Because R is an integral domain Frac(R) = F is a field and R \to F is injective
#groups-rings-fields
dim widget
If there was an injection R^n \to R
Then we would get an injection F^n \to F which is impossible by linear algebra
novel parrot
from dimension counting?
dim widget
The reason that F^n \to F would have to be injective is because localization preserves injections.
dim widget
from dimension counting?
Yes from dimension counting you see that this is only possible when n = 1 or 0.
delicate orchid
They have a basis 
dim widget
A free module is just $R \oplus R \oplus \dots R$ for some set of factors.
cloud walrusBOT
Topos_Theory_E-Girl
novel parrot
A free module is just $R \oplus R \oplus \dots R$ for some set of factors.
but could be infinite
dim widget
Basically it means that, as wew is saying, there is a set of generators with no relations.
novel parrot
infinite factors
dim widget
Yes some set of factors, not necessarily finite.
The problem with ideals is that (a, b, ...) always has the relation ba - ab = 0.
delicate orchid
Another way of thinking about freeness is that each element can be uniquely expressed as some combination of some fixed (basis) set of elements
novel parrot
Another way of thinking about freeness is that each element can be *uniquely* ex...
that is my intuition
delicate orchid
The problem with ideals is that (a, b, ...) always has the relation ba - ab = 0.
So this wouldn’t be free as ab = ba - two different ways of writing the same element (one with a being a scalar, and one with b being a scalar)
formal ermine
I just think of freeness as being like a vs
delicate orchid
The things me and topos are saying are equivalent so go with whatever
novel parrot
ok and free modules with finite basis of size n are always isomorphic to R^n as R modules, right?
delicate orchid
Yus
coral spindle
My face when Z is isomorphic to 2Z
novel parrot
My face when Z is isomorphic to 2Z
is it not true?
south patrol
As Z modules sure lol
novel parrot
As Z modules sure lol
Modules do not contain a multiplicative identity, hence 2Z not containing 1 doesnt matter right
obsidian loom
I know everyone has a different way to multiply cycles so it will probably be hard to help me
but I went wrong somewhere and I got: (1,2,7)(3,5)(4,6)(8)
I know 3 goes to 5, and then when f is applied, 5 doesn't move
so why is it (3,5,8)
lavish nexus
5 goes to 3 then goes to 8
obsidian loom
5 goes to 3 then goes to 8
ok but aren't you supposed to always start with the lowest possible number
like since (1,2,7) is a cycle the next cycle starts with (3,..)
lavish nexus
g changes 5
obsidian loom
lavish nexus
g sends 5 to 3
obsidian loom
Ok lets assume what you say is true
5 gets sent to 3, then 3 to 8
so it will just be a transposition (5,8)
obsidian loom
this topic is so confusing for no reason
obsidian loom
If I'm not mistaken this just breaks my lecturer's method of multiplying cycles
Would you mind sharing how you do it?
lavish nexus
take the smallest number and write it down
apply g then f
find out the number and write it down
apply g then f to this number
find out the number and write it down
etc until this number is one of your previous numbers
close the parenthesis
choose the next smallest number that has not been written down
do the same thing again
until you run out of numbers
obsidian loom
choose the next smallest number that has not been written down
ok right now im at 3
applying g to 3 gives 5
then applying f to 3 gives 8
obsidian loom
f doesn't move 5
lavish nexus
it does nothing exactly
so 3 goes to 5
now write down (35 and apply g to 5 and f to whatever the result is
obsidian loom
it makes sense now
im not exactly sure where my confusion was but its gone
thanks for your patience
lavish nexus
south patrol
Modules do not contain a multiplicative identity, hence 2Z not containing 1 does...
Indeed a Z-module is just an abelian group equivalently
And 2Z and Z are clearly isomorphic as groups
If you wanna think of it in a more natural way lol
novel parrot
And 2Z and Z are clearly isomorphic as groups
yes
formal ermine
And 2Z and Z are clearly isomorphic as groups
ong?
coral spindle
feels weird to me that R can be isomorphic to principal ideals as R modules
I pointed out that Z is iso to 2Z in response to this
It should not be surprising anymore
coral spindle
If you think of fields as commutative rings with no nontrivial ideals, it becomes clearer imo
It's interesting to me personally that the equivalent for noncommutative rings is so different: noncommutative rings with no nontrivial two-sided ideals are matrix rings over division rings
Like, you'd really want it to be just division rings, but no...
No
formal ermine
iff it's projective
coral spindle
Almost never
novel parrot
stupid question sorry
coral spindle
Bruh now it looks like I'm talking to myself
delicate orchid
take any cyclic group over Z as a counter example
formal ermine
related to the above topic (kind of): are there any equivalent conditions for all nonzero ideals of a ring to be isomorphic to a power of the subring of that ring as the subring modules? e.g. this is true for the ring of integers of a number field: every non zero ideal is isomorphic to Z^n as Z mods where n is the degree of the number field
coral spindle
yup I expect so
delicate orchid
I'd do smith normal yeah
or just bash them together like a caveman until they became linearly independant
novel parrot
how do i find a basis for N?
delicate orchid
what have you tried
delicate orchid
anything would be a good start but one mo
$N = \langle (X^2, X^3, X^5+X), (1, X+1, X^2+1), (X, X, X) \rangle$
cloud walrusBOT
Wew
novel parrot
yeah
delicate orchid
or just bash them together like a caveman until they became linearly independant
then do this 
coral spindle
Wew is suggesting you use Gaussian Elimination
novel parrot
oh
novel parrot
for the second part
delicate orchid
2nd part has to be smith normal surely
novel parrot
Yeah
it does
but
im wondering
do i need to use the linear independent list in the second part
or use original matrix
both should work right?
delicate orchid
both should work yeah, they span the same space
novel parrot
ill get an isomorphic cokernel?
novel parrot
Wew is suggesting you use Gaussian Elimination
i can only elemntary row operations right?
not column
seems to be Linearly independant already
if im not mistaken
this shit is painful
obsidian loom
Does a group G have a subgroup for every factor of |G|?
