0+0=0
#groups-rings-fields
void cosmos
solar glacier
zero plus itself is 0
high cypress
Y’all are on another level
solar glacier
lol im preparing for an alfebra qual
high cypress
I understand now 🤓
solar glacier
algebra*
winter yew
lol im preparing for an alfebra qual
qual?
solar glacier
phd qualifying examinations
high cypress
WHATTT
solar glacier
once u get in u need to take 2 or 3 depending on ur institution
winter yew
I can’t even get past the first question
First two videos: https://www.youtube.com/playlist?list=PLBY4G2o7DhF0JCgapYKrqibGaJuvV4Gkb
celest cairn
What’s the splitting field for $x^4 - 2$ over $\mathbb{Q}$? Is it $\mathbb{Q}(\sqrt[4]{2}, i \sqrt[4]{2})?$
cloud walrusBOT
Sapphire Gaming
void cosmos
yes
solar glacier
is the correct way of proving xy and yx have the same order as eachother in any group the following
(xy)^k=e_G implies
x(yx)^{k-1}y=e_G
thus $(yx)^{k-1}=x^{-1}y^{-1}=(yx)^{-1}$
cloud walrusBOT
MyMathYourMath
solar glacier
then hitting both sides by a yx we have that yx to the k is e_G
woeful sage
can you have vector spaces over finite fields 
prisma ibex
unique valve
I am a vector space over myself
solar glacier
so anyone who wants to check my 33 abstract solutions let me know 🙂
theyve all been written out just need verifying.
tribal moss
"Just".
solar glacier
lol what? theyre all finished
tribal moss
Verifying homework solutions is time-consuming, tedious, boring, and not particularly rewarding work. There's a reason universities need to pay graders to do it.
solar glacier
question on the proof that if H is normal and K is just a subgroup then HK is a subgroup
so i let $x,y \in HK$ then $x=h_1k_1, y=h_2k_2$
cloud walrusBOT
MyMathYourMath
MyMathYourMath
MyMathYourMath
MyMathYourMath
MyMathYourMath
solar glacier
how do i show this
cloud walrusBOT
MyMathYourMath
solar glacier
Verifying homework solutions is time-consuming, tedious, boring, and not particu...
this is something im doing on my own time in prep for preliminary exams
tawny pine
this is something im doing on my own time in prep for preliminary exams
doesnt have to be hw, checking sols is generally tedium
solar glacier
doesnt have to be hw, checking sols is generally tedium
Who knows maybe someone will find it fun
I would
barren sierra
you think you would
and then you'll realize many parts suck
God how I wish I could force student to typeset and/or be able to remove points for poor handwriting
solar glacier
10$ cool lol and when i get paid which is this tuesday
can someone help me on the above proof
if H is normal and K is a subgroup then HK is a subgroup
i know I need to show xy^-1 \in HK for x\in H, y \in K
solar glacier
**MyMathYourMath**
then this
barren sierra
ok thank god
solar glacier
ok i got this then its easy
barren sierra
I forget the reason why that version isn't necessarily true
cause ofc "I can't show it's true" isn't enough
tawny pine
or is it H subgroup, K normal?
it doesnt matter which one is normal
barren sierra
_> i can't get the proof to work
obsidian sleet
the subgroup test doesnt care which one is inverted
the proof is symmetric u just invert the other guy
barren sierra
oh duh
edgy plume
is a field an abelian group under "multiplication"
barren sierra
0 has no inverse in nontrivial fields
so not quite
the intuition is kind of like that but you're better off not really thinking about fields as just two groups mashed together
edgy plume
noted, thank you!
agile burrow
I'm trying to show if H is a finite index subgroup of G, then G is of type FP_n if and only if H is of type FP_n for 0 <= n <= infinity.
The forward direction is straightforward, as we may regard a (partial) projective resolution for Z over ZG of finite type (a projective resolution where each projective module is finitely generated) as a (partial) projective resolution for Z over ZH. Since H is a finite index subgroup, each projective module is still finitely generated when viewed as a ZH-module.
For the other direction, I want to use an argument along the following lines: start with the augmentation map ZG -> Z, view it as the start of a free resolution over ZH. Then we can extend it to a (partial) free resolution of finite type over ZH. Is there some way I can lift this to a free resolution of finite type over ZG by using induced modules and the fact that H has finite index in G?
formal ermine
the roots of $x^3 - 3 \in \bQ[x]$ are $\sqrt[3]{3}$ and $-\frac{\sqrt[3]{3}}{2} \pm \frac{1}{2}i3^{\frac{5}{6}}$. is the splitting field $\bQ(\sqrt[3]{3}, \sqrt{3}, i)$? cuz like we can get the $3^{\frac{5}{6}}$ with cbrt(3) * sqrt(3)
cloud walrusBOT
rectangle cube
rustic crown
but you can't get sqrt3 :p
the splitting field is Q(cbrt3, i sqrt3)
or said in another way, Q(cbrt 3, omega)
celest cairn
Why does $\mathbb{Q}(\sqrt[4]{2}, \sqrt[4]{2}i) = \mathbb{Q}(\sqrt[4]{2}, i)$?
cloud walrusBOT
Sapphire Gaming
formal ermine
but you can't get sqrt3 :p
ah right. thanks
rustic crown
uwu?
pastel cliff
smoky ivy
im having struggles understanding this bijection at the bottom
i don't see how it actually maps into the fiber of f over P
rustic crown
show both inclusions
like since Q is prime in S_[P], it's inverse image epsilon^-1(Q) will be a prime in S/I which doesn't contain U
now the further inverse image, pi^-1(epsilon^-1(Q)) will not contain any elements outside P (else epsilon^-1(Q) will contain something in U, and it would be sad), so it is contained in P
similarly, since epsilon^-1(Q) is an ideal in S/I
the final inverse image pi^-1(epsilon^-1(Q)) will have to contain P
as elements of P die under that map
rustic crown
deft ferry
how to show that a group of order 4 or less is abelian?
tribal moss
Just wing it for each of the possible orders.
deft ferry
hn
tribal moss
1 is easy. 2 and 3 are primes, and you should already know that groups of prime order are abelian. Then there's only order 4 left.
tribal moss
Do you know they're cyclic?
tribal furnace
proof by exhaustion
tribal moss
Lagrange's theorem shows that.
deft ferry
hn
deft ferry
Do you know they're cyclic?
i just know the definition of cyclic groups
deft ferry
proof by exhaustion
that groups of prime order are abelian?
tribal furnace
that groups of prime order are abelian?
