btw
#groups-rings-fields
scarlet estuary
im not familiar enough with algebra textbooks to respond.
winter thorn
okay
chilly ocean
every vector space is a group right?
rustic crown
under +, yep
chilly ocean
but only an abelian group?
rustic crown
yep, that's how we define vector spaces. abelian group with some cool action of a field
chilly ocean
I saw a meme that said a function is the same as a homomorphism between two sets, that's not correct right? like for example you can have a function from a set with only the + operation to a set with only the • operation and that wouldn't be a homomorphism
stark sigil
Well they said sets, not groups !
chilly ocean
Well they said sets, not groups !
they did but a mapping is only a homomorphism when the domain and the codomain are the same algebraic structure right?
stark sigil
Using the word homomorphism for the category of sets is pretty weird here since there is no algebraic structure to preserve; hence, homomorphism is a synonym of function
chilly ocean
like a vacuous truth?
stark sigil
Sort of, the conditions for a function aren't completely trivial still
chilly ocean
Using the word homomorphism for the category of sets is pretty weird here since ...
but still you can find a counter example, like a function that maps stuff from an algebraic structure to a different algebraic structure? you're still mapping from a set to another set but it cannot be a homomorphism of sets
or am I just being stupid?
stark sigil
A group is a set plus a binary operation
A set is a set, nothing else
A homomorphism of groups is a function between sets compatible with the binary operation
A homomorphism of sets is a function between sets.... and that's it
The notion of homomorphism between sets doesn't change just because they happen to be groups
$\Hom_\Set$
cloud walrusBOT
Icy001
Compile Error! Click the 
reaction for more information.
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cloud walrusBOT
Icy001
stark sigil
The default preamble has \Grp bolded hmmmmmmm
past temple
why is Artin recommended over Gallian?
i personally really dislike artin's book
the first chapter of the book is about multiplying matrices
and in the second chapter he condenses like
basically all of elementary group theory into the span of like 20 pages
comparing artin to dummit and foote
artin basically covers all of the content in the first 5 or so chapters of dummit and foote in just one chapter lmao
his combined approach of abstract algebra and linear algebra in one text also didnt rly work for me
his textbook feels like
if u took a text in linear algebra
and a text in abstract algebra
that are each self contained
and then u combined them into one book with the odd chapters being from the linear algebra text and the even oens being from the algebra text
winter thorn
yeah, I didn't really enjoy him condensing cyclic groups into 2 pages of content.
His concise writings may appeal to someone who has a foot in, in the door of abstract algebra.
But to a beginner like me, he was incomprehensible at times.
past temple
yeah artin can be a pretty solid reference
but as a first textbook in algebra like
its pretty terrible in my opinion
winter thorn
I'm finding Gallian way more readable.
Though his second chapter introducing what symmetry is wasn't as good as I'd like it to be.
past temple
ive personally never read gallian
winter thorn
what did you start with?
past temple
dummit and foote
hm
im reading through a pdf of gallian right now
he doesn't cover group actions??
winter thorn
I think group actions get covered later?
I'm at chapter 3 and he has covered the basic definition of a group, order of a group, subgroups, cyclic groups, centralizers of a group and an element etc.
so I dont know
past temple
doesnt seem like it
maybe its in the special topics section
yeah he has like
a two page section on group actions in chapter 29 lmao
group actions are like
arguably one of the most important topics in an undergrad algebra class
winter thorn
I guess I'll have to refer to Artin for that
hopefully I'm erudite enough for that when the time comes.
waxen hedge
I saw a meme that said a function is the same as a homomorphism between two sets...
In general, a set set doesn't have an operation (+,*,..) !
hidden haven
Sets are algebraic structures and set functions are structure homomorphisms in the sense of model theory
Homomorphism usually refers to structure homomorphism rather than any morphism
I mean morphism and homomorphism aren't synonymous
chilly ocean
In general, a set set doesn't have an operation (+,*,..) !
containment is an operator
sorta
thorny fiber
Hey hey, I was wondering if someone would agree to take time with me at looking into my algebra course from top to bottom ? It would be done regularly during the same time, and would be presented like a talk, but with any interaction being let possible. How does that sound ?
thorny fiber
The goal would be to at least look at the half of it (the last subject of the half being cycles)
thorny fiber
It's basic algebra, and these "talks" can really last how much be decide it would last
hidden haven
containment is an operator
Relation 
Operators are functions
Unless you mean operator from 2^S x 2^S → 2 lul
hidden haven
Oh like elements?
chilly ocean
ye
im jokingly saying this tho
because ik ppl get in their feels with this nuance
thorny fiber
just ping when
I guess we could really start now if we want to lol
chilly ocean
,
lavish nexus
yeah
stark sigil
How about the ideal (2, (1+sqrt(-5))) over Z[sqrt(-5)]
Ok I needed a refresher on the definition of a simple module because that is not a simple module
next obsidian
If M is a simple module over commutative ring D and Tor M = 0
Yes
Any simple module is cyclic aka generated by a single element
Proof: if it isn’t then it has a proper cyclic submodule
Let it be generated by x
Then the map A -> M given by a -> ax is surjective by assumption
It is injective by TorM = 0
Bijection
next obsidian
I think
You can prove something amazing from this
A has to be a field here
Because you’ve shown that A is a simple module which means it has no ideals
Other than A or (0)
So the existence of this module is a ridiculously strong condition on the ring
lavish nexus
right
that's why for this problem they got another thing called indecomposable
so if it's indecomposable and torsion free it is a free module of rank 1 over PID
instead of a field
very interesting
I did not realize this earlier
winter thorn
the path to the order of the multiplicative group mod n goes through the Euler totient function, right?
rustic crown
goes through? it's in some sense the definition of phi(n)
lavish nexus
yeah coprime to n is equivalent to having an inverse (Z/nZ)*
winter thorn
I need to prove that a group G of even order has to have an element whose order is 2.
