I think it was a good idea in any case, and there are still details to work out
#groups-rings-fields
coral spindle
white yoke
My supervisor told me that it follows from a known result... but it does not seem known haha
Idk what an idempotent is
topaz solar
X^2 = X
rustic crown
det is sleepy
delicate orchid
there's a canonical choice for idempotents for characters here - but again it's over Z 
coral spindle
Sadge
Yeah seems like a tough problem
So I guess if we're deciding to abandon that
white yoke
I'm using a result by Milnor to prove that projectives are free over Z[Q_8] and to prove this, showing projectives are free over Z[C_2xC_2] is one of the steps
coral spindle
Oh well, if you have that projectives are free over Z[Q_8]
I think you should try to interpret Z[Q_8] as a Z[C_2xC_2]-module
topaz solar
Well it’s a step to show that
coral spindle
Or perhaps the other way around, lol
in any case I think this should at least reduce the problem significantly
white yoke
Oh well, if you have that projectives are free over Z[Q_8]
No no, I want to prove it for Z[C_2xC_2] and use that to prove it for Z[Q_8]
coral spindle
Oh well nevermind then
delicate orchid
interesting they're free over Q_8 too
coral spindle
Yeah
delicate orchid
wonder if it has something to do with Z[G] with G Hamiltonian
white yoke
What does Hamiltonian mean
white yoke
Ah cool
coral spindle
Honestly I have no other ideas at this present moment than to just calculate the ideal structure of Z[C_2 x C_2] and see what that gets you
delicate orchid
yeah, honestly it shouldn't be too bad (I pray)
u got ur.. uhhh ... (0) and uhhh (p) for p prime and uhhh... uhh
coral spindle
There are some interesting details in the ideal structure, but I don't think it should be too bad either
Very interesting question indeed
topaz solar
Do you even have any weird ideals in that
white yoke
For example to prove projectives are free over the Hurwitz integers, I proved it is euclidean which means it is a principal ideal ring so projectives are free. If L are the integral quaternions and H are the hurwitz integers, then L/2H is a local ring, over which we know projectives are free. I was looking for some kind of result like this that can be applied.
Because something my supervisor said implied that there is such a condition here
coral spindle
Do you even have any weird ideals in that
There are certainly some mistakes that would be easy to make in determining the structure
coral spindle
For example to prove projectives are free over the Hurwitz integers, I proved it...
Perhaps you can prove that this is a PID. I wouldn't be surprised.
delicate orchid
wait can we not formulate Z[C_2xC_2] as a subring of the hurwitz integers
topaz solar
There are certainly some mistakes that would be easy to make in determining the ...
I would imagine it’d do something, but showing it’s PID seems reasonable
white yoke
wait can we not formulate Z[C_2xC_2] as a subring of the hurwitz integers
are you sure it's a subring?
white yoke
wait can we not formulate Z[C_2xC_2] as a subring of the hurwitz integers
By hurwitz integers I mean this btw https://en.wikipedia.org/wiki/Hurwitz_quaternion
delicate orchid
yeah yeah ik dw
white yoke
ahhh maybe not, I forgot ij = -ji
yeah i was thinking its not abelian
but PID might be a shout
delicate orchid
yeah see if you can show it's a PID
topaz solar
Let me know if you find any weird ideals
delicate orchid
yes please do so I can post them to weirdideals.com
white yoke
yes please do so I can post them to weirdideals.com
Bruh i almost thought this was real
topaz solar
Trolled
delicate orchid
Ayo can you give it a lattice order now that I think of it?
I thought of looking at it like a lattice
virtual character type beat
topaz solar
If you can, I know of a paper on a Structural Theory of(for?) Lattice Ordered Rings
Probably not quite relevant though
Since it’s looking for l-ideals and f-rings
(This is not my wheelhouse)
Well if we had a lattice ordering on Z[C2 x C2], if we made any of the group elements positive we’d have an issue combining them I think
Maybe
delicate orchid
it wouldn't play nice with multiplication I don't think
molten viper
So, I'm working through homework, and I've done part b, so I know R/I x S/J is a domain, what facts would be relevant to actually finishing the proof? I is an ideal in R, J is an ideal in S
delicate orchid
what properties would this ordering need
topaz solar
We can obviously get 4 copies of Z type of ordering
But I don’t think we can order elements of G here wrt 0
topaz solar
what properties would this ordering need
Monotone multiplication by positives
delicate orchid
I thought so, that definitely doesn't play nice if you want the group elements positive
wet zodiac
ok im thinking of the next problem (describe all groups that contain no proper subgroup)
topaz solar
Well, if x <= yg, and 0 < g, we’d have xg <= yg^2
coral spindle
ok im thinking of the next problem (describe all groups that contain no proper s...
Hint: cyclic subgroups
wet zodiac
im guessing it’s all cyclic groups where |G| is greater than 2 and is prime
coral spindle
where |G| > 2?
wet zodiac
yes
topaz solar
That 0<g isn’t strict as a necessary condition for multiplication but since it’s appended it’s clearly not 0
coral spindle
Does the cyclic group of order 2 contain a proper subgroup?
coral spindle
u don't say
topaz solar
Well, if x <= yg, and 0 < g, we’d have xg <= yg^2
g^2 = 1, so we get x <= yg iff xg <= y
coral spindle
Perhaps you can prove that this is a PID. I wouldn't be surprised.
I no longer believe that it is a PID, unless I'm simply being very silly
delicate orchid
so then it’s for all cyclic groups with a prime order
yeah, plus one more
topaz solar
Though how do you supremum 1 + g mmmm
wet zodiac
ok what other group has every subgroup disobey closure
topaz solar
I no longer believe that it is a PID, unless I'm simply being very silly
What kind of ideal kills it?
wet zodiac
coral spindle
What kind of ideal kills it?
Choose some generator g of a cyclic subgroup, then I think (2, 2g) is a counterexample
delicate orchid
ok what other group has every subgroup disobey closure
I don't know what this means but you don't need it to find the missing one from your set
topaz solar
Well in particular we get C_2 x C_2
delicate orchid
you guys can work with these ideals in your head so fast 
wet zodiac
you can always find a subgroup that contains the identity and inverse of whatever elements you have
topaz solar
But if you have a cyclic element wouldn’t you be able to cycle around to get 1?
coral spindle
you guys can work with these ideals in your head so fast <:KEK:58624087735835034...