I mean the groups of order 4 case
tribal furnace
For 2 and 3, as time cube has already said, it follows from Lagrange's theorem and the fact that cyclic groups are Abelian
tribal moss
You can also brute-force the possible group operations on a set with 2 or 3 elements. Lagrange's theorem will be handy for the 3 case too.
deft ferry
with 3 elements i suppose g = {e,a,b}, where e is neutral. In this case, b need to be the inverse of a, right?
formal ermine
yes, there's only one group of order 3
formal ermine
(it's isomorphic to Z/Z3 fwiw)
tribal furnace
(it's isomorphic to Z/Z3 fwiw)
Since when do you write Z/3Z? 
deft ferry
with 4 elements
deft ferry
hn, ok
formal ermine
(at least that's how I did it back when I had to do this as a homework)
south patrol
eh you can do it by counting too
okay tbf maybe I am just aassuming you can use lagrange there thoguh, if you can't then a table works ofc
formal ermine
he doesn't know lagrange
south patrol
sure
deft ferry
(at least that's how I did it back when I had to do this as a homework)
i'm trying to apply rules on the four elements so that this implies that the group is abelian
formal ermine
also wouldn't you need prime square order
formal ermine
i'm trying to apply rules on the four elements so that this implies that the gro...
sure
deft ferry
i'm trying to apply rules on the four elements so that this implies that the gro...
in this way, g = {a,b,c,d}, let a be neutral. So, b, c and d needs to have inverse
formal ermine
something you might instantly see:
we know there are 2 groups of order 4, one of them being cyclic. you therefore only have to find the other one (which is not cyclic) and verify that it is also abelian
south patrol
(in fact, if every element in a group G has order 2, then G is abelian and if G is also finite then G is isomorphic to (Z/2Z)^n for some n >= 1)
tribal moss
(vacuously because there's always an element of order 1 != 2)
next obsidian
(in fact, if every element in a group G has order 2, then G is abelian and if G ...
this is true even for not finite
It becomes an F_2 vector space
south patrol
Indeed ye I realised after
next obsidian
And so you just replace n with some cardinal
south patrol
Since the way I'd prove it is to use vector spaces anywa6
tribal moss
Counterexample: the group of finite or cofinite subsets of N under symmetric difference is not (Z/Z2)^n for any n.
chilly radish
For finite you can appeal to classification
Crushing a cockroach with a sledgehammer
deft ferry
and i get other question
next obsidian
Counterexample: the group of finite or cofinite subsets of N under symmetric dif...
This is infinite tho, in which case it’s (Z/2Z)^(+)alpha for some cardinal alpha
deft ferry
how to prove that there is an n positive interger that a^n = e?
next obsidian
Is your group finite?
rustic crown
next obsidian
There’s some i,j such that a^i=a^j and i ≠ j
Or you get infinitely many elements
Now take a difference of i and j
formal ermine
how to prove that there is an n positive interger that a^n = e?
a different proof: let x be an element out of your group (x in G). let k be its order (k = |x|). by lagrange kl = |G| for some integer l, so you have x^|G| = x^(kl) = (x^k)^l = 1^l = 1
is $\bZ_2(2) = \bZ_3$?
cloud walrusBOT
rectangle cube
next obsidian
Literally wtf is this notation
solar glacier
whats $\Bbb{Z}_2(2)$? @formal ermine
cloud walrusBOT
MyMathYourMath
next obsidian
Bro wtf is YOUR notation. \Bbb{Z}?
formal ermine
,, \left(\bZ/2\bZ\right)(2)
cloud walrusBOT
rectangle cube
next obsidian
Literally what is that
elder wave
formal ermine
is the field extension $\bZ_3/\bZ_2$ simple
cloud walrusBOT
rectangle cube
next obsidian
That’s not a field extension!
chilly radish
Characteristic...
formal ermine
but $\bZ_2 \subset \bZ_3$
cloud walrusBOT
rectangle cube
chilly radish
Who is the order 2 element in Z/3?
next obsidian
This isn’t even a subgroup inclusion
tribal moss
a different proof: let x be an element out of your group (x in G). let k be its ...
Huh? The task was to prove that every element has a finite order, which you seem to be assuming as known when you say "... let k be its order".
solar glacier
lol
south patrol
What is $Z_2$ and what is $Z_3$ in ur notation
cloud walrusBOT
potato
formal ermine
What is $Z_2$ and what is $Z_3$ in ur notation
Z/2Z
solar glacier
z2 like @next obsidian said isnt a subgroup of z3
south patrol
okay maybe you are just going off of like {0,1} being contained in {0,1,2} or something
and constructing them that way which is leading to confusion
chilly radish
Think about the cardinality a field extension of a finite field would have to have. Think of vector spaces as someone above said
formal ermine
Huh? The task was to prove that every element _has_ a finite order, which you se...
with order here I mean the order of the cyclic subgroup generated by x, which is obviously finite as the group itself is finite
next obsidian
Or even just Lagrange to see that Z/2Z can’t be a subgroup of Z/3Z
south patrol
Usually one does not think of them as such and if you do then this is kinda besides the point as these inclusions aren't subgroup inclusions
next obsidian
Because 2 doesn’t divide 3
formal ermine
z2 like <@146419489083228160> said isnt a subgroup of z3
this is field theory
south patrol
oof
formal ermine
huh
formal ermine
**rectangle cube**
why is this false
solar glacier
as a subgroup?
formal ermine
as fields
solar glacier
byt LaGranges theorem the order of Z2 would need to divide the order of Z3
its not even a subgroup let alone a subfield
chilly radish
with order here I mean the order of the cyclic subgroup generated by x, which is...
