ok so now I have that x*conj(x) = -d and conj(x) = c - x
#groups-rings-fields
restive star
delicate bloom
now just go through and check that the inverse trick you'd use in complex numbers works here as well
delicate bloom
might be clearer to show that (a+bx)*(a+b conj(x)) is in F
then you can just divide that element of F over and put it with the part on the right
restive star
would that prove it's a field?
delicate bloom
that would directly construct the inverse...
restive star
ah
but wouldn't either way work?
actually I'm just ending up with (a^2+2abx+b^2 x^2)/(xc-x^2)
delicate bloom
you should have x+conj(x) = c and x*conj(x) = -d when you expand out your denominator
meaning it will have only elements of F
restive star
so I went back and did it your way, by expanding that expression. And got a^2 + abc - abx + abx + b^2cx - b^2x^2
but I have no way of dealing with that x^2
oh nvmind
delicate bloom
I don't know what you're doing
restive star
I see my mistake
delicate bloom
lol ok cool
(a+bx)*(a+b conj(x)) = a^2 + ab(x+conj(x)) + b^2 x conj(x) = a^2 + abc- b^2d
restive star
yeah I was dumb and didnt use that x conj(x) was -d
delicate bloom
divide (a+bconj(x)) by that and the RHS is just 1
so that means you have your inverse
delicate bloom
yeah you're welcome
chilly ocean
yeah you're welcome
I'm sorry, but where may I post a question regarding elementary differential forms?
next obsidian
I think that's something for #point-set-topology
rustic portal
Oh hm
Formula sheet that I made recently
It's just copies of the theorems from the book but I think it looks nice :P
final pasture
what's a domain ? 🤔
rustic portal
yea
i didnt include any definitions that i didnt need or i had enough practice w/ cause limited on space
rustic portal
whoopsies
yeah
wrote this in a pinch cause i had 2 finals on thursday and needed it for yesterday
cause my final was yesterday :P
delicate bloom
I feel like if you have to consult a formula chart for that you're in trouble
rustic portal
yeah i didnt use that
it was mostly just the other formulas tbh
i didnt actually use the formula sheet at all, wasn't needed
i had enough practice with everything thankfully
yeah thats true
it helped to make because there was a theorem i misinterpreted
rewriting that kinda made me realize :P
this class was tough as hell though... lots of practice near the end saved me though
group actions 
yeah theyre super cool
just have a lot of weird things that took a long time for me to understand
final pasture
they're just acting cool 
rustic portal
hahaha
im sure the concept wouldve come easier to me if not for COVID :p
we kinda race through proofs and the choppy connection she has didnt help
next obsidian
OOF
final pasture
What I've read: "race through proofs"
What I imagine: "Ok so we have exactly 17 seconds for the proofs of the Sylow theorems, y'all ready ?"
rustic portal
wait thats really cool
next obsidian
Basically just numbers dictate that p-groups be nice lol
Yeah aditya do the class formula
Actually fuck it I’ll prove it rn
Any non trivial conjugacy class has size a multiple of p
This is cuz the stabilixsr has to divide p
rustic portal
I feel a bit weird given that my introduction to some really cool algebraic topics was during a time that made it pretty difficult to learn or atleast grow some significant passion with
lots of these topics are like "holy shit thats cool i wanna know more"
next obsidian
And then the size of the conjugacy class is |G| / |stabilizer|
Then class formula tells you that |G| = |Z(G)| + sum |conjugacy classes|
Now reduce mod p
The left is 0
And all the sizes of conjugacy classes become 0
So |Z(G)| must be 0 mod p
But it’s at least 1
Since the identity is there
So it has to be at least size p
Gg
next obsidian
It’s just numbers
and they don’t lie
AND THEY SPELL DEFEAT FOR YOU AT SACRIFICE
latent ingot
proving the class formula is the janky part of that i think
next obsidian
I don’t think it really is, you just note that conjugacy classes partition G
Since they’re all orbits
latent ingot
i had some problem with it
next obsidian
Fair
latent ingot
it does make sense though
next obsidian
I mean on a first pass I didn’t really get it either
But if u sit on it for a while it’s kinda like
“Oh huh, right”
IMO
latent ingot
like you're just counting the elements
rustic portal
thats most of this class for me
latent ingot
and taking the center out
next obsidian
Yeah
latent ingot
really obvious, then the proof like, had some useless step iirc
next obsidian
Eh
Maybe it’s like
Conjugacy class trivial iff in center
Because you kind of write it in terms of like
Singleton conjugacy classes + everything else
Then you collect all the singletons to make Z(G)
rustic portal
chad
next obsidian
Rip
latent ingot
or maybe i'm misremembering
next obsidian
Honesty
It might just be that u weren’t as good
At the time you saw it in class
Like no shade
This has been my experience
next obsidian
If A is a DVR and K its residue field, then suppose that B is a valuation ring of K-bar which dominates A. I want to show that B is a non-discrete dimension 1 valuation ring. I can show it's non-discrete, but I can't show that it's dimension 1. I know that it's equivalent that B's valuation is Archimedean which is what I'm trying to show, but I can't figure it out
It's equivalent to show that for any a,b in m_B, there's an n such that a^n/b is in m_B
but idk how to do that :(
fringe dawn
What is this notation supposed to mean?
I am not familiar with the notation $\mathrm{Ann}_M$
cloud walrusBOT
PseudoRndNbr
fringe dawn
only $\mathrm{Ann}_R$ where $R$ is a ring.
cloud walrusBOT
PseudoRndNbr
rustic crown
guessing from the context, it should be just all the elements in M that die when you multiply by p.
fringe dawn
I assumed as much, I just never saw anyone use the subscript to denote the module instead of the underlying ring.
rustic crown
Ann_R gives you an ideal, which is a submodule of R, while Ann_M gives you a submodule of M
i'm seeing this notation for the first time as well 😛
fringe dawn
thanks, appreciate the help
rustic crown
yw, i didn't do a thing though 😶
vestal snow
Can someone help me formalize a category theory argument?
chilly ocean
Everytime I see your name I read Have a Banana at first.
vestal snow
Wdym?
