Oh
#groups-rings-fields
opal osprey
sturdy marsh
sure...
opal osprey
End(V) is also a ring, right?
sturdy marsh
it is
sturdy marsh
it's a k-algebra
opal osprey
So you think about End(V) as a ring in this example
sturdy marsh
yes
opal osprey
So it's a ring homomorphism from k[x] to End(V)
opal osprey
Haven't really studied much of polynomial rings yet.
sturdy marsh
alright
opal osprey
So I don't know how we could naturally define this homomorphism.
sturdy marsh
Try to do it in the reverse order
assume we had a map from f:k[x] ---> End(V)
so we have f(x) in End(V)
say f(x) = T
then what is f(x^2)?
remember it's a ring map
sturdy marsh
T^2, but yes
sturdy marsh
so what is f(p(x))?
sturdy marsh
where p(x) is a polynomial in x
yes
so the image of 'X' seems to determine the homomorphism
i.e. you can recover the map if I told you where X went
as if you know where x went, you know where all polynomials in x go
and that's the entire ring
so now, say I gave you a linear map T: V---> V
opal osprey
This somehow works kind of like a basis for a vector space.
sturdy marsh
can you define a 'special' map from k[x] ----> End(V)
sturdy marsh
This somehow works kind of like a basis for a vector space.
Exactly!
That's a great observation
opal osprey
can you define a 'special' map from k[x] ----> End(V)
So I want a ring homomorphism k[x] --> End(V) that takes x to T : V --> V?
sturdy marsh
Yup!
opal osprey
Idk, can't you just let f(x) = T, for instance? And then define f(p(x))=p(f(x)) for every p(x) in k[x]?
sturdy marsh
yup
opal osprey
Also, just a silly question
A k-algebra is a vector space equipped with a bilinear product, right?
So the bilinear product in end(V) would be just linear map composition?
Or is it another bilinear product I can't see?
sturdy marsh
an R-algebra is almost the same thing as an R-module
except there's a multiplication on the module
and R acts in a way compatible with the multiplication
you might want to try interpreting it as a map from R to an endomorphism ring again
but be careful about the kind of endomorphisms
and the the properties the map should have
opal osprey
Ok so
I kind of get this example down conceptually
But now
I am wondering why is this example important
I guess that what you are trying to say to me
sturdy marsh
there's a structure theorem for modules over k[x]
or really modules over a PID (k[x] is a PID)
opal osprey
Is that R-module homomorphisms have some similar properties as linear transformations, i.e vector spaces homomorphisms.
sturdy marsh
and you can recover a lot of the canonical forms for matrices using the strucure theorem
opal osprey
What do you mean by Structure Theorem?
sturdy marsh
classification of finitely generated modules over a PID
opal osprey
That's already a motivation for me
sturdy marsh
ideals of the ring are modules over the ring
so you can study ideals using module theory
really, just read d&f
they have some very detailed examples
including the vector space + linear operator example
opal osprey
really, just read d&f
It's just that I got to track down some few easy examples and motivation before actually working stuff out.
sturdy marsh
it might also help to try out some linear algebra problems, assuming knowledge of canonical forms for matrices
you'll see that they are very powerful
opal osprey
Yeah, I've already studied a bit of canonical forms.
But tbh
Besides the applications of the Jordan Canonical form to study ODEs
I've never really found any use of them
So I might be a little bit rusty
sturdy marsh
well, cayley-hamilton is a one-liner assuming RCF (or JCF in the algebraically closed case)
and a good part of math is either turning things into linear algebra or exploiting whatever "linear algebra" you have
opal osprey
and a good part of math is either turning things into linear algebra or exploiti...
Yeah, I kind of have got this idea in mind already.
But the problem with canonical forms is that I don't really get how much of information you can get about a linear transformation/matrix by looking at it's canonical form.
sturdy marsh
so do problems
inland bough
can someone explain what the localization of a commutative ring by a prime ideal is?
i'm reading stuff online and it's not making sense
this doesn't make sense if S is a prime ideal
inland bough
ohh gotcha and since it's a prime ideal it's complement is closed under mult
tysm literally was so confused by that
latent anvil
np
opal osprey
There's another thing I'd like to ask
Which is
What are some important results about Commutative Rings which are not true to non-commutative rings?
Is it a big loss?
latent anvil
yes
Localization is much weirder and harder to work with for noncomm rings, for example
there's not a good notion of a prime ideal
opal osprey
I'd always liked to see a list comparing both structures.
dim escarp
Does monoid(or free monoid) attain many uses?
latent anvil
it would be a long list
prisma ibex
@opal osprey Say that two rings are Morita equivalent if their categories of modules are equivalent. Any two commutative rings are isomorphic iff they are Morita equivalent, but this is far from the case for non-commutative rings (which is why the notion of Morita equivalence is actually interesting in this setting)
opal osprey
Oooh
This example is actually cool
Because I've asked before how can you study rings by looking at its R-modules.
So it kind of gives somewhat of an answer to the question when it comes down to Commutative Rings.
prisma ibex
Yup! In the non-commutative case it gets quite interesting, in a lot of cases you can study modules over some ring you're interested in by finding some Morita equivalent ring whose modules are easier to understand
a really stunning example of this is the following: say you have a finite dimensional k-algebra A over a field k. Then A is Morita equivalent to a path algebra kQ/I where Q is some quiver, kQ is its path k-algebra, and I is some two-sided ideal of kQ
but the structure theory of kQ/I-modules is substantially more tractable: most of the relevant computations can be done combinatorially and much of the structure is inherited from graph-theoretic properties of the quiver Q
next obsidian
How in the fuck would you show rings are Morita Equicalent lol
Also I know you can talk about a more general notion of Morita Equivalence but I have no idea which categories that’s for lol
prisma ibex
usually you prove Morita equivalence by constructing a suitable progenerator
If you have two rings R and R' and additive functors F:RMod->R'Mod and F':R'Mod->RMod then these are equivalences iff there is a balanced (R',R)-bimodule R such that P_R and P_R' are finitely generated projective generators and we have natural isomorphisms F=P_R\otimes_R- and F'=Hom(P_R',-)
in this case R and R' are Morita equivalent iff R'=End(P_R) and R=End(P_R')
usually these things can actually be constructed/computed explicitly, so this is to say that Morita equivalence is something you can usually check explicitly by constructing certain bimodules with nice properties
next obsidian
Weird wtf okay
dim escarp
I'd like to ask again, Does monoid(or free monoid) attain many uses?
