Yeah
#groups-rings-fields
bleak abyss
Alright so
kindred rivet
So is thst just the definition or what?
i'll do my best to understand, but my ignorance is hindering
bleak abyss
Yeah
cloud walrusBOT
Daminark:
stone fulcrum
Looks a lot like getting a differential?
Don't mind me I'm not smart enough for this
cloud walrusBOT
Daminark:
kindred rivet
Kaynex we’re both not smart enough for this, maybe a 3rd mole will be
kindred rivet
Or 1st or 2nd
bleak abyss
I'm not sure if I'm smart enough for this either I'm just winging it
Uh oh
(From the errata)
OH FUCK
This lengthens the argument a lot
Prob no commentary I'll just get through this fast I need to be done with things soon
vestal snow
Can someone tell me why M (x) A_m/mA_m = M_m/mM_m?
bleak abyss
FINALLY GOT THROUGH THAT DAMN PROOF
@vestal snow I think you can just do it in steps? Like first handle localization and then quotients?
pure crest
Hey guys
what does it mean when [E:F] = 1?
is there any way that i can prove that E = F?
scarlet estuary
in what context is this notation arising
pure crest
Degree of an extension feild
pure crest
i was thinking that i can argue that E is a vector space over F,
haha 😂
and use a basis
I can say that 1 is in E since its a field
and it takes elements x from F
so 1*x is in E
pure crest
right thank you
scarlet estuary
we call such an extension a "trivial extension" btw
pure crest
ohh i see
it seems rarely talked about
i cant even find a wikipedia article on it haha
i wonder what is the most obscure mathematics there is
solemn rain
bleak abyss
Based on the URL I feel like those two notions counter each other a bit
Unless he's asking for like, one person knows a given theorem is super useful in proving stuff but mathematicians are sleeping on it
pure crest
i was thinking more in the lines of a subject people study, but only has like 10,000 posts on google
not just one theorem
that would kinda be lonely
vestal snow
@bleak abyss Could you explain how exactly though?
bleak abyss
Alright lemme think
bleak abyss
Now if you tensor with M you get
vestal snow
Ah got it
bleak abyss
Yup
vestal snow
Thanks
vestal snow
Sloth, M isn't necessarily flat
So in the last step, when we tensor with M, we might not get an isomorphism
bleak abyss
No but it's right exact
vestal snow
I don't know what right exact means...
Does it mean that it preserves everything except the injection in the first arrow?
bleak abyss
Yup
vestal snow
I still don't see how that gives us the required isomorphism
Yes
So that gives us that M (x) k(m) is isomorphic to the M (x) A_m mod the homomorphic image of M (x) mA_m
Ahh I see now
We then expand M (x) mA_m to M (x) m (x) A_m
cloud walrusBOT
Daminark:
bleak abyss
Whoops
vestal snow
And then using properties of tensor product, that equals mM_m
I guess we still need to check the commutativity of the isomorphisms till this point, but doing that is painful
pure crest
can i say that $\mathbb{Q}(\sqrt{2})(\sqrt{1 + 3\sqrt{2}}) = \mathbb{Q}(\sqrt{1 + 3\sqrt{2}})$
cloud walrusBOT
boat:
pure crest
im trying to find the degree of $\mathbb{Q}(\sqrt{1 + 3\sqrt{2}})$, i was thinking that i can divide it into two subsets $ \mathbb{Q}(\sqrt{2})(\sqrt{1 + 3\sqrt{2}}) \subset \mathbb{Q}(\sqrt{2}) \subset \mathbb{2}$
cloud walrusBOT
boat:
pure crest
but i am not sure if i can say that the above statement are equal
elder valley
Yeah you can do that. I would just compute the minimal polynomial personally
elder valley
How are you computing it?
pure crest
$a = \sqrt{1 + 3\sqrt{2}}$ then $a^2 = 1 + 3\sqrt{2}$, then $(a^2 - 1)^2 = 18$
cloud walrusBOT
boat:
pure crest
but then this will always give a root
elder valley
Yeah that's it. So you get a degree 4 polynomial and just have to show it's irreducible
What do you mean?
pure crest
ohh
right
i am so dumb
you are right it is irreducible!
but then how can i argue that it is a simple extention over Q?
elder valley
What's the definition of simple extension?
pure crest
Rational roots theorem
to show that there are no linear factors
then i just have to show there are no quadratic factors
elder valley
Yeah, right on
You might be able to make some argument about there being the intermediate subfield Q(sqrt(2)) to help with irreducibility, but I'm not sure how that argument would go
pure crest
i am just curious as to what separates a single extension to a more complicated extension
elder valley
Seems there should be a connection there. Maybe someone else knows
Well every finite extension of Q is simple so things don't get complicated in that sense here
pure crest
what is the difference between $Q(\sqrt{2})(\sqrt{3})...(\sqrt{n})$ and $Q(\sqrt{1} + \sqrt(2) + ... + \sqrt{n})$
cloud walrusBOT
boat:
elder valley
It's called the primitive element theorem
Actually you now know that the degree is at most 4. You know there's the subfield Q(sqrt(2)) which has degree 2, and so the degree of the top field is a multiple of 2 by the tower theorem. But the top field can't be degree 2 since it's not equal to Q(sqrt(2)), so it must be 4
smoky cypress
Let $\alpha$ be an automorphism on a finite group $G$ and let $I=\brc{g\in G:\alpha(g)=g\inv}$. If $|I|>\frac34|G|$ then $G$ is abelian; if $|I|=\frac34|G|$ then $G$ has an abelian subgroup of index 2
cloud walrusBOT
Whoever:
smoky cypress
How would I go about showing this
Using earlier exercises we know that $II=G$, so every element $g$ can be written as $xy$ where $\alpha(x)=x\inv$ and $\alpha(y)=y\inv$
cloud walrusBOT
Whoever:
smoky cypress
But I don't see how 3/4 comes into play
golden pasture
Your goal is to eventually show |Z(G)|>|G|/2
||try considering the set I\cap gI for some g in I||
||what is a lower bound for |I| ||
pure crest
how can i find the irreducible polynomail of $\sqrt{3 +2\sqrt{2}}$
cloud walrusBOT
boat:
pure crest
i had a similar question last night, but this time the method i used about doesnt work
i actually do get a reducible polynomial
1 - 6 x^2 + x^4
which is
burnt pond
You can rewrite $\sqrt{3 +2\sqrt{2}}=1+\sqrt{2}$, thus the minimal polynomial is $(x-1)^2=x^2-2x-1$
cloud walrusBOT
Ceana:
smoky cypress
(x-1)^2 is not x^2-2x-1
burnt pond
yes forgot the -2 there
pure crest
wait isnt $3+2\sqrt{2} = 1 + \sqrt{2}$
cloud walrusBOT
boat:
elder valley
relevant article: https://en.wikipedia.org/wiki/Nested_radical
pure crest
nvm
that was a stupid statement
not your haha
mine
its obviously not 1 + sqrt(2)
becuse the factors are different
but what you said is correct, i just made an error in my calculations
burnt pond
Do it like this: $3+2\sqrt{2}=2+2\sqrt{2}+1=\sqrt{2}^2+2\sqrt{2}+1=(\sqrt{2}+1)^2$
cloud walrusBOT
Ceana:
pure crest
ohh damn
that makes senes!