Is the right direction to assume that there is no element in G such that a = a^-1 for all a in G?
rustic crown
there is always the identity of the group satisfying that
winter thorn
a =/ e ***
rustic crown
yep that's the right approach
winter thorn
okay
dull root
So I am having trouble finding all invariant subspaces of a given representation. Let S_3 be represented by the 3x3 permutation matrices over a two element field {0,1}. How do I go about finding all possible (nontrival) invariant subspaces. It is easy to see the one dimensional subspace with (x,x,x) as a basis is invariant and in fact, this is the only nontrivial 1 dim subspace, since any 1 dim subspace is just 2 vectors since we are over F_2, but I am not sure how to find others
dull root
I have seen Maschke's theorem, but I don't think that applies here since we are over F_2
dull root
So, a couple questions that I need help clarifying. It is sufficient to find all dimension 2 invaraint subspaces, since this space is dim3 over F_2 right, and we already found all dim 1 invariant subspaces
Then is there an easy way to show the existence/non-existence of a dim2 invariant subspace?
With Maschke, this is clear, but we can't apply it here
hot lake
{(0,0,0) ; (1,1,0) ; (1,0,1) ; (0,1,1)} is invariant
dull root
So you took (1,1,0) ; (1,0,1) as your basis
hot lake
I didn't choose a basis
dull root
How did you come to that subspace?
hidden haven
It's usually the complementary subspace when maschle does apply
Set of vectors with coefficients summing to 0
hot lake
I picked a random element of F2^3 and looked at the smallest invariant subspace containing it
and you basically only need to look at (1,0,0) (1,1,0) and (1,1,1) to get all the possibilities
dull root
So I see if an invariant subspace contains (1,0,0), the subspace itself is the whole space. If it contains (1,1,0), the smallest one looks like the one you posted and if it is (1,1,1) the smallest subsapce is ((1,1,1),(0,0,0))
Then it also looks like the subspace W_1 generated by (1,1,0) forms a direct sum with the subspace generated by (1,1,1)
Even so, we were not allowed to apply masche since the Char(F_2) divides the order of S_3
Then is it true that Maschke truly fails when we look at F_3? Because the whole argument applies; we only need to find the smallest invarant subspace containing (1,0,0), (1,1,0), and (1,1,1). It is clear the smallest invariant subspace containging (1,0,0) is the whole space itself. Where this differs from F_2, is that the smallest invaraint subspace with (1,1,0) is also the whole space. Then the only nontrivial invariant subspace is the one with (1,1,1)
hot lake
no there are others
(1,2,0)
for representation of Sn as permutations, you can always get an invariant subspace {(x1...xn) | x1+...+xn = 0}
that will always be a (n-1) dimensional subspace
but not necessarily orthogonal to the 1-dimensional subspace
for example for S3 over F3, it would contain the subspace generated by (1,1,1)
I don't know if <(1,1,1)> has a complementary invariant subspace in that case
dull root
So I see that if (a,b,c) where a+b+c = 0, then we can find an invariant (n-1) dimensional invariant subspace which contains (a,b,c), but now I'm confused about when do we know we have found all invariant subspaces. It was easy to understand in the F_2 case
hot lake
well then you have to look at invariant subspaces generated by x where you pick one x in each orbit of the action of S3 on F3³
dull root
Oh, I see. I also think that the subspace we have is not a orthogonal complement to the 1 dim subspace (1,1,1), since (2,2,2) is in both. So is that enough to conclude that the space CAN NOT be written as a direct sum of two subspaces that are both invariant, showing Maschke is not necessarily true when char(F) divides the order of the group
woven obsidian
I wanted to show that X^4+1 is irreducible over Z_p for any prime p
I did it by showing that $$X^4+1$$ has a root $$\alpha$$ in $$F_{p^2}$$ such that the minimal polynomial of $$\alpha$$ has degree less than 2 and divides $$X^4+1$$
cloud walrusBOT
AoiKunie
woven obsidian
The exercise also asks me to show that X^4+1 is either a product of 4 linear factors or 2 irreducible quadratics (irreducible over Z_p). But I cannot seem to exlude the case of 2 linear factors and 1 irreducible quadratic
hot lake
how did you show it had a root in Fp² ?
woven obsidian
If p > 2
I factorized it directly in Z_2
Because then the order of the multiplicative group of Fp^2 is divided by 8
hot lake
ah then why not look at where in that cyclic group are the 4 roots, and where are the elements of Fp
woven obsidian
Ah right
I only showed there was one root
But by elementary group theory I guess there are more...
Ah right, there should be eulerphi(8)=4 of them
And they all lie in the same cyclic subgroup, so if one of them is in the small field then all of them are
cloud walrusBOT
det
cloud walrusBOT
det
rustic crown
right
next obsidian
But…
rustic crown
is it true if k = Q and discriminants of E and F are coprime?
rustic crown
no, the product
next obsidian
rustic crown
so our profs scared us in the quiz
next obsidian
I don’t know field theory
rustic crown
rustic crown
lol idk enough grammar to understand that
next obsidian
The point is
It says “if X”
Then just ends
There’s no like, “then Y”
The then is after a period
rustic crown
ah, right it should be a comma
rustic crown
they converted the quiz to an assignment
next obsidian
So I am of no help
rustic crown
they converted the quiz to an assignment
and asked us to assume the extensions are galois
but i still wanted to know what we can do in general
next obsidian
¯_(ツ)_/¯
rustic crown
yea, this problem lies in the intersection of field theory and number theory... but i thought my question belonged here
discriminant of a finite extension F over Q is defined very weirdly. you look at the ring of integers O_F. This turns out to be a free abelian group of rank [F:Q]. so we can take it's Z-basis. if {sigma_i} are embeddings of F into Q bar, and {alpha_j} is an integral basis, then disc(F/Q) = det(sigma_i(alpha_j))^2
i was wondering if we could do some normal closure stuff and use that if one of the extensions is galois then nice stuff happens.
uncut girder
I don’t know field theory
U suck
thorn flint
Hi can someone help me with this? I’m confused what they want for 1a
waxen hedge
@rustic crown we had this exercice in my number theory class (sorry it's in French), it may be relevant for you
rustic crown
oh arigato. but i don't think it helps me completely... i'm trying to understand what will happen if we drop the [L:Q] = [K1:Q][K2:Q] and instead pick up K1 intersect K2 = Q.