Saying the cyclic subgroup generated is finite is equivalent to a priori assuming the order is finite...
At this level you can't assume that kind of thing
formal ermine
he assumed that the group itself is finite, and a subgroup of a finite group must always be finite, by definition
solar glacier
the only subgroups of Z3 are the trivial and Z3 itself
next obsidian
You’re the one that asked why it isn’t
formal ermine
K/F if F is a subset of K
chilly radish
....
next obsidian
You’re fucking kidding me
rustic crown
.<
solar glacier
You’re fucking kidding me
lets at least try and keep the language G rated lol
chilly radish
he assumed that the group itself is finite, and a subgroup of a finite group mus...
One would, in this case, have to prove that the cyclic subgroup generated by an element is something that is even well defined
In any case lagrange is too strong here
formal ermine
fair
chilly radish
If they are tasked with proving every element of a finite group has finite order, they don't know lagrange's theorem
formal ermine
sorry for annoying you guys with this question
next obsidian
> K/F if F is a subset of K
This definition doesn’t even say anything about the field structure
There’s no way this is your definition, you’re misinterpreting a definition
formal ermine
Definition 4.1.1. A field extension L/K is an inclusion K subseteq L of two fields
that's what's written in my lecture script
formal ermine
wdym
chilly radish
K is a subfield of L
coral spindle
No, it's not.
chilly ocean
You got to show it's minimal such positive integer
south patrol
eh a few words make it fine though
||o(yx) <= o(xy) by this and then "by symmetry"||
xd
south patrol
This isn't enough, because all you showed is that the order of yx is less than k
south patrol
I've not actually read the proof lol
It seems fine other than that you seem to be assuming k = 4 notationally
solar glacier
I forgot the ….
coral spindle
a^k = 1 does not mean that the order of a is k...
south patrol
This isn't enough, because all you showed is that the order of yx is less than k
ye this is what i mean
But yeah after correcting it to this, you're like a few words away
solar glacier
So I’ve shown order of yx is at most k
chilly ocean
First equality is also bad
solar glacier
I meant to include a …. In the parenthesis
so I assume there is some l<k such that the order of yx is l?
And find a contradiction ?
A little lost here
chilly ocean
You proved that ord(xy) <= ord(yx)
solar glacier
oh now do the other way
chilly ocean
That is, assuming ord(yx) is finite
solar glacier
im assuming its finite order
chilly ocean
solar glacier
sorry but for the other direction i expand (yx)^k
and deduce that
$(xy)^{k-1}=y^{-1}x^{-1}$
cloud walrusBOT
MyMathYourMath
solar glacier
i e. xy has at most order of yx
cloud walrusBOT
MyMathYourMath
solar glacier
since $yxyxyx...yx=y(xyxyxy...xy)x$ $k$ times
cloud walrusBOT
MyMathYourMath
stoic rose
does this suffice for a proof that the orders of xy and yx are the same
Another way to see that is to note that xy and yx are conjugate.
And conjugation is a group automorphism, hence preserves order.
solar glacier
yeah but im trying to do it the classical direct way
deft ferry
how to prove that any subgroup of a cyclic group is cyclic?
i guess that any element of this subgroup has form a^n, so from definition, that cyclic
but i'm not sure
chilly ocean
"G is cyclic" means that there is a g in G such that every element of G is of the form g^n for some n
you are given such a G and a subgroup H
solar glacier
how to prove that any subgroup of a cyclic group is cyclic?
thats a big proof
chilly ocean
you need to find an h in H such that every element of H is of the form h^n for some n
solar glacier
it uses the division algorithm no?
next obsidian
thats a big proof
Not really
It follows by some very elementary element counting and the most basic of NT
chilly ocean
i encourage you to try a few examples of cyclic groups and subgroups out before doing a proof
solar glacier
does this proof suffice for that
chilly ocean
solar glacier
i typesetted this myself
chilly ocean
come on
smoky ivy
it uses the division algorithm no?
yeah, quite easy with that
deft ferry
to show that a Group G that has no subgroup of its own is cyclic, I can show by saying that if the belongs to G, any power of G must belong to G and that therefore the set of powers of a is a cyclic subgroup of G, and as by hypothesis G has no subgroup of its own, then the G itself is cyclic?
next obsidian
to show that a Group G that has no subgroup of its own is cyclic, I can show by ...
Yeah this works
Notably: you have to allow {e} and G as subgroups, so you mean no non-trivial subgroups
And then you take a ≠ e
(Or I guess e if the group is trivial lol)
solar glacier
i get it with this
You mean with the division alg right
south patrol
not sure what you mean by "subgroup of its own"
deft ferry
not sure what you mean by "subgroup of its own"
proper subgroups
sorry about my english
south patrol
sure dw
deft ferry
You mean with the division alg right
ye, k = mq + r
deft ferry
and m is your smallest integer such that x^m \in H
yes
deft ferry
where k is your...?
k is any interger that x^k \in H
solar glacier
and you claim H equals the subgroup generasted by x^m
solar glacier
k is any interger that x^k \in H
yes
south patrol
tbf you don't really need much of a division algorithm
well
i guess you do but it's simple
deft ferry
bros
how can i prove that if G has no proper subgroups, G is cyclic and of order p prime
that is cyclic i can prove
but
with order prime i cant
some hint?
south patrol
You'd wanna say proper, non-trivial subgroups
Okay, so uh
Do you know anything about the subgroups of Z/nZ?
