ye we chilling
#groups-rings-fields
pastel cliff
rancid sleet
The solutions to a problem state that this group action is faithful. But when I work it out, I find this case that contradicts the fact that the group action is faithful. I've also checked that my group element belongs to GL2(R). Is there a hole in my counter example or is the solution wrong?
dim widget
The solutions to a problem state that this group action is faithful. But when I ...
A group action on a set X is faithful if for all g \in G there is some x \in X such that gx != x
Just having one fixed point is not enough to say that the action isn't faithful
dim widget
Yep that is equivalent
rancid sleet
Because in that definition of a faithful group action I have
dim widget
yes that doesn't imply that
Because you haven't shown it for all x
just (3, 4) for some specific matrix
viral lynx
anyone here could help me w a galois question quickly regarding splitting fields of resolvant cubic of a quartic?
rancid sleet
Sorry for the stupid question but doesnt giving a counter example mean that the statement is not true for all of x
viral lynx
Let f ∈ Q[x] be an irreducible degree 4 polynomial, with splitting field K/Q. Let g be the resolvent cubic of f , and let M/Q be the splitting field of g. Show that K = M (α) for any root α ∈ K of f .
i've been bashing different cases of whether g has a unique root, and so forth but can't seem to get anywhere
dim widget
Sorry for the stupid question but doesnt giving a counter example mean that the ...
think about your statement 2 carefully
viral lynx
we know they have the same discriminant, so if the discriminannt is a square, we have that splitting field M/Q has to be A_3, and K is contained in A_4 but that really doesn't tell me why adjoining a root gives us enough
we also know that M=Q(a1a2 + a3a4, a1a3 + a2a4, a1a4 + a2a3) but i can't seem to algebraically arrive at why adding an extra root gives us K=M(ai)
rancid sleet
think about your statement 2 carefully
Ahh is (for all x in X statement) different from (statement for all x in X)?
dim widget
we know they have the same discriminant, so if the discriminannt is a square, we...
So what does the discriminant tell you about the degree of the cubic extension vs degree of the quartic extension?
Not even thinking about the subgroups necessarily
dim widget
Yeah it means the cubic extension is either trivial or degree 3 if the disc is a square, otherwise it could be dihedral
viral lynx
yeah with you so far
viral lynx
and V_4 yes
dim widget
okay V_4 isn't transitive though if you're saying what I think you're saying
viral lynx
oh wait yes sorry
dim widget
So anyway
viral lynx
wait (1), (12)(34), (13)(24), (14)(23)}. is isomorphic to V_4 is transitive no?
dim widget
uhh
i don't think that's a subgroup?
maybe I'm missing something dumb let me multiply cycles
okay hmmm
Anyway I don't think it's very important for what I was going to say
Let's add V_4 to the list and that should be all
viral lynx
okay
dim widget
So the point is that if the discriminant is a square we are reduced to the examples contained in A_4
And we know that the cubic resolvent generates an extension of degree either 1 or 3
if the cubic resolvent generates an extension of degree 1 then we know that 3 doesn't divide the order of the galois group
viral lynx
yeah
because there would be a 3cycle and that would contradict Q being the fixed field
go on
dim widget
So we reduce to only the groups of order 4
C_4, and V_4
This means that the extension obtained by adjoining a root of our quartic is just degree 4.
viral lynx
right okay i get it
dim widget
Anyway you play the same game if the discriminant is not a square
viral lynx
thx
dim widget
no worries
static temple
let's say D8 for now
D8? Isnt it too soon for us 2?
pliant raptor
I would like to show that $l(M \otimes N) \leq l(M) l(N)$, where $l$ is the length of a composition series of a module. I know that the tensor product is right exact, but I can't find an exact sequence that will help me here to solve the problem
cloud walrusBOT
ImHackingXD
coral spindle
I would like to show that $l(M \otimes N) \leq l(M) l(N)$, where $l$ is the leng...
I think you may be able to get one via the tensor-hom adjunction. Do you have a result about the length of Hom(M, N)? Just a suggestion.
grand cliff
Seems like a super basic q, but If g is in Nx where N is a normal subgroup (and therefore Nx is a coset).
Is g^-1 in Nx?
For the examples I can think of, it holds: e.g. Alternating subgroup of symmetric, Cyclic subgroup of dihedral
lethal dune
no this is not true in general
lethal dune
the examples you gave is for index 2 subgroup, where it turns out to be true
grand cliff
Maybe its coz index 2
yeah lol
dang internet caused my message to lag after yours
now i just look like a loser
lethal dune
how are you sure that my internet didn't lag 
@grand cliff you can say something more if that's true, you'll get x ∈x'N so x² ∈ N which reduces to [x]² =[N] in G/N so every elery element has order 2 in G/N from that you have G/N ≅ ⊕Z/2
grand cliff
huh cool
grand cliff
I think you can say something more if that's true, you'll get x ∈x'N so x² ∈ N which reduces to [x]² =[N] in G/N so every elery element has order 2 in G/N from that you have G/N ≅ ⊕Z/2
dannngg
just came late
internet lag again :|||
lethal dune
you can produce these groups easily by taking representation of N in Gl(n, F_2) and creating a semidirect product
grand cliff
Okay, maybe you can help me understand what this object is supposed to be then:
Imagine I have a group G with a normal subgroup N.
(g_1, g_2, ...., g_n) where we have g_i g_j^{-1} is in N for all i,j.
I was thinking that the elements g are in the normal cosets, but clearly they can't be since tuples like this are supposed to form a group.
Because I am trying to prove that (g_1^-1, g_2^-1, ...., g_n^-1) also satisfies the property g_i g_j^{-1} in N.
lethal dune
subset of normal coset
grand cliff
hmm okay
lethal dune
$g_i^{-1}(g_j^{-1})^{-1}=g_i^{-1}g_j= (g_j ^{-1} g_i)^{-1} \in N$ as $ x= g_j g_i^{-1} \in N \implies x^{-1} \in N$ as N is a group
cloud walrusBOT
grand cliff
That was a bit embarrassingly easy, but thank you
dim widget
^average european internet enjoyer 
elder wave
European but not German internet enjoyer*
lethal dune
I miss 780mbps days
grand cliff
||I heard there was an ancient Chinese technique...||
delicate orchid
there is an ancient Chinese proverb
grand cliff
Why does this seem like such a confrontational way to introduce the CRT
*intimidating
novel parrot
because it is probably lol
grand cliff
Just look up a yt video applying the CRT
novel parrot
so i need to find an element mod 30 such that mod 2 its 1, mod 3 its 2 and mod 5 its 3 ?
grand cliff
it's usually taught in a first algebra/pure math class at a uni
novel parrot
so i need to find an element mod 30 such that mod 2 its 1, mod 3 its 2 and mod 5...
is this basically it?
if i understand correctly
hmm i so i guess its 23 + (30)Z
delicate orchid
Why does this seem like such a confrontational way to introduce the CRT
cause it's in a pretty general form
novel parrot
am i correct
delicate orchid
but yes you just apply the isomorphism
delicate orchid
am i correct
see if it satisfies the equations!
delicate orchid
well there we go
delicate orchid
is this just asking if 6 is a quadratic residue in Z/11Z, chat
novel parrot
isnt it asking if 6 is a square in Z/11Z
tribal moss
Yes: "quadratic residue" is a traditional fancy term for "square in Z/nZ".
