they're the smallest field extenstion such that all the roots of a given polynomial are contained in it
#groups-rings-fields
delicate orchid
equivalently, it's the smallest field extenstion such that the polynomial factors into linear thingies
rocky cloak
Alright, so thanks to tropos we assume f(1) = 1.
Then we want to show f(z) = z^n . So if f(z) = g(z)/h(z) we want to show
g(z) - z^n h(z) is identically 0.
This is a polynomial of degree say m, so it's enough to show that it vanishes on m+1 points.
Let w be a primitive pth root, with p big enough so that the field generated by the coefficients of f doesn't contain primitive pth roots of unity, and there are more than m+1 primitive pth roots.
Assume f(w) = w^n (can we prove this).
And let s be a Galois automorphism permuting pth roots of unity, say s(w) = w^k. Then f(s(w)) = sf(w) = (w^n)^k = (w^k)^n . So f equals z^n on more than m+1 points, hence they are equal.
@frigid lark
somber sleet
AHHHHHH okay, I get it
Okay this was really easy
That explanation was fire, thx again
delicate orchid
yeah everything is easy once you understand it - it's understanding it that's hard! 
uneven fossil
mr robot is a great show
hasty lance
If more context, definitions are needed I will provide
delicate orchid
just include the left/right thirds of the text
tribal moss
Alright, so thanks to tropos we assume f(1) = 1.
Then we want to show f(z) = z^...
Where do you get n from? I've argued that if there is an n at all, it has to equal f'(1), but to begin with we don't know such an n even exists, or for that matter that f'(1) s an integer.
hasty lance
just include the left/right thirds of the text
Sure, will send now
wooden ember
while you're here tropos is it at all useful?
coral spindle
If you adjoin an algebraic element you get a finite extension. If you repeat this finitely many times, you still get a finite extension. A splitting field of a polynomial is formed by adjoining all of the finitely many roots, hence a finite extension.
hasty lance
tribal moss
while you're here tropos is it at all useful?
I've been trying to figure out an argument along those lines, but I can't quite wrap my head around it yet.
rocky cloak
Where do you get n from? I've argued that if there is an n at all, it has to equ...
Yeah, shoot. n depends on m and m depends on n...
tender wharf
stop spamming my god
delicate orchid
tropo can you do something about this guy
tepid tartan
Dont spam pls
coral spindle
Cmon you're encouraging this thing
tender wharf
not even abstract algebra...
tepid tartan
Cmon you're encouraging this thing
My bad
tribal moss
Muted while I figure out if there are any extenuating circumstances.
delicate orchid
ok, did you mis-calculate the sign of one of the permutations?
that's the only way I can see a numerical error coming into play
hasty lance
Oh, I'll send what I've done.
wooden ember
I've been trying to figure out an argument along those lines, but I can't quite ...
cause now that we know this we get f(w)=w^n(w) for each root of unity where n(w) is an integer depending on w no?
hasty lance
I didn't write it all out explicitly, I tried being clever 
delicate orchid
a grave blunder
this is cool though I've never really thought about the exterior algebra like this
elder wave
this is a good addition to my low res reaction pic collection
wooden ember
this is cool though I've never really thought about the exterior algebra like th...
i prefer to do algebra inside
delicate orchid
no problem boss
delicate orchid
I agree with 2pi_4(f_i (x) blah blah blah (x) f_l) so far
oh I see I slightly misunderstood the multiplication
is f_i (x) f_j (x) f_k = D_ijk a standard thing?
hasty lance
Yep, that's the determinant function
tribal moss
cause now that we know this we get f(w)=w^n(w) for each root of unity where n(w)...
Oh, especially if we can show that n(w) is less than the order of w when the latter is high enouh.
wooden ember
well you can choose it to be no?
tribal moss
Ah yes of course.
If the existence of n(w) holds in the first place, which still feels slippery to me.
delicate orchid
ngl you've kind of convinced me it should be 12
delicate orchid
did you spot it
delicate orchid
LETS FUCKIN GOOOOOO
I bet they forgot the 2 infront of the D_ijk that's my fan theory
hasty lance
Lol ye
delicate orchid
also with this much errata maybe find a different source lol
hasty lance
I dunno if they released a newer edition
I have the physical copy back home, but I still remember there being a 6 in that
tribal moss
I follow you as far as [K(f(w)) : K] divides [K(w) : K] which is at most the order of w, but why can't the order of f(w) be larger than that?
wooden ember
gah i see your point
darn
can't even get a nice rational power from that cause n/phi(n) -> infty if im not mistaken (tbf rational powers would be dodgy anyways if we're working all around the unit circle)
tribal moss
Hmm, but we can conclude that the order of f(w) is at most the order of w, times the highest order of a root of unity in K.
rocky cloak
Wait, so if n(w) exists, we can choose n(w) such that |n(w)| < the order of w divided by 2 (i.e. letting it be negative half the time). Then if one can find p such that phi(p) > p/2 + m, where m is the maximal degree of g and h in f = g/h.
Then my counting argument should go through.
wooden ember
Hmm, but we _can_ conclude that the order of f(w) is at most the order of w, _ti...
could you elaborate a bit?
tribal moss
I thought I could conclude that, but now it eludes me again. :sigh:
tribal moss
Wait, so if n(w) exists, we can choose n(w) such that |n(w)| < the order of w di...
It's not clear to me we have an argument for n(w) existing yet.
Can we prove in general that if L/K is an extension and x in L is a root of unity, then the minimal polynomial of x over K is of the form x^k = a?
wooden ember
Can we prove in general that if L/K is an extension and x in L is a root of unit...
this isn't true though
just take L=Q
i mean K=Q
tribal moss
Oh right, there are easy counterexamples to that.
rocky cloak
Okay, how about this. It's it true that the only roots of unity in K(w) are of the form aw^k where a is root of unity in K? If so, you have
f(w)/a = w^k
And the argument would go through I think
wooden ember
we do have phi(o(f(w))) <= phi(o(w)) so surely we should be able to say they arent too far from each other
wooden ember
Okay, how about this. It's it true that the only roots of unity in K(w) are of t...
i dont think this works: Q(i,sqrt(3))/Q(sqrt(3)) has a primitive 8th root of unity
tribal moss
6th, you mean?
rocky cloak
I see, I'm out of ideas then
Maybe someone actually has to read this book on Diophantine geometry that is referenced in the problem
idle roost
What was the original problem?
tribal moss
Here's the original problem.
wooden ember
we're probably missing a simple step
i mean tbf it's a galois theory textbook and weve barely used any
somber sleet
guys I'm so confused, if I have a field of characteristic p, why does it follow that the the polynomial ring over the algebraic closure has characteristic 0
rocky cloak
guys I'm so confused, if I have a field of characteristic p, why does it follow ...
It has characteristic p, so it doesn't follow at all.
somber sleet
yyeah right, I thought it too
I do not understand a proof we did in the lesson and it is in german
ebon heron
Anyone needing help can reach my dm
formal ermine
I do not understand a proof we did in the lesson and it is in german
send it
somber sleet
it's about the statement that a field of characteristic p>0 is perfect iffthe frobenius endomorphism is bijective
this is the proof, and it's exactly what my professor said since I listened to the lecture again which is registered
formal ermine
Anyone needing help can reach my dm
that's not how this server works
what is your definition of a perfect field
somber sleet
when every irreducible polynomial over K is separable
formal ermine
what part are you confused about
somber sleet
in the 4th row of the "=>"-direction
the first implication
that K has characteristic p implies that in (K bar)[X] p=0
oh no wait, this does exactly mean that it has characteristic p right?
rocky cloak
Yes
somber sleet
omg, these mistakes...................
somber sleet
I was so confused for a moment haha, thank you guys
it's always these kind of things which make me struggle
somber sleet
another easy question for you: why is every injective homomorphism f:K->K, where K is a finite field, surjective?
formal ermine
another easy question for you: why is every injective homomorphism f:K->K, where...
every field homomorphism is injective btw
chilly ocean
what can you say about an injective function from a finite set to itself
chilly ocean
every injective function from a finite set to itself is ||also surjective||
coral spindle
||This fact characterises finite sets btw. A set X is finite when every map f : X → X is surjective iff it is injective. I just really like this fact||
chilly ocean
every field homomorphism is injective btw
ideal fact
delicate orchid
ideal fact
moderators ban this individual
fervent gyro
I know that being a finitely generated module isn't a local property in general, but is there a name for a ring such that it is the case? Have rings with such a property been studied? I ask because that would cause the property of having finite global dimension to be a local property too
coral spindle
When you say 'local' do you mean that it is inhereted by its submodules?