Say Z/6Z, what are the subgroups
deft ferry
Do you know anything about the subgroups of Z/nZ?
hm, no
next obsidian
You'd wanna say proper, non-trivial subgroups
proper
non-trivial
Cmon man, G is a trivial subgroup
south patrol
Fair
south patrol
Well I would usually take it to mean {e} but ig that's more if they just say trivial group rather than subgroup
In any case it's clear from context ig
next obsidian
Yah
south patrol
owo
agile burrow
Mfw when the field extension is normal and separable
Idk I never learned that stuff
agile burrow
Better call Galois
south patrol
lol
Yeah so uh
my course defines finite galois as spl fld of a sep poly
which is more based
bruh
agile burrow
Hm
Oh I guess if you have finite extension then you can view it as splitting field of like the product of minimal polynomial of generators or something
Something along those lines
thorn delta
mine defined galois as its fixed field being the base field 
next obsidian
Oh I guess if you have finite extension then you can view it as splitting field ...
Apply primitive element
To get a generator for which the ext is the splitting field of, poly is separable
Hmmm
agile burrow
primitive element theorem is epic
next obsidian
Actually now thinking
Okay yeah you have to use extra thm
Or no uhh
Okay idk I think no work
agile burrow
sad
next obsidian
Or maybe it does
agile burrow
happy
next obsidian
Chmonkey
Okay I think it does
So L contains all conjugates
Of alpha
Say L = K(alpha)
This cuz normal
We know that m_alpha is separable cuz separable extension
Then we just need to know dat L = splitting field
So we need to know it minimal
But it has degree deg(alpha)
So it minimal
Maybe that works?
next obsidian
With the sauce!
next obsidian
You know what’s cool
In char 0
That actually for deep reasons makes a lot of AG in char 0 tick
That u can take Galois closure
agile burrow
L
agile burrow
next obsidian
Walter
agile burrow
yes but i am trying to learn more group coho
south patrol
Char 4 is cool
south patrol
Char 8 extensions are more interesting
next obsidian
Go play some fortnite
agile burrow
too late, i grew up
south patrol
There is a bijection between Char 8 field extensions and abelian non-solvable groups
agile burrow
char 1738
south patrol
which is a rich area of study
delicate bloom
wth is a characteristic 8 field
chilly ocean
a field where 1 added to itself 8 times is zero, duh
agile burrow
common examples include
chilly ocean
F_un
next obsidian
common examples include
and
barren sierra
8 is prime for prime values of 8
Therefore there are fields of characteristic 8
qed
agile burrow
Let G be a group, N a perfect normal subgroup. If we have a partial projective resolution P_2 -> P_1 -> P_0 -> Z -> 0 over ZG where each P_i is finitely generated and projective over ZG, I can see that applying (-)_N (taking coinvariants under N) preserves exactness. However, I don't see why (P_i)_N is projective over ZN.
Hmm, I don't think it's true that taking coinvariants preserves being projective. However, if the partial projective resolution is in particular free (which I can do), then I can view each P_i as a free ZN-module. Then the coinvariants are free abelian groups of the same rank as the free ZN-modules
next obsidian
Have u tried Chmonkey theorem?
agile burrow
realized I was being dumb, I'm constructing a projective resolution of Z over Z[G/N]
next obsidian
Maybe Chmonkey theorem was the right thing to do actually
agile burrow
in which case taking coinvariants of free ZG-modules under N literally gives me free Z[G/N]-modules
thanks chmonkey
agile burrow
I'm reading that it's unknown if there exists a group of type FP which is not of type FL. This is equivalent to asking if there exists a group G of type FP such that the integral group ring ZG has a finitely generated projective module which is not stably free. Interestingly, there are examples if you drop the FP condition, one being the integral group ring over Z/23Z. I find it neat because there are a number of results which we can prove for groups of type FP that we can't prove for groups of type FL.
I'm reading all of these from a book published in 1983 and I know some of the conjectures stated in this book have been solved by now. Does anyone know if there has been interesting progress regarding this question? Feel free to ping me if you have an answer or just want to talk about the problem lol
next obsidian
I'm reading that it's unknown if there exists a group of type FP which is not of...
What is FP and FL
agile burrow
FP means Z admits a finite length projective resolution over ZG where each projective module is finitely generated
FL means Z admits a finite length free resolution over ZG where each free module is finitely generated
agile burrow
L for libre
next obsidian
Definitely
next obsidian
P for projectif
next obsidian
I'm reading that it's unknown if there exists a group of type FP which is not of...
Wait wtf
This seems…
Like wtf
I mean I guess I don’t know counterezamples to this for just resolutions over whatever ring
But like that’s kinda kooky
agile burrow
Yeah, supposedly the answer is "certainly yes" if you work over arbitrary ring
Which I'll think about more when I'm not sleepy
But the equivalence is kinda neat, the idea being that if G is FP, then G has finite cohomological dimension n (meaning cohomology vanishes in degree greater than n). Then you can construct a partial free resolution of length n - 1, but you can't guarantee that the n-th term is free, only that it's projective.
If P is stably free then you can do the Eilenberg trick or whatever it's called to replace it with a free module, proving the group is FL
rustic crown
good luck uwu
formal ermine
ok so two questions:
- what's the connection between purely inseparable (idk the translation) and separable. they seem so arbitrary but yet are named similarly
- why do we need galois extensions to be separable? like, what's the background behind this?
tbh the second question might not be the smartest to ask right now
cuz all we've done in lecture was show that Aut_K(L) = Hom_K(L, K bar) and define Gal(L/K) := Aut_K(L)
oh also another question
- is there an easy way to finding the galois group of an extension
formal ermine
let L/K be an extension and char(K) = p > 0. a in L is called purely inseparable if a^(p^m) in K for some m
could be translating it badly
coral shale
ok im not too sure about this - but moving to the next
So say your irreducible polynomial has a squared factor. Eg.
