
thorn delta
right. it just wasn't clear until 5 minutes ago i could do rank(R^n) = nrank(R) 
golden pasture
actly that isnt true if you aren't given invariant basis number
youll need to prove it from the condition - R=M+N implies N=0 or M=0
thorn delta
I showed that R^n = R implies that n = 1. i.e. any basis for R has one element
thorn delta
there is a theorem that says two free modules with the invariant dimension property have the same rank iff they are isomorphic. that's probably important
golden pasture
we haven't shown R has invariant dimension property have we tho
only showed R=R^m -> m=1
but thinking of examples of R without IBN is a pain
rustic crown
but thinking of examples of R without IBN is a pain<:scream:797705821881499648>
its true for commutative R... and I don't like non-commutative R 😛
thorn delta
Suppose X is a basis for R and |X| = n. Then R = R^n. But we showed n must be 1, i.e. any basis for R has one element.
My professor constructed a ring with IBN in class lol. pretty pathological stuff
golden pasture
its true for commutative R... and I don't like non-commutative R 😛
consider End(k^\infty) :)
but yea
next obsidian
Without IBN?
golden pasture
consider End(k^\infty) :)
prob one of the nicer examples tho
thorn delta
Let $S$ be a ring. Set $F = \bigoplus_{i=1}^\infty S$. Let $R = \operatorname{Hom}_S(F,F)$. Then $R \simeq R\oplus R$.
golden pasture
yesh
cloud walrusBOT
kxrider
rustic crown
yep
golden pasture
i still cant convince myself that if
R=R^m -> m=1
then
R^m=R^n -> m=n
rustic crown
i missed the earlier discussion... but we can do it directly... right?
golden pasture
not sure tbh
rustic crown
if R^n = R^m ... then pick a maximal ideal and do some stuff to get isomorphism of vector-spaces (R/M)^n = (R/M)^m.... dim is a thing now so m = n
thorn delta
R is not commutative lol
rustic crown
i missed the earlier discussion... but we can do it directly... right?
should have read the actual question... my bad
golden pasture
isok i thought R was commutative at first then i realize it's too trivial for this
thorn delta
This is what I'm saying: $R \simeq R^m \implies m = 1$ shows that the rank of $R$ is 1 has a free $R$-module. Suppose $R^n \simeq R^m$. We have $\operatorname{rank}(R^n) = n$ and $\operatorname{rank}(R^m) = m$, and by a theorem, isomorphic free modules with IBN have the same rank, so $m=n$.
cloud walrusBOT
kxrider
golden pasture
is the rank rlly well defined tho?
like isnt the rank being a thing a consequence of IBN
thorn delta
the previous exercise was this:
golden pasture
yea you need IBN
thorn delta
not of F \oplus G. Just F and G
thorn delta
hmm, i see what you mean
wind parrot
Yeah in general, for any positive integers n and k, you can find a ring R such that R^n is iso to R^(n+k) and n,k are the least integers with this property
So you can find a ring with R^n iso R implies n=1, while still having R^2 iso R^3 for instance
cyan marten
Is the invariant dimension property only about free R-modules of finite rank?
Or is it the statement "R^I = R^J implies |I| = |J|" for general index sets I, J?
simple valley
"why" is it that the only finite subgroups of C* are cyclic?
my thought was to consider the torsion subgroup C*_T which turns out to be U(1)_T which is essentially Q/Z
and then in Q/Z it just so happens that a product of two cyclic subgroups is cyclic, so all finitely presented subgroups of Q/Z are cyclic
but the question suggests that there is a generalization of this, but I don't think my approach is much generalizable
am I missing an elementary proof
rustic crown
"why" is it that the only finite subgroups of C* are cyclic?
my first thought would be that any element must have norm 1 else there would be infinitely many elements... and next the angle made would be a rational multiple of 2pi otherwise again we have infinitely many elements... putting everything together and picking the element with least positive argument (this exists since its a finite set) that should be a generator
simple valley
sure
"norm 1" is repersenting C* as U(1) x R^*_+
and R^*_+ has no finite subgroups
rustic crown
what's U(1)?