thank you very much
thank you for the article as well Auvera, i never really looked into roots so much
plucky flicker
Hey! Is not that phi supposed to be from R^k to only R? Why R^l? I mean in the third row it is written that f_1, ..., f_k is in R, and then he just throws there that R^l.
in this sense f_i would be l tuples of polynomials in R, but he beggin to present f_i as simple polynomials in R
okay i guess i've got it, if phi : R^k -> R, its kernel is the first syzygy module, but we could ask for the generators of R^k, because in fact it is finitely generated, so we ask for the kernel of the pi: R^s -> R^t homomorphism, which is the second syzygy and so on
elder valley
If $F \subset F(\alpha) \subset F(\beta)$ are field extensions of finite degree, can anything be said about the relationship between the minimal polynomials of $\alpha$ and $\beta$, aside from $\deg(\alpha) \leq \deg(\beta)$?
cloud walrusBOT
Auvera:
elder valley
i'd expect not because for a fixed beta you can vary alpha over (potentially infinitely many) different primitive elements, which in turn yield different minimal polynomials
next obsidian
I don’t really think you can say much as you said
cloud walrusBOT
deiknymi0:
elder valley
as a start you can separate into cases based on the characteristic of R, then look at the action of f on the additive subgroup <1>. the map f sends this subgroup to 1, which gives at least |<1>| many distinct solutions to the polynomial x^pq-1
can probably hit it with some field theory at that point, with the field being the fraction field of R
oh i messed up, f acts trivially on <1>
so the polynomial is actually x^pq-x
vale flint
the thing is that this showed up as an exercise for a section that just introduced what zero divisors and integral domains are. up to this point, neither of the notions of ring characteristics or fractional fields have been defined
i don't know how to prove that there is exactly one integral domain for the map $f(a) = a^6$ even (is it like Z2 or something?)
cloud walrusBOT
deiknymi0:
elder valley
p and q are arbitrary positive integers?
vale flint
p and q are distinct primes, sorry i should have clarified at the start
burnt pond
I might have overlooked somthing, but for$f$ to be a ringhomorphism you need $(a+b)^{n}=a^{n}+b^{n}$ for all $a,b \in R$ But this is only true if $n$ is a prime and $R$ has prime characteristic.
cloud walrusBOT
Ceana:
next obsidian
I’m not sure if that’s only true
In that case
It certainly is true in that case
But requiring being an integral domain
Means prime characteristic
But I think that if n is a multiple of the characteristic
Then it’s fine right?
Or is that also just false, hmm
burnt pond
If you expand $(a+b)^n$ with the binomial theorem and use that $R$ is integral then for the equation above to be true you need $\binom{n}{k} \equiv 0 \mod n$ for all $0<k<n$. But this is equivalent to $n$ being prime
cloud walrusBOT
Ceana:
next obsidian
Huh, I guess so
But then wouldn’t f(a) = a^{pq} never be a homomorphism?
In an integral domain
Also wait, why does that imply n is prime?
Is there a specific term you look at?
elder valley
i'm not seeing how you got that congruence. maybe i'm slow
next obsidian
Expand (a + b)^n
Actually I also don’t quite see it now that I think about it
Why can’t the different powers cancel each other out
elder valley
yeah
burnt pond
Ok. Slowly. So $$(a+b)^n=\sum_{i=0}^n \binom{n}{k}a^{i}b^{n-i} = a^n+b^n+\sum_{i=1}{n-1}\binom{n}{k}a^{i}b^{n-i}$$
cloud walrusBOT
Ceana:
next obsidian
Yes
elder valley
yep
burnt pond
So now you know that $R$ is integral , so neither $a^i$ nor $b^{n-i} is zero, when $a,b \neq 0$.
next obsidian
Yeah
cloud walrusBOT
Ceana:
Compile Error! Click the 
reaction for details. (You may edit your message)
next obsidian
Also I can come up with a counterexample
Take any char 0 integral domain
And just do (1 - 1)^7 or something
Isn’t that still the same as (1)^7 + (-1)^7?
vale flint
can we go down to the case $pq=6?$ is there no work-around the freshman's dream thing from above?
cloud walrusBOT
deiknymi0:
next obsidian
Actually still not true
Tske Z/2Z
Then a -> a^2 is an automorphism
Do it twice tells you a -> a^4 is too
Even though that’s actually the same map
burnt pond
and 4=2*2 not coprime
next obsidian
But you said
That n needs to be prime
And equal to the characteristic
And I said I thought multiples of the char can work
I think this shows that powers of the char can work
Because you can do frobenius over and over
burnt pond
Yeah true
next obsidian
But perhaps it’s only powers
burnt pond
It even works when $n$ is a charmichael number
cloud walrusBOT
Ceana:
elder valley
can we go down to the case pq=6? is there no work-around the freshman's dream thing from above?
@vale flint well the map f is multiplicative on R for any commutative ring R, so really the only interesting aspect to look at is the homomorphism on the additive group. so that leads to looking at (a+b)^n = a^n + b^n
next obsidian
Honestly a classification seems really really hard
I would be absolutely stunned if you can get a nice classification of them
elder valley
i mean i think you can solve it with field theory but deiknymi0 said this problem was posed before learning that
next obsidian
Do you suspect the thing has to be a finite field?
I’m not really even sure how to leave the land of integral domains
vale flint
aren't all finite integral domains fields?
next obsidian
Why does this need to be finite tho?
