it is honestly
#groups-rings-fields
scarlet estuary
maiden ocean
a secret one
latent anvil
Aww
scarlet estuary
the worst part
latent anvil
Yes
scarlet estuary
of the album
latent anvil
Post in suggestions sloth
maiden ocean
i believe in u nami use ur power for good
scarlet estuary
like i respect that spotify underpays artists and its an intentional choice and blahblah
but fuck it
also we already have a gay channel
latent anvil
YES
maiden ocean
does this mean we have permission to post htere instead
we should have like advanced-lounge or something and then only people with the advanced role can see it because
latent anvil
Category theory is gay (cringe)
maiden ocean
statistically i like the people with that role
latent anvil
commutative algebra is gay (based)
maiden ocean
@latent anvil twink is not a slur but know that if i ever call u a twink i meant it as a slur
latent anvil
if you called me a twink it would mean nothing
scarlet estuary
we should have like advanced-lounge or something and then only people with the advanced role can see it because
discord user advocates for segregation, 2020
latent anvil
i do not look like a twink
maiden ocean
yea nami between the cool people and the lame people
latent anvil
even when I feel positive about how i look i do not think I look like a twink
latent anvil
lmao
maiden ocean
one sec let me post something in ivory
latent anvil
hmm have u considered they r hot
chilly ocean
ivory
maiden ocean
sad
south storm
What are you guys discussing?
maiden ocean
homosexuality
chilly ocean
||bears > twinks||
maiden ocean
same thing
south storm
chilly ocean
bye
chilly ocean
gay theory
scarlet estuary
i dont get it
maiden ocean
ill tell you when ur older
chilly ocean
latent anvil
Like the word gay
scarlet estuary
oh sorry in canada we pronounce it "ked"
latent anvil
Ahh gotcha
maiden ocean
honestly shamrock my mind usually jumps to ketamine before "gay"
latent anvil
I'm ged
chilly ocean
time to learn K-theory
latent anvil
I really want to
chilly ocean
mat437 time
latent anvil
ttera learn k theory with me
Dami and I were going to this summer
But then I was burnt out
chilly ocean
there is a chance ill take a course that talks about it next semester
latent anvil
im taking a bundles course
which will presumably get close to it
or at least set me up to read and understand a k theory book on my own
scarlet estuary
damn can you teach me
maiden ocean
burnt out shamrock stops doing math burnt out moth has a complete mental breakdown and stops functioning for 2 days before going back to normal we are not the same
chilly ocean
scarlet estuary
im pretty sure only 1 person understands thomason trobaugh
latent anvil
namington do you have the power to grant honorable?
chilly ocean
im jealous sham
scarlet estuary
and it had 2 authors
maiden ocean
does one of the authors understand or is it like
a 50/50 deal
each 50% understands
chilly ocean
oh nami are you ready for another annoying toronto question
scarlet estuary
well one of the authors is only a coauthor because he visited the writer in a dream
maiden ocean
ohhh its that book
latent anvil
@maiden ocean burnt out shamrock can get pretty bad lol
scarlet estuary
namington do you have the power to grant honorable?
technically i CAN but im not supposed to
latent anvil
oh okay
scarlet estuary
@chilly ocean ?
chilly ocean
mat437 with george, do you know anything about it?
maiden ocean
shamrock a few days ago i got outed to my mom for the fourth time and shes pretending nothings happening again lmao
scarlet estuary
no im replying to tterra
maiden ocean
extreme levels of cope
latent anvil
what is the thought process
how do you
yeah
oops guess my kid is just really unlucky
maiden ocean
i asked for hormones and she was like "ill think about it" and then went back to deadnaming me 30 minutes later
chilly ocean
george's mat436 is so... peculiar, im considering taking his followup
latent anvil
lmao
latent anvil
I wish I could say that it gets better re:parents
but I don't think it does a lot of the time
maiden ocean
i do not think they are ever gonna budge so im gonna stop trying and just wait it out
latent anvil
yeah :/
maiden ocean
its not worth the mental energy
scarlet estuary
i figured he'd ask me to grade for it at some point
or something
but he likes having a lot of control over it ig
maiden ocean
its mostly just annoying for paperwork reasons and also no hormones
hmmmm
HMMMMMM
chilly ocean
"grade for it" as in... what? 