I guess I could ping helpers as well
<@&286206848099549185>
nova plank
Well, all rings with identity, groups, etc are monoids and those are useful
carmine fossil
I guess the person meant monoids that aren't groups
nova plank
I guess so too, but they didn't specify 
golden pasture
not rlly
i dont rlly see it being investigated to a super significant extent but there are defo people to study all these stuff
like there are books on really painful structures like near semirings or smt
nova plank
I know someone who studies near semirings. Looks boring af.
dim escarp
Oh, so ppl could study, like, monoids as well
Someone talked in terms of free monoid with sets as elements, I was baffled at their term usage
I guess that kind of stuff is also kind of valid
prisma ibex
monoids can be useful yes
prisma ibex
So I mean to begin with monoids are completely natural much like groups are
there's a general notion of monoid objects in different categories
monoid objects in sets are classical monoids, which are a fairly general thing
but if you interpret this in other categories you get lots of other familiar algebraic objects
e.g. monoid objects in the category of Abelian groups are rings in the usual sense
prisma ibex
free monoids specifically axiomatize words and concatenation
they come up in computer science fairly often for instance
dim escarp
Yep
I mean
It was like CS ppl trying to talk abt math
And they used terms like "list of set S"
That made me got disgusted
prisma ibex
I mean that's more or less accurate
dim escarp
I know it is fairly useful in CS, but idk if it is worthy in math
prisma ibex
I mean there are lots of structures in math that in part have this kind of structure of like words + concatenation
lots of algebras are built out of this
dim escarp
Interesting
prisma ibex
e.g. algebras associated to quivers (directed graphs) where the monoid structure is given by concatenation of paths in the graph
prisma ibex
in the case where your paths are not reversible yea
dim escarp
I guess I (who doesnt even know much) was just behaving like gatekeeper then
prisma ibex
no I think this is a fair question
dim escarp
I mean I was being cynical about using notation like "list of set"
Guess I was just gatekeeping towards them talking about math
prisma ibex
yea I mean "list of set" is maybe not the most precise way to say this but it's roughly the right idea
dim escarp
Time to time I get the urge, as I feel like they are abusing words. (Perhaps they even are not doing that, but)
I need to tone myself down with this urge.
marble whale
I'm having some trouble comprehending factor groups. I have Z + 2/5 and I'm supposed to figure out it's order in (Q/Z, *) where (Q, +) and (Z,+). (Z, +) is said to be the normal subgroup of (Q,+)
marble whale
I understand how orders work in Z_n but I don't really have good samples of how factor groups operate
carmine fossil
(Z+2/5)+(Z+2/5)=(Z+4/5)
in general, (a+Z)+(b+Z)=(a+b)+Z
And (a+Z)=(b+Z) iff a-b is in Z
marble whale
hmm that looks neat, wouldn't that mean that the ord in this case is undefined as adding 2/5 any number of times will not result in 0
carmine fossil
Z=2+Z
marble whale
Hmm this is rather hard to comprehend
(Z+2/5)^a would probably be Z+2*a/5 if I've understood this correctly
but the notation (Q/Z,*) doesn't tell me what the e of (Q/Z, *) is and if it even has one
marble whale
to reverse Z + 2/5 in * I'd need Z-2/5
depends on how the group works. Here things seem to work by addition and thus it would be Z+0 as it doesn't do anything in (Z+a)*(Z+0)
ah so the identity can't be constructed with (Z+2/5)^c where c >= 1 which is something I'm used to
since exponent notation works in a strange way with these algebra notations, (Z + 2/5)^0 would be the only way to get the identity and it would mean that (Z+0*(2/5)) is the identity
meaning that order is 0
marble whale
Yes as Z + 5*(2/5) = Z + 2 but what's significant about it
oh it's the first integer that comes up, is that what Q/Z means?
Ah this seems rather simple now when I understand the meaning of Q/Z
chilly ocean
Can someone help me understand this definition of the “thickness” of a subfield?
I’m not sure what tr and d mean.
wind parrot
transcendence degree probably
thorn delta
this just follows directly from the fact that C is a k dimensional subspace of Fn. There are k vectors in a basis for C and there are |F|^k distinct linear combinations you can make out of basis vectors in C
well, they are distinct assuming the vectors in C we are taking linear combinations of are linearly independent
np
final dove
I want soe help with representation theory. I'm trying to understand the Frobenius-Schur indicator. I don't understand why it's non-trivial to find a G-invariant bilinear form for an irreducible representation r of a group G. Doesn't the usual "weyl inner product" automatically give us an inner product? (which is a symmetric bilinear form)?
sturdy marsh
As they used the word "usual", they might be talking about the averaging trick
final dove
@sturdy marsh yeah, that's the one I mean
I got the issue; the problem is that inner products on C aren't bilinear; they're linear in the first argument, and conjugate linear in the second
so the existence of an inner product doesn't tell us anything about the existence of a bilinear form
I learnt that the type of form an inner product on C is called "sesquilinear"
are there other interesting examples of sesqulinear forms?
prisma ibex
yes, a lot of natural examples of sesquilinear forms are (skew)-Hermitian forms and there are many sources for these
e.g. inner products on complex Hilbert spaces in general
Hermitian forms show up pretty naturally in Kähler geometry for instance
final dove
I see.
chilly ocean
A field is rigid if its only automorphism is the identity map. What kinds of field extensions of $\mathbb{Q}$ are or are not rigid? Finite extensions, algebraic extensions, transcendental extensions, or what?
cloud walrusBOT
lugita15
uncut girder
Q(cube root of 2 is rigid)
Q(alpha) where alpha is an algebraic number and Q(alpha ) does not include any conjugates of alpha is a rigid extension
chilly ocean
@uncut girder Can infinite-degree algebraic extensions of Q be rigid? Can transcendental extensions of Q be rigid?
chilly ocean
@uncut girder Yes to both?
chilly ocean
@uncut girder What about algebraic extensions of infinite degree? Can they be rigid?
chilly ocean
@uncut girder OK, maybe I should turn my question around: what types of fields can NOT be rigid?
uncut girder
(bro im so good at galois theory )
if you have at least one nontrivial automorphism
uncut girder
any splitting field of a rational polynomial with degree >1 is not rigid
chilly ocean
@uncut girder In any case, here is my situation. I am working on a research problem studying a certain field, and I’m trying to figure out its properties. One of the few properties I know it has is that it’s rigid, so I was hoping I could use that to derive other properties about it.
chilly ocean
what is the field?