(also i can read the math, so it doesn't matter if it's French 
)
lethal dune
Hi can someone help me with this? I’m confused what they want for 1a
R* means the set of units?
thorn flint
So I need to show that R* subset of {f(x)/g(x) ... }, which are set of units?
And the set of units {f(x)/g(x) ... } subset of R*? @lethal dune
lethal dune
you need to show that a inverse will exists iff f(0) neq 0
what about the function g(x)/f(x)? think about it
thorn flint
well if we have this function then it will produce a unit since the two functions cancel each other to equal 1
plucky flicker
Hi! What do we mean by the complex conjugate of a character? The character of a representation is a function from G to the base field, so how can one interpret the complex conjugate of such a map?
It means that for every g \in G, we take the complex conjugate of the trace of its representation matrix
sharp sonnet
it only makes sense if your base field is a subfield of the complex numbers
plucky flicker
it is actually the complex numbers
sharp sonnet
in general however you can take the dual character
and stuff like the orthogonality relations will still be true
oh, so yeah
you just take the complex conjugate of the trace, yes
plucky flicker
So we talk about characters of representations, right? And for every representation of G we have such a function from G to F
sharp sonnet
yes
plucky flicker
thanks
And if I want to show that the complex conjugate of a character of a given representation is a character as well, I have to find another representation for which the complex conjugate will be its character?
vocal wolf
Hey everyone! We're in an abelian category and im hoping to use axioms only. So I want to show that if this sequence is exact:
\begin{tikzcd}
0 \arrow[r] & A \arrow[r] \arrow[r, "f"] & B \arrow[r, "g"] & C
\end{tikzcd}
\
then $f=ker g$. I already have:
\
\begin{tikzcd}
& & & coker f \
0 \arrow[r] & A \arrow[r] \arrow[r, "f"] \arrow[d, shift left] & B \arrow[r, "g"] \arrow[ru] & C \
& im f= ker g \arrow[u, dotted, shift left] \arrow[ru] & &
\end{tikzcd}
\
$A\rightarrow ker g$ exists cause $fg=0$. I know that if we can find $A\leftarrow ker g$, then $f$ satisfies the kernel univ prop and we're done, but i cant see where this arrow will come from. Any hints?
cloud walrusBOT
Brian485
plucky flicker
i already hinted at what representation to check
This is so trivial or I am missing something. So given phi : G -> GL(V) and its character chi : G -> C such that for every g \in G chi(g) = trace(phi(g)). I want to show that chi' : G -> C, chi'(g) = complex conjugate of chi(g) is also a character. But we know that for every g \in G the complex conjugate of chi(g) is just chi(g^-1), so chi'(g) = chi(g^-1) for all g.
so the representation can be alpha : G -> GL(V) such that for all g alpha(g) = phi(g)^-1
sharp sonnet
is that really a homomorphism though
plucky flicker
alpha?
sharp sonnet
yes
alpha(gh) = phi(gh)^{-1} = (phi(g)phi(h))^{-1} = phi(h)^{-1}phi(g)^{-1} = alpha(h)alpha(g)
sharp sonnet
you need to look at the dual vectorspace, that is Hom(V, K), where K is your base field
plucky flicker
ohh
sharp sonnet
then if phi is in Hom(V, K), you can define a "linear action" from G as phi*g phi \circ g^{-1}
something like that anyway
this should work
plucky flicker
so the trick is simply that the complex conjugate of the given character will be the character for a representation of G over the dual vector space
sharp sonnet
yes
i think this direction is also better to look at
its natural to ask whats the character of the dual representation
i.e. G acts linearly in a natural way on the dual vector space, what is its character?
(this way you also get a more general character theory that doesnt require you to work over C)
plucky flicker
thanks 😄
vocal wolf
**Brian485**
bump
plucky flicker
chi(g) = chi(g^-1) if and only if g and g^-1 are conjugate?
where chi is a character
We know that characters are constant on conjugacy classes, but is the other direction true?
robust pollen
We know that characters are constant on conjugacy classes, but is the other dire...
No, look at the character of the trivial representation of a group in which there is an element g that is not in the conjugacy class of g^-1
plucky flicker
Yeah, thanks
So how to prove this: chi(g) is real if and only if g inverse is a conjugate of g. I proved the <= implication
hidden haven
isn't the trivial rep again a counterexample? or is there more context?
because chi(g) is real for any trivial rep
and any g
plucky flicker
For the other direction we know that chi(g)=chi(g inverse)
Here is the problem: Let g \in G. Show that chi(g) is a real number for all characters chi if and only if g inverse is a conjugate of g.