I want to show that Ker(F(g))/Im(F(f)) is isomorphic to F(Ker(g)/Im(f)) if F is an exact tensor and gf=0
So here's the idea
Since F is exact, it commutes with kernels and cokernels
fringe dawn
gf = 0 implies you have a short exact sequence
vestal snow
gf = 0 implies you have a short exact sequence
So I have the sequence 0 -> A -> B -> C -> 0
Not necessarily
rustic crown
(i'm not sure if the above works... the definition of exact i know is, preserving kernel and cokernel, does that imply it also preservers all limits and colimits?)
vestal snow
My bad
fringe dawn
Well what categories are you working in?
fringe dawn
I think it's pretty much by definition
vestal snow
Yeah but I am having trouble formalizing it
fringe dawn
0 -> A -> B -> C -> 0 is exact, since gf = 0
Exactness of F implies that
0 -> F(A) -> F(B) -> F(C) -> 0
is exact
fringe dawn
C is chosen such that the sequence is exact.
rustic crown
you can't choose C right
fringe dawn
Ok sure so let me not call it C lol 😄
0 -> A -> B -> B/im(f) -> 0
Something like this?
vestal snow
Yeah
fringe dawn
Which then implies that
0 -> F(A) -> F(B) -> F(B/im(f)) -> 0
is exact
fringe dawn
Yeah right, f needs to be a monomorphism.
fringe dawn
I would probably break down my argument into 3 parts. Image is preserved, kernel is preserved and quotient is preserved.
First one follows from F being a functor
Second one requires some sort of argument via a short exact sequence and then using exactness of the functor
And the third one will involve a short exact sequence that involves the homology
vestal snow
$
\begin{tikzcd}
& B/f(A) & & & \
A \arrow[r, "f"] & B \arrow[u, "\pi", two heads] \arrow[r, "g"] & C & & \
0 \arrow[r] & Im(f) \arrow[r, "i", hook] \arrow[u, "k_\pi", hook] & Ker(g) \arrow[r, "\pi'", two heads] \arrow[lu, "k_g"', hook] & \frac{Ker(g)}{Im(f)} \arrow[r] & 0
\end{tikzcd}$
Bottom row is exact
All the verical/diagonal maps commute with F due to preservation of kernels and cokernels
So I have proved that F(Im(f))=Im(F(f))
cloud walrusBOT
Have a Banana, Bitch
vestal snow
and F(ker(g))=ker(F(g)
All that remains is to show that the quotient of these two also commutes with F
That is, F(ker(g)/im(f))=Ker(F(g))/Im(F(f))
fringe dawn
0 -> Im(f) -> Ker(g) -> Ker(g)/Im(f) -> 0 is a short exact sequence and so is
0 -> F(Im(f)) -> F(Ker(g)) -> F(Ker(g)/Im(f)) -> 0
by exactness of F
fringe dawn
And then you are done.
fringe dawn
Unless I'm missing something
Exactness tells you that F(Im(f))/F(Ker(g)) = F(Ker(g)/Im(f))
And then you use F(Im(f)) = Im(F(f)) and F(Ker(g)) = Ker(F(g)) and you are done.
And this works because F(Im(f)) -> F(Ker(g)) is the inclusion map, right?
Or rather Im(f) -> Ker(g) is the inclusion map (let's say i)
And then F(i) is the map F(Im(f)) -> F(Ker(g))
vestal snow
Okay one question
fringe dawn
And preserving monomorphism is a limit property
as long as your functor preserves pullbacks it also preserves monomorphisms.
vestal snow
We know that F(ker(g)) is isomorphic to Ker(F(g)) and F(Im(f)) is isomorphic to Im(F(f)), but how do we know that these isomorphisms convert into isomorphisms of the short exact sequence?
fringe dawn
Do we actually need that?
We use exactness of the functor to go from the short exact sequence 0 -> Im(f) -> Ker(g) -> ... -> 0
to the short exact sequence
0 -> F(Im(f)) -> F(Ker(g)) -> ... -> 0
Then we use exactness of this sequence to show that
F(Ker(g))/F(Im(f)) = F(Ker(g)/Im(f))
And then we use the isomorphisms F(Ker(g)) = Ker(F(g)) and F(Im(f)) = Im(F(f))
vestal snow
Yes, but we want to show that Ker(F(g))/Im(F(f)) = F(Ker(g)/Im(f))
fringe dawn
Yeah but why would you want to show that the two short exact sequences are fully isomorphic?
vestal snow
I'm just saying that the problem isn't done yet
fringe dawn
Let's number what we have shown:
- F(Ker(g))/F(Im(f)) = F(Ker(g)/Im(f))
- F(Ker(g)) = Ker(F(g))
- F(Im(f)) = Im(F(f))
vestal snow
Yes
fringe dawn
By 1: F(Ker(g)/Im(f)) = F(Ker(g))/F(Im(f))
By 2: F(Ker(g)/Im(f)) = Ker(F(g))/F(Im(f))
And then by 3 we finish
I don't understand what you think is missing
I'm not saying that you're wrong, I genuinely don't see it atm.
vestal snow
Okay so
I think you are saying that A = B, C = D implies A/C = B/D
which is not true unless the isomorphisms behave nicely with each other
fringe dawn
Yeah I understand, but exactness of the functor gives us another short exact sequence
Hmm
I think you need additivity of the functor
I see what you mean, but I am pretty sure this is guaranteed to behave nicely because of the context we are in
But you are right, there's something missing to fully justify the conclusion
vestal snow
I see what you mean, but I am pretty sure this is guaranteed to behave nicely be...
which is exactly why I was having trouble formalizing it 🙂
fringe dawn
My gut says additivity of the functor might help
Since then you can represent the functor as tensoring by a module
I have to get back to reading this paper that I'm trying to wrap my head around 😄
I hope you made some progress.
fringe dawn
But yeah, I would look into additivity or something like this.
And your intiution about colimits/limits isn't far off
Since that's sort of along the lines of whether the functor will preserve monomorphisms.
vestal snow
If F was tensoring with a flat module what would you do then?
vestal snow
If B is an A algebra, can we say anything about $(M\otimes_A B) \otimes_B (L \otimes_A B)$?
cloud walrusBOT
Have a Banana, Bitch
untold sapphire
I feel like it should be an A Algebra?
untold sapphire
Since there is a copy of A inside right
vestal snow
Assuming A is not commutative
untold sapphire
Oh
vestal snow
So the tensor products are abelian groups
untold sapphire
Not sure then
oblique river
Does that tensor product even make sense? Im assuming M and L are right A-modules?
vestal snow
Yes
old lava
uh do you mean B is not commutative
pretty sure A has to be commutative
for anything to be an A algebra
no?
oblique river
So then when you tensor on the right with B you get two right B-modules
vestal snow
My bad
oblique river
But i dont think you can tensor two right B-modules together
vestal snow
One of them is right and the other one is left
oblique river
So yeah step 1 is to figure out exactly which actions M and L have on which sides by B and A and then write the order correctly
delicate hawk
is there any way for me to check my answer
i found that it was 24
but im garbo @ lin alg so idk
untold sapphire
Maybe try substitution to get some clearer relations between the generators?