It requires some knowledge of logic to describe it, are you familiar with ultrapowers?
chilly ocean
no
OK, well basically the field is obtained by taking an infinite product of Q with itself, and then taking a quotient of that.
uncut girder
whats the quotient?
if its rigid all that tells you is that galois theory isnt interesting in this case 
delicate bloom
the real numbers and p-adic fields are rigid and both come from completions, I wonder if constructing a field with infinitely many extensions by completion is always rigid
chilly ocean
whats the quotient?
That’s the part that’s hard to describe. But suffice it to say that it’s quotient by an equivalence relation that says that two elements of the infinite product are equivalent if they agree on “most” of their components. Where the definition of “most” is what requires some background in logic.
delicate bloom
ah, nevermind, you could just do a finite extension before completing
vestal snow
Is anyone here doing Arizona Winter Semester?
vestal snow
Nice!
uncut girder
are you?
vestal snow
Yeah
uncut girder
nicenicneincie
prisma ibex
I missed the deadline to apply but also a lot of the things in the semester are things I've already studied a lot in the past
should have applied as a helper or something though
uncut girder
they are probably posting all the lectures online anyway as the have done in the past
prisma ibex
oh yea of course
vestal snow
I missed the deadline to apply but also a lot of the things in the semester are ...
I think it would mostly be stuff you've done already
prisma ibex
yea exactly
I'm kinda sad that the planned topic was moved to next year
although the timing might be better for me
more time to learn more Langlands stuff
uncut girder
im very happy they have this undergrad workshop
uncut girder
whats that
vestal snow
Check the Zulip server
chilly ocean
Suppose $F$ is a rigid field of characteristic zero and infinite transcendence degree. Then is $F$ necessarily “large” in the below sense?
cloud walrusBOT
lugita15
chilly ocean
compact needle
It's false for the real numbers
Maybe I'm wrong, I didn't notice smooth in the hypothesis
My counter example is not smooth at the R-rational point
compact needle
I haven't any idea if it's true. A search for references on the subject has only turned up proofs of other classes of fields as being large might land you an annals paper. If you actually want to know, you might try overflow or directly emailing people who study large fields.
next obsidian
I was thinking about applying for Arizona Winter but it's too NT for me given what I know at the moment I felt
Proving the exactness of the snake lemma in an abelian category is hellish
fierce perch
Without freyd mitchell you mean?
golden pasture
Is anyone here doing Arizona Winter Semester?
yeeee
golden pasture
Without freyd mitchell you mean?
:screamin:
carmine fossil
1 generates a subgroup of G
If you take the set of all elements in G,You will find the subgroup generated by the set is the Group itself
H should also be closed under the operation
marble whale
if I have a group G = {Z_13 \ {0}, *} and its subgroup, H ={1, 3, 9}. Would the factor group (Z_13\ {0}/H, •) be
{1,2,4,5,6,7,8,10,11,12}?
From what I've understood, in (Z_a/H, *) reaching either a or any value in H will result in the identity of *. And as 1 is the identity of • in this case, the set of values is the same as in Z_13{0} {3,9}
hot lake
nova plank
The factor group consists of the cosets of H. What are the distinct cosets of H?
marble whale
I have not yet found an understandable explanation of what / means in the factor group notation and that there was what I got from the only example I've understood this far.
nova plank
xH is just the set you get by multiplying all the elements of H by x. So you can just manually work out what the cosets are.
All the elements that are in the same coset, will give the same coset when you multiply it with the elements of H.
Actually do it once to see that, then in the future you will know, once an element appears in a coset, you don't need to multiply by it because you will just get the same coset.
marble whale
Well, I've now taken 1,3 and 9 and started to multiply them by integers starting from 1 to 13. Multiplication by 2 and by 5 seem to be the same, just different order, same with a few other multiples. Going on after 13 should be pointless since mod13.
nova plank
I did not say to multiply the integers by all the elements of Z_13. I said to multiply the elements of Z_13 by the elements of H. For example, if H = {1,3,9} then 2H={2,6,18}={2,6,5}
Then do 3H, 4H, 5H, up to 12H.
marble whale
Hmm that's what I thought I did. I got 1H = {1,3,9}, 2H = {2,6,5},... ,12*H={12, 0,0}
nova plank
Oh, okay, maybe you just explained it in a way that confused me.
The last one is definitely wrong
marble whale
Perhaps I explained it poorly
nova plank
You should have four distinct cosets, and they should all appear three times. Can you tell me the 3 different answers?
marble whale
a moment, I probably did some mistakes here
{1,3,9}, {2,6,5}, {4,12,10},{7,8,11} they all appear 3 times.
nova plank
That looks right
marble whale
values in cosets seem to be unique and this is a bit surprising but perhaps it was to be expected
nova plank
I'm not sure what you mean by that. Each element appears in exactly one coset. So we say the cosets partition the group.
marble whale
that's just the thing, if Z_13 would have been Z_12 or something else, I would have had different results. This coset partitioning would probably not have happened at least not as neatly as this.
vestal snow
I was thinking about applying for Arizona Winter but it's too NT for me given wh...
I'm doing it and I can guarantee you know more than me
marble whale
Thanks a lot @nova plank. I was stuck on this for hours. I'd repay you somehow but I guess I'll just tell you that my other cat is named Luna and so you can imagine that I named my cat after you or something.
nova plank
Haha
next obsidian
I'm doing it and I can guarantee you know more than me
No I literally know like 0 NT haha
vestal snow
Did your algebra class not do stuff like Gaussian integers when doing the section on PID, UFD, and EDs?
chilly ocean
I haven't any idea if it's true. A search for references on the subject has only...