Nah I got it wrong
wet zodiac
otherwise the premise of G being a group is incorrect
topaz solar
you guys can work with these ideals in your head so fast <:KEK:58624087735835034...
1000 computations/instant and their all wrong
delicate orchid
Nah I got it wrong
yeah but you noticed you've got it wrong, something I can only kinda see by now
topaz solar
Notice the incorrect their
delicate orchid
you multiply by the other generator for a little trolling right
topaz solar
It’s C_2 x C_2 though
delicate orchid
topaz solar
So the other generator shouldn’t really ruin that?
delicate orchid
yeah then I can't see why that ideal is principle at all 
, is it just (2g)?
coral spindle
Choose some generator g of a cyclic subgroup, then I think (2, 2g) is a countere...
But I think that a similar idea does work. If g is the generator of the first C_2, and h the generator of the second, then I think (2g, 4h) would be an example of a non-principal ideal. I reckon that would work, but I'm going to now go to sleep.
wet zodiac
ok that problem wasnt that bad
coral spindle
But I think that a similar idea does work. If g is the generator of the first C_...
I am returning to say that this is just (2g) thanks for listening
delicate orchid
ah now I'm starting to get the hang of these
coral spindle
It's probably a PID so WORK ON IT !!!!!!!!!!!!
delicate orchid
NO!!!
topaz solar
But I think that a similar idea does work. If g is the generator of the first C_...
Yeah you can’t get (1) out of that but h = gh * g
So (2g)?
delicate orchid
let $C_2 \times C_2 = \langle a, b\rangle$ cause I said so
Take an ideal $(x_1a^{i_1}b^{j_1}, x_2a^{i_2}b^{j_2})$ is this always principle?
cloud walrusBOT
Wew Lads Tbh
topaz solar
Well I think you want x1, x2 not coprime so you don’t get (1) in there
delicate orchid
yeah that bit is trivial, so we can assume they're not coprime
wet zodiac
whats a PID
delicate orchid
then can we immediately reduce one of them to gcd(x_1, x_2)?
topaz solar
ab * b = a, and ab * a = b
So you should be able to get the bigger one from the smaller by that
delicate orchid
whats a PID
a ring where every ideal is generated by one element
delicate orchid
ab * b = a, and ab * a = b
ok yeah so just like
topaz solar
So you should be able to get the bigger one from the smaller by that
If you have x1 a then you also get x1 b and x1 ab
And x1 ofc
delicate orchid
take some element in the ideal, since this mf is abelian we can just automatically say that xg in the ideal for some g in C_2xC_2 then ALL xg is in that mf
and since x1 is in there
that mess ^ = (x_1, x_2) = boom
topaz solar
so any ideal containing two+ terms from Z[G]\Z can be reduced to one term?
delicate orchid
I believe so
topaz solar
Wait that’s a tad fast
delicate orchid
wait maybe not general G
abelian, probably
thinkin bout my character decomp again 😋
delicate orchid
then why do you say it's a tad fast?
topaz solar
Well what about reducing 1+g kinda thing
delicate orchid
oh yeah, lol
delicate orchid
if an ideal is contained within a principle ideal does it have to be principle?
google time!
delicate orchid
probably... NOT!!!
topaz solar
(X) contains any other cursed ideal of x terms
quiet pelican
probably... NOT!!!
Every ideal is in (1)
delicate orchid
oh my god who CARES
south patrol
And then by considering quotients etc we get more general ones
delicate orchid
ok so... epic fail
topaz solar
But anyway
Any non principal ideal has at least 1 generator term in Z[G]\Z
G being our cyclic product
Since Z PID
How does 2+4g look combining it with something
topaz solar
so can you do something similar for x+yg, a+bh
delicate orchid
Completely general sums are scary though, we should probably just try and consider them instead of a billion cases
hot lake
you should be able to complete the table by carefully using all the orthogonality relations (columns and rows), the fact that the three 2-dimensional representations are related by tensoring with the 1-dimensional ones, and looking at the few choices of eigenvalues for (1 & 1 // 0 1)
topaz solar
Completely general sums are scary though, we should probably just try and consid...
I mean if you can reduce these two terms that should net you anything
Except maybe (n, x+yg)
delicate orchid
I mean if you can reduce these two terms that should net you anything
I’ll take your word for it
delicate orchid
I… don’t see how that solves (n, x+yg)
topaz solar
you could also consider a+b+ab too but that should be a tad nicer
topaz solar
I… don’t see how that solves (n, x+yg) <:KEK:586240877358350341>
Well h is either 1, a, b, ab
delicate orchid
WHERES THE H
topaz solar
so can you do something similar for x+yg, a+bh
Here
topaz solar
There’s like very few cases, it’s just picking 2 numbers from 1 to 4
delicate orchid
This is agonising there has to be a quicker path
topaz solar
Wait no, that specific case relied on it being g on both parts
That indeed is a different case, but solving that (2, 2) case solves (1, 2)
Since we can take one as the identity gg
delicate orchid
Sure
topaz solar
There’s like very few cases, it’s just picking 2 numbers from 1 to 4
I mean, every term can be written as like 4 elements of Z
There’s probably some dumb sane way to solve this immediately
chilly ocean
Why on earth E1j belongs to Lj
hot lake
every coefficient is 0 except one at the intersection of column i and row 1
so every coefficient is 0 outside of column i, so it is in Li
yeah the wording is a bit ambiguous there
pastel cliff
blunder...
blunder
pastel cliff
wew wanna verify some proofs for me 
i'll give you a cookie
i think p^{-1} is wrong but i couldnt be assed when i wrote this
south patrol
Slight typo when you say "we see that m = sum_{\alpha=1}^{i} ..." since this should be aminus, but i assume you made a typo lol
But yes, this is perfect otherwise lol
Note that the proof is literally the same as for vector spaces
when you prove rank-nullity
Just you don't have to show the independence bit lol
pastel cliff
Slight typo when you say "we see that m **=** sum_{\alpha=1}^{i} ..." since this...
yes typo
ty for pointing that out
i also had thiis
which felt way too easy
south patrol
ty for pointing that out
np
That solution doesn't really work
The union of two linearly independent sets needn't be linearly independent
(Otherwise we could just union two bases)
Hint: every spanning set contains a basis
south patrol
Well
my hint is unnecessary actually
Try the following lemma, which I assume you know: if B is linearly independent and v isn't in the span of B, then B u {v} is linearly independent
Eh i mean i guess there are many ways to do this lol
topaz solar
That lemma always felt like basically the definition 
pastel cliff
Hint: every spanning set contains a basis
well this would just end the proof there no? i think the spanning set i have is at least right
topaz solar
Eh i mean i guess there are many ways to do this lol
Become HoTT/type theorist/proof theorist and consider if they are equivalent proofs 
topaz solar
well this would just end the proof there no? i think the spanning set i have is ...