A ring for which every f.g. module has only f.g. submodules is Noetherian.
hidden haven
local usually means inherited by all localizations at primes I think
coral spindle
Feel free to ignore me then lol
delicate orchid
local usually means inherited by all localizations at primes I think
finally, somebody ACTUALLY explains what local means
formal ermine
local usually means inherited by all localizations at primes I think
NOO
why< did y<ou tell him
you are supposed to gatekeep it
delicate orchid
tbf that only accounts for like
formal ermine
$10 if someone can name a local property that does not hold for maximal localizations
delicate orchid
one of 1266 different usages of the word
formal ermine
$10 if someone can name a local property that does not hold for maximal localiza...
as in like
hidden haven
you are supposed to gatekeep it
Did you consider that I might have intentionally given him wrong information 
formal ermine
most local properties are A has P iff A_p has P for every p iff A_m has P for every m
I have yet to see one where the last iff is false
delicate orchid
a similar idea holds for groups btw, in my area we deem something a "local" property if it can be gleamed by looking at some \pi-elements for some set of primes \pi
formal ermine
🤓
delicate orchid
most local properties are A has P iff A_p has P for every p iff A_m has P for ev...
you have no fuckin right to nerd react me after posting this
delicate orchid
that's standard for a set of primes 
formal ermine
JUST SAY "pi" LIKE A NORMAL HUMAN
formal ermine
I feel offended
delicate orchid
that would be because I offended you
delicate orchid
14 years of age... grandpa..
formal ermine
ok time to actually do math
I've been trying to do math for the past like 2 weeks but video games are too distracting
toric geometry is killing me
celest furnace
I've been trying to do math for the past like 2 weeks but video games are too di...
Which one
chilly ocean
I know that being a finitely generated module isn't a local property in general,...
Being finitely generated is local tho?
celest furnace
hypixel skyblock
So true
formal ermine
my friend let me borrow his hyperion so I've just been doing f7 runs without an end
chilly ocean
Or what is your definition of local?
celest furnace
Hyperion? What’s that?
formal ermine
the best weapon in the game
it's a sword that lets you teleport forward and make a big explosion
it's worth 2.1 billion coins
the best player right now has like 100 billion
fervent gyro
I mean local in the sense that a property being true for a module M is equivalent to it being true for all localizations of that module, and also equivalent to all localizations in the primes, or in the maximals (viewed respectively as modules over the localized rings)
formal ermine
this
formal ermine
oh no timo
elder wave
most local properties are A has P iff A_p has P for every p iff A_m has P for ev...
I do know a counterexample to this
formal ermine
moderator moment 🤓
formal ermine
krull?
elder wave
yes
formal ermine
interesting
elder wave
obv it's true for maximal ideals because it only has one
but fails when localising at (0)
because that's the fraction field and has dimension 0
formal ermine
but that's not a local property then no?
formal ermine
yeah but I'm looking for a local property that is not maximal local
chilly ocean
huh
formal ermine
I'm looking for an A and P such that A_m has P for every m does not imply that A has P
and that P is a local property of A
chilly ocean
That's not gonna work
elder wave
when taking A_p has P ... as definition of local
formal ermine
why?
chilly ocean
As that's the definition of a local property
formal ermine
m is a maximal ideal
elder wave
yes
chilly ocean
yes
elder wave
formal ermine
so if A has P <=> A_p has P for every prime ideal p, we can always conclude that A_p has P for every prime ideal p <=> A_m has P for every maximal ideal m?
wait no
that's phrased wrong
uh
let me think
chilly ocean
Oh
formal ermine
like
idk how to phrase this properly
most local properties are: A has P iff A_p has P iff A_m has P
what I'm asking is if we can just remove the last iff because it will always be true anyway
I think no but I'm looking for a counterexample
elder wave
you're asking for a property s.t. A_p has P <=> A has P but A_m has P is not equivalent to A having P?
chilly ocean
Yes that's equivalent if I'm not stupid
elder wave
it should be by looking at the maximal ideal that contains p
chilly ocean
Because if R has P for all maximal ideals, then Rm has P and then localise that at p and you got Rp
formal ermine
wait why is (R_m)_p = R_p?
chilly radish
wait why is (R_m)_p = R_p?
Properties of tensor products
formal ermine
wut
hidden haven
Maps from R_m_p to S are maps from R_m to S that map things outside p_m to units which are maps from R to S that map things outside both p and m to units
chilly radish
If you localise twice you're not gaining any nrw information
hidden haven
But things outside p and m are the same as things outside p
chilly radish
Just use the universal property to get a map and show it's S
an iso
formal ermine
Because if R has P for all maximal ideals, then Rm has P and then localise that ...
but how do we know that R_p has P
hidden haven
So homs out of both things are isomorphic functors
elder wave
but how do we know that R_p has P
that's what you asked it to
chilly radish
So homs out of both things are isomorphic functors
Me when the yoneda embedding reflects my isomorphism
hidden haven
Maps from R_m_p to S are maps from R_m to S that map things outside p_m to units...
Btw this same proof works for the third isomorphism theorem
formal ermine
Just use the universal property to get a map and show it's S
an iso
what is S
chilly radish
A typo
delicate orchid
lets draw the triangle
cloud walrusBOT
wew ladz
delicate orchid
yup this looks like a universal property to me
chilly radish
doesn't it just characterize j
The definition of the localisation needs the information of S^-1R AND j
You need to know both who your ring is and how R relates to it
chilly radish
Ok but your ring embeds inside a certain way
delicate orchid
yup this looks like a universal property to me
to show this characterises S^-1R we do the standard thing of setting T to be some different localisation (S^-1R)' and then the two maps induced by the universal properties must be inverses and you get that the localisation is an isomorphism
chilly radish
If you embed it differently then it won't act like you want it to
You want your ring to embed in such a way that respects the ring operations
Among other things
(generally there might be different embeddings tho)
formal ermine
hm
chilly radish
But you have to specify who your ring is inside the big ring
So when you do the fraction do you do this implicitly
By setting elements of your ring to be r/1
formal ermine
to show this characterises S^-1R we do the standard thing of setting T to be som...
why must they be inverses
how do you show that
hidden haven
The identity map is makes the triangle
R → S^-1 R
↓
S^-1 R
commute when you put it in the diagonal
chilly radish
All together now
delicate orchid
why must they be inverses
this is a standard argument for universal properties, you always have the identiy map from an object to itself, so if we have two maps like in the pic, they must compose to give the identity as there is only one map from S^-1R to S^-1R. This is assuming the embeddings are the same of course nah
hidden haven
The identity map is makes the triangle
R → S^-1 R
↓
S^-1 R
commute when you put ...
