My gut's telling me I can partition GL_2(F_p) into p+1 sets that each conjugate the subgroup generated by (1,1,0,1) into one of the p+1 subgroups, but that's what led me to start this discussion in the first place lol. No clue if there's even a way to find those sets without just exhaustively searching through the group
#groups-rings-fields
noble hedge
dim widget
My gut's telling me I can partition GL_2(F_p) into p+1 sets that each conjugate ...
I think there is a conceptual way coming from the Bruhat decomposition, that's what I would do
But that is beyond a first course in group theory.
Let B = upper triangular matrices
then $Gl_2(F_p)/B(F_p) = P^1(F_p)$ is the space of lines through the origin in $F_p^2$
cloud walrusBOT
Topos_Theory_E-Girl
dim widget
because your sylow subgroup U has normalizer B, this also classifies conjugates of U.
So for each line in $F_p^2$ there is a conjugate of U
cloud walrusBOT
Topos_Theory_E-Girl
dim widget
You can then see which elements of Gl_2(F_p) take the line x = 0 to all of the other lines.
noble hedge
Why is B(F_p) normal to GL_2(F_p)?
noble hedge
So then how did you get that quotient group?
dim widget
It's not a group, it's just a quotient space
noble hedge
Ah okay
noble hedge
So it's just a set of cosets then?
dim widget
Yes
noble hedge
neat
dim widget
but it has a nice geometric interpretation
dim widget
yes, of course. you can also use geometry to see that there are p+1 lines
noble hedge
This is pretty cool, thanks!
dim widget
no worries!
lethal dune
probably they meant R or treating X as the spec R
south patrol
what does "quotient field" here mean?
synonym for fraction field
I remember being confused by that at first like lol what maximal ideal are we quotienting by
dim widget
I remember being confused by that at first like lol what maximal ideal are we qu...
especially confusing because here it's a quotient field of a quotient
south patrol
Lol
lethal dune
qquotient
formal ermine
thanks
why do we need the last part though?
isn't the last map just an embedding
dim widget
isn't the last map just an embedding
yes it is
it's just explaining that every point in Spec(R) comes from a k-point for k large enough
dim widget
like 15(7) = 1 and 15(5) = 0?
yes but in that case the prime ideals are maximal
formal ermine
yeah
a) are all of them just closed cuz C is a field so quotiening will always give us a splitting field i.e. all of the ideals are maximal (except of course 0, and just x but that's also just C)?
b) does K[x]_(x) here mean anything special? or is it just localization at (x)?
rustic crown
yea all points except 0 are closed. and yea that means the localization at the prime (x)
formal ermine
yea all points except 0 are closed. and yea that means the localization at the p...
thanks
so for b
by comm alg the prime ideals of it are all prime ideals of K[x] that are contained in (x)
because K[x] is a pid, as K is a field
rustic crown
right
formal ermine
(x) supseteq (r) so x | r
so it's all irreducible polynomials except the constant ones but without 0
as in like
Spec R = 0 + irreducible polynomials - constant polynomials
is that correct
wait we don't even need to remove the constant polynomials cuz they're just units
right
so it's just Spec R = irreducible polynomials + 0?
rustic crown
yea and what can you say about irred poly divisible by x
formal ermine
oop
rustic crown
so yea Spec R = (0) and (x)
formal ermine
I'm dumb
rustic crown
nu >.<
formal ermine
thanks tho
rustic crown
rustic crown
a) are all of them just closed cuz C is a field so quotiening will always give u...
also notice in general, as point correspond to prime ideals, the closed points would correspond to maximal ideals
in the second case, R was local, so can have only one closed point
formal ermine
also notice in general, as point correspond to prime ideals, the closed points w...
ye
formal ermine
in the second case, R was local, so can have only one closed point <:eeveeKawaii...
ohhhhhh
btw hi det
rustic crown
hewwo illu
formal ermine
how's life
rustic crown
tired
rustic crown
yea, it was on last wednesday and i prepared it only on the last two days 
so tired since then
formal ermine
what was it on?
rustic crown
wanna stay home and do nothing >.<
riemann roch for surfaces and two applications of it
formal ermine
oo nice
rustic crown
it was fun
formal ermine
wanna stay home and do nothing >.<
I was forced to do that for the last 4-5 days
now I'm getting back on the math grind
rustic crown
I was forced to do that for the last 4-5 days
i want a week's worth of weekend
formal ermine
get fever, ez solution
rustic crown
nu, but then exercise sheets pile up
formal ermine
I also get wednesday and friday off
formal ermine
wednesday cuz oral final exams for the people at my school
and friday because my school just said "hey have a day off"
5 day weekend 💪
rustic crown
just what i want
rustic crown
get fever, ez solution
i did have a tiny fever today, and it just made the sunday worse >.<
rustic crown
formal ermine
I have an english exam tomorrow though
first two lessons
I have yet to start learning for it
formal ermine
thanks
lapis trench
I'm trying to prove that if $G$ is a finite $p$-subgroup and if $|G|>p$ for some prime $p$, then $|G/G'|>p$ where $G'$ is the commutator subgroup of $G$.
formal ermine
why does the zero ring have no prime ideals?
lapis trench
I've shown that $[G,G']=G'$ and I believe this is supposed to help
lapis trench
why does the zero ring have no prime ideals?
A prime ideal has to be a proper subset of the ring if I recall correctly
lapis trench
And there would be no proper subsets of the zero ring
formal ermine
shouldn't the zero ideal be a prime ideal cuz quotening you get F_1
lapis trench
I think there are certain useful properties that prime ideals have which are useful
But I forgot them
lapis trench
I've shown that $[G,G']=G'$ and I believe this is supposed to help
I'm trying to use a characterisation of central series to solve this
But have not had much success so far
coral spindle
I'm trying to prove that if $G$ is a finite $p$-subgroup and if $|G|>p$ for some...
The easiest way that I can think of doing this is using the fact that finite p-groups are nilpotent, after which this is a simple corollary.
coral spindle
I've shown that $[G,G']=G'$ and I believe this is supposed to help
This is typically not true, if I'm not terribly mistaken.