simple valley
the unit circle
chilly ocean
Hah, *_+ is a funny face
rustic crown
oh okie
rustic crown
Hah, \*_+ is a funny face
lol
simple valley
anyway that's a proof yeah, but that doesn't really explain "why"
what property of C^* makes this happen
what is a generalization of this statement
rustic crown
i think its true that any finite subgroup of group of units of a field is cyclic
because say N(d) were the number of elements of order d, so either N(d) = 0, else there is atleast one element, call it g. now g^0, g^1, ..., g^(d-1) must be all the roots of x^d - 1=0 and now any element of order d satisfies that and so its easy to see that only phi(d) of them have order d. So N(d) = 0 or N(d) = phi(d). We can surely say N(d) <= phi(d). Now say the cardinality of the group was n, so each order would divide n and we can take the sum of N(d) over all positive divisors of n.
n = sum_ {d | n} N(d) <= sum_{d | n} phi(n) = n
the first one follows because each element is counted once... and the second is a pretty famous identity which can proved by noting that there are phi(d) elements of order d in the group Z/nZ
But this forces each N(d) = phi(d)... in particular there are precisely phi(n) >= 1 elements of order n.
This is an elementary proof... i'd say. But you can always use the structure theorem of finite abelian group to make the work easy...
rustic crown
uwu
simple valley
not super comfortable with fields
rustic crown
rustic crown
not super comfortable with fields
oh so the property of fields or rather integral domains we used was a polynomial of degree m has atmost m roots
golden pasture
Is the invariant dimension property only about free R-modules of finite rank?
finite i believe but for infinite you can prob jus do some cardinality argument?
cyan marten
finite i believe but for infinite you can prob jus do some cardinality argument?
Good point, but that doesn't work if R is really big.
golden pasture
yea
cyan marten
Suppose R^n ~ R^m, with n >= m, and we know that R^k ~ R iff k = 1. I thought of this argument, but I can't find the flaw in it:
R^n / R^(m - 1) ~ R^(n - m + 1)
and R^n / R^m is trivial, but also isomorphic to (R^n / R^(m - 1)) / (R^m / R^(m - 1)). So R^(n - m + 1) ~ R, and thus n - m + 1 = 1, so n = m.
golden pasture
im somewhat uncomfortable with the quotienting tbh
it is like saying R+A=R+B implies A+B
cuz right
there can be different modules such that M/N = M'/N i believe
(cuz it's just a extension problem and jus compute size of Ext^1_R(N,M/N))
R is noncommutative
cyan marten
Like, R^m can be embedded in R^n in several ways, not necessarily as a direct summand?
golden pasture
(cuz it's just a extension problem and jus compute size of Ext^1_R(N,M/N))
is this even well defined for noncommutative rings
huh actly
cyan marten
is this even well defined for noncommutative rings<:monkey:705409501423206420>
No idea
golden pasture
with this perspective it seems to work
ok idk noncommutative rings are screwey
i feel like it shouldnt work
but i see no reason why
cyan marten
I think the "R^n / R^m is trivial" bit is inaccurate because we only know that R^n / f(R^m) is trivial, where f : R^m --> R^n is an isomorphism.
But how can f(R^m) sit in R^n?
golden pasture
ok i see the problem is right
cyan marten
R(1, 1, 1,...., 1) is a copy of R in R^n, but is not a direct summand, I think.
golden pasture
Suppose R^m = R^n, m>n and suppose n is minimal.