novel parrot
oh
delicate orchid
yeah and then I guess this is a contradiction for some reason 
I refuse to think about number theory for longer than 2 seconds at a time
novel parrot
its supposed to be algebra
tribal moss
The multiplicative group of Z/11Z is cyclic of order 10, so 6 is a square iff its order is 5.
tribal moss
That's probably easier, yes.
novel parrot
whats the answer?
none of them squared are 6
whats the contradiction though lol
a^2 = -5b^2 mod 11
supposing b^2 is not 0 mod 11
rustic crown
right
delicate orchid
its supposed to be algebra
what do you think number theory is
novel parrot
why must there be c such that c^2 = -5 mod 11
tribal moss
If b is nonzero modulo 11 then it is invertible. Divide a^2 = -5b^2 by it twice ...
novel parrot
there's 11 cases, check them
yea, its not true
novel parrot
If b is nonzero modulo 11 then it is invertible. Divide a^2 = -5b^2 by it twice ...
ah
delicate orchid
If b is nonzero modulo 11 then it is invertible. Divide a^2 = -5b^2 by it twice ...
ok, so there isn't one... therefore ^
yeah now you've got it
and the tensor plopducts
delicate orchid
exact sequences are rather funny
but it's far more fun when they're not quite exact because you can do (co)homology on them
novel parrot
i want to explore more real world applications
novel parrot
is it
lethal dune
summer path
wow that emoji is cursed
dim widget
okay what have you tried?
novel parrot
not much, no idea really how to start
but M = Z^m
so a basis would have size of m
dim widget
What if m = 2
novel parrot
it would be isomorphic to Z^2
dim widget
it would be isomorphic to Z^2
ok well so how would you prove the statment in that case
how would you try to build a basis of that form?
novel parrot
it would be isomorphic to Z^2
if we said a basis is {m_1, m_2}
then a_1m_1 + a_2m_2 generates every element
then apply the endomorphism?
novel parrot
ok well so how would you prove the statment in that case
not too sure ..
dim widget
Well can you find one element that might be part of a basis?
novel parrot
phi(m_1) ?
dim widget
what is m_1?
novel parrot
a basis element
dim widget
Okay so are m_1 and \phi(m_1) linearly independent? Why?
novel parrot
linearly independant
am_1 + bphi(b_1) = 0 => (a-b)m_1 + (a+b)phi(m_1) = 0
cant happen unless a and b are both 0
novel parrot
yeah it is circular
novel parrot
am_1 + bphi(b_1) = 0 => (a-b)m_1 + (a+b)phi(m_1) = 0
i applied the endomorphism
using the fact that phi^2 = -1
rustic crown
instead of adding the two, maybe you can do something more interesting 
dim widget
@rustic crown is here 
rustic crown
dim widget
Soon you will be saying q d
novel parrot
instead of adding the two, maybe you can do something more interesting <:catThin...
am_1 = bphi(m_1) ?
dim widget
I will leave you in det's capable hands
novel parrot
I will leave you in det's capable hands
thnak you
rustic crown
lemme just write m and n = phi(m) >.<
so you have am + bn = 0 and applying phi an - bm = 0
novel parrot
yeah
rustic crown
you don't understand both m and n together
so a good idea is to remove one of them from existence :p
novel parrot
so you have am + bn = 0 and applying phi an - bm = 0
thats what i had
then i added them
rustic crown
so you have am + bn = 0 and applying phi an - bm = 0
use your linear equation knowledge to solve for m :p
like multiply the first eqn by a and second by b and then subtract
that way you don't have to deal with that n anymore
novel parrot
i see
rustic crown
so yee, what do you get
novel parrot
(a^2 + b^2 ) m = 0
rustic crown
it's a + but yea :p
novel parrot
oops lol
novel parrot
by assuming it was free, we dont need to care about torsion free ness ?
rustic crown
yep
novel parrot
since both are equiavalent
novel parrot
but only for f.g over pid right
rustic crown
anyway, you can see this is a bit complicated to do by hand >.<
so far you only get that m and phi(m) are linearly indep
still need to make a choice for m such that they basis
a different approach is to take the help of the pid Z[i]
(i hope you could see that some Z[i] stuff was lurking around, phi^2 = -1, a^2+b^2 showing up, etc)
novel parrot
(i hope you could see that some Z[i] stuff was lurking around, phi^2 = -1, a^2+b...
hm no not really
phi^2 + 1 = 0 in End(M)
rustic crown
right
novel parrot
but then what
rustic crown
so like in general, if you have an R-module M and an endomorphism phi, then you can make M also a R[x] module! (if R is non-commutative, then you need ensure that phi and r * id commute, i.e. phi is R-linear)
because an R-mod structure is just a ring map R --> End(M) and you can extend this to R[x] --> End(M) by telling what x goes to
so here you can define a Z[x] mod structure on M, such that x acts by phi, and since phi^2=-1, this map factors through Z[x]/(x^2+1) -->End(M)
which gives you a Z[i]-module structure
rustic crown
if you quickly chase around these isomorphism, you see that the induced action is just (a+bi) * m = am + bphi(m)
rustic crown
but how is that useful?
if you can show it's a free Z[i] module then you get your m_j
from there you can show that {m_j, i*m_j} as j varies would be a Z-basis
and i*m_j = phi(m_j)
give it try
novel parrot
i see
its more complicated than i thought
a z basis for Z[i] would be {1, i} yeah?
rustic crown
yep
rustic crown
its more complicated than i thought
right... but think about it for a moment. don't you find it weird that the rank of this M is forced to be even somehow? so first showing it's a rank n Z[i]-module is a nice approach because it would make it a rank 2n Z-module
novel parrot
right... but think about it for a moment. don't you find it weird that the rank ...
i see
south patrol
Hm what is the centre of the ring of upper triangular matrices? Idk why I can't seem to find much online
Maybe I'll just play around with it lol
lethal dune
south patrol
Why aha
Basically this is cause this ring seems to come up in exams often in the context of rep theory oop
dim widget
Hm what is the centre of the ring of upper triangular matrices? Idk why I can't ...
The center is just the scalars.
coral spindle
If I'm not mistaken, the (nonzero) scalars are also the centre of the group of nonsingular upper-triangular matrices :)
dim widget
If I'm not mistaken, the (nonzero) scalars are also the centre of the *group* of...
unless you're over F_2 lol
slim kayak
How do you transform {x^2,xy,y^2} into a nielsen-reduced set? Is there some sort of general algorithm I can try to follow?
south patrol
If I'm not mistaken, the (nonzero) scalars are also the centre of the *group* of...
Lol
south patrol
The center is just the scalars.
Yee sure thanks
I assume we can just run the exact same argument with like matrices with only one 1
I do wonder if given a matrix A it's easy to find a specific B such that A and B don't commute
But ig that argument shows you can just take it to be a particular element of like the canonical basis lol
dim widget
I assume we can just run the exact same argument with like matrices with only on...
wait what are you trying to do?
coral spindle
I think you can set B to be the matrix that sends a basis element v_i to v_j and everything else to zero, yes
dim widget
oh I see what you're saying
I was thinking of looking at the upper triangular matrices k[B] as a module over the diagonal matrices k[T]
and showing that every diagonal matrix has some strictly upper triangular matrix which it doesn't commute with unless the diagonal matrix is scalar
delicate orchid
unless you're over F_2 lol
isn't this still technically true lol
delicate orchid
it just so happens that the only scalar over F_2 is the identity
white oxide
I’m a dead man
delicate orchid
good luck boss
sonic coral
hopefully annoying the server did me well
same but it did more than wonders for my grade
you’ll do great
coral spindle
I have my algebra final now
Best of luck!
coral spindle
isn't this still technically true lol
No! For example over F_2 the group of 2x2 upper triangular matrices is Abelian.