I mean H <= G and |H| is a factor of |G|
formal ermine
Does a group G have a subgroup for every factor of |G|?
if it's abelian sure
otherwise not necessarily
e.g. A_4 has no subgroup of order 6
or more generally
for n \geq 5 A_n can't have a subgroup of index 2 as that would mean it's no longer simple
obsidian loom
I see
formal ermine
if G is not abelian you have the sylow theorems
obsidian loom
I'm trying to prove "If G has an even order, then G has an element of order 2"##
I think the intended way is proof by contradiction
but I'm trying to do it directly
by lagrange
delicate orchid
It’s easier to prove it directly
obsidian loom
How? I can only show the element is of even order
delicate orchid
But not by lagrange
delicate orchid
How? I can only show the element is of even order
It’s even order, so you can split the elements into pairs
There’s a particular pairing which is useful here
obsidian loom
each element with its inverse?
delicate orchid
Yes, exactly
hollow tartan
I think this is the 1st sylow theorem
delicate orchid
There’s at least one element that doesn’t pair up nicely
south patrol
Sylow
formal ermine
it's simpler than sylow
obsidian loom
I'm kinda lost
delicate orchid
Yeah it’s just cauchy, of which I’m trying to guide someone through a proof of
obsidian loom
is the left over element the identity?
delicate orchid
It is, yes
obsidian loom
So how does that show there's an element of order 2
formal ermine
I think what wew meant to say was that you're supposed to pair every element in G \ { e } with its inverse
delicate orchid
Because there’s an even number of elements, and all but one can be paired up with their inverses - there must be a second element that can’t be paired too
delicate orchid
I think what wew meant to say was that you're supposed to pair every element in ...
No, I didn’t
You can’t pair them up, that’s an odd number
delicate orchid
Then why “correct” me?
obsidian loom
Because there’s an even number of elements, and all but one can be paired up wit...
holy fuck
and that element is b^2 = e
delicate orchid
Yes! Since that element can’t be paired with its inverse it must be equal to its own inverse - cause everything else is in a pair (or is the identity…)!
delicate orchid
No worries - this is one of those proofs which is very memorable but near impossible to come up with
For what it’s worth, I think you can do this with Lagrange as well but I can’t remember it and it’s not nearly as cool
hollow tartan
I was learning mobius transformations, and then wondered how many group actions are there? I know there's a library website for groups/rings/fields has everything been discovered about group actions or its just another tool that is not "perfect"
hollow tartan
novel parrot
for finitely generated module over a PID
we have this, "M is torsion free if and only if M is free"
why is that true?
chilly ocean
the quotients R/(a_i) will give you torsion elements if they are not trivial
pastel cliff
,av tterra
cloud walrusBOT
chilly ocean
pastel cliff
kilroy
novel parrot
the quotients R/(a_i) will give you torsion elements if they are not trivial
will those kill the free part?
novel parrot
the quotients R/(a_i) will give you torsion elements if they are not trivial
a_i is a torsion element of R/(a_i)
but a_i is not a torsion element of R
can you explain?
chilly ocean
can you explain?
no
cloud walrusBOT
ActiveChapter
novel parrot
any m can be written as (r_1, \cdots, r_n, r mod (a))
how does this get killed by a ?
bleak abyss
What does an element of the direct sum look like?
novel parrot
What does an element of the direct sum look like?
tuples?
delicate orchid
(0, \cdots, 0, 1) is very clearly torsion
delicate orchid
free => flat :packwatch:
delicate orchid
How could it be torsion as a vector space
You just showed that free modules are torsion free
delicate orchid
I’m not smart but thanks
novel parrot
How would i do it though? M if finite dimensional over k => M is torsion module over k[X]
i did the other direction
rustic crown
other is false >.<
f.g. torsion => f.d.
(but i assumed u had that implicitly >.<)
novel parrot
other is false >.<
wym?
charred crescent
i got a perfect score on my final. thanks for all of your help this semester y'all!
pure laurel
Hello, I hope this is the right channel for representation theory. I am doing an exercise where I should write down a complete basis of symmetric rank 2 tensors, and then use this basis to express the representation of generators of the rotational subgroup of D6 (i.e. rotations of $2\pi /6$ about the $z$-axis.
I have constructed a basis of symmetric rank 2 tensors by just using
$E{ij} \equiv \frac{1}{2}(e_i \otimes e_j + e_j \otimes ei)$
but I can't figure out how to figure out how to write the representation of the rotation in this basis. I know the fundamental representation of the rotation and I know how my symmetric basis tensors transform under this rotation giving me 6 equations $E'{ij}$ as a linear combination of the untransformed $E_{ij}$ but I'm really stuck here. Any pointers? Should I use the fact that you can decompose rank 2 tensors into a Kronecker delta, an anti symmetric part and a traceless symmetric part?
cloud walrusBOT
Solklar
pure laurel
The question specifically asks "a complete basis of second rank symmetric tensors under rotations", I am unsure of what effect the "under rotations" has on the sentence before
dim widget
The question specifically asks "a complete basis of second rank symmetric tensor...
Under just the rotations coming from D_6?
delicate orchid
you sound like a physicist, no offense
pure laurel
Yes exactly only the rotation subgroup
pure laurel
you sound like a physicist, no offense
Haha you got me
delicate orchid
you can define a group action of D_6 on the two basis vectors as it acts on R^2
novel parrot
just torsion doesn't imply f.d.
but its torsion over k[x]
and i showed that its f.d over k
rustic crown
like so like consider infinite product of k[x]/(f)
it's torsion, as f kills everything
so but f.d. over k
but if you assumed f.g. over k[x], then you good
dim widget
Haha you got me
Wait also are you starting with a 3-dimensional representation where D_6 acts trivially on the z axis?
pure laurel
you can define a group action of D_6 on the two basis vectors as it acts on R^2
I know how the fundamental rep of the rotation looks like, I am just unsure of how I can write it in a basis of symmetric tensors
novel parrot
but if you assumed f.g. over k[x], then you good
oh i did
rustic crown
are you allowed to use structure thm? >.<
novel parrot
yes
rustic crown
nice :3
novel parrot
im not sure about the reverse direction xd
M being f.d over k => M is torsion over k[x]
pure laurel
Wait also are you starting with a 3-dimensional representation where D_6 acts tr...
Yeah so the fundamental representation is $\begin{pmatrix}1/2 & -\sqrt{3}/2 & 0 \ \sqrt{3}/2 & 1/2 & 0 \ 0 & 0 & 1\end{pmatrix}$ but I have no idea how I could write this in a basis for symmetric tensors, since it's not even symmetric itself
cloud walrusBOT
Solklar
rustic crown
M being f.d over k => M is torsion over k[x]
f.d. implies f.g. so you can use structure thm 
delicate orchid
yeah I agree with you, that matrix generates the other reps
novel parrot
yes so M = k^n
dim widget
Yeah so the fundamental representation is $\begin{pmatrix}1/2 & -\sqrt{3}/2 & 0 ...
The tensors are living in the tensor algebra for the representation, and I'm confused what you mean by write this in a basis for symmetric tensors?
novel parrot
not sure where to get next
dim widget
They just want you to find the invariants in Sym^2V where V is that representation
rustic crown
not sure where to get next
show that free part is 0, so the remaining must be torsion
pure laurel
The tensors are living in the tensor algebra for the representation, and I'm con...
So the matrix I wrote is the "fundamental" representation using cartesian basis vectors, the question asks "Write down a complete basis of second rank symmetric tensors under rotations
and use this basis to express the representation of the generators of the rotational
subgroup" and I'm super confused
dim widget
So the matrix I wrote is the "fundamental" representation using cartesian basis ...