A basis, but you need one containing A
So you need a statement that all elements of A are in it ofc
pastel cliff
topaz solar
What was that one theorem/lemma about uhhh interchange or something
This seems really similar to it (if not the same but I’m not recognizing it)
pastel cliff
Try the following lemma, which I assume you know: if B is linearly independent a...
since X = {b_1 ... b_m-n} U A is a spanning set of V and A is a linearly independent subset, then the subset of X given by A U {b _i \in B s.t. b_i not in the span of A} is a basis of V...?
that feels a bit ehh
topaz solar
Well B = {bi} spans yes?
pastel cliff
wdym
i meant that as like
take the basis elements of b that are not in the span of A and just union them to A
topaz solar
Well if B is a basis, could kinda just step by step take the first one that’s independent of A until you get a basis I think yeah
pastel cliff
since X = {b_1 ... b_m-n} U A is a spanning set of V and A is a linearly indepen...
ig a problem w this is that it mght not necessarily span V?
pastel cliff
yeah that makes sense
topaz solar
You could make it way more strong of an argument by saying some dim span A < dim span B = V kinda thing
Something something induction and appending a_n+1 in V\span A
pastel cliff
topaz solar
Well I’d say, instead of “all such b,” something like “iterating this process”
The union {b1 ... bn-m} u A stated as spanning A before showing you can do that is odd
pastel cliff
going back to my q from way before tho, imma assume this is almost enough?
maybe im just being lazy
but do i really need to get into universal property stuff to show that this is true
like univ. property of free module and coproduct
topaz solar
Universal property of product is that if we have B<-X->A then we get a unique arrow X->AxB
However direct sums are also the direct product for finitely many, right
Wait I think I mixed that up with product oop
Shh
south patrol
That isn't quite what the universal property says
south patrol
However direct sums are also the direct product for finitely many, right
And yes that is true for abelian categories
They "admit finite biproducts"
topaz solar
Coproduct is B->X<-A gives a unique A+B->X
topaz solar
And specifically these arrows are important in that they commute with the injection/projection
south patrol
That's part of the statement yes
topaz solar
So X->AxB->A is the same as straight to X->A
south patrol
Ye
topaz solar
Reverse for A+B
pastel cliff
hmmmm
me thinks this will be a "clearly" and "as shown in class" proof
(i havent been to lecture in 2 weeks)
topaz solar
Any element of N \oplus M should basically be a sum of elements of N & M right?
I mean think about the projections $N \leftarrow N\oplus M\to M$
cloud walrusBOT
Sharp
sonic coral
am i missing something or is this wrong? The group generated by a is H=<a>={e,a,a^2}, and after checking gHg’ for all g in D6 but not in H, gave me something back in H
sonic coral
gotcha overlooked that
i thought it had to be normal is a^2 was in H since everything in D6 but not in H had order 2
smoky cypress
Is there anything interesting to be said about rings that have composite characteristics
topaz solar
well they're certainly not fields
sonic coral
I realized why I was having trouble. i have a handout where i filled in a cayley table of D6, but the a and the b are swapped in the definition of them…
topaz solar
gotta be careful about symbols like that and remember sometimes they're only used within that scope
sonic coral
it was just so subtle that i overlooked it. i’m just gonna treat this at <b> since it’s easiest looking at the cayley table
chilly ocean
Is there anything interesting to be said about rings that have composite charact...
They're not integral domains
I think Z/nZ for composite n is a prime example here
smoky cypress
Well yeah
south patrol
A prime example
smoky cypress
But is there anything to say about them
chilly ocean
Maybe we could decompose such a ring if the characteristic is nm with gcd(n, m) = 1
smoky cypress
Maybe like a classification?
chilly ocean
I think R = mR + nR would be a direct sum then?
I'm sleepy but the only thing I'd expect is being able to reduce it to prime powers
Rings with char being a prime power I mean
solar glacier
I think Z/nZ for composite n is a prime example here
Z mod p is in fact a field even and if p is composite then it’s not even an integral domain cause it has zero divisors
Also different copies can be concatenated if the integers are coprime
rapid junco
Let X be a subspace of a vector space Z. Show that there exist linear
operators S and T in L(Z) with Range T = ker S = X.
How would one appraoch this
simple valley
you should be able to complete the table by carefully using all the orthogonalit...
Figured but that shit sucks man
I was hoping for a less painful way
white oxide
for 41 (underlined in blue), would this fail the condition of closure of a subgroup? for take sigma, tao in this set. then sigma(tao(b)) = sigma(c) where c is in B. however, with the given information, we only know that sigma(b) is in B and not sigma(c). is that what they mean by particular? or does that mean for any element in B
in other words, does the set described in 41 contain all permutations such that sigma(x) is in B for all x in B, or just sigma(b) is in B just for one element b in B?
tender wharf
i mean you can check it
white oxide
oh right oops
topaz solar
b -> b’ in B, but b’ -> a in A
white oxide
b -> b’ in B, but b’ -> a in A
yes, but how do we know whether it's in B or in A - B?