We have some submodule M isomorphic to R^{n-1} in R^n, giving us R^n/M=R
say f:R^n->R^m is a isomorphism, we have
f(R^n)=R^m and f(M)=R^{n-1} otherwise contradicting minimality, then f(R^n/M)=R^{m-n+1} which cannot be equal to R unless m=n
seems legit
unless the quotient isnt well behaved
then
hm but this proof doesnt work if M=R^n tho as it implies all rings as IBN
yh i think the quotient may be more screwy
tho could we use the minimality condition
my brain cant into noncom rings
simple valley
@rustic crown thats slick
rustic crown
yep... its one of my favorite proofs...
simple valley
How do you come up with this, I have no intuition for algebra
old lava
do every exercise in d&f
carmine fossil
do every exercise in d&f
You can skip the computations tho
Like "find the grobner basis of this ideal generated by 14 elements "
golden pasture
carmine fossil
If you actually try to compute this without knowing this is the minimal basis,That would probably take like 100 steps
carmine fossil
old lava
so, we'll see when we get there
cyan marten
I don't think d and f expects you to compute this.
Maybe with a computer? Or does that trivialize it?
Dummit himself has apparently did some work in computational mathematics.
old lava
I remember one d&f question where they explicitly ask you to write a program to do it
two actually
next obsidian
Maybe with a computer? Or does that trivialize it?
Dummit himself has apparently...
I think a lot of group theorists have. Especially those involved with the classification
carmine fossil
Maybe with a computer? Or does that trivialize it?
Dummit himself has apparently...
I mean, That would just give you a grobner basis. You would still have to reduce it to a reduced basis
cyan marten
I mean, That would just give you a grobner basis. You would still have to reduce...
Well, maybe you can rely on a computer only partially (expanding polynomials and so on).
carmine fossil
Ye
old lava
if there's a bunch of computational stuff in the section
I'll just code up a python program
that's what I did for Sylow
most of the boring calculations part was done by python
carmine fossil
1/3 rd of the exercises are computations
old lava
nice
cyan marten
Over a domain, we can always write M = M' + Tor(M), right?
Tor here is the submodule of torsion elements.
stark sigil
Looks false
If the domain isn’t a principal ideal domain
You might be able to find a counterexample looking at modules over Z[x]
next obsidian
I mean
If you aren't specifying it's a direct sum, then sure
Take M' = M
if you want an actual direct sum then this sounds fake
sturdy marsh
If the domain isn’t a principal ideal domain
it isnt even true over a PID, you need a finite generation condition
for finitely generated, torsion free implies free
but you only get torsion free implies flat in the general case
but if you're willing to work with fin gen stuff, then the PID condition can be weakened to dedekind
cyan marten
I see
narrow ibex
If ( R ) is a ring with identity, not necessarily commutative, and ( x \in R ), what does the notation ( xRx ) mean? Naively I'd think all elements of the from ( xrx ) for some ( r \in R ), but I just wanted to make sure.
cloud walrusBOT
g_ijoe
cyan marten
whats the identity subgroup? no mention of this until now
The group {1}
unique juniper
i see ty
vestal snow
If \( R \) is a ring with identity, not necessarily commutative, and \( x \in R ...
I have seen that notation also used for the ideal generated by the set xrx for r in R
which ends up being finite sums of elements of the form xrx
narrow ibex
I have seen that notation also used for the ideal generated by the set xrx for r...
oh ok that makes more sense, I guess it parallels the notation ( RxR ) as finite sums of elements of the form ( rxr )
cloud walrusBOT
g_ijoe
next obsidian
wouldn't RxR just be equal to (x)?