I found a solution on stackexchange, but => this part is not correct:
He says that chi(g) = chi(g^-1) hence g and g^-1 are conjugate
hidden haven
Isn't there some result that says that characters span the space of class functions on a group
I forgot the conditions for that
hidden haven
Then that should be enough
given 2 elements of the group
not in the same class
if all characters agreed on them, all class functions would agree on them
oh wait I misread
sharp sonnet
you can also use the second orthogonality relation
hidden haven
Is G finite?
hidden haven
I think you might be able to say that character of g inverse is conjugate of character of g
hidden haven
phew I am just learning this stuff rn so I am very worried that I might be saying wrong stuff lol
sharp sonnet
Isn't there some result that says that characters span the space of class functi...
if G finite, you just need algebraically closed field and characteristic not dividing order of G (you can deduce this from the orthogonality relations)
if G not finite i have 0 idea lmao
but orthogonality relations prove slightly stronger result, so there is easier way i guess
sharp sonnet
thats the first one
its one of the statements i remember bcs it lets you compute character tables
plucky flicker
Well I have an idea
Suppose g and g inverse are not conjugate, and chi_i (g) = chi_i (g^-1) for all g, and chi_i. Since g and g inverse are not conjugate by the second orthogonality relations we get that sum_{i=1}^k chi_i(g)^2 = 0. Is that contradicting to something?
well then chi_1 (g)^2 + ... + chi_k(g)^2 = 0 for all g in G, but this can be true iff chi_i(g) = 0 for all g since all the chi_i-s are real
so {chi_i} can't be an orthonormal basis (??)
It seems legit for me, can you find any mistake in the proof?
ehh, here I supposed that every character is irred
so this won't work in general 😦
or does is? well from the second ort rel we get that for all characters chi(g) = 0 for all g
so this is true for the irred ones
contradicting the first ort rels
sorry, the second is also for irred characters, so this proof is wrong
or maybe not.. damn I'm confused 
sharp sonnet
all characters are sums of irred characters so it should suffice to consider those?
plucky flicker
well I don't really need to use that, if I'm not mistaken I just supposed that for all characters of G and for all g in G chi(g) is real, so this is true for the irred ones
applying the second ort relations to g and g inverse we get that for all irred characters chi(g) = 0 for all g
and then it contradicts the first ort relations
sharp sonnet
works i guess
you can also just plug it directly into it
using $\chi(g) = \chi(g^{-1}$ for all irreducible characters $\chi$ you get that
$$\sum_{\chi \text{irred}}\chi(g)\chi(g) = \sum_{\chi \text{irred}}\chi(g^{-1})\chi(g)$$
is not zero
cloud walrusBOT
Lochverstärker
sharp sonnet
as g is conjugate to itself
so again by second orthogonality relation, g and g^{-1} are conjugate
(this in fact works if you replace g^{-1} by any element h)
(this also requires G finite)
plucky flicker
I'm guessing that G is finite, we're dealing with representation theory of finite groups
thanks for your time 😄
chilly ocean
The set of all matricies with real entries form a ring right?
rustic crown
you need to fix some n and look at n x n matrices
rustic crown
yee
chilly ocean
cool ty
rustic crown
it's a non-commutative ring with identity to be precise
waxen hedge
it's even a R-algebra 
robust pollen
you need to fix some n and look at n x n matrices
Why?
rustic crown
multiply a 2x2 matrix with a 3x3
robust pollen
Can one not put some weird graded structure?
rustic crown
but you still need to be able to multiply any two elements right
all matrices form a category
objects are natural numbers and yes 0 is a natural number
robust pollen
Ofc 0 is a natural number, who thinks otherwise?
rustic crown
Hom(m, n) = n x m matrices
rustic crown
Ofc 0 is a natural number, who thinks otherwise?
there are people
like my number theory prof
robust pollen
But yeah, I was thinking more like graded ring, but that doesn't work, at least not with the usual definition of graded ring
rustic crown
yea, graded rings are rings first
chilly ocean
Ofc 0 is a natural number, who thinks otherwise?
it's not a natural number 
N = {1, 2, 3....}
rustic crown
robust pollen
N = {1, 2, 3....}
That S U C C s
hidden haven
Bruh just say natural numbers with unity or without
I personally believe that natural and natral is a good distinction
rustic crown
unity? 
rustic crown
I personally believe that natural and natral is a good distinction
ohh this is cute
robust pollen
Did you just say "natral"
hidden haven
Natral
rustic crown
naturals are clearly more natural than natrals
rustic crown
hidden haven
Brilliant compromise that solved absolutely nothing
south patrol
Based
robust pollen
Natral nmbers
hidden haven
But at least I can get an nlab page on me
chilly ocean
let's all just agree to disagree and use $\mathhbb{Z}^+$ and $\mathbb{Z}^+ \cup {0}$
cloud walrusBOT
neamesis
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
south patrol
British version na'ral numbers {1,2..}\{t}
robust pollen
British version na'ral numbers {1,2..}\\{t}
Na, should be "noombahs"
rustic crown
Det is a set truther
nice, i'm someone i don't know
chilly ocean
what's a set truther?
robust pollen
Ababou holy sht, totally forgot about that absolute king
rustic crown
whats ababou
robust pollen
NJ Wildberger reaccs only
chilly ocean
But I expect 🤡s to exist
those are the people who think that N should contain 0
Lol
chilly ocean
I'm jk btw
rustic crown
i've accepted i'm cringe 😌
hidden haven
rustic crown
scarlet estuary
if H isnt normal, G/H is not well defined
at least not in the sense of having a usual group structure
so presumably its hidden in the "proposition" it mentions.
regal carbon
N⁰ means non zero integers= whole numbers right?
robust pollen
N⁰ should really be denoted N, since the natural numbers are {0, 1, 2, 3, ....}
regal carbon
If a group satisfies associative property then it definately satisfies commutative property right?
robust pollen
no
tropic spade
Pretty sure my counterexample was bad lol.