Also what you said is true
You can even get away without assuming it’s commutative and prove it’s a field
vestal snow
Does this not imply P^(n) = P^n?
next obsidian
No
Because P^n isn’t prime
It doesn’t get preserved
If you knew P^n were prime then yes, but really this is looking at phi^-1(phi(P^n)) which one general is simply contained in P^n
vestal snow
Oh okay
next obsidian
Where phi sends x -> x/1
elder valley
f being a homomorphism means that all elements of <1> satisfy x^pq-x, which has at most pq distinct roots. so char(R)=|<1>| <= pq. you can extend f to an injective map on the field of fractions F of R. if R is finite then f is an automorphism of a finite field, and so must be a power of frobenius
does that make sense?
although like you said if R is a finite integral domain then it's a field, so you can immediately get that f is a power of frobenius
i'm not certain what that says about R in terms of p and q though
for example you can have R=Z/2Z and pq=anything, which implies f is the identity map = frobenius^0
vale flint
@elder valley sorry i'm new to this, what do you mean by <1>? trivial subring? i appreciate you writing this up. it's a bit too advanced for me, but \i'll work up to it in a few days though
elder valley
The additive subgroup generated by 1
pure crest
How do i find $Gal(E/Q)$ if E is the spiting field of $f(x) =x^4 - 10x^2 + 1 $
solemn rain
what did you try
elder valley
Is that supposed to be degree 2?
solemn rain
and yea i think you meant x^2-10x+1
cloud walrusBOT
boat:
pure crest
sorry i mean degree 4
pure crest
ig?
solemn rain
i guess
pure crest
$Q(\sqrt{2}+\sqrt{3})$ is a splitting feild of f(x)
cloud walrusBOT
boat:
pure crest
or perhaps $Q(\sqrt{2}, \sqrt{3})$ but they are equivalent
cloud walrusBOT
boat:
pure crest
but i am not sure how to proceed further
elder valley
Any automorphism of E fixing Q is determined by it's action on sqrt(2) and sqrt(3)
And you can reason the possible options for where those can be sent
pure crest
but all the roots invovle both sqrt(2) and sqrt(3)
elder valley
You can forget about f now basically, since you have E=Q(sqrt(2),sqrt(3))
pure crest
right
but isnt g in GAL(E/F), such that if x is a root of f(x) then g(x) is a root as well?
solemn rain
you can think of Q(sqrt(2),sqrt(3)) as a Q-vector space
elder valley
Yes. Apply that logic to the minimal polynomials of sqrt(2) and sqrt(3) though
elder valley
It doesn't now that you have the splitting field. That's what I said above
but isnt g in GAL(E/F), such that if x is a root of f(x) then g(x) is a root as well?
@pure crest the f in this statement can be any polynomial, not just the one you started with
elder valley
Pretty much
pure crest
once you have the splitting feild, you can consider the isomorphism in it that fixes F
and works on sqrt(2) and sqrt(3)
but i also learned that if f(x) = g(x) h(x), then the if x is a root of g(x) then phi(x) (an automorphism) must be a root of g(x) and not h(x)
why does this happen if the function doesnt matter?
elder valley
Because automorphisms permits the roots of polynomials in general. So it permutes roots of g and permutes roots of h separately
The fact that f is the product of g and h is kind of irrelevant to that
pure crest
ohh i see
elder valley
The full statement I think is: if E/F is a galois extension and f(x) in F[x] splits, then for any automorphism phi in Gal(E/F) we have f(phi(r))=phi(f(r))=0 for any root r of f. Meaning that phi(r) is also a root of f
E need not be the splitting field of f there
elder valley
No worries. For me it was a bit overwhelming at first
pure crest
are you a phd student?
elder valley
Formerly
pure crest
ohh i see
solemn rain
is there a linear algebra approach
thinking of phi ( an element in the galois group ) as a linear transformation
then computing basis elements
?
pure crest
makes sense that you are so good at this, i am in fourth year and i am struggling with it
honestly i feel like i learned very little over the 4 years haha
elder valley
is there a linear algebra approach
@solemn rain in the example we had above if you take {1, sqrt(2), sqrt(3), sqrt(2)sqrt(3)} as the basis, then phi is determined by it's action on sqrt(2) and sqrt(3). You can express phi(sqrt(2)) and phi(sqrt(3)) as linear combinations of the basis elements and solve for the coefficients using linear algebra
You'll have 2 equations and 8 unknowns
pure crest
so i would have 4! possible
and i would have to elimiate any that maps sqrt(2) to sqrt(3) by using the fact that it maps a root to a root?
elder valley
It's easier than that that approach will work but you can use an easier method
What are the minimal polynomials of sqrt(2) and sqrt(3)?
elder valley
Right. So the roots are sqrt(2), -sqrt(2) for one and sqrt(3), -sqrt(3) for the other
So a single phi has to send sqrt(2) to either sqrt(2) or -sqrt(2), and similarly for sqrt(3)
pure crest
why are we able to make that claim?
elder valley
The full statement I think is: if E/F is a galois extension and f(x) in F[x] splits, then for any automorphism phi in Gal(E/F) we have f(phi(r))=phi(f(r))=0 for any root r of f. Meaning that phi(r) is also a root of f
because of this
elder valley
You take f=x^2-2 and r=sqrt(2) in that statement
pure crest
so once i have the spiltting field, i can choose any polynomial
like here we chose x^2 - 2
let r be a root of x^2 - 2
solemn rain
the roots of the polynomial that splits in the field of polynomials are permuted by the elements of the galois group
elder valley
so once i have the spiltting field, i can choose any polynomial
@pure crest yes, as long as it's coefficients are in F and its roots are in E
pure crest
then 0 = phi(0) = phi( r^2 - 2) = phi(r)^2 - 2 ==> phi(r) = +- sqrt(2)
elder valley
Yes
elder valley
Yeah
pure crest
thank you very much
i was under the impression that it has to be with just F
since f(x) is irreducible
thats why i was so confused as to how to calculate it
How did you find grad school?
i feel like i am too dumb for it
at least grad school level mathematics
i want to go into something less abstract
but maths all i know
elder valley
I thought it was great fun. A few classes were pretty tough but overall not bad. Many of my colleagues disagreed though and definitely did not find it fun
pure crest
did you always like mathematics?
elder valley
By the way, our approach up there significantly reduced the number of possible phi out of the original 4!