A basis for Sym^2V is just the usual symmetric tensors (it's a 6 dimensional vector space) and you can write out the action on simple tensors explicitly.
delicate orchid
They just want you to find the invariants in Sym^2V where V is that representati...
right, NOW I get it
dim widget
Then the invariants are just solving for the 1-eigenspace of the generating rotation.
rustic crown
M being f.d over k => M is torsion over k[x]
a direct proof woudl be to say consider the elements {m, xm, x^2m, ...} as these are infinite, they must linearly dependent over k. so a certain linear combi must be 0, this gives you a nice non-zero polynomial which kills m
pure laurel
A basis for Sym^2V is just the usual symmetric tensors (it's a 6 dimensional vec...
I found out how this representation of the rotation acts on the symmetric tensors constructed by just using symmetric, i.e. how the basis tensors $E_{ij} = \frac{1}{2}(e_i\otimes e_j + e_j \otimes e_i)$ transform under this rotation, is that what you mean by action on simple tensors?
cloud walrusBOT
Solklar
dim widget
One invariant tensor that you know exists is \sum_i v_i \otimes v_i since that's invariant under any rotation.
dim widget
I found out how this representation of the rotation acts on the symmetric tensor...
e_1 transforms as it does inside of v
So if $\sigma$ is the rotation then $\sigma(e_1 \otimes e_2) = \sigma(e_1) \otimes \sigma(e_2)$
cloud walrusBOT
Topos_Theory_E-Girl
pure laurel
Yeah so I have 6 expressions, one of them is for example
$\sigma(E_{11}) = \frac{1}{4}(E_{11} + 3E_{22}) - \frac{\sqrt{3}}{2}E_{12}$
which I got from doing that. What I'm confused about is how I go from these expressions to expressing the representation in this new basis. Is it supposed to be a 3x3 matrix?
dim widget
So Sym^2V is a 6-dimensional vector space so if you tried to do this in the most doctrinaire way it would be a 6x6 matrix
novel parrot
a direct proof woudl be to say consider the elements {m, xm, x^2m, ...} as these...
i understand this
cloud walrusBOT
Solklar
dim widget
There are however, certain subspaces that you can throw out and restrict to a smaller subspace where the eigenvalue problem is easier.
novel parrot
show that free part is 0, so the remaining must be torsion
how would i show this though? $M \cong K[x]^m \oplus K[x]/(f_i)$
cloud walrusBOT
ActiveChapter
novel parrot
showing that m = 0
dim widget
Because V = W \oplus 1 where 1 is the trivial representation (the z axis)
coral spindle
I was learning mobius transformations, and then wondered how many group actions ...
Not everything is known for group actions. For example, take the notion of a primitive action. Classifying these for finite groups is equivalent to finding all maximal subgroups of a finite group. Computational evidence suggests this is a hopeless problem in general, and as for the finite simple groups, the maximal subgroups of the monster group are still unknown.
pure laurel
So Sym^2V is a 6-dimensional vector space so if you tried to do this in the most...
Ah I see, thank you! Is there any easy way of going from the expressions I have to writing down the 6x6 matrix? (Sorry this is ELIphysicist)
coral spindle
I was learning mobius transformations, and then wondered how many group actions ...
Maybe it should be mentioned, since you don't seem to be aware: transitive group actions correspond to subgroups up to conjugacy, and every action is a 'direct sum' (whatever that means) of transitive group actions, so in some sense it is fair to say that every group action is 'known'.
rustic crown
how would i show this though? $M \cong K[x]^m \oplus K[x]/(f_i)$
oh that may not be true...
you have M = k[x]^r ⊕k[x]/(f1)⊕k[x]/(f1) ... ⊕k[x]/(fn)
each of k[x]/f_i is f.d. but notice that's not the case with k[x].
so only way M is f.d. is if r = 0
chilly ocean
"M f.d." = manifold
pure laurel
Because V = W \oplus 1 where 1 is the trivial representation (the z axis)
Is this related to how you can decompose a rank 2 tensor into a kronecker delta, anti symm part and symmetric traceless part?
delicate orchid
Maybe it should be mentioned, since you don't seem to be aware: transitive group...
did someone say the burnside ring I think I heard someone say the burnside ring
👂 🤚
dim widget
Is this related to how you can decompose a rank 2 tensor into a kronecker delta,...
Not exactly
but the basis for Sym^2 is related to this
pure laurel
But let's see if I understand the question correctly, they want me to find out how this representation acts on the space of symmetric rank 2 tensors?
dim widget
But let's see if I understand the question correctly, they want me to find out h...
Yes so they want you to write out how a rotation acts on the basis E_{i, j}
novel parrot
each of k[x]/f_i is f.d. but notice that's not the case with k[x].
wouldnt k[x] be f.d over itself?
trivially
since we are considering M as a k[x] module, and M is f.g as a k[x] module
pure laurel
Yes so they want you to write out how a rotation acts on the basis E_{i, j}
Ahhhh okay now I got it finally I think, so if I express the symmetric subspace as a 6x1 "vector" I'm looking for the 6x6 matrix that represents the rotation? Because that I can do
dim widget
Ahhhh okay now I got it finally I think, so if I express the symmetric subspace ...
Yep exactly
rustic crown
wouldnt k[x] be f.d over itself?
right, but we're talking about dimension of M over k.
pure laurel
Yep exactly
Thank you!!!
dim widget
Ahhhh okay now I got it finally I think, so if I express the symmetric subspace ...
There are smarter ways of doing this but I think it's good to practice just doing it by hand
rustic crown
right, but we're talking about dimension of M over k.
f.d. as a k-vector space implies it's f.g. as a k[x]-module
pure laurel
There are smarter ways of doing this but I think it's good to practice just doin...
Yes for now it will do lol, thanks again (!!)
novel parrot
right, but we're talking about dimension of M over k.