All groups satisfy associativity but not all groups are abelian though.
robust pollen
If a group satisfies associative property then it definately satisfies commutati...
what makes you think that?
smallest counterexample: S3
chilly ocean
the identity element e needs to satisfy a * e = e * a = a for all a in the set. the operation * is commutative so it's enough to check a * e = a. the first part of that paragraph shows that e = 1 works
i dunno what properties G1 - G4 are but supposedly one of them is your set having an identity element
regal carbon
Can I write like this?
It's confusing right?
I don't know how they have arrived to b=-a+2
hot lake
they explain a little in the analysis paragraph
to find the identity they write a * e = a and solve for e
regal carbon
Yes
hot lake
to find the inverse of a, they write a * b = 1 and solve for b
regal carbon
But it's a+b-1=1
I don't know how they arrived here
The last line of the paragraph
regal carbon
a*b= 1 ?
regal carbon
I'm Solving, sir
regal arrow
Quick Q
what's a curve over a field?
Also, for a curve over a field K, what does it mean for the curve to have a point with "K-coefficients" .__.
For K-coefficients, does it mean coordinates in K?
stark sigil
Sounds like an algebraic geometry question!
A curve over a field k is a curve with a morphism to Spec k :)
A "point with K-coefficients" sounds like the concept of a K-point, which is a morphism from Spec K to the curve!
regal arrow
uhhh
is there a less complicated answer
this book literally assumes you don't even know multivariable calculus
regal arrow
Sounds like an algebraic geometry question!
I figured so, but also the third message in my reply to your answers.
civic tapir
I’m revisiting some abstract algebra and I’m having a hard time with this. I’ve gotten to if the order of b is less than 36, then e=a^m b^q, for some 1<=m<=11,1<=q<=2, but I’m having a hard time proving this is impossible. Additionally, I’ve avoided using the hint because I’m not sure why I can only consider factors of 36
chilly ocean
I’m revisiting some abstract algebra and I’m having a hard time with this. I’ve ...
lagrange’s theorem
chilly ocean
I’m revisiting some abstract algebra and I’m having a hard time with this. I’ve ...
Why are you considering m and q in a range?
civic tapir
Given that Lagrange’s theorem comes after this problem in the book, I’m assuming that’s not allowed in the proof. So I’m trying to do it algebraically, which led me to those ranges
chilly ocean
Well a is a cube root
with order 12
ab isnt identity because that contradicts orders
similarly with ab^2
if you consider a^11 multiplying by b gives a^14=a^2
similarly a^17 for b^2
you can show similar happens with a^10
no nvm this is wrong
past temple
so this is the claim that i want to prove
and my first thought was
constructing this mapping
and showing that its a bilinear bijection
am i going in the right direction here?
fossil shuttle
it's not going to be a bijection
in general
the tensor product isn't isomorphic to the cartesian product
You should definitely construct that map beta and check it's bilinear.
That's a good start.
Then by the universal property you get the desired map
which you want to prove is an isomorphism
The way I recommend you do this is as follows:
let $Z$ be an arbitrary $A$-module and let $f : A/I \times M \otimes Z$ be a bilinear map. Show that $f$ extends uniquely along $\beta$ to give a map $M/IM\to Z$
cloud walrusBOT
diligentClerk
fossil shuttle
then consider the case $Z = A/I\otimes M$ and $f$ the identity, this gives the desired inverse to $\beta$
cloud walrusBOT
diligentClerk
past temple
gotcha
wait
what do you mean by "f extends uniquely along beta"?
@fossil shuttle
thorn delta
I think this pops out when you consider
0 -> I -> A -> A/I -> 0
regal carbon
@past temple use fii note on playstore so you can write math easily
next obsidian
You have to prove M (x) A = M via multiplication
Then it does follow from
0 -> I -> A -> A/I -> 0 by right exactness
next marsh
hey! im super confused on this problem i think ive confused myself im not sure where to go
next obsidian
In general, what’s the degree of the extension Q(a)/Q?
Can you give me a different way to get that number in terms of a?
prisma thunder
Suppose $\mathrm{char} , F$ does not divide $\mathrm{Aut}_FL$ and that ${\beta_1, \dots, \beta_n}$ is a basis for $L$ over $F$. Let $H$ be a subgroup of $\mathrm{Aut}F L$. Let
$$\gamma_j = \sum{\sigma \in H} \sigma(\beta_j)$$
Show that $F(\gamma_1, \dots, \gamma_n)$ is the fixed field of $H$.
cloud walrusBOT
eM
prisma thunder
I was able to show $L^H \subseteq F(\gamma_1, \dots, \gamma_n)$ but haven't found a clear way for the reverse inclusion.
cloud walrusBOT
eM
prisma thunder
I would have probably brute forced it but I don't think that's what we want to do.
rustic crown
Notice that each gamma_j is fixed by the action of H. The sum is just permuted.
next obsidian
Gotta love sums over the entire everything which stays stable over an automorphism
By this I mean like doing some sum over all g in G then you end up doing a sum over all g^-1 in G (Maschke’s theorem or something does this.)
Love that stuff
next marsh
Can you give me a different way to get that number in terms of a?
hmm dfjklhasdf im completely spacing
next obsidian
I mean so
What’s the degree of Q(i) over Q
Hint: this doesn’t require anything about Q in particular. I could just as well have asked what is the degree of F(a)/F for any field F
Well, I actually have to drive so I’ll instead just say stuff
In general the degree of F(a)/F is the degree of a’s minimal polynomial over F
In your specific case, a is the rea root of a cubic polynomial
First determine if a can be a rational number, if it is then Q(a)/Q is degree 1, they’re the same field
If not, then a’s minimal polynomial divides the cubic polynomial of the problem
So a isn’t in Q, so its minimal poly has degree > 1, and it satisfies a deg 3 poly so it has degree <=3
If it was degree 2, then the cubic polynomial factors like
f(x)(x-c) where a is a root of f(x) and f(x) has degree 2
But now one of the other two roots, b or c (I wrote it as c when I factored) has to be a root of a linear poly
So that c is in Q
But… c is a complex number 😨
So a’s monomial poly can’t be degree 2, so it’s degree 3
Thus Q(a)/Q is degree 3
Once you adjoin another root it will end up having to be degree 2 over Q(a), so Q(a,b)/Q is degree 6 (this is because Q(a) ≠ Q(a,b) since the latter has a complex number, but the former has only real numbers. It has to be degree 2 since this is contained in Q(a,b,c) which has degree at most 3! = 6)
Then it follows that c is also in Q(a,b,c) since the splitting field of a deg n poly has degree at most n!
