F(a) is the smallest field containing F and a
#groups-rings-fields
rocky cloak
coral spindle
Yes
rocky cloak
(and that's always a field)
coral spindle
Yes
static yew
Maybe I was thinking of Q[x] vs Q(x)
It ain't easy keeping all this straight
That's it
Going back to basics now
Everything in terms of Peano axioms
Countable rings only. Only 1 has a multiplicative inverse
Fox only, no items, final destination
delicate orchid
Countable rings only. Only 1 has a multiplicative inverse
the -1 in question
crystal turtle
-1 crying rn
delicate orchid
could be char 2 albeit
static yew
Ok I might give -1 a pass
static yew
could be char 2 albeit
As a software engineer I am partial to char 2
delicate orchid
anyway just make sure whatever you're quotienting by is generated by a minimal polynomial and we're all good
static yew
Minimal? Is that the same as irreducible?
delicate orchid
it's the polynomial of smallest degree that has something as it's root
they are irreducible though, see if you can prove it
south patrol
Peano
crystal turtle
Piano
static yew
they are irreducible though, see if you can prove it
a is a root of a polynomial h(x) IIF evaluating h with x=a results in 0 (additive identity) right?
delicate orchid
I think ryxiann knows
crystal turtle
Yes
delicate orchid
although the proof I have in my head doesn't really use this fact
anyway if f is a minimal polynomial for some a and it isn't irreducible then it can be factored as f = gh, since f(a) = 0 then either g(a) or h(a) = 0. But since deg(f) = deg(g)+deg(h), both g and h have a stricly smaller degree than f while having a as a root, contradicting the fact that the degree of f is minimal
static yew
they are irreducible though, see if you can prove it
Ok discord mobile is too tough to type this on
But I'm thinking: if h(x) is not irreducible, h(x) = f(x)g(x) for non constant f,g (so deg f, g > 0)
If h(a)=0 then f(a) = 0 or g(a) = 0
Which means that if h is not irreducible then there is another polynomial with Lower degree (f or g) that also has a as a root
delicate orchid
I'll give you the benefit of the doubt that you didn't read my message
static yew
I'm doing all this in F[x] btw. No idea if it translates to G(x)
static yew
G is psi(F) where psi(x) is the letter adjacent on the stupid onscreen keyboard
delicate orchid
I feel a bit uneasy calling elements of F(x) polynomials
static yew
Good, me too
static yew
Kinda like how C has no well defined ordering?
delicate orchid
well you can definitely still talk about the degree of things in there
hmmm
no perhaps it does still work
static yew
R -> C gets you more algebraics but "less than" is nonsense
Fraction fields are fields but now "minimal " is nonsense
delicate orchid
it could still work I might have been too hasty as per usual
whatever they are they're definitely not "irreducible" though, cause we're in a field
dim widget
Can we call this channel the “Halliday Memorial Channel (RIP 5ever in our hearts 💙🦋💙)”
delicate orchid
you spent 6867 hours searching for the butterfly emoji
delicate orchid
the department refuse to pay me
crystal turtle
Based
dim widget
based department
delicate orchid
yeah I have concluded it does not make sense because I cannot make sense of it
you can talk about the minimal polynomial of something in F(x) though
which is in F(x)[y]
dim widget
If everything is minimal… no one is 🥺
static yew
If everything is minimal… no one is 🥺
static yew
If everything is minimal… no one is 🥺
Just stick to prime fields
No polynomials
delicate orchid
just realised that F(x) will have the same universal property as F[x] but for fields
the only nice construction in the category of fields
crystal turtle
Yes
static yew
Universal property? Like Jurassic Park? 🙂
delicate orchid
no dear
static yew
Technically, a universal property is defined in terms of categories and functors by means of a universal morphism
I think this is beyond me
crystal turtle
Fields don't exist
static yew
Fields don't exist
False. F_p^k exists and can be explicitly constructed.
R and C, on the other hand...
delicate orchid
False. F_p^k exists and can be explicitly constructed.
"ermmmm aktually if u check the rules... 🤓 "
crystal turtle
(F_p)^k is not a field
delicate orchid
rip Q
crystal turtle
delicate orchid
(F_p)^k is not a field
no you dinkleberry
static yew
(F_p)^k is not a field
Brb defining the operator for G, the group of all fields
static yew
Yeah whatever did Q do to you?
After careful consideration Q gets to exist
delicate orchid
here's a little thought for you stevie
static yew
(for the prime 0)
0 isn't prime
It has pi and e as factors (pi×e×0)
delicate orchid
the dyslexia strikes again
delicate orchid
0 isn't prime
It has pi and e as factors (pi×e×0)
(0) is a prime ideal in Z sweatie
delicate orchid
here's a little thought for you stevie
what if we took the set of all sequences in Q and made it into a ring by multiplying and adding them together
and then what if we said that two sequences were the same if they approached the same value
I wonder what we would get
static yew
Sounds like some nonsense Dedekind would say
dim widget
The padics
delicate orchid
the 0-adics
rocky cloak
Sounds like some nonsense Dedekind would say
Actually I think Cauchy would say it
crystal turtle
P-adics my beloved 
static yew
n-adic digits are 0..n-1
0-adic
...-1
delicate orchid
completion w.r.t. a non-prime ideal... I'm spooked solid quite frankely
static yew
what if we took the set of all sequences in Q and made it into a ring by multipl...
Only if that set is countable, which means no infinite sequences 🙂
dim widget
In base 1: 0.000000000… = 1.0 = 1.00 = … = 1.00000000… = 2
rocky cloak
completion w.r.t. a non-prime ideal... I'm spooked solid quite frankely
All rings are integral domains moment
delicate orchid
Only if that set is countable, which means no infinite sequences 🙂
... Q is countable
crystal turtle
Only if that set is countable, which means no infinite sequences 🙂
What
delicate orchid
what if we just completed with respect to (1) huh what would you do then HUH???
static yew
Q is countable but any set of_sequences_ of elements of Q that contains countably infinite infinitely-long sequences is uncountable as per Cantors diagonal construction
delicate orchid
any set
chooses a set containing a single sequence
nothing personell kid
static yew
Then that set does not contain countably-infinite infinitely-long sequences
Okay I gotta get back to work
coarse kestrel
List the rational numbers between 0 and 1.
For each unique rational number construct a list of integers between 0 and 9 that are the digits of the rational number to the right of the decimal, using infinite 0's as needed.
This creates a countably infinite list of infinitely long sequences of a subset of rational numbers.
The conclusion of Cantor's diagonal argument doesn't work, or is at best blurry, when applied to rationals.
ivory trail
Q is countable but any set of_sequences_ of elements of Q that contains countabl...
the diagonal argument is for the set of all sequences, not for sets of sequences
at least what you're going for here
dim widget
The diagonal argument is when you convince each member of a group of people to argue with themselves.
So person i argues with person i, j with j, etc.
ivory trail
i thought it's when you argue something countably many people disagree with
static yew
the diagonal argument is for the set of all sequences, not for sets of sequences
True, though wew started with the set of all sequenced
static yew
i thought it's when you argue something countably many people disagree with
Inb4 arguing with aleph-one people
dim widget
The incidence matrix of the argument is the nxn identity matrix
crystal turtle
top tier shitpost
red quartz
hmm when did this exist
crystal turtle
always
long nebula
Since a couple hours ago!
red quartz
ok idk if this is the right place to ask but why is [H,K] a subgroup of G if H and K are subgroups of G
dim widget
You’re thinking of #1100503863586455632
long nebula
ok idk if this is the right place to ask but why is [H,K] a subgroup of G if H a...