If [G, G'] = G' then all nilpotent groups are Abelian, which is false.
lapis trench
Oh, I proved it under the additional hypothesis that G/G' is cyclic
lapis trench
The easiest way that I can think of doing this is using the fact that finite p-g...
Ok, I understand that they are nilpotent but can't see how it can be a simple corollary
coral spindle
Oh, I proved it under the additional hypothesis that G/G' is cyclic
I'm doubtful of this but I can't think of a counterexample off the top of my head.
coral spindle
Ok, I understand that they are nilpotent but can't see how it can be a simple co...
It is a corollary of the equivalence of the two characterisations of nilpotent groups: one in terms of successive centres, and the other in terms of the lower central series
lapis trench
Ok thanks, I'll try to look at that
delicate orchid
I'm doubtful of this but I can't think of a counterexample off the top of my hea...
it's true for nilpotent groups - I'm presuming G' is [G,G]?
I ain't scrolling up to check!
formal ermine
it's true for nilpotent groups - I'm presuming G' is [G,G]?
yes
delicate orchid
sorry, it isn't true I was a lying FRAUD
lapis trench
I think it also needs the assumption that G/G' is cyclic
I proved it under that assumption
Unless I made a mistake
coral spindle
Could you share the proof?
I just realised you can prove that G' =/= G for p-groups using character theory, so that's cool :)
delicate orchid
yurrrrr
lethal dune
I just realised you can prove that G' =/= G for p-groups using character theory,...
can you share
coral spindle
Yes!
delicate orchid
you can basically show that each p-group must have a non-trivial, non-faithful linear character I believe - unless the p-group is C_p and then you cry
coral spindle
^
Well we don't need the faithfulness
you just show there is at least one nontrivial linear character, after which you're done :)
It's not too hard to see (and well-known) that the linear characters of a finite group are in bijection with the irreducible characters of its Abelianisation (that is, G/G')
Now if we know that the degree of any irreducible character divides |G|, and the sum of the squares of the degrees is |G|, then we know that there must be at least p distinct linear characters of G
delicate orchid
you just show there is at least one nontrivial linear character, after which you...
this isn't true, consider C_{p^k} - you need the non-faithfulness
but yes for non-abelian
coral spindle
So we're done! We know G/G' is nontrivial, so G' is not G
I think you're mistaken Wew
delicate orchid
yeah, I am
or am I
like, obviously there are normal subgroups of C_{p^k} which is a far easier argument for it not being perfect
coral spindle
It's also... Abelian!
delicate orchid
abelian alschmelian
oh right I misread what you were saying
I am indeed Miss. Taken
coral spindle
Oh no she's taken :,(
delicate orchid
Mr. Taken will not be pleased
lethal dune
by linear character, do you mean 1 dimensional character?
lethal dune
ah cool proof
can you not generalise it to nilpotent groups (finite) as they are product of p groups
delicate orchid
I am indeed Miss. Taken
OR AM I????!!!
I just had a different proof in my head
in fact I had a different statement in my head
prisma ibex
1-dimensional character
delicate orchid
what's the issue here n"g"roupoid are you going to cry because we didn't say degree
prisma ibex
yes 
lethal dune
🍑
coral spindle
If you don't say degree, u don't get ur degree
lethal dune
not a rep theorists so idc
delicate orchid
I've got my degree and I say dimension because characters and C[G]-modules are literally the same thing to me
the distinction interrupts my workflow
a most heinous act
coral spindle
Ah well you see, I'm a pedant
lethal dune
coral spindle
um actually, cartoon frogs cannot shoot me
lethal dune
ϵ<0, cry
coral spindle
ew gross, use \varepsilon
lethal dune
0 ε 1
coral spindle
now that is truly unforgivable
coral spindle
can you not generalise it to nilpotent groups (finite) as they are product of p ...
So I've been thinking about this, and frankly I don't know. I'll look into it!
dim widget
I just realised you can prove that G' =/= G for p-groups using character theory,...
You can also use elementary group theory: by the Sylow theorems |G| = p^n thus G admits a subgroup of order p^{n-1}, but subgroups of index p where p is the smallest prime dividing |G| are normal.
You could also prove the subgroup of order p^{n-1} thing by induction without the sylow theorems
dim widget
can you not generalise it to nilpotent groups (finite) as they are product of p ...
It's true by definition for nilpotent groups, this is a proof that p-groups are nilpotent essentially!
chilly ocean
i know math upto calc, and am familiar with proofs at a basic level, can i work on dummit and foote?
coral spindle
I would say to just start reading. If you're not up to it you'll know soon enough.
chilly ocean
ty
toxic zephyr
say we have a cyclic group of order M. then, if k divides M, will there always exist an element of order k? i suppose taking the generator g^(M/k) would work? how could i cite this fact (need this in the context of number theory)?
prime sundial
why do you need to specifically cite it
it sounds like something easy enough to prove, or almost direct enough to warrant no proof
south patrol
Yes, this is a standard fact and you could just say that lol
In fact, slightly more interestingly, there is exactly one subgroup of order k for each k dividing M (and any subgroup of a cyclic group is cyclic so this entails your thing too)
fickle zealot
If I have $p(x)$ such that $G(p, F)$ is solvable, is there a (easy(ish?)) sufficient condition or like a way I can reasonable check that the radical tower $F \subset F_1 \subset F_2 \dots \subset F_m$ has $F_m$ \textit{exactly} the splitting field of $p$ over $F$ and not just $F_m$ contains the splitting field?
cloud walrusBOT
56
frigid lark
If I have $p(x)$ such that $G(p, F)$ is solvable, is there a (easy(ish?)) suffic...
I might be wrong, but I think assuming F_m/F is Galois is enough
fickle zealot
F / Q?
tropic moth
Which abstract algebra books are you guys reading?
frigid lark
F / Q?