So Q(a,b,c)/Q is also degree 6
I have to drive now, so hopefully you can use this to work out the details necessary to fill in whatever gaps
chilly ocean
fields can be groups under both of their operations but rings(that don't have a multiplicative inverse) can only be groups under only one of their operations right?
rustic crown
um not quite
fields aren't groups under multiplication
non-zero elements of a field form a group under multiplication
you have to throw away the elements which don't have multiplicative invereses
in the case of fields, that just a single 0
but in case of rings, you might have to throw many more elements
each ring has at least 1 invertible element, namely 1
if the ring was Z, and you throw away all the elements that don't have multiplicative inverses, you're left with {1, -1} and that's a group under multiplication
chilly ocean
fields aren't groups under multiplication
ah right so field - {0}
chilly ocean
you have to throw away the elements which don't have multiplicative invereses
right okay
chilly ocean
if the ring was Z, and you throw away all the elements that don't have multiplic...
oh wow
thanks
rustic crown
this is called the group of units of a ring
chilly ocean
what is called group units? the elements that have an inverse?
cloud walrusBOT
det
rustic crown
so for fields k, the group of units k* = k \ {0}
next marsh
ohh that makes sense thank you
thorn flint
Hi can someone help me understand how to figure out 1b?
golden pasture
Hi can someone help me understand how to figure out 1b?
if b(0)=0 what does this tell you about the x^0 coefficient of b
thorn flint
@golden pasture if b(0)=0, then we have the zero polynomial and some constant right?
golden pasture
uhhhh
no?
try writing $b(x)=\sum_{i=0}^db_ix^i$, what would $b(0)$ be in this case
cloud walrusBOT
ari 亲
golden pasture
hence if b(0)=0, what does this tell you about some factoring b(x)
thorn flint
just 0?
thorn flint
yea
golden pasture
b(0)=?
thorn flint
1
golden pasture
yea
thorn flint
We're not just plugging x = 0? wouldn't it just be b_i(0)^j?
golden pasture
uhhhhhh
thorn flint
the sum of those
golden pasture
sum of b_i0^i yea
thorn flint
yea so why is it not 0?
golden pasture
b(x)=1+2x+3x^2+4x^3 rite
what happens in this case
cloud walrusBOT
ari 亲
golden pasture
so you can factor
thorn flint
So then this would be the a(x) that they are asking about
thorn flint
Thank you @golden pasture
regal carbon
When it comes to identity property ae = ea=a. I didn't understood how this relates to ab = a+b-1. And the explanation is like at the bottom analysis line.
ae=a+e-1=a which follows e=1
Be substituted e on the place of
b why?
And I don't know why took b=-a+2 by solving a*b=a+b-1=1.
I don't know why he ended up there. It's goddamn confusing.
The core thing is to prove a * e=e *a=a. But I'm not understanding what's happening over there.
Will anybody eloberate the step???
robust pollen
You define the operation * on Z by setting a * b = a + b - 1.
Then you ask: Does this operation have a unit, that is, does there exist an element e such that a * e = a = e * a for all a in Z?
Since * is commutative, it suffices to just try to solve a * e = a.
Now, simply write it out a * e = a + e - 1 and this should equal a, so the choice e = 1 does the job, since a * 1 = a+1-1=a for all a.
next you ask: Does every a in Z have an inverse with respect to the operation *? What does that mean? Well, you have to find an integer b such that a*b = e (remember that * is commutative, so you get b*a = e for free).
We have seen above that e = 1, so now expand a*b = e: it simply is a + b - 1 = 1. And now you solve this for b, to find b = 2 - a
regal carbon
You define the operation `*` on `Z` by setting `a * b = a + b - 1`.
Then you ask...
This is beautiful peace of art! thankyou so much.
robust pollen
Hey, so a while ago I asked the following question. Suppose that $A$ is a ring, and that $P$ is a generator in the category of right $A$-modules.
(This means that for every right $A$-module $M$ there is some set $I$ and an epimorphism $P^{(I)} \to M$, hence "generator").
The question/task is now to show that the functor $P \otimes_A -$ reflects short exact sequences.
cloud walrusBOT
expectTheUnexpected
robust pollen
And I have absolutely no idea. I feel like this question is not even well-posed: I am not told what the source and target of the functor are. Do we go from A-bimodules to right A-modules? Or do we go from left A-modules to abelian groups?
I would guess the former, since we cannot say anything about the generating properties of P when we're talking about left modules (since a priori, the categories of left and right modules can be quite different).