And so do the composites, but such a map has to be unique
chilly ocean
**wew ladz**
how does this work? like what do I have to tell the bot to draw that? do I just need to write latex code and it does that? ;-;
formal ermine
how does this work? like what do I have to tell the bot to draw that? do I just ...
click latex then copy paste the code
formal ermine
ye
chilly ocean
thanks
barren sierra
q.uiver is fire
summer path
im sure there's some people out there that just type it out by remembering the commands
chilly radish
It has some limitations but there's like 50 others similar tools with different limitations that you can also use
formal ermine
ok I think I understand it let me just repeat it to make sure I didn't get anything wrong
in the triangle we have
then we set f = id so j circ g = id and thus they are both isomorphisms?
so R_m is isomorphic to (R_m)_p
how wrong is that
formal ermine
ah frick
delicate orchid
the composition of the straight dotted maps must be equal to the curved one because there is a unique map from S^{-1}R to itself by the universal property
fuck I drew them backwards oh well
it's symmetric
you can apply this argument to anything that has a universal property
formal ermine
fuck I drew them backwards oh well
the bottom one is supposed to be the other way around?
delicate orchid
yeah imagine the dotted arrows are going the other way - this diagram is still correct it's just upside down
formal ermine
both?
delicate orchid
all three if you want
I don't really know how to explain it in any simpler terms
uhhh
formal ermine
why must g circ g' = id
delicate orchid
because, by the universal property
chilly ocean
because it's unique
wraith cargo
couldn't it be any isomorphism?
delicate orchid
so by uniqueness they must be the same map
formal ermine
but the uni prop just says something about the triangle
not about other stuff you put in no?
delicate orchid
I just joined two triangles together 
wraith cargo
sounds like goddamn category theory to me
formal ermine
I feel stupid right now
wraith cargo
I feel stupid right now
bruh I'm reading abt sets rn and I feel stupid af so I know ur pain
chilly ocean
The identity map and the composition of the maps make the diagram commute. The universal property says there is exactly one map which makes the diagram commute
wraith cargo
ZF didn't know dick why did they come up with so many axioms
formal ermine
The identity map and the composition of the maps make the diagram commute. The u...
but what diagram? doesn't the universal property just talk about one single triangle and not that whole diagram?
chilly ocean
Yes, apply the universal property to the outer triangle
formal ermine
ahhhhh
delicate orchid
the sinister third triangle
formal ermine
okay that's what I was missing
I don't see how we can use this to show that R_m_p = R_p
delicate orchid
no clue
wraith cargo
I don't see how we can use this to show that R_m_p = R_p
note that all the non units in R_m_p are those contained in p and same story for R_p
I think you can use a similar argument to show that the units are all the same
this is a rough idea but I think it would work sorta
without any fancy machinery
chilly ocean
white oxide
how can I see that $F(y_1, \dots y_n)$ is separable over $E$?
cloud walrusBOT
okeyokay (analysis is cool)
frigid lark
Are all the y_i distinct?
If so you have a polynomial, specifically f in E[X] which each y_i is a root of, and that f has no double roots
Hence each generator of F(y_1,...,y_n) is seperable
Over E
Hence the extension is seperable
frigid lark
If so you have a polynomial, specifically f in E[X] which each y_i is a root of,...
Hence for each y_i, Irr(E, y_i, X) won't have double roots
rocky cloak
Presumably f is assumed to be seperable right before this
chilly ocean
Are all the y_i distinct?
yeah
white oxide
Hence each generator of F(y_1,...,y_n) is seperable
oh i thought you had to show that every a in F(y_1, ..., y_n) is separable
but i suppose the y_i generate that field so that makes sense
i guess i should try to prove that formally then lol
frigid lark
Try first with one generator
And then apply seduction induction
Well that argument would only hold for finite field extensions
white oxide
oh yea ig applying seduction makes sense
ye
my guess is that i would have to use the homomorphism property
frigid lark
Have you defined the separable degree of an extension?
white oxide
no i don't think so
frigid lark
How was a seperable extension defined?
white oxide
they only defined it for finite extensions, but if ${E: F} = [E: F]$
cloud walrusBOT
okeyokay (analysis is cool)
white oxide
where ${E: F}$ is the number of isomorphisms of $E$ onto a subfield of the algebraic closure of $F$ fixing $F$
cloud walrusBOT
okeyokay (analysis is cool)
formal ermine
to hopefully drop a handle
OMG GUYS I DROPPED SOMETHING THAT'S LIKE 400 MIL AND HANDLE IS 600 MIL
LET'S GO
IT'S 0.01% CHANCE
frigid lark
Alr, good enough, you can also consider {E : F} to be the number of embeddings of E into some algebraic closure of F that fix F
void cosmos
beat your wife to it
frigid lark
Try first with one generator
Spoiler for this ||if a generator, b, is seperable, let f = Irr(F, b, X), then you can embed F(b) into F^a over F by sending b to any root of f||
white oxide
Spoiler for this ||if a generator, b, is seperable, let f = Irr(F, b, X), then y...
oh yea im def gonna come back to this an exercise just doing some reading rn
frigid lark
Okeyokay
void cosmos
why is every submodule now free of rank 0 or 1
wouldnt this imply the Gi+1/Gis are free
why does it follow for every other ssubmodule
ivory trail
finally, somebody ACTUALLY explains what local means
actually locally free and stalkwise free are different
void cosmos
for notation F_i is just the submodule generated by all x_j such that j < i
x_i is the basis for the free module F and G is a submodule and we want to show that G is free aswell
given the ring is a PID
delicate orchid
actually locally free and stalkwise free are different
chilly ocean
yes but that's not a local property
ivory trail
and then they're related to projective in some annoying way
white oxide
why is every submodule now free of rank 0 or 1
ayy is this hungy
void cosmos
yea it just appeals to the noobies like me
and the exercisesa re not terribly hard
anyways can u help with the proof?
white oxide
oh no i just got to section 2 of that chapter i believe lol
(just learned about exact sequences from section 1)
srry
ivory trail
yes but that's not a local property
maybe not in the sense of "local property" that most local things are, but "locally free" has the word "local" in it
chilly ocean
yes but that's a different local
ivory trail
and @delicate orchid said "what local means" not "local property" so i'm just trying to be clear
because i spent like half an hour on deciphering this shit myself
chilly ocean
Yes that's what local means in algebra
ivory trail
that's what "local property" means in commutative algebra
chilly ocean
Yes
ivory trail
but "local" can mean a bunch of different things
chilly ocean
Yes...
chilly ocean
But context clearly matters
ivory trail
so what is the issue?
chilly ocean
idk u tell me ;-;
barren sierra
dumb queston, but subgroups of imprimitive groups are imprimitive right?
It seems true to me
like just from the other angle
if a permutation group is primitive, then none of it's elements can fix a non-trivial partition
void cosmos
why is every submodule now free of rank 0 or 1
nvm this turns out to be stupid
celest cairn
Let $\zeta_{3}$ be a primative 3rd root of unity. I was using the fact that $\Phi_{3}(x) = x^2 + x + 1$ in order to find the minimal polynomial of $2 + \zeta_3$ over $\mathbb{Q}$, and got $a^2 - 3a + 3 = 0$. Is this correct?
cloud walrusBOT
Sapphire
glossy crag
Let $\zeta_{3}$ be a primative 3rd root of unity. I was using the fact that $\Ph...
The minimal polynomial will be \Phi_3(x-2), which I can't be bothered to calculate tbh.
Yes, looks like it.
If you require justification, x\mapsto x-2 is an automorphism of Q[x], therefore the polynomial \Phi_3(x-2) is irreducible. It is also monic and has \zeta_3+2 as a root, therefore it is truly the minimal polynomial of \zeta_3+2 over Q.
The same procedure works for any algebraic element over any field for exactly the same reason: if f(x) is the minimal polynomial of \alpha over K, then 1/c^nf((x-d)/c) is the minimal polynomial of c\alpha+d over K.
frigid lark
Okay, how about this. It's it true that the only roots of unity in K(w) are of t...
This should be true. Let n be maximal such that $\zeta_n$ is a primitive n'th root of unity in K. And let w be a primitive m'th root of unity. Write n = da, m = db, where d is the gcd of n and m, and a,b are coprime. Then WLOG let $\zeta_n = \zeta_{dab}^b$ and $w = \zeta_{dab}^a$. Then there exists some integers, r,s such that rb + sa = 1. Hence $\zeta_n^rw^s = \zeta_{dab}$. Hence $K(w) = K(\zeta_{dab})$, thus all of the roots of unity in $K(w)$ are contained in $\mathbb{Q}(\zeta_{dab})$, and we just showed that we can write any root of unity in $\mathbb{Q}(\zeta_{dab})$ say $\zeta_{dab}^c$ as $\zeta_n^{rc}w^{sc}$
cloud walrusBOT
parrottea
loud frost
hello! does anyone know of a good source for solutions for pinter's abstract algebra?
tribal moss
This should be true. Let n be maximal such that $\zeta_n$ is a primitive n'th ro...