Wait, I misread, F_m/F
Actually, I think F_m should have the n'th primitive root of unity where n is coprime to p (the char of F) and np^s = [F_m : F]
What I'm saying isn't iff, I'm just working with theorem 2 from section 7 from chapter 6 of Lang
frigid lark
I might be wrong, but I think assuming F_m/F is Galois is enough
And for this you only need F_m/F to be separable, as it is already Normal
wooden ember
Something I never really explicitly noted when I was doing this stuff but that I’ve noticed in looking at the universal coefficients theorem: is it true that for finitely generated abelian groups $A$,$B$ we have $$\operatorname{Ext}^1_\mathbb{Z}(A,B)\cong \operatorname{Tor}1^\mathbb{Z}(A,B)\cong T(A)\otimes\mathbb{Z} T(B)$$
cloud walrusBOT
𝓛ittle ℕarwhal ✓
wooden ember
Where T denotes the torsion part of that group
rustic crown
doesn't look true to me. like for B = Z/nZ, then Ext1 is A/nA and Tor1 is A[n], and last thing is T(A)/nT(A)
for A = Z, first is different from other two
for A = Q/Z, second is different from the other two
wooden ember
yeah it didnt sound true to me either but i cant figure out what's going wrong with my argument
im probably doing something clearly wrong from me misremembering properties
but like, if I write A and B as a direct sum of a free part and a bunch of cyclic groups
since everything is finite i can distribute my direct sums, any Ext/Tor group with a free factor disappears and im left with a sum of Ext(Z/nZ,Z/mZ) and Tor(Z/nZ,Z/mZ) for various m,n, but we know those are equal to Z/(n,m)Z at order 1 no?
rustic crown
you can probably show Ext0 and Tor1 are same, but Ext1(free, something) doesn't die in general.
rustic crown
🐟
cloud walrusBOT
aabb
wooden ember
okay so at least the iso on the right is correct yeah?
it feels weird there wouldnt be some nice dual iso for Ext
I know Ext^1(A,Z)=T(A) for A finite gen so torsion should be involved surely
rustic crown
okay so at least the iso on the right is correct yeah?
it's involved but i think you can only say Ext0 is Tor1 and Tor0 is Ext1
wooden ember
it's involved but i think you can only say Ext0 is Tor1 and Tor0 is Ext1
i cant believe ive already frogot so much of this stuff it's barely been a semester
lethal dune
wooden ember
Basic question but how does it follow that A/a is Noetherian ring?
ring is noetherian iff it's noetherian as a module over itself
rustic crown
it's involved but i think you can only say Ext0 is Tor1 and Tor0 is Ext1
This is prolly wrong >.<
Ill check the details after I return home.
tropic moth
Which abstract algebra books are you guys reading?
Anyone?🥲
lethal dune
I've seen a hom_tensor like adjunnction for tor and ext
prolly Ext¹(Tor_1(A, B), C)= Ext¹(B, Ext¹(A,C)) for modules over PID but I'll have to check
lethal dune
ring is noetherian iff it's noetherian as a module over itself
No I'm asking how does it follow that A/a is Noetherian A module then it's Noethreian as A/a module.
probably I'm missing something trivial
frigid lark
Anyone?🥲
I read Lang, I've know some people in here read/recommend Dummit and Foote, Aluffi, Fraleigh, and maybe some others
rustic crown
i cant believe ive already frogot so much of this stuff it's barely been a semes...
i forget stuff after 1 month of not being in touch with it >.<
tropic moth
average Herstein enjoyer
Its so short, is it enough as a starter?
fickle zealot
as an intro it's more than fine at least imo
tropic moth
I have dummit and foote but it has way more content than herstein
wooden ember
No I'm asking how does it follow that A/a is Noetherian A module then it's Noeth...
That’s not what they’re saying. They’re saying A is a noetherian A module, so A/a is a noetherian A module, and thus A/a is a noetherian module over itself
lethal dune
Yeah but how does that follow?
wooden ember
Yeah but how does that follow?
Last or first implication?
lethal dune
Last
wooden ember
Last implication is cause it’s naturally an A/a module since a annihilates it
Like if (a_1,…,a_n) generates an A submodule, then it also does so as an A/a submodule since a is in the annihilator
lethal dune
Ok it’s trivial lol
tropic moth
I say this as an ugrad tbf
Im doing postgrad
fickle zealot
ah lmao
elder wave
don't multipost
long nebula
Sorry I told them to come here for future questions, probably should've clarified not for the same question haha, my fault
zenith hollow
why are groups considered as the repeated use of bijective functions? i get that you take 2 elements and assign them a new one
but do you do that on repeat for it to become a group?
delicate orchid
multiplication by a group element is bijective function from that group to itself
although ngl your question doesn't really make sense, I presume that is what you were asking
long nebula
Yeah, you can treat any group as a set of bijections (in the finite group case, this is Cayley's theorem)
delicate orchid
they do not know about S_\infty
long nebula
(real numbers, +, *) is a field
zenith hollow
wouldn't that mean to be a Ring it would need to meet the condition of distribution law?
But a+(b*c) is not a+b x a+c
long nebula
But a+(b*c) is not a+b x a+c
a*(b+c) is ab + ac though
zenith hollow
Oh does your textbook list multiplication before addition
nope i was just wondering if the reversal would be a Ring
zenith hollow
that explains why they wrote 1 as the identity element
in the field
of (R, + , *)
long nebula
0 is the additive identity and 1 is the multiplicative identity
zenith hollow
ye but R,+ is a group anyway
long nebula
Yeah, (R,+) is a group and (R\{0},*) is a monoid
In fact (R\{0}, *) is also a group
Because R is a field
long nebula
Because you can't divide by zero
So you consider only the nonzero elements
For a field
zenith hollow
huh
long nebula
Sorry, (R,*) is a monoid and (R\{0}, *) is a group
zenith hollow
the field only has + and *?
long nebula
Sorry what's your question?
zenith hollow
Sorry, (R,*) is a monoid and (R\\{0}, \*) is a group
yeah here i mean why without the 0
zenith hollow
ohhh
long nebula
But for R to be a field, we require that every nonzero element of R has a multiplicative inverse
Which turns (R\{0}, *) into a group
zenith hollow
thx now i know all the definitions really well ^^
zenith hollow
Die Menge der Äquivalenzklassen bildet mit der Addition die Restklassengruppe modulo n. (Z/nZ, +)
correct?
zenith hollow
Ah ok ^^ sometimes i don't know the english vocab for it
(Z/primeZ, + , *) is even a field apparently
i guess that's why primes are so important in general
lethal dune
in fact if Z/nZ is a field iff n is prime
zenith hollow
btw i hear the word permutation a lot lately
what topic exactly covers permutations? or are they omnipresent?
lethal dune
groups
zenith hollow
groups
wait fr?
i am sure i heard it elsewhere as well
does it also occur at graphs for example?
long nebula
does it also occur at graphs for example?