OK thanks
next obsidian
I touched on it two years ago
All I know is like when does a prime factor in Gaussian integers and it’s about your residue class mod 4
And then 2 is weird
But that’s the extent of it lol, I don’t know the slightest lick of ANT
sturdy marsh
Chmonkey you could probably still follow whatever is being taught at the winter school
with no trouble
prisma ibex
^
thorn delta
what is the winter school? Don't most people have regular classes and stuff while this is going on?
vital quail
What's the 🅱️ínter school can hsers apply
vital quail
They really did me dirty like that
Speaking of which I mean I don't actually mind much about that but are there any such things going on this summer
"Arizona summer school" mayhap
next obsidian
Oh haha
I looked at the topics and went
“Ah this looks like real NT shit I don’t know this stuff”
latent anvil
Yeah chm I was thinking of applying
And we have the same amount of NT background
(decided not to because I have enough going on this quarter and it would be concurrent with my classes)
drowsy echo
Maybe not correct place for question but I doing a groups course and is question in my specimen test I don't understand
next obsidian
what's a specimen test lol
chilly ocean
biology server in #old-network 
drowsy echo
Let S be non-empty set. Which of following does not give rise to a partition of S.
a)Injective function with codomain S
b)The subset {(s,s)|s in S} of S x S
c)equivalence relation on S
d) None of above
i think a) or d).
my thought: if injective function is bijective, inverse can be use.
if not, union of {s in im(F)|s in S} and {s not in im(F) | s in S}
next obsidian
This isn't really group theory, but the set in b) isn't even a subset of S
drowsy echo
it is subset of S x S
next obsidian
I guess it's hard to figure out how you're interpreting "gives rise" to a partition
scarlet estuary
what is meant by "specimen test"?
next obsidian
I mean if you take that to be an equivalence relation you get a trivial partition of S into singletons
drowsy echo
what is meant by "specimen test"?
practise exam for real exam next week
scarlet estuary
ah.
next obsidian
Idk like I think you could argue that each of them gives rise to a partition
scarlet estuary
yeah this question is kinda vague
next obsidian
like for a) you can get a partition by two sets
im(f) and the complement
for b) if you interpret that defining an equivalence relation you get the partition of S into singletons
and c) just gives rise to a partition by equivalence classes
But it isn't precise what it means by "gives rise" so like...
scarlet estuary
S partitions itself trivially
next obsidian
I don't like that this is multiple choice lol
drowsy echo
i do not either
next obsidian
Yeah idk what to say
c) just
in no world could that not give rise to a partition
for a) and b) it's too vague IMO
drowsy echo
yes, i agree
drowsy echo
the empty function injective yes?
next obsidian
trivially yup
drowsy echo
does this give rise to partition
scarlet estuary
the most consistent interpretation of this question i have is that each is describing a set of ordered pairs
next obsidian
then yeah, it gives you the partition into just S
scarlet estuary
so in (a) we're actually looking at graph(f)
next obsidian
ah
scarlet estuary
and thus all 3 are relations
next obsidian
I guess that makes sense
scarlet estuary
in which case (a) doesnt necessarily give rise to a partition
scarlet estuary
but its worded really vaguely
next obsidian
it couldn't unless f is a self-map
scarlet estuary
its hard to tell just from this context what theyre asking for
next obsidian
like you're not even a subset of the right thing
drowsy echo
i think give rise to is synonymous to 'can always be used to create a partition'
drowsy echo
in which case (a) doesnt necessarily give rise to a partition
why is this?
drowsy echo
the most consistent interpretation of this question i have is that each is descr...
and ordered pairs of?
scarlet estuary
but "can always be used" is very vague :/
like im interpreting these parts as all describing relations (functions ofc just being a specific type of relation)
so the function in (a) is a set of pairs {(x, f(x)) | x in domain}
and obviously (b) is a set of pairs (s, s) as described, and (c) is just an equivalence relation, so also a set of pairs describing elements related by it
and if we interpret this as asking us to partition the set from the corresponding relation, as per https://en.wikipedia.org/wiki/Partition_of_a_set#Partitions_and_equivalence_relations, (a) may not necessarily give rise to such a thing
since if the domain of f is a strict subset of the codomain with strictly smaller cardinality, it will never partition the set in that way
but
its hard to tell if this is the interpretation theyre looking for
since again the way they worded it is really vague
if they allow ANY interpretation, even a generous one
then we could add an extra "if it's not in the relation then just partition it into its own class" and then (a) is fine
i really dont like how this question is phrased/formatted, in any case.
fierce perch
L a finite extension of K then it is Galois iff L \otimes_K L is isomorphic to L^n for some n. Is there something similar for infinite?
light tusk
semisimplicity
@fierce perch I think you're talking about the stuff in ch 8 of Milne's Notes: https://www.jmilne.org/math/CourseNotes/FT.pdf
surreal basalt
Is it correct to say that two homotopic chain maps might have non homotopy-equivalent cokernels?
I’m trying to gain intuition on why the homotopy category is not abelian
junior furnace
I think this falls under abstract algebra but let me know if this is the wrong channel: suppose (i, j, k) are an orthonormal basis for R3, and P(i, j, k) is a polynomial over i, j, and k. I'm trying to find the polynomials for which P(Ri, Rj, Rk) = P(i, j, k) for any (proper) rotation matrix R. For example, I think P(i, j, k) = i^2 + j^2 + k^2 and P(i, j, k) = ijk + jki + kij - ikj - jik - kji both have this property. Does anyone know of an existing answer to this question, or how I could go about trying to identify all of them?