Narwhal pointed out a counterexample to
all of the roots of unity in K(w) are contained in Q(zeta_dab)
namely K=Q(sqrt3), and w=i
which gives n=2, m=4, dab=4, zeta_dab=±i, but K(i) contains all the 12th roots of unity.
frigid lark
Okay, I see the last conclusion is wrong
tribal moss
I think your argument shows that zeta_dab is in K(w), but not that it generates K(w).
Wait, perhaps it does.
frigid lark
No K(w) is contained in K(zeta_dab), but not all roots of unity in K(w) are of the form zeta_dab^s
i.e. not all roots of unity in K(w) are contained in Q(zeta_dab)
delicate bloom
we still on the same problem from earlier
tribal moss
Yes.
delicate bloom
my strategy of napping and hoping someone else solves it before I wake up didn't work so well
frigid lark
same
tribal moss
Narwhal also pointed out that we probably haven't applied enough Galois theory for the exercise to be about that.
delicate bloom
what book is this from
frigid lark
Lang
delicate bloom
oh ok, cool I have that, maybe skimming the section will give some ideas for tools to throw at it at least
frigid lark
Maybe
I tried
It doesn't smell like anything present in chapter 6
Maybe something to do with characters
but that's pushing it
tribal moss
Can we find some property of K that allows us to bound the ratio between "highest root of unity in K(w)" and the order of w?
frigid lark
We could get a not so useful bound just by considering [K(w) : Q]
Then the highest root of unity, say zeta_n cannot have phi(n) > [K(w) : Q]
then as phi(n) > sqrt(n/2), we have n < 2[K(w) : Q]^2
tribal moss
What I hoped for was a bound of n/ord(w) that didn't depend on w.
But I've no idea where to look for that.
delicate bloom
I might have an idea
since we know |f(z)|=1 when |z|=1, I think that should mean the derivative of g(t)=|f(e^it)| should be 0
which might just be like looking at f'(z)=0 when |z|=1
so if we write $f(z)=c \frac{\prod_i (z-\alpha_i)}{\prod_j (z-\beta_j)}$ and do logarithmic differentiation it'll force a condition on what the roots/poles have to be
cloud walrusBOT
merosity
delicate bloom
we already know the alpha and beta can't be roots of unity, so they must lie inside or outside the circle
carrying out the differentiation I think gets us $$\sum_i \frac{1}{z-\alpha_i} = \sum_j \frac{1}{z-\beta_j}$$
cloud walrusBOT
merosity
delicate bloom
for all |z|=1
whether this is right or wrong, I don't think this will be enough either since we have some f(z) earlier that map the unit circle to itself that weren't wz^n
might be something much simpler than this actually, almost falls out by symmetry
if we have $f(z)=c \frac{\prod_i (z-\alpha_i)}{\prod_j (z-\beta_j)}$ then $$1 = |f(z)|=|c|\frac{\prod_i |z-\alpha_i|}{\prod_j |z-\beta_j|}$$ if you look at the individual distances, and take $\alpha_i$ somewhere inside the circle that isn't the center, then $|z-\alpha_i|$ will be lopsided as you move around and will have to be accounted for
cloud walrusBOT
merosity
tribal moss
Does that lead towards an argument that excludes, say, f(z)=2(z-½)/(z-2)?
delicate bloom
whether this is right or wrong, I don't think this will be enough either since w...
that's what I had in mind when I made this comment
tribal moss
Yes, I hoped the post after that somehow suggested a way around the problem.
delicate bloom
it might be enough to narrow it down to f of the form that's a product of (az-b)/(bz-a)
celest lily
Considering Clifford Algebras are generated by a quotient of the tensor algebra and can contain spinors, what exactly defines these spinors as sufficiently "isomorphic" to regular tensors so that they don't exceed the size of the tensor algebra?
rocky cloak
We could get a not so useful bound just by considering [K(w) : Q]
The more general statement is indeed proved in the book Fundamentals of Diophantine geometry as theorem 7.3. I don't have time to read it now, but it is there
frigid lark
thx
coral steeple
I'm not exactly sure what I have to show here. My best guess was to define $d=\prod_{p\mid a,b}p$ and show that $a\mathbb{Z}+b\mathbb{Z}=d\mathbb{Z}$, but I can't figure out how to do this without needing to re-prove the theorem from the book (for all integers a,b there is a unique d so that aZ+bZ=dZ). Am I showing what's being asked for? Is there a better way to do it?
cloud walrusBOT
person2709505
tender wharf
is there a specific way your book defines gcd
coral steeple
I think it says that gcd(a,b)=d (the d in the parentheses in my last comment)
It also gives that d is a common divisor of a,b as a corollary
tender wharf
I think it says that gcd(a,b)=d (the d in the parentheses in my last comment)
hm I believe if you show aZ + bZ = dZ you can then use the book's theorem to conclude that since d is unique it is the gcd
i don't think you'll need to reprove it in this case
coral steeple
I tried doing that but as far as I could tell, showing aZ+bZ=dZ reduces to showing that if p,q are prime then there are integers r,s so that rp+sq=1, which I think is most conveniently shown by proving bezout's identity. Maybe I misunderstood?
(The part I'm stuck on is showing dZ \subseteq aZ+bZ)
coral steeple
That's ${z\in \mathbb{Z}:z=ax+by\textrm{ for some }x,y\in\mathbb{Z}}$
cloud walrusBOT
person2709505
tender wharf
is gcd being a linear combination given to you
or bezout's lemma as it's called
if it is given just go ahead and use it and then you're done
coral steeple
I assume our proof will have to depend on the fundamental thm of arithmetic somehow
tender wharf
sure you can use 2.6 then
coral steeple
So the idea is we show $\prod_{p\mid a,b}p=d$?
cloud walrusBOT
person2709505
tender wharf
yeah pretty much that's your job
frigid lark
The more general statement is indeed proved in the book Fundamentals of Diophant...
For anyone interested the specific proof is given in thm 6.4
wooden ember
it might be enough to narrow it down to f of the form that's a product of (az-b)...
Yeah I was thinking about how to reduce to just working with blaschke products yesterday but I couldn’t get a handle on the poles
frigid lark
For anyone interested the specific proof is given in thm 6.4
oh 6.1 is enough as well
wooden ember
Yeah I was thinking about how to reduce to just working with blaschke products y...
Nvm just realized can’t reduce to this anyways since you could absolutely have poles within the unit circle, but I guess might be able to still say all poles are in the unit disk or outside to reduce to that case up to a reciprocal
One thing we do know is zeroes and poles are related by a circle inversion if I’m not mistaken
Yeah so I think you can kinda do the same trick as with blaschke products to collect poles and zeroes and get an explicit form for f in terms of a product of disk automorphisms to some integer powers
I realize this is probably what y’all were basically already saying
coral ivy
HI
glossy crag
Reposting old question.
glossy crag
Reposting old question.
I've partly solved the first question, but not entirely, would appreciate some help.