I'm not sure what you mean by graphs (graph of a function? graph theory?) but permutations are usually studied under group theory (permutation groups)
You can also study them in combinatorics (how many permutations?)
zenith hollow
yeah i mean graph theory
long nebula
Oh I mean permutations are a very natural/broadly applicable concept but I can't think of any specific connections between permutations and graph theory right now
long nebula
Sure
elder wave
translate it
zenith hollow
i guess i should just post a picture of the german text somewhere else
it's hard to read this way
rain oxide
does it also occur at graphs for example?
A graph isomorphism is a structure preserving permutation on the vertices
zenith hollow
A graph isomorphism is a structure preserving permutation on the vertices
So when we got this graph for example can we calculate the number of triangles, (each edge is a triangles edge) with permutation ?
rain oxide
I dunno
zenith hollow
with the help of the adjacency matrix i meant
that was basically the (German) task i mentioned earlier
except i dunno whether you can do it with permutation
warped viper
what can we use the ring of fractions for?
like has this thing any use?? why am i learning this
chilly ocean
the algebraic construction of the rationals from the integers seems important enough to generalize, no?
embedding integral domains into fields is nice
more generally localization is extremely important
to say the least
warped viper
the algebraic construction of the rationals from the integers seems important en...
yeah definitely but that is an example of a ring of fractions
i was thinking like how can we use the ring of fractions within the field of algebra, like connect it to flatness or noetherian rings
i know that if A is noetherian then its ring of fractions also is, and from that and from the ring i've managed to construct a new neotherian ring using an homomorphism, so i was happy to see that you can construct new noetherian rings from the ring of fractions
but then i realised that i could just build a neotherian ring from a noetherian ring, without needing the ring of fractions
i also know that the ring of fractions is flat but i don't know what i can deduce from that
my teacher said they played an important role in flatness but i've been trying to find a link for a week and i can't so i'm starting to wonder if this was a missunderstanding
dim widget
All fields are noetherian, noetherian-ness of a ring has nothing to do with its field of fractions being noetherian.
warped viper
All fields are noetherian, noetherian-ness of a ring has nothing to do with its ...
"ring" of fractions
i don't wanna abandon my research on the basis that i found nothing
so i'm still searching for some use in flatness like my teacher said
dim widget
What do you mean by ring of fractions? Are you just talking about localization of a ring?
warped viper
What do you mean by ring of fractions? Are you just talking about localization o...
yes basically
when you use Ore's conditions instead of a commutative ring, sources call it the ring of fractions
dim widget
If you are familiar with localization of a commutative ring that should motivate why we care about localization of noncommutative rings don’t you think?
warped viper
the motivation you are talking about are local rings?
like how the localization of a commutative ring over a set whose complement is a prime ideal is a local ring?
the reason i worked on noncommutative localization actually was that the construction of "the category of fractions" is very very very similar to the ring of fractions in the non commutative case, even though they are very different in nature, so i went from the construction of the ring of fractions in the integral case (talked about the construction of the rationals with this part) then the commutative case (talked about local rings here) then the non commutative case, which i used to making the category of fractions easier to understand
but still, i have constructed all these things, but it feels pointless somehow, the connections i saw with algebra in general were in chapter 1 and 2, and i felt like i needed some more motivation for the next chapters (Ore and categories)
that motivation i still haven't found
obsidian loom
Hello, I need some help
since tao is applied first, sigma*tao = sigma right?
If so, isn't it already written as a "product of disjoint cycles"
what more do they want me to do
formal ermine
since tao is applied first, sigma*tao = sigma right?
no?
obsidian loom
oh right tao actually moves them around by 1
fervent gyro
This exercise seems super easy, but I'm absolutely blocked when trying to prove it. It seems similar to the prime avoidance theorem, but similar techniques don't seem to work here to prove it
slim kayak
That's a bit confusing. Can't you repeatedly use this result in order to show that any ideal is irreducible?
coral spindle
Well the exercise is false as stated! Take I_0 to be any prime ideal, and set a = I_0.
Clearly there's some info missing.
fervent gyro
The exercise is used by Jason McCullough in his Introduction to Commutative Algebra notes to prove this next lemma
slim kayak
How is Ass (hehe) defined? It seems like this is meant to be the prime avoidance lemma
next obsidian
Associated primes
For the purposes of this proof you can use prime avoidance, the stronger version where 1 (you can actually even get away with 2) are not prime
fervent gyro
I see, thanks for the help
next obsidian
Yeah idk how to see the result
You can try to avoid the dumb obvious fail by saying there’s more than two ideals there, and then I feel like you can do a prime avoidance type argument, but you did say that it wasn’t seeming to work
Okay I think you can mimic prime avoidance
Suppose that there’s more than 1 ideal here
The case of n = 1 is standard, you don’t even need proneness of either ideal
slim kayak
The way it is stated it is false so no wonder the prime avoidance type arguments fails. For the time being it seems reasonable to just assume that this is meant to be the prime avoidance lemma. I am unsure what weaker hypothesis you would want to add instead
next obsidian
Hmm
Okay my hypotheses are that n > 0 and that you’ve removed redundancy, so the union is not equal to I_0, I think this might work
I guess I’ll just try and write a proof
n = 1 is standard
Hmmm
Too tired, and writing the proof actually seems insanely annoying kekw kekw kekw
slim kayak
Imaging writing a proof: enjoyable, based. Actually putting the pen on the paper: Annoying, cringe.
fervent gyro
It definitely looks like the standard prime avoidance lemma works just fine. For the purposes of my work I can simply show any paper that proves it, so no need to elaborate a proof here, thanks
next obsidian
I was trying to figure out what modification of the exercise is true, if you can really move to subsets and only one prime ideal. I have ideas but don’t care enough to hash them out
formal ermine
been thinking about this for the past two days and still can't come up with anything. do you have a hint for me?