topaz solar
well consider S_3 and B = {2, 3}
b = 3
If it fails here then clearly it’s not (necessarily) a group in general
The classic disproof by counterexample
3 -> 2 is fine
white oxide
but that's not in B
sigma(3) = 2 \in B
but we can define tao(2) = 1 which is not in B
topaz solar
Doesn’t matter if 2->1 because the only condition is 3 goes to {2, 3} since the later permutation could send 3->3
gg
white oxide
Doesn’t matter if 2->1 because the only condition is 3 goes to {2, 3} since the ...
oh yeah right
topaz solar
So $\Phi(\sigma)\land\Phi(\tau)$ but $\neg\Phi(\tau\circ\sigma)$
cloud walrusBOT
Sharp
topaz solar
This kind of compatibility with composition/multiplication is what you’re looking for, and similarly with inverses and identity
white oxide
but we can define tao(2) = 1 which is not in B
wait but since tao is in this set doesn't it have to map to some element in B
like
wait
nvm
topaz solar
wait but since tao is in this set doesn't it have to map to some element in B
Tau 3 = 3
That’s all we need
white oxide
But both work independently
How is (12) in this set, because if I'm understanding correctly $(12) = \begin{pmatrix} 1 & 2 & 3 \ 2 & 1 & 3 \end{pmatrix}$ so $\sigma(2) = 1 \notin B$
NVM
topaz solar
But b = 3
white oxide
i'm stupid
ignore that lmfao
How is (12) in this set, because if I'm understanding correctly $(12) = \begin{pmatrix} 1 & 2 & 3 \ 2 & 1 & 3 \end{pmatrix}$ so $\sigma(2) = 1 \notin B$
cloud walrusBOT
okeyokay
topaz solar
\sigma(3)=3
topaz solar
wait why can't we take b = 2
b is arbitrary, so if it fails for one choice it doesn’t hold for the general case asked in the question
b is fixed tho
white oxide
ohhh
okay i was a bit confused by that part
in terms of letting b be one particular element
white oxide
lmao i slightly remember that from last semester
so in this case b is 3
and it's fixed
for all permutations
such that any permutation sigma(3) must be in b
nvm i got it
brain not working
as usual
tranquil plover
Hi guys, I have a question concerning group theory. I am just a beginner in the field, and I would like to know if there are certain games that can be represented using finitely presented residually finite groups. I am doing a small project on games (can be any type of game, like the rubik's cube) and I wanted it to be related to group theory. Any help is very much appreciated!
summer path
not sure if this helps at all, but https://en.wikipedia.org/wiki/Rubik's_Cube_group
i also recall there's something you can do with card shuffles, though I don't quite remember the details anymore
rustic crown
a younger sibling of rubiks cube is the 15-puzzle (or n^2 - 1 puzzle if you wanna generalize). here you have 15 tiles in a 4x4 grid and like rubiks cube it's a permutation puzzle. it can be easily shown that you can solve the puzzle if and only if the given permutation is even 
topaz solar
well I mean a key detail is that you can reverse the moves, so if you can do that you can probably treat it as a group
formal ermine
a younger sibling of rubiks cube is the 15-puzzle (or n^2 - 1 puzzle if you wann...
sgn?
rustic crown
yee
formal ermine
icic
rustic crown
it's quite easy to show you can't reach odd permutations
rustic crown
,av derpZ
cloud walrusBOT
formal ermine
matching pfps with @chilly ocean
rustic crown
tender wharf
formal ermine
how do I show that k[a,b]/sqrt(a-b^2) isn't isomorphic to k[x]
sqrt is radical here btw lol
rustic crown
is it not?
k[x] --> k[a, b]/(a-b^2) given by sending x --> b should be an iso, the right is like k[sqrt(a)]
formal ermine
sqrt is radical here btw lol
.
topaz solar
I mean does radical of (a-b^2) add anything
rustic crown
it doesn't, because prime ideals are radical if k is a field or something
if k has nilpotents, it might be weird
formal ermine
k is algebraically closed here
Oh wait lmaoo
I messed up the polynomial
it's ab - 1
not a - b^2
🤦
rustic crown
only units in k[x] are (non-zero) constants
but that's not the case for k[a,b]/(ab-1)
geometrically, k[x] is a line while the other thing is line without origin
topaz solar
which even with the radical added in there you'd still get ab=1 in the quotient
no idea if the radical should be in there with this new polynomial
not that I'd expect there to be anything extra
rustic crown
yea (ab-1) is also prime lol
topaz solar
I don't get the geometric interpretations though
topaz solar
yea (ab-1) is also prime lol
add the radical to confuse them anyway
rustic crown
k[x] are the ring of functions defined on a line
formal ermine
yea (ab-1) is also prime lol
why is it irreducible?
rustic crown
why is it irreducible?
because it's primitive and linear? think of it as a polynomial in (k[a])[b]
formal ermine
ah I see
formal ermine
geometrically, k[x] is a line while the other thing is line without origin
how do you see this?
rustic crown
okie, so when you remove the origin on the line, the coordinate would always be non-zero
formal ermine
yes
rustic crown
so functions on a line without origin are polynomials in your coordinate x, but you're also allowed to divide by x
so the ring is k[x][1/x]
topaz solar
yeah still not getting the line w/o origin idea but I see a line
formal ermine
or is it wrt zariski topo
because A^1 is a line
I don't think what I'm saying makes any sense lol
rustic crown
okie det was being sloppy
topaz solar
Geometric intuition really does make everything worse frfr 
rustic crown
k[x] is the ring of functions of an affine line
formal ermine
Geometric intuition really does make everything worse frfr <:bleakkekw:913344117...
why is there no algebraic geometry without the geometry
rustic crown
you're not ordering the functions to put them on a line lol
formal ermine
no I don't get it
elder wave
why is there no algebraic geometry without the geometry
If only there was something like that
topaz solar
why is there no algebraic geometry without the geometry
develop a syntactic theory then (there probably already is one tho)
elder wave
(Commutative algebra)
elder wave
Integers
formal ermine
zero set
elder wave
rustic crown
rustic crown
zero set inside what lol
formal ermine
nvm it
formal ermine
if k[x] is a line then I see how k[a, b]/(ab - 1) is a line without the origin
south patrol
Good for you
formal ermine
but I still don't see how k[x] is a line
south patrol
Jk
formal ermine
Good for you
pomeano
formal ermine
(x - a)
no?
ah I think I see it?
so we look at spec k[x]
and THAT is a line (kind of)
cuz prime ideals are x - a
so our a is the "point" on the line
chilly ocean
Is there any particular reason why in the definition of a prime ideal, we require the ring to be commutative?
rustic crown
nope
but iirc you define prime ideals a little differently for non-commutative rings
if a and b are two ideals in your ring, such that ab is contained in p, then either of a or b must be contained in p
chilly ocean
I see
rustic crown
but yea they're more useful in commutative setting
i have never thought about primes in non-commutative rings lol
novel parrot
guys, for part 4. F_2[x]/<f> where f is any degree 3 polynomial, even just X^3 gives a field of 8 elements right
simple valley
guys, for part 4. F_2[x]/<f> where f is any degree 3 polynomial, even just X^3 g...