If $g:A\otimes_K L\to\operatorname{End}L(V)\subset\operatorname{End}K(V)$ is another isomorphism, then using Skolem Noether ($A,B$ simple $K$-algebras, $A$ central and finite-dimensional, $B$ finite-dimensional, $f,g:B\to A$ homomorphisms $\implies\exists u$ with $g(b)=u^{-1}f(b)u$) we get a unit $s\in\operatorname{End}K(V)^\times$ such that $z^g=s^{-1}z^hs$ for all $z\in A\otimes_KL$. The same applied to $g^{-1},h^{-1}$ (now considered as $L$-algebra homomorphisms because $A\otimes_KL$ is $L$-central) yields a $t\in A\otimes_KL$ with $x^{g^{-1}}=t^{-1}x^{h^{-1}}t$ for all $x\in \operatorname{End}L(V)$. Let $v\sigma$ be the units such that $x^{g^{-1}\sigma g}=v^{-1}\sigma xv\sigma$ for all $x\in\operatorname{End}L(V)$. The map $a\otimes\lambda\mapsto a\otimes\lambda^\sigma$ is a $K$-algebra automorphism, so $t^{\sigma}$ is again a unit, likewise $t^{\sigma h}$. Using Skolem-Noether twice we see $x^{g^{-1}\sigma g}=((t^{-1})^\sigma x^{h^{-1}\sigma}t^\sigma)^{g}=s^{-1}(t^{-1})^{\sigma h}x^{h^{-1}\sigma h}t^{\sigma h}s$ for all $x\in\operatorname{End}L(V)$. The element $r=t^{\sigma h}s$ is a unit, so $v^{-1}\sigma xv\sigma=r^{-1}u^{-1}\sigma xu_\sigma r$, so $u_\sigma r=v_\sigma a_\sigma$ for $a_\sigma\in L^\times$. Now if not for the pesky $r$ I'd be done, is there any way to get rid of it or show $r\in L^\times$?
cloud walrusBOT
leave_no_norm
formal ermine
let m be a maximal ideal and p a prime ideal, why is pm = p
coral spindle
Bet this is nakayama
formal ermine
wait no nvm
chilly ocean
m = 2Z, p = 3Z
coral spindle
Yeah nvm I guess
formal ermine
hmmm
coral spindle
CONTEXT!!!!!!
coral spindle
Is A local
formal ermine
no
coral spindle
Then this is false innit, by a similar example to Terra's
m = 3Z and p = 5Z this time
formal ermine
wait why is (R_m)_p = R_p?
here
coral spindle
Well actually the same example works lmao
formal ermine
that was yesterday
coral spindle
Wise wtf
formal ermine
but didn't we just give a counter example
wraith cargo
but didn't we just give a counter example
u didn't tho
formal ermine
tterra did no?
wraith cargo
I mean I assumed when you asked the question you meant that p is contained in m
coral spindle
CONTEXT!!!!!!
wow
formal ermine
wow
oh
the context was like
we were talking about local properties
and how they're always given as A has P <=> A_p has P for every prime p <=> A_m has P for every maximal m
I asked if the last iff is always true anyway
or if it's only true for some local properties
then nine pointed out that if A has P <=> A_p has P and R has P for all maximals, then R_m has P <=> R_m_p has P and then said that R_m_p = R_p
wraith cargo
note that there's a one to one correspondence of prime ideals of R_m and prime ideals of R that don't intersect R-m
So if you're taking (R_m)_p I assume that p there is actually something interesting in R_m
wraith cargo
well we wanna show it for every p no?
for two maximal ideals (R_m)_m' is 0 when m =/= m' so it doesn't always hold
this is what TTerra essentially said
formal ermine
wait so what did @chilly ocean mean
wraith cargo
wait so what did <@456226577798135808> mean
I think they meant in the case where p is contained in m
chilly ocean
y'all are really stretching the meaning of my original statement lol
i didn't mean anything deep i literally only gave a counterexample to "mp = p"
chilly ocean
Yeah that only works if p is contained in m
Pretty sure I said that in my message as well tho
formal ermine
oh so the maximal local => local local thing is not true?
wraith cargo
Eisenbud actually taught me something 😎
My day is getting brighter
chilly ocean
oh so the maximal local => local local thing is not true?
What?
It is
So
So if R has P <=> R_p has P for all primes p, then you want to show this is eq to R_m has P for all m.
One direction is easy. Now if you want to show that R_p has P, you choose a maximal ideal m that contains p, and then you have (R_m)_p = R_p, so R_p has P
coral spindle
you choose a maximal ideal m that contains p
and there the riddle is solved
chilly ocean
CONTEXT!!!!!!
wraith cargo
zorn my beloeved
chilly ocean
WE FINALLY GOT IT
coral spindle
I'm so proud of this community 🥹
formal ermine
abs alg's greatest moment
tribal moss
The roots-of-unity thing?
formal ermine
wait but how do you show that pm contains p lol
chilly ocean
It doesn't, but that's not what you need to show
formal ermine
then how do we use the proposition
chilly ocean
A_p means A[A\p^-1]
formal ermine
oh I forgor
untold turret
field k has algebraic closure K. B is a finitely generated k-algebra. Suppose we're able to extend the inclusion map i:k -> K to i:B -> K. is this enough to say B is an algebraic extension of k?
untold turret
oh! suppose instead we have a nontrivial extension of the inclusion map? maybe it works in this case?
coral spindle
OK extend the inclusion map by sending x to 1
untold turret
hm... right. thanks
but there is some way to infer that B is an algebraic extension from the given premises here, right?
coral spindle
Let B = k(x).
untold turret
so the homomorphism extension must be an injection
coral spindle
If the extension is an injection, this is the equivalent of saying "If B is a subfield of the algebraic closure of k, then is it a subfield of the algebraic closure of k?"
untold turret
what has me a bit confused is, I've previously encountered the definition of a subfield to be that of a field that is an actual subset of the parent field, not anything given by a map
coral spindle
It is a mild abuse of terminology to say that "is a subring" rather than "is isomorphic to a subring" but this has the same effect.
untold turret
oh I see! this makes sense, if it is isomorphic to an algebraic extension (subring of algebraic closure) it might as well be one too
coral spindle
being isomorphic to an algebraic extension is the same as being an algebraic extension
Yes you edited your message to be more accurate
untold turret
i see, thanks a lot for clearing this stuff up!
chilly ocean
Hello I have a problem
formal ermine
Hello I am a problem
chilly ocean
Let p be a prime and the set M=Z_p × Z_p. We define the operation "o": for every (a,b), (a',b') in M, (a,b) o (a',b')= (aa'-bb', ab'+a'b). We know (M,o) is a commutative monoid. What is the number of invertible elements? I saw that (1,0) is the identity of M.
I am stuck
Any hint?
ivory trail
is this just Z_p[i]
tribal moss
Yes. Which means that its properties depend on whether Z_p already contains a square root of -1 or not.
If there is a b such that b² == -1 (mod p) then (b,1) o (b,-1) = (0,0) so those elements certainly cannot be invertible.
But if there's no such b, then the usual rule for division of complex numbers works.
coral spindle
It's not quite Z_p[i].
For example if p=5 then since 2 is a square root of -1 already, Z_5[x]/(x^2 + 1) is just Z_5 again
Or well ok let me amend that
what I said was wrong
It is what I wrote down, but Z_5[sqrt(-1)] as written would be just Z_5 still in that particular case.
simple valley
hmm so say I have G acting transitively on X, and I consider the componentwise action on X^n
then the orbits will correspond to essentially partitions of n (the set of n elements)?
the "equality structure" of X^n will be preserved
no wait that's not necessarily true is it, these can in turn fall apart into smaller orbits
is there a notion of when this does happen? like when the action on X^n is "as transitive as possible"
thinking about the space of bezier cubics... as functions R -> R^2 they have an action by Aff(R^2) (on the image), but also an action by Aff(R) (on the parameter)
the space of bezier cubics is almost everywhere 4*2-dimensional (determined by 4 control points), and Aff(R^2)xAff(R) is 6+2 dimensional
though we know that the action is not transitive because in the "general" case there are self-intersecting and non-self-intersecting beziers and these two classes are preserved by this action
I wonder if the orbits are discrete however
white oxide
how would you know that we have to find an extension of the rationals of degree 2 prior to knowing what ${\rho_0, \rho_1, \rho_2, \rho_3}$ fixes?
cloud walrusBOT
okeyokay (analysis is cool)
white oxide
like say for example we wanted to find the degree of the fixed field of ${\rho_1, \rho_2, \rho_3, \rho_4}$ without knowing what each of them fix, but we know the subgroup diagram
cloud walrusBOT
okeyokay (analysis is cool)
tribal moss
is there a notion of when this does happen? like when the action on X^n is "as t...
"As transitive as possible" sounds like you basically want G to be the product of symmetric groups on each orbit.
simple valley
what do you mean by a group "being" another group on an orbit?
tribal moss
(Disclaimer: I haven't absorbed all the details of your bezier curve example)
I meant G would need to be isomorphic to the product of each orbit's symmetric group, in order if we want to be able to predict the orbits in X^n from the orbits in X.