Wait
im proving that M is f.d over k => M is torsion over k[x]
rustic crown
yep
novel parrot
so as k[x] module, M = k[x]^r ⊕k[x]/(f1)⊕k[x]/(f1) ... ⊕k[x]/(fn)
but you are saying that k[x] is not f.d ?
novel parrot
over k yes
rustic crown
(you usually dont' use the word "dimension" unless the thing is a vector space)
for general free modules, people prefer the word "rank"
novel parrot
k[x] over k[x] has dimension 1?
rustic crown
yea rank 1 :p
novel parrot
im super confused
M = k^n as k module
and M = k[x]^r ⊕k[x]/(f1)⊕k[x]/(f1) ... ⊕k[x]/(fn) as k[x] module
rustic crown
a little imprecise, but okie >.<
novel parrot
i didnt understand how you forced r = 0
rustic crown
so isomorhpism of k[x] modules also gives you an isomorhpism as k-vector space
just forget that multiplication by x is a thing
so the left side is finite dimension over k
if r was non-zero then the right side would be infinite dimensional
so you're force to have r = 0
and now the remaining part k[x]/(f1)⊕k[x]/(f1) ... ⊕k[x]/(fn) is a torsion k[x]-module
infact the single polynomial f = f1 f2 ... fn kills the entire thing at once!
uwu?
novel parrot
not yet 
rustic crown
an isomorhpism is a k[x] linear bijection
so in particular it's a k-linear bijection
as k is a subring of k[x]
right?
like maybe we can look at the same situation over abelian groups.
finite abelian group if and only if torsion and finitely generated
rustic crown
if A is a finite abelian group, then it is finitely generated. so by structure theorem,
A = Z^r ⊕Z/a1 ... Z/an
novel parrot
so k^n = k[x]^r ⊕k[x]/(f1)⊕k[x]/(f1) ... ⊕k[x]/(fn) as k-modules, yeah?
rustic crown
but since A is finite and Z is infinite, r must be 0
rustic crown
so k^n = k[x]^r ⊕k[x]/(f1)⊕k[x]/(f1) ... ⊕k[x]/(fn) as k-modules, yeah?
yee
novel parrot
uwu
rustic crown
novel parrot
thank you 🙂
rustic crown
np
rotund aurora
So I sps this is like the chain rule
but shouldnt the left be DF(x_1,..., x_n) and not DF(X_1,...,X_n) ?
dim widget
but shouldnt the left be DF(x_1,..., x_n) and not DF(X_1,...,X_n) ?
I think this notation is just bad, LHS should be something like $(D(F(x_1, \dots, x_n)))$
cloud walrusBOT
Topos_Theory_E-Girl
rotund aurora
rustic crown
also the end should be (D(X_i))(x_i)
rustic crown
but shouldnt the left be DF(x_1,..., x_n) and not DF(X_1,...,X_n) ?
oh nvm, you're right... i assumed D is a derivation on k[X1, ..., Xn] i think they mean it's a derivation on R
so yea only the left needs to be little x_i instead of big X_i
.<
@rotund aurora >.< gomen
rotund aurora
gomen?
rotund aurora
I dont speak anime
rustic crown
south patrol
d-det chan, p-please keep speaking in weebiness
rustic crown
ryu
south patrol
just to check uh
to show F(X) has trivial centre provided |X| >= 2, does it suffice to say like
(thinking in terms of classes of words) take a reduced word w such that its class [w] is in Z(F(X)); suppose w is non-empty, so it starts with some letter x. Pick y in X, not equal to x. By hypothesis, yw and wy are equivalent words. But yw is reduced by construction, so, by uniqueness of reduced words, wy must be reduced too (or it would reduce to smth shorter than wy). But then again by uniqueness, this means yw = wy as words - impossible, since only one of them starts with y
rustic crown
yee
terse rose
I’m a bit confused on the final part
What’s the motivation of defining the binary operation like that? The markscheme didn’t show what the homomorphism actually is, I presume it’s T( (a,b)+H)=a?
rustic crown
(thinking in terms of classes of words) take a reduced word w such that its clas...
if i had to write a really precise answer that's what i would have written. but i would think about it more like this: in yw the letters y and x always appear in that order. your reduction rules don't change this so it can never be transformed to wy where somehow x is now first.
rustic crown
What’s the motivation of defining the binary operation like that? The markscheme...
that's a general construction in algebra called "quotienting". you take a nice subgroup, and declare that to be 0, to get a fun new group. for example in Z if you declare 5Z to 0, then you get the group {0, 1, 2, 3, 4} under addition mod 5
south patrol
if i had to write a really precise answer that's what i would have written. but ...
Yup sure
Thanks
I remember thinking like hm I wonder if there is a purely categorical proof of this
But yeah idk lol
Actually maybe that is true lol so like
pastel cliff
is an algebraic extension necessarily finite
south patrol
Nope
pastel cliff
i know finite implies algebraic
south patrol
Say $x,y$ are distinct elements of $X$. Take your favourite non-abelian group $G$, with elements $r,s$ that don't commute. There's a map $F(X) \to G$ sending $x \mapsto r$ and $y \mapsto s$. Absurd.
Hehe okay that is cute
south patrol
i know finite implies algebraic
Yeah
So I think the funniest way to create a counterexample is like
pastel cliff
trying to show this
south patrol
What is the biggest algebraic extension of Q you can think of (that's contained in C)
hint: it's tautologial
south patrol
trying to show this
Oh this can just be done by hand
pastel cliff
isnt it just Qbar
south patrol
But yeah probably the easiest example to explicitly show is like
Okay maybe there are easier examples, but for me it'd be uh
take Q and adjoin all the (2^n)-th roots of 2
Lol
pastel cliff
chilly radish
Say $x,y$ are distinct elements of $X$. Take your favourite non-abelian group $G...
Wait why r u assuming that there's an element of X in the centre
south patrol
Wait
Yeah okay that was dumb
Good point
Lol
aiukgyahbglyhb
Okay so my original ting worked better
pastel cliff
trying to show this
wb this then tho
chilly radish
You need find a group that allows you to send some word w to an element r and some letter y to an element s
south patrol
Yeah
south patrol
Yeah so maybe the wordy one is cool
chilly radish
wb this then tho
Use the defn of being algebraic
south patrol
Hug
chilly radish
You don't need any finiteness conditions
delicate orchid
Is Q bar / Q finite?
yup
chilly radish
F is algebraic in K
What does 'F is algebraic in K' mean
pastel cliff
sorry wait im doing the dumb
det appears
wait
nvm
but what info do we get about L from K being algebraic
rustic crown
where did det go
i realized shin and potato already discussed whut i said >.<
rustic crown
but what info do we get about L from K being algebraic
nothing, that's why you're asking stuff being algebraic over L... if both K and L were alg over F, then you could also say that KL is alg over F
pastel cliff
wait im not sure i get that
i was thinking if E was algebraic over K
then L would have to be algebraic over F since it's like an intermediate to an algebraic extension E/F
chilly radish
Hint: ||KL=L(K)||
rustic crown
then L would have to be algebraic over F since it's like an intermediate to an a...
right... but we only need to show KL is alg over L, not necessarily over F.
which is why we don't need any assumptions about L
rustic crown
Hint: ||KL=L(K)||
my hint would be more like ||consider the set of elements in E which are alg over L, by magic this is a field. it contains K and L||
chilly radish
Hmm that also works
I was thinking ||just use the fact that extension generated by alg elements is alg||
pastel cliff
havent looked at the last two hints but
pastel cliff
Hint: ||KL=L(K)||
does this make KL just automagically algebraic?