Like it certainly is contained in this, and the other one follows by letting some of the r be equal to 1
Oh wait, we want r to be the same r on both sides
oblique river
if someone wrote RxR I wouldn't assume that the r had to be the same on both sides
carmine fossil
I mean, There's the map where you map everything to 0_S
cloud walrusBOT
squirtlespoof
carmine fossil
no
it shouldn't be from a ring with the abelian group,generated by 1_R
that would just be trvial hom
because phi(a)=phi(1)+phi(1)... some m times=0_S
because everything gets mapped to 0,yes
sorry,mb
you are correct
yes
simple valley
what's the non-awful way of showing S_4/V_4 = S_3
next obsidian
What in the shit is V_4
carmine fossil
Z_2 x Z_2
viscid pewter
K4
next obsidian
Surely there’s more than one copy of V_4 in S_4
next obsidian
So I don’t like how the problem is stated but I guess you can swing it to make it work in any case
simple valley
there's only one normal V_4 in S_4
simple valley
modulo Aut of S_4 ofc
next obsidian
I feel like you should be able to show there’s no element of order 6
Just look at the image of every order 6 element in S_4
Then you’re only left with C_6 or S_3 as your options for what the group is
So if you can rule out C_6 you’re done
simple valley
what are the order 6 elements of S_4 
simple valley
I think there isn't
next obsidian
You’re done
carmine fossil
Is there a group action way of doing this?
simple valley
That’s easy as hell
actually how do you do that
do you just play sudoku with the cayley table?
?
next obsidian
You can reduce to the case there’s 2 elements of order 3 and 3 of order 2 quickly
As long as you know internal direct products
This is for showing a non-C_6 one is S_3
simple valley
You can reduce to the case there’s 2 elements of order 3 and 3 of order 2 quickl...
how?
next obsidian
Assume it’s not C_6
You have an element of order 3 and an element of order 2 by Cauchy
simple valley
yes
next obsidian
Subgroup of order 3 is normal by being index 2
simple valley
yes
next obsidian
If you had two subgroups of order 3 you only have room for one of order 2
But then order 2 is normal so you get that the group is the internal direct product of the subgbrpip of order 2 and 3
So it’s C_2 x C_3 = C_6
Contradiction
simple valley
huh
next obsidian
So you have 1 subgroup of order 3 = 2 elements of order 3
And 3 of order 2
1 of order 1
You can act on the subgroups of order 2 by conjugation
This has to be transitive because if it’s not the orbits are either size
1,1,1 or 1,2
simple valley
hold up
simple valley
But then order 2 is normal so you get that the group is the internal direct prod...
what
next obsidian
You have
Identity
And 4 elements of order 3
So there’s only one element of order 2
So 1 subgroup of order 2
So normal
wind parrot
The action of S4 on the corners of a square gives rise to a permutation of these 3 figures, by permuting the corners and leaving the lines intact (for instance, for the permutation (12), if you switch corners 1 and 2 in the first figure you get the 3rd, if you do it in the 3rd you get the 1st and the 2nd gives itself), so S4 acts on the set of these 3 figures, giving a group hom S4 to S3. Then you can check manually that the only permutations that fix everything are {(), (12)(34), (13)(24), (14)(23)} = V4, so that is the kernel of this homomorphism. So you get an injection S4/V4 to S3, which is surjective by cardinality considerations
simple valley
by "1 subgroup" here you mean 1 mono C_2 -> G
next obsidian
What
I mean any subgroup of order 2 is C_2
There’s an isomorphic copy of C_2
you decompose the group as H x K for H order 2 and K order 3 so this is iso to C_2 x C_3 = C_6
simple valley
yeah in other contexts when we say 1 subgroup we mean modulo automorphisms
simple valley
no?
next obsidian
there’s only room for a single subgroup
simple valley
in this case yes
next obsidian
There’s only 1 element of order 2
Act by conjugation on the 3 subgroups of order 2
It’s transitive because if it weren’t one of them is fixed aka is normal
Which we know isn’t possible because then we’d be C_6
I think this is almost enough
This gives a map from your group of order 6 to S_3 which has image at least size 3
Because of transitivity
You need to just show that it isn’t size 3 by either showing it’s faithful
Or exhibiting a 4th permutation
simple valley
hold on I lost you
next obsidian
At which point?
simple valley
you decompose the group as H x K for H order 2 and K order 3 so this is iso to C...