It's a group of order 2!
dim widget
isn't this still technically true lol
Then (1 1 0 1) is in the center
south patrol
and showing that every diagonal matrix has some strictly upper triangular matrix...
Noice yes
I meant that like the normal way I know to prove that Z(M_n(k)) = diagonal is like
multiply by the matrices with exactly 1 non-zero entry
delicate orchid
Then (1 1 0 1) is in the center
............. WTF
south patrol
Me when uhhhh
delicate orchid
true though
delicate orchid
No! For example over F_2 the group of 2x2 upper triangular matrices is Abelian.
oh 2x2
coral spindle
............. WTF
It's not that surprising, if you realise there are only two elements in that group :)
delicate orchid
I thought these mfs were nxn
there's probably still some non-trivial memer that's in there but
w/e
2 isn't a prime anyway
coral spindle
I have not thought about the general case 
south patrol
what i'm actually tryna do is uhhh
dim widget
I thought these mfs were nxn
for nxn over F_2 the center is still Z/2
south patrol
show that the ring of upper triangular matrices over a field has non-zero Jacobson radical lol
But I shall think
dim widget
it's Id + \delta_{i, n} \delta_{j, n}
delicate orchid
I'm spooked solid
south patrol
Basically my aim was to look for central nilpotents, which fails, though I've heard like you can do fun stuff with like
Strictly upper rtiangular matrices lol
coral spindle
I'm spooked solid
you were previously liquid??
delicate orchid
$I_n+\delta_{in}\delta_{jn}$
cloud walrusBOT
Wew
delicate orchid
you were previously liquid??
I was a more sinister second option
novel parrot
dim widget
Didn't you ask this before?
novel parrot
i may have, i just wanted to check my answer
didnt completely understand the last time
south patrol
In other words, show ||R^n/I^n is iso to (R/I)^n|| right
novel parrot
so in M/IM if a_1m_1 + ... + a_nm_n = 0 <=> a_i are all in I ?
which means they are 0 in R/I
dim widget
so in M/IM if a_1m_1 + ... + a_nm_n = 0 <=> a_i are all in I ?
yes you want to show this
novel parrot
by definition of IM its true right
only 0 if its a finite sum of ab (a in I b in M)
next obsidian
Are you familiar with the tensor product?
novel parrot
is it needed here 🥹
delicate orchid
freeness of M implies is why that implication is two directions
jesus brain
freeness of M is why that implication is in two direcitons
there we go
delicate orchid
yes
novel parrot
M/IM if a_1m_1 + ... + a_nm_n = 0 => a_i in I
delicate orchid
no, other way
novel parrot
i know this without free ness
delicate orchid
you had it right
novel parrot
how?
because if a_1m_1 + ... + a_nm_n = 0 that means that a_1m_1 + ... + a_nm_n in IM
and that implies that a_i are in I
delicate orchid
pick a minimal generating set for M, if M isn't free there will exist some linear dependence in that generating set
if there isn't a linear independence, that generating set is a basis and M is free
novel parrot
because if a_1m_1 + ... + a_nm_n = 0 that means that a_1m_1 + ... + a_nm_n in ...
but how is this not true
night bridge
Am I missing something really obvious here? It's 1 mark so it should be a very simple explanation but I'm not seeing it
(sorry for the sideways image)
delicate orchid
how do you determine if a set of elements is linearly independent
there is a non-trivial linear combination of them that equals 0
brb need to go put my turkey dinosaurs in the oven
wraith cargo
Am I missing something really obvious here? It's 1 mark so it should be a very s...
Hint: First isomorphism theorem
novel parrot
But is it not true that $\sum a_im_i \in IM $ iff $a_i \in I$
cloud walrusBOT
ActiveChapter
wraith cargo
But is it not true that $\sum a_im_i \in IM $ iff $a_i \in I$
This is true
novel parrot
This is true
so am i not QED?
wraith cargo
Let me see what you were proving lol
novel parrot
hi 
wraith cargo
I think that you're conflating it being a free R module and a free R/I module
dim widget
so you know like wew was saying that if $a_i = 0$ that $\sum_i a_i \bar{m}_i = 0$
cloud walrusBOT
Topos_Theory_E-Girl
night bridge
Hint: First isomorphism theorem
ok so that gives me that G1/ker f1 = im f1 but i dont see how that helps (i dont remember ever learning any relations between isomorphisms and normal sub groups)
i know that it would be normal if f1 is surjective but idk how to show that with the info given
dim widget
**Topos\_Theory\_E-Girl**
if a_i = 0 mod I that is
wraith cargo
ok so that gives me that G1/ker f1 = im f1 but i dont see how that helps (i dont...
The quotient isn't the important part of the theorem here
novel parrot
I think that you're conflating it being a free R module and a free R/I module
but am i not correct tho?
wraith cargo
What does the theorem tell us about the kernel
dim widget
but am i not correct tho?
Okay so in the other direction how were you trying to prove this?
novel parrot
havnt proved the other direction
night bridge
nothing that ive been taught
stating the first isomorphism theorem is the very last thing in my course
wraith cargo
nothing that ive been taught
stating the first isomorphism theorem is the very l...
It tells us that the kernel is a normal subgroup of the domain
So now that you have that you're set
night bridge
oh ok ye, did know that
we just didnt derive it from this theorem
dim widget
havnt proved the other direction
Okay so for the other direction: $\sum_i \bar{a_i}\bar{m}_i = 0$ iff $\sum_i a_i m_i \in IM$ iff $\sum_i a_i m_i = \sum_j i_j m_j$ for $i_j \in I$. Because M is free this means that $a_i \in I$ so we see that $\bar{a_i} = 0$.
cloud walrusBOT
Topos_Theory_E-Girl
novel parrot
$\sum \overline{a_i}m_i \in IM => a_i \in I$
cloud walrusBOT
ActiveChapter
dim widget
$\sum \overline{a_i}m_i \in IM => a_i \in I$
err what does $\sum_i \bar{a}_i m_i$ mean if $\bar{a}_i \in R/I$ and $m_i \in M$?
cloud walrusBOT
Topos_Theory_E-Girl
dim widget
$M$ is not an $R/I$-module
cloud walrusBOT
Topos_Theory_E-Girl
dim widget
perhaps thats where i was mistaken
read the argument I gave and see if it makes sense to you
night bridge
So now that you have that you're set
oh shit ye thats really obvious, thanks
thats a mark i shouldnt have lost...
novel parrot
read the argument I gave and see if it makes sense to you
the last part doesnt
if M is free, how that implies that a_i is in I
you are rewriting the m_j in terms of the elements in the free set?
dim widget
if M is free, how that implies that a_i is in I
because there is a unique way to write a = \sum_i a_i m_i in terms of the m_i
The thing that can happen if M is not free is
Well let's give an example
We can look at Z^2/Z where the copy of Z is diagonal (n, n)
novel parrot
**Topos\_Theory\_E-Girl**
you are rewriting the m_i and m_js in terms of the basis elements right?
dim widget
you are rewriting the m_i and m_js in terms of the basis elements right?
those are the basis elements
the m_i
novel parrot
how do you know?
dim widget
That's part of the proof
First we pick a basis for M
we know that M surjects on to M/I by definition
So the $\bar{m}_i$ are a spanning set for $M/I$
cloud walrusBOT
Topos_Theory_E-Girl
dim widget
We are trying to show that actually they're a basis
novel parrot
i see
dim widget
But in M
there is a unique way to write any element as \sum_i a_i m_i
the a_i are uniquely determined
long oracle
Hello, I am new here, so I don't know if this is the right place to ask this question. Is the tensor product of two vector spaces the set of all bilinear forms on the Cartesian product of the vector spaces with the entries reversed? Let $E_1$ and $E_2$ be two finite dimensional vector spaces over $\mathbb{R}$. Then is it true that tensor product $E_1 \otimes E_2$ is defined as the set of all bilinear maps from $E_2 \times E_1$ to $\mathbb{R}$?