The only other ones I have found have been derivative of those two in some way
delicate bloom
what does it mean to multiply basis vectors
junior furnace
It's distributive and associative and that's it
like (i + j)(j + k) = (ij + j^2 + jk)
delicate bloom
what's the difference between a rotation matrix and a proper rotation matrix?
junior furnace
Just clarifying that improper rotations don't count
delicate bloom
what's an improper rotation?
junior furnace
A rotation that flips the orientation so its determinant is -1 instead of 1
delicate bloom
only thing that jumps out at me is it requires cyclically permuting i,j,k as an obvious criteria
junior furnace
Yeah, that was where I was too but it's insufficient
delicate bloom
and I suppose you also have keeping one fixed while turning one into the other negates the other
junior furnace
For example it fails on P(i, j, k) = i^4 + j^4 + k^4
If you take $P(\frac{1}{\sqrt{2}}(i + j), \sqrt{2}}(i - j), k)$
oops
$P(\frac{1}{\sqrt{2}}(i + j), \frac{1}{\sqrt{2}}(i - j), k)$
cloud walrusBOT
NotASemicircle
junior furnace
Then your i^4 term is 1/2 instead of 1
junior furnace
There are fourth degree answers, though. Like $P(i, j, k) = ijij + jiji - ijji - jiij + ikik + kiki - ikki - kiik + jkjk + kjkj - jkkj - kjjk$
cloud walrusBOT
NotASemicircle
sturdy marsh
Is it correct to say that two homotopic chain maps might have non homotopy-equiv...
should be true
have you tried to find an example?
anyway, derived cats are rarely abelian
next obsidian
The homotopy category isn't the derived cat yet
It's only triangulated, but the same statement holds
sturdy marsh
well yeah it's almost
wind steeple
it's not triangulated, why ? the suspension functor isn't an autofunctor
next obsidian
hmm??
isn't the homotopy category triangulated?
I thought the whole point is you get a triangulated category and then you localize it to get your derived category
wind steeple
uh are talking about hTop or the homotopy category of chain complexes ?
in hTop there's no desuspension in general
latent anvil
homotopy category of chain complexes
At least I assume so, seeing as this started by talking about cokernel of chain homotopic chain maps
surreal basalt
@sturdy marsh I tried, but I’m kinda slow with computations
homotopy category of chain complexes
@latent anvil yep
latent anvil
I think you can take a contractible complex
which is nonzero
Wait no lol
Nvm lmao
surreal basalt
@next obsidian> isn't the homotopy category triangulated?
@next obsidian yes, although I dont think that itself implies no cokernels (something about homological dimension)
Man I should really understand how to quote properly
cloud walrusBOT
Mecejide
surreal basalt
Notation for what?
sly crescent
Symmetry groups
I was wondering whether the plus sign was supposed to be superscript.
wind steeple
oh yeah it is triangulated then but it is not interesting since it admits cokernels and kernels ig
languid meteor
hey all, can I assume that since I can factorise 17 = (3+2i)(3-2i) that the only maximal ideal which contains 17 is (3+2i). [Since 3+2i is irreducible in Z[sqrt(-2)] and its a PID]
could there possibly be any other ideals?
the other factorisations should be associates of (3+2i)
languid meteor
oh thats true, sorry. So we have two then
its not possible to have any others right? Since 17 factors into those two elements
sly crescent
Is there a simple way to find the largest common subgroup of two groups?
next obsidian
its not possible to have any others right? Since 17 factors into those two eleme...
argument makes sense to me. A maximal ideal containing 17 must contain one of those two, but then it’ll be equal to ideal generated by that irreducible
uncut girder
cause i wanted to show that factorization was in Z[2i]
uncut girder
yeah
uncut girder
a prime number is a sum of two squares iff its 2 or 1 mod 4
next obsidian
oh that's only for primes
uncut girder
also a composite number is a sum of two squares iff all its prime factors are 2 or 1 mod 4
wait i said that wrong
anyway the ideal generated by 17 in Z[sqrt(-2)] splits completely as
(1 + 2\times 2i)(1 - 2\times 2i). The two ideal factors are distinct because 17 does not ramify in Z[sqrt(-2)]
if using the fact that ramified primes divide the discriminant, and the discriminant of Z[sqrt(-2)] is -2*4
if you dont have access to this fact then you need to prove 1 + 2\times 2i and 1 - 2\times 2i are not associates in another way
next obsidian
Just iterate all the units
there's only 4
err... in Z[2i] there's only 2
you could like embed into Z[i] if you know the units there already
vestal snow
Is there a good description of the function field corresponding to an elliptic curve?
prisma ibex
I mean you can put the elliptic curve into Weierstrass form and write down the function field that way
red imp
Let $G=S_n$ and $\Omega$ be the set of $k$-subsets of ${1,2,\dots,n}$ with $k<n/2$. Let $G$ act on $\Omega$ by ${a_1,a_2,\dots,a_k}^g={a_1^g,a_2^g,\dots,a_k^g}$.
I've shown that this action is transitive, but the next part of the question is to find all orbitals of $G$ in its action on $\Omega$. Is there not only a single orbit since the action is transitive?
cloud walrusBOT
woven delta
Hint: If you fix a k and look at all the k element subsets of {1,2,...,n} then the action is transitive on those subsets. On the other hand if you look at all of the k element subsets where k can range there is more than one orbit
red imp
I thought so! Maybe my teacher worded the question badly
(b) made me assume that k was fixed
woven delta
yeah I guess there was just a mistake in the writing
red imp
thanks Liquid
also the rank of G would be n-1 right since it's generated by 2-cycles (12), (13), ..., (1 n)
woven delta
I'm not sure what rank you're referring too
red imp
me neither lol
is there one specific to permutation groups
oh right he defines the rank as the number of orbitals, not the number of generating elements
👍 thanks again
nova plank
The hint is pretty clear. Where are you stuck?
next obsidian
Is this an exam?
paper flint
even if not attempted
nova plank
Well, why not? Jk
nova plank
Anyway, do say where you are stuck
next obsidian
I wouldn’t worry about the subgroup bit, it’s completely irrelevant
Do as the hint says
It tells you which theorem to apply, and then you can use a result which tells you the order of a quotient of groups
carmine fossil
Hint:||Divisibility||
cloud walrusBOT
Mecejide
sly crescent
This is really cool
junior furnace
What does it mean?
sly crescent
What does it mean?
The symmetry group of the 24-cell dual compound is isomorphic to the direct product of the symmetry group of the cube with itself.
junior furnace
The symmetry group of the 24-cell dual compound is isomorphic to the direct prod...