Since after I started college yes
pure crest
i like mathematics, but i always hated analysis
its kinda my worst subject
i am afraid that i wont be able to get in with my bad analysis marks
elder valley
We should stay to abstract algebra topics in this channel though, in case people have questions and don't want to interrupt the off topic conversation
pure crest
Oops sorry
elder valley
No worries
vale flint
@elder valley I'm sorry if I'm being really stupid but I have no clue what your argumentation was from before. First, how does f being a homomorphism that takes x -> x^{pq} mean that all elements of additive subgroup generated by 1 satisfy the equation x^{pq} - x? secondly, why does that let us say something about Char(R)?
i'm thinking something about the Frobenius endomorphism, but that's about fields of prime characteristic, so none of that's applicable here. if it is, why?
elder valley
for example f(3)=f(1+1+1)=f(1)+f(1)+f(1)=1+1+1=3, but also f(3)=3^pq, so 3^pq-3=0
so all elements satisfy x^pq-x, but this polynomial has at most pq roots. since we found |<1>|=char(R) many solutions, we have char(R)<=pq
vale flint
wait but that isn't true in general...? $f(x+y) = (x+y)^{pq}$ while the homomorphism gives us $f(x+y) = f(x) + f(y) = x^{pq} + y^{pq}$... and then if we're operating modulo pq, then this is a tautology?
wait well i guess i see what you're saying, but then wouldn't it be $(x+y)^{pq} - (x^{pq} + y^{pq}) = 0?$ but aren't we taking everything modulo $pq$??
vale flint
nevermind i'm being dumb
elder valley
yeah i'm not sure what you're saying xD
i'm specifically looking at x=y=1 basically
vale flint
well yeah we can do that since we can decompose each element of R as a sum of 1_R... now i understand sentence #1 of your argument, thank you
elder valley
not each element necessarily. like <1> doesn't have to equal R
vale flint
why not...? 
elder valley
for example if R is real numbers then <1> is only Z
in a finite field of size p this will be true
or in R=Z this is true. but those are pretty much the only times it works
i dunno if this whole argument is even useful so it might be a big waste of your time
pure crest
that would be true only in Z_p right?
elder valley
maybe i misunderstood but i thought the statement was that <1>=R
so that's true when R=Z
vale flint
what did you mean by "power of frobenius"? i clearly don't have the same understanding as you, since i'm thinking frobenius endomorphism... 😕
but that can only be extended to p^k, not at all to two distinct primes like we're asked... so i guess the problem remains unsolved?
maybe we could decompose f as the composition of two individual frobenius endomorphisms and try to prove something with that? but then we get a field of a non-prime characteristic!?
elder valley
yeah power of frobenius means p^k for some k. but yeah i'm not sure where to go from pi^k=f, where pi is frobenius
essentially you get r^(s^k)=r^(pq) for all r in R, where s=char(R)
this was assuming R was finite, so you can maybe get some modular equation with |R|
chilly ocean
hello guys
I wanted to ask something owo
If the determinant is a multilinear form, can it be represented as a tensor?
like, is the determinant an element of T_{n}^{0}V?
owo, I think I answered myself
burnt pond
you did
chilly ocean
lol
burnt pond
for a orthonormal basis the determinant is then represented by the levi-civita symbol
chilly ocean
oh, that's nice to know <3
vestal snow
How do I prove the existence of minimal elements in the set of closed sets that are not the finite union of irreducibles
I tried the standard trick of zorn's lemma and considered a chain C_i
intersected all of it to get another closed set C
However, I don't know how to show that C itself cannot be written as a finite union of irreducibles
burnt pond
Why are you intersecting the C_i?
vestal snow
To get a maximal (minimal) element
I'm defining $A \leq B$ as $B \subseteq A$ in the set
cloud walrusBOT
The formerly edible banana:
burnt pond
Whats your definition of a noetherian topological space?
vestal snow
Descending chain condition on closed sets
burnt pond
use that on your chain
vestal snow
Can I do that?
Chains are referring to different things here
Perhaps I should call it the descending sequence condition
Because in one use of chain we can have an arbitrary index and in the other we cannot
burnt pond
Okay. Let's a chain $C_ 1 \leq C_2 \leq C_3 \leq ...$.Then $C_1 \supseteq C_2 \supseteq C_3...$
cloud walrusBOT
Ceana:
vestal snow
Hold on
burnt pond
So a maximal element with respect to $\leq$ is a minimal element in the descending chain of closed sets
cloud walrusBOT
Ceana:
vestal snow
A chain (in this context) need not be countable
All we need is that it is totally ordered
next obsidian
Just suppose there wasn't minimal elements. Then you can just keep getting a descending chain of closed sets S1 > S2 > S3 >...
But this contradicts the descending chain condition
For the Noetherian topological space
vestal snow
Wait
next obsidian
oh wait
This is really dumb
Noetherian is ascending
lmao
Artinian topological space bullshit
Umm
Wait am I being dumb
vestal snow
Noetherian is descending chain of closed sets
next obsidian
That is noetherian right
vestal snow
You're right lol
next obsidian
yeah so that should work right?
vestal snow
Yup
next obsidian
brainlet moment
burnt pond
wait hows your argument different from mine, what am i missing
next obsidian
I just didn't argue it in terms of Zorn's or chains or something
tbh I just skipped over that and said "pretend there isn't"
Cool fact: A group with finitely many subgroups is finite
pure crest
If E is a field extention of F, and if K is a subfeild of E, then must $ F \subset K$ or $ K \subset F$
cloud walrusBOT
boat:
next obsidian
you can have neither occur
take like algebraic closure
Think like
Q-bar
as an extension of Q(sqrt(2))
and then the subfield like, idfk
Q(2^1/3) or something like that idk
burnt pond
Q(sqrt(2),sqrt(3)) is a field extension of Q(sqrt(3)). And Q(sqrt(2)) is a subfield.
pure crest
ohh damn
you guys are right
i am trying to prove the fact that if [E:F] is a prime, then is K is a subfield, then K = E or K = F
i tried to make the arguement that if K is a subfeild of E, then F \subset K \subset E
but i dont think i can make that
next obsidian
I mean
this is just multiplicity
[E:F]
= [E:K][K:F]
so one of them is 1 the other is p
if [E:K] = 1 then E = k
pure crest
right
elder valley
you have to assume F is a subfield of K though
pure crest
but i need F \subset K \subset E
next obsidian
Oh right
uhhhh
Is it Galois?
lol
Actually that again only says stuff for intermediary ones
next obsidian
I think you need to assume more then
Namely that it's an intermediary subfield
Because as stated Ceana did have a counter example
burnt pond
It works when F is the prime field of E. Maybe that’s the missing assumption?
elder valley
it doesn't need to be though. that just forces K to be an extension of F. assuming F contained in K should be the assumption with most generality that makes it work
woven delta
Do the elementary row operations generate GL_n(Z)?