The ring will be a field iff f is irreducible. But yes it is always 8 elements.
simple valley
Factors of f are zero divisors
novel parrot
How to do this? gcd(a_i) = b => (a_i) = (b)
I only thought this was true in euclidean domain
if and only if
simple valley
Extended euclidean algorithm gives you a linear expression of b in terms of a_i
rustic crown
we in pid
rustic crown
How to do this? gcd(a_i) = b => (a_i) = (b)
do you remember the def of gcd?
simple valley
Oh wait gcd has a different meaning then
novel parrot
Extended euclidean algorithm gives you a linear expression of b in terms of a_i
yes but in pid do we still have this?
i thought that first
simple valley
I'll let det take this over
rustic crown
yes but in pid do we still have this?
yee, not quite... there is a notion of dedekind-hasse valution, but that is "algorithmic" as euclidean division
novel parrot
do you remember the def of gcd?
yes d is gcd of a and b is d | a and d | b and if any other d' does this. d' | d
rustic crown
yee, not quite... there is a notion of dedekind-hasse valution, but that is "alg...
which iirc was, if b is non-zero then you can find p and q such that size of ap-bq is smaller than size of b
but can't force p = 1
rustic crown
yes d is gcd of a and b is d | a and d | b and if any other d' does this. d' | d
yeap
novel parrot
we need to use dedekind hasse ?
rustic crown
nah nah
novel parrot
oh
rustic crown
so we'll verify directly using the def
formal ermine
so we look at spec k[x]
is tis correct
novel parrot
yeah
rustic crown
and if some c divides all a_i then c divides b
rustic crown
then it's more or less done
rustic crown
since pid, the ideal (a1, a2, ..., an) has to be principal
say it is (b')
then b' is the gcd
novel parrot
oh
rustic crown
and uniqueness of gcd implies b and b' are associates
novel parrot
so since gcd(a1, .. an) = b => (a1 .. an) = (b') <= (b)
then we can show b' and b only differ by units
right?
rustic crown
yep
novel parrot
thnx
in showing that Q is not free module
i should say that Q has a basis
so every p/q = r1a1 + .. + r_n a_n ri is integer and ai is rational
i can clear all the denominators of the ai
so now get xp/q = linear combination of only integers
but q may not be cleared basically
elder wave
in showing that Q is not free module
i'm not sure what you're trying here
elder wave
oh
novel parrot
but q may not be cleared basically
so the contradiction is that linear combinations of integers can equal a rational number
elder wave
i'm not sure what you're trying to do tbh
novel parrot
so i want to show that Q is not a free module as a Z-module
So i assume that Q is free, meaning it has a basis
say Q = <a1, .. an>
then every p/q = some Z-linear combination of those ai
uniquely
since ai are rational, we can clear all those denominations of ai by multiplying
say ai = pi / qi
we multiply by q1q2q3..qn
so q1q2..qn p / q = linear combinations of only integers
so q1...qn p / q is an integer
always
but q can be such that q doesnt divide q1..qnp
so remains rational
?
hope this makes sense haha
idk where i went wrong with this
elder wave
i feel like this doesn't work but i can't spot a mistake rn
novel parrot
elder wave
novel parrot
what would you have done?
elder wave
show that Q isn't cyclic as Z module
because for two rationals x,y you always find integers a,b s.t. ax+by=0
so the only possible rank would be 1
hm let me think
novel parrot
say Q = <a1, .. an>
i guess on way its not correct is here
the basis may have infinite lenght right
not finite
elder wave
yeah but your linear combination would still be finite
elder wave
no they show that it doesn't include a certain element
novel parrot
oh
elder wave
they show that it would have to be cyclic
therefore generated by one f
so Q is equal to Zf
but that does not include avg(f,2f) which is rational
kind jacinth
can I get feedback for this proof:
Proposition 5.2 Let σ and τ be two disjoint cycles in SX . Then στ = τ σ
simple valley
what is this even saying?
kind jacinth
was trying to show that if we start from left we would get a1->a2->...->ak and then ak->a1
simple valley
the permutation $\begin{pmatrix}1 & 2 & \dots & k \ a_1 & a_2 & \dots & a_k\end{pmatrix}$ is not the cycle $(a_1a_2\dots a_k)$
cloud walrusBOT
mniip
simple valley
the two-row notation $$\sigma = \begin{pmatrix} x_1 & \dots & x_k \ y_1 & \dots & y_k\end{pmatrix}$$ usually means $\sigma(x_i) = y_i$
cloud walrusBOT
mniip
kind jacinth
ok in that case instead of 1 2 ... k id have them numbered as a1 a2 ... ak
simple valley
if the top and the bottom row are the same that would be an identity permutation
not a cycle
summer path
I think it is probably just easier to write out where each term is sent to explicitly
simple valley
yeah but what was written suggests that there's some fundamental misunderstanding
summer path
I suppose that's also true
kind jacinth
yeah but what was written suggests that there's some fundamental misunderstandin...
i do see what the issue is in having top and bottom numbered the same now that i think about it. a1->a1 ak->ak which would mean they are all fixed
simple valley
I'm also noticing that you said $\sigma = (a_1 \dots a_k)$ but then that $\sigma(b_l) = b_1$?