I ask this question since I know that $ V \otimes V^* $ is the space of all bilinear forms on $V^* \times V$, if $V$ is a finite dimensional vector space.
cloud walrusBOT
apoorvpotnis
lethal dune
It doesn’t matter which one you put first as its over commutative ring
What definition of tensor product are you using
novel parrot
the universal property says that bilinear maps on E1 x E2 are in bijection with linear maps on E1 (x) E2
long oracle
I am using the following definition. If $V$ is a vector space over a field $K$, then a $(p,q)$ tensor $T$ on $V$ is a multilinear map $T: V^* \times \cdots V^* \times V \cdots V \rightarrow K.$ The space of all such tensors is denoted by $V\otimes \cdots V \otimes V^* \otimes \cdots \otimes V^*$.
cloud walrusBOT
apoorvpotnis
long oracle
I am learning this in the context of differential geometry.
lethal dune
No the entries are not reversed
delicate orchid
that raises a good point why tf did they write it backwards
lethal dune
Double dual thing maybe
delicate orchid
hmmmm
white oxide
same but it did more than wonders for my grade
haha same
yo is the only way to find all irreducible degree 3 polynomials in Z3[x] by just trial and error and finding all polynomials who don’t have zeros lmfao
or could you use Eisenstein in some way
That question was annoying af just did trial and error for like 30 minutes
south patrol
yo is the only way to find all irreducible degree 3 polynomials in Z3[x] by just...
Yeah that's probably the easiest way unless I'm mistaken
Like there are only 9 (assuming monic)
dim widget
yo is the only way to find all irreducible degree 3 polynomials in Z3[x] by just...
It's like finding primes
south patrol
and then some casework eliminates some clear ones
lethal dune
googling up is the easiest
dim widget
googling up is the easiest
I like the new name
south patrol
Eisenstein won't really work since tht's for stuff over Z (or more generally in UFDs i guess)
lethal dune
Cool result
white oxide
That’s annoying
coral spindle
yo is the only way to find all irreducible degree 3 polynomials in Z3[x] by just...
You can restrict to monic polynomials and then just check for roots. This means there are precisely 9 polynomials to check.
You can then also remove the obvious cases where the constant term is 0, meaning you have only 3^2 - 3 = 6 polynomials to check
so in fact this isn't a large search at all
sonic coral
why can you get away with removing the constant term of 0
coral spindle
Imagine a polynomial x^3 + ax^2
sonic coral
something like checking polynomials of degree 3 in Z2 makes sense
coral spindle
can you spot a root?
coral spindle
I did make a mistake though, I accidentally was counting degree 2 polynomials rather than degree 3. My method still works, but we need to check 3^3 - 3 = 24 instead, which is significantly more.
Wait, bad counting!
This is 3^3 - 3^2 = 18 :)
Wow I really can't do arithmetic
sonic coral
for my final, i had to write down a field with 9 elements, something like that isn’t too bad
sonic coral
it gets annoying quickly
white oxide
But shit was so annoying
lethal dune
white oxide
This is 3^3 - 3^2 = 18 :)
Wait why -9
coral spindle
Because we can remove the cases where the constant term is 0, as I explained above
sonic coral
my favorite was looking for irreducible degree 2 over Z2, since you need a constant term and an odd numbered amount of terms in general
coral spindle
I got like 8-9 irreducible polynomials
I just checked by hand using the method I outlined, and I get 8 irreducible polynomials, so it looks like you got that right
white oxide
YESSS
ok that was the one question I was worried about
well there were some others and I used results proved beforehand so hopefully that should be fine
sonic coral
how many students were in your class?
white oxide
Like one of them was prove that this polynomial that was irreducible over Z3 when quotiented out by Z3[x] was a field
and I was like bruh it’s maximal so it’s a field
hopefully that’s ok
sonic coral
i was the only one, my exam was 2 weeks ago ish
sonic coral
no i just graduated undergrad
white oxide
ohh ok
coral spindle
This question makes no sense, but in the first place this is the wrong channel to ask in, as this has nothing to do with algebra. You might get better luck in a channel actually about differential geometry, such as #diff-geo-diff-top.
coral spindle
Great :|
Please don't spam your question in several channels.
I think rather than mathematics not making sense, the blame lies elsewhere :)
coral spindle
I will never understand the emojis in this server
rustic crown
chilly ocean
the most important one to understand is 
rustic crown
.<
coral spindle
I get that one 
chilly ocean
although it's fallen out of fashion recently
coral spindle
I get too
rustic crown
whut about
rustic crown
coral spindle
To make up for this, I will share an interesting result in representation theory
but I shall not explain why it holds
cloud walrusBOT
Boytjie
rustic crown
nice font
coral spindle
Ty I like it too
lethal dune
south patrol
Hm
solar glacier
so i get how to prove if f:R \to S is a ring homomorphism and I is an ideal in R then f(radI) \subset rad(f(I)) but how do you show if the map is onto and I contains the kernel of f then you get equality? Heres what i have so far
let $f: R \to S$ be surjective ring homomorphism, we w.t.s. $\sqrt{f(I)} \subset f(\sqrt{I})$ let $b \in \sqrt{f(I)}$ then $b \in S$ and there is a $k \geq 1$ such that $b^k \in f(I)$ and by surjectivity there is an $a \in I$ such that $\phi(a)=b^k$
cloud walrusBOT
MyMathYourMath
solar glacier
i dont see where it uses the fact that I contains the kernel of f
coral spindle
You have surjectivity from R, not from I.
You do not know that there exists such an a in I, but rather only one in R.
solar glacier
so there is an element of R that hits b
hits b^k namely
so if i can show i get hit by an element of I im done as I \subset radI
so what i had assumed proved it basically but i was incorrect to assume a is pulled from I but rather from R
white oxide
Why can a composition series not have any further refinement? I get it's because all the subgroups have to be maximal, but can't we regard a subgroup H as a maximal subgroup of X or a maximal subgroup of Y?
idk if that makes sense
like for example say we have a composition series H0 < H1 < ... < Hn
and for Hj < H{j + 1}
Why can't we insert a subgroup Hk in between them
Like Hj < Hk < H{j + 1}
because then Hj is maximal in Hk so that should be fine right
and also Hj is not a subgroup of H{j + 1}
or is it
oh wait i'm stupid
H1 a subgroup of H2 a subgroup of H3 implies H1 is a subgroup of H3 right
solar glacier
H1 a subgroup of H2 a subgroup of H3 implies H1 is a subgroup of H3 right
yes this is true
white oxide
lol i have yet to encounter a book which defines permutation multiplication of $\sigma\tau$ as $\tau\bigl(\sigma(x)\bigl)$, all the books i've read mention warnings of other books doing it that way
cloud walrusBOT
okeyokay
agile burrow
Why can a composition series not have any further refinement? I get it's because...