That is really cool
thorn delta
what does the first isomorphism theorem tell you here?
next obsidian
Use Lagrange’s theorem
It tells you that the size of G/H = [G:H]
And the latter is just |G| / |H|
thorn delta
theta is any homomorphism. it has a kernel N = ker(theta), which is a normal subgroup, and the first isomorphism theorem just says that theta induces a well-defined isomorphism from G/ker(theta) to the image of theta given by that formula.
so, the elements of G/ker(theta) = G/N are cosets Ng for g in G
does that help kinda?
np, so how can you apply it to your problem? The homomorphism in question would be f \circ i
thorn delta
you have an isomorphism from G/ker(f circ i) to im(f circ i) = f(A). Now think about how the orders of G, ker(f circ i) and f(A) fit together here
And lagrange says that |Ng| divides |G|?
kind of. lagrange says that the order of any subgroup of G divides the order of G. N is a subgroup of G, and |N| = |Ng|, so yes, |Ng| divides |G|.
thorn delta
so you've got |G|=|f(A)| |ker(foi)|.
should be |A| = |f(A)||ker(foi)|
thorn delta
first iso gives you |A/ker(foi)| = |f(A)|
and lagrange gives you |A|/|ker(foi)| = |A/ker(foi)|
I think slim was using G for a generic group at one point and that got mixed up with A lol
np
next obsidian
What if it wasn't?
errr
Okay so here's a way to think about it
consider f|_A: A -> H
what's the kernel of this?
You can write it in terms of stuff that show up
f restricted to A
you can consider this as f\circ i
if you like
it's the same function
so what's its kernel?
by definition it's the stuff in A which die via f|_A which is just the same thing as f
right
that's ker f|_A
and ker f = {g in G| f(g) = e_H}
so if you die via f|_A
1: you're in A
2: you're in ker f
right?
does it go the other way?
if 1 and 2 are satisfied are you in ker f|_A?
okay
so this says ker f|_A = ker f \cap A
Okay so A being a subset of ker f
is equivalent to saying that ker f\cap A = A
does that make sense?
intersection
$\ker f\cap A$
cloud walrusBOT
Chmonkey 2.0
next obsidian
so do you believe this?
This is a general set theoretic thing
prove that $S\subseteq T$ if and only if $S\cap T = S$
cloud walrusBOT
Chmonkey 2.0
next obsidian
anyway, I need to go, but the point of this is
once you get that, you get the following
Oh shit
I think
I did it backwards
Okay
$\ker f|_A = \ker f\cap A$. We wanted to show that $A\subseteq \ker f$ which is equivalent to $\ker f\cap A = A$
Oh wait
OKay
cloud walrusBOT
Chmonkey 2.0
next obsidian
So to show $A\subseteq \ker f$ we can just show that $\ker f|_A = \ker f\cap A = A$
cloud walrusBOT
Chmonkey 2.0
next obsidian
But $\ker f|_A = A$ simply says that everything dies under $f|_A$
cloud walrusBOT
Chmonkey 2.0
next obsidian
so suppose that it were not the case that everything died under $f|_A$
cloud walrusBOT
Chmonkey 2.0
next obsidian
then $|f|_A(A)| > 1$
cloud walrusBOT
Chmonkey 2.0
next obsidian
By Lagrange's theorem $|f|_A(A)|$ divides the order of $H$, but this leads to a contradiction because by part 1 you know that $|f|_A(A)|$ has to divide $|A|$
cloud walrusBOT
Chmonkey 2.0
next obsidian
The contradiction comes from the assumptions you started with for part (b)
anyway gtg
try to work out all the details
carmine fossil
A=ker(f|_A)
Ok, That's not direct
But, Since we are dealing with finite groups,|A|=|ker(f|_A)| and ker(f|_A) is a subset of A, that's true
dim escarp
Is it normal that I forgot most of what I learned in intro class after a month or two
golden pasture
depends on familiarity sometimes (at least me) still needs to open certain texts to like check again
carmine fossil
You shouldn't forget things like first iso(if group theory was an intro class)
chilly ocean
as long as you remember why you care about groups
red imp
is there a way I can see whether GL(3,R) acts on R^3\{0} primitively?
red imp
me? no but I know linear algebra relatively thoroughly
oh really
I'm only now learning about group actions
carmine fossil
Well,He doesn't say that explicitly
dim escarp
You shouldn't forget things like first iso(if group theory was an intro class)
I forgot this
meh
carmine fossil
idk if this is helpful, but let GL(3,R) act on R^3 by:
A.v=T(v) where T is the corresponding linear transformation,given a choice of basis
dim escarp
Oh no I even forgot what maximal subgroup means
red imp
it's a subgroup such that if $H\leq K\leq G$ then either $K=H$ or $K=G$
cloud walrusBOT
red imp
where K is another subgroup
so like the only subgroup it's contained in is the entire group
red imp
I guess it's worth noting that the first action is transitive but I'm not sure where to go from there
okay I ended up submitting the quiz and the top option was correct, but I'd still like to know exactly how we can tell
golden pasture
I forgot this
it occurs often enuf for other structures that you can easily rederive the group version someday
dim escarp
Oh, thank you
golden pasture
Oh no I even forgot what maximal subgroup means
same for this as well
dim escarp
okay I ended up submitting the quiz and the top option was correct, but I'd stil...
Now that I see what it is, I think it is imprimitive because of presence of a block
red imp
yeah I couldnt spot the block though
dim escarp
Did you spot it now
red imp
no
dim escarp
Perhaps looking at what is preserved through linear transformation would help
dim escarp
What is linear transformation after all?
red imp
is that a rhetorical question
thorn delta
y=mx+b
dim escarp
I mean literally
red imp
what's the block you saw
dim escarp
You'll find out if you formalize what linear transformation is
red imp
I know what a linear transformation is lol
dim escarp
I mean, would you explicitly write it out
Hm strange, it is inevitably linked with the block
red imp
I found it now, you could have just said "rays"
red imp
@ivory dust whip out a rectangular prism shaped object and see if you can spot the symmetries that generate the group
from flipping my phone around I think there are 2 of them
ivory dust
If its in the form of a parrallelogram-based prism i see only 2
the identity, and a rotation of 180 around Z axis
The question kinda confuses me
if its a rectangular prism, shouldnt it be 180 around each axis?