How are you defining [H,K]?
If not as a subgroup of G
long nebula
commutator
presumably it's the subgroup generated by [h,k]?
delicate orchid
yeah so
long nebula
otherwise it doesn't form a subgroup in general
red quartz
like for example [h,k]^{-1}=[k,h]
delicate orchid
all elements of h are in G, all elements of k are in G
red quartz
but if H is not K then what happens
long nebula
the product of commutators isn't necessarily a commutator
red quartz
how is it a group tho
red quartz
the product of commutators isn't necessarily a commutator
oh
long nebula
The set of commutators of H and K is not necessarily a subgroup
red quartz
but [G,G] is a subgroup?
long nebula
[H,K] should be defined as the subgroup generated by commutators of the form [h,k]
delicate orchid
yeah sorry I presumed you were taking <[h,k]>
long nebula
but [G,G] is a subgroup?
The way you've defined it, no
red quartz
rt
long nebula
The way it's usually defined is that [G,G] is the subgroup generated by commutators
red quartz
The way you've defined it, no
oh
delicate orchid
for subgroups you define [H,K] to be the group generated by <[h, k] : h in H, k in K>
long nebula
The smallest example where the set of commutators of G is not a subgroup though is of order uhhhh
95?
cloud walrusBOT
solar vessel
bro thought 95 was prime
long nebula
My bad, 96
delicate orchid
wait no no no I don't believe it I simply do not believe it
yeah there we go
all groups of order 95 are abelian
delicate orchid
rancid
delicate orchid
I presumed it wasn't a manual search
what's the actual morphisms involved in those semidirect products
I wanna see the character table
long nebula
Yeah, backstory is I was about to assign the problem "show that the set of all commutators isn't necessarily a subgroup" in the problem set for the algebra summer study group
And I didn't realize that the smallest example was that
ivory trail
i thought it's when you argue something countably many people disagree with
I should've said "annoy countably many people at once"
long nebula
ivory trail
or generally annoy everyone at once
delicate orchid
Yeah, backstory is I was about to assign the problem "show that the set of all c...
the free group? lol
long nebula
I was gonna use the free group but I couldn't figure out how to prove that something wasn't a commutator
So I was just like, screw it, and took the problem off the problem set
delicate orchid
well the free group is nice because all commutators just look like commutators lmfao
long nebula
The free group on four variables should work tho
delicate orchid
or as I like to call it, the free group on two variables
delicate orchid
yeah the free group on 2 elements is way bigger than the free group on 4 elements
long nebula
as a subgroup?
He just meant the free group on two variables is a better example
teal vessel
so understanding check: The normalizer of a set A in the group G is the set of all elements in G such that gAg^-1 is an automorphism, right? in other words, acting on the set is at most a shuffle of the set (since set order doesn't matter).
long nebula
so understanding check: The normalizer of a set A in the group G is the set of a...
Yes
If H is normal in G, then the normalizer of G is G
Also by "gAg^-1 is an automorphism" I think you mean that "gAg^-1 = A"?
The former doesn't make sense to me
long nebula
guys how is pi^2 algebraic in R?
It's algebraic over R since it's the root of x - pi^2. It's not algebraic over Q since pi isn't.
solemn dew
oh okay i think i got it
so if a number is in that field, it us algebraic over that field if it has solutions in that field
teal vessel
algebraic means that it is a solution to a polynomial with coefficients in the field in question.
solemn dew
yes okay
but look at this:
says alpha is not in F
but isnt pi^2 in the field R?
teal vessel
it is, that's strange
that equation also doesn't make sense
because that would say that pi^4=pi which is false
little shadow
stupid questions: when we say a known group (G) is a subgroup of a totally different group (H) does it mean that its isomorphic to a subgroup of H? like this for instance (in this case its the reserve)
teal vessel
Also by "gAg^-1 is an automorphism" I think you mean that "gAg^-1 = A"?
thinking of f:A → A f(a)=gag^-1
not a group isomorphism, so yeah, confusion abounds I suppose
crystal turtle
stupid questions: when we say a known group (G) is a subgroup of a totally diff...
equivalently, G would be (isomorphic to) a subgroup of H iff there is an injective homomorphism G --> H
rocky cloak
yeah the free group on 2 elements is way bigger than the free group on 4 element...
It's even bigger than the one with (countable) infinitely many generators
little shadow
equivalently, G would be (isomorphic to) a subgroup of H iff there is an injecti...
yes thats makes sense. I'm just a bit confused since ive never seen something like "D_8 is a subgroup of Z/16n) or something like that
crystal turtle
because it doesn't really matter how we label the elements, so much as the structure that the group operation has. So instead, we should consider up to isomorphism, which is why they are calling it a "subgroup" even if, technically, the elements are not in the larger group
limber sequoia
thinking of f:A → A f(a)=gag^-1
but that is a group isomorphism?
f(ab) = gabg-1 = gag-1gbg-1 = f(a)f(b)
oh the set is outside of G
cool cool
teal vessel
I'm overloaded on terms atm lol. I spent way too long in set theory, so this group theory stuff can be strange for me to keep in group theory terms
little shadow
because it doesn't really matter how we label the elements, so much as the struc...
i see, thanks for clearing it up! 😁
limber sequoia
I'm overloaded on terms atm lol. I spent way too long in set theory, so this gr...
same i just started and it's an overload
and my course notes really don't go in depth!
solemn dew
try learning this on your own
long nebula
Yeah, I guess because specifically it's talking about stuff that's not in the field. Anything in the field is automatically algebraic over that field though.
long nebula
not a group isomorphism, so yeah, confusion abounds I suppose
yeah, I guess saying that "conjugation by g restricts to a bijection on A" would be better 👍
delicate orchid
This wording is confusing me, do we just want the minimal polynomial of sqrt(2)+sqrt(3)
solemn dew
i think so
topaz solar
looks like it
ivory trail
i would just take the 0, 1, 2, 3, 4th powers and then come up with a polynomial from that
delicate orchid
Yeah that’s the trick
south patrol
You can just kind do it by hand, like find min poly over Q(sqrt 2) first
Which is clear
delicate orchid
Then Q(sqrt(3))
upper pivot
I would multiply all the conjugates but thats probably a bit dumb
like (x+sqrt(2)+sqrt(3)(x-sqrt(2)+sqrt(3))(x+sqrt(2)-sqrt(3))(x-sqrt(2)-sqrt(3))
and start a expandin'
ivory trail
true
upper pivot
shouldnt be too bad oof
delicate orchid
like (x+sqrt(2)+sqrt(3)(x-sqrt(2)+sqrt(3))(x+sqrt(2)-sqrt(3))(x-sqrt(2)-sqrt(3))
Based tbh
upper pivot
I liked what rho said tho lol, seems smarter
void cosmos
R is a ring , A is an R-module , N and K are submodules of A
pi is the natural projection map to A/K
is pi(N) = N / K = (N+K)/K ?
delicate orchid
n+k+K = n+K
solemn dew
void cosmos
suppose {a1,a2,...,an} generate M/mM ( R is a ring local ring with M being its maximal ideal )
(those a_is are equivlence classes but too lazy )
now we wish to show that these a_is generate M aswell
consider N = (a1,a2,...) in R
then pi(N) = (N+M/mM)/M/mM = N/mM = M/mM ( since they generate M/mM)
---> N/mM / (N+M/mM/M/mM) = 0 --> M = N+mM
so N = M by nakayamas lemma?
delicate orchid
Yeah I buy it
molten viper
So here's the context, I've got a local ring A, and its maximal ideal is m. I have some x \in m idempotent, I wish to prove x = 0. I can show that 1-xy is a unit for all y \in A. So I believe the idea is picking the right y to show some contradiction if x \neq 0. Would like a hint, please
warm gate
Hi groups, you cant take infinite products in a group can you?
dim widget
@warm gate you can if you give it some extra structure like a topology
warm gate
Ahhh so you need more structure
that soothes me a little
my question was gonna be
if i have a group G that is generated by countably infinitly many generators , is there a smallest group that contains G and like all the products with an infite amount of different generators in the product
delicate orchid
Hi groups, you cant take infinite products in a group can you?