Do you understand this now?
next obsidian
I read a group theory book which defined f•g(x) to be g(f(x))
I think it’s not uncommon in group theory
Because of the group action thing
I’ve only ever seen it in group theory things
sonic coral
yeah my professor told me ab how some disciplines write compositions and multiplication backwards
whereas algebraist use right multiplication and cosets ect
white oxide
Do you understand this now?
yup
so im reading hungerford rn and he's defining the generalized associative law, is that essentially to say that the associativity of any semigroup extends to any product $(a_1a_2 \dots a_n)$?
cloud walrusBOT
okeyokay
next obsidian
yes
it just says that you can put the parentheses in anyway you want that makes sense and they're all the same
white oxide
ah that makes sense thank you
ok dumb question but what does he mean by [resp: x], does it mean it also applies to x?
for example
If $G$ is a group [resp. semigroup, monoid] and $a \in G$...
cloud walrusBOT
okeyokay
next obsidian
it means that they'll later give another statement with (resp.) which modifies the claim
like here's a really really dumb example
Let x = 1 (resp. 2), then x + 1 = 2 (resp. 3)
the structure is almost exactly the same, but just is slightly different for the different cases
so they use resp. to reuse the framework, and then include the relevant minor modification
white oxide
ohh i see so like if x was equal to 2 then 2 + 1 = 3
in other words you can replace it
next obsidian
yes
little root
What are some good examples of finite groups? Like good examples to work through to build intuition
For instance S_3 is nice since it’s the smallest non abelian group
lusty marlin
What are some good examples of finite groups? Like good examples to work through...
S_n, D_n (dihedral groups), the icosahedral group (actually isomorphic to the alternating group A_5) etc.
lusty marlin
What are some good examples of finite groups? Like good examples to work through...
What 'sort' of examples do you want though?
little root
What 'sort' of examples do you want though?
Idk tbh, I just feel like I know a lot of theorems but not a lot of examples
Like a lot of groups I come across are either too simple (like cyclic groups, or anything abelian) or too complicated and unstructured (idk like pretty much any finite group)
The dihedral groups feel like the only example that’s somewhere in the middle that I’m somewhat comfortable with
lusty marlin
Like a lot of groups I come across are either too simple (like cyclic groups, or...
'Unstructured' as in?
little root
Like it’s not very good at helping me understand how other groups might behave
Or at least I don’t have a good enough grasp of group theory to see how
lusty marlin
I see
lusty marlin
Like it’s not very good at helping me understand how other groups might behave
I think just going through the exercises of an appropriately chosen textbook would help with this issue.
little root
Any suggestions?
lusty marlin
I used 'Algebra' by Artin for group theory, I personally found it very nice, stuff was explained well and the exercises really helped to strengthen my understanding.
Although I'm just an undergraduate myself, so you can take my opinion on this with a pinch of salt...
frigid lark
I used 'Algebra' by Artin for group theory, I personally found it very nice, stu...
Artin Tate 👀
~~Sorry Lang referenced something with Artin - Tate, so I thought it was one person ~~
south patrol
Lol
wooden ember
aight am i being really dumb or is this clearly false (K=Q(sqrt{d}))
what would be the correct statement?
formal ermine
what is X
jaunty forum
Can anyone help me with this problem:
If p is a prime. Consider the polynomial ring F_{p}[X] in the variable X over the field with p elements F_p. Let L := F_{p}(X) = Frac(F_{p}(X)) (an infinite field with characteristic p > 0) and let A: L -> L; a |-> a^p be the Frobenius homomorphism. We set K := im(A) = A(L) which is a subfield of L. Show that [L:K] = p
wraith cargo
If p is a prime. Consider the polynomial ring F\_{p}[X] in the variable X over t...
So the question is
Show that [F_p(x): F_p(x^p)]=p
Hint: consider the minimal polynomial of x in F_p(x^p)[T]
jaunty forum
So the question is
Show that [F_p(x): F_p(x^p)]=p
Well I think it's saying to show that [Frac(F_{p})(X):im(A)] to be p; or is that the same thing here?
wraith cargo
Well I think it's saying to show that [Frac(F_{p})(X):im(A)] to be p; or is that...
it's the same thing
Also note that here when I say F_p(x) I mean Frac(F_p[x])
jaunty forum
Also note that here when I say F_p(x) I mean Frac(F_p[x])
What exactly is Frac? I fail to recall
Can't find in my notes either
jaunty forum
Hint: consider the minimal polynomial of x in F_p(x^p)[T]
Okay, so perhaps I'm wrong but would the minimal polynomial be T^p - x^p; x is obviously a root here and it's irreducible since T^k - x^p won't have x as a root for k < p
wraith cargo
You can also use eisenstein
but that works
and that immediately gives you the degree
jaunty forum
Okay, how would I write it more as a formal argument; that's one of the main things I struggle with, especially in Analysis!
wraith cargo
idk like
To find the degree we wanna find the degree of the minimal polynomial with coefficients in F_p(x^p) of x
You can then say it's T^p-x^p and use either of these arguments to argue that it's irreducible
And from that you immediately get the result
jaunty forum
Okay nice
The second part of the question is showing that each a in L\K is not separable over K
wraith cargo
T^p - X^p = (T - X)^p
this isn't true
in our field
x doesn't exist as an element
Ah wait
are you talking abt the second part of that problem
sry
opal osprey
Suppose $V$ is a vector space over a field $\mathbb{K}$ (of not necessarily finite dimension) and let $X : V \rightarrow V$ be a nilpotent operator. Is it then true that the adjoint map $[X, \cdot] = \text{ad}(X) : \mathfrak{gl}(V) \rightarrow \mathfrak{gl}(V)$ is also nilpotent?
cloud walrusBOT
MisterSystem
dim widget
Suppose $V$ is a vector space over a field $\mathbb{K}$ (of not necessarily fini...
What is your definition of nilpotent operator that includes the case of finite dimensional vector spaces? Is it 0 \neq Ker(X) \cap Im(X^k)?
opal osprey
Nilpotent for me just means that there exists some $k \geq 0$ such that $T^{k} = \underbrace{T \circ \cdots \circ T }_{k , \text{times}} = 0$
dim widget
Ah okay good
cloud walrusBOT
MisterSystem
opal osprey
also, idk if you are aware, but adjoint here in the sense of lie algebras
so ad(X)(Y) = [X,Y] = XY-YX
dim widget
Yeah I got that part 🙂
dim widget
also, idk if you are aware, but adjoint here in the sense of lie algebras
Okay I think there is a proof that works in general
For a second I tried to reduce to the finite dimensional case but it didn't really work out
So Ad_X acting on End(V) is just the difference of two linear operators on End(V)
X* and *X which we can call l_x and r_x
opal osprey
oh, and these commute
opal osprey
so we can apply the binomial theorem
and taking (l_x-r_x)^2n makes everything vanish
dim widget
No worries!
cloud walrusBOT
MasakaBakana
wraith cargo
I don't believe that is a finite extension
dim widget
Yes it is a finite extension
wraith cargo
Oh really hmmmm
cloud walrusBOT
MasakaBakana
formal ermine
ah I thought you meant over Fp^n
dim widget
Yes that is true! You can give a basis explicitly, or you can verify that it is separable and prove that the galois group is Z/nZ
wraith cargo
Does that hold generally for Galois extensions
That if L/K is Galois, then L(t)/K(t) is Galois?
Wtf Wikipedia tell me that K(x)/K is a finite extension??
dim widget
That if L/K is Galois, then L(t)/K(t) is Galois?