Or any standard operations on the matrix
viscid pewter
the order is 1: a = b = c = 1 
untold sapphire
Ah lol
viscid pewter
i'm just twiddling it rn and it looks like like order of b is at most 6, order of c is at most 12
but there might be simplifications, i'm going to try some things
order of a is at most 6
i don't see how to get any better, might be missing something
untold sapphire
That doesn't account for the words which are products of a single generator
Aren't"
rustic crown
just write the presentation matrix and then do the smith normal form
in more details, you have the map
$\mathbb{Z}^3 \to G$
cloud walrusBOT
det
rustic crown
given by sending e1 -> a, e2 -> b, e3 -> c
this will be surjective
now what are the relations among these genrators?
the column vectors (2, 4, 4), (-6, 6, 12), (10, -4, -16) all go to zero and generate the relations
so we have the sequence
$\mathbb{Z}^3 \to \mathbb{Z}^3 \to G \to 0$
cloud walrusBOT
det
rustic crown
Where the first map is given by
$\begin{bmatrix}2 & -6 & 10\ 4 & 6 & -4\4 & 12 & -16\end{bmatrix}$
cloud walrusBOT
det
untold sapphire
Right yeah so basically calculate the syzygy modules
rustic crown
the group is the cokernel of this map
so choose different basis so that this map looks nicer
untold sapphire
Ok but that's basically the same as performing row operations
rustic crown
yea
untold sapphire
Idk what this person's background is
rustic crown
but this is like one of the nicest ways to think about it, if not directly bashing.
delicate hawk
this module chapter has been tough for me since idk much lin ag
untold sapphire
But det is the most correct
untold sapphire
but this is like one of the nicest ways to think about it, if not directly bashi...
Yeah for sure
rustic crown
sage: A = matrix(ZZ, [[2 , -6 , 10], [4 , 6 , -4], [4 , 12 , -16]])
sage: A
[ 2 -6 10]
[ 4 6 -4]
[ 4 12 -16]
sage: A.smith_form()
(
[ 2 0 0] [ 1 0 0] [ 1 -1 -1]
[ 0 6 0] [ 0 1 -1] [ 0 3 -2]
[ 0 0 12], [ 2 0 -1], [ 0 2 -1]
)
sage: ```so sage says this map looks like diag(2, 6, 12) in some weird basis
which means the image is 2Z oplus 6Z oplus 12Z
and if you quotient this out, you get the group (Z/2Z) + (Z/6Z) + (Z/12Z)
oblique river
Ok but that's basically the same as performing row operations
Row and column operations
Because you can change the basis of both copies of Z^3
untold sapphire
Yep , my bad
rustic crown
yea so the order of the group is actually 2*6*12 and not 24
delicate hawk
is there anyway i can read on this? section 14.5 on artin was supposed to be about this but its been completely useless
untold sapphire
I like Aluffi but some people hate it
But his chapters on Linear Algebra are really good
Either that or just read Lang
rustic crown
(i thought his linear algebra was one of the thing on the bad side 😶 )
untold sapphire
Fair enough , to each to their own
rustic crown
but i really loved that proof of structure theorem
i have seen people prove it by purely using smith normal forms over PIDs, but that proof is kinda ugly
if you can't compute gcds in pids there is very little point of an algorithmic proof
but that makes a lot of sense over Euclidean domains
delicate hawk
so @rustic crown did you just do row operations on that 3*3 matrix till you reach a pseudo row echelon form ?
untold sapphire
Yeah that is nice, I haven't studied PID stuff in a while
rustic crown
so <@!671237809029251103> did you just do row operations on that 3*3 matrix till...
nah, you are allowed to do both row and column operations, but be careful, you are only allowed to multiply and divide by units
the diagonal matrix you get in the end is called the smith normal form
do row and column operations until (1,1) is the only non-zero entry in the first row and first column and make sure that (1,1) entry divides all others. now continue inductively
storm meteor
first time asking here, it would be helpful if you could guide me to the answer 😅
old lava
f'(0) = 0 means that there is no linear term, i.e., A contains polynomials of the form b_0 + b_2 x^2 + b_3 x^3 + ... Since gcd of 2 and 3 is 1, any integer can be written as an integer linear combination of 2 and 3. Also, every integer >= 2 can be generated by positive lin combinations of 2 and 3 (prove this if you need to). Therefore, every power in b_0 + b_2 x^2 + b_3 x^3 + ... can be generated by products and sums of x^2, x^3 and constants from k.
so the map is surjective
storm meteor
i need to hit my head on a wall
i can feel my brain cells leaving me, should sleep, thanks for the heads up, ill try that again
delicate hawk
@rustic crown by "only allowed to multiply and divide by units", you mean i cant do something like $R_1 \rightarrow R_1 + \frac{7}{2}R_2$ even if multiplying the elements in $R_2$ by $\frac{7}{2}$ keeps everything integer?
cloud walrusBOT
Fiwam
untold sapphire
i can feel my brain cells leaving me, should sleep, thanks for the heads up, ill...
Geometry is hard
rustic crown
<@!671237809029251103> by "only allowed to multiply and divide by units", you me...
nope! you can only do R1 --> R1+3 R2
but when i said not multiply by unit i mean you can't do something like R1 --> 1/2 R1
we just want all the operations to be invertible, so each operation is mutiplication by some integer matrix that is invertible
old lava
untold sapphire
just ask
old lava
that's certainly a way to define semidirect products
cringe category pilled definition though
cloud walrusBOT
expressive group
oblique river
i'm not sure what you're trying to show here
next obsidian
What are you assuming?
You haven’t made clear where you’re starting from, it sounds as if you’re trying to show any surjection splits which isn’t true
Where semidirect product means the SES splits?
Factors uniquely as KH, what a trash way to phrase it monkaS
What book is this
tfw
Just find a book that doesn’t treat it in a stupid way like this
You should learn what an internal semidirect product is
And then it becomes obvious
Then..........
It’s hard, idk wtf they expect you to do
I’m not even a priori sure how you show that G/H is K lol
what map do you use?
Wut
You need a map G -> K
To use the first iso
You want to use first iso to show that G/H is K right?
Uh
Sure whoops
You still need a map G -> H then
What SES
All you know is that G = KH “uniquely”
Whatever that means
Right?
Wutttttttt
You don’t get maps from G into subgroups
For free
Cuz...
Like you can just find a counterexample
Like
old lava
Factors uniquely as KH, what a trash way to phrase it monkaS
based and correct
next obsidian
Like okay
Your map is identity on H
And 0 everywhere else?