Yeah watch this
g_1 * g_2 * g_3 * g_-
delicate orchid
if i have a group G that is generated by countably infinitly many generators , i...
Is this another “free group of rank 2” moment chat
warm gate
well? is it?
delicate orchid
Just take the group all “infinite” products generate I suppose
But I believe this will be the entire group
ivory trail
So here's the context, I've got a local ring A, and its maximal ideal is m. I ha...
x(1-x)(1-x)^-1
delicate orchid
let a be one of these “infinite” products and g some element, then ga is in this subgroup
And thus so is gaa^-1 = g
warm gate
surely its more than th3 entire group
molten viper
x(1-x)(1-x)^-1
thanks frendo
delicate orchid
How could it be more than the group if the group is closed under multiplication
warm gate
Wait i mean the new group that has infinite products is bigger than the one that doesnt.
night onyx
How could it be more than the group if the group is closed under multiplication
a larger group with a copy of G in it, which somehow also supports infinite products
delicate orchid
Right so we are putting something topological on it
And then taking the limit points
warm gate
Do we have to
delicate orchid
How else could we possibly talk about a convergence without a topology
delicate orchid
And then taking the limit points
In which case it’s equal to the group if and only if it’s compact
topaz solar
The other option would be formal products, like sequences N->G or something quotiented out appropriately ig
warm gate
Yh thats more what i was thinking along the lines of
topaz solar
But that’s uh
delicate orchid
Taking like a R[[X]] kinda deal
topaz solar
Annoying
delicate orchid
Ay ay ay
warm gate
Is it messed up
delicate orchid
Yeah this is why we need to quotient by stuff
topaz solar
So now you want finite length things are easy to deal with, but what if I have some infinite sequence, it’s a bit more troublesome to quotient these all out
Obviously we’d want things that look the same to be the same, but if I were to take, say, our original group countable
Then if I enumerate the whole group in a different order
What then
Consider this for finite non commutative groups
Like uhhh, S_4 or smth?
Does the order I take the product matter?
Or maybe S_3
warm gate
hm
topaz solar
suddenly everything should get way more annoying since also $\prod a^{(-1)^n}$ would show up
cloud walrusBOT
The eternal Sharp
topaz solar
$a a^{-1}a a^{-1}a a^{-1}a a^{-1}...$
cloud walrusBOT
The eternal Sharp
rustic crown
topaz solar
sure, we can apply our little reducing rule and move finitely many at a time
topaz solar
but you can't move infinitely many at once
red quartz
if G=[H,H] then is G a normal subgroup of H
warm gate
Yeahhh.... hmmm....
red quartz
cuz i got $hh_1h_2h_1^{-1}h_2^{-1}h^{-1}$
cloud walrusBOT
xoonkckskskkckskckxkcks
red quartz
so this has to be in [H,H]
delicate orchid
but you can't move infinitely many *at once*
DYEL bro?
topaz solar
@warm gate so basically you've got issues in things being normal once you take infinite products. That above product is exactly 1-1+1-1+1... in R which we know doesn't converge
warm gate
yea
so its kind of evil... ic...
thats why we hwve the rules... and topology to talk abt convergence otherwise... makes sense!
red quartz
if G=[H,H] then is G a normal subgroup of H
bump
topaz solar
DYEL bro?
I meant just with the expected sequence equivalence classes being like uh a_n being equivalent to multiplying two adjacent terms a_n a_n+1 to like (a_n * g) (g^-1 a_n+1)
topaz solar
thats why we hwve the rules... and topology to talk abt convergence otherwise......
ofc, you can consider these as formal products, but don't expect them to be sensible
warm gate
thx for entertaining my daydreams sharp and wew
delicate orchid
if G=[H,H] then is G a normal subgroup of H
Let x be in [H, H] then x^-1ux = uu^-1x^-1ux = u[u,x] which is in [H,H]
delicate orchid
A normal subgroup is invariant under conjugation in the larger group
red quartz
ye
delicate orchid
So no
topaz solar
A normal subgroup is invariant under conjugation in the larger group
they want [H, H] normal in H
delicate orchid
Oh right I got my letters backwards who gives a shit
topaz solar
you conjugated u by x in [H, H]
delicate orchid
I’m drunk you all got the point
crystal turtle
based
red quartz
**xoonkckskskkckskckxkcks**
i got htis but idk
topaz solar
thats why we hwve the rules... and topology to talk abt convergence otherwise......
There's some ways to do it without full on considering topology but basically just don't do that lmao (G^\mu, orders, other such notions of converging)
long nebula
how does this follow
What have you tried so far
delicate orchid
I think the easiest way to see this is to consider the quotient right
red quartz
What have you tried so far
idk i didnt try to prove that fact
long nebula
Try it
red quartz
**xoonkckskskkckskckxkcks**
but i tried this for the normal subgroup
long nebula
The one direction is probably easier, in the commutator subgroup => can be written as a product which when rearranged gives the identity
delicate orchid
I think the easiest way to see this is to consider the quotient right
Nah just showing it directly is easier
red quartz
The one direction is probably easier, in the commutator subgroup => can be writt...
yea
red quartz
**xoonkckskskkckskckxkcks**
if arbitrary expressions like this rearange to the identity then how do you prove its in [H,H]
delicate orchid
If you rearrange them and it’s the identity it means that for each g in that expression you need g^-1 to be in there as well
red quartz
ye but it could be very far away
delicate orchid
Doesn’t matter you just have to rearrange it
Oh
I get what you mean, they have to be next to each other to be in the commutator
That’s the trick you have to figure out
topaz solar
hollow mica
There’s a model theoretic proof too which is cool
topaz solar
Where’s this from?
a random presentation of ultraproducts in algebra
delicate orchid
Model theory is scary albeit
topaz solar
There’s a model theoretic proof too which is cool
this is model theoretic
3.5 assumes CH, which isn't quite necessary
hollow mica
Oh the proof I’m familiar with shows that a counterexample in C would imply a counterexample in a finite field
topaz solar
it holds without it
topaz solar
Oh the proof I’m familiar with shows that a counterexample in C would imply a co...
this is exactly that
holds for all finite fields -> Los -> holds in C
hollow mica
Oh is Los that
topaz solar
yes
$\prod \mathcal M_i/U \models \phi ([\mathbf x_i])$ iff ${i: \mathcal M_i \models \phi (\mathbf x_i)} \in U$ or wtv
cloud walrusBOT
The eternal Sharp
topaz solar
a random presentation of ultraproducts in algebra
I was looking up something to see if I could see any ultra group stuff since Harper got me thinking 
topaz solar
$\prod \mathcal M_i/U \models \phi ([\mathbf x_i])$ iff $\{i: \mathcal M_i \mode...