.
next obsidian
This is if you have a normal subgroup of order 2
Then by looking at that and a normal subgroup of order 3 you show the group is iso to C_6
So we know the subgroups of order 2 can’t be normal
This is an internal direct product
simple valley
so the way I see it, if H is generated by h and K is generated by k, if they're both normal then h and k commute, and hk has order 6
next obsidian
Yes
simple valley
ok
next obsidian
So from here we have the group as one subtbroip of order 3 and 3 of order 2
simple valley
actually is that true
next obsidian
Yes it is
simple valley
seems to be only true if K is order 2
next obsidian
K is order 2
simple valley
yes but not in general
next obsidian
More generall if you have two normal subgroups H and K
Such that the product of their order is |G|
And their intersection is trivial
Then G = H x K
This is what’s called an internal direct product
You get that G = HK and you can exhibit an isomorphism from HK and H x K
You just send elements of the form (h,k) to hk
simple valley
how do you get that h and k commute?
next obsidian
Two normal subgroups which are disjoint commute
Because if h is in H and k is in K
next obsidian
Yes
simple valley
cool
next obsidian
So act on the 3 subgroups of order 2 by conjugation
Looking at orbits you must have a single orbit of size 3
Because if not there’s an orbit of size 1
And then you get a normal subgroup of order 2
Which is not okay
simple valley
ok
next obsidian
So this gives a map G -> S_3 and the image is a transitive subgroup
So it’s either size 3 or size 6
simple valley
hold up
next obsidian
So it either looks like <(123)> or S_3
simple valley
how does this give a map G -> S_3?
next obsidian
You’re acting on a set of size 3
next obsidian
So we want this to be surjevtive / injectivr
simple valley
transitive subgroup?
next obsidian
A subgroup of S_n where if it acts on {1,...,n} the action is transitive
Basically we relabel the 3 subgroups as being subgroup 1,2,3
And the image of this map is the set of permutations of 1,2, and 3
So we showed it acts transitvely
Meaning it’s a transitive subgroup
simple valley
weird name, is there a connection to other transitivities?
next obsidian
¯_(ツ)_/¯
So we want to show our map G -> S_3 is injective / surjective
Because for finite sets of the same size that’s equivalent to being bijective
simple valley
sure
next obsidian
So you have to either show that the action on the subgroups is faithful which would be injectivity
Or show that it’s not just generated by a 3-cycle because then the image can’t be <(123)> so it must be S_3
I think it’s easiest to do the not 3-cycle thing
Because if your three order 2 subgroups look like
{e,a_i} for i = 1,2,3
Then conjugating the subgroup {e,a_2} by a_1 should just swap this with the subgroup {e,a_3}
And it fixed {e,a_1}
rustic crown
are you still discussing S4/V?
rustic crown
won't it be easier to show that its not commutative?
rustic crown
the commutator [(12)V, (23)V] = (132)V
next obsidian
But we’re almost done classifying groups of order 6 anyway so
rustic crown
did you classify like D_2p?
simple valley
Then conjugating the subgroup {e,a_2} by a_1 should just swap this with the subg...
why not the identity?
simple valley
a_1 fixes <a_1> in place, why not fix the other two?
simple valley
yeah
next obsidian
So
a_1a_2a_1^-1 = a_1a_2a_1
If this were equal to a_2 then we would get that
a_2a_1 = a_1^-1a_2 = a_1a_2 which says the two elements commute.
And then a_1a_2 by them commuting has order 2
Because it can’t be identity
And by squaring since they commute we get identity
simple valley
yes
next obsidian
And from there this is equal to either a_2 or a_1
Err I guess it could be a_3 but then you get an issue in that
Okay so it couldn’t be a_2 or a_1 because by multiplying by a_1 or a_2 we get one of a_1 or a_2 is identity
So it has to be a_3
Fuck, this is annoying this should have been a lot easier
simple valley
if a_1 a_2 a_1^-1 = a_2 then <a_2> is normal and we deduced that's not the case
next obsidian
Not necessarily
simple valley
oh?