cloud walrusBOT
mniip
kind jacinth
so it should be sigma(b1)=b2
simple valley
but that's still false
the correct notation (and that's just what this is, a different notation) would be
$(a_1 a_2 \dots a_k) = \begin{pmatrix} a_1 & a_2 & \dots & a_{k-1} & a_k \ a_2 & a_3 & \dots & a_k & a_1\end{pmatrix}$
cloud walrusBOT
mniip
kind jacinth
**mniip**
this would represent the first cycle, and if i want to do the permutation would it be
$(a_1 a_2 \dots a_k b1 \dots b_k) = \begin{pmatrix} a_1 & a2 & \dots & a{k-1} & a_k \ a_2 & a_3 & \dots & a_k & a_1 b_2 \dots & b1 \end{pmatrix}$
cloud walrusBOT
tom310
kind jacinth
no not like that
i mean after ak id add the b1 ... bl
but this doesnt take into account the things that are fixed
simple valley
$(a_1 \dots a_k) \cdot (b_1 \dots b_k) \ne(a_1 \dots a_k b_1 \dots b_k)$
cloud walrusBOT
mniip
simple valley
but yeah as suggested above, all this is really asking is to write out how a cycle acts on elements of X, and verify that the two ways to compose them yield the same action
$\sigma(\tau(x))$ vs $\tau(\sigma(x))$ where you consider cases of whether $x$ appears in the $a_i$ cycle or the $b_i$ cycle or neither
cloud walrusBOT
mniip
kind jacinth
so how id write it in permutation form?
like something like the above using both cycles
simple valley
so how id write it in permutation form?
this is correct yes
kind jacinth
i thought it would be a1 a2 .... ak b1 b2 ... bl
a2 a1 b2 b1
simple valley
sure but that sort of requires the knowledge of what you're trying to prove
tribal moss
You can write either $$\begin{pmatrix}1&2&3&4&5&6\2&4&1&3&6&5\end{pmatrix}$$ or $$\begin{pmatrix}1&2&4&3&5&6\2&4&3&1&6&5\end{pmatrix}.$$
In this notation the order of the columns doesn't matter.
cloud walrusBOT
Troposphere
simple valley
say, how have you used the fact that the cycles are disjoint?
because this certainly isn't true if the cycles aren't disjoint
kind jacinth
say, how have you used the fact that the cycles are disjoint?
because σ and τ are given to me as disjoint
oh wait sorry ur asking me to apply whats given
summer path
I guess tangentially related now: but if the confusion here is regarding how to compute permutations, I think it may be a good idea to just write down some examples of products of (non disjoint) cycles explicitly and just compute them
kind jacinth
**mniip**
ok using some examples i see now why this is the case only for disjoint.
because if i have lambda = (1 2 3) and theta (2, 5) and try to do theta(lambda(2)) = 3 but lambda(theta(2)) = 5
i will try proving this now
is it good now?
the exclamation is that its not always the case (the last element will map to the first element) but this can be solved by using mod
should be a(i+1) mod k
warm crow
For a commutative ring with unity, it is true that every maximal ideal is a prime ideal. but is this true if we remove the unity?
actually this is an exercise from Gallian, I have to find a maximal ideal which is not a prime ideal, given only a commutative ring
I thought it was a typo at first
woven obsidian
4Z in 2Z perhaps?
warm crow
ill try on my own, but thanks for the hint
just wanted to make sure if the question was okay
kind jacinth
is part (b) like part (a) but i use induction?
and my base condition would be what I did in part a?
south patrol
Yes
Exactly
Well
You can start the induction with n=1 but you'll need a) for the inductive step
kind jacinth
alright will give that a try thanks!
kind jacinth
You can start the induction with n=1 but you'll need a) for the inductive step
would it make sense to start at n=1 though? considering lcm requires 2 numbers?
south patrol
Lcm of a single number is just itself
Well or ig you can take n=2
But I see no issue taking n=1
void cosmos
proving Q(sqrt(2),sqrt(3),sqrt(5)) is of degree 5 over Q
is it enough to just consider an element
and keep like recursing untill u find out the basis?
like ik how it works but is that an enough proof?
so Q(sqrt(2),sqrt(3)) (sqrt(5))
so elements of them form a+bsqrt(5) where a is in Q(sqrt(2),sqrt(3)) and so is b
and keep unfolding this
hence we get some linear combination
say thats our basis and thast it?
south patrol
That can't be of degree 5
void cosmos
8
south patrol
I'm not really what exactly what your method is
void cosmos
oh
so i need to prove linear independance aswell?
it doesnt follow from anything XD
?
south patrol
I would personally use Galois theoretic considerations but lol idk if that is what you are allowed to use
void cosmos
im not
south patrol
so i need to prove linear independance aswell?
Uh I mean yes
If it follows from smth then you are still proving it
void cosmos
what about
just saying this field is isomorphic to Q[x]/(x^2-2)(x^2-5)(x^2-3)
or is that even not true
lmao
void cosmos
yea
south patrol
What you've written is isomorphic to the product of Q(sqrt(2)) x etc
void cosmos
this wont be an irreudicible anyways
ig
isnt showing linear independance
like trivial?
south patrol
Linear independence of what
void cosmos
say thats our basis and thast it?
the elements i would get
south patrol
Why should it be trivial on the face of it
void cosmos
have no idea lmao
south patrol
Okay so there are a couple of ways to do this ig
First is uhh
Do you have any bounds on the degree
First try to find an upper bound
Lower bound is easier
void cosmos
ig its either 4 or 8
south patrol
so Q(sqrt(2),sqrt(3)) (sqrt(5))
Well basically try this
south patrol
ig its either 4 or 8
Yes
And so really you just need to check uh
Well what do you need to check
void cosmos
that its not 4
south patrol
Well sure min poly
But equivalently
Just need to check if, say, sqrt(2) is in Q(sqrt(3), sqrt(5))
warm crow
For a characteristic of a ring R, char(R) = additive order of unity if unity is in R. So for the ring Z[i] / <2+2i> = {0 + I, .... , 7 + I}, where I = <2+2i>, isn't the characteristic the additive order of 1 + I, which is 7?
void cosmos
nvm
warm crow
For a characteristic of a ring R, char(R) = additive order of unity if unity is ...
.