(the row operations are swapping 2 rows, adding one row to another, multiplying a row by a unit in Z)
scarlet estuary
every invertible matrix is a product of elementary matrices, yes.
hot lake
it works but it probably gives you stuff over Q
scarlet estuary
wait fuck i read GL_n(Z) as GL_n(F)
smoky cypress
In Jacobson page 76, it gives the formula $\operatorname{Stab}axa\inv=a(\operatorname{Stab}x)a\inv$, where $\operatorname{Stab}x$ is the stabilizer of $x$ in a set $S$ which the group $G$ acts upon. But $xa\inv$ doesn't make sense since group action acts on the left side
How should I interpret this formula
cloud walrusBOT
Whoever:
hot lake
it's Stab(ax) instead
smoky cypress
Oh ok makes sense
charred pewter
scarlet estuary
have you tried anything? what's holding you up?
scarlet estuary
no worries
lavish sigil
Let $G$ and $H$ be groups, and let $X$ be a set. If $\theta:H\times X\to X$ is a group action and if $f:G\to H$ is a group homomorphism, then we have a group action $\phi:G\times X\to X$, defined by $\phi(g,x)=\theta(f(g),x)$. If we view $\theta$ as a homomorphism $H\to\mathrm{Sym}(X)$ (where $\mathrm{Sym}(X)$ is the group of permutations on $X$), then $\phi$ is just the composition of $\theta$ and $f$. But is there any other widespread notation or name for this action $\phi$ in terms of $\theta$ and $f$?
cloud walrusBOT
gustavn64:
elder valley
it looks something like a pullback
smoky cypress
It claims here that $G$ acts on $S$ imprimitively if and only if there is a proper subset $A$ of $S$ such that $|A|\geq2$, for any $g\in G$ either $gA=A$ or $gA\cap A=\varnothing$. The only if direction is pretty obvious but I can’t see why the if direction is true
cloud walrusBOT
Whoever:
smoky cypress
Oh I see the “cosets” of A should be a partition that works
smoky cypress
@next obsidian Jacobson’s basic algebra
next obsidian
Lol wut
knotty mason
Jacobson’s basic algebra. nice
vestal snow
What are the prerequisites for Qing Liu's book on Algebraic Geometry?
olive mirage
you definitely want to know a good bit of Ring Theory. I'd highly recommend significant previous experience with algebraic geometry (SHafarevich and Eisenbud/Harris), as well as some content from Riemann Surfaces.
It is probably the textbook I have spent the most time studying @vestal snow
woven delta
Lol
olive mirage
I really don't think people should learn algebraic geometry without learning complex geometry. Without that piece, it makes no sense.
vestal snow
Isn't Qing Liu's book an introduction to Algebraic Geometry?
olive mirage
Liu's book is designed so that if you understand it you can do research with it. It technically has "introduction" in the title, but I would not touch it without a year of serious algebraic geometry study
(it is a very good book, and if I had a vote people would use it instead of Hartshorne)
elder valley
do you also feel then that hartshorne is not a good intro book?
bleak abyss
Hmm, I guess it depends on how much complex geometry
Do you think you need like, Kahler metrics or something?
next obsidian
do you also feel then that hartshorne is not a good intro book?
@elder valley yes he has expressed that
I’m also don’t think it’s a good intro book but I am a slave to the structures my forefathers put in place
E.g. slave through that book
olive mirage
Hartshorne is not intended as an intro book, but I don't think it is a good book for becoming an algebraic/arithmetic geometry researcher either. But of course there is a lot of AG out there, I'm sure there are some things you could want to do that it would be helpful with
(and I should be clear, there is nothing wrong with studying the foundations of AG, a la Hartshorne. It is fascinating stuff. It just doesn't really help you do research, because AG is geometry and the foundations are algebra)
Of course I feel that way because I was taught by people who feel that way. I'm sure there are others who feel differently. We benefit from diverse viewpoints.
knotty mason
why does a conjugacy class size divide the order of the group?
elder valley
the size of the conjugacy class is the index of the normalizer, which divides the order of the group
it's a direct consequence of the orbit stabilizer theorem i think
nova plank
@olive mirage Recommended intro and prerequisites for that intro?
Sorry, if you've already said and I missed it
olive mirage
I think the ideal path into algebraic geometry is something like:
- Miranda
- Shafarevich
- Eisenbud/Harris
- Liu or Vakil
nova plank
Prereqs for Miranda?
olive mirage
Complex Analysis mostly.
nova plank
Thanks
smoky cypress
May I just say that Sylow theorems are pretty cool
nova plank
You may.
vestal snow
@olive mirage My advisor said that he felt Hartshorne does not do a good job of teaching you how to do problems in AG.
olive mirage
haha that's what my advisor told me too
vestal snow
We might have had the same advisor since you're from Arizona too
nova plank
Dox yourselves and find out
olive mirage
My identity is not super hidden haha
I once stumped @vestal snow's advisor with a problem from my abstract algebra exam haha
nova plank
Are you Miranda and that's why you're plugging your own book? /j
olive mirage
haha I wish.
olive mirage
It was "GIve an example of an element of order 6 in A_7 or prove none exists"
olive mirage
I think it is a bad question because it is a bit too easy to write a "proof"
but in my experience undergrads do better than grad students, and grad students do better than PhDs at this question haha
vestal snow
Does no such element exist?
olive mirage
we are in the #groups-rings-fields channel, I'll let folks think about it (-;
knotty mason
are those in A_6 though?
vestal snow
But I don't remember if no. of transposes in A_n was required to be even or odd
knotty mason
even
smoky cypress
even
knotty mason
🤔
vestal snow
It's in A_7 right?
knotty mason
looks good
another technique that might work
|A_7| = 7!/2 and 2 and 3 divide this
so by cauchys theorem you get elements of those orders
vestal snow
Ah slick
knotty mason
nothing beats a concrete example though
vestal snow
And then the subgroup generated by them intersects trivially
I just did the partitions trick
smoky cypress
Wait but is the group abelian?
vestal snow
No
smoky cypress
Yeah so you have an element of order 2 and 3, but does that guarantee an element of order 6?
smoky cypress
But if the group is abelian then it should work right?
vestal snow
Yes
smoky cypress
Yeah that's what I thought
vestal snow
As long as one of the elements is in Z(G), it should work I think
smoky cypress
Makes sense
vestal snow
Or if you want to make it as general as possible
One is in the centralizer of the other
golden pasture
as long as x^2=e, y^3=e and xy=yx
vestal snow
Oh yeah Ari's right
olive mirage
I recall having my mind blown when I realized that if x had order 2 and y had order 3 then xy could have infinite order
chilly ocean
cna u give an exmaple
smoky cypress
chilly ocean
what is this link for
smoky cypress
What can the order of xy be in a group
chilly ocean
ure asking in ableian groups? since this link is about abelian ones
smoky cypress
The answer didn't assume the group is abelian
chilly ocean
what would be the case when ordx=2 ord y=3 and ord xy=infty?
and in the answer I think its only for finite ord
smoky cypress
True
wind parrot
you can take < x,y | x^2=y^3=e >
chilly ocean
Ok yeah I guess
olive mirage
haha yeah that would be my example
vale flint
going back to the A_7 question, can someone please explain why (12)(345)(67) works? would (123)(45)(67) work as well?
smoky cypress
Yeah
Those are conjugate elements
and conjugate elements have the same order
Well actually they're conjugate in S_7, and not necessarily in A_7
next obsidian
The order of a product of cycles is just the lcm of the orders of each cycle
Since they commute
(When the cycles are disjoint)
Since they commute
@next obsidian actually this isn’t enough, but you can just show it manually
chilly ocean
@smoky cypress the reason it works the reason that having an order-6 element g implies order-2 and -3 elements is because you can take powers: g^3 and g^2. But there's no reason for it to work the other way around.
smoky cypress
?