since K is algebraic
pastel cliff
i know that is a is irreducible then F(a)/F is algebraic
chilly radish
Wdym by an irreducible element
pastel cliff
sorry not irreducible, algebraic
but it would also be the root of an irreducible polynomial over F
that's unrelated just reminding myself
chilly radish
Ok in this case maybe det's hint would work better for you
pastel cliff
i still wanna know 
like if it was FK that would be enough but we dont have any info about L
chilly radish
You don't need info about L
Well, except that it contains F
What I was getting at is that an extension generated by algebraic elements is algebraic
Do you know this?
pastel cliff
i thought i did, but im getting mixed up
oh wait
im dumb
F doest matter
it's algebraic over L
then my confusion is, K is algebraic over F, what about those elements of L that are not in F, we dont know if they're algebraic over L do we?
rustic crown
pastel cliff
K/F algebraic doesnt give information about elements of K over L
i know im not thinking straight rn but bear w my silliness pls 
rustic crown
it doess because F[x] is a subring of L[x]
a in K satisfies f in F[x] then it also satisfies f in L[x]
pastel cliff
oh
that's why shin said use definition of algebraic
if an element of K satisfies a polynomial in F
then adjoining that to L doesnt change that
since F is in L so the same polynomial will be in L when it's the "base field" of KL/L
pastel cliff
thank you det
for this question, how do we think about F(a)[x]
asking bc i already think of F(a) as F[x]/(m_a(x))
wraith cargo
for this question, how do we think about F(a)[x]
polynomials with coefficients in F(a)
and you can imagine F(a) as polynomials in a with coefficients in F that cut off at some power
coarse kestrel
Take something like $\mathbb Q(\sqrt2) = {a+b\sqrt2\mid a,b\in \mathbb Q}$ so the coefficients in your polynomial look like $a+b\sqrt2$.
It's the same for polynomials with complex coefficients since $\mathbb C \cong \mathbb R(i)$.
cloud walrusBOT
Zybikron
frigid lark
and you can imagine F(a) as polynomials in a with coefficients in F that cut off...
That's a bit deceptive, F(a) denotes the smallest field containing F and a, and while your statement is true when a is algebraic over F, it isn't true when a is transcendental, in that case F(a) is the quotient field of F[a], viewing a as a letter
wraith cargo
That's a bit deceptive, F(a) denotes the smallest field containing F and a, and ...
In no way is it deceptive
It literally says in the problem that a is algebraic
frigid lark
Sorry, I thought you were giving a definition for F(a)
pastel cliff
im not sure why the answer here to the first part is correct
or well i dont see how this implies irreducibility
pastel cliff
i got that
different question: Suppose $\mathbb{K}/\mathbb{F}$ is a field extension and that $v,u \in \mathbb{K}$. If $v$ is transcendental over $\mathbb{F}$ and algebraic over $\mathbb{F}(u)$, show that $u$ is algebraic over $\mathbb{F}(v)$.
is there a way to connect the polynomial in F(u) whose root is v to like
cloud walrusBOT
not sebbb not stμ₂dying
pastel cliff
the minimal polynomial of u in F(v)?
actually can i somehow use that a finite extension is algebraic
dim widget
actually can i somehow use that a finite extension is algebraic
Can you TeX the whole question?
dim widget
Oh I thought this was related to the MO post
pastel cliff
oh no this is different
pastel cliff
or well i dont see how this implies irreducibility
this was the last thing about the MO posst
dim widget
Okay well so the goal is just to show that [F(u, v):F(v)] is finite
pastel cliff
oh?
dim widget
Yes that's what it means to show algebraicity
pastel cliff
oh that makes more sense after the edit
dim widget
hmmm okay
pastel cliff
iss F(v) finite?
not sure if ik anything about adjoining transcendental elements
ive just been thinking abt transcendental as not algebraic
pastel cliff
uhhhhh
well v transcendental
i mean im not sure what a transcendental extension really means
though i guess you gave a hint, it's "not algebraic"
dim widget
though i guess you gave a hint, it's "not algebraic"
So because F(u) is transcendental, what do the subextensions of F(u) look like/
white oxide
you can determine the order of an element of a factor group by simply computing it's order in G right?
oh wait
until it is in the factor gorup
group
tender wharf
not necessarily
for example if we have the factor group G/H and x is in H then xH has order 1 but x might not be the identity in G
white oxide
for example if we have the factor group G/H and x is in H then xH has order 1 bu...
ah that makes sense
pastel cliff
Well can you say anything about the degree of F(v) over F(v) \cap F(u)?
v should be algebraic over that intersection right
sorry for disappearing topos :(
you can check #「the-obsidian-twink-jesus」 for why i was MIA
tender wharf
i haven't seen you in the server for a while
me inactive
white oxide
me inactive
:(
pastel cliff
for K/F what does it mean to say that K is generatedd over F by a finite number of separable elements
does this just mean like
K = F(a1, ... an)
next obsidian
Where each one is separable
chilly ocean
is anyone good at algebra 2
next obsidian
Wrong channel
frigid lark
From Lang Chapter 6 section 10, is $Tr_G$ well defined for $#(G)$ infinite
cloud walrusBOT
Parrot Tea
long nebula
until it is in the factor gorup
yeah if G is a group with identity element e, then
(xH)^n = eH in G/H
iff
x^n is in H
lethal dune
From Lang Chapter 6 section 10, is $Tr_G$ well defined for $\#(G)$ infinite
There’s no infinite sum
novel parrot
what does this mean?
what does the matrix of the map look like?
diagonal matrix with only a's ?
rustic crown
nu
pick a Q-basis of k
Now multiplication by a of any basis element is a Q-linear combi of other basis thingies
That's the matrix
So its a matrix with entries in Q, not k
ofc is a was in Q to begin with then its exactly diag(a,...,a)
novel parrot
That's the matrix
yes
when it says det(a) and tr(a), that means of the matrix right
north sand
well this is a different way to define what tr and det are. and yes it fits together with the matrix
dim widget
how would we define the discriminant of a polynomial?
Why do you want to know this? It is an interesting question but irrelevant to the picture that you posted.
novel parrot
because there are questions asking to find the discriminant of a polynomial lol
dim widget
because there are questions asking to find the discriminant of a polynomial lol
What degree?
novel parrot
mth degree
dim widget
There is a general definition but it might not be right for this question.