You’re stipulating that for any subgroup H
That the complement of H union {0} is a normal subgroup
Consider S3 and C3 inside it
Then you have a size 4 normal subgroup of S3
The point I think
Is that an element of G can be uniquely written as
kh for k in K, h in H
And maybe you do some stupid ducking kh-> h
But it’s not clear at all
That this is a group Hom
It isn’t abelian
Why is that a group Hom tho
Like, I don’t think that this is actually correct
Because this is a semidirect product you’re twisted a bit
Or something like that
Or maybe that is it, but like it’s not clear that
kh(k’h’) will have hh’ in the like... H slot
cloud walrusBOT
expressive group
oblique river
I'm just rejoining but I'm confused at the confusion here
the definition is "K normal and G = KH uniquely"
cloud walrusBOT
expressive group
oblique river
we can form G/K and then show that's iso to the subgroup H just by using the unqiueness
next obsidian
By?
oblique river
restrict the surjection G --> G/K to H --> G/K
which is injective (H cap K = {e}, proved from uniqueness) and surjective (since everything in G is of the form kh)
next obsidian
Okay, but now you need to figure out what the map G -> H is
And find a section of it
next obsidian
Then you need the inverse of H -> G/K
oblique river
no you don't lmao
next obsidian
Or you just say you replace H with G/K
oblique river
we've found an isomorphism H --> G/K which is canonical coming from the original data
we also have a (canonical) G --> G/K which means we have a (canonical) G --> H
next obsidian
At the end of the day I don’t like these definitions, and I think this is more confusing than its worth
old lava
I think the more "direct" definition of semidirect products is better, and then showing that a splitting exact sequence is a semidirect product is nice
rather than defining it using exact sequence memery
that just seems like you define it that way because you can
oblique river
okay, so we have two different definitions and either way the first step is to prove that they're equivalent
in terms of how people actually use thsi stuff in nature, it's very rare that you actually are given something which looks like KH
instead, you have some SES which you can prove splits, and therefore you know that it's of the form KH
yes -- if G = KH uniquely and K is normal in G, we can form the quotient G/K. Restrict the quotient map G --> G/K to the subgroup H, this gives us a map H --> G/K. this map is both injective (b/c of unique decomposition) and surjective (because everything in g is a product of something in K and something in H) and therefore an isomorphism
this gives us a surjection G --> H with kernel K and which clearly has a splitting, namely, just send H back to itself as a subgroup of G
what do you mean "the same thing as"?
the map H --> G/K came from the map G --> G/K
yes, they are both surjective
think of an abelian example G = Z x Z and K = Z x {0}. Let H = {0} x Z. Then the quotient map G --> G/K can be restricted to the subgroup H, and that restriction H --> G/K is also surjective (and even an isomorphism; both sides are just Z)
sure, although it might be more obvious if you use right cosets of K instead of left cosets of K
untold sapphire
everything is impossible until it's obvious!
oblique river
the idea is that K(kh) = Kh
every right coset of K is determined by some element of H
and in a unique way -- if Kh = Kh' then h = h'
that's why G/K is iso to H
yes but now you understand it better :P
untold sapphire
I think Aluffi does this in a really nice way for groups
oblique river
G --> Aut(K) --> Aut(K)/Inn(K)
K is in the kernel of this composition
and so G/K also maps to Aut(K)/Inn(K)
this is basically the idea of the first isomorphism theorem
yes
whenever you have G --> G' and K is in the kernel, you get G/K --> G'
if K is equal to the kernel, that new map is injective
but as long as K is in the kernel you get that map and the image is the same
chilly ocean
cardinality of A copies of H maybe
smoky verge
A's a set, so I'm unsure what the implication is there
chilly ocean
all sets of same cardinality are isomorphic so
oblique river
maybe "direct sum indexed by A" then
smoky verge
and A = {1 .. n}
oblique river
sure, the groups H^N, H^Z, and H^Q are all isomorphic to each other, but they still look different in how you write them dwon
if someone said "countable direct sum" I would imagine the first group, not the third one
chilly ocean
does that difference ever matter?
fiery berry
can someone give me a hint how i can prove that x^6 + x^3 + 1 is irreducible over Q ?
oblique river
it's about how you communicate the information
there is a surjective map H^Q --> H^N
given by "restrict the function from Q to N"
that function isn't an isomorphism -- it's very very far from it
chilly ocean
there is a surjective non injective from H^N to H^Q too
oblique river
yes lmao i know that, they're both abstractly isomorphic to each other
I'm saying that you might be in some setting where you are considering functions from Q to H and functions from N to H in the same problem
carmine fossil
can someone give me a hint how i can prove that x^6 + x^3 + 1 is irreducible ove...
Note that over C,x^6+x^3+1 factorises as (x^3-w)(x^3-w^2) where w is a third root of unity
oblique river
and so being able to write down "the natural map H^Q --> H^N" is much better than trying to describe that map as some map H^N --> H^N
carmine fossil
If suppose that were reducible over Q,you end up with another factorisation in C
But C[x] is a UFD => contradiction
chilly ocean
you mean natural map H^N to H^Q?
chilly ocean
right im stupid forget what i said lol
past temple
if I have a tower of extensions K/L/F
and K/F is Galois
is it true that K/L is Galois?
oblique river
my point is that groups are more than just abstract things, they often come with some kind of interpretation or meaning in relation to each other, and it's nice when our notation remembers that
for example, if you went around using N to refer to the rational numbers with your justification being "well technically they have the same cardinality so I might as well" then people would look at you funny
fiery berry
Note that over C,x^6+x^3+1 factorises as (x^3-w)(x^3-w^2) where w is a third roo...
is there an easy way to see that such a factorization exists ?
oblique river
or, in this case, suppose that H were actually a ring, then H^N and H^Q could also be made into rings, and they wouldn't be isomorphic as rings
so while they might be abstractly isomorphic as abelian groups, there are plenty of reasons to not conflate the two
carmine fossil
is there an easy way to see that such a factorization exists ?
if you take x^3=a that expression becomes a^2+a+1
Which is (a^3-1)/(a-1)
chilly ocean
wait why are they not isomorpic as rings?
oblique river
because N and Q aren't isomorphic as groups
well sorry N isn't a group, but semigroup or whatever
chilly ocean
Isnt H^S just functions from S to H with coordinatewise operations
oblique river
that's not the ring structure I had in mind
I was thinking of it as a graded ring
where the grading was given by S
chilly ocean
oh idk about those
oblique river
but yes if you did it all coordinatewise then they would be the same
chilly ocean
is it similar to monoid rings?