But I now don't think I was right about that, if n is small compared to the size of the orbits.
simple valley
the exact property I had in mind was that given $(x_1, \dots, x_n) \in X^n$ and $(y_1, \dots, y_n) \in X^n$ such that $\forall i j, x_i=x_j \iff y_i=y_j$, there exists $g$ with $\forall i, gx_i = y_i$
cloud walrusBOT
mniip
simple valley
this seems to be equivalent to the definition where given $x_i \ne x_j$ and $y_i \ne y_j$ we have $\forall i, gx_i = y_i$, except the degenerate cases where $|X| < n$
cloud walrusBOT
mniip
tribal moss
Yeah -- that's the same as saying the action of G is n-transitive, I think.
simple valley
well
k-transitive for all k<=n
an action is always trivially n-transitive if n>|X| but not so much with my thing
but if n<=|X| then n-transitivity implies n-1-transitivity
tribal moss
Okay, point.
simple valley
okay yeah but bezier curves
tribal moss
I suppose your Bezier curves are infinite curves, not with definite start and end points?
simple valley
I'm thinking of maps f:R -> R^2
they have "start" and "end" points as much as they have f(0) and f(1)
by considering the image of f instead, you're implicitly quotienting by the symmetries of the argument space, and I don't want to do that quite yet
tribal moss
Okay, so transforming the argument space gives you something you consider a different curve, not just a different parameterizaton of the same curve.
simple valley
in practice bezier curves are quadruples of control points
reparameterization is a symmetry that acts on this space
tribal moss
I suspect it would be enlightening to pretend we're working over a finite field so we can count instead of trying to imagine dimensionality ...
white oxide
how would you know that we have to find an extension of the rationals of degree ...
anyone?
simple valley
curves over finite fields I am scared
tribal moss
It seems to be the most important distinction is between curves whose four control points are collinear and ones where they affinely span the entire plane.
(Which is a distinction preserved by all of the transformations).
simple valley
well, there's also the distinction between self-intersecting and non-self-intersecting curves
tribal moss
Oh, and degenerate curve where all the control points coincide.
simple valley
the Aff(R^2) will act rather differently depending on how many points are unique yes
white oxide
so my book (fraleigh) defines a separable extension as a finite extension E of F such that [E: F] = {E: F}, where {E: F} is the number of isomorphisms of E onto a subfield of the algebraic closure of F keeping F fixed. How would you define a separable extension in the infinite case, where E is an infinite extension of F?
tribal moss
The self-intersecting-or-not distinction shows that we'll have more than one orbit, but I don't think it directly influences the size of the orbits.
A special kind of non-self-intersecting curves, though, are the ones of degree 1, since they are the one where the Aff(R^2) and Aff(R) actions overlap, so their orbits are smaller.
Hmm, this can also happen for degree-2 curves. For all of them, though?
tribal moss
This classification becomes complicated and might not even be helpful.
Let's heuristically assume that most curves are in general position and have enough distinguishing features (self-intersections, inflection points or the like) that a parameter transformation cannot look like an affine output transformation.
next obsidian
so my book (fraleigh) defines a separable extension as a finite extension E of F...
E is separable if given any F-algebra A, A (x)_F E is reduced.
Equivalently, in the case that E is finitely generated as a field extension, there exists a “separating transcendence basis” {x1,…,xn}< E which is a transcendence basis and such that E/F(x1,…,xn) is separable algebraic.
This requires you to say what a separable algebraic extension is, in the finite case one definition is what you said, but the “proper” one IMO is that each element is separable, meaning that its minimal polynomial has no repeated roots
I don’t feel like saying anything more, so if you have more questions please consult Google or someone else
simple valley
Here's my line of thought: if Aff(R) x Aff(R^2) has finitely many orbits, then every bezier curve can be produced by reparameterizing and affinely moving one of these finitely may "primordial" bezier curves
for lines and parabolas this is obviously true
however there's a certain sense in which a parabola is on equal grounds to a curve where 2 control points coincide
but these are obviously in different orbits
tribal moss
For a finite field k of size p, the order of Aff(k²) is p²·(p²-1)·p(p-1). And it acts faithfully on general-position curves.
Aff(k) has order p(p-1), so the order of Aff(k²)×Aff(k) is p²(p²-1)p(p-1)p(p-1).
Huh, this means that there actually isn't room for more than one "general curve" orbit in the finite case when p is large.
So much for speculating that the finite-field case would be helpful.
white oxide
E is separable if given any F-algebra A, A (x)_F E is reduced.
Equivalently, in...
oh ok thanks for typing tha tout
next obsidian
I guess I can say one last thing
(Mostly) nobody fucking cares
It’s rare you deal with this, so you can mostly forget it
A lot of people will never have to care about separability in fact because everything is separable in char 0
simple valley
ok so here's a possibly important fact: the space of self-intersecting curves forms an open subset
and e.g. curves where P1=P3 lie on the boundary. Parabolas also lie on the boundary
maybe we can move to #math-discussion cause this is leaving the realm of abstract algebra
white oxide
how would i show that $\varphi$ is a representation? i'm trying to show that $\varphi_{\sigma_1\sigma_2}$ and $\varphi_{\sigma_1}\varphi_{\sigma_2}$ agree as functions on the standard basis, i.e. $e_{\sigma_1\sigma_2(i)} = e_{\sigma_1(i)}e_{\sigma_2(i)}$ for all $i \in I$, but so far I haven't got anywhere
cloud walrusBOT
okeyokay (analysis is cool)
white oxide
because say $\sigma_2(i) = j$, then it sums up to showing $e_{\sigma_1(j)} = e_{\sigma_1(i)}e_j$
cloud walrusBOT
okeyokay (analysis is cool)
rotund dragon
@white oxide how are you getting e_sigma1(i) e_sigma2(i)
group operation for GL_n(C) is given by matrix multiplication so phi_sigma1 phi_sigma2 corresponds to a composition of the permutations
white oxide
is there any non-painstaking way to check that $\rho$ is a homomorphism? i don't want to check all combinations of elements of $S_3$ lol but it seems like that's the only way since the representation was defined in terms of the generators of $S_3$
cloud walrusBOT
okeyokay (analysis is cool)
white oxide
or well i guess the point is that i only have to check it for the 6 elements of S_3 expressed as a combination of the generators?
idk
rotund dragon
or well i guess the point is that i only have to check it for the 6 elements of ...
ye
cloud walrusBOT
okeyokay (analysis is cool)
white oxide
because i don't know it's a homomorphism yet
chilly ocean
define it on generators and extend in such a way that it is a homomorphism
void cosmos
yo if A is a free R-module then its torsion free obviolusly right?
cuz it would contradict independance?
white oxide
define it on generators and extend in such a way that it is a homomorphism
Ok I’ll try that thx
chilly ocean
you know how in linear algebra you can define a linear map by saying what it does on a basis? similar idea
it's probably a good idea to think about the general question: given a presentation of a group, when does a function on the generators to another group extend to a homomorphism?
warm ember
i odnt understand
is p a prime of Z or of A
or what
and what is the multiplication in pa
multiplication in the ring A or in Z
tender wharf
is p a prime of Z or of A
p is the same prime as char A
tender wharf
multiplication in the ring A or in Z
$pa$ is just $\ob{a + a + \cdots + a}{p times}$
cloud walrusBOT
tender wharf
characteristic?
tender wharf
characteristic of a ring R, denoted char R is the smallest positive integer n such that for all elements r in R, nr = 0
warm ember
also does anyone know how this is proved cuz i have no idea
tender wharf
first show
warm ember
ye for two
void cosmos
can someone explain the implication that choosing this r gives us rX is contained in F?
the proof is easy for me after that
warm ember
oh it's a meme role
wait wdym
tender wharf
I found it funny that pre university has access to the advanced channels just by the advanced role
vast quiver
can someone explain the implication that choosing this r gives us rX is containe...