Also, a reminder: "reflects short exact sequences" means if $ 0 \to P \otimes_A K \xrightarrow{P \otimes f} P \otimes_A M \xrightarrow{P \otimes g} P \otimes_A N \to 0$ is exact, then $0 \to K \xrightarrow{f} M \xrightarrow{g} N \to 0$ was already exact
cloud walrusBOT
expectTheUnexpected
robust pollen
I guess since all categories considered are abelian it would be enough to show that tensoring with a generator reflects epis and monos?
oak grove
I know its not the hardest problem but I still need some help on this guy 
I get the roots should be $\pm \sqrt{ \pm i}$
cloud walrusBOT
jan Niku
oak grove
how can i be sure that I write my field in a minimal way though? I have a lot of doubts that $\bQ ( \sqrt i, \dots)$ is the best way to write it
cloud walrusBOT
jan Niku
rustic crown
notice that splitting field will contain i as well as (1+i)/sqrt(2) = sqrt(i)
oak grove
right, it has to contain some things like i and also 1/i etc
how does that inform you about what the minimal choice would be
rustic crown
Q(sqrt(i)) = Q(i, sqrt(2))
rustic crown
wait is your question like you don't wanna write all 4 roots?
oak grove
i guess i dont really understand what all these roots are hitting in the context of the field
so im not sure how to write them
i assume ive written all four roots?
or you mean in the extension
rustic crown
Q(sqrt(i)) is enough because a = sqrt(i) then other roots are just a, a^3, a^5, a^7
oak grove
weird
but do you capture everything you need?
or i guess you dont know for sure that 1/i is in it
you just need -i
rustic crown
what's your definition of Q(something)?
usually it's smallest field containing Q and something
and Q[something] is smallest ring containing Q and something
oak grove
well idk my working definition in my head is its q, then starting with some base power of some element not in q, all elements needed to capture all orders of that element and something in Q until you enter back into q
so capturing all actions you could make with the element youre adding
you dont need for sure that you have a multiplicative inverse i guess
or i guess just sqrt i to some power is its own inverse
so it doesnt matter anyways
this problem is so confusing
so you'd just note that since Q(sqrt i) is a field it'd have -sqrt i
how can you be sure itd have -sqrt -i
it must be picked up in some power of sqrt i?
rustic crown
sorry, i'm not sure what you mean by action exactly
oak grove
like operation
rustic crown
how can you be sure itd have -sqrt -i
it's sqrt(i)^3
like draw these 4 things on the complex plane
they'll form vertices of a rotated square
oak grove
im not really sure how to be honest
i was looking this weekend
i get they look like a square
i guess thats the thing to notice
rustic crown
oak grove
x^a + 1 has roots that look like a a-gon on the complex plane
youre saying this is enough to inform a person that sqrt i is sufficient
i think?
rustic crown
yep, because if you look at the 45 degree angle, all the other angles are multiples of it
rustic crown
yep, a, a^3, a^5, a^7
oak grove
why odd?
rustic crown
even things correspond to roots of x^4-1 = 0
rustic crown
nah, a^2 = i
oak grove
oh duh you need 3 to get from the first to second
okay
sorry for the dense brain I appreciate your help
🙇♂️
rustic crown
also in case of algebraic extensions, inverse of any non-zero element is a polynomial in the element itself
rustic crown
it basically means that every element satisfies some polynomial over Q
oak grove
im gonna have to think on that one
rustic crown
i think you're a little confused in giving a basis and a generator
so Q(cbrt(2)) is generated by a single cbrt(2), but a basis for this over Q is {1, cbrt(2), cbrt(4)}
oak grove
right
since cbrt(4) isnt sufficiently captured by cbrt2 and some element of Q
you need an additional power
rustic crown
if you're only allowed to add, then it's not... but when we write Q(a) you're allowed to multiply as well
so Q(cbrt(2)) contains cbrt(2)^2
remember it is smallest field contains Q and cbrt(2), so of course we can multiply any two elements in the field or add them
rustic crown
also in case of algebraic extensions, inverse of any non-zero element is a polyn...
to just get the idea of this without the words, say a is non-zero and it satisfies some polynomial f(x) over Q. Since a is not zero, you can assume that constant term of f is not 0. because otherwise f(x) = g(x) * x^n with g having a non-zero constant term.
so notice g(x) = x * h(x) + b, where b is a non-zero rational
plug in x = a to get 1/a = (-1/b) * h(a)
is that what you were looking for?
rustic crown
**jan Niku**
if you want to find a basis for this, say a = sqrt(i), then {1, a, a^2, a^3} is a basis. The other roots since a^4 = -1, the roots are {a, a^3, -a, -a^3}
another nice basis is
{1, sqrt(2), i, sqrt(2)i}
oak grove
sorry @rustic crown i have class so i had to check out suddenly but i appreciate it 🙇♂️ ill take a look at it
rustic crown
okie
plucky flicker
Hey guys! Do you have any idea how to start the 11th problem? Is there enough information? Don’t we need the size of conjugacy classes?
,rotate
cloud walrusBOT
robust pollen
the first row is easy 
trim grove
How can i show that $\sqrt{-5}$ is prime in $\mathbb{Z}[\sqrt{-5}]$ without using norm, any hint?
cloud walrusBOT
Algebra
robust pollen
Hey guys! Do you have any idea how to start the 11th problem? Is there enough in...
One way to "solve" the exercise is to compute some character tables with sage. This is my way, since I don't know character theory. I hope your way will be different.
chilly ocean
How can i show that $\sqrt{-5}$ is prime in $\mathbb{Z}[\sqrt{-5}]$ without usin...
Assume sqrt(-5) * (a+bsqrt-5)= xy for x and y in the ring. If sqrt-5 doesnt divide x and y, then a+bsqrt-5 has to divide x, y or xy
this is also a domain so you can cancel stuff I guess
rustic crown
you can also show the quotient is an integral domain
Z[sqrt(-5)] =~ Z[x]/(x^2 + 5)
the ideal (sqrt(-5)) on the left corresponds to the ideal generator by x on the right
so (Z[x]/(x^2+5)) / ((x, x^2+5)/(x^2+5)) =~ Z[x]/(x, x^2+5)
(x, x^2+5) = (x, 5)
Z[x]/(x, 5) =~ Z/5Z
Z/5Z is field so done
robust pollen
beautiful 
trim grove
so (Z[x]/(x^2+5)) / ((x, x^2+5)/(x^2+5)) =~ Z[x]/(x, x^2+5)
How this can you please explain?
rustic crown
okie so we have a ring Z[sqrt(-5)]
so we'll get the surjective map Z[x] --> Z[sqrt(-5)]
the kernel is (x^2+5)
trim grove
Okk
rustic crown
now what's the inverse image of the ideal (sqrt(-5))?
it's the ideal generated by (coset of) x on the left, okie?
that has to be the ideal (x, x^2+5)
trim grove
now what's the inverse image of the ideal (sqrt(-5))?