oblique river
a monoid ring being something like R[G]?
fiery berry
if you take x^3=a that expression becomes a^2+a+1
i see, couldn't i immediatley deduce now that it must be irreducible since x^2 + x + 1 is irreducible over F_2 ? the substitution a = x^3 shouldnt bother here right ?
fiery berry
reduction mod 2
chilly ocean
so its like formal sums of formal multples rg?
oblique river
that's what R[G] is
chilly ocean
yea
oblique river
a G-graded ring is any ring R which has a decomposition as a direct sum of R_g for g in G
fiery berry
I don't see how F_2 is useful here
yea you are right, thank you
oblique river
such that R_g*R_h lands inside R_(gh)
untold sapphire
Miller & Sturmfels have a whole chapter on this
really fun stuff
with cool links to matroid theory
fiery berry
i see, couldn't i immediatley deduce now that it must be irreducible since x^2 +...
i just want to quickly add that im pretty sure now that this argument is wrong, it would have been something like this: Suppose that f(x) = x^6 + x^3 + 1 is reducible, then f(x) = g(x)h(x) => f(x^(1/3)) = f' = x^2 + x + 1 = g(x^(1/3))h(x^(1/3)) => contradiction since f' is not reducible over Q. However i think what's wrong with this is that g(x^(1/3)) mustn't be in Q[x] so this is probably not correct. However, one can prove irreducibility of f via a substitution x -> x+1 ( this is fine, since it doesn't change the degree of the polynomial) and then use eisenstein criterion
carmine fossil
Well, It's actually f(x)=(x^3-w)(x^3-w^2) can be factorised further into product of irreducibles not in Q[x]
And if we find a factorisation into irreducibles in Q[x],we will have two different factorisations
fiery berry
Well, It's actually f(x)=(x^3-w)(x^3-w^2) can be factorised further into product...
yea this would probably also work, i just now realized that the argument from me above which i thought was right is just wrong in general
carmine fossil
Hold on,I don't think this approach is correct
f(x) could factorise as g(x)h(x) in Q[x] which ultimately factorises into the product as above in C[x]
chilly ocean
what does a product like this mean if the group doesn't have to be commutative? is it a mistake in the question?
prisma ibex
No
I agree in general this is a suspect notation
But in this case it’s fine
Implicit in this notation is the fact that the order doesn’t end up mattering
smoky verge
(in this case, doesn't it mean that any ordering results in f?)
chilly ocean
oki makes sense
restive star
When trying to show something is a minimal polynomial for some given x, is it enough to show it has x as a root and is irreducible?
restive star
the definition I saw was that it is the polynomial of lowest degree with it as a root
I figured the two statements were equivalent, but just had to double check
thanks!
chilly ocean
it follows from K[x] being a PID i think
rustic crown
Yea that's corrected in the errata
next obsidian
@chilly ocean
rustic crown
Just write it as a product equaling identity with last two elements that do not commute, this equation wound then tell you that their commutator is identity
chilly ocean
i tried finding a counterexample but im bad with noncommutative groups lol
next obsidian
Actually hold on
So first of all, I think the errata does fix it
But maybe it’s true regardless
Oh fugg
Let f the only order 2 thing
For any y in G, let
z = yfy^-1
Then z^2 = yfy^-1yfy^-1 = yffy^-1 = yy^-1 = e
So z^2 is either identity (impossible) or order 2
So z is order 2, so z = f
But this says that f commutes with y
No nvm, I am still on “not defined”
Hurb
I was thinking you can then always write the prodict of all elements as
f*product of rest
But it’s product of rest that’s not well-defined
I think you can cook up an example by taking
C2 x non abelian group of order 27
rustic crown
I'm saying the equation can be written as Cab=Cba=e then C=b'a' which gives b'a'ba=e which can't be true
next obsidian
Cuz then the only order 2 thing is (1,e)
Wait so wut
Are you saying that if it was well-defined G has to be commutative?
It looks like you’re saying b and a commute
rustic crown
Yea
rustic crown
This will be true for any a, b
rustic crown
lol
rustic crown
Yea
next obsidian
Yagga
restive star
I just wasted a long time trying to prove a certain polynomial is irreducible and just realized eisenstein applies lmao
oblique river
No
Assuming that L and K are both galois over F, then Gal(L/F) is a quotient of Gal(K/F)
Gal(K/L) is a subgroup of Gal(K/F)
past temple
ok
past temple
I'm making a presentation in latex
is anyone familiar with how to change the colors of these boxes
latent ingot
like, the latex?
unique juniper
x^3 - 2 is seperable right?
unique juniper
well Q
past temple
if its over Q its always separable right? or am i being dumb
unique juniper
my main confusion is
we obviously dont have roots to x^3 - 2 in Q, but thats not a requirement to be separable right?
past temple
no
a polynomial is called separable if it doesnt have any repeated roots in a splitting field
unique juniper
a polynomial is called separable if it doesnt have any repeated roots in a split...
OHHH
right
past temple
let a be the cube root of 2
then the roots of ur polynomial is a times the three powers of the third root of unity
those are all distinct
past temple
in general though if you know that the polynomial is over Q, its separable
unique juniper
i havnt seen that yet
unique juniper
XD
yeah once you said char Q = 0 i realised XD
i thought this would have been 3 ...
am i wrong?
sturdy marsh
yes
unique juniper
aw
sturdy marsh
try factoring it
unique juniper
3 factors
sturdy marsh
what are they
unique juniper
$(x-\frac{-1+\sqrt{-3}}{2})(x - \frac{-1-\sqrt{-3}}{2})(x-1)$
cloud walrusBOT
Yes
sturdy marsh
yeah this stuff is tricky
unique juniper
and im kinda rushing through everything
but
wouldnt [Q{c}:Q] = 3 ? c is 3rd root of unity
next obsidian
No
x^3 - 1 isn't irreducible
right?
I think it factors as like (x - 1)(x^2 + ax + b)
Probably?
am I stupid
maybe
unique juniper
oh
sturdy marsh
chmonkey probability theorist confirmed
sturdy marsh
yes
next obsidian
what is that ummm
Oh right
lmfao difference of cubes
I know this formula
Chmonkey
Anyway, does that make sense? If x^3 - 1 was irreducible then yeah it would be deg 3
but it isn't D:
next obsidian
:D
I guess the other way to see it is that of the third roots of unity, one of them already lives in Q
next obsidian
I don't think you want -1
Or well
yeah that will work
you can just add 1/2 to kill that bit off
unique juniper
yeah
next obsidian
:D
next obsidian
past temple
i'm having trouble wrapping my mind around internal direct products
basically, if you have two normal subgroups of H, N < G
with HN = G
and hn = nh for each h in H, n in N
and the intersection of H, N is {e}
then we write that G is the internal direct product of H and N right?
how does this definition extend to infinite direct products?
is it even possible to?
carmine fossil
How are you defining infinite direct product?