<@&286206848099549185>
void cosmos
Just need to check if, say, sqrt(2) is in Q(sqrt(3), sqrt(5))
this is as hard as proving linear independance for me
or is it easy?
coral spindle
Why does it suffice to prove if G is finite ablien it has a cyclic tower?
Because if we have an Abelian tower, we can refine it to a cyclic tower in that case.
we simply insert the necessary subgroups into the tower
kind jacinth
Prove that the order of an element in Sn equals the least common multiple of the lengths of the cycles in its cycle decomposition.
why does the LHS entail the RHS?
formal ermine
does A iso B imply spec A homeo spec B?
coral spindle
Yes
formal ermine
ok so k[x] and k[a,b]/(ab - 1) aren't isomorphic because the latter has a "hole" at the origin?
so it can't be homeomorphic
kind jacinth
Prove that the order of an element in Sn equals the least common multiple of the...
why would it not hold in the case they are not disjoint (i guess i can see how that is true for disjoint case)?
cunning dust
why would it not hold in the case they are not disjoint (i guess i can see how t...
Take for example (1 2) and (1 2) (or in general a cycle and its inverse)
frank cosmos
Because if we have an Abelian tower, we can refine it to a cyclic tower in that ...
why does he add e at the end
frank cosmos
Because if we have an Abelian tower, we can refine it to a cyclic tower in that ...
How?
south patrol
Suppose H is a normal subgroup of G with G/H abelian. Then there's a subgroup L of G/H of order p, which lifts to a subgroup K of G such that [K:H] = p. Then we are done by induction
frank cosmos
lifts?
south patrol
So this is the correspondence theorem basically
I mean that the image of K in G/H is L
frank cosmos
what is correspondence theorem?
south patrol
So basically it states that given a group G and a normal subgroup N, there is a bijection between subgroups of G containing N and subgroups of G/N
frank cosmos
oh ok
frank cosmos
is this the same thing as G/H isomorphic with (G/K)/(H/K)?
south patrol
It has many pleasant properties, like preserving normality etc
No, that is one of the isomorphism theorems
frank cosmos
are you talking about the one that says G/H injects into f(G)/f(H)?
er, f^-1(G)/f^-1(H) injects into G/H
im trying to find this correspondence theorem
frank cosmos
im not sure i get it still
does cyclic refinemet mean adding cyclic groups into the tower or adding subgroups such that the resulting tower is cyclic?
south patrol
The latter
frank cosmos
thats what i thought
south patrol
(Adding subgroups of G such that all the quotients are cyclic groups)
frank cosmos
yep
so lets say I have an abelian tower
G>...>G_n then I know that G/G_1 is ablian and thus has a cylic tower
how do i use this to refine G>...>G_n into something cyclic
south patrol
Okay so we only need to worry about each step
So we may as well just assume you have G > H
Now G/H is abelian and has a tower of prime cyclic groups
By induction on size of the group, say
But now the correspondence theorem says that uhh
frank cosmos
prime?
south patrol
Our tower G/H = K0 > K1 > K2 > ... > 1 that we have with Ki/K(i+1) cyclic of prime order
Yes, we can do that
Like cyclic groups of prime order
frank cosmos
why can we do that
south patrol
Oh I mean like that is by induction on size of group
Uhhh let A be an abelian group
If A is prime cyclic we are done
Otherwise, take an element of order p for some p to form a non trivial subgroup B
Then A/B has order smaller than A, hence a tower with prime cyclic quotients
By inductive hypothesis
frank cosmos
Does Lang skip all this in the proof?
south patrol
But then the correspondence theorem tells us that that gives us a tower A = A1 > A2 > .. > B with prime cyclic quotients
frank cosmos
That i sent earlier
south patrol
Oh I didn't know you sent one earlier lol
frank cosmos
oh ok lol
frank cosmos
here
south patrol
Wait lol lang gives the proof I've been describing
frank cosmos
my issue is with the statement "suffices to prove...tower"
the proof after is pretty clear to me
although I didn't convince mysef using the correspondence theorem i think
Oh wait I guess I did, just didnt call it that
south patrol
Ah okay sure
Okay so the point is like
Let's assume every abelian group has a cyclic tower
And now suppose you have H a normal subgroup of G with G/H abelian
Well then G/H has a cyclic tower (by assumption) and we may write it as
G/H = L0 > L1 > ... > Ln=1
Then by the correspondence theorem, that gives us a cyclic tower uh
G = G0 > G1 > ... > Gn = H
[This bit is actually yes due to the isomorphism of (G/K)/(H/K) with G/H if you look at what the correspondence theorem does]
Do you follow me so far?
frank cosmos
yeah i get all of this
oh shit
i get it now
I take each joint in the tower, take the quotient which is abelian, apply the sufficient statement thingy, correspondence theorem, and boom I have a cyclic tower
😎
i was trying to write this out on my own and failed miserably 
frank cosmos
I take each joint in the tower, take the quotient which is abelian, apply the su...
what was throwing me off was him adding the e at the end
but that was just so we can get canonical^-1(e)=G'
phi sends G' to e in G/X
south patrol
I take each joint in the tower, take the quotient which is abelian, apply the su...
Yup yeah
Using correspondence like this is a super useful ideal in general, by the way
You can sometimes "chop up" a group (or some other object) G into G/H and H (as you hopefully see what I mean in this example) and work with the smaller objects G/H and H
lethal dune
this is as hard as proving linear independance for me
say LD the √2=a√3+b√5 then squaring both sides derive a contradiction
void cosmos
whats LI
lethal dune
LD* linearly dep
south patrol
Hm that isn't quite enough
Since Q(sqrt(3), sqrt(5)) is a 4 dimensional vector space over Q, not 2 as your hint seems to suggest
You need like 1, sqrt3, sqrt 5, sqrt 15 ig
Which is why it gets hard eek
lethal dune
oh I assumed they are trying to show √3, √5 don't generate √2
lethal dune
You need like 1, sqrt3, sqrt 5, sqrt 15 ig
it's harder now but doable with brute force
south patrol
Indeed
This q seems surpsignly annoying lol
But I did one like this a while back and it ended up being ok
I'll try to find it at some point lol
void cosmos
it sucks ass
lethal dune
probably we can show x²-2 is irred over Q(√3, √5)
probably some norm argument to show 2 is prime
next obsidian
…
south patrol
Eh that's equivalent tho right lol
next obsidian
Show it has no root
south patrol
Indeed
lethal dune
I mean how do you know √2 is not in Q(√3, √5)
next obsidian
You write down sqrt(2) = asqrt(3) + bsqrt(5) and then square
south patrol
Lol
lethal dune
lol
void cosmos
im not following you guys at all tbh
south patrol
Cause ryu said that earlier but isn't that wrong
next obsidian
2 = a^2•3 + b^2•5 + absqrt(15)
south patrol
As you need sqrt 15 and 1 as well
next obsidian
What?