simple valley
oh right
next obsidian
Oh right
So duh
Okay
You get an order 4 subgroup
{e,a_1,a_2,a_3}
We could just as well have applied this argument to conjugating {e,a_3}
And then we’d get the same thing that the product a_1a_3 = a_2 and they commute
And likewise we can conjugate {e,a_3} by a_2 and her that a_2a_3 = a_1
Like everything I said is symmetric in the a_i
Since we know the exact same stuff about each of them
simple valley
doesn't really matter what a3 is, if a1 and a2 commute we get a subgroup of order 4 anyway no?
simple valley
ok
next obsidian
And that’s a contradiction
simple valley
I think that concludes the proof
next obsidian
Yes
Blech
That’s a bit more involved than I liked
The real good way to finish this is with semidirect products
simple valley
what bothers me is you swept 3 pages of arguments under "easy as hell"
next obsidian
I guess yeah
simple valley
and I get that it's to do with intuition
next obsidian
Well with semidirect products it does become really easy
simple valley
but like where does it come from 
next obsidian
But I tried to come up with an argument on the spot not using them
Which is kind of sadchamp
So like the easy proof using semidirect product says
Take normal subgroup of order 3 and subgroup of order 2
Call these H and K repsectiely
Then you get that G is the semidirect product of H and K and thus you look at maps from K -> Aut(H)
We know K = C_2 and H = C_3 so this becomes maps
C_2 -> C_2
Of which there are only 2
So there’s at most 2 isomorphism classes of groups
And then S_3 and C_6 fit the hill since they are not isomorphic
As one is commutative and the other isn’t
But I guess this uses a lot more theory which doesn’t fall under “easy as hell” unless you know all the theory
My B
Also det had a much easier proof of he S_4/V_4 thing
As did someone else using troop actions
carmine fossil
The action of S4 on the corners of a square gives rise to a permutation of these...
.
old lava
you might want to note that this is exhaustive for classifying groups of order 6 because by cauchy, a subgroup of order 3 always exists and such a subgroup is of least possible prime index, so it will always be normal
next obsidian
That’s what I started with
carmine fossil
tbh, Semidirect products are pretty OP
simple valley
I had to stretch my brain to remember how to prove that a subgroup of index 2 is normal today
rustic crown
group actions are so OP
I still marvel at the proofs of sylow theorem and that any subgroup of index p where p is the smallest prime dividing the order of G is normal
old lava
group actions are indeed op
chilly ocean
group actions are neat
next obsidian
Tfw I didn’t use them the entire first time I did group theory
chilly ocean
too bad it only took a mini course on lie groups for me to appreciate them
next obsidian
Tfw I did up to semidirect products and still didn’t reLize they’re good
It was during my second algebra run I was like
old lava
semidirect products are so fun
viscid pewter
i need to do more group theory, i was literally just getting into group actions and i heard that's where all the fun is
scarlet estuary
group actions are the real reason we care about groups
honestly
like some people enjoy the study of groups for their own sake and they make nice prototype algebraic structures
to build off of when defining rings or w/e
simple valley
I don't get group actions 
scarlet estuary
or algebrass over [blah]
simple valley
like you get a map from G to Aut(X) ok
simple valley
what do you do with the maps
simple valley
next obsidian
This helps a lot when
X is finite cuz then it’s just S_n
Sometimes you can figure out the permutation representation of the action
So you can start building the subgroup of S_n with elements
simple valley
might as well act on a basis of sorts and feed it to lapack
next obsidian
Lol
simple valley
guess I'm not seeing it yet
next obsidian
I found it useful to try and compute automorphism groups
Because they act naturally on the group by just applying them lol
You know a good deal of info about how many could exist because they need to send things of a given order to other things of that order
Etc
So if you have partial information about a few automorphisms you make by hand you can use that to do stuff inside of S_n
¯_(ツ)_/¯
simple valley
how do you make a group out of that tho
say |Aut(C_n)| = phi(n)
what the automorphism group is though I have no idea
next obsidian
Well okay for C_n it’s maybe not as useful
But for like Aut(Q_8) it was useful
Idk, it’s just a way to get free maps
Also here’s an insane thing Brofibration told me
The problem was to show a group of order 1722 can’t be simple
1722 can be done without sylow (you just need cauchy ig). Let the group act on itself, look at the map to Sn. The element of order 2 is an odd cycle (as 1722 = 2(2k+1)). So composing with sgn, you get a surjection onto Z/2 which gives you a normal subgroup.