f = g^2 h where g, h are linears
Now when you take K[x] / f, you still get a field right
as f is irreducible
formal ermine
yes
coral shale
but now you cannot distinguish between 2 of the roots
i edited - g, h gotta be linear
so say (x-a)^2(x-b)
If it was separable
(x-a1)(x-a2)(x-b)
You would have extra automorphisms by mapping a1 to a2 and viceversa
but setting a1 = a2, you've lost these
formal ermine
we haven't connected roots to the galois group or w/e yet
coral shale
so the galois group gets smaller so number of possibly homomorphisms decrease
formal ermine
I think it's too early for me to ask this question
coral shale
mm
formal ermine
but thanks anyway!
coral shale
hmmm
formal ermine
I'll come back to this sooner or later
formal ermine
- is there an easy way to finding the galois group of an extension
what about this
rustic crown
nope, finding it is usually hard
you can do it for small degree polynomials quite easily
but in general, it is probably undecidable lol idk tho
and about your first question
purely inseparable extensions are like the literal opposite of separable ones
for separable extension you know that an irreducible polynomial will not have any repeated roots when you go into some bigger field
but for purely inseparable, all the roots clump together into a single thing
formal ermine
but for purely inseparable, all the roots clump together into a single thing
they do?!
rustic crown
yep
pick any element alpha
it's minimal polynomial is like x^(p^n) - a
so this factors like (x-alpha)^p^n
coral shale
oh so purely inseparable means galois group is trivial
coral shale
uh, ok
rustic crown
you phrase that using separable degree
given a field extension F/k, one defines [F:k]_s = number of extensions of k --> k bar to F
this is also multiplicative like the usual degree
and a finite extension is separable if and only if [F:k]=[F:k]_s
formal ermine
it's minimal polynomial is like x^(p^n) - a
might be a stupid question, but why?
rustic crown
oh because alpha will satisfy alpha^p^n = a in k for some n
so alpha is a root of some x^p^n - a
formal ermine
ahhhhhhhh
rustic crown
and since this factors like (x-alpha)^p^n, then if you choose n to be minimal such that alpha^p^n lies in k, then the polynomial will be irreducible in k[x]
formal ermine
why does it factor like that?
rustic crown
hehe :p
this is the reason why inseparability only happens with positive char
there is no such thing when char is 0
formal ermine
negative char exists?
rustic crown
lol i just mean 0 vs positive :p
formal ermine
ah
rustic crown
anyway, one last piece of information :p
formal ermine
~~what if you're french 
~~
rustic crown
given any field extension F/k, you can define the subset F^sep of F given by elements which are actually separable over k, i.e. F^sep = {alpha in F : alpha satisfies some polynomial with no repeated roots}
then F^sep is automatically a field extension of k
formal ermine
when do we use lowercase letters for field
rustic crown
upto you
formal ermine
given any field extension F/k, you can define the subset F^sep of F given by ele...
yes we covered this
rustic crown
yea so the separable degree i defined above is then just [F:k]_s = [F^sep : k]
formal ermine
yip
rustic crown
and in particular htis means separable degree divides the actual degree, so one can also define an inseprable degree to be [F:k]_i = [F:k]/[F:k]_s
rustic crown
formal ermine
ok so for the galois group
rustic crown
yee?
formal ermine
I have the field extension Q(cbrt(3), isqrt(3))/Q
that is the splitting field of x^3 - 3
rustic crown
okie
rustic crown
okie, so first lets confirm that this is a galois extension
formal ermine
yes I already did that
rustic crown
which is clear in this case as you gave it as a splitting field of a separable polynomial
so now you wanna find the group of automorphisms which fix Q
rustic crown
one nice thing about using generators is that you only need to keep track of where these generators go, the whole automrophism will be determined by them
so say f is an automorphism fixing Q, what can you say about f(cbrt 3)?
formal ermine
(acbrt(3)+bisqrt(3))/(ccbrt(3)+disqrt(3)) ?
or wait lol
I'm overcomplicating this
rustic crown
yep
:p
use that f is an automorphism
the image of cbrt 3 isn't an arbitrary element!
it has to satisfy a really nice polynomial
formal ermine
the minimal polynomial of cbrt3?
rustic crown
yee exactly
formal ermine
why
rustic crown
since (cbrt3)^3 - 3 = 0
if you apply f to this and use that it is a map of fields and also fixes Q, you see
f(cbrt3)^3 -3 = 0
this is common slogan "roots go to roots"
rustic crown
and if moreover the extension is simple, "a root can go to any other root!"
(but use the letter n for exponents >.<)
right so image of cbrt 3 is another root of x^3-3
so has only 3 options
rustic crown
similarly image of isqrt(3) has to be a root of x^2+3
and you have 2 such options
so this actually gives you an upperbound for the number of such f
since f is determined by where it sends cbrt3 and isqrt3
there are at most 6 such things
for that you need to be a little careful
formal ermine
I misread what you said
rustic crown
it is true, but one needs to verify that you can make both these 3 choices and 2 choices independently
so instead of doing it directly, it's nicer to just use some theory
for example, if p is an irreducible separable poylnomial over k, then the galois group of the splitting field field of p would be a transitive subgroup of S_{deg p}
this is because, just like we proved "roots go to roots", each automorphism defines an injective function on the set of roots
formal ermine
transitive subgroup?
rustic crown
yee, coming to that >.<
rustic crown
this is because, just like we proved "roots go to roots", each automorphism defi...
this is automatically a bijection because of cardinality reasons
oblique river
@spice whale yes
rustic crown
so it shows that it's a subgroup of permutation group on the roots
formal ermine
hi buncho
spice whale
ok thanku buncho
oblique river
hi
rustic crown
hewwo
oblique river
i was answering a question alison asked in another channel
spice whale
oblique river
hi det
spice whale
I'm doing some rep theory exercises
rustic crown
so it shows that it's a subgroup of permutation group on the roots
and yea, a transitive subgroup of S_n is a subgroup such that for any 1 <= i, j <= n, there is a permutation in the subgroup which sends i to j
spice whale
from artin
oblique river
(alison, i think this is only true over char 0 though)
spice whale
(alison, i think this is only true over char 0 though)
he's only doing complex reps so this is fine
rustic crown
and yea, a transitive subgroup of S_n is a subgroup such that for any 1 <= i, j ...