Elements of form h_1 h_2 h_3... Where h_i in H_i such that only finitely many h_i are not identity is one possible defn,I think. Is that how you are doing it?
past temple
i think thats the direct sum, right?
by direct product i mean h1 h2 h3... where hi can be any element in H_i
carmine fossil
You know how you extend it to more than 2 subgroups?
past temple
like not rly lol
carmine fossil
Like if G=HKN ,(all 3 subgroups are normal) where H inter K, H inter N and N inter K are all {e} then G=HxKxN
past temple
ummm
i guess all you need then is the commutativity property
idk how you'd express it for three subgps tho
compact needle
As you "go to infinity" here, the more obvious thing you'll end up generalizing is the "direct sum"
past temple
i think so but for the problem i'm doing i have to consider the infinite direct product
the next problem deals with the infinite direct sum
carmine fossil
Same criteria will apply,I think
past temple
anyways yeah @carmine fossil i'm not sure how to generalize the commutativity thing for H,K,N
carmine fossil
All subgroups should be normal and intersection of any 2 subgroups should be Identity
Prove HK is normal in G and HK inter N={e}
And then write HKN as HK x N
=H x K x N
carmine fossil
Yes
past temple
still kinda struggling on how it generalizes to infinity tho
wait, im reading the wikipedia page
so is there no way to generate an infinite direct product from normal subgroups of G?
how is an infinite direct product even defined then?
other than just saying like
(h1,h2,h3,..)
oblique river
That’s it
If all the G_i are the same then you can also define it as the set of functions from I into G
carmine fossil
Is it true that $(HK \cap N)=(H \cap N)(K \cap N)$ where H,K and N are all normal subgroups of a group G
cloud walrusBOT
What's NDY?
storm meteor
an old question paper i found, would appreciate any help
The first part is a direct application of the previous question which was to show that A/m is a finite extension of k. I'm stumped at the second part
where A is a finitely generated k algebra
cloud walrusBOT
Have a Banana, Bitch
hidden haven
Is it true that $(HK \cap N)=(H \cap N)(K \cap N)$ where H,K and N are all norma...
Nope, take D_n = < r,s | r² =1, s^n =1, rsr = s^{-1} >
H = < rs >
K = < s >
N = < r >
The right side is trivial, the left isn't
I think n might need to be even
Yeah right side is trivial only if n is even
carmine fossil
Prove HK is normal in G and HK inter N={e}
I am trying to prove HK cap N if trivial is H cap N,H cap K and K cap N are trivial. All three are normal subgroups of a group G
hidden haven
Oh all 3 normal
Forgot about that
Wait all 3 are normal in my example
But will think about the specific case
hidden haven
I am trying to prove HK cap N if trivial is H cap N,H cap K and K cap N are triv...
Doesn't that example work for that too
Unless I messed up checking normality
carmine fossil
Like if G=HKN ,(all 3 subgroups are normal) where H inter K, H inter N and N int...
So,Is this not true?
hidden haven
Yeah I think so
carmine fossil
What extra condition do you need for G to be a direct product?
hidden haven
Direct product of H, K and N?
carmine fossil
Yes
If H inter K is 1 and H and K are both normal you can deduce hk=kh for all h,k
hidden haven
Yeah
hidden haven
What extra condition do you need for G to be a direct product?
I can't think of any nicer condition than H intersection K should be trivial + HK intersection N should be trivial + HKN should be the whole group
Actually Z² also works as a counterexample to the earlier claim
Take the 3 normal subgroups to be the 3 order 2 subgroups it has
Product of any 2 is the whole group
hidden haven
an old question paper i found, would appreciate any help
For the real case, with $f$ defined as in the problem, but to R, notice that $I = (x-f(x), y-f(y))\subset m$. Then $m=m+I$, so $\bR \cong A/m = A/(m+I) \cong (A/I)/(m/I)$. Since $A/I \cong \bR$, $m/I$ must be trivial
Complex case will be very similar, you just quotient by (x-f(x)) first, then quotient by that quadratic in y, and you get C already, so there can't be anything left to quotient by
cloud walrusBOT
Moldilocks
sturdy marsh
**Have a Banana, Bitch**
@vestal snow this is true for any exact functor (which in this case is tensoring with a flat module). Show that the kernels and images are preserved.
alternatively, spectral sequence of a double complex
next obsidian
Lmfao
chilly ocean
whether it is "all 0" or "not all 0" i am sure you get it. (personally i am fairly sure it is written correctly but)
just some annoying negating mathematical statements
unique juniper
all 0 tells me that a_i = 0 for all i
chilly ocean
"no nontrivial solution"
chilly ocean
maybe it is an issue of parsing english 🙂 there are too many negations, and maybe this sentence is not clear to begin with
unique juniper
"not all 0" says that some a is not 0 and the sum will be 0, so we have a non trivial solution to this. "all 0" says that it can only be equal to zero when all a's are zero?
chilly ocean
i think he is saying that a nontrivial solution would be "not all 0", and that we are saying there should be no nontrivial solutions
chilly ocean
i think this is just written weirdly
obsidian sleet
yeah it is a little weird to parse but a professor of mine wrote like this and he meant that at least one was nonzero
the equation is written as an example of a nontrivial solution
pain
i guess it is easier to see this as saying the only solution is the trivial solution where all the ai are 0
looks like linear algebra lmao but it looks different what is this?? im new to math
sharp sonnet
linear characters
group homomorphisms from a group G to L^\times for some field L
at least i think so
sharp sonnet
(they were introduced by dirichlet from a group to C^\times to study number theoretic stuff)
(there is a slightly different definition of character now in the context of representation theory)
unique juniper
oh cool
boreal pecan
Hello 🙂 I have one question that I just cant find in books I have. What this A^B notation means? I think its probably group of some homeomorphisms but need to know what exactly :/
boreal pecan
Ohh, thats why I couldn't find it in algebra books 🤣
I was like searching in books and online and couldnt find it. Asked one profesor I am working it, she told me she remembers seeing it but doesnt remember what it is haha
Thank you 😄
summer geyser
Does anyone have an idea on how to do part b of this question? (also just to make sure as I don't think "root depth" is common term, I'll post all defintions momentarily)
This is the definition of root depth:
And this is 25.8:
sqrt(K_(i-1)) is simply the set of all square roots of the respective field. so K_i is constructed by adjoining all these square roots.