next obsidian
For all intents and purposes the +1 on the y-part means nothing, just shift by -1 and it’s basically t -> (t^2,t^3)
Now it’s pretty clear the minimal relation is (t^2)^3 = (t^3)^2
There’s your polynomial
zenith hollow
But 0 doesn't have an inverse in a field
ohh whenever it hits 0 it has no inverse
so basically 0 + 3*x has no inverse
north sand
in Z/3Z those are all the same thing
it's only Z/3Z and Z/2Z in which all multiplicative inverses are the same element again. that doesnt happen for other fields
south patrol
@chilly radish lol long shot but do you know anything about rings with like AeA = A
for e a (central) idempotent
I can't find much online but thought someone might know the terminology better like you lol
Might learn some noncomm alg
😼
earnest sky
Hi guys, was wondering if anyone could help with this question
summer path
delicate orchid
for e a (central) idempotent
Surely if it’s central AeA = Ae = A => e = 1
Maybe the last implication isn’t right - woke up 3 mins ago
Did u mean eAe? Cause I’ve seen that expression way more
south patrol
@delicate orchid No, for example consider the case of A = C[G] and e the principal idempotent, right
delicate orchid
Considering
delicate orchid
thinking
I’m not saying full => 1 I was thinking full + central => 1, but obviously the principle idempotent is central
Yeah ok now I can see how it’s full
Interesting
chilly radish
<@102705166754549760> lol long shot but do you know anything about rings with li...
Not really no
Sorry
south patrol
Np lol wasn't sure if it was a common class of rings
delicate orchid
Np lol wasn't sure if it was a common class of rings
are you asking for rings where every idempotent is full
delicate orchid
what do you want from me you silly billy...
south patrol
Okay idk lol
dw
I am sleeps
I was computing char table of A5 and it seems painful unless you use the fact it acts as group of rotational symmetries of icosahedron sadge
delicate orchid
I am sleeps
it's 1pm boss
A5 isn't that bad imo imo
like you have a very obvious action on 5 points
south patrol
doesn't that just gib u standard rep?
formal ermine
it's 1pm boss
it's 2 pm
south patrol
german moment
delicate orchid
doesn't that just gib u standard rep?
yur so there you go there's one of them
south patrol
lol
formal ermine
german moment
you live in the wrong timezone
delicate orchid
there's two conjugacy classes that don't square to themselves, so there is a pair of conjugate irreducibles
dunno if you know about that life hack
delicate orchid
ok so we have 60 = 1+16+2x^2+y^2
now you pull out your handy python script...
south patrol
and also unique rep fo dim 5 so it's real
i mean yes i worked out that lol
but ye
delicate orchid
ok so why u crying about eye deez potents
delicate orchid
simple diff
south patrol
Lol
left estuary
I did end up learning a lot of ring theory and all the prerequisites so I’m glad I did this and I look forward to continuing learning this subject when I can but for now it’s my birthday and I’m about to graduate so I will put stuff on hold for now
Thanks to everyone who helped me and showed me the right track
valid night
Is it true for a group $G$ that for $g \in G$ if $g^{d}h = hg^{d}$ for all $h \in G$ then $gh=hg$?
cloud walrusBOT
Kroros
valid night
If so how would I prove it?
delicate orchid
see what happens if you set h = d
valid night
Do you mean h=g?
delicate orchid
no
valid night
Oh $g^{d}$ means g repeated d times, sorry for the confusion lmao
cloud walrusBOT
Kroros
delicate orchid
no it's ok, mb
in which case I don't think this is true
t^2 is in the centre of D_8 but t isn't
delicate orchid
turns out it is true if you're conjugating though 
valid night
Lmao
zenith hollow
stuff like this can't be solved in R?
lethal dune
gilded frost
is there a difference between a group and a module over Z?
fervent gyro
Z-modules are assumed to be commutative (I don't know if it's been studied without this assumption)
gilded frost
right sorry i meant abelian group
fervent gyro
Ah, yeah, then none
fervent gyro
I believe modules were created to generalize abelian groups, among other things like ideals of commutative rings and vector spaces
When you have a group it's kind of intuitive how you can define a product by an integer by repeatedly applying the group operation
south patrol
Z-modules are assumed to be commutative (I don't know if it's been studied witho...
This doesn't really make sense like
You don't usually call a module commutative/noncommutative in itself
fervent gyro
I mean commutative as in the underlying group is commutative
south patrol
Sure I mean that is just modules in general but ye
I do now wonder if there is some more general thing but i guess then you've got smth more messy
chilly radish
I do now wonder if there is some more general thing but i guess then you've got ...
Closest thing that I've seen that is applicable is a group being acted on by another group such that the action is compatible with the group multiplication
You see this when you wanna look at extensions with nonabelian kernel
delicate orchid
oh kinda like a biset
feral umbra
I can't quite follow the proof of the following lemma (from Gromov's paper on polynomial growth)
I have the following questions:
- If there is a hom with infinite image, how does it follow from Tits' thm that we find a \Delta' \subset \Delta (of finite index) s.t. [\Delta',\Delta'] \subset \Delta' has infinite index? It seems like it follows directly from the thm that \Delta itself is nilpotent, but I might be mistaken.
- Why does it follow that \Gamma is almost (or virtually) solvable? We have shown that we can find \Delta' (of finite index) s.t. [\Delta',\Delta'] is almost nilpotent, however, this commutator subgroup has infinite index in \Gamma so isn't of much use directly.
- Why is it sufficient for \Gamma to be almost solvable, rather than solvable to apply the thm by Milnor-Wolf?
Here are the lemma's referenced: (note the induction is on the growth)
cloud walrusBOT
Joe 1
devout slate
how can I show that Q(√a,√b)= Q(√a+√b) , where a≠b and neither a or b is a perfect square? I tried to show that Q(√a,√b)⊂ Q(√a+√b) and Q(√a,√b)⊃ Q(√a+√b) to prove that they are equal. But I am having trouble in showing Q(√a,√b)⊃ Q(√a+√b)
formal ermine
that direction should be the trivial one lol
slim kayak
how can I show that Q(√a,√b)= Q(√a+√b) , where a≠b and neither a or b is a perfe...
Um, did you mix-up the inclusion there?
rustic crown
that direction should be the trivial one lol
illu i got a holiday tomorrow as well
slim kayak
how can I show that Q(√a,√b)= Q(√a+√b) , where a≠b and neither a or b is a perfe...