latent anvil
be safe with hormones. not saying they're bad for people under 18, just that taking meds without talking to a doctor is unsafe
maiden ocean
i am hatching a plan
latent anvil
like do your research (i don't need to tell you this but i will worry)
chilly ocean
yeah
chilly ocean
ok thanks nami
maiden ocean
will update you in n months
scarlet estuary
delicate bloom
haha no I'm good I was going to say stay safe that's all
wear your mask I mean
😬
chilly ocean
sorry mero, you need to offer a ring before you can go
delicate bloom
7-adic integers
chilly ocean
put a ring on me
delicate bloom
1-adic integers
chilly ocean
ring of functions
latent anvil
oh hell lyeah
chilly ocean
ring of smooth functions on manifold
chilly ocean
latent anvil
agoomer brain
delicate bloom
brb getting a sheave of stalk of germs or some nonsense
latent anvil
lmao
scarlet estuary
chilly ocean
is this an AG meme im too smooth-manifold brained to get
latent anvil
oh umm
so ttera
let ring mean commutative ring with identity
basically the starting point of schemes is
chilly ocean
as all rings are
maiden ocean
virgin seven rings vs chad seven rings in hand
latent anvil
think of every ring as a ring of functions on some space
and by "think of" I mean "actually construct a """"space"""" which makes this true"
If A is a ring, Spec A is a """"space""" on which the global functions are A
this is sort of the point of affines schemes
and then schemes are things which locally look like this
anyways
chilly ocean
i see
chilly ocean
||fartshorne||
delicate bloom
you can think of primes as being points in the complex plane
and field extensions correspond to covering spaces
latent anvil
Primes meaning primes in C[x, y]?
chilly ocean
kinda reminds me of that one weird approach to smooth manifolds where a smooth structure is given by specifying the smooth functions in it
delicate bloom
no, literal prime numbers
latent anvil
Smooth Manifolds and Observables
chilly ocean
yeah that one sham! my difftop prof recommended it, i skimmed it and it looked super neat
latent anvil
it is sick
would recommend
you can characterize a lot of stuff nicely in terms of functions
like obviously vector fields = derivations on functions
but also like, you know how an injective smooth immersion i : X -> Y determines the smooth structure on X? Like really it's a condition on a continuous map of topological spaces X -> Y
latent anvil
if you want to take an injective continuous map X -> M where M is a manifold (maybe X is also a topological manifold of the same dimension?), you can characterize when this is secretly a smooth immersion in terms of the sheaves
chilly ocean
sheaves 
delicate bloom
I see why you're simping for TTerra to be honorable now @latent anvil 😏
chilly ocean
lol
latent anvil
so that I can sheaf pill him??
maiden ocean
honorable is unnecessary for sheaf pilling
scarlet estuary
vector bundles are what sheaves want to be
maiden ocean
just kidnap
latent anvil
lmao
chilly ocean
vector bundles are neat
scarlet estuary
half of algebra is figuring out how to think of [x thing] using vector bundle intuition
scarlet estuary
the other half is lamenting your failure to do so
latent anvil
lol
maiden ocean
merosity heart eye reacted because we share a practical appreciation for criminality
delicate bloom
amen to that
latent anvil
all i understood was that i was being bulli
maiden ocean
shamrock will you send it to me in n years when i want it
delicate bloom
we don't always disagree, but when we don't, it's beautiful
chilly ocean
Why are you sending textbooks, just libgen it
latent anvil
presumably it will be out
maiden ocean
the book is not out yet
latent anvil
@chilly ocean unpublished draft version
chilly ocean
Oh
latent anvil
well I will
chilly ocean
I guess I would not do that personally
just spreading knowledge
it'd end up on libgen anyways, why not speed up the process?
leaden finch
anyoone good with cartesian coordinates for rings?
wind steeple
Hey, why for a group representation pi : G -> Vn we have the formula khi_pi(g²) = khi_S^2V(g) - khi_A^2V(g) ?
where S^2V is the symetric subspece of V tensor V, and A^2V the antisymetric one
and khi are characters
glossy yoke
wind steeple
oh ok lol, eigen values are lambda^2 where lambda is a eigenvalue of g
those of S^2V(g) are lambda_ilambdaj with i<=j and A^2V(g) are the same with i<j
ok thx
but is there a more elegant proof ?
for instance, khi(g)^2 = khiS2(g) + khiA2(g)
because V tensor V is S2V sum A2V
light tusk
See Serre's finite rep book, the end of ch 1 and beginning of ch 2
A lot of the details I think end up being what you just mentioned
He calculates the sums of products of eigenvalues, just keeping track of alternating/symmetric.
wind steeple
Is this formula general ?
light tusk
uhh... i think the main things going on is finite dimensional and algebraically closed scalars
otherwise, are you asking about the group? i'd have to get back to you on that, lol.