Hmmm
Okay well in general K[x] is a euclidean domain, and the discriminant is (up to a sign) the gcd(f(x), f'(x)) where f'(x) is the derivative.
novel parrot
oh
dim widget
You can turn this into the problem of finding a p(x) of degree < n and a q(x) of degree < n-1 such that p(x)f'(x) + q(x)f(x) = 0 which turns this into a linear algebra question as you can make a (2n-1)^2 square matrix out of the coefficients of f, f' which has a 0-eigenvector iff f, f' have a gcd. This is the discriminant.
dim widget
oh
Alternatively if you know the roots of $f(x) = \prod_i (x - a_i) $ then the discriminant is $\prod_{i < j} (a_i - a_j)^2$
cloud walrusBOT
Topos_Theory_E-Girl
novel parrot
i see
novel parrot
i see
meager jolt
Let $f(x) \in K[x]$ have splitting field $L$. Is $\textrm{Aut}(\tfrac{L(\mu_n)}{K(\mu_n)})$ a subgroup of $\textrm{Aut}(\tfrac{L}{K})$?
cloud walrusBOT
Jeeves
meager jolt
($\mu_n$ is a primitive root of unity)
cloud walrusBOT
Jeeves
meager jolt
This is the first part of a question and I've done all the later parts, so I know that this can't be hard to show 😢
dim widget
This is the first part of a question and I've done all the later parts, so I kno...
Have you tried drawing a square?
dim widget
with K, L, K(\mu_n), L(\mu_n) as vertices
meager jolt
To show a correspondence between the automorphisms?
dim widget
yea
meager jolt
I've thought about it but I got hung up on how the roots of unity may interfere with the roots of f
And so 'hypothetically' the roots may permute differently
dim widget
They might, f could be the m cyclotomic polynomial for m|n
Why don't you draw the square
meager jolt
sure 👍
dim widget
frigid lark
dark stirrup
Just to check, suppose you have a cyclic subgroup with order n, does the order of every element in the cyclic subgroup divide n?
meager jolt
yup
meager jolt
So it's just because we can restrict any L-automorphism in L(mu_n) to a K-automorphism in K(mu_n)?
dim widget
So it's just because we can restrict any L-automorphism in L(mu_n) to a K-automo...
And to one of L
And if both of those are trivial then the automorphism is trivial
meager jolt
It took a while for me to 100% to convince myself but I get it
One more thing - do we use the fact that any of these extensions are Galois?
meager jolt
Ah because of the fixed fields ofc
dim widget
You could probably prove the statement if K(\mu_n) wasn't galois? not really sure though, it would be a bit different at least.
meager jolt
Every Galois theory question I get stuck on just requires faith in embeddings that I need to develop
thanks for the help 🙂
south patrol
I think the key point is that L/K is Galois right, or am I doing stuff differently lol
dim widget
Yeah I guess in general this line of reasoning shows that Gal(LK/F) = Gal(L/L \cap K)
if L/F is galois
south patrol
Poggies
dark stirrup
I’m a bit confused on the final part, how does it know that a is a power of c and commutes?
dim widget
I’m a bit confused on the final part, how does it know that a is a power of c an...
The notation is horrible
But they're just saying if ac^n = c^m then a is a power of c
c commutes with its powers obviously
dark stirrup
I see thanks
Is there a general approach when trying to find the size of a subgroup?
dim widget
No
Two elements in a group can generate a group of almost any order given some obvious restrictions.
If the group is abelian then it is trivial to see the size of the group they generate
otherwise hard.
Just throw all of the standard theorems at it and hope something helps
kind temple
hmm. i think there is some process to find presentations of groups…
i remember seeing something about it but i can’t find anything on it right now
dim widget
hmm. i think there is some process to find presentations of groups…
If they are finite there is obviously an algorithm, but that's not what was asked.
pastel cliff
the difference between endomorphisms and automorphisms is that the former isn't necessarily a bijection right
delicate orchid
yes
elder wave
Endo -> source=target, auto->endo and iso
delicate orchid
an endo maybe epi and mono but an auto must be epi + mono, iso
pastel cliff
epi(c)
elder wave
I assume you’re working in some category where bijection makes sense
pastel cliff
yeah field automorphisms
elder wave
Could’ve gone for the mega pedantry right there
delicate orchid
Could’ve gone for the mega pedantry right there
all categories are concrete
pastel cliff
all rings are domains
delicate orchid
u wot
pastel cliff
i was trying to do a silly but did it wrong
if we say K is generated over F by a finite number of separable elements
that's not enough to say K is finite is it
dim widget
if we say K is generated over F by a finite number of separable elements
It doesn't make sense to talk about elements being separable unless they're algebraic
any fg algebraic extension is finite
dim widget
Yeah I guess you could say transcendental elements are separable
That would be natural
then the answer is no
i guess the right definition of separable is doesn't satisfy an irreducible polynomial with repeated roots
pastel cliff
oh in that way it's kinda somewhere between being transcendental and algebraic
24 hr galois speedrun is so fun
not impossible though
elder wave
sounds possible
oblique river
Does “inseparable” mean “not separable”, or does “separable” mean “not inseparable” 
rotund aurora
Math has to be flawed
chilly ocean
I have no idea how i can show it. My first ide was to show, that e=yxxy and e=xyyx but that doesn't help at all. Can someone give me a hint?
dim widget
I have no idea how i can show it. My first ide was to show, that e=yxxy and e=xy...
You actually get x^2y^2 = e
and yx^2y = e
rustic crown
(except e :p)
chilly ocean
because xy=x-1y-1 is true for all x and y so its also true when y=e ?
Im a bit confused
but that makes sense
Thanks, I think I get it now
sand parrot
How would I go about proving that if A and B are subgroups of (Z_n, +), then |A| = |B| <=> A = B? If it's true at all. I feel like I just need a nudge in the right direction and hopefully I haven't been trying to prove something that's false
dim widget
How would I go about proving that if A and B are subgroups of (Z_n, +), then |A|...
It is true.
Maybe a hint that will help: try to show that every subgroup is cyclic, then count the orders of elements.
sand parrot
there's no need to prove that Z_n under addition is cyclic, right
dim widget
I guess that's basically the definition of Z/nZ
But if you feel like it's not obvious from your definition it couldn't hurt to prove it
chilly radish
Does anyone have a cute application/demonstration of the chinese remainder theorem (the modular arithmetic one ofc)
lethal dune
solving system of modular equations?
chilly radish
Sure but like
I was wondering if there's some neat way of using it that I can show my class
dim widget
I was wondering if there's some neat way of using it that I can show my class
It's useful for counting large numbers of things precisely
Human memory is very fallible in storing numbers with more than 6-7 digits so it's better to do multiple passthroughs with modular arithmetic
Not a very sexy application though, but militaries have used it to count supplies for ages.
chilly radish
Huh, that's interesting
dim widget
This was generalized to mathematical computing eventually. In parallel processing it's a common speedup if you have lots of independent cores
For the polynomial one partial fractions and lagrange interpolation
There are also the cryptographic applications of CRT
Like secret sharing amongst k people.
of a codeword N less than p_1...p_k
chilly radish
I know RSA implementations use it also
dim widget
yeah
lethal dune
Don't remember RSA using CRT
chilly radish
I don't think there's really anything I can show them within this timeframe, i'll probably just end up reformulating it for them in terms of solving systems of modular equations and do an example since they saw it as a statement about a group isomorphism
dim widget
Don't remember RSA using CRT <:stare:826920038623346729>
very common in decryption algos for RSA
sand parrot
Maybe a hint that will help: try to show that every subgroup is cyclic, then cou...