Anyways, if you wanna go the elementary route note that Q(sqrt(a)+sqrt(b)) must contain sqrt(ab) and hence it contains a sqrt(b) + b sqrt(a), from there you can extract either sqrt(a) and sqrt(b).
formal ermine
illu i got a holiday tomorrow as well <:eeveeKawaii:556858499930259471>
yoooooooooo
lesgo
I got uni tomorrow
lecture at 2 pm
have to wake up early
lethal dune
i still have to go on friday
Don’t ig
slim kayak
As long as you have the script or recordings, yeah
rustic crown
but it so fun to attend it live
slim kayak
but it so fun to attend it live <:eeveeKawaii:556858499930259471>
Incomprehensible statement for a commuter
lethal dune
Lol
rustic crown
commuter? 
lethal dune
commuter? <:catThink:526983414226747393>
Ig their commutator is 0
slim kayak
Yeah, i stan'ed algebraic topology back and then still do
rustic crown
det wanna learn topowogy too >.<
elder wave
real
rustic crown
awgebwaic uwu
elder wave
ooh i didn't realize you meant pointless topology as in the subject
slim kayak
det wanna learn topowogy too >.<
do you need like a license to start or?
formal ermine
have to wake up early
@elder wave I usually wake up at like 3-4 pm on holidays / weekends
lethal dune
Pointed pointless topology
elder wave
i found it funny
rustic crown
do you need like a license to start or?
me wanna learn everything >.< and me slow 
formal ermine
me wanna learn everything >.< and me slow <:sad:532803514892812288>
you already know so much
rustic crown
awgebwa is
formal ermine
where does p-adic analysis go? #advanced-analysis?
lethal dune
I find point set cringe now
slim kayak
Points? Never heard of them
rustic crown
i've done more point-set stuff after AG than in actual point set top course 
formal ermine
where does p-adic analysis go? <#576511214025310260>?
@elder wave mr moderator
lethal dune
It’s just “hey did you know that if you X and Y then Z holds and here’s an random space as counterexample”
slim kayak
Is that this weird set stuff? I thought that trend had finally passed
formal ermine
it's algebraic number theory all over again
slim kayak
Whats a CW complex? It's where the arrows start and end
rustic crown
(me forgot rep theowy >.<)
formal ermine
I'll have a go joe
lethal dune
Got reps exam in a couple weeks lol
I got an interview coming 
south patrol
Gl
Actually can you just link the sheets
i can probs find them
lol
Dw
Oh okay
Wtf are these exams lmao
lethal dune
Wtf are these exams lmao
Yeah lmao
willow geyser
yeah i did
south patrol
Anyway I'll have a go at this q anyway lol
I like how uh
Some of this is more advanced than us
Some is stuff I did in first year physics
Weird lol
We didn't do Mackey's theorem actually
which is that
Using pi for a character bruh
rustic crown
me looked at that a few weeks ago and already forgot
agile burrow
walter forgets too
rustic crown
walter
delicate orchid
State mackey's theorem
:breakfast: just keep writing down intersect symbols and subscripts until you get bored
summer path
my rep theory class straight up just didn't do induced reps
agile burrow
det
rustic crown
me looked at that a few weeks ago and already forgot <:kongouDerp:48398839209944...
i refuse to read that >.<
delicate orchid
I would not uhhh denote it like that haha
lethal dune
<:kongouDerp:483988392099446794>
Same
formal ermine
<:kongouDerp:483988392099446794>
it's almost as bad as analysis theorems
south patrol
Oh wait we did this i think lol
delicate orchid
wait
how do I write it
there's a nicer way of phrasing it than this ^ using an inner product
summer path
yeah, but the prof was running out of time and went for a bunch of nice applications instead; they were fun 
ill take rep theory again later and probably learn it then
thanks for reminding me
det
rustic crown
tubu
delicate orchid
this, apparently
no idea where I got this from it was just in the archives (randomrepshite.tex)
rustic crown
cute notation
delicate orchid
yeah I use le arrows
rustic crown
no idea where I got this from it was just in the archives (randomrepshite.tex)
(what's with the last e? in 'randomrepshite')
delicate orchid
the word "shite"
rustic crown
time for dictionary again
formal ermine
zufälligescheiße.tex
rustic crown
rustic crown
yeah I use le arrows
whut do you use for coinduction?
formal ermine
coinduction?
formal ermine
induction in Z^op 
agile burrow
Lol
delicate orchid
right that's roughly what I excpeted
the honest answer is I literally never do that
agile burrow
Wew's groups are finite
rustic crown
Wew's groups are finite
oh right, they were magically iso
lethal dune
Ugly
delicate orchid
I wonder if there's some hom-tensor adjunction nonsense to do with this that you can pull into something more tangiable
like they do with frobenius reciprocity
south patrol
Okay yeah yikes lol
But the way you split up ur sum isn't legit in ur method
Like by the same logic 1 = 1 * 1 = (0 + 1)(1 + 0)= 0 * 1 + 1 * 0 = 0
delicate orchid
oh now this looks interesting
lethal dune
lol
south patrol
But yeah the answer they've given in the next bit implies like
< πl, πk> = l + 1 if l <= k right
Oh I have an idea perhaps
hm
delicate orchid
you need to find the number of subsets in X_r that are not subsets of the set of fixed point of a given permutation, that should be the value of the character
what this mofo actually is is anyone's guess
no, sorry
that are subsets
this is a third year question?
where tf do you go
right I should have guessed
having a fucked course is better than having no course at all
formal ermine
only in ohio
delicate orchid
my uni did not have a rep theory course or much algebra at all
they had 2 algebra courses for my masters year
I was on an integrated masters course so idk for that one
for PhD uhhh different since you're from cambridge but you'll want a 2-1 probably
delicate orchid
yeah potato I think ur initial guess doesn't make sense btw
delicate orchid
we can all agree on the degree at least, I hope
delicate orchid
how to find the number of doodads that fix a set though that is the question
south patrol
The problem it's the other way round like
Number of sets fixed by the doodad
which is bruh
delicate orchid
lemme just
delicate orchid
$\pi_r(\sigma) = |{Y \in X_r \colon \sigma Y = Y}|$
what fuckin error u cry baby
cloud walrusBOT
Wew
delicate orchid
this is (n-r)!P(n-r) no?