plus all the reflections
red imp
I thinkkkkk that rotations around two of the axes are the generating elements, and rotations about the 3rd axis can be composed out of those two other rotations
oh yeah if you wanna include reflections then I guess that will add some too
ivory dust
Are reflections in R3 about a plane? as opposed to a line
red imp
nah you can pit a kebab skewer through it and spin it around
ivory dust
But also could be a plane?
things in R3 just mind fuck me 😂
im legit playing with a tissue box rn
the "pit a kebab" analogy and spinning it would be a rotation tho
not a reflection
correct?
ivory dust
A reflection just swaps coordinates
red imp
yeah so maybe 5 generating elements of order 2? I'm a bit of a group theory noob so don't take my word for it
ivory dust
Ahh so I was confused about the question, its actually a rectangular prism
no slants
which means you can rotate it 180
in 3 axes?
red imp
yeah but I think that rotating in the z axis is the same as rotating in the x and then the y axis
ivory dust
whats the difference bw rotating in x axis and y axis
red imp
like if you draw them like this
spinning it around the x axis and then spinning it around the y axis is the same as spinning it around the z axis
red imp
yeah that's what I'm thinking
ivory dust
as they are included/redundant
and the reflections I think i know
its in the planes: xy-plane (z=0), zx-plane (y=0), and zy-plane (x=0)
and no diagonal reflections
red imp
sounds decent though I'm not an expert, also not entirely sure what your teacher wants when they say "describe"
ivory dust
probably gnna write in set notation w variables
representing rotations or reflections
somehow
ill figure it out
😂
so the rotations all together are
scarlet estuary
perhaps they want a group presentation?
scarlet estuary
though you might be able to find a group its isomorphic to
which sounds like itd also suffice
ivory dust
{e, R90, R180, R270, H, Diagnol 1, Diagnol 2, V}
red imp
oh he wants literally all the elements then 
scarlet estuary
...ew
red imp
I was just thinking about using the generators to characterise the group
ivory dust
but my textbookk writes
like variables a,b
for reflection or rotation
then a, a^2, a^3, e, b, ba ba^2, ba^3
red imp
do all elements of this group have order 2
ivory dust
UHhh whats order again
red imp
as in like if you do them twice you get identity
ivory dust
Yea!
Wait no
not rotations
for squre
square*
just reflections
rotations are order 4(?)
red imp
for the prism though I think it might be the case
ivory dust
Oh i see
red imp
that means you wont have so many elements in the group I guess
scarlet estuary
||the rotations are the klein 4-group, and together with the reflections, we find an isomorphism between the symmetry group of a prism and the semidirect product of the klein 4-group with the cyclic group of order 2||
idk if thats meaningful to you
scarlet estuary
okay, ignore that then lmao
ivory dust
and obviously i know what isomorphism is
red imp
I'm on my second algebra course and only just now seeing semidirect products
ivory dust
I think its order 8
identity, rotation by 180 in 3 axes as pivots, and reflections of 4 faces
red imp
that sounds right, if what Namington said was right
ivory dust
reflection around xy, xz, zy and origin*
yea
ok i got it
thanks guys
ahh actually would you be able to reflect it around the origin? ie center of the prism
if not itd be order7
i basically swapped diagnol vertices
red imp
yeah thats the same as reflecting it in all 3 planes
scarlet estuary
by the way, as a heuristic: this group certainly cant be of order 7 since groups of prime order are cyclic
which means theyre generated by any of their elements
i.e. youd be able to get any possible element by just applying the same reflection or rotation over and over
which obviously isnt true
so we know the order of the group, at the very least, must not be a prime number.
red imp
I've done the <= part of this proof but cant really see how to do the => part
i.e. I need help showing that if they are permutationally isomorphic then there exists that automorphism
next obsidian
by permutationally isomorphic
is this saying like there's a G-equivariant isomorphism H -> K?
If so, I think you want to assume you get a map from H -> K an isomorphism which is equivariant
hmm
red imp
next obsidian
Oh huh
okay
so it's like an augemented equivariance
so this says like
f: G_1 -> G_2
red imp
we also use input-on-the-left notation so I think that changes what is meant by "right cosets"
next obsidian
and say g: S -> T
then for any x in G_1
you have
g(xs) = f(x)g(s) right?
like it's almost like it's G-linear but up to the isomorphism of G_1 to G_2
I guess the notation is a lot different
so maybe you aren't familiar with what I'm writing
red imp
yeah not entirely haha
next obsidian
Like
yeah okay
I translated
and yeah it's the same thing
so I thikn
literally
so like the permutationally isomorphic thing
gives you an isomorphism from G -> G right?
red imp
yep
next obsidian
and a function rho:H -> K
I think that iso G -> G should be the automorphism of G
at least at a first glance
like
red imp
and I think I want to show that phi = rho
next obsidian
no
rho is not defined on G
but like
consider this
okay so
take h in H
then you know that H^h = H right?
like err
red imp
yeah
next obsidian
(eH)^h
okay
so by assumption
Then
right okay maybe you need to shift phi a little bit
okay so sorry
rho(eH) = gK for some g
right?
red imp
yeah
next obsidian
then you know that rho((eH)^h)) = rho((eH)) = gK
but also by the like equivariance here
rho((eH)^h) = rho(eH)^phi(h) = (gK)^phi(h)
so (gK)^phi(h) = gK
red imp
hmm let me write it out 1 sec
next obsidian
err, I'm doing left cosets not right but whatever
red imp
okay yep I'm following so far
next obsidian
okay so now consider the following
make a new phi' where phi'(x) = g^-1phi(x)
this is also an automorphism of G
Apply this to an h in H, we knew that (gK)^phi(h) = gK, so it follows that
(gK)^phi'(h) = (gK)^{g^-1phi(h)} = (gK)^{g^-1})
err
I guess do g^-1 on the right
red imp
it's not obvious to me that that is a homomorphism
next obsidian
????
it's composing with the multiply by g map
or am I in the wrong here
oh wait I am
hmmmmmmmmmmmmmmn
Okay I almost had something
lmfao sadge
okay hold up
I swear this is almost right
Maybe I don't need to shift
sorry about that lmao
okay I got it
I think
red imp
all good haha
next obsidian
sad boys okay so
SO!
waaaay back to the beginning
rho(eH) we want to show this is eK
to do so
note that rho((eH)^e) = rho(eH)
but by equivaraince
Wait ACK!