Thanks, just finished it.
barren sierra
Anyone have good reference texts on finite group theory?
Stuff about actions and representations and the like
elder wave
@delicate orchid
sand parrot
I have another one
Let (G, *) be a group and a, b from G with order n. Let a * b = b * a. Prove that n is divisible by the order of a * b
Do I just use the Lagrange theorem here?
barren sierra
Yea probably
delicate orchid
Anyone have good reference texts on finite group theory?
Representations and Characters of groups - Liebeck, James. Fulton Harris is a classic also
barren sierra
You're going to want to compute the order of a and b in terms of the order of a and the order of b
Actually no you don't
But yes Lagrange
white oxide
can somebody give me a hint for this question? i feel like i have to use the fact that 33 can be written as 11 x 3 where 11 and 3 are primes
sand parrot
Actually no you don't
How would I proceed if I don't?
wraith cargo
can somebody give me a hint for this question? i feel like i have to use the fac...
what does it mean if they are prime
i.e. what does it imply
white oxide
well that they're cyclic
but doesn't a cyclic subgroup of G not imply that G is cyclic
wraith cargo
but all the subgroups are cyclic
wraith cargo
ag true
wraith cargo
I am not good at group theory lol sry
barren sierra
How would I proceed if I don't?
just find a relevant subgroup
wraith cargo
tbh this question sounds like it could possibly use like sylow p-subgroups
barren sierra
can somebody give me a hint for this question? i feel like i have to use the fac...
off the top of my head this is some Sylow bs
white oxide
I am not good at group theory lol sry
lol all good
white oxide
off the top of my head this is some Sylow bs
oh shit
haven't gotten to that yet
ig makes sense since this is a practice final from another algebra class
barren sierra
maybe there's a way without Sylow
white oxide
i stg this was a gallian problem tho
barren sierra
but off the top of my head that's how you do it
white oxide
like if G has order pq where p and q are primes
then show that G is cyclic
hm ok
oh nah
the proof was to prove that every proper subgroup of G was cyclic
barren sierra
like if G has order pq where p and q are primes
it is
white oxide
if |G| = pq where p and q are primes
barren sierra
and you prove that using Sylow
white oxide
not to prove that G itself is cyclic
barren sierra
in the case that p doesn't divide q - 1 at least
novel parrot
math is fake
white oxide
oh huh must have missed it then
white oxide
the proof was to prove that every proper subgroup of G was cyclic
in which case then it's easy
white oxide
but off the top of my head that's how you do it
don't know what a centraliser is sadly
or the direct product theorem
also how much did Cauchy contribute to algebra
formal ermine
wtf is this
barren sierra
Cauchy very important
white oxide
i know he was huge in analysis but the only mention i've seen so far is him noticing that R[x]/<x^2 + 1> was iso to C
at least that's all i've seen mentioned in the texts that i've been reading
formal ermine
and every group is embedded into S_n
white oxide
wtf is this
idk
barren sierra
ok so the centralizer of $A \leq G$ is denoted $C_G(A)$ and it's ${ x \in G \mid \forall a \in A,~ xa = ax}$
cloud walrusBOT
Spamakin🎷
barren sierra
so all elements of G that compute with all elements of A
idk what they mean by direct product theorem
white oxide
so all elements of G that compute with all elements of A
ohh ok
white oxide
idk what they mean by direct product theorem
yeah me neither
can i get a hint for this question? trying to show inclusion in both directions, farthest i've got is letting {1, c, c^2, ... c^n - 1} be a basis for F(c) and expressing x in F(c) as a linear combination of this basis, don't know if i'm on the right track tho
(in terms of showing right including left)
rustic crown
whut's your def of F(c)?
(say the extension was E) then is it the smallest subfield of E containing both F and c?
chilly radish
idk what they mean by direct product theorem
They surely mean that if A,B are normal, intersect trivially and |AB|=|G| then G is iso to the direct product of A,B
white oxide
(say the extension was E) then is it the smallest subfield of E containing both ...
oh yes true
hm ok i'll think about that thx
pastel cliff
is there a torsion z module with annihilator 0
lethal dune
Yes
pastel cliff
i saw that
lethal dune
i saw that
shhh 🤫
rustic crown
(saw whut >.<)
lethal dune
I’ll never tell
pastel cliff
so it's gonna be some infinite direct sum right
dim widget
What’s your second favourite injective Z module
(Q/Z)^2
lethal dune
That’s illegal
pastel cliff
pastel cliff
howd you do that
lethal dune
Did you get the counter?
pastel cliff
not sure yet
sonic coral
LOL
lethal dune
not sure yet
Already floating around
karmic moat
ruined…
delicate orchid
wow actually proving it properly rather than just going "le character table"... cringe
slim kayak
Does N2 not imply N3? If I choose v=v1,w=v2 then I can represent v and w as reduced products pa and a^{-1}q, so then vw is equal to pq, the latter being reduced too since I could just extend a accordingly if the first letter of q and the last letter of p are inverses. But from N1 I'd get that |vw|=|pq|>|v|=|pa| and |pq|>|w|=|a^{-1}q| i.e. |p| > |a| and |q|>|a|. So in a product at most less than half the world could reduce. So I could write any u in U as l(u)m(u)r(u) with m(u) being the untouched letters. But then |uvw|=|l(u)m(u)m(v)m(w)r(w)| would hold if u and v as well as v and w reduce each other. But the right hand side of that inequality then has length |l(u)m(v)| + |m(w)r(w)| so it would be automatically satisfied.
(Oh. My translated version changed the "greater or equal than" into a "greater than" on which I based the argument... that might be why the argument works although that means N2 is ever so slightly weaker than making N1 into a strict inequality which is a bit funny)
rustic crown
(does he know)
whut does it mean 
dim widget
whut does it mean <:NervousSweat:789206248419491900>
He doesn't know :pepepoint:
chilly ocean
<:shiver:934961643643871332>
what was the original message?
delicate orchid
what doesn't he know...
rustic crown
the name i gave to my qed eevee 
delicate orchid
gulp
pastel cliff
Already floating around
Z/p1 + Z/p2 + ...
south patrol
hm is this argument fine like uhhh
Say I'm given the character table of some finite group G and I wanna find the char table of its centre Z(G)
I guess the point is given g in Z(G) and some irred character χ of G, well g just multiplies by some scalar λ(g,χ) on the corresponding irrep
And that irrep will split into 1 dim irreps of Z(G) upon restriction and we know how g acts on each of those lol
So we can use the char table to deduce all the characters of Z(G) which come from restrictions of irreducible characters of G
But each irrep of Z(G) is a direct summand of the restriction of an irreducible character (using like induction, say)
So we are done
I imagine that may be overcomplicated though? aha
coral spindle
really cool
smoky cypress
Say you have a base field k=Z/p or Q, and fields K,L both contain k, K is algebraically closed field and infinite transcendence degree, and L is generated by finitely many elements over k, then does there exist an embedding L -> K?