I never said anything about the existence of an element of order 6 implies the existence of elements of order 2 and 3
chilly ocean
maybe I misread something
smoky cypress
Yeah so you have an element of order 2 and 3, but does that guarantee an element of order 6?
This is what I said
next obsidian
If they commute 😏
smoky cypress
What's the most efficient way of showing $\Aut(S_3)=S_3$?
cloud walrusBOT
Whoever:
solemn rain
next obsidian
Let $B \subseteq A$ where $A,B$ are integral domains, $A$ is a finitely generated $B$-algebra, and there are finitely many primes of $A$ lying over $0$. Then Frac$(A)$ is a finite extension of Frac$(B)$.
We know that $A = B[a_1,\dots,a_n]$, and it follows immediately that Frac$(A) = \text{Frac}(B)(a_1,\dots,a_n)$. We show that this is an algebraic extension, implying that the $a_i$ actually generate Frac$(A)$ as an algebra, but then by Zariski's lemma Frac$(A)$ is a finite extension of Frac$(B)$ as desired.
Suppose that some $a_i$ is transcendental over Frac$(B)$, then Frac$(B)[a_1,\dots,a_n] = C$ has transcendence degree $d > 0$ over Frac$(B)$. By Noether Normalization, there exist elements $\gamma_1,\dots,\gamma_d$ algebraically independent over Frac$(B)$ such that $C$ is finite over Frac$(B)[\gamma_1,\dots,\gamma_d]$, so in particular it is integral over this subring. As polynomial rings over a field have infinitely many primes, this subring has infinitely many primes, and then by the lying over property (going up) it follows that $C$ has infinitely many primes. We can view $C$ as the localization of $B[a_1,\dots,a_n]$ at the set $B\setminus{0}$, so in particular there are infinitely many primes of $B[a_1,\dots,a_n] = A$ which intersect $B$ only at $0$, which is to say there are infinitely many primes which lie over $0$. This contradicts our hypothesis, so that we conclude all $a_i$ are algebraic, and as argued before Frac$(A)$ is a finite extension of Frac$(B)$.
@vestal snow
:(
Rip the bot not doing it
This was a proof which used these sorts of arguments a lot which I sucked at before but got better at
cloud walrusBOT
Chmonkey:
vestal snow
Cool
elder valley
lol
solemn hollow
buncho watch out
smoky cypress
Suppose $G$ is a finite group, $\alpha$ is a group automorphism, and $I=\brc{g\in G:\alpha(g)=g\inv}$. If $|I|=\frac34|G|$ then $G$ has an abelian subgroup of index 2
cloud walrusBOT
Whoever:
smoky cypress
It's an exercise that I can't prove lol
chilly ocean
I kbow a person who solved it but is on differebnt math server
chilly ocean
I added him to friends but he mightve blocked me by accident and i checked he left the othrr server too but ill see what i can do
sonic current
hello everyone! I'm trying to prove that I_a is a subuniverse of the lattice R.
to show that it is closed under V operation is easy; I'm mostly having trouble with the ∧ operator
steady axle
is that spanish
sonic current
Portuguese
steady axle
oh i see
sonic current
So (x ∧ y ) v a, for all x, y, must be equal to a
and I can't rely on distributivity, of course
I think I have three possible cases for x ∧ y: it's either x, y, or an element z such that z ≤ x and z ≤ y
how can I argue that z ≤ x, z ≤ y, and since x v a = a and y v a = a, then z v a = a
since x v a = a, then x ≤ a; similarly, y ≤ a
but if z ≤ x and x ≤ a, then z ≤ a
cool!
burnt pond
@smoky cypress do you want a full proof for this or only a hint?
timid hull
Suppose $G$ is a finite group, $\alpha$ is a group automorphism, and $I=\brc{g\in G:\alpha(g)=g\inv}$. If $|I|=\frac34|G|$ then $G$ has an abelian subgroup of index 2 . I'm really curious to see the sol for this problem ... I think $I$ generates the entire group $G$
cloud walrusBOT
WhyamIsohot?:
burnt pond
almost correct
timid hull
Basically every element a of G can be written as a=xy where x,y are in I ...
@burnt pond Can u say your sol ?
thanks!
solemn rain
whats H
chilly ocean
Let G be a group and f be an automorphism. Let H = { x : f(x) = x^-1 }. Assume |H| > 3/4 |G|. Prove that G is Abelian and f(x) = x^-1 for all x in G
burnt pond
The proof for the original question is almost the same. For any $x,y \in A$ we have $xy \in A$ iff $xy=yx$. Since $|A|=\frac{3}{4}|G|$, $G$ can't be abelian, otherwise $A$ is a subgroup of $G$. Now let $x \in A$ be any element which is not in the centralizer if $G$. We have
$A\cap xA \subset C(x)$ and $|A\cap xA|\ge\frac{1}{2}|G|$ by inclusion–exclusion-principle.
Thus $|C(x)|=\frac{1}{2}|G|$ and $C(x)=xA\cap A$. Last thing to show is that $C(a)$ is abelian. Use $xy \in A$ iff $xy=yx$ for that.
cloud walrusBOT
Ceana:
shy bluff
How does an infinite direct product of groups work?
Like I'm reading the wikipedia page
why can we take the direct product of an infinite number of groups?
Or like in general, why can we go and do things like take an infinite product?
solemn rain
why not
shy bluff
hrm
Idk
It just seems weird to me to go and use infinity
Anyawys, how do you show that something's an orbit?