I think the general form is like the Weyl Character Formula?
wind steeple
Uh sorry, by general I meant, is the formula doesn't depends on the base field the representation is. In fact it is bc one can compute the trace with eigenvalues in an algebraic closed field
But I was looking for a proof without using eigenvalues, idk why, it seems ugly to me to compute eigenvalues because for the formula khi(g)^2 = khiS(g) + khiA(g) we don't have to compute eigenvalues
light tusk
There's Dummit & Foote section "Tensor Algebras, Symmetric and Exterior Algebras", but they don't explicitly calculate much tensor products in the representation chapters...
I think however going that route, you mostly have to keep track of the characteristic of the field; D&F introduces the symmetrization, which involves a coefficient in the form of 1/k!
snow cliff
I'm a bit confused by the process of trying to find all ring homomorphisms from one ring to another. From what I gather on math stack exchange, two things seem to be important in this examination: 1) preserving identities and 2) idempotent elements (which seem similar to generators)
I'm not really certain why 2 in the above matters and it just seems to work and as a result it seems like symbol pushing with the facts needed for a ring homeomorphism (f(ab) = f(a)f(b) and f(a+b) = f(a) + f(b) for a,b in R, assuming our map f is from the ring R to the ring S). I think 1) makes sense since a homomorphism should preserve structure and so the identity from R should be sent to the identity to S.
maiden ocean
@snow cliff idempotent means that e^2 = e right?
so... apply a ring homomorphism f to e^2
snow cliff
@maiden ocean why? This is the part i don't seem to understand
It seems to work, but I don't understand why it does
chilly ocean
A clock is just mod12 right?
viscid pewter
yes, except 0 is replaced by 12
maiden ocean
dan
chilly ocean
dan
maiden ocean
i do not think this question belongs here maybe in #elementary-number-theory
unless it is going to connect to some D_6/D_12 memery
chilly ocean
idk it was just a random question.
But yeah isnt a clock a D_12?
Because its cyclic
thorn delta
D_12 aint cyclic
chilly ocean
Z_12?
maiden ocean
also clocks do not really have a group structure
chilly ocean
but there's no 0
viscid pewter
0 is replaced by 12
light tusk
Are you suggesting my clock is disconnected?
cinder bone
is the subgroup of a normal subgroup normal?
carmine fossil
No
cinder bone
K intersect G would be a subgroup of N tho right
Na=aN for all a in G -> Na=aN for all a in K as K<G
cinder bone
normal in K
so like
K intersect N would be a subgroup of N* sorry my bad
didnt mean G
maiden ocean
counterexample D8 = <a, b | a^4 = b^2 = e>
cinder bone
halp pls
viscid pewter
i am also very tired and therefore being pedantic
so i don't screw something up
e is normal in D8 right
or
viscid pewter
and then with <b>?
viscid pewter
okay but that's normal in <b>
it's not normal in D8 but i think they weren't asking for that
maiden ocean
i know but they just asked "is the subgroup of a normal subgroup normal"
which is a common question
so im giving a counterexample : p
viscid pewter
wait joshua what were you asking again
were you asking whether the intersect is normal in the normal subgroup or normal in the whole group
maiden ocean
they have a more specific exercise but they prefaced it with
is the subgroup of a normal subgroup normal?
viscid pewter
in what
maiden ocean
i presume they mean the whole group lmao
cinder bone
i dont get it
cinder bone
arent i supposed to show (K ∩ N)a = a(K ∩ N) for all a in K
maiden ocean
ok yes break down what that means one step further
cinder bone
the left and right cosets are the same
maiden ocean
yes
and what this means is that
if i take any a in K cap N
and k in K
kak^{-1} is in K cap N
cinder bone
im so lost lol
chilly ocean
N is normal in G
cinder bone
ok maybe i can try another q
so i have G1 = Z', H = 3 Z', and G2=S3
and have a function from Z' to S3 that is a homomorphism
namely f(n) = (13) if n is odd and = (1) if n is even
what should i do next
tryna show f(H) is not normal in S3
trying to give a counterexam
cinder bone
i know f(H) = {f(h) | h in 3Z'}
maiden ocean
just say G is not cyclic
scenic sage
but he has to show it
maiden ocean
idk if theyre asking for proof or phrasing lol
uhhh
im not really sure what ur asking
cyclic means its generated by 1 element
yeah so do u want a proof
is that what you meant by "say" here lol
or are you confused on terminology lol
cloud walrusBOT
scenic sage
,w 7*5 mod 18
cloud walrusBOT
scenic sage
,w 17*5 mod 18
cloud walrusBOT
scenic sage
,w 13*5 mod 18
cloud walrusBOT
scenic sage
,w 11*5 mod 18
cloud walrusBOT
maiden ocean
that is all of them duke
scenic sage
mod 18
and modular multiplication is well defined so it is enough to check those ones < 18
well you can
but 35 is the same as 17 mod 18
so if i am not dumb by exhaustion we showed that your group is cyclic
maiden ocean
id say that |Z_18*| = 6 so if it was cyclic it would be isomorphic to Z_6
and its probably not that hard to show that no such isomorphism exists
less annoying then computing a bunch of products
maiden ocean
injective group homomorphisms preserve the order of elements
i imagine this kills it pretty easily
scenic sage
@maiden ocean Z/18* is literally isomorphic to Z/6
steep hull
(Z/nZ)* is cyclic iff n=1,2,4,p^k, or 2p^k for some odd prime p and positive integer k.