carmine fossil
Doesn't that work for literally any even order group? Find an element x of even order and consider the map:G->Z/2
Which maps x to (-1) and identity to 1 giving you a surjective map
oblique river
Why would that always extend to a homomorphism?
oblique river
It's also just not true, A_n is simple for all n >= 5
but its order is even
it is true if |G| = 2m for some odd number m (I think?)
golden pasture
as an example consider the group of 2D rotations with it's action of the 2D plane
or the group of translations
or maybe symmetric group S_n on the set of n elements
unique juniper
$\varphi : G \times A \rightarrow A$
cloud walrusBOT
Yes
unique juniper
phi is my group is action here? what are the properties saying?
chilly ocean
$\varphi(g_1, \varphi(g_2, a)) = \varphi(g_1g_2, a)$ and $\varphi(1,a) = a$.
cloud walrusBOT
(T*(Terra), -dτ)
chilly ocean
if you want to write things in terms of phi
chilly ocean
unique juniper
golden pasture
unique juniper
chilly ocean
what if i started talking about differential geometry? flows are just local group actions it's relevant
unique juniper
idk that
chilly ocean
something something solutions to autonomous ODE gives you something called a flow and it has the same properties as a group action
but it may not be defined everywhere
unique juniper
nice
chilly ocean
hence "local"
celest brook
the euclidean algorithm is just matrices
latent anvil
What the fuck is petthedan
Or pethesully
I hate it
@chilly ocean answer for your crimes
chilly ocean
golden pasture
nice
latent anvil
Hate this
woven delta
Is there an emote server?
latent anvil
I think tertea has a personal one
uncut girder
Smfh
latent anvil
Tterra category theorist???
carmine fossil
Tterra category theorist???
Always has been
latent anvil
@chilly ocean you're a category theorist right can you help me with this diff top/homological algebra thing?
Check #category-theory
chilly ocean
Is there a complete classification of countable fields of infinite transcendence degree over Q?
next obsidian
Cant that not exist for set theoretic reasons?
Like you get a copy of Q[infinitely many variables] inside of it which is at least as big as an infinite product of Q
next obsidian
Actually idk if that’s as big as an infinite product of Q
bleak abyss
@next obsidian I think you should write it as the union of things in the first n variables
Like Q(x_1) \cup Q(x_1,x_2) \cup ...
So if each is countable...
next obsidian
Ah I guess so
bleak abyss
But yeah I'm not aware of such a classification unfortunately
next obsidian
I am very much disliking that Q[infinitely many] is smaller than Q[[x]]
bleak abyss
I guess the reason it's painful is that you tend to think of an element of Q(infinitely many) as big when in reality it's a polynomial
So just finitely many terms
next obsidian
Wait Dami hold up
bleak abyss
Holding down
next obsidian
This is for countable transcendence degree
Only
So that reduces the classification a bit
Since uncountably many variables literally already gives you uncountably many elements
bleak abyss
Yeah
next obsidian
bleak abyss
Like you're literally just taking any algebraic extension of Q(infinitely) as far as I'm concerned
next obsidian
Yeah
bleak abyss
But it seems like knowing that boils down to understanding all irreducible polynomials over Q no?
And that feels hard
cyan marten
What abelian groups can be endowed with field structure?