and this is easy to see in our case, given any two roots of p, say alpha and beta, there is map k(alpha) --> k(beta) which fixes k and sends alpha to beta. now you just extend this to an automorphism of the whole extension
rustic crown
ig you can replace this with some concrete calculations
but usually people would wanna avoid them :p
rustic crown
and this is easy to see in our case, given any two roots of p, say alpha and bet...
i'll just end this lol.
so in our case, we'll have a transitive subgroup of S_3, and since complex conjugation is also an automorphism, by some simple group theory you see it's all of S_3
okie so for the more computational approach
formal ermine
is the galois group always a subgroup of a symmetric group
rustic crown
first i would say it's easier to pick a different set of generators
coral shale
is the galois group always a subgroup of a symmetric group
all groups are
rustic crown
use t = real cbrt3, w = primitive cbrt of 1
formal ermine
is the galois group always a symmetric group
rustic crown
so you want Gal(Q(t, w)/Q)
so before i began this stuff about transitive subgroup and all that
we proved that roots go to roots
formal ermine
why primitive cbrt of 1
rustic crown
i don't want you to pick cbrt 1 = 1 lol
coral shale
is the galois group always a symmetric group
its fairly easy to construct an example for C2 x C2
rustic crown
like w = (-1 + sqrt(-3))/2 or its square
formal ermine
but don't we have Q(cbrt 3, isqrt 3)
rustic crown
what's the example of C_n in your mind?
coral shale
idk 
rustic crown
but don't we have Q(cbrt 3, isqrt 3)
they're the same fields lol
coral shale
ok, C_n for some n?
formal ermine
they're the same fields lol
huh
coral shale
looking at C_p
formal ermine
ah
rustic crown
looking at C_p
the standard example is the cyclotomic extension Q(mu_n)/Q which has galois group U(n) = (Z/nZ)*
rustic crown
ah
like the only benefit of this is that you can write your automorphisms explicitly
f(t) = t * w^a and f(w) = w^b
here a varies in {0, 1, 2} and b in {1, 2}
because t * w^a are roots of x^3-3 and w, w^2 are roots of minimal polynomial of w, which is x^2+x+1=0
coral shale
is the galois group always a symmetric group
anyways, so i think this relates to the inverse galois problem, and its thought that all finite groups are galois groups
rustic crown
now you would need to explicitly check that these are indeed field maps
rustic crown
anyways, so i think this relates to the inverse galois problem, and its thought ...
(finite)
it's false that all groups are galois groups
even if the base is not Q
because galois groups are profinite
formal ermine
anyways, so i think this relates to the inverse galois problem, and its thought ...
it has not been proven yet?
rustic crown
and so they naturally carry a topology, which is compact, hausdorff and totally disconnected
coral shale
no
rustic crown
but that means you could put a finite haar measure on it, which further tells you that it can't be countable
so galois groups are never countable :3
coral shale
ic
rustic crown
it has not been proven yet?
it's known for a huge class of groups ig
like all finite abelian groups can are galois groups over Q
S_n is also possible
A_n too then
solvable i guess are also proven
idk more tho
rustic crown
now you would need to explicitly check that these are indeed field maps
anyway, once you do this, it's easy to see that galois group is non-abelian and has order 6, so S_3
rustic crown
spice whale
not sure how to use frobenius reciprocity in the proof of (b)
I've got it from the definition of induced rep
south patrol
i imagine the point is that you can use orthogonality to decompose the representation
oblique river
@spice whale I think the idea is to show that the characters of ind(res R) and R + R' are the same. write them out as class functions, and order the conjugacy classes so that those which make up H come first (just for simplicity)
you want to show that each entry of these vectors is the same
next obsidian
Omg Buncho
oblique river
so take the inner product with e_i
and for ind(res R) use frobenius reciprocity and for the other side, just write out the character of R + R' based on the character of R
and if you sort that out I think that should work?
hi chm
how's it going?
oblique river
are you ready for them?
next obsidian
But neither are any of my classmates
oblique river
ez
next obsidian
It’ll be gg ez
oblique river
haha
next obsidian
I proved a neat thing
oblique river
oh yeah?
next obsidian
If you have a domain where every prime is principal, it’s a PID
This isn’t that hard actually, fg primes => Noetherian, Principal Ideal Theorem => dim 1, then it’s a Dedekind domain
All ideals are products of prime then you’re 
The cool part is you can drop “domain”
oblique river
huh
next obsidian
And the proof is geometric, you take the irr components of Spec A and then you argue they’re connected components, basically because they can’t intersect or else the point of intersection has too large a tangent space
It has to be dim > 1, but the (co)tangent space is like m/m^2
So that contradicts m being principal
So all the irr components are disjoint so they’re connected components, so actually A breaks apart as the product of its irr components
Then you can show it for them separately
It’s pretty sweet
agile burrow
Good job chmonkey
next obsidian
Thx Walter
agile burrow
And good luck on finals
agile burrow
Give me like 3 semesters
next obsidian
Ok
next obsidian
It’s nice!