false
I'm thinking about this as taking each permutation in 1-row, it fixes Y if and only if none of the numbers in the 1-row form are in Y, and thus there are n choose n-r of them - and then we have P(n-r) different ways of partioning this to get different cycle structures
nope this is still false, if EVERY number in the 1-row form is in Y it also fixes Y
it's a good start though
is the correct statement "fixes Y if and only if all of the numbers are in Y, or none of the numbers are in Y"? I think so
yeah it does
I'm not sure how yet but I can feel that it does
delicate orchid
is the correct statement "fixes Y if and only if all of the numbers are in Y, or...
let N(sigma) = X-Fix(sigma) be the set of points acted on non-trivially by sigma, then a pair (Y, Z) is in your Cartesian product if and only if (Z,Y) \subseteq N(sigma)^2 or (Z,Y) \subseteq Fix(sigma)^2
south patrol
I think I've got an answer by doing it the other way round lol
south patrol
Oh maybe that is what is being done lol
delicate orchid
OH SHIT YEAH
once you've got it in one set you can swap out enumerating over the subsets to enumerating over the permutations can't you?
or am I smoking weed again
ah no you'd have to iterate over pairs of subsets
igojwperk'sdgwrosp[gikjag[
delicate orchid
basically what joe did on line 3
delicate orchid
mf got the python script online
yeah I mean i could check out S_4
@south patrol which symmetric group u working on
south patrol
Oh I was doing general case
I think you can do it by counting in a different way and considering how two sets overlap but seems painful
So I reckon there's a smarter way
In particular, noting that pi_r is a constituent of pi_s for r <= s (by the next part lol)
delicate orchid
In particular, noting that pi_r is a constituent of pi_s for r <= s (by the next...
literally cheating but also this is to be expected
wait is it
wtf do you have for the degree
|X_r| right
delicate orchid
ok then how can r <= s imply it's a consituent when n choose n is 1
south patrol
We are dealing with k,l <= n/2
south patrol
lol
south patrol
fudnamental group
delicate orchid
NO you fool
what topology she got on? in those subsets?
pi_2 of S_4 is (6,2,0,2,0) with pi_2(12) = pi_2((12)(34)) = 2
interesting
delicate orchid
let N(sigma) = X-Fix(sigma) be the set of points acted on non-trivially by sigma...
my idea about this seems correct @south patrol
south patrol
Inch resting hm
delicate orchid
well, maybe not about the pairs, but if you just take either Y or Z it's definitely true
cause I think if Z is in N(sigma) and Y is in Fix(sigma) or the other way around then you're fine
it's when the set of "non-fixed" points intersects Y but doesn't contain it, then it cannot be fixed under the action
hold on a minute
boss
@south patrol boss
south patrol
helo
delicate orchid
would it be easier to work out the orbits instead
delicate orchid
then burnside's lemma that mofo
ah it's not so simple
could still orb-stab it but
not sure
we're definitely missing something here like idiots
delicate orchid
if sigma contains only k-cycles with each k > r then pi_r(sigma) = 0
there's something
there will be a way to calculate the inner product without knowing the character values explicitly
thing is I don't care anymore I need to know what these values are
delicate orchid
I'm gonna keep legit working on this cause it seems tangentially related to a tangent of my actual research stuff lol
that's my justification for getting nerd sniped
south patrol
Lol
This stuff seems much more advanced tho
LOL
@willow geyser
Old problem sheet question
delicate orchid
lets see if i'd have ever gotten this
south patrol
Oh yes
You can find the orbits
(A,B) and (C,D) in same orbit iff | A \cap B| = |C \cap D|
delicate orchid
=> obvious right
no - I don't see what A and B have to do with each other
delicate orchid
well, no it isn't - I had to think about it... a grave sacrifice
south patrol
You can then find # of orbits
and basically like
Apply Burnside to that action i think
Well you can like
delicate orchid
yeah just find number of orbits LOL?!?
south patrol
Just count in a different way lol
delicate orchid
I'm gonna get these character values irregardless
south patrol
Well what we are doing is like
delicate orchid
wait is this for the inner product or the actual values
south patrol
That can be computed w burnside
delicate orchid
it's just one fix
south patrol
Yeah
delicate orchid
right
delicate orchid
what a bunch of bullshit
delicate orchid
anyway
south patrol
AAnyway I assume this is the sort of thing where like
It's okay if you've already done it
delicate orchid
I had the right steps I just couldn't be arsed
south patrol
Cause that's what our exam questions are like
Like you do enough past papers and you know roughly what to do ig
well that is all exams
lol
delicate orchid
(A,B) and (C,D) in same orbit iff | A \cap B| = |C \cap D|
minus this observation which I guess helps
languid saddle
can someone remind me of the name of the group whose elements are kind of like the permutation group, but it's like you're drawing lines from n elements on the left to n elements on the right
and the lines are like a string, they can twist around each other
south patrol
r these braid groups
delicate orchid
yeah those are braid groups
languid saddle
i've no clue i just remembered someone telling me about the basic definition and i thought it was super neat
delicate orchid
Fundamental group of space of n points on R^2 right
you mean R^2-a few points?
my memory of this is foggy
south patrol
no
Dw this is basically uh like
Ordered configuration space of n points on the plane so the points are n-tuples of points where each is distinct
But the idea is like uhhh
You can think of a braid (with the twisting) as a class of paths those n points can take
Which is cool
delicate orchid
I'll just google the definition of configuration space
south patrol
Llol
south patrol
Ordered configuration space of n points on the plane so the points are n-tuples ...