here ya go in a categorical formulation
delicate orchid
**The eternal Sharp**
mathcal M?! I love the irreducible marks 
topaz solar
Mod(C) is exactly pretopos functors C->Set btw
delicate orchid
Mod(C) is exactly pretopos functors C->Set btw
Thanks, now it’s crystal clear to me
I find it funny how we’re talking about model theory in here and row echelon reduction in #advanced-algebra
upper pivot
since we are talking about model theory type stuff. I started reading about R*, v cool
Was thinking, what is Gal(R*/R)
topaz solar
upper pivot
lmfao nice
topaz solar
As best as ive got atm, but likely horrible
upper pivot
transedental galois groups seem disgusting
apperantly even for K(x1,x2,x3)/K are unknown oof
topaz solar
It can be proved that if \kappa is the smallest \omega measurable cardinal then any group of cardinality < \kappa is an ultrapower
nice
topaz solar
apperantly even for K(x1,x2,x3)/K are unknown oof
how bad can it be tho
upper pivot
oh interesting
upper pivot
topaz solar
ofc there's trivial ultrapowers lol
but take your countably complete ultrafilter on \kappa and it's \lambda complete for any \lambda < \kappa
so it's nonprincipal but spits the original back out
upper pivot
(this is group of automorphism P^k, by AG magic, there is an equivalence at looking at their function fields and cover)
topaz solar
Was thinking, what is Gal(R*/R)
Anyway, this is 100% gonna be independent of ZF at least since ultrafilters, and likely ZFC since cardinalities being screwy without GCH?
upper pivot
yeah i believe that
topaz solar
But uhh, with GCH, it's likely as horrible as it can get
upper pivot
In my field, R= specific factor, the question on wether ultrapowers of R are all iso is equiv to CH
topaz solar
So, # of nce ultrafilters on \kappa is 2^2^\kappa
GCH says all saturated
so all iso
(GCH equivalent to all these holding for all k, according to Ultra)
which I'd believe
topaz solar
so all iso
I don't believe these isomorphisms are necessarily natural
But if you place all the 2^2^k things down with isos so it's a groupoid, then Gal(R^\mu, R) < loops in this space, uhh \pi_1(X, R^\mu)
upper pivot
this sounds cool. Ill come back after finishing goldbring kek and read this fully
topaz solar
no idea how many really keep R fixed per se, but these iso's are likely the interesting ones
since they should only scrable the nonstandard elements
upper pivot
right but this is already interesting, bc it says atleast about Gal(R*/Q)
and we know for Gal(R/Q) this is trivial
upper pivot
all field autos fix Q yeah
topaz solar
fixing R is a little screwier
but, these saturation induced isos should all commute with the diagonal so
it fixes Q, so it fixes the orders and all
so moving R is not easy
(These isos all preserve field and order stuff I think)
But the infinitesimals and infinites can kinda be moved around essentially freely
And the sizes of them depends on your size of ultrafilter,
So picking \kappa is important anyhow
upper pivot
actually I think all of these should fix R now that you say it. Here is why:
you preserve order, bc sigma(x)^2 =sigma(x^2), and Q is order dense in R
topaz solar
N*/~ where f ~ g iff f-g finite is a dense linear order and is as saturated as your ultrapower, so k saturated
south patrol
Tbf this looks more like model theory so go to #algebraic-geometry
warm gate
I was looking up something to see if I could see any ultra group stuff since Har...
harper got you thinking? :o thats me :D
topaz solar
actually I think all of these should fix R now that you say it. Here is why:
ye there's no way you're screwing up the order because thats exactly the structure you want to preserve
ultrapower as ordered fields
upper pivot
its the same arguement as to why Gal(R/Q) is trivial basically
topaz solar
ye
topaz solar
N*/~ where f ~ g iff f-g finite is a dense linear order and is as saturated as y...
R*/R is not trivial though I believe, since we can take this and kinda freely slide around the galaxies
upper pivot
right i believe that
upper pivot
I need to get comfortable with all this hyper logical stuff oof, seeing so many cardinalities kinda made me dizzy lol
topaz solar
the size is k dependent, but for any a < k we should be able to make a saturation argumenr
Starting with just countable ultrapowers then
N*/~ should be 2^N saturated
I think
Assuming GCH
because I am afraid of the cardinals in general without at least CH here
so for any countable collections X, Y with X < Y around, we can split em X < c < Y
actually scratch that I have a brilliant idea
topaz solar
right i believe that
If I'm right in my assumption that any sort of "rotation" of the galaxies works, since densities and such, then injective monotone function N*/~ -> N*/~ satisfying f[0] = [0] should have at least one corresponding element in Gal(R*,R)
moving galaxies is moving Z* = moving Q* and then we would just want to say Gal(*R, *Q) trivial
upper pivot
oh wow interesting i didnt even think of that
topaz solar
but since it'd be auto on a.e. of the copies it's trivial
upper pivot
Gal(R*/Q*)
Does this follow from transfer principle? it does I think, write \sigma aut -> sigma = id
and it transfers
topaz solar
it's trivial a.e. so ye
Here's the technical point: are galaxy rotations always extendable
2: are they uniquely extendable
- is if the idea works, 2) is if we only give a partial description
Verifying my saturation argument is also important
but yeah you could 100% make a small paper on this
upper pivot
yes sharp john collobaration???
topaz solar
yeee lets goooooo
upper pivot
nice give me a week to finish goldbring
topaz solar
but like you could actually do that if you felt like it
topaz solar
nice give me a week to finish goldbring <:opencry:586078614865444865>
nah fam that why you drop it on the grad student RA
coral shale
what if we banish yellownames from here
upper pivot
bro I am the grad student RA 😭
coral shale
then we can talk about our baby algebra
summer path
then you'll never see det again
topaz solar
then we can talk about our baby algebra
I meant me but true
coral shale
then you'll never see det again <:kongouDerp:483988392099446794>
not if i banish myself too
summer path
coral shale
i dont trust you to talk only about baby algebra
coral shale
summer path
can't wait for april fools for this channel to be actually baby algebra
topaz solar
I don't know how to write things and you actually know people
topaz solar
<:wg:1130147597428998184>
that was a categorical formulation, less horrible looking than logic
coral shale
topaz solar
you guys are just at a level below baby
the infinitesimal algebra
red quartz
hmm so how do you prove [G,G] is a normal subgroup of G?
or should i just assume this
delicate orchid
I gave you the proof earlier but just mixed up u and x
south patrol
Lol
delicate orchid
could you not just unswap them in your head?
south patrol
uwu
topaz solar
But swapped yourself
red quartz
could you not just unswap them in your head?
btuh
south patrol
btuh
red quartz
how is u[x,x] in [G,G]
dim widget
A better question is why is [DG, DG] normal where DG is [G, G]
red quartz
ok lemma find that message
south patrol
A better question is why is [DG, DG] normal where DG is [G, G]
Yoneda.
dim widget
@dim widget I guess even better is to prove that [N, N] is normal whenever N is normal in G
rocky cloak
Quick solution: conjugation is an automorphism QED
dim widget
I am curious about the proof writing technique that compels one to mention that conjugation is an automorphism but not any other part of the argument
delicate orchid
it's called being based u wouldn't understand
south patrol
The class of groups for which this holds is weakly saturated and
delicate orchid
SATURATED???
rocky cloak
I am curious about the proof writing technique that compels one to mention that ...