we have two cases. If $x\in S$, then $rx \in F$ because $F$ is the free submodule generated by $S$. If $x\in X-S$, then we have the $r_x x = -\sum_{i=1}^k r_i x_i \in F$, so in particular $rx \in F$ because $rx$ is a scalar of $r_x x$.
cloud walrusBOT
thejoesully
next obsidian
Oh fugg
I either misread, or you edited the question
I thought you were asking why rX < F means rA < F lol
void cosmos
ohh
so thats why they included all the product of r_i
yea that was samrt
thank you i got it
chilly ocean
If o(a) =m and o(b)=n , gcd(m,n)=1 , and ab=ba then I have to show that o(ab) = mn
Now in this
Only thing remains to show is
mn | o(ab)
And that is where I can't think anything
What's the idea?
tender wharf
If o(a) =m and o(b)=n , gcd(m,n)=1 , and ab=ba then I have to show that o(ab) = ...
what is o
chilly ocean
Order of an element
tender wharf
you use contradiction
chilly ocean
How
chilly ocean
That mean
tender wharf
it follows quite fast
chilly ocean
o(ab)=( mn)(p) + r
tender wharf
can you write your work down on a piece of paper or a latex document somewhere else
chilly ocean
Yes it is
tender wharf
now notice w > 0
chilly ocean
It's r
chilly ocean
So we have (ab)^ r = e??
tender wharf
yeah you should be able to finish the rest yourself
coral steeple
I'm not sure where to start with this. It's simple to see that for any x,y in G the isomorphism has to conjugate xy by y, but this doesn't exactly specify a bijection. For what it's worth there's an easy homomorphism (just map everything to 1)
For reference
prime sundial
can you think of any unary operation that “flips” xy
coral steeple
Conjugation by y...?
prime sundial
fwiw conjugation by any element is an automorphism
coral steeple
Oh yeah wait
prime sundial
but i don’t think that suffices here since it’s specific to the y you pass in
i had something different in mind
think of some matrix actions that swap the matrices
prime sundial
yes
coral steeple
Ok yeah I got it that shouldn't have taken that long
prime sundial
the name of the group is also a nice hint
solemn dew
Any hints on 2?
Can I show closeness with two $$\phi_{1}$$ and $$\phi_{2}?$$ And show that the product of those two are in G, thus t he closeness is G and because of this closeness shows it is a automorphism?
and then try to do the same with inverses?
cloud walrusBOT
jonathan100
prime sundial
the product of those shouldn’t be in G
it should be in Aut(G)
do you remember what the group operation is in Aut(G)
solemn dew
I don't, what is it?
prime sundial
composition of maps
prime sundial
“group under composition of automorphisms in terms of maps”
your goal is to show the composition of two inner automorphisms is an inner automorphism
think about what element that would be a conjugation by
i.e. if you do conjugation by g and then conjugation by h, that’s the same thing as conjugating by something else
prime sundial
you used h twice
each phi should be conjugation by a different element
it doesn’t necessarily have to be different, but it should be arbitrary and so you should use g and h (or some other 2 distinct letters)
trim stirrup
Let G be a nonempty set and let * be an associative binary operation on G. Assume that for any element a,b in G, we can find x in G such that ax=b, and we can find y in G such that ya=b.Show that (G,*) is a group.
prime sundial
Let G be a nonempty set and let * be an associative binary operation on G. Assum...
what have you tried
prime sundial
i can’t tell what happened in that third equality
trim stirrup
I tried to equate the two equations
prime sundial
i agree, i just don’t understand how those outer terms got there
solemn dew
I'll write it more clearly
prime sundial
I tried to equate the two equations
i would begin with the axioms axioms a group
the operation is associative
trim stirrup
we have closure and associativity
trim stirrup
identity element and inversibility
prime sundial
what happens if you set a = b?
trim stirrup
if a=b then x and y must be the identity
prime sundial
let’s rephrase that
if a = b then there is some x such that
ax = a
and such that ya = a
you somehow need to show x = y
trim stirrup
both x and y are equal to a*a^-1
trim stirrup
because ax=a for all a in G, doesn't that mean that there exists a right identity element in G? and doing the same with ya=a, we get that there exists a left identity element in G. now setting a=x in the second equation we get that yx=x but yx=y, so x=y. is this correct?
sacred gust
because ax=a for all a in G, doesn't that mean that there exists a right identit...
Why x is the same for all elements in G?
"ax=a for all a in G" it isn't given. What's given is that for each a in G there is some x such that a x = a.
summer path
Ok I didn't realize there was more context
But what you can do to modify your proof is to fix your first x such that ax=a for some a, and try to work things out from there
tender wharf
sleemo why did you delete that, this is the right channel
trim stirrup
"ax=a for all a in G" it isn't given. What's given is that for each a in G there...
let's say we have ax=a with x fixed. from the second equation, ya=b, we get that y(ax)=ya => (ya)x=b, so bx=b. we proved that the x satisfying ax=a also satisfies bx=b. so we can choose the same x for all elements in G. correct?
prime sundial
i feel like you are now mixing and matching the x's
hm actually it looks fine, i think i misunderstood
that suggests that x is a right identity for all x yes. the rest of the identity proof should be straightforward
do you know how to approach the inverses?
trim stirrup
we can set b=e and from ax=b we get that ax=e, so x is the inverse of a?
prime sundial
right inverse, but yes
trim stirrup
if we denote the right identity element with er and the left identity element with el, then er=el*er=el, so we have an identity element in G
trim stirrup
yes
prime sundial
werent we doing inverses
trim stirrup
sorry
trim stirrup
from the second equation, we set b=e so ya=e
prime sundial
so how can you characterize y
trim stirrup
y is the left inverse of a
and we have to prove that the left inverse and the right inverse are equal
chilly ocean
Are you OK from here on out? @trim stirrup
trim stirrup
I think I am
chilly ocean
Good luck
trim stirrup
I'll figure it out
prime sundial
it's quite similar to the other proof we did
sacred gust
Goodjob
trim stirrup
thakns @prime sundial for helping
worthy sonnet
sleemo why did you delete that, this is the right channel
I thought it would be better in help
Let G3 be the symmetry group of a cube.
(a) Determine the order of G3.
(b) Consider the subgroup H3 < G3 generated by the three reflections on the planes x1 = 1/2, x2 = 1/2, x3 = 1/2. Determine the order of the group H3.
(c) Let S(P) be the stabilizer of a vertex P of the cube. What is the order of S(P)?
(d) Prove that every element of G3 can be written as h·s, where h ∈ H3 and s ∈ S(P). Hence, G3 = H3 · S(P).
(e) Recalculate the order of G3.
But I got no answer so yea, let me repost it^^
tender wharf
where are you stuck
flat treeBOT
What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
a) How does one do this in their head? Like aren't there pretty many?
b) What is meant with the plane x1=1/2? Is this the plane where x2=0 and x3=0 and it's just parallel to the other axis?
c) Isn't this every element of G, such that for example the reflections go through the vertex P and the rotation axis go through that point? But there is no rotation axis such that the whole cube stays the same when it goes through a vertex p, so I guess S(P) is only id and bunch of reflections?
d) no idea how to prove this
e) also no idea
delicate orchid
a) Who says you have to do it in your head
b) it’s the plane (1/2, x_2, x_3)
c) the group is generated by reflections, but they’re not the ones given
d) take an arbitrary element and show it’s a product of those two elements, not much else to it
e) worry about it when you get there
worthy sonnet
b) Do I just think of every possible way, those reflections can be arranged. e.g.: x1 o x2, x1 o x3, x1 o x2 o x3 ....?
c) I don't get it. How is there more than 1 reflection?
teal vessel
G3 being the symmetric group of a cube? well, how many ways can you arrange the faces with only rotations and possibly reflections? (the book I use doesn't allow reflections for the symmetric group, yours may differ).
try thinking of it in terms of one face and the surrounding vertices
worthy sonnet
ok so for c i found 3 reflections now and I guess the identity
and rotations of k*90°?
tender wharf
another way is to orb stab it
celest cairn
On my last line, did I distribute everything correctly?