Yes it is generally true.
If your t is a transcendental
dim widget
Wtf Wikipedia tell me that K(x)/K is a finite extension??
Based wikipedia
lethal dune
crime
against humanity
(I remember lang using k for non commutative ring with unity)
lethal dune
wraith cargo
I was questioning my existence for like the last 10 minutes
Makes a lot more sense lol
jaunty forum
T^p - X^p = (T - X)^p
how does this prove the result?
lethal dune
try expanding
smoky ivy
are there fields where this doesnt hold
chilly ocean
it is true over all fields
tribal moss
The result is a polynomial identity in the matrix entries and lambda. Polynomial identities that hold in R hold not only in every field, but in every ring.
lethal dune
even for non commutative ones?
oblique matrix
Around a month ago, I asked whether in a poset (X, <=) one could obtain generalizations of suprema and infima by taking the collection of upper bounds of a subset A then taking an infemum or taking the collection of lower bounds of a subset A then taking the supremum. The answer given was no, and I tried working on it but I wasn't able understand why.
Here's what the question is
Any help would be appreciated, and if anyone knows what changes when the maximum or minimum operator is replaced by the maxima/minima operator which gives the set of maxima/set of minima I would appreciate any help/advice
fast stratus
Let H be a subgroup of G. How to prove S is normal… i’m able to prove its a subgroup of G
delicate orchid
a subgroup is normal if and only if xH = Hx for all x in G, it’s basically automatic in this case
smoky ivy
it is true over all fields
is GL_n(K) dense in M_n(K) for all fields?
well, it depnds on the topology tho
i guess
smoky ivy
XY=YX holds in R, after all.
but why does this XY = YX hold
tribal moss
Any help would be appreciated, and if anyone knows what changes when the maximum...
Prove that A if necessarily a subset of LU(A). Now if LU(A) has a maximum x, then x is >= everything in A, so x in U(A). On the other hand this implies that x is a minimum of U(A).
chilly ocean
when K is Q or R or C the choice of topology is clear. when it's not any of these, you should have the zariski topology in mind
tribal moss
but why does this XY = YX hold
Because multiplication of real numbers is commutative.
smoky ivy
Because multiplication of real numbers is commutative.
wait, but i thought we are considering X, Y to be matrices
they dont necessarily commute
tribal moss
No, it was just an example of a polynomial identity that is true in R, but is not true in arbitrary (not-necessarily-commutative) rings.
smoky ivy
are there fields where this **doesnt** hold
i was talking about if X, Y in K^{n x n} for K arbitrary field
smoky ivy
when K is Q or R or C the choice of topology is clear. when it's not any of thes...
how do you show that
like that it holds for all fields
tribal moss
You were talkiing about A, B -- my X and Y are just single elements of the field.
rotund aurora
unless Im smoking
lol the result was on equivalence of norms, and the base field has to be complete
🤦♂️
smoky ivy
You were talkiing about A, B -- my X and Y are just single elements of the field...
oh well, i guess, but what does that have to do with my question
sorry
oblique matrix
Prove that A if necessarily a subset of LU(A). Now if LU(A) has a maximum x, the...
wow I'm dumb - thanks for pointing that out
tribal moss
det(AB-λI) is a certain polynomial in 2n²+1 variables, namely the n² entries in A, the n² entries in B, and λ.
det(BA-λI) is another such polynomial.
The coefficients of both polynomials are integers and don't depend on which field we get the values of the variables from.
But we know the two polynomials always have the same value when the inputs are real.
That means all the coefficients must be the same.
^ @smoky ivy
smoky ivy
det(AB-λI) is a certain polynomial in 2n²+1 variables, namely the n² entries in ...
"The coefficients of both polynomials are integers and don't depend on which field we get the values of the variables from.
But we know the two polynomials always have the same value when the inputs are real."
Could you expand on that?
tribal moss
Which of them?
smoky ivy
Why are the coefficients integers and how do you know you get the same value when the inputs are real
and why are we talking about the input being real
tribal moss
The coefficients are integers because each entry of AB is an integer polynomial in the 2n² variables; then each entry of AB-λI is an integer polynomial in all 2n²+1 variables. The determinant of an n×n matrix is an integer polynomial of its entries (see e.g. the Leibniz formula for determinants), and the composition of integer polynomials is again an integer polynomial.
The case you already know is that det(AB-λI)=det(BA-λI) when A and B are real matrices and λ is a real number -- that is, the two polynomials in this equation produce the same results for any values you plug into the 2n²+1 variables.
smoky ivy
ohh, i get it
i see
but why does it not depend on where you get the values of the variables from
tribal moss
Look at the Leibniz formula -- it doesn't say anything about which field you're working in, just gives you some +1 and -1 coefficients that work in all fields.
For example for a 2×2 matrix, the determinant is ad-bc no matter what the field is.
slim kayak
The coefficients are integers because each entry of AB is an integer polynomial ...
Can't that be rephrased into the proof of the statement not utilizing any properties of the reals beyond being a commutative ring and general defining results like the leibniz formula?
tribal moss
It is useful to work in field of characteristic 0 that has 2n²+1 elements that are algebraically independent over Q. Plugging them into the identity guarantees that the determinant of A will be nonzero (because it is a polynomial in those transcendentals), and thus it is invertible. And then AB = A(BA)A^-1 so AB and BA are similar and have the same characteristic polynomial ...
It doesn't have to be R in particular -- it could be Q(x1,x2,x3,x3,.....,xenough).
(Characteristic 0 makes sure all integer polynomials are distinguishable by their values, so results will indeed generalize to arbitrary rings).
slim kayak
I am supposed to find two distinct automorphism $\alpha$ and $\beta$ from $Aut(F_{3})$ such that their they belong to distinct equivalence classes in $Out(F_{3})$, such that the induced maps from $\alpha$ and $\beta$ on $F_{3}/[F_{3},F_{3}]$ are identical.
Making the simplifying assumption that beta is equal to $\gamma$ not in $Inn(F_{3})$ composed with alpha leaves me with having the condition that $x^{-1} g x x \gamma(g) x^{-1}$ being an element of $[F_{3},F_{3}]$ for all g.
Nothing comes to mind there and I am thinking there might be a better way to approach this.
cloud walrusBOT
Kerr
tribal moss
Is F_3 the field with 3 elements? That doesn't have any nontrivial automorphisms.
slim kayak
Oh whoops, it's the free group of three elements here.
Generated by three elements*
[G,G] being the commutator here
tribal moss
Ah.
delicate orchid
obvious maps to try are the ones that just swap the generators around, I presume they don't work?
tribal moss
How about the ientity and some nontrivial permutation of the generators?
Conjugation in the free group preserves the net number of each generator in the reduced word, so there is no inner automorphism that takes one generator to another.
slim kayak
obvious maps to try are the ones that just swap the generators around, I presume...
Hm, yeah but it isn't that obvious when that looks like an element of the commutator of the free group if ever.
delicate orchid
wait what do you mean
elements of the commutator look like products of wvw^-1v^-1 for some words w, v
tribal moss
if you have the free group on generators {a,b,c}, then there's an obvious homomorphism to Z^3 that sends a to (1,0,0), b to (0,1,0), c to (0,0,1). Since the codomain is abelian, any inner automorphism will be invisible to this homomorphism.
if you have an automorphism that sends a to b, they map to different elements of Z^3 -- namely (1,0,0) and (0,1,0) -- so that automorphism cannot be inner.
slim kayak
wait what do you mean
Well, I need two automorphism alpha and beta so that alpha inverse composed with beta isn't an inner automorphim and that alpha^{-1](g)beta(g) is always an element of the commutator.
delicate orchid
right I see now
slim kayak
And I am unsure if just permuting elements and identity would work when choosing the word g in a bad way
Might help to notice that for N the commutator F_3/N is just isomorphic to Z^3, so I am more or less looking for an alpha and a beta such that them acting on Z^3 is identical but the way they act on F_3 differs by a non-inner automorphism.