Like usually an orbit of x is just the result of the group action on said x
But this question asks me to show that something is an orbit instead?
elder valley
find an x such that the thing you're looking at is the orbit of x
shy bluff
But I'm given said x?
elder valley
so show everything is in the orbit of that x
shy bluff
I dont' think I'm understanding
elder valley
give some more detail about what you're trying to show and maybe i can explain better
i'm not sure what you're confused about
shy bluff
I'm trying to show that if N is a nilpotent 2x2 matrix, then for each i = 0 or 1, the set of nilpotent matrices of rank i is an orbit under conjugation by elements GL_2
So to me that means that for every g in GL_2, we have to have that gNg^-1 is in the orbit of either N with rank 0 or N with rank 1?
elder valley
yeah that's phrased a bit poorly. i can see why you're confused
i assume by "an orbit" it means "the orbit of N". otherwise N has no purpose
shy bluff
like the way that I'm understanding it is that we have some sets N_i such that N_i is made up of all the nilpotent matrices of rank i
And they want us to show that all of N_i is ... an orbit? Under conjugation?
elder valley
yeah that's probably the best way to interpret it. seems like N in the statement is irrelevant
maybe someone else has a better idea
elder valley
it means you can find a matrix M such that N_i is the orbit of M under conjugation by GL_2
shy bluff
As in you can find M such that MN_iM^-1 is ... in O(N_i)?
wouldnt' that M always exist...?
elder valley
no, you have the group GL_2 acting on N_i. so you should get that N_i = GL_2 * M, where * is the group action. so every N in N_i should be able to be written as N=G * M * G^-1 for some G in GL_2
the * in the last sentence is matrix multiplication, sorrry
elder valley
basically you're showing that all nilpotent matrices of rank i are similar
shy bluff
thank you 
WEll
all nilpotent matrices of rank i would have the same characteristic polynomial
ok thank you
smoky cypress
@chilly ocean that's not quite the problem, what I asked for was |H| equals 3/4|G|, not greater than
@burnt pond thank you, I almost got the proof except I didn't notice that G cannot be abelian
next obsidian
It just seems weird to me to go and use infinity
@shy bluff infinity here is sort of not capturing the whole picture
You can do it with respect to any indexing set, of arbitrary cardinality. So like you could assign to each natural number, a group which forms one part of the direct sum/product
And you could even index it by R if you chose (you’d want to well-order it), but it comes down to infinite ordinal / cardinal stuff
That’s why when you treat it rigorously instead of describing it as tuples you describe it as the set of functions {f: I -> \cup G_i| f(i) in G_i}
smoky cypress
Let say L/K is a finite degree field extension, then is it necessarily the case that L is isomorphic to K[x]/(f(x)) for some prime f in K[x]?
oblique river
in characteristic 0, yes
this question is basically equivalent to the primitive element theorem
err, answered by
smoky cypress
If K has characteristic 0 then this is true?
oblique river
yes
smoky cypress
Interesting ok
oblique river
if the extension is separable then it's true
knotty mason
I don't think it's true is it?
it needs to be a normal extension
oh wait I get what you're saying about primitive element, for Q(2^{1/3}) you'd use the minpoly of 2^{1/3} + zeta_3 right?
oblique river
yeah, in general, you'd just use the minpoly of the primitive generator
vestal snow
Can someone explain what transcendence degree is?
From what I understand, the tr deg A over k is the "largest" polynomial ring over k that one can embed into A
vestal snow
If $k=K$, how does that give us that $x_i - a_i \in m$?
cloud walrusBOT
The formerly edible banana:
oblique river
X_i - a_i is 0 in K
because the image of X_i is equal to a_i
and therefore X_i - a_i is in the kernel
vestal snow
Abusing notation?
oblique river
no
X_i - a_i is an element of the polynomial ring
the X_i all live in k[X1, ..., Xn]
(sorry)
when I said "X_i - a_i is 0 in K" I mean the image under the map we're given
vestal snow
Wait so a_i = x_i + m you mean, right?
oblique river
K is equal to k
so therefore the image of X_i is some element of k
call that element a_i
then X_i - a_i is an element of k[X1, ..., Xn]
and its image is 0 under the map k[X1, ..., Xn] --> k given by "quotienting by m"
I guess you can write a_i = X_i + m
vestal snow
This might be a bit dumb, but are we treating = and "is isomorphic to" as the same here?
oblique river
this all depends on how you've been stating things
in the proof you pasted, it refers to a previous theorem and says "therefore K = k"
or something like that
I'm treating K and k as equal to each other
which is stronger than just isomorphic to
oblique river
if you assume k is algebraically closed, then it follows that K is literally equal to k
vestal snow
We should get x_i + m = something in k
oblique river
yes
vestal snow
So we call that something a_i
oblique river
you don't have to call it X_i + m though
oblique river
I just personally find that carrying around the +m isn't always helpful, but ymmv
as I phrased it above, "the image of X_i in k is a_i"
we have a map k[X1, ..., Xn] --> k given by quotienting by m
let a_i be the image of X_i
vestal snow
Ahh this is confusing me a bit lol
oblique river
X_i is only an element of the polynomial ring
and we can talk about its image in K (which is equal to k)
a_i is an element of k
which we can also think of as an element of the polynomial ring as just a constant
X_i - a_i is an element of the polynomial ring
vestal snow
Oh
oblique river
it's not an element of K
vestal snow
OH
vestal snow
But x_i - a_i + m = 0 in K
vestal snow
Therefore x_i-a_i is in the kernel of the map from K[x_1,...x_n]
oblique river
yes
vestal snow
which is m
oblique river
always help a fellow banana
vestal snow
Haha yeah
@oblique river Now that I write it down, aren't we implicitly assuming that a_i (as a polynomial) gets mapped to a_i (as an element of K=k[x_1,...,x_n]/m)
Like consider x_i - a_i (the polynomial)
oblique river
yes -- that's automatic
these maps are all k-algebra homomorphisms
which preserve elements of k
vestal snow
Oh okay
That makes sense
Though I don't see it mentioned in the statement of the theorem
Is that something we're supposed to assume?
implicitly
oblique river
hmm no I think it can be deduced, let's see
so I think it comes down to the fact that K is literally k
I'm goign to call the polynomial ring R cuz i'm tired of writing it all out haha
vestal snow
So R = k[x_1,...,x_n]/m?