Hello
scenic sage
1->5->7->17->13->11 just make isomorphism 1 to 0, 5 to 1, etc.
we just showed that there is isomorphism with cyclic group
and if it is then original is cyclic
maiden ocean
i would explain but currently i am lazy and tired
scenic sage
because isomorphism says us that two groups differ only by notation
maiden ocean
also @steep hull have you read hatcher/bredon/tam dieck or whatever there r like 50 books
scenic sage
because isomorphism says us that two groups differ only by notation
@scenic sage i mean this is condensed essence of isomorphism
but anyway, you can just show by exhaustion that all coprimes to 18 in Z/18 can be obtained from 5
yw
cinder bone
help wit dis? 😮
i was recommended by teacher to find surjective homomorphism function
which i did
for f(A) = det(A)
f: GL(2,R) -> R \ {0}
but not sure where to go next
i guess the quickest way would be to show nAn^-1 is an element of N for n in N
chilly ocean
i think det(AB)=det(A)det(B) and det(A^{-1})=det(A)^{-1}, right?
then this should make the problem easy
cinder bone
what would ido after that?
chilly ocean
just as you were saying, nAn^-1 is an element of N
chilly ocean
just think a little more
scarlet estuary
every element in a group generates itself
light tusk
Every element g of a group G generates a cyclic subgroup C. By construction, g is a generator for C.
Woohoo!
next obsidian
It’s unique by virtue of being biggest
If theres another thing which Is the largest the core and that other thing of subsets of each other
light tusk
H = gHg^-1 for all g in G would mean H is normal in G. Did you mean all g in N?
leaden finch
can someone help me with well define and homorphism stuff?
golden pasture
not sure what you mean but when we say well defined we mean that the thing we're looking at is actually a function
in the sense of take in a value give out another value
homomorphism is like a structure preserving morphism
cuz like
homo(same)-morphism(type/shape)
wait this isnt med school nvm
but like say for groups
you have multiplication
so you want f(ab)=f(a)f(b)
or say rings with addition and multiplication, so you want f(a+b)=f(a)+f(b) and f(ab)=f(a)f(b)
leaden finch
yeah that stuff im not sure how to apply the definition into various problems
golden pasture
or vector spaces/modules with
f(a+b)=f(a)+f(b), f(ka)=kf(a)
icic
try like what maps are homomorphism from the group of real numbers with addition as group operation to itself
how about the group of real numbers with addition to the group of positive reals with multiplication
pallid ember
you mean how to make sure your homophism is well defined? @leaden finch
leaden finch
pallid ember
which part
leaden finch
part a
golden pasture
essentially can i actually compute this
so the usual issue that pops up here would be like
leaden finch
well i need help on each part lol
golden pasture
[0]=[3]=[6]=...
so does the function really give the same output for each possible choice of a?
obsidian path
I'm doing question d
I can just check all of the properties for ring and field, but that would be very time consuming
are there other ways I could solve these?
I found this example that might prove useful to solve d
but I'm not sure cause I'm not comfortable with field extensions
help?
golden pasture
i mean you dont rlly need to write out explicitly right🤔
obsidian path
wdym?
like because we know addition and multiplication is associative for the reals ?
and addition is commutative
golden pasture
something along the lines
like you can see multiplying sqrt(2,3 or 6) would lead to something inside
stone fulcrum
You should check all the properties, but most of them boil down to "because Q is, then Q[√2] is"
obsidian path
I'm worried that's not rigorous enough..
viscid pewter
it's completely rigorous
stone fulcrum
Now's a good time to ensure you can identify a group's inverse
viscid pewter
you can prove it
golden pasture
you dont need to care too much about rigour
a simpler way to see is Q(sqrt(2)) is field
Q(sqrt(2))(sqrt(3)) is field
stone fulcrum
Again, worth checking to be sure you can do all the properties, and that you know what a ring and field are
obsidian path
true..