Which intuitively would just be a permutation of the generators 🤔
coral spindle
I think $\alpha \colon a \mapsto ab^{-1};\ b \mapsto bc^{-1};\ c \mapsto ca^{-1}$ is an automorphism, right? If you define $\beta$ similarly but e.g. send $a \mapsto b^{-1}a$ I have a feeling this would give you what you're looking for, but this is a guess.
cloud walrusBOT
Boytjie
coral spindle
My idea is that these are the same on the Abelianisation, but they exploit the noncommutativity.
I think I might be off on this being an automorphism here, but I think this general idea should work.
slim kayak
Thanks, I think I got it now since I subconsciously glanced over the fact that the commutator subgroup is normal and tried to think solely in terms of products of commutators of pairs of elements 💀
Let alpha be the identity and beta permute x1 with x2, beta isn't an inner automorphism since $\beta(x_{3})=x_{3}$.
Then alpha and beta clearly have different equivalence classes and given any words g I can write $g=u(x_{1} x_{2})v$ and with $v^{-1} x^{-1}{2} x^{-1}{1} x_{2} x_{1} v$ being inside $N = [F_{3},F_{3}]$, so I can keep switching out any instances and x1 and x2 in $\alpha(g)$ until I have $\beta(g)$ and therefore $\alpha(g) N = \beta(g) N$
coral spindle
Way easier example. Nicely done!
cloud walrusBOT
Kerr
tribal moss
Well, I need two automorphism alpha and beta so that alpha inverse composed with...
Ah, sorry didn't notice the second condition. By the way, the projection into G/[G,G] is exactly the map into Z^3 I mentioned.
(Oh you noticed that already)
slim kayak
Yeah I know, although it did help remind to consider what I was "actually" doing
tribal moss
How about alpha=id and beta inverting each generator separately?
slim kayak
It acts differently on the generators of Z^3 so that wouldnt give me the equivalence on the induced maps.
For example beta(x^{-1}_{1}) would map to (-1,0,0) instead of (1,0,0)
tribal moss
Was
that alpha^{-1](g)beta(g) is always an element of the commutator
a misstatement of what you need?
slim kayak
No, doesn't seem like it. Why?
frigid lark
how does this prove the result?
I have no idea if this was answered but just imagine I had money to get discord nitro and change my name to x
Any automorphism of L over K must take x (hey that's me)
to another root of X^p - T^p, which can only be x (me again)
As x (me) generates L over K, we only have one automorphism
tribal moss
Because with alpha=id and beta inverting each generator separately, we would have alpha^{-1](g)beta(g) in [G,G] for all g.
slim kayak
The induced maps sending g to alpha(g)N and beta(g)N are equal iff alpha^{-1}(g)beta(g)N=N, i.e. said product is inside N already.
slim kayak
Because with alpha=id and beta inverting each generator separately, we _would_ h...
Wait, is x^{-2}_{i} in [G,G]?
oh...
tribal moss
I may be confusing myself. [G,G] is the kernel of the projection into Z^3, isn't it?
tribal moss
Ah.
slim kayak
It's meant to be the inverse of the element alpha(g) 😅
tribal moss
How about (flailing wildly now)
alpha = id
beta(a) = b^-1ab
beta(b) = b
beta(c) = c
slim kayak
bab?
tribal moss
Whoops, fixed
If this beta is inner, then it would be conjugation by something that commutes with both b and c ...
slim kayak
The projection map makes the induced map of beta send b^-1 a b to (0,-1,0)+(1,0,0)+(0,1,0)=(1,0,0) so that works too
Ah... I realized that I assumed that all words would already have pairs of x1 and x2, beta(x1)=x2 and that gets mapped to (0,1,0) instead of (1,0,0).
tribal moss
Yeah, I see that permute-the-generators wouldn't satisfy that criterion I missed.
slim kayak
Okay... if I invert x3, then add the inverse of x3 onto x1, swap x3 inverse back to itself, then map x2 to x2x3 I'd get:
beta(a)=ac^-1
beta(b)=bc
beta(c)=c
Wait no...
tribal moss
I'm confused, why are you speaking both about x1 x2 x3 and a b c?
slim kayak
Old habits die hard and then I copied your notation afte the middle since it was more conveniant
tribal moss
Ah, I was afraid they meant different things and I had missed something.
slim kayak
beta(a)=b^-1 c^-1 a b c
beta(b)=b
beta(c)=c
seems like it should work
tribal moss
Why didn't beta(a) = b^-1ab work?
slim kayak
Since in any instance of beta(a) the stuff on the left and right cancel out, but it is barely not an inner automorphism due to the order
slim kayak
Why didn't beta(a) = b^-1ab work?
For the first condition beta can't be an inner automorphism, or well rather alpha and beta must differ by a non-inner automorphism
Oh.. I realize what you did there now
tribal moss
Oh, I see, it's easier to argue it's not inner that way (?)
slim kayak
Yeah...
No I thought you wanted to check what happens with an inner automorphism
Yeah your example always worked
Thanks for the help
kindred quest
anyone know what the 2 . means in this exact sequence?
chilly ocean
multiplication by 2
kindred quest
also confused about how this is exact. Wouldn't exactness require that the kernel of Z/2Z -> 0 is 0?
chilly ocean
it is the map from Z to Z which sends n to 2n
chilly ocean
also confused about how this is exact. Wouldn't exactness require that the kerne...
it would require that the kernel of that is the image of the map Z -> Z/2Z
kindred quest
yeah, makes sense considering it as the 2n map. I thought it was the identity and was confused
chilly ocean
you might be mixing up exactness at the second Z with exactness at Z/2Z
idk but i am glad to have cleared things up
lethal cipher
Silly question, but are ideals closed under addition?
chilly ocean
please go look at the definition of an ideal
lethal cipher
I'm reading Dummit and Foote and it is only saying it is closed under left/right multiplication.
chilly ocean
i do not think that that is the case
lethal cipher
Ah, it has a little note after the definition that says I must also be closed under subtraction.
chilly ocean
being closed under addition is in one way or another part of the definition
lethal cipher
Yep, I'm with ya. That's what I thought. The argumnt I had for 4a) was relying on it being closed under subtraction, so I had to make sure.
chilly ocean
where in 4a did you need to do subtraction?
slim kayak
Silly question, but are ideals closed under addition?
Closed in what sense?