frigid lark
For this you can and should use a calculator/software
Like some mindless computation isn't gonna make you better at math, it's just kinda wasting your time
summer path
If you just have a computation problem, go to #prealg-and-algebra
tender wharf
spicy
solemn dew
solemn dew
I can't find a way to show that $$(\phi * \psi *\phi) (g) = \phi(g) $$
cloud walrusBOT
jonathan100
chilly ocean
that's not what you have to show though
take a careful look at the definition of a normal subgroup
brittle hare
ah your handwriting is a little hard to decipher
chilly ocean
you have to show that conjugation of an inner automorphism by any automorphism is again an inner automorphism
solemn dew
Ahh I found it, only have to show that it is a subset of it
So my proof is enough I think
chilly ocean
the second line is wrong. you're not showing that conjugation by psi fixes phi, you're showing that conjugation by psi sends phi to another inner automorphism
given phi in Inn(G) and psi in Aut(G), you must prove that psi o phi o psi^{-1} is in Inn(G)
the rest of the proof is actually correct in this regard
minus a small typo at the end of the third line
you correctly showed that, if phi is conjugation by h, then psi o phi o psi^{-1} is conjugation by psi(h)
chilly ocean
"because psi(h) is in G" is irrelevant
strictly speaking it's not (psi o phi o psi^{-1})(g) that's in Inn(G), but psi o phi o psi^{-1}
this looks better though
solemn dew
Can you prove this without the "(g)"?
chilly ocean
uhh yeah
chilly ocean
wait
no i misunderstood you
you have to show one way or another that psi o phi o psi^{-1} acts on G by conjugation by psi(h)
solemn dew
yes
chilly ocean
computing its action on an arbitrary element g of G is one way to do so
it is also probably the most direct way to do so
delicate orchid
Does it even make sense to chain mods like that? Thinking about the correspondence theorem, the ideals of Z mod p are exactly the ideals contained in (p), which is just (0) - which is to be expected as Z mod p is a field
Even implicitly embedding back into Z from Z mod p doesn’t guarantee that x is some kth power so I don’t see why this holds at all
static yew
Trying to grasp field extensions and algebraic closures of prime fields
If F2-bar is F2 union F4 union F8 onto infinity
Is the element x+1 in GF(4) the same as the element x+1 in GF(8)?
This doesn't really sound right to me but I don't have a good picture in my head, and I cant see how an infinitely large GF(2^k) could eventually have an element where x^2 + x + 1 = 0 in F2
hidden haven
As written, that is not a construction of F2bar. To take a union of sets, the sets must first be embedded in some larger common set, but that larger set is what you are trying to construct.
In any case, F4 cannot be embedded into F8.
static yew
So the x+1 in GF(4) and the x+1 in GF(8) are not the same?
hidden haven
What is x
static yew
Gf(4) has four elements
0 1 x x+1
hidden haven
Are you constructing those fields as F2[x]/(quadratic or cubic)?
static yew
So x is the transcendental(indeterminate?) Used to construct the field
hidden haven
No x is not transcendental. It satisfies x² + x + 1 = 0
static yew
Hrm does it matter? If all fields of the same finite size are isomorphic
hidden haven
No but I was just asking what you meant by x
hidden haven
Ye so the x in F4 and the x in F8 are different things entirely
The weird part isn't that F4 doesn't embed into F8, but that it does embed into, for example, F16
static yew
Hrm so while F2-bar is the union of a copy of all GF(2^k)
The actual definition of the operators in F2-bar is not that of any of the subfields it's made from
hidden haven
Operators? You mean elements?
static yew
If f2 bar is a field it has two binary operators
hidden haven
You can construct it from these finite subfields, just that you have to be more careful and can't just say that it's a union since the bigger set doesn't exist yet
Oh ye the operations on F2bar extend those of the finite ones
In the sense that all of these finite ones embed into F2bar
static yew
And (x+1)[F4] + (x+1)[F8] equals an element in F2bar
But none of the source fields has an operator that works on both
hidden haven
Yeah but also you have to be careful when you say x+1 of F4
F4 does embed into F2bar but not uniquely
static yew
You can construct it from these finite subfields, just that you have to be more ...
Ugh. Wikipedia literally says
For a finite field of prime power order q, the algebraic closure is a countably infinite field that contains a copy of the field of order qn for each positive integer n (and is in fact the union of these copies).
hidden haven
So x+1 can have different meanings there
hidden haven
Ugh. Wikipedia literally says
> For a finite field of prime power order q, the a...
Yeah it's true but not a definition
static yew
So then the x+1 in the copy of F4 is the same x+1 in the copy of f8?
I'm not sure if F4 is a true subfield of F2bar
hidden haven
There's no the copy, but in no choices of copies can they coincide because that would give you an embedding of F4 into F8 and that isn't possible
static yew
One sec
hidden haven
F4 can't embed into F8 because then F8 would be a vector space over F4 but the cardinalities don't allow that
hidden haven
I'm not sure if F4 is a true subfield of F2bar
It is a subfield but not in a unique way
You can swap x and x+1 in F4. This is an automorphism
So if you take an embedding of F4, you can apply this automorphism and then the embedding to get a different embedding
static yew
so in F_2-bar
there is an element a corresponding to (x+1) from GF(4)
and an there is an element b corresponding to (x+1)from GF(8)
and a!=b
hidden haven
Though it is true that this new embedding will have the same image (this is true for all choices of embeddings because all finite char 2 fields are normal extensions of F2)
hidden haven
so in F_2-bar
there is an element a corresponding to (x+1) from GF(4)
and an t...
Yeah sure
static yew
if a == b then you could embed F4 in F8 which doesn't work
hidden haven
Yes
static yew
and what I'm really trying to get to is
there are four elements in F2-bar that correspond to 0,1,x,x+1 from GF(4)
is the structure of those four elements equal to that of GF(4)? I can't think of how to describe it
basically I guess those four elements are isomorphic to GF(4) under F2-bar's definitions of+ and -
white oxide
how would we know that we're trying to find an extension of Q of degree 2 prior to knowing what rho0, rho1, rho2, and rho3 fix?
hidden haven
there are four elements in F2-bar that correspond to 0,1,x,x+1 from GF(4)
is t...
Yes
And in 2 different ways
You're having a hard time here because you've given them the same name
white oxide
how would we know that we're trying to find an extension of Q of degree 2 prior ...
is it because of the one-to-one correspondence in the fundamental theorem of galois theory?
hidden haven
If you call the elements in F2bar y and y+1, then maybe it'll be easier to say what you want
static yew
so that means GF(4) is a subfield of F2-bar
hidden haven
so that means GF(4) is a subfield of F2-bar
Yes
static yew
are there any finite fields that have subfields?
reading my book, it likes to focus on Q and e.g. Q(sqrt2). Q is obviously a subfield of Q(anything)
imma start doing Q(i) to upset people
hidden haven
F4 has a subfield F2
hidden haven
More generally F_(p^m) embeds into F_(p^n) if and only if m | n, and in that case there are m different embeddings, all with the same image
static yew
wait doesn't m|n mean 'm divides n'? aka mk=n for some positive integer k
hidden haven
how would we know that we're trying to find an extension of Q of degree 2 prior ...
Do you know the index of H_1 in the whole group? The fundamental theorem also says that the degree of an intermediate extension is the same as the index of its aut group
static yew
but you said F4 doesn't embed F16? or did I misunderstand
hidden haven
It does
static yew
err embed into
hidden haven
Not in F8
static yew
F4 _does_embed into F16
hidden haven
Yep
white oxide
Do you know the index of H_1 in the whole group? The fundamental theorem also sa...
ohhh right
thanks!
static yew
oh Imisread what you said
F4 doesn't embed into F8 (3/2 not in Z) but it embeds into F16 (4/2 is in Z)
hidden haven
Yes
static yew
but if I named the elements of F16 0,1,x,x+1,x^2,x^2+1,...
the subfield of F16 that is isomorphic to F4 is not 0,1,x,x+1?
hidden haven
No
static yew
that's a 'no it's not' or 'no you're an idiot'