Just assume that we have already proven that any automorphism preserves the derived subgroup
Makes life much easier
Tbf it actually feels easier to prove that any automorphism preserves derived subgroup, than it does proving it for conjugation.
night onyx
hmm so how do you prove [G,G] is a normal subgroup of G?
An important part of the proof is that if a subgroup H is generated by a subset X, so that H = <X>, then gHg^{-1} = <gXg^{-1}>. Since the commutor is generated by elements [a,b] for a,b in G, it's enough to show g[a,b]g^{-1} is also a commutor in order to show that [G,G] is normal.
south patrol
Tbf it actually feels easier to prove that any automorphism preserves derived su...
inventor's paradox?
alpine island
Oh this is a cool new channel. Glad it exists.
delicate orchid
t'other is t'new one
coral shale
cool
dim widget
@rocky cloak yes it is no easier for conjugation than anything else of course. Morally this has to be true because N semidirect product Aut(N) is also a group
delicate orchid
It is infact my favourite group
delicate orchid
yeah
long nebula
nice give me a week to finish goldbring <:opencry:586078614865444865>
LOLLLLLLLLLLLLLL
long nebula
yes sharp john collobaration???
Wait I wanna see it
long nebula
<:wg:1130147597428998184>
yeah exactly, baby algebra
topaz solar
Wait I wanna see it
I had some ranting in algechill
hollow mica
what does “galaxy” in this channel refer to lol
crystal turtle
sharp's rants
topaz solar
what does “galaxy” in this channel refer to lol
Copies of Z in nonstandard models of PA
It’ll always look like N+AZ for some dense linear order w/o endpoints A
topaz solar
not really
hollow mica
So who came up with it
hollow mica
Hmm
formal ermine
NO
void cosmos
okay good
tender wharf
what happened to aa
crystal turtle
whaat
formal ermine
what happened to aa
toxic zephyr
finally, an algebra channel that isn't too big brained for me. at least I hope.
waxen ibex
How is the map f : G→G/N is surjective homomorphism where G is a group and N is a normal subgroup of G
ashen heron
you haven't fully defined the map
waxen ibex
a→aN a∈G
toxic zephyr
well if you have $aN\in G/N$, then it has a preimage doesn't it?
cloud walrusBOT
nixxy nilpotent (abbott hater)
toxic zephyr
@waxen ibex
waxen ibex
I forgot what was I confusing at, no question now.
solemn garden
can't wait for april fools for this channel to be actually baby algebra
What will #prealg-and-algebra be then
frigid lark
Combinatorics
distant summit
Idk if this counts as "advanced", but I wanted to try proving this and would like someone to check. I have used $\gamma=\gcd(a,d)$.
$ord(g)=d \iff g^d=e$. Consider $g^a$ to the power of $d/\gamma$: $$ (g^a)^{d/ \gamma} =g^{da/\gamma}.$$
But, by definition, $\gamma | a$, so $da/\gamma=dk, k\in\mathbb{Z}$. Hence $$(g^a)^{d/\gamma}=(g^d)^k=e.$$ To show that $d/\gamma$ is the least such integer, suppose $\exists \theta \in \mathbb{Z}, \theta \leq d/\gamma$ such that $(g^a)^{\theta}=e$. This means that $(g^a)^{\theta}=g^{da/\gamma}$, so $a\theta=da/\gamma$. But this means that $\theta=d/\gamma$. So $d/\gamma$ is the least such integer, and therefore it is the order of $g^a$.
cloud walrusBOT
Douglas
distant summit
I'm a little unsure about $a\theta=da/\gamma$, because there might be some modular stuff I'm forgetting about
cloud walrusBOT
Douglas
dim widget
@distant summit the last step of your proof doesn’t work. g^x = g^y doesn’t mean x = y
summer path
obviously a sylow problem but i don't know what to do after finding n_3 = 1, 7, or 49 and that P_7 = Z/7Z x Z/7Z or Z/49Z; but clearly i'm going to have P_7 = Z/49Z for both cases..
dim widget
So this is not really a sylow question necessarily.
I can show a different solution and we can think about how to use the sylow theorems after if you like?
dim widget
@summer path very important theorem: if a subgroup H has index the smallest prime which divides the order of G, then H is normal.
(It’s a fun exercise to prove it, it’s “follow your nose” but very useful once you know it)
So the subgroup generated by g of order 49 is normal
Alternatively you could note that by the sylow theorems the number of subgroups of order 49 has to be a divisor of 3 and also be 1 mod 7
So it must be just 1 of them
Now by Cauchy you also know that the map G \to Z/3 induced by taking this quotient has to have a section
Indeed G has an element of order 3, which does not lie in our subgroup of order 49 bc of its order
So then G is a semidirect product: Z/49 \to G \to Z/3
Thus all such G are determined by a Hom(Z/3, Aut(Z/49))
distant summit
<@711934347934040125> the last step of your proof doesn’t work. g^x = g^y doesn’...
Because of modular arithmetic?
I was thinking that $g^x=g^y \implies g^x g^{(-y)}=g^0 \implies g^{x-y}=g^0 \implies x=y$
cloud walrusBOT
Douglas
distant summit
Oh actually, in saying x-y=0 I'm assuming my conclusion aren't I?
Yeah I think the modular arithmetic thing makes sense, like if $g^a=g^b$ and $g^n=e$, then you say $a \equiv b (\mod n)$ I think, but not necessarily $a=b$
cloud walrusBOT
Douglas
dim widget
@distant summit yes exactly
@summer path anyway so the conclusion is that there are 3 maps from Z/3 to Aut(Z/49), one is trivial and two are non trivial. And the two nontrivial ones give isomorphic groups
distant summit
<@711934347934040125> yes exactly
Hmmm, not sure how to "patch" the proof.
I know I could say $a\theta=\frac{ad}{\gamma} \mod d$, but obviously $\frac{ad}{\gamma}$ is a multiple of $d$ so is congruent to zero $\mod d$. So if $a\theta=0 \mod d$ then $a\theta=dn, n\in \mathbb{Z}$
cloud walrusBOT
Douglas
summer path
Thus all such G are determined by a Hom(Z/3, Aut(Z/49))
i follow until here, i don't really understand how you reach the conclusion that there are 3 of them
but given that there are 3 of them, i think it would make sense that one sends 1 to some automorphism \sigma and 2 to \sigma^2, and the other sends them to swapped ones?
dim widget
@summer path it’s just because Aut(Z/49) = (Z/49)* = Z/6 \times Z/7
And yes the point is that the two non trivial homomorphisms differ by the automorphism of Z/3
dim widget
@distant summit if g^ab = 1 then d|ab. So the smallest nonzero b such that d|ab is….?