I will do that because I have to do the rest of the questions
I'm just trying to get shortcuts where I can cause its very lengthy.. :'')
obsidian path
o-o'
golden pasture
tho id imagine it may be tricky to state them if currently you're at proving rings and field section?
obsidian path
yeah we've just begun
stone fulcrum
If P is irreducible. You'll get there soon enough though
obsidian path
but we can use whatever we want if we find it I guess
maybe
alright thanks everyone!
the help is encouraging (:
leaden finch
@golden pasture how do we know which number maps?
golden pasture
by plugging into the function given?
f([a])=([2a],[3a])
then check if it is well defined
maiden ocean
umm
do you mean automorphism
ohhh
uh i dont think so
i think there exists an isomorphism like that
uh i think E and E' have to be splitting fields of the same polynomial
the context i know this from is that if u have an isomorphism phi of a field F -> F', and some f in F[x], then this extends to an isomorphism of E -> E' where E and E' are the splitting fields of f and phi(f)
so you take the special case where phi is the identity
and see that if E and E' are two splitting fields of the same polynomial f there is an isomorphism between them fixing F
can you give more context
like whats the base field and what are our polynomials
ah
right
i think ur right for Q
im not sure if this works for any base field
yea
haha
hope that helped : )
chilly ocean
Anyone know where I can find practice problems with solutions?
Maybe prelim problems (do solutions usually get posted for these?)
Or random university course websites that post homework solutions
south storm
What‘s an isomorphism between all left triangular invertible n by n matrixes and all right triangular invertible n by n matrixes? (Under matrix multiplication)
Are they even isomorphic? It would really make sense if, but I’m not sure.
thorn delta
for square matrices, left/right invertibility is equivalent to invertibility
south storm
Yes I was talking about left and right triangular invertible matrixes forgot to mention
Sorry for that
Fixed it now
next obsidian
Uhhh
Hmm
Wait
Left triangular matrices dont form a group
They don’t have the identity
Or err
What operation do you want to endow them with?
viscid pewter
they said multiplication
next obsidian
Multiplication then you can’t
Oh I see haha
Yeah left triangular matrices dont have an identity
south storm
They do, left triangular are of form $\begin{pmatrix} a&0&0 \ b&c&0 \ d&e&f\end{pmatrix}$ quite sure, maybe mixed up the name one second
viscid pewter
so wait is
[1 0]
[0 1]
not left triangular
cloud walrusBOT
27182818284tropy:
south storm
Nope got it right
next obsidian
Hmmmm
south storm
Yes I tried that
next obsidian
I mean...
south storm
And tried transpose over anti diagonal
Doesn’t work either
And any combination of them
next obsidian
Okay so cheat the following way
thorn delta
row operations?
next obsidian
Define multiplication on left triangular
To be backwards
:^)
Then trace works
I feel like maybe they aren’t isomorphic
viscid pewter
they surely have to be
south storm
Like why wouldn’t they be
viscid pewter
okay so you've tried the transpose
stupid question: what if you flipped it the other way
next obsidian
Then transpose will work haha
viscid pewter
along the other diagonal
next obsidian
They said they tried it
south storm
Yes I said I tried that too
next obsidian
Also
south storm
$(AB)^\tau=A^\tau B^\tau$
next obsidian
If you do that you still get left triangular matrices no?
south storm
And yeah the problem is
next obsidian
Still
cloud walrusBOT
27182818284tropy:
next obsidian
Left triangular and upper triangular matrices aren’t a group right??
They aren’t all invertible
Or am I dumb
thorn delta
i think they said "invertible left/right
cloud walrusBOT
27182818284tropy:
south storm
How does one even disprove something isn’t isomorphic
south storm
No doesn’t
south storm
$(AB)^T=B^TA^T$
next obsidian
It flips the operation so it actually works
south storm
It isn’t even a hom
next obsidian
Yes but G^op
next obsidian
Listen to me!!!
south storm
$((AB)^T)^\tau=(B^T)^\tau (a^T)^\tau $
next obsidian
G^op takes a group G
south storm
That‘s literally what I said
cloud walrusBOT
27182818284tropy:
next obsidian
Makes a new group where the objects are elements of G
And you define g•_op h = h•g
south storm
Sounds like G to me
cloud walrusBOT
27182818284tropy:
next obsidian
You flip the order of operation
south storm
Alright
next obsidian
This is an iso of lower triangular and upper triangular^op
south storm
Oh right and then show they aren’t isomorphic?