south patrol
It's a 4d v space
lethal dune
Hm that isn't quite enough
^
south patrol
Not 2
next obsidian
Oh sure you have sqrt(6) too
void cosmos
in Q(sqrt(2),sqrt(5)) looks like @next obsidian
south patrol
And 1
lethal dune
whole point was to avoid pain
void cosmos
what about some counting argument
next obsidian
Honestly
void cosmos
like use tower theorem
next obsidian
I don’t think it’s pain man like
rustic crown
what we proving
next obsidian
2 is really really small
rustic crown
det had a nice nap
next obsidian
When you square a lot of stuff becomes positive
void cosmos
the problem is to prove it
this is just the problem
so the professor knows its not trivial
next obsidian
So you force like constant^2 = 0
void cosmos
it wouldnt be a problem
south patrol
I reckon just square a relation and use linear independence of the basis
rustic crown
ooh walter
lethal dune
some sort of norm argument to show (2) prime in Q(√2,√3)
south patrol
Well
lethal dune
maybe?
south patrol
The ring of integers is annoying enough to conpute
I mean one can find formulae for it
But then you might as well assume sqrt 2 isn't in Q sqrt 3 sqrt 5
void cosmos
yo why cant we just say this is the same as Q[sqrt(2)+sqrt(3)+sqrt(5)]?
south patrol
Lol
rustic crown
ah, you mean 2O_K >.< nvm 🙈
south patrol
I mean we know it'll be smth like that by primitive element theorem
But how do we know it helps
Lol
Okf
Okay also the problem I did was a special case where o used another trick
Namely showing that 2^1/4 isn't in Q(sqrt 2, sqrt 3)
south patrol
B sqrt 2 + C sqrt 3 + D sqrt 5 where B = 10000, C = 295858 and D = 383838
rustic crown
wait so we proving sqrt(2) is not in Q(sqrt(3), sqrt(5))?
void cosmos
ok guys if i see this problem on the exam i am just going to skip it
rustic crown
ahhh nuu
void cosmos
wait so we proving sqrt(2) is not in Q(sqrt(3), sqrt(5))?
main problem is proving that Q(sqrt(3),sqrt(5),sqrt(2)) is of degree 8 over Q
void cosmos
whats a trace argument
rustic crown
yea
rustic crown
i'll show sqrt(pn) is not in Q(sqrt(p1), ..., sqrt(p_{n-1}))
lethal dune
next obsidian
I’m pretty sure you can show this extension is Galois
rustic crown
yea but i don't think it might help much
void cosmos
I’m pretty sure you can show this extension is Galois
cant use that
summer path
i think at some point they said they can't use galois arguments
rustic crown
Namely showing that 2^1/4 isn't in Q(sqrt 2, sqrt 3)
in this case, it helps a lot
next obsidian
But if you know it’s Galois you have three degree 2 intermediate subfields
summer path
trace
rustic crown
trace
next obsidian
And no order 4 groups has more than 3 subgroups of index 2
So sqrt(2) can’t be in there
void cosmos
is it the one from ANT?
rustic crown
idk lmao
void cosmos
then define it 😠
rustic crown
given a in F, you look at the linear map F --> F given by multiplication by a and take it's trace lol
void cosmos
ok
rustic crown
anyway
lethal dune
how about taking character Z_+: → Q(√2,3,5) by
n → √2ⁿ, √3ⁿ, √5ⁿ, √15ⁿ, 1 and and say they're LI
rustic crown
so say sqrt(p_n) = sum a_S sqrt{p_S} where S varies over subsets of {1, 2, .., n-1}
if you apply T
you get 0 = [F:Q] * a_{empty}
and you can exploit this more
oh btw p_S = prod of p_i for i in S
multiplying by 1/sqrt(p_S) and then applying T gives you 0 = [F:Q] * a_S
so you get all a_S are zero
which gives sqrt(p_n) = 0
oh i did use transitivity of trace
or more specifically
that tr_{F/k}(a) = tr_{k(a)/k}(a) * [F:k(a)]
you can prolly verify this by hand
lethal dune
hard to read so many {}s
lethal dune
how about taking character Z_+: → Q(√2,3,5) by
n → √2ⁿ, √3ⁿ, √5ⁿ, √15ⁿ, 1 and an...
feels like this should work 
maybe no
rustic crown
whats a trace argument
basically the idea is to write it as a linear combi, and trace helps you grab the individual coefficient of a certain term and say that's zero. this because trace of sqrts is 0 while trace of 1 is non-zero
maybe it looks eww when i write it out lol :p
void cosmos
peak math
this is easy but i need to understand the trace
more
ohhh
its the same
yea
super cool
rustic crown
rustic crown
empty set
void cosmos
and let T = tr_{F/k}
.
rustic crown
so say sqrt(p_n) = sum a_S sqrt{p_S} where S varies over subsets of {1, 2, .., n...
i used the subsets to index the sum
oh sure
rustic crown
that tr\_{F/k}(a) = tr\_{k(a)/k}(a) * [F:k(a)]
will you let me use this 🙈
next obsidian
If they can’t use Galois theory I don’t think they can use trace methods
void cosmos
yes but its esaier to build
like i can literally whip it off on the paper
i think
and just define shit
next obsidian
rustic crown
you can bash out some galois theory too, you don't need the whole power anywya
next obsidian
Show that 2 doesn’t ramify in Q(sqrt(3),sqrt(5)) by computing the discriminant KEK:
rustic crown
like by induction show you have an automorphism which conjugates just one of the squareroots