I've done (a) by constructing the minimal polynomials of each x_i, however I have no clue how to do (b). Really the only thing we've discussed about elements being in what K_i is this theorem below:
However it's only discussing K_1, not any others. I was thinking about maybe showing x in K_i/K_i-1 iff Gal(K(x)/K) os abelian of exponents 2i? However I have no idea if this is true or not, let alone how to go about proving it, actually I don't really know how to prove just the theorem above.
rustic crown
i think a simple induction will do the job
summer geyser
Yeah I thought about this, I can show that x_i is in either Q_i or Q_i-1 (assuming x(i-1) is in Q_i-1) but not sure how to prove otherwise
summer geyser
Do you have a hint as towards how to prove that? I've been thinking about it for a while but not sure how
the structure of Q_i isn't entirely clear to me so its hard to do like a direct proof similar to something like sqrt(5) not in Q(sqrt(3))
rustic crown
(oh i was still thinking, i don't have a proof yet, maybe i'll say when i'm completely done)
summer geyser
ah alright I see thanks
rustic crown
okay i have an idea, but i still need a little work to show it, maybe you can help there
summer geyser
perhaps, I'm not that good at abstract algebra though :P
rustic crown
so if i can show that that degree 4 polynomial is irreducible, then i'll be done because then the galois group of the polynomial would have exponent 4.
cloud walrusBOT
det
summer geyser
hm wait how this it being irreducible imply that the galois group is exponent 4? and how does this say x_(i+1) is not in Q_i (sorry)
rustic crown
The galois group will be Z/4Z
rustic crown
and for the other, i'm using this theorem with K = Q_{i-1}
summer geyser
oh wait yeah of course you can just replace K with those
rustic crown
The galois group will be Z/4Z
you have already shown this because you know the galois group over Q(x_{i-1}) itself is Z/4Z
and if the polynomial is irreducible, then galois group will have size atleast 4
I meant this follows from part (a)
summer geyser
sorry just digesting all of this
I can't quite get my head around it being Z/4Z
what rules out the klein-4 group?
rustic crown
$\text{Gal}(\mathbb{Q}(x_i)/\mathbb{Q}) = \mathbb{Z}/2^i\mathbb{Z}$
cloud walrusBOT
det
rustic crown
This is what you have shown.
rustic crown
$\text{Gal}(\mathbb{Q}(x_{i+1})/\mathbb{Q}(x_{i-1}))$ is a subgroup of $\mathbb{Z}/2^{i+1}\mathbb{Z}$ with order 4.
cloud walrusBOT
det
summer geyser
yeah I see this argument now, thanks. just gotta go back to (a) and show its cyclic
cloud walrusBOT
det
rustic crown
but yea, once we show that the polynomial is irreducible, we should be done.
summer geyser
hm I see, alright I need to read through it all to fully grasp it
but thanks for the help thus far, was really helpful!
urban acorn
are there any interesting examples of abelian groups where all endomorphisms commute?
cyclic groups are like that
including finite ones and Z
and I'm not sure about it, but I think these constitute the only finite examples
I'm also fine with imposing extra structure and only considering endomorphisms which respect it
like, R as a topological group has multiplication by constants as it's endomorphisms
hidden haven
are there any interesting examples of abelian groups where all endomorphisms com...
I found this https://mathoverflow.net/questions/35355/criterion-for-an-abelian-group-to-have-a-commutative-endomorphism-ring
urban acorn
thanks, by the title it's exactly what I was looking for
rustic crown
yep
For principal ideals, notice that (a)*(b) = (ab)
(2)^n = (2^n) in Z
yep
i didn't get that
what are you canceling?
hidden haven
i think det was saying yes to the "am i wrong" part
rustic crown
i think det was saying yes to the "am i wrong" part <:hmmCat:831508801646755850>
yes lol
rustic crown
urban acorn
literally 2Z
rustic crown
is that 69th root?
strong valve
420
rustic crown
you can show by eisenstein that x^69-69 is irreducible by taking p = 3
cloud walrusBOT
Carla_
chilly ocean
oopsp
urban acorn
yeah okay I'm confident the degree is 69
strong valve
sinful mirage
how can I show that the intersection of the two is zero?
Let $a \in \bigwedge^2 V \implies a=\sum c_{ij} \frac{1}{2} (e_{i} \otimes e_{j}-e_{j} \otimes e_{i} )$. Now let's act with F on this element:$ F(a)=\frac{1}{2}\sum c_{ij}( F(e_{i} \otimes e_{j} - e_{j} \otimes e_{i}) )=\frac{1}{2} \sum c_{ij}( F(e_{i} \otimes e_{j}) - F(e_{j} \otimes e_{i}) )=\frac{1}{2} \sum c_{ij} ( e_{j} \otimes e_{i}- e_{i} \otimes e_{j})=-\frac{1}{2} \sum{c_ij} (e_{i} \otimes e_{j})=-a$
then similar proof works for $a \in Sym^2 V$ and $a=-a$ is true iff $a=0 \implies Sym^2V \cap \bigwedge^2 V={a|a=0}={0}$
cloud walrusBOT
ProphetX
sinful mirage
this is my trial but i'm not sure whether this is ok 
cloud walrusBOT
ProphetX
unique juniper
what does "minimal number r of nonzero b_i" mean exactly?
like smallest b_i that is not 0?
urban acorn
no
each solution consists of an (n+1)-tuple
and any entry is either 0 or nonzero
so each solution has some amount of nonzero entries, which is a natural number (including zero)
so you choose one such that no other solution contains less nonzero entries than it
in particular, in this case apparently the minimal number of nonzero entries is r
the missing context makes it annoying to read
urban acorn
alright
unique juniper
to confirm, the solution such that any other solution will have more 0 entries
alright
ty
brisk ferry
Been stuck on this for a while and asked for help in linear already
how do i jordan decompose this thing
chilly ocean
shouldnt this be in #linear-algebra
brisk ferry
tried asking it there, but they could not answer it
brisk ferry
i did the ker(A-eigen I)^3 but the nullspace given is not compatible on what i got with wolfram
I get 3 vectors for the p in pjp^-1
2 seem correct but one is 2x too small
next obsidian
Do you know anything about this map?