next obsidian
And I think something is iso to it’s opposite IFF abelian or some shit??
south storm
But how to do that?
next obsidian
The point is if lower and upper were iso
We get upper iso to upper op
Which sounds mega fake
south storm
I mean proofs
south storm
Look at what I wrote
next obsidian
Doesnt flip the order of multiplication
south storm
People look at what I send
next obsidian
Just show upper triangular matrices aren’t iso to their opposite
south storm
Yeah how that
viscid pewter
i swear they really should be isomorphic tho
next obsidian
WAIT
south storm
Send proofs 👀
next obsidian
I’m dumb
south storm
We dumb
south storm
Oh right
next obsidian
So this is the iso
south storm
Wait so they are isomorphic
south storm
Alright could you please write everything out explicitly I really wanna be sure so I can double check if it‘s right.
next obsidian
No lol
south storm
Still thanks
Why LOL?
Huh alright
Anyways thanks for helping, wondered about this for a while
next obsidian
I don’t mean to be rude but I really don’t feel like writing a random proof for a problem I don’t have any real interest in lol. All the ideas are here so you can just formalize them and it should work
south storm
No no it‘s fine. Was over the top to ask to write it out, I’ll do it myself.
south storm
@next obsidian wait so what was the isomorphism between right_op and left?
next obsidian
Just map something to its inverse
south storm
Wait but both are closed under inverses
next obsidian
Yeah but this goes G to G^op
So you send g to g^-1
This is a bijevtion
And it’s a homomorphism because of how multiplication in G^op is defined
south storm
Yes
next obsidian
That’s just transpose
south storm
Wait that won’t work
next obsidian
It’s a Hom because right^op has backwards multiplication
south storm
Since it will map back
next obsidian
?
south storm
Wait I got the isomorphism between G and G op
next obsidian
Send M to M^T
next obsidian
I legit don’t know where you wrote the iso of G and G^op
next obsidian
Sorry so G’ is the left triangular?
next obsidian
And G is the right triangular
south storm
Yes
south storm
No that doesn’t work since it will still remain a right triangular matrix
next obsidian
G^op still are right triangular matrices
south storm
Not under transpose no
next obsidian
When you do the transpose?
next obsidian
Okay so
next obsidian
The matrix
$$
\begin{bmatrix}
1& 0\
1 & 1
\end{bmatrix}
$$
south storm
Yes left triangular
next obsidian
When you take the transpose you get an upper triangular matrix
south storm
No
next obsidian
WhatV
south storm
That‘s not how transpose works
cloud walrusBOT
Chmonkey:
next obsidian
The transpose gives you 1,1;0,1
south storm
Wo what‘s the isomorphism between T and T dot?
Because I used the transpose there,
And the „anti transpose“ wouldn’t work
next obsidian
What is T and T dot? I can’t read your notation. The point is left triangular is iso to right triangular op via transpose
The transpose of a left triangular matrix is right triangular
next obsidian
I can’t read it dude
south storm
Ohhh
Thanks!
Yup that worked perfectly, sorry for me not understanding first. Thanks again!
cinder bone
so im trying to prove gng^-1=n_1 for all g in GL, n in SL, and some n_1 in SL
and i have g*n
maiden ocean
What is the determinant of the product of two matrices
cinder bone
oh the product of the determinants
cinder bone
and the inverse determinant is 1/d?
maiden ocean
right
cinder bone
oh
maiden ocean
so
maiden ocean
yeah
cinder bone
ok beautiful
for thisone
i have xy in H
xhx^-1 in H
and yhy^-1 in H
for all h in H
nvm
i can plug y(xy)y^-1?
cinder bone
it works right ?
maiden ocean
ya
cinder bone
so coming back to this one
gHg^-1=g^-1Hg=H right?
or are they distinct elements of H and the notation is throwing me off
chilly ocean
what do you mean distinct elements of H?
gHg^-1=g^-1Hg=H right?
that's the definition of a normal subgroup, yes
cinder bone
like gHg^-1 means that ghg^-1 = h* for some h* in H
is this the same h* as g^-1hg=h*
for all h in H
and all g in G
chilly ocean
no, this would require that ghg^-1=g^-1hg, or g^2h=hg^2 (h commutes with every square in G)
carmine fossil
Solution:
||Show k^-1h^-1 k h is in H int K||
chilly ocean
OK WHAT IS THIS
too dense for me can someone try to explain it to me
there is no more context what are the n_i's
I don't get the summations, I suppose its like the only way to write a m-degree polynomial?