@summer path because Hom(Z/3, Z/6\times Z/7) = Hom(Z/3, Z/6) = Hom(Z/3, Z/3) = {0, Id, inv}
Where inv is inversion composed with the previous homomorphism
distant summit
<@711934347934040125> if g^ab = 1 then d|ab. So the smallest nonzero b such that...
d|ab means ab=dk and d≤ab, so d/a ≤ b, i.e. b must be at least as great as d/a
dim widget
@distant summit what is the smallest number divisible by d and a?
summer path
oh, ok i thought that for a second, then didn't know what to do with the Z/7
but i guess it makes sense that you can kind of throw it away since 3 doesn't divide it
thanks tteg 
i will now go enjoy my 6am dinner and then wonder why the sun isn't up when i get up again
dim widget
@summer path no worries, good luck with your nocturnal lifestyle 🙂
distant summit
<@711934347934040125> what is the smallest number divisible by d and a?
ad works
i think there are special cases, but if a,d shared no prime factors in common then ad would be the smallest number divisible by both a and d i think
dim widget
Yes but even if they do share prime factors there is a nice expression for their least common multiple
In terms of their greatest common divisor
distant summit
Yes but even if they do share prime factors there is a nice expression for their...
well ik that w the gcd you take the common prime factors and the "multiplicity" will be the minimum, e.g. gcd(p^3 q^1, p^5 q^2)=p^min(3,5) q^min(1,2)=p^3 q^1, but im not sure abt the lcm
crystal turtle
field
solemn dew
but no bells ring of what it can be
F_25 means what then?
A field with 25 elements?
solemn dew
okedoke thanks
warm wyvern
there's a field with p^k elements for all positive k and prime p
is there only one such field?
up to isomorphism ofc
formal ermine
is there only one such field?
yeah
warm wyvern
interesting
warm wyvern
tf is a splitting field
crystal turtle
Galois moment
formal ermine
crystal turtle
tf is a splitting field
smallest field for which a polynomial splits into linear factors (equiv: if the poly has degree n, it has n roots counting multiplicity)
one can show that splitting fields are unique up to isomorphism, they always exist, and in the case of x^(p^k)-x, any field of order p^k is a splitting field for this poly
thus, they are unique
iirc at least, this is the proof
I don't know of a nice description for the elements of them lol
But for primes p not congruent to 1 mod 4 (I think?), one can use gaussian integers mod p to get a field of order p^2
coral shale
tf is a splitting field
bonk
dim widget
In general Q(zeta_{p^n - 1}) mod p is F_{p^n}
coral shale
tex it
dim widget
no u
crystal turtle
Interesting, didn't know
crystal turtle
No just some elements duh
dim widget
just the integral elements, or just the elements with no denominators divisible by p
If you reduce those rings mod p you get F_{p^n}
night onyx
tf is a splitting field
in general polynomials over a field don't have roots in that field (like x^2 + 1 in R[x]), a splitting field of a polynomial is the smallest field containing the roots of that polynomial (so in the case of x^2 + 1 in R[x] the splitting field would be C)
coral shale
whats Q adjoined with
dim widget
e^{2\pi i/(p^n - 1)}
indigo ridge
is there always a homomorphism between two rings if not then what about the zero mapping?
coral shale
ok
dim widget
is there always a homomorphism between two rings if not then what about the zero...
in some texts ring homomorphisms have to sent 1 to 1, in other texts the 0 homomorphism is acceptable
warm wyvern
smallest field for which a polynomial splits into linear factors (equiv: if the ...
I see 
crystal turtle
is there always a homomorphism between two rings if not then what about the zero...
Yes if you allow zero map, no if you require 1 maps to 1
warm wyvern
in general polynomials over a field don't have roots in that field (like x^2 + 1...
interesting
dim widget
It depends if you only allow "rings with identity maps" or just "rng maps"
warm wyvern
rng
dim widget
this right
It should be Z[x] not Q[x]
rocky cloak
whatever iso is
$\cong$
cloud walrusBOT
jagr2808
crystal turtle
is there always a homomorphism between two rings if not then what about the zero...
example: if you require 1 going to 1, then there is never a map from the zero ring to another nonzero ring, since that map would have to send 0 to both 0 and 1 simultaneously
cloud walrusBOT
indigo ridge
It depends if you only allow "rings with identity maps" or just "rng maps"
what do you mean by rings with identity maps?
dim widget
Maps that have to send the identity to the identity
so they preserve the structure of "having an identity"
crystal turtle
Maps that have to send the identity to the identity
different authors take different conventions on that, which is why it depends
multiplicative
dim widget
No the multiplicative identity, any ring has an underlying abelian group, and homomorphisms of groups always have to send the identity to the identity
indigo ridge
a ring has the property that it is an ableian group with respect to addition though not multiplication
why would we assume the ring in question would have a multiplicative identity
dim widget
why would we assume the ring in question would have a multiplicative identity
It is sometimes part of the axioms of a ring.
When it is it is often the case that authors require that ring homomorphisms send that identity to the identity of the target
crystal turtle
some authors require a multiplicative identity, some don't
indigo ridge
oh
crystal turtle
lmao
open sluice
crystal turtle
https://imgflip.com/i/7urdak
i don't want to
dim widget
https://imgflip.com/i/7urdak
Mfw there is no ring of compactly supported functions on R
sad
open sluice
the price we have to pay
coral shale
rings do contain 1
formal ermine
https://imgflip.com/i/7urdak
they do
coral shale
rngs may not
formal ermine
non unital rings don't exist
crystal turtle
true
crystal turtle
https://www.youtube.com/watch?v=ST42VjR0LYk
skill issue
static yew
a ring has the property that it is an ableian group with respect to addition tho...
I was about to say, I thought 2x2 matrices formed a ring
Actually any transitive closure of n×n no?
crystal turtle
n×n matrices form a unital ring yes
unkempt stream
granted F is a field, the polynomial p(x) is irreducible iff (p(x)) is maximal in F[X] (a PID) correct?
static yew
I'm trying to look up "unital ring" and all I'm finding is crap from a Japanese novel series
unkempt stream
unital- with unity
coral shale
sao uwu
unkempt stream
i hope you bite your tongue violently
coral shale
who.
unkempt stream
You
coral shale
formal ermine
granted F is a field, the polynomial p(x) is irreducible iff (p(x)) is maximal i...
yes
unkempt stream
also a polynomial need not split entirely over a single field extension correct?
i.e not solely over the injection into the splitting field
coral shale
what
formal ermine
yeah
unkempt stream
trying to think of an example
unkempt stream
i’m trying to nail down the correspondance between splitting polynomial & fixing/stabilizing automorphisms
static yew
n×n matrices form a unital ring yes
Taking it a step further... what is the weakest structure the matrix elements must belong to?
Obviously a field will work, fields work everywhere
If your elements are from a semiring then the resulting structure will be at most a semiring cuz some matrices won't have additive inverses
Can my matrix elements be elements of a ring and I still get a ring?
unkempt stream
it’s isomorphic to the endomorphism ring over the ring’s direct sum with itself N times
Afaik
static yew
it’s isomorphic to the endomorphism ring over the ring’s direct sum with itself ...
I'm not 100% convinced that this isn't nonsense
indigo ridge
when there is a homomorphism between two rings, I always assumed this was to show that there are inherent similarities between these two rings (aside from the fact that they're both rings) or there just happens to exist a function that 'preserves' the ring
coral shale
wat
unkempt stream
I'm not 100% convinced that this isn't nonsense
should be a module endomorphism
static yew
Wouldn't such a function be an isomorphism?
coral shale
theres a copy in either direction
unkempt stream
Wouldn't such a function be an isomorphism?
?
formal ermine
it’s isomorphic to the endomorphism ring over the ring’s direct sum with itself ...
which is isomorphic to $$R^{\oplus N} \tensor \left(R^{\oplus N}\right)^\vee$$
cloud walrusBOT
static yew
"There just happens to be a function that preserves the ring"
unkempt stream
wtf are you talking about my man
Granted I have to ask that to myself on a hourly basis
warm wyvern
rings do contain 1
depends on your definition
static yew
What jayz said
warm wyvern
some maniacs (like john) don't include that requirement
indigo ridge
"There just happens to be a function that preserves the ring"
so this?
unkempt stream
To be fair I actually kinda pulled rhe endomorphism thing out of my ass
i never formally proved it, but it seems right
warm wyvern
which is isomorphic to $$R^{\oplus N} \tensor \left(R^{\oplus N}\right)^\vee$$
tf is ^\vee?
static yew
I was hoping you wouldn't remember that
crystal turtle
What
wraith cargo
bruh I thought that said Z/12
static yew
But then, what does it mean to contain 1?