Also it made me like
Think back to this result I proved before
If A is a Noetherian ring such that A_p is an integral domain for all p, then it’s the product of A/pi over the minimal primes
I did it using a lot of algebra before, but the geometric proof is way easy
Again you just want to show the irr components are disjoint so they’re the connected components
But if irr components intersect at p, then A_p isn’t an integral domain, because it has more than one minimal prime
So you’re just done
spice whale
<@279884747017289728> I think the idea is to show that the characters of ind(res...
i see
oblique river
geometry so ezpz
next obsidian
Literally
spice whale
this is so much more work than just using definition of induced rep
oblique river
I don't think that the "using frobenius reciprocity" remark was meant as a hint to get you started
i think the purpose was as a directive to force you to practice using that theorem
because it's true in much more generality than what you proved in the previous problem
spice whale
the thing is like
part c requires use of frobenius reciprocity
oblique river
maybe part b was meant to be a warm-up for that then?
spice whale
or at least is made much easier with it
spice whale
maybe part b was meant to be a warm-up for that then?
you use part B's result
that's about it
next obsidian
Occasionally ppl also write down some nonsense if all other theories as to why they wrote something down fail
I remember there was some problem in my analysis class where literally everyone went “that hint was useless, I just did X”
oblique river
i mean i think the purpose of exercises isn't just to compile a list of true facts
it's to give you a little sandbox to try to understand how something works
formal ermine
how do I find a normal extension that isn't separable
chilly radish
You're gonna need to start with a non perfect field
So no finite fields and no char 0 fields
formal ermine
You're gonna need to start with a non perfect field
what is a perfect field
formal ermine
perfekter Körper
don't know what that means either
formal ermine
google search results are personalized and most of the time when I google stuff for aa I just get nonsensical results
grizzled narwhal
formal ermine
chilly radish
Protip right there
formal ermine
oh it's called vollkommener körper in german
next obsidian
Anyway, I think F_p(t) over F_p(t^1/p) works?
formal ermine
at least that's how my prof calls it
next obsidian
This is not separable, it’s the canonical example of one
And I’m pretty sure this expresses F_p(t) as the splitting field of t [whoops I meant of F_p(t^1/p) over F_p(t)]
next obsidian
No this is backwards
formal ermine
lemme try showing that it's separable
formal ermine
No this is backwards
uhhh
next obsidian
It’s F2(t) over F2(sqrt(t))
Or wait
Think
You’re right
I’m being crazy lmfao
So sure, normal because degree 2
And it’s not separable
Namely, what’s the min poly of sqrt(t)?
Then find the two roots of that
formal ermine
X^2 - t
next obsidian
Right
formal ermine
sqrt(t) and -sqrt(t)
next obsidian
But in char 2 this is the same as?
formal ermine
(X - sqrt(t))^2?
formal ermine
I don't quite understand why -sqrt(t) isn't a root, doesn't it fulfill X^2 - t?
chilly radish
(The important characterisation of perfect fields btw is either char 0 or char p + every element is a p-th root. In particular finite fields are perfect)
next obsidian
I don't quite understand why -sqrt(t) isn't a root, doesn't it fulfill X^2 - t?
This is the same as sqrt(t)
chilly radish
The field is char 2
next obsidian
Because -1 = 1
next obsidian
This is the canonical example of non-separable extensions btw
You should memorize this example
formal ermine
I will
it's so annoying
my course hasn't introduced the F_n notation and it feels weird to use it
rustic crown
F_p >.<
south patrol
owo
formal ermine
Z/n is cursed
chilly ocean
H\G/K
rustic crown
I also like to write Z/Zn or nZ\Z
.<
barren sierra
Z/nZ for additive, Z_n for multiplicative
case closed
I won't take any questions
chilly ocean
Z_n for integers modulo n
next obsidian
Is Z_p = Z/pZ, the p-adics, or S^-1Z where S is {1,p,p^2,…}
And if you’re dumb, Z_(p)
rustic crown
lol
chilly ocean
Why not Z_nZ
agile burrow
Hi det
coral shale
A3 C3 Z_3 Z/3Z Z/(3) 
tribal furnace
I saw the S3
coral shale
what S3
tribal furnace
The group of order 6 that definitely is not isomorphic to F_3
coral shale
lie
tribal furnace
Also not a group of Lie type or a Lie group
next obsidian
Chmonkey
agile burrow
Commutator subgroup of GL_2(F_2)
tribal furnace
50% monkey, 49% chair, 1% 
rustic crown
chain complexes in monkey 🙈
agile burrow
Computing chmology
tribal furnace
Also not a group of Lie type or a Lie group
wait, it might be a group of Lie type, idk group theory
A_1(2) = SL(2,2) = S3 says Wikipedia
coral shale
{z/n: z in Z}
south patrol
that's why i said basically always lol
i think if it's in an algebraic context it's more clear it means Z/nZ
chilly ocean
$\mathbb{Z}_{n\mathbb{Z}\ \mathbb{Z}}$
cloud walrusBOT
Blitz
coral shale
barren sierra
$n_\mathbb{Z}$
cloud walrusBOT
Spamakin🎷
chilly ocean
Perfect!
coral shale
the hell
cloud walrusBOT
Spamakin🎷
quaint minnow
can i ask a question on field theory here?
next obsidian
quaint minnow
ah aiight
next obsidian
I scared them off 
quaint minnow
for 2 algebraic numbers a and b which is true
cloud walrusBOT
Frru1ty
next obsidian
I don’t think this requires anything about algebraicity
quaint minnow
huh
next obsidian
But you can get every one, the first is guaranteed
quaint minnow
cool
delicate bloom
for 2 algebraic numbers a and b which is true
Easy example of case 2: a=sqrt(2)+1 and b=-sqrt(2)
shell agate
and for an easy example of case 3 you can take a=b=0.
formal ermine
Let $K$ be a field and $\operatorname{char}(K) = p > 0$. Let $f \in K[X]$ be an irreducible polynomial. Show that $f$ is not separable iff there exists a $g \in K[X]$ with $f = g(x^p)$.
cloud walrusBOT
rectangle cube
formal ermine
I can easily show the $\impliedby$ direction by using the fact that $f$ separable $\iff f' \neq 0$
cloud walrusBOT
rectangle cube
elder wave
if f is not seperable then it's formal derivative is 0, that means that every monomial in f is of the form a_r * X^(rp)
formal ermine
if f is not seperable then it's formal derivative is 0, that means that every mo...