sharp sonnet
it's a polynomial in m variables, the n_i are the highest power with which z_i appears
it's not the only way, technically you can write it as a single sum
or replace the n_i by infinity
not sure if that helps in parsing
@chilly ocean
chilly ocean
yeye idk
I never understand double+ summations
but its just a standard polynomial right
sharp sonnet
you can think of it as a single sum $\sum_{(v_1, \dots, v_m) \in\bN^m }A_{v_1, \dots, v_m}z_1^{v_1}\dots z_m^{v_m}$, where all but finitely many of the $A_{v_1, \dots, v_m}$ are zero
cloud walrusBOT
Lochverstärker:
sharp sonnet
ye, just a polynomial
light tusk
You can think of it in terms of set builder notation also, like the multi-index sum above, over a specified subset of the indexing variables
$\sum_{(v_1,...,v_m) \in I} A_{v_1,...,v_m}z_1^{v_1}\cdots z_m^{v_m}$, where $I = { (v_1,...,v_m) \in \bN^m| 0 \leq v_i \leq n_i \text{ for each } i, 1 \leq i \leq m}$
cloud walrusBOT
Apopheniac:
sharp sonnet
i wanted to tex that at first but decided against it due to being lazy
light tusk
Lol I thought about giving up...
stone fulcrum
That one is completely computational. Just compute f(ab) and f(a)f(b)
That is assuming part a did come out to "well defined"
chilly ocean
Hey guys, I need to do a project on a group theory topic. What's something fun and not too challenging to look into?
besides a rubiks cube
maybe you can apply some group theory to another topic you're interested in 
technically a group theory topic 😉
chilly ocean
classify finite groups
viscid pewter
jfc
maiden ocean
groupoid moment
@chilly ocean shamrock did his analysis project on groupoids
lmfao
i read the paper
chilly ocean
maybe ill ask him for it later
From what I understand ISBNs and Credit Card numbers are types of groups. Is that true?
pallid ember
right so then parts b and c should be easy
leaden finch
if it turns out not well define do we still prove homo/isomorphism?
chilly ocean
no
thorn delta
to show that f is well defined, you just have to show that for any a,b with a = b mod 3, that f(a) = f(b).
@leaden finch
pallid ember
if its not well defined then its not even a function @leaden finch
and you need a function for a homomorphism
thorn delta
im going to ask a question, but feel free to interrupt if you still need help sunshine
Let $M$ be a maximal ideal of $\bC[x_1, \dots, x_n]$ and let $\pi_1$ be the composition of the inclusion of $\bC[x_1]$ into $\bC[x_1, \dots, x_n]$ with the canonical map $\bC[x_1, \dots, x_n] \to \bC[x_1, \dots, x_n]/M = K$. Assume that $\ker \pi_1$ is non zero. Artin says that since $K$ is not the zero ring, $\ker \pi_1$ is not the whole ring. Why is this?
obviously, if K is the zero ring, then the kernel would be all of C[x_1]. But i don't see why the converse is true
cloud walrusBOT
kxrider:
thorn delta
i.e. i don't understand the underlined part here
wait hmm, if ker(pi_1) = C[x_1] then 1 is in the kernel and since ker(pi_1) is contained in the kernel of the canonical map, the kernel of the canonical map contains 1, so it is all of C[x_1, ..., x_n]. Okay, i think i answered my question
chilly ocean
@leaden finch no, he meant that for one argument you can't have two different values
(otherwise it wouldn't be a function)
snow cliff
is x a generator of the ring R[x]?
latent anvil
it is true that R[x] is (freely) generated as an R-algebra by x
mint gulch
If I have an algebraic extension $K(\alpha)/K$ and see $K(\alpha)/K$ as a $K-Space$ I have the following linear transformation
$$\phi : K(\alpha) \to K(\alpha)$$ $$ \beta \to \alpha \beta$$
I need to prove that $\alpha$ is an eigenvalue of $\phi$, but, $\alpha$ doesn't belong neccesarily to $K$ then, by definition, it can be an eigenvalue
Any hint about the problem?
cloud walrusBOT
Enigsis:
chilly ocean
Hm, that's weird
I guess you can say that if this map does have an eigenvalue, then it must be alpha and alpha must be in K
mint gulch
The thing is that, I have to prove that the carasteristic polynomial of that transformation is the same as the irreducible polynomial of alpha
Then, I have to prove that $\alpha$ is an eigenvalue (Solution to the polynomial)
But, the thing is, that alpha doesn't belong to K necessarily
For example $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$
cloud walrusBOT
Enigsis:
chilly ocean
The characterisitic polynomial is (t-alpha)^n right?