That is the definition
delicate orchid
way easier for me to parse in notation
south patrol
Okay sorry so like
delicate orchid
so yeah it IS R^2 minus a few points you LIAR
south patrol
tht's ordered & then unordered is when u have the natural action
lol
No it's like
R^{2n} minus a few points
Few meaning a lot
delicate orchid
wtf surprise n
slim kayak
Fundamental group of space of n points on R^2 right
With what edges?
south patrol
R^2 minus n points is just S1 \wedge itself with n summands
delicate orchid
yeah I thought that's what you were referring to lol
join they circles
wait yeah how tf could that be a braid group lmfao
delicate orchid
le abelian braid group has arrived
delicate orchid
ok I know one of them is
south patrol
wedge of circles doesn't usually have abelian
delicate orchid
two counting the trivial group
delicate orchid
ermmm so anyway uhh so like nayway uh huh
slim kayak
Nice weather today
south patrol
I need to practice more reppies
rustic crown
cute name
summer path
algae 
rustic crown
commie and homie algae
coral spindle
Reppies, Gal Wah Fearie, Algae number theory
guh-loys feery
upper pivot
learnt about loop braid groups in a conference i went to kek
and how they are connected to yangs mills operators that conserve charges
(i didnt really understand a lot of it!)
slim kayak
- I am a little bit unsure about the first bit. Is it because that one then factors the composition to get the extension of phi?
- Um, why isn't that a loss of generality? How does it connect to the case where k[x] isn't a field, or is k[x] as a finitely generated ring for some reason always a field?
wraith cargo
1. I am a little bit unsure about the first bit. Is it because that one then fac...
For q 2 they tell you why in the first paragraph
slim kayak
Yeah but isn't that only for the special case when k[x] is indeed a field?
wraith cargo
No
Because they tell you you can extend it to the general case
Usually wlog means you restrict it to a special case but you can get the general case from that special case
summer path
the entire first paragraph is explaining why it is without loss of generality; also you have that \mathfrak{M} is a maximal ideal, so k[x]/M is a field
i think that might be the part you're forgetting(?)
wraith cargo
Essentially the idea is that if you prove it for when k[x] is a field you can just apply the theorem to \phi \circ \sigma^-1
And get back the general case
slim kayak
What exactly is the general case here, k[x] is not necessarily a field?
summer path
the general case is the hypothesis given by the theorem, i.e. k[x] is a finitely generated ring over k, so not necessarily a field
slim kayak
Like, it goes "If k[x] is a field and since k[x]/m is a field, as is sigma k [...] then one obtains an extension" as far as I can see
I don't get how that relates to the k[x] is not a field case, since I can't pass the argument to k[x]/m since I'd have to use the (left) inverse $\sigma^{-1}$ of the quotient map, which only exists if the quotient map is already an injection, hence an isomorphism. But then k[x] would be a field.
cloud walrusBOT
Kerr
wraith cargo
I'm not exactly sure what the issue here is hmmmmm
slim kayak
I don't understand how the "k[x] is not a field" case is covered
wraith cargo
I don't understand how the "k[x] is not a field" case is covered
Okay so
Take \phi \circ \sigma^-1
That's a homomorphism
From some field into L
slim kayak
Where sigma inverse only exists if k[x] is a field : (
wraith cargo
That is not true
delicate orchid
sigma inverse exists because the quotient map is surjective
or is that right inverese
oh well!
slim kayak
Left inverse is injective
delicate orchid
darn!
slim kayak
And quotient maps are surjective, so a left-invertible quotient map is a ring isomorphism which doesn't really work from a non-field into a field?
slim kayak
That is not true
The "pullback" to the general case works like this, no?
The homomorphism $\phi \sigma^{-1}$ from $\sigma k$ to L extends to $\sigma k[\sigma x_{1}, ... ]$, one then factors this extended homomorphism at the image of $\sigma^{-1}$ which is just k[X] to obtain the extended homomorphism from k[x] to L.
cloud walrusBOT
Kerr
solar glacier
1. I am a little bit unsure about the first bit. Is it because that one then fac...
is this out of Lang
i noticed the Nullstellensatz font lol
"theorem of zeros"
slim kayak
Oh okay, I think I see now. Since sigma restricted to k is an isomorphism to $\sigma k$ this works, the $\sigma^{-1}$ is meant to be restricted to k or rather $\sigma k$
cloud walrusBOT
Kerr
solar glacier
i have a question whenever this ones finished lol
formal ermine
what is your question on
formal ermine
what's rad2
slim kayak
radical ideal of (2)?
solar glacier
showing $3+\sqrt{2}$ is irreducible in $\Bbb{Z}[\sqrt{2}]$
cloud walrusBOT
MyMathYourMath
formal ermine
gaussian integers 
solar glacier
gaussian integers <:skullpods:979290331499606066>
theyre so fun!
and the norm is 11
so it its reducible = ab for non units a,b
then N(a)N(b)=11
so WLOG say N(a)=1, N(b)=11
Then a is a unit
and we are done
formal ermine
correct
solar glacier
cool just checking
slim kayak
Yeah, maybe you wanna verify that only units has norm 1 with an extra sentence but otherwise this is pretty much done and dusted
solar glacier
and the trick of G abelian iff G/Z(G) cyclic is noting all elements of G can be written as x^nz where x is generator for G/Z(G) and z \in Z(G) and some n \in Z^+
formal ermine
now... what are the units in Z[sqrt(2)] 😉
solar glacier
as G/Z(G) is cyclic it has a single generator
coral spindle
and the trick of G abelian iff G/Z(G) cyclic is noting all elements of G can be ...
IFF? NO!
Oh wait
yeah you're good
solar glacier
now... what are the units in Z[sqrt(2)] 😉
id have to sit and compute this out
formal ermine
id have to sit and compute this out
oh god please no
rustic crown
boytjie
coral spindle
You don't say
solar glacier
hi det
rustic crown
hewwo mmym
formal ermine
when illu sleep?
I was gonna sleep like an hour ago but I wanna read more p-adic analysis 
solar glacier
I was gonna sleep like an hour ago but I wanna read more p-adic analysis <a:wahh...
want a text i did on it
slim kayak
oh god please no
Fr, just apply the same norm argument to an arbitrary unit a+bsqrt(2)
delicate orchid
I was gonna sleep like an hour ago but I wanna read more p-adic analysis <a:wahh...
oh have you finished real analysis then
solar glacier
oh have you finished real analysis then
u mean inequality theory lol
formal ermine
formal ermine
want a text i did on it
sure
white oxide
sorry to interrupt but is the set of all nonconstant polynomials under multiplication an example of a semigroup that is neither a monoid or a group?
solar glacier
undergad analysis is inequality theory
rustic crown
measure theory is algebra
formal ermine
I have to hold a presentation on p-adic exp and ln