I think they are precisely direct sums of C_p and C_infinity (aka Z).
cyan marten
Yeah
carmine fossil
How can the abelian group under R,i.e.( R ,+) be seen as a direct sum in that way?
cyan marten
Hmmmm..
next obsidian
You’re right
next obsidian
This would be for fields generated by one element
Under addition
So like... as a Z module
Which is just as an abelian group bwahahahaha
cyan marten
No, C is generated by two elements and is isomorphic to R.
next obsidian
Not over Z
I meant that your classification
Holds for fields generated by 1 element over Z
cyan marten
Which is just as an abelian group bwahahahaha
Exactly. I should've said copies of C_p or copies of Q.
next obsidian
I’m not sure how R is a direct sum of Q though
cyan marten
Let K be any field, and F its prime subfield. Then K is an F-vector space, so is isomorphic as an abelian group to copies of F. Now F is either Q or C_p.
next obsidian
I’m really sleepy lol
Oh right
Now I don’t like that, I am going to instead simply not think about R being a Q vector space and instead go to sleep
😪😪
carmine fossil
Choice intensifies
cyan marten
lol
cyan marten
Let G be the absolute Galois group of the rationals. Is G isomorphic to G/H for any finite subgroup of it, H?
molten silo
What does U(R) actually mean?
old lava
the set of units of R
molten silo
So for the reals what would that be?
old lava
+- 1
mb
for the reals, it's everything
but 0
for integers it's +- 1
check the axioms
old lava
yes
next obsidian
The only case it is is for the 0 ring
Since then it’s an empty ring
Is that even a ring?
No it can’t be
sharp sonnet
zero ring
next obsidian
Yeah so I stand by my statement
next obsidian
Wait
next obsidian
I guess in the case of the zero ring
sharp sonnet
0 is a unit in the zero ring
next obsidian
0 is a unit
sharp sonnet
an element with multiplicative inverse
cyan marten
Let G be the absolute Galois group of the rationals. Is G isomorphic to G/H for ...
I normal H with finite index*
What I really want is whether the absolute Galois group of K, where K/Q is a finite extension, is isomorphic to the Galois group of Q.
cyan marten
It never will have 0
How about {0} U R*? (R* the group of units).
next obsidian
Compute an example
cyan marten
Ohh.. We will have troubles with 1.
golden pasture
I normal H with finite index*
What I really want is whether the absolute Galois ...
cyan marten
So the answer is strongly negative, because both Q and K are algebraic number fields?
That is, does the definition of algebraic number field include Q?
golden pasture
yesh to both
obsidian path
is it fair enough to say that, because this is a homogeneous system with fewer equations than unknowns, we have infinitely many solutions therefore there must be a solution where the coeffs are not all 0?
rustic crown
yep
celest brook
modern algebra
main quiver
modern algebra iff y = mx + b
cyan marten
I feel like this should be trivial, but it wasn't for me:
Suppose A, B are nonisomorphic groups of the same order. Then A^2 and B^2 are nonisomorphic.
(A, B finite)
I believe the statement is correct, however.
obsidian path
yep
wait, wouldn't it be a problem if the infinite solutions were just by one variable?
yeah nevermind, I don't think thats enough because it has to be a quadratic curve
rustic crown
wait, wouldn't it be a problem if the infinite solutions were just by one variab...
you have a 5x6 matrix, so it can't have rank 6 rank, which means kernel is non-trivial. This gives you non-trivial coefficients whose quadratic curve satisfy the given 5 points....
even if all our 5 points were same... in this situation, the rank reduces even further but this just means that dimension of the kernel increases which just amounts to saying that there are multiple conics passing through that given point.
obsidian path
but you'd need to define a linear map first right?
sorry nvm lol, i get your point
just thinking of a more formal grader level proof
thanks tho
i got it
rustic crown
i think this is formal enough... but okie 
rustic crown
you have 5 equations and 6 unknowns
rustic crown
okie uwu
rustic crown
obsidian path
my one question is tho
how do u define the linear combination
you put the vector {A, B, C, D, E, F} and multiply with matrix?
rustic crown
yep