vestal snow
Okay
oblique river
so like, think of the composition of maps k --> R --> K (= k)
this map must be surjective
vestal snow
Yes
oblique river
because if not, then that would give us a finite field extension K of k
oblique river
k is supposed to be algebraically closed
and so shouldn't have any finite field extensions
oblique river
yeah -- the map k --> R --> K is an isomorphism
and we can just use that to label elements of K
like, when we call something in R/m "a" where a is an element of k
we mean the image of a under this map k --> R --> K
which is also just the image of the constant polynomials
vestal snow
Oh okay
oblique river
here's another way to say it -- a field might have different ways of labeling elements, but the labels aren't part of the field itself
vestal snow
Got it
oblique river
for example let's say we could write down all of the complex numbers, and someone asked each of us to point to the number i in that list
you might point to "i", and I might point to what you call "-i"
all this means is that there is a field automorphism of C which sends i to -i (and we know what it's called -- complex conjugation)
this isn't a problem, it just means one of us should speak up and say "no, we're labeling things this way so that we're on the same page"
vestal snow
Got it
oblique river
that's kidn of what is happening here -- we're just saying like "okay, once and for all, when we talk about k, here is how we're labeling things"
vestal snow
So we are defining an embedding of k into K in the way that a_i + m = "a_i"
oblique river
yeah
the embedding of k into K given by a_i --> a_i + m is surjective
and so we might as well call a_i + m as a_i
oblique river
yeah
vestal snow
One last question
oblique river
this proof could be rephrased to be a little more precise, but at the end of the day I think because it really just comes down to naming conventions, it's not really a huge issue
vestal snow
If k -> R -> K (=k) is not surjective
What do you mean when you say that we get a finite field extension
Is R the finite field extension you are referring to?
oblique river
no
R is a polynomial ring
definitely not a field
the image of k under hte composition k --> R --> K
is just k
and so K is a finite field extension of "the image of k under this map"
but k doesn't have any proper finite field extensions
therefore the map must be surjective
oblique river
it's algebraic and generated by finitely many elements
shy bluff
Given that $G$ is a divisible abelian group and $H$ is a proper subgroup of $G$, how do I show that $G/H$ is infinite?
cloud walrusBOT
Liria ^(;,;)^:
shy bluff
I've looked at the arguemuent fo rQ/Z but I don't really see how that works?
vital quail
This is the difference between an edible and not banana
shy bluff
Pardon?
shy bluff
For all g in G, there exists some integer n and some y in G such that ny = g
in a group
Err wait
Yea for every n
sorry
every postiive n*
so I want to prove that for every positive n, and for every g in G, there exists some y in G such that ny = g
undone bay
I think this should be a sketch of a proof @shy bluff (I'm sorry if I'm misunderstanding something here, not very familiar with divisible groups).
Divisible groups are necessarily infinite, demonstrated as follows: let $g \in G$, let $y_n \in G$ denote the element such that $ny_n = g$, then show $y_n \neq y_m$ iff $n \neq m$, thus ${y_n ; | ; n \in \mathbb{N} } \subset G$ is infinite. \
So it suffices to show the quotient of a divisible group is divisible. Let $g \in G$: given that $ny_n = g$, can we find $z_n \in G/H$ such that $n (z_n H) = gH$? As $gH$ is a generic element of $G/H$, this would ensure divisibility of $G$.
cloud walrusBOT
craft:
latent anvil
@undone bay i don't think yn ≠ ym is true necessarily. Take g = 1/2 + Z in Q/Z. Then 3 g = g so y3 = y1
I mean you don't even have unique yi's necessarily
next obsidian
Sham this is mega overkill but
Is there a reason injective Z module must be infinite
Like a good one
latent anvil
I mean my proof that divisible abelian groups are infinite is the classification of finite abelian groups
There's probably a much more direct way to see it
next obsidian
Lmfao nice
undone bay
@latent anvil good catch, yea my way of showing divisibility -> infiniteness isn't right
but im pretty sure that fact does hold
undone bay
Let $G$ be a non-trivial divisible group with finite order $n$, then let $G \ni g \neq e$. By hypothesis, $\exists y \in G$ such that $ny = g$, but we also know $nh = e, \forall h \in G$, thus $g = e$ and the group is trivial (a contradiction).
cloud walrusBOT
craft:
next obsidian
Maybe you can still get some finiteness via some sort of idea similar to that
Something like only for at most finitely many n
Or something
next obsidian
Hahaha thx man
undone bay
yeah sorry I don't mean to be trampling on your proof I just wanted to fix what I said and not leave all that incomplete.
latent anvil
No you're fine
Your proof also works for "divisible" nonabelian groups
Which is sweet
undone bay
shamrock pointed out that the hint I included for demonstrating that all divisible groups are infinite was misguided. all divisible groups are indeed infinite, but you have to prove it a different way (one way being the new proof I gave a few messages back) @shy bluff
shy bluff
The one about how it's trivial?
undone bay
yep. basically I showed that a finite divisible group must be the trivial group {e}, so divisible groups are either that (which we really don't care about) or infinite
undone bay
yeah that fact nh = e isn't obvious. basically, Lagrange's theorem tells us that the order of a group must be divisible by the orders of each of the subgroups. One subgroup of a group G is the subgroup <h> = {h, 2h, 3h, 4h, ...}. For a finite group, this set has to "end" at some point: there must be an "m" such that mh = e, and we start over ((m+1)h = h, etc). If m is the smallest integer for which this happens, then the subgroup <h> has order m. Because m must divide n (the order of G), then nh = cmh = c(mh) = ce = e.
you get infinite-ness because assuming finiteness requires your group to be trivial, so there are no non-trivial divisible groups
latent anvil
Nope
knotty mason
Q(2^{1/3}, 3^{1/3}) is a finite extension
but we can't apply primitive element theorem
latent anvil
Primitive element theorem requires the extension be separable
vestal snow
Is k[x,1/x] = k(x)?
chilly ocean
No
elder valley
no. there's no way to get other denominators on the left one
vestal snow
Lmfao nice Godel
Cool that's what I thought
Does this fact give an alternate proof to this theorem?
if r>0 and A is a field, then you can embed k(x) into A
and then we can use that k(x) cannot be obtained by adjoining finitely many elements to k
shy bluff
Say that I have 2 disjoint cycles, t = (1, 2) and s_i = (i, i + 1... n)
What does <t, s_i> look like?
Or like how do generators work with disjoint cycles?
next obsidian
Pretty sure it should just look like t^ns_i^m
For n between 0 and the order of that
Since they’ll commute
shy bluff
oh so it works the same way
next obsidian
Yeah normally you have to worry about the order in which you multiply
But disjoint cycles commute so
next obsidian
Like
shy bluff
For S_n
next obsidian
The element with which you’re
Taking the power of
I mean you could go higher
But then you’ll loop back around
shy bluff
Oh like the order of t and the order of s_i?
shy bluff
So like for t it's between 0 and 2 and then for s_i it's just (n - i)?
next obsidian
I guess
You can afford to make one of those strict
On either side doesn’t matter
Otherwise you include making it the identity twice
It’ll just be the length of the cycle
shy bluff
oh ok
next obsidian
For its order
shy bluff
And what about the order of the thing generated by <t, s_i>? Usually the order is just the lcm right? But does that still work for this?
next obsidian
Yes