dsfjoisdfjl;asd
okay give me a second this is almost the right idea
I swear
red imp
no hurry hahaha I appreciate it man
next obsidian
rho((eH)^h) = rho(eH)^phi(h) = (gK)^phi(h)
okay so headed back to this
this little paragraph
I had yet to mess anything up
we showed that for any h in H, that (gK)^phi(h) = gK
red imp
yep I have it written up on my whiteboard here and it all seems solid
red imp
yeah
next obsidian
but unless g is in K, or equivalently gK = eK
gK isn't a subgroup
for example it won't have the identity
so really phi(H) < K
red imp
yeah that makes sense
next obsidian
so now you're basically done
by just swapping K and H and phi,rho with inverses
you show that phi^-1(K) < H
so that phi(H) = K and phi^-1(K) = H
and then you're done
red imp
aren't they the same statement
red imp
ohh okay
next obsidian
the statements phi(H) = K and phi^-1(K) = H are the same
but you need to know phi^-1(K) < H
but this is like... just formal basically
the literal same exact proof works
it's symmetric
red imp
1 sec im just trying to comprehend how we know we know we're done here
next obsidian
ah well phi is an automorphism of G
and you just showed that phi(H) = K
which is exactly what you wanted
red imp
so if phi(H)<K and phi^{-1}(K)<H then we can deduce that phi(H)=K?
next obsidian
ah
so take k in K
do phi(phi^-1(k)) = k
so the element phi^-1(k) is in H (by phi^-1(K) < H)
thus k is in phi(H)
red imp
ohh I think I see it
next obsidian
this is just like
a bijection thing
if you have f: A -> B a bijection
then for S < A
even if f(S) < T
for T a subset of B you don't know that f restriction to a bijection of S and T
this is what you were noting
but if you can show also that f^-1(T) < S
then f does restrict to a bijection of S and T via the argument I gave
red imp
niice yes I think I've got it now
next obsidian
Yee
red imp
thanks a heap Chmonkey
red imp
this warning was given under the definition so it's probably a rebranding of something that's commonly seen
cloud walrusBOT
Mecejide
warm fiber
Let p and q be primes. The equation x^4 - px^3+q = 0 has an integer root. Find the sum of p and q.
carmine fossil
The integer root has to be 1,q,-1 or -q
sly crescent
Ok, I got it. It’s the projective special linear group of degree two over the field of nineteen elements.
chilly ocean
So in this video https://www.youtube.com/watch?v=RYPt7kGdo7s, he says that "the orthogonal group gives a group action on R^3 that preserves the function x^2+y^2+x^2," which makes perfect sense. But then he says "x,y,z are linear functions of R^3 so x^2+y^2+z^2 is a polynomial on R^3". What is a polynomial on R^3?
bleak abyss
@chilly ocean Do you know ring theory at all?
bleak abyss
So if A is a ring you have the polynomial ring A[x]
A polynomial on R^3 just means R[x,y,z]
bleak abyss
Yeah
R^3 being the domain when you think as functions, rather than as the coefficient ring
Also nice I know folk who work on syzygies
I vaguely know what they are but I have no idea what's the underlying geometry
chilly ocean
Nice
But also later on
He says that "if V is a vector space and we have a polynomial function from V to a field k"
this part
also doesn't quite make sense
bleak abyss
So the naive thing to say is that V is k^n
chilly ocean
That would make sense
bleak abyss
That's a bit unsatisfying since it's non-canonical
But like
Let's say v_1,...,v_n is a basis for V
And w_1,...,w_n is another
Then if you give me a function f:V->k which is a polynomial in the coordinates v_i
Then you just precompose with a change of basis matrix, it'll still be a polynomial in the coordinates w_i
chilly ocean
Oh I see
bleak abyss
So I guess you can just say "Pick any basis" here
chilly ocean
So the polynomials on V with this definition is independent of the choice of basis
bleak abyss
Yup
chilly ocean
Ok I'm watching the video and I feel like I have a lot of questions
So I might ask more
bleak abyss
Sure. I might go to bed soonish but while I'm awake I'll answer
chilly ocean
Ok I finished this video and it's quite nice I would say
bleak abyss
Dope
chilly ocean
The exact sequence nicely encode the relations of the invariant ring and the relations of the relations
Or syzygy
I guess he really didn't need to say vector space
He could've just said a group acting on k[x1,...,xn] where k is a field with char 0
viscid pewter
how exactly do you use coxeter groups/reflection groups to show things about polyhedra and hyperpolyhedra and such? like what would an example of a nice property be that's proved by group theory
viscid pewter
bugger, polytopes, that's the word
google's saying i should probably look more closely at group actions
sly crescent
how exactly do you use coxeter groups/reflection groups to show things about pol...
Removing an unringed node gives you a facet of the polytope.
sly crescent
Removing an unringed node gives you a facet of the polytope.
If it’s a valid polytope, that is.
viscid pewter
uhhhhh
what's an unringed node
alternatively where can i read about this proof and related stuffs
sly crescent
@viscid pewter oops. I thought you were talking about Coxeter diagrams.
viscid pewter
well that's just the exechamp
nova plank
Long division
What's the remainder when you divide x^3 by x^2+x+1
Uh, actually your equivalence classes look wrong?
Oh, I guess not
But it's strange to use [x^2] instead of [x+1]
Yep
Np
Yeah, you have to do long division. Sometimes it's pretty easy to see what it is without actually doing it, but normally you just have to do it.
thorn delta
let $\mathfrak m$ be a maximal ideal of an integral domain $R$ and let $R_{\mathfrak m}$ be the localization of $R$ at $\mathfrak m$. Then $R_{\mathfrak m}$ is a subring of the quotient field of $R$.
cloud walrusBOT
kxrider
thorn delta
Is there even anything to prove here? R_m being a subset of the quotient field is immediate and assuming we know the ring of fractions is a ring.... thats it
Making sure Im not misunderstanding something
next obsidian
There is something
Well...
Eh I guess sort of but not really
The only thing I can think of is the fact that in a non-integral domain the map from a ring to a localization need not be injective
But to have a quotient field you need to be an integral domain
gritty adder
Technically you also need to show that the map is well-defined
next obsidian
I suppose so