The set itself {0} does not contain 1
It does have a multiplicative identity, but that's 0
crystal turtle
"I said some bullshit and hoped you wouldn't catch it" 
wraith cargo
But then, what does it mean to contain 1?
The set itself {0} does not contain 1...
it means a multiplicative identity other than 0
wraith cargo
yeah...
That's called the 0 ring
static yew
Where 2 is the addition of the multiplicative identity to itself
static yew
F2 be my favorite field
Well. Base field, anyway. Things are more interesting with its extensions
coral shale
But then, what does it mean to contain 1?
The set itself {0} does not contain 1...
1 is defined to be the element that goes 1a = a1 = a
smh
static yew
What if I only have left and right 1s
crystal turtle
Cope
coral shale
then those rnt 1s are they ...
static yew
Cuz multiplication is noncommutative?
crystal turtle
Bros just making up arguments now
coral spindle
But then, what does it mean to contain 1?
The set itself {0} does not contain 1...
It means there is a multiplicative identity
static yew
I don't actually care that much. None of the rings I work with have separate left and right 1s
Crud. Gotta go
coral shale
I don't actually care that much. None of the rings I work with have separate lef...
GL_n
well ig not
but cmon smh
teal vessel
Given that they just gave this as a definition at the beginning of te section, I'm assuming they want me to prove this as a special case of a kernal, right?
oh no wait, those elements are in reverse order, nvm
cool, not hard, I just suffer from can't-read disease
I've heard it's similar to allergic-to-going-on-islands disease
just left multiply g inverse and right multiply g onto a, and then recognize that for all g in the centralizer, gag^-1=a, so make the replacement, and get (g^-1)a(g)=(g^-1)(gag^-1)(g)=a, therefore the centralizer may be described with either order.
teal vessel
that moment when you have to break down the definition in words to make sense of it:
"the group of all elements in G that commute with (the set of all elements that commute with all of G)"
seems quite obvious when put that way.
in other words, every member of Z(G) is commutative with any element of G, therefore, the set of elements in G that commute with members of Z(G) must be every member thereof.
coral spindle
What if I only have left and right 1s
Suppose there is a left unit a and a right unit b. Think about the product ab. What do you conclude?
static yew
I conclude that either a=b, or one of the two does not exist (i.e. a ring can have at most one multiplicative identity, be it right, left, or both)
wait a minute
that means that salesman who was offering a spear that could pierce any armor and a shield that could block any attack... 🤔
crystal turtle
huh
teal vessel
we're talking rings, right? isn't it the case that, supposing there is a left identity but not a right one (using the jargon loosely), then what would the right inverse of the left identity be?
oh, wait, mult. inverses needn't exist in rings
I'm dumb
should probably wait until I get there
coral spindle
I conclude that *either* a=b, or one of the two does not exist (i.e. a ring can ...
Correct!
static yew
oh, wait, mult. inverses needn't exist in rings
you want multiplicative inverses, go find yourself a nice field
if you want cyclic resign yourself to prime powers
teal vessel
in proving that for any A⊆B⊆G where G is a group, C(B)≤C(A); is it sufficient to show that C(B)⊆C(A) and then rely on the fact that centralizers are both subgroups of G to get the rest of the way? I don't think there's any way for a group to be a subset of another, and grouped under the same operation, and yet not be a subgroup. There's no extra elements that could break closure or inversion, and if it's a group under the same operation, identity must be in there.
static yew
I don't think there's any way for a group to be a subset of another, and grouped under the same operation, and yet not be a subgroup.
Isn't that literally the definition of a subgroup?
- the set is a subset of the parent group's set
- the group operation is the same
- it is a group (i.e. closed)
teal vessel
I'm paranoid about my own fallibility, as evidenced by thinking about multiplicative inverses in rings.
I also prefer to think out loud where I can get bullied when I'm wrong.
crystal turtle
yeah that works
teal vessel
Like I always say: I'm stupid, just fast
if you brute force all the ways you can fail faster than it takes to succeed the first time, then is there really a difference?
static yew
okay that's because because, unless the ring is a field, you've got a bit of a twist going on:
- there is a "multiplicative group" which is the subset of elements that, under the * operator, form a group (closed, identity, all elements invertible)
- That subset of elements is not necessarily closed under addition
- That subset of elements is (unless it's a field) a strict subset of R\{0}
teal vessel
wait a second.... my memory banks don't have enough threads..... the symmetric group 3 is isomorphic to the dihedral group of order 6.... so I can check centralizers via generators instead of by hand ...! that makes calculating the centralizers of every subset of S3 much easier.
oh, my can't-read disease is showing up again. It's just saying "each element" not "each subset"
I was about to say, 64 checks would be quite tedious to do by hand, even if I were separating into cases.
teal vessel
oh, interesting thought. Shouldn't the centralizer of any subset H of G be the intersection of the centralizers of all elements of H?
i.e. the set of all elements that commute with h1 AND h2 AND h3 etc.
formal ermine
yeah
teal vessel
in that case C_s3(s3)={1} because that's the only element in all single-element centralizers.
rocky cloak
in that case C_s3(s3)={1} because that's the only element in all single-element ...
In general C_G(G) is just the center yeah
teal vessel
Yeah, I was just keeping notation across the mini lemma
summer path
would it be correct to say that in the inner case you have H some somegroup and N some normal subgroup such that H n N = e and HN = G, and then you can give the natural action H ⤵️ N to get the map H --> Aut(N)
but then in the outer case you remove the requirement for N to be normal and just have H,N any (sub?)groups, in addition to some hom H --> Aut(N), and you define some new group N x|_phi H using the map, which need not be G? Though if H,N need not live inside the same big G, then it doesn't really have any reason to produce a semidirect product equal to G?
dim widget
@summer path I always find the distinction pretty superficial
The only difference between the inner and outer picture is whether you think of the two groups as being “a priori” embedded in some G or after the fact.
In the second case you could set G = N x| H and then it would be an instance of the first case
unkempt stream
I still don't quite comprehend the two different "views" of Galois Theory
or specifically galois extensions
One important fact is that [F : F^G] = |G| for a finite subgroup G of F's automorphism group, and F^G is the fixed group. I'm pretty sure this holds for all characteristics
But the special case where the largest field being fixed by G is the one it stabilizes means it's a galois extension
which implies it's seperable & normal, but I don't quite see how those automorphic properties imply these two properties
formal ermine
wat
unkempt stream
An extension A over B is Galois iff B is the largest subfield of A fixed by the automorphisms of A that fix B
but also this same property implies the extension is separable and normal
idk that direction
warm ember
how do you prove this
south patrol
What is (1)
warm ember
root extension
dim widget
What is (1)
It's the ideal generated by (1) in K
warm ember
bruh no
south patrol
Ah ok
warm ember
ye
dim widget
They're saying K as in (1) to mean that K is the same as (1)
warm ember
this one
dim widget
pretty basic algebra
warm ember
where each extension is a root extension
dim widget
bc K is a field
warm ember
bruh
south patrol
Okay I assumed there was valuable context cause the lemma statement uses notation which we couldn't see
formal ermine
pls halp this is my alt
what's your main
red quartz
ye this is the context sorry
this
rocky cloak
idk that direction
Let G be a group of automorphisms of A, and B = A^G. Consider a in A, and let a1, a2, .., an be the elements in the orbit of a. Then a is a root of (x - a1)...(x - an). Notice that that action of G just permutes the factors, so this polynomial is fixed by G, and hence had coefficients in B. Since this polynomial is seperable and a was arbitrary A is seperable.