"monomial in f"?
elder wave
as in components of the sum that make up the polynomial
this describes how each summand must look like for the derivative to be 0
elder wave
yw
formal ermine
is there any simple criterion for when an extension is cyclic
rotund aurora
In general no I think
But you have things like this
See Kummer theory and Artin-Schreier theory
A main problem I think is that F<K cyclic and K<E cyclic does not imply F<E cyclic
This is what Clark says:
@formal ermine
tribal niche
We've defined the free group F(X) of a set X to be the set of words in the symbols x, x^{-1}, as x ranges elements of X
alongside the universal property on X: let G be a group and φ : X → G be a set map. then (G, φ) has the property if for any other pair (G', φ'), there is a unique group homomorphism f : G → G' such that fφ = φ'.
How would I prove the following?
If (G, φ) has the universal property on X, then G is isomorphic to F(X) via the identity on X
I know that there exist maps a : F(X) → G and b : G → F(X) such that ai = φ and bφ = i (by the universal property applied to (F(X), i) and (G, φ), where i here is the inclusion map)
And by uniqueness, I know that ba is the identity on X.
The hint is to show that ab is also the identity to conclude that a and b are isomorphisms, but I'm not sure how this works
white grotto
is a symmetric matrix just stretching? i just checked the singular value decomposition for it and it shows that a symmetric matrix is just a rotation then stretching then rotating back(the inverse of the first rotation).... then isnt it just strectching?
i mean i know it isnt
but i dont see how it cannot be
tribal niche
The hint is to show that ab is also the identity to conclude that a and b are is...
this is the picture I have in my head
<@&286206848099549185>
white grotto
is a symmetric matrix just stretching? i just checked the singular value decompo...
nvm i got it
tribal moss
The hint is to show that ab is also the identity to conclude that a and b are is...
Do you agree that if ab and ba are both the identity, then a and b are isomorphisms?
The rest of it is just to repeat your argument for "ba is the identity" the other way around. There's not really any particular property of the official free group involved, just "if two groups both satisify the same universal property, then they're isomorphic".
formal ermine
<@391270706186420224>
thanks
formal ermine
is there something similar to this but for char 0?
tribal moss
And by uniqueness, I know that ba is the identity on X.
(I hope you mean "ba is the identity on F(X)" here).
tribal niche
Do you agree that if ab and ba are both the identity, then a and b are isomorphi...
Re the first part: yes, since then they're both invertible.
I'm unsure of the repeat your argument bit, and I'm trying to work through the details right now: how do you know that the unique homomorphisms formed in the reverse case are still b and a, and not something completely different?
chilly radish
is there something similar to this but for char 0?
Yes, see kummer theory
tribal moss
You get the morphisms a and b once and for all by applying the universal property of G and F(X). They are ones you have before you start asking whether they are each other's inverses.
Hmm, perhaps I'm skipping a few steps here.
Have you proved yet that F(X) itself together with the inclusion from X into F(X) has the universal property?
tribal niche
Yes, I'm assuming above that (F(X), i) has the universal property
My thoughts are: given (G, φ) has the universal property, treating (F(X), i) as our other pair, you know there's a unique homomorphism b : G → F(X) such that bφ = i. Given (F(X), i) has the property, treating (G, φ) as the other pair, you know there's a unique homomorphism a : F(X) → G such that ai = φ, and from this you can show that ba is the identity.
I don't see how you can easily reverse this though
tribal moss
That looks right. But what you're saying is completely symmetric, you can just swap all references to (F(X),i) with (G,phi) and vice versa, and then your argument word-for-word proves that ab is the identity too.
The fact that a and b are the unique solutions to ai = φ and bφ = i ensures that it will be the same maps you're talking about.
tribal niche
When we "swap all references", don't the equalities ai = φ and bφ = i remain unchanged? How do you prove ab is the identity using those two?
rotund aurora
is there something similar to this but for char 0?
Highly doubt it
rotund aurora
Highly doubt it
Note that we are using the characteristic of the field very strongly. That extension is of degree char K
tribal niche
Ok yeah, ab is then the identity on G (not the identity on F(X), which was a source of confusion), so a and b are inverses
tribal moss
You also need to swap a and b, of course. Then after all the swapping "ai = φ" and "bφ = i" have turned into each other(!), but the conclusion "ba is the identity" has become "ab is the identity".
tribal niche
Great; thanks so much
Actually, sorry to bother you again, but how do we then know that the isomorphism from G to F(X) is the identity on X? If G was e.g. generated by X and φ was the inclusion, then sure, but I'm not sure why this is true in general (because all you know is that a(x) = φ(x) for x in X)
Perhaps the exercise should read "G is isomorphic to F(X) via φ on X?"
tribal moss
the isomorphism from G to F(X) is the identity on X
This claim doesn't seem to make sense to me.
X is not necessarily a subset of G.
On the other hand, since phi = ai and i is an injection, we know that phi must be an injection too, so each x in X corresponds to an element of G.
I'm not sure how to interpret "via the identity on X" in the exercise; I'd suggest ignoring it.
tribal niche
> the isomorphism from G to F(X) is the identity on X
This claim doesn't seem t...
It doesn't make sense to me either: the claim in my notes is formulated as "If (G, φ) has the universal property on X, then G is isomorphic to F(X) via the identity on X", but I suspect the correct version involves φ (and yeah, φ injects into G in 'the same way' as i injects into F(X))
tribal moss
I don't think you can make it mean anything more interesting than what ai = φ and bφ = i already say.
sweet echo
Are Poisson algebras defined to be commutative algebras with a Poisson bracket? Looking at an article that doesn't impose commutativity but then claims $\text{ad}{a^{-1}} = -a^{-2}\text{ad}{a}$, which I can't seem to prove without commutativity
cloud walrusBOT
ΣAC
chilly ocean
some people require commutativity, some don't
maybe whatever you're reading has a comment about its conventions somewhere?
maybe it follows the conventions of some of its sources which might?
sweet echo
The textbook written by the same people that wrote the paper that this article references does impose commutativity
sweet echo
I was just wondering if the identity holds in the non-commutative case, either it doesn't and they should have said commutatitive, or I just can't figure it out
chilly ocean
i'll stare at it once i'm caffeinated and see if it's possible to prove without commutativity