Is it linear, is it a ring hom, etc
This is an automorphism of C
As a ring
Its inverse is itself
chilly ocean
and it fixes R
next obsidian
And tht, yeah
Just look at what it does restricted to R
It does nothing
If you want it as a homomorphism C -> C, yes
But if you want to show it’s a homomorphism restricted to R, no
Since on R it’s just identity, so it just... is a homomorphism
chilly ocean
did you show it
next obsidian
Just compute it
It isn’t gonna be hard
Well it depends on what you were claiming is an automorphism
If you’re claiming it restricted to R, then sure
But maybe what you wrote was claiming that complex conjugation is an automorphism
cloud walrusBOT
squirtlespoof
next obsidian
Yeah but you can just say it’s identity, and I don’t see how that isn’t enough to say it’s an automorphism
plucky hare
Hi there! So I am working on homework problem for my abstract algebra 1 class, and I seem to have gotten stuck.
Here is the problem:
I am working on part b. Here is what I have so far:
next obsidian
You can find an inverse to phi
Pretty easily
It basically comes down to “undoing” x goes to 5 - 2x
Then you just need to show both of them are ring homomorphisms
plucky hare
I got part a done already. I am stuck on part b, since I'm not sure how to prove that it isn't both onto and one-to-one
next obsidian
I think it’s not onto, but maybe it’s one-to-one
The fact you have a squared is very suspicious
It might not even be one-to-one honestly
Nvm it totally is
plucky hare
That's what I was thinking too. We have an example from class that states the square makes it not onto, but the example was for a problem in the rationals. I wasn't sure if it changed since we are working in the reals.
next obsidian
Show that for example
The polynomial f(x) = x
Is not in the image
Because you’re replacing x with 5 - x^2
You can never have an odd power in the image
For one-to-one show that it f(x) is mapped to 0, it was 0 to begin with
My suggestion is to look at the highest power, nothing should be able to kill it
i.e. if you had
f(x) = a_nx^n + lower degree stuff
When you replace x with 5 - x^2 you should get
a_nx^2n + lower degree stuff
And so the leading term will remain non-zero
plucky hare
That makes a lot more sense. I think that's what my professor was trying to explain in the notes, but it didn't quite make sense. Thank you for that!
warped bane
hey everyone,
Is it trivial that: Span(f1,f2,f3) + Span(e1,e2) = Span((f1,f2,f3,e1,e2) ?
scarlet estuary
what do you mean by +?
warped bane
normal addition?
scarlet estuary
okay sure, vector space addition
then the statement is true
U + V for subspaces U, V of the same vector space is defined as {u + v | u in U, v in V}
elements of span(f_1, f_2, f_3) are of the form a_1f_1 + a_2f_2 + a_3f_3
warped bane
even tho Span(f1,f2,f3) and Span(e1,e2) are not necessarily supplementary?
scarlet estuary
elements of span(e_1, e_2) are of the form b_1e_1 + b_2e_2
so Span(f1,f2,f3) + Span(e1,e2) = {(a_1f_1 + a_2f_2 + a_3f_3) + (b_1e_1 + b_2e_2)}
and of course this is just the span of the 5 vectors
supplementary in the sense of having a well-defined direct sum? thats fine if theyre not
warped bane
so Span(f1,f2,f3) + Span(e1,e2) = {(a_1f_1 + a_2f_2 + a_3f_3) + (b_1e_1 + b_2e_2...
got it
scarlet estuary
you just change your coefficient
if f_1 = e_1 for example, then a_1f_1 + b_1e_1 = (a_1+b_1)f_1
or if you prefer, (a_1 + b_1)e_1
doesnt really affect anything
scarlet estuary
caveat: this is, of course, assuming your spans come from the same vector space over the same field
otherwise addition doesnt really make sense
warped bane
yeah sure
wraith obsidian
@warped bane This also very elegantly follows from the characterization of span S being the smallest subspace containing S.
Namely, if S={f1,f2,f3} and T={e1,e2}, we have the following for every subspace W containing both S and T:
- it must contain S, hence span S. - Also it must contain T, hence span T.
- since it contains the last two, it also must contain span S + span T, as this is the smallest vector space containing both.
- but span S + span T already is a vector space containing S and T
- since we have shown that such a W containing S and T must already contain span S+span T, the latter must be the minimal subspace containing S and T, and this is the characterizing property of span S\cup T.
All what we need here is that span is what's called a closure operator, so the proof automatically works for „subgroup generated by“, the convex hull, „ideal generated by“, …
(closure operators are fucking neat)
Note also that $V+ W$ is just a fancy way of writing $\operatorname{span} (V\cup W)$
cloud walrusBOT
lux
boreal pecan
Guys if we have 2 abelian groups A and B how to interprate "componentwise group operations" on set of all function from B to A?
rustic crown
I don't think you need B to be an abelian group here, just a set would work. You look at the set A^B that is set-functions from B to A and then define an operation on these function pointwise. That is, for f and g in A^B define f+g in A^B by (f+g)(b) = f(b) + g(b)
So A^B acquires an abelian group structure!
I should stress the + in f(b) + g(b) is the operation of A
boreal pecan
Thanks 😄 I just needed to know how to think about this component wise part 😄
chilly ocean
think of it the same as products
rustic crown
You usually don't call this component wise though.... i have heard the word "point-wise" a lot more.
But yea, you can look at a function from f:B --> A by a tuple (f(b))_{b in B}
boreal pecan
Yea, I mean script that prof. gave me have all details left out so I am on my own hahaha So I have to ask here to clerify terms since my native isnt english 😄
rustic crown
I don't think you'll need anything significantly more than the definition. f splits over K and K = F(roots of f)
i is primitive 12th root of 1?
hot lake
I don't think that's correct
it should be 12 yeah but your explanation is a bit wrong
but in general finding the splitting fields is pretty non obvious
cloud walrusBOT
squirtlespoof
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hot lake
yeah but I don't see how you go from "a*b is in the splitting field" to "a and b are both in the splitting field"
rustic crown
do you see that $\sqrt[6]{3}e^{i\pi/6}$ and $e^{i\pi/3}$ are both there and these 2 elements are enough to generate all roots of that polynomial?
cloud walrusBOT
det
chilly ocean
it makes me feel yucky when i can't generate a field like Q(purely real, purely imaginary) 😞
rustic crown
yea lol
hot lake
it's 6
cloud walrusBOT
squirtlespoof
rustic crown
its the quotient of first 2 roots
notice that $\alpha = \sqrt[6]{3}e^{i\pi/6}$ then $\alpha^3 = \sqrt{3} e^{i\pi/2} = i\sqrt{3}$
cloud walrusBOT
det
rustic crown
i just meant look the roots are alpha*beta^k for k = 0, 1, 2, 3, 4, 5 where beta was the second element
the splitting field contains alpha and alpha*beta so must contain beta as well. But as soon as it has alpha and beta, we can generate all roots so the splitting field is Q(alpha, beta)