Save argument also shows normality.
crystal turtle
Was it supposed to be F_0=F
crystal turtle
Both ends cannot be K, everything in the chain would be K then
red quartz
o well i didnt make this handout
F=K_0\subseteq K_1\subseteq...\subseteq K_s=K
i think it shoudl be this
dim widget
But what are the properties of the K_i
are they just random extensions
can they all be equal?
Are they, perchance, generated by adjoining a single root of an element in K_{i-1}?
dim widget
What is n? Are n and a indepentent of i?
red quartz
you add an nth root each time
n_i
tribal moss
If there are finitely many of them, you can always choose n to be the same each time. :-p
dim widget
wiat what im confoos
this is clear
red quartz
ok i found the proof in dummit and foote
round hull
d/f slash fiction
frigid lark
In part b is n divisible by 2 distinct primes?
Cause if n = pqs, for primes p and q, I can show that both p and q are in (1 - zeta), the ideal generated by 1 - zeta, and if p and q are distinct, I am done.
lethal dune
wym by both p and q are in 1 - ζ
frigid lark
they are in the ideal generated by 1 - zeta
lethal dune
how so?
frigid lark
say a is a primitive p'th root of unity, then a = z^k for k a multiple of n/p. But then (1-a)/(1-z) = (z^k - 1)/(z - 1) = 1 + z + ... + z^(k-1), is in Z[zeta]. Thus 1 - a is in (1 - z). This applies to all primitive p'th roots of unity, taking the product of (1 - a) over all p'th roots of unity yields p
Last claim is given by excercise 19 of Lang ch6
also z is zeta, cause I am lazy
rocky cloak
In part b is n divisible by 2 distinct primes?
My calculations suggests it's not true for n=9, so probably 2 distinct primes yeah
frigid lark
Oh thanks for that
delicate bloom
I think there's a nice proof for the general case
something like, since $\Phi_{p^k}(x)=\Phi_p(x^{p^{k-1}})$ we have $\Phi_{p^k}(1)=p$ and that makes $$p=\prod_\zeta (1-\zeta)$$ and then thinking about the prime ideal generated by $1-\zeta$ basically breaks it from being a unit
cloud walrusBOT
Merosity
unkempt stream
Let G be a group of automorphisms of A, and B = A^G. Consider a in A, and let a1...
is the group action on the roots faithful?
i also don’t know how seperability is involved here in terms of the action
rocky cloak
is the group action on the roots faithful?
No, the group action need not be faithful. But you don't take the product over the elements of the group, just the elements of the orbit. So the roots are distinct by construction
frigid lark
Is this for a splitting field of a polynomial?
rocky cloak
That is what you will end up constructing yes
frigid lark
Oh rip, I misread faithful as transitive
I was gonna say the action has to be transitive
unkempt stream
though I haven’t looked much into positive characteristic fields
rocky cloak
Well nothing here depends on the characteristic
unkempt stream
Let G be a group of automorphisms of A, and B = A^G. Consider a in A, and let a1...
isn’t this just Artin’s Lemma
frigid lark
No, the group action need not be faithful. But you don't take the product over t...
Wait, doesn't it need to be faithful, cause if s is an automorphism of the splitting field of a polynomial, and s acts trivially on the roots, then s will act trivially on the splitting field, apply the Galois correspondence, the dimension of the splitting field is finite, we see that s must be the identity
rocky cloak
Wait, doesn't it need to be faithful, cause if s is an automorphism of the split...
The action on A is obviously faithful, since G is a subgroup of the automorphism group. But the action on the orbit of a specific element a need not be
rocky cloak
isn’t this just Artin’s Lemma
Don't think so. Artins lemma says [A:B] <= |G|, right?
unkempt stream
i suppose that could be used for a proof that for finite G [F :F^G] = |G|
I thought it’s a strict equality
rocky cloak
Yeah, the strict equality is true
frigid lark
The action on A is obviously faithful, since G is a subgroup of the automorphism...
Yeah, I didn't read the construction too specifically
rocky cloak
Don't know if people also call that Artins lemma
unkempt stream
i wondered if I could prove it using the polynomial ring F[X_1…X_|G|] more simply than the crazy linear alg argument that Artin provides
and how we can extend the automorphisms into the polynomial ring by fixing the indeterminates or by permuting them
rocky cloak
i wondered if I could prove it using the polynomial ring F[X_1…X_|G|] more simpl...
I guess if you can prove that F/F^G has a primitive element, then it follows by just noting that G is determined by where said primitive element is mapped
unkempt stream
I guess if you can prove that F/F^G has a primitive element, then it follows by ...
how do I prove it has a primative element
rocky cloak
Yeah, maybe that is just as tricky
rustic crown
you have to use something tricky at some point, the two possible routes are with artin's lemma or with the primitive element theorem.
you've already proven that F/F^G is separable and normal, and also that any alpha in F has [F^G(alpha) : F^G] <= |G|
so take any intermediate field E/F^G which is finite, it's also separable, so by primitive element theorem E = F^G(alpha) and thus [E:F^G] <= |G|
this has to imply [F:F^G] <= |G|
as any finitely generated (algebraic) extension is actually finite
yea so since F/F^G itself is finite, F = F^G(alpha) and now
G is contained in Gal(F/F^G) which has size equal to number of distinct roots in F of minimal poly of alpha over F^G. by separable and normal this is same as [F:F^G]
therefore |G| = [F:F^G]
and Gal(F/F^G) = G
det hasn't thought much about the linear algebra trickery that goes into Artin's lemma
but the proof of primitive element theorem can be made a lot cuter and feel less like a trick once you introduce separable degree
frigid lark
Primitive element theorem, best theorem
rustic crown
it's a simple theorem 
frigid lark
I've never seen simple be used in an algebraic context
frigid lark
Oh, never heard of that
south patrol
Simple extension, simple module, simple group
south patrol
simple-icial set
frigid lark
Is there interesting stuff with simple extensions?
Cause in (finite) Galois stuff there probably isn't any point
Cause everything is
rustic crown
idk you introduce it at the very start cause they're the easiest to deal with
and a decent bit of work is required until you finally prove finite+separable=simple
so it deserves a name
frigid lark
Well Lang has reserved the word simple to condescend his readers
Okay that's enough
unkempt stream
and Gal(F/F^G) = G
pain
unkempt stream
det hasn't thought much about the linear algebra trickery that goes into Artin's...
it’s honestly more intuitive than that approach
teal vessel
new dumb realization of the day: when computing single element centralizers, the centralizer of an element is the set of all elements whose centralizers include it. i.e. C_D8(s)={1,s,sr^2,r^2} which means that s is in the centralizers of 1 (obviously), itself (again trivial), sr^2, and r^2.
this is in essence saying "the set of all elements which commute with this one is the same as the set of all elements who's list of commuting elements include this one"
delicate orchid
if x commutes with y then y commutes with x yes