dim widget
Say you have a base field k=Z/p or Q, and fields K,L both contain k, K is algebr...
yes
And the hypotheses are unnecessary
K just needs to be algebraically closed.
rustic crown
and L/k just algebraic is enough, no need for f.g.
dim widget
and L/k just algebraic is enough, no need for f.g.
so potentially transcendental
smoky cypress
Yeah
rustic crown
oh oop
dim widget
Okay then you do need at least as many transcendentals as L has.
smoky cypress
I'm not sure how this works
The hint says to first embed the transcendence part then the algebraic part
I'm not really sure what that means
rustic crown
pick a trancendence basis B, define k(B) --> K and extend to L
summer path
Many colored people >.<
neat aspen
Can someone help me with my benchmark I have tomorrow
dim widget
I'm not really sure what that means
So essentially you can pick an embedding as det says and devissage to the case that the base field is k = k(B), and L is algebraic over k
rustic crown
devissage
neat aspen
Can anyone help me#
summer path
You're going to have to be more specific
neat aspen
Who
summer path
You
neat aspen
Oh
rustic crown
smoky cypress
Oh wait I never learned the transcendence degree thing 
smoky cypress
So for any field extension it's always generated by transcendence elements then algebraic elements?
neat aspen
dim widget
So for any field extension it's always generated by transcendence elements then ...
Yep, if you think about it those are the only two options
summer path
Can someone help me with my benchmark I have tomorrow
What is it on
dim widget
Either you satisfy a polynomial or you don't
neat aspen
What you mean.
rustic crown
So for any field extension it's always generated by transcendence elements then ...
if you have a field extension F/k you can use zorn to get a maximal algebraic indep set B. then by maximality F/k(B) is algebraic
summer path
What is your work on
dim widget
I just need help with my wok
Don't call people on discord without asking my dude
neat aspen
I misclicked.
summer path
If it's on percent portions, go to #❓how-to-get-help
white oxide
by this exercise do they mean same up to isomorphism or literally the same splitting field?
algebra moment
rustic crown
literally same
white oxide
hm ok thx
rustic crown
by that i mean: L is a splitting field of f if and only if it's also a splitting field of f(a+x)
rustic crown
anyway, whenever you say "the" splitting field, there are two possible scenarios. (a) you're only talking stuff modulo some isomorphisms (b) you have fixed a big enough field (for example an algebraic closure) and are you talking about splitting field embedded into that big field
white oxide
anyway, whenever you say "the" splitting field, there are two possible scenarios...
wdym by (b)
rustic crown
so like when you work with finite extensions of Q, since you know that C is a thing, you can always talk about subfields of C where your given poly splits
and this way you always have a unique such thing
like given f in Q[x], take F = Q(roots of f in C)
white oxide
oh right
rustic crown
by this exercise do they mean same up to isomorphism or literally the same split...
both the statements here would be equivalent. because if you show the polynomial splits in one, it would also split in the other via the isomorphism.
white oxide
both the statements here would be equivalent. because if you show the polynomial...
ye but they wouldn't be literally the same tho right
rustic crown
literally same
(so i should have said, "doesn't matter" here, but i said "literally same" to not complicate life further :p)
white oxide
ah i see
rustic crown
ye but they wouldn't be literally the same tho right
the notion of a splitting field is defined only up to isomorphism. you can say F is "a" splitting field of f over k. but not "the" apriori. but once you show that any two splitting fields of a polynomial are isomorphic, then you're good.
so yea like both Q(sqrt2) and Q[x]/(x^2-2) are splitting fields of x^2-2 over Q. but they're not equal.
white oxide
for the problem i just wrote $f(x) = b(x - a_1)(x - a_2)\dots(x - a_n)$ for $a_j \in E$ and $b \in F$ where $E$ is a splitting field for $f(x)$, then i plugged in $x + a$ and got $f(x + a) = b(x + a - a_1)\dots(x + a - a_n)$, but the smallest field containing $b$ and $a - a_j$ is $E$ itself, is that right?
cloud walrusBOT
okeyokay
white oxide
the notion of a splitting field is defined only up to isomorphism. you can say F...
ohh ok gotcha
rustic crown
by this exercise do they mean same up to isomorphism or literally the same split...
depending on what you treat "literally same" vs "isomorphic" you'll parse this statement in different way.
white oxide
the notion of a splitting field is defined only up to isomorphism. you can say F...
ohh ok that makes so much sense thanks!!
rustic crown
**okeyokay**
yee that's good
white oxide
yessss thanks!!!!
white oxide
i know boytjie was explaining this earlier to me, but just to be clear; we can simply define an isomorphism $\phi: \mathbb{Q}[\pi] \to \mathbb{Q}[x]$ in the obvious way, by just replacing each $\pi$ by $x$ - therefore the field of quotients $\mathbb{Q}(\pi) \simeq \mathbb{Q}(x)$, because $\pi$ is transcendental over $\mathbb{Q}$ and the denominator never equals 0?
cloud walrusBOT
okeyokay
white oxide
and hence every element of Q(pi) has the preceding form?
stone fulcrum
Yes. This is a way to note that pi has no algebraic relationship with Q
white oxide
can somebody explain to me the sentence where it says at most 4, i'm struggling a bit to see why
i mean they explain it
but it's not clicking
oh
is it because x^4 - 3 is irreducible and of minimal degree
pastel cliff
but adjoin an element that is a root of a degree four min polynomial => deg 4 extension
same
8am
white oxide
luckily field extensions and what not won't be on it
pastel cliff
,ti
cloud walrusBOT
The current time for stμ₂dying is 08:53 PM (EDT) on Mon, 08/05/2023.
white oxide
oh shitt
pastel cliff
12 hours
white oxide
you got this
pastel cliff
i hope so
white oxide
,ti
cloud walrusBOT
You haven't set your timezone! Set it using the interactive timezone picker with ,ti --set.
white oxide
,ti - set
cloud walrusBOT
cloud walrusBOT
white oxide
oops
unkempt pond
can somebody explain to me the sentence where it says at most 4, i'm struggling ...
It just says that since 4th root of 3 is a root of x^4-3, it's minimal poly over Q(cube root 2) is at most degree 4. Hence degree of 4th root of 3 is at most degree 4 over that field.
pastel cliff
a splitting field of a polynomial over Q is just Q adjoined the roots of that polynomial right
@white oxide how's your prep going
white oxide
<@418474799208071180> how's your prep going
it’s going well, ended up doing some real analysis tonight cuz I’m feeling decent about it
i will say I’m very nervous tho
Like extremely
pastel cliff
yeah mee too