mint gulch
Not necessarily
For example, I found that, with the example above is (x- root(2))(x + root(2))
That is x^2 - 2, the irreducible polynomial of root(2)
chilly ocean
I guess I have not done much algebra in a while. Why can't I take a random basis of K(alpha), then the matrix of this transformation is a diagonal of all alpha? Then this implies characteristic polynomial is (t-alpha)^n
(sorry if I'm wasting your time 🙂 )
mint gulch
Don't worry hahah
I did that
But, the charasteristic polynomial isn't neccesarily (t-alpha)^n
The example above shows that
chilly ocean
Ah sorry, I was being a dumbass
mint gulch
@open torrent Do you know Spanish?
Let me write it down
Given a simple algebraic extension $L := K(\alpha)/K$ and the linear transformation on the K-space.
$$L \to L$$
$$\beta \to \alpha \cdot \beta$$
Prove that the characteristic Polynomial
$\phi(x) = det(x \cdot I - T)$ coincide with the irreducible polynomial of $\alpha$
cloud walrusBOT
Enigsis:
mint gulch
T is the matrix of the linear transformation
But, the roots of the characteristic Polynomial aren't the eigenvalues?
Oooh
I see
I see
But, If I Proof that \alpha is an eigenvalue I prove that is a root?
Yes
Yes
I see
I have to prove that is a root directly then
Ok, thanks
thorn delta
In an integral domain R, an element c is irreducible if and only if (c) is maximal in the set of all proper principal ideals of R. If R is a commutative ring and R^2 = R, then a maximal ideal is prime.
In a non-PID, being "maximal in the set of proper principal ideals of R" is not the same as being a maximal ideal, right?
next obsidian
Yeah, if you know what height is a principal prime ideal will be height 1 so any prime ideal containing it is either it or not principal
To illustrate a specific example, k a field take the ring k[x,y]. Then (x) and (y) for example are maximal principal ideals which are not maximal, as (x,y) contains both. This actually holds for any ideal (f(x)) with f(x) irreducible as it's a height 1 prime
Also I should expand a bit on the first message, a priori if we have a prime ideal (f) we don't know any principal ideal containing it is prime so I can't make the conclusion I did. But suppose (g) contains (f), then we can write f = ag and by looking at irreducible components (we need UFD for this at least, so maybe you can find a counterexample where our ring isn't a UFD) either a is irreducible or g is irreducible. In the former case g is a unit so (g) is the entire ring, that's bad. In the latter case g is irreducible so (g) was actually prime.
carmine fossil
How do I show (Z/pZ)^k cannot be generated by less than k generators?
maiden ocean
as a group or as a ring
carmine fossil
As a group
maiden ocean
@carmine fossil i think if we treat (Z_p)^k as a k dimensional vector space over Z_p
carmine fossil
Ok,That works
next obsidian
Depending on the structure of the group you can use theory to find them
Examples being like S_n, but in general I don't thinks o
thorn delta
for D_{2n}, im pretty sure its still just manual computation, even if its works out to not be that bad
leaden finch
can someone help me with subrings
maiden ocean
leaden finch
can someone check my work for this one
thorn delta
computing all the conjugates is not a herculean task here.
thorn delta
do u define rings with a multiplicative identity?
leaden finch
hmm i think so
thorn delta
assuming rings have identity for you, you just need to mention that the subring contains the multiplicative identity (the identity matrix).
other than that, looks good
leaden finch
oh i think the mult identity is for integral domain
thorn delta
what is this part for?
leaden finch
oh to check if it has zero divisors
maiden ocean
hurb
leaden finch
im not sure if it has a zero divisor
maiden ocean
this is not what zero divisors mean
thorn delta
that doesn't have anything to do with being nonempty. its enough to say the zero matrix is in the set, so its nonempty
thorn delta
oh wait, are you combining parts a,b,c or something?
leaden finch
oh shoot, yes this was only for part b
maiden ocean
that should be done after part a
thorn delta
this is the same picture xd
thorn delta
what is M(R)?
pallid ember
that notation usually means set of matrices with real entries
thorn delta
other than the other spot where you wrote M(R) instead R, the proof looks good
pallid ember
you probably meant to write S instead of MR there
leaden finch
yes , i think i should write s
thorn delta
ah shoot yea. wherever you have [matrix] \in M(R) it should be [matrix] \in S.
leaden finch
oh okay
thorn delta
because ur showing that S is nonempty and closed under the operations, not M(R)
small wind
nice handwriting
leaden finch
thanks lol
thorn delta
u almost have 3