Within themselves?
chilly ocean
lethal cipher
Wording sorry. Yes, I wanted to make sure elements of I were closed under addition (ultimately subtraction)
delicate orchid
your wording was fine
slim kayak
Ideals are by definition additive subgroups of your ring so yeah
Maybe they just said subgroups, although with that they mean that in relation to the additive structure since in general the multiplicative structure fails to be a group
chilly ocean
brother dummit and foote is one of the most recommended algebra textbooks of all time i do not think that they describe ideals as mere additive subgroups and nothing else
white oxide
yo how am i suppose to find a common factor of this polynomial and it's derivative (which is a criterion for multiple zeros) if i can't even tell if this polynomial is reducible or not
wraith cargo
yo how am i suppose to find a common factor of this polynomial and it's derivati...
me when derivative
white oxide
huh
wraith cargo
you can take the derivative
wraith cargo
that's the condition for multiple roots
wraith cargo
i'm trying to find the common factor
why tho
white oxide
white oxide
so all you have to do is show that a function is differentiable?
that seems a lot stronger
than saying they have a common factor
white oxide
a zero of f
also i thought this only works for polynomials of degree 2 or 3
slim kayak
brother dummit and foote is one of the most recommended algebra textbooks of all...
hm?
wraith cargo
also i thought this only works for polynomials of degree 2 or 3
why would it work only for those??
white oxide
a polynomial f(x) of degree 2 or 3 in F[x] is reducible iff it has a zero in F
white oxide
why would it work only for those??
idk you would think that they would make the statement a lot stronger
a polynomial f(x) in F[x] is reducible iff it has a zero in F
is a lot stronger so why wouldn't they just say that
wraith cargo
idk
What book are you reading Gallian?
I'm sort of not a fan of his book he does stupid stuff all over the place
(at least he did in the group theory section)
white oxide
gallian and fraleigh
wraith cargo
a polynomial f(x) in F[x] is reducible iff it has a zero in F
well I mean in this case you can easily factor out (x-a)
Since if x-a is the minimal polynomial for a is a is in F
And since the minimal polynomial divides any polynomial that has a as a root
It's easy to see it's reducible
white oxide
well I mean in this case you can easily factor out (x-a)
Since if x-a is the min...
i mean over Z3 there's no zeros
wraith cargo
oh yeah
OK so here's what I'd do
assume alpha is a root of your polynomial
and alpha is in some extension of Z3
now take the derivative of your polynomial
and show that alpha is also a root of your derivative
white oxide
ah ok i'll try that thanks
oh
well i mean like the derivative is just 0
lmao
in Z3
wraith cargo
yeah
slim kayak
You could go a bit further by factoring out the largest power of (x-a) to obtain some g(x) which doesn't have a as a root. Then look at its derivative and conclude that it can only have a common factor with f, if the exponent of (x-a) is larger than 1.
wraith cargo
nahh it's ok ur not dumb lol
white oxide
well thanks anyways!
white oxide
You could go a bit further by factoring out the largest power of (x-a) to obtain...
ahh ok
sonic coral
a polynomial f(x) in F[x] is reducible iff it has a zero in F
this is equivalent to being irreducible if f(x) can’t be written as the product of two other elements of F[x] each having positive degree?
white oxide
this is equivalent to being irreducible if f(x) can’t be written as the product ...
i'm not sure tbh, idek if that statement is true
sonic coral
well writing it for for reducibility got me points on an exam so i hope it’s true, i’m just trying to think about how to get one from the other
white oxide
well writing it for for reducibility got me points on an exam so i hope it’s tru...
yeah same
well i said deg 2 or 3 polynomials which i know is true
also what is the point of this question, isn't the splitting field for just this polynomial just Z3 adjoined with the complex roots of the polynomials
wraith cargo
also what is the point of this question, isn't the splitting field for just this...
nop
white oxide
nop
y not
oh because those aren't all the potential zeros?
i'm so confused i thought a splitting field was the smallest field containing F and all of its zeros in an extension E of F
wraith cargo
OK forget my comment I'm too tired for this rn lol
wraith cargo
so it is?
See the thing is that in Z3 x^2+x+1 and x^2+x+2 are the same polynomial
but they have different roots when viewed with coefficients in Z
white oxide
See the thing is that in Z3 x^2+x+1 and x^2+x+2 are the same polynomial
wait huh
what do you mean
as in the same function
wait no they wouldn't even be the same function
cuz 1 is not equal to 2 in Z3
white oxide
LOL all good
wraith cargo
still
x^2+x-1 has real roots
vs. x^2+x+2 has complex roots
OK so this problem can get a bit tricky
but maybe a hint
is that the splitting field
is a finite field
of the form
F_{3^n}
for some n
light moon
Forgive me for this question but what is a splitting field
chilly radish
Smallest field extension over which the polynomial splits into linear factors
wraith cargo
some authors define it as any field containing all the roots of some polynomial*
chilly radish
Those authors are wrong and bad
wraith cargo
This is how my prof taught us lol
chilly radish
Usually it's nice to be able to talk about THE (up to iso) splitting field
wraith cargo
ah wait
I am dumb
OK my prof did not define it like that lmao
I have just always read it like that because my prof always said something is a splitting field
guess I missed the lecture where he originally defined it lol
I am extra dumb :(
white oxide
I am extra dumb :(
if you're extra dumb then i'm extra extra dumb :(
what?? why is [Q(2^1/3, 3^1/4):Q(2^1/3)] = 4? i thought the minimal polynomial for 2^1/3 was x^3 - 2 which is of degree 3??
i'm so fucking confused holy
oh
Q(2^1/3) refers to the coefficients
wait but like why can't you apply that logic to [Q(2^1/3, 3^1/4):Q(3^1/4)]
chilly ocean
there are too many fucking symbols in this paragraph lol
white oxide
actually holy shit
like why can't we say that [Q(2^1/3, 3^1/4):Q(3^1/4)] is at most 4 since 3^1/4 is a zero of x^4 - 3 in Q(3^1/4)
wraith cargo
wait but like why can't you apply that logic to [Q(2^1/3, 3^1/4):Q(3^1/4)]
you can
white oxide
bro this dude thinks he's an ancient egyptian or smt
mfs writing in hieroglyphics bro chillll
summer path
you're reacting to your own message
empty rose
what?? why is [Q(2^1/3, 3^1/4):Q(2^1/3)] = 4? i thought the minimal polynomial f...
Q(2^1/3, 3^1/4) and Q(2^1/3) both contain 2^1/3, the thing we're adding there is 3^1/4
empty rose
like why can't we say that [Q(2^1/3, 3^1/4):Q(3^1/4)] is at most 4 since 3^1/4 i...
here we're adding 2^1/3, since that's not in Q(3^1/4)
hollow tartan
What is the geometric intuition behind the fact that field F with matrices with entries (a b -b a) 2x2 is isomorphic to complex field C? Is it because the product of two complex numbers is similar to what a matrix does when multiplied? Roughly speaking very similar to what the Jacobian does
chilly ocean
complex numbers multiply complex numbers by scaling and rotation. such matrices as you wrote act on R^2 by scaling and rotation
white oxide
Q(2^1/3, 3^1/4) and Q(2^1/3) both contain 2^1/3, the thing we're adding there is...
ah ok thanks
what? isn't the subset of G just e?
oh wait
it's a homomorphism
so it doesn't have to be injective
i have no clue what that has to do with theorem 13.15 tho
like bruh isn't that just the kernle
kernele
kernel
because if gx = x for all x in X then \sigma(g) is surely the identity permutation
<@&286206848099549185>
agile burrow
Yes, the subgroup of elements which act trivially is sometimes called the kernel of the action because it is the kernel of the corresponding homomorphism into the symmetric group
white oxide
okay thank you
i don't fucking know why he didn't just write kernel wasted me like 30 minutes glossing over it lmfao
novel parrot
$\mathbb{Z}[\sqrt{-11}]$
cloud walrusBOT
ActiveChapter
novel parrot
Are the units 1 and -1 ?
glossy crag
Do direct sums of (nonabelian) groups ever come up in important ways?
novel parrot
try to prove it
i already did, just wanting to confirm my answer