Why to check inverse image?
rustic crown
one way to see this is via the correspondence theorem,
Z[x] --> Z[x]/(x^2+5)
ideals of the right correspond to ideals of the left containing (x2+5)
oh because
we want to find the kernel of the composite
Z[x] --> Z[sqrt(-5)] --> Z[sqrt(-5)]/(sqrt(-5))
if you're new to working with maps, it might feel weird or hard, but eventually this feels very comfortable
you just use the isomorphism theorems without thinking
This statement below is very useful for concrete example, but think of it like just using the isomorphism theorems
cloud walrusBOT
det
prisma thunder
Let $L/F$ be Galois with Galois group $G$. For $\alpha \in L$, show that $L = F(\alpha)$ if and only if the images of $\alpha$ under $G$ are distinct.
cloud walrusBOT
eM
prisma thunder
What does it mean for the images of $\alpha$ to be distinct? I'm not sure if I'm understanding the question. Any automorphism of $L$ fixing $F$ acting on $\alpha$ are all distinct from one another?
cloud walrusBOT
eM
cloud walrusBOT
det
cloud walrusBOT
eM
prisma thunder
Exactly
chilly ocean
unless those are the same for some power
prisma thunder
But the images are distinct
chilly ocean
yeah but f can be equal f^2
next obsidian
It means that g(alpha) for every g in the automorpjism group are different
This is I think equivalent to saying that G acts freely on the orbit of alpha
rustic crown
faithfully?
rustic crown
what was free action i forget
rustic crown
i remember something like free groups act freely on a tree lol
prisma thunder
yeah but f can be equal f^2
I guess. Identity automorphism as an example.
hidden haven
free action is G acting on a bunch of disjoint copies of G
left adjoint to forgetful 😎
rustic crown
OH
hidden haven
equivalently an action is free iff gx = x implies g = 1 for any g and x
trivial stabs
prisma thunder
Same lol
hidden haven
faithful is weaker
next obsidian
But I mix up faithful and free
prisma thunder
O_o
hidden haven
gx = x for all x implies g is identity
rustic crown
ah right
hidden haven
free is
gx = x implies g is identity for any x
chilly ocean
ohhhhhhhhh
next obsidian
Well in our case we act only on 1 orbit
rustic crown
Ohhhhhhhhh
next obsidian
So it’s the same
hidden haven
Ohhhhhhhhhhhh
chilly ocean
Please be mature in the advanced channels.
next obsidian
Joe papa 😂
chilly ocean
blocked
rustic crown
wait, doesn't Sn act on {1, ..., n} faithfully and transitively?
hidden haven
wait I might be wrong then
next obsidian
I was wrong 
chilly ocean
....
prisma thunder
F*ck
next obsidian
rustic crown
hidden haven
ye faithful + transitive to free isn't true
next obsidian
hidden haven
F
next obsidian
,w 69!
cloud walrusBOT
next obsidian
Yeah
next obsidian
rustic crown
yeo i remember this
hidden haven
didn't you block him
rustic crown
70! > 10^100
next obsidian
It’s the first one
chilly ocean
Yes I did.
chilly ocean
prime minnow
Hey guys! Do you have any idea how to start the 11th problem? Is there enough in...
$$\begin{pmatrix}
1 &1 &1 &1 \
? & -1 & ? & ?\
2 & 0 & \frac{\sqrt{5}-1}{2} & ?\
? & 0 & ? & ?
\end{pmatrix}$$
$<C1,C2> = 0$ gives
$$\begin{pmatrix}
1 &1 &1 &1 \
1 & -1 & y & a\
2 & 0 & \frac{\sqrt{5}-1}{2} & b\
x & 0 & z & c
\end{pmatrix}$$
You can obtain with the norm of the second column : card(G) is even
If $card(G) = 2n$, $\frac{\sqrt{5}-1}{2}$ is sum of two unity root of order 2n since the subrepresentation is of dimension $2$
The aim is to find $2n$
$e^{ik\pi/n} + e^{il\pi/n} = \frac{\sqrt{5}-1}{2}$
Since the sum is real and not zero : $k = -l$
$2cos(k\pi/n) = \frac{\sqrt{5}-1}{2}$ and $k\pi/n = 2\pi/5$ which gives for instance $n = 10$
cloud walrusBOT
Boomer
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
prime minnow
multiples of 10 can work but we will try with n = 10
1+1+4+x² = 10k maybe you can show by solve the equation in x that the only integer possibility is when x = 4 and card(G) = 10
trim grove
What will $\frac{\mathbb{C}[x]}{\langle x^2+1 \rangle}$ will look like? Any hint
cloud walrusBOT
Algebra
rustic crown
do you know CRT?
trim grove
Yes
rustic crown
awesome!
trim grove
Here (x-i) and (x+i) are its factor
rustic crown
yep, and notice these are coprime!
CRT says, that ring is isomorphic to the product of two quotients
trim grove
Hmm how to check the are coprime in complex? Or i can say one doesn't divide other that's why they are co prime?
rustic crown
linear polynomials are irreducible over fields, so you just need to check they are distinct. since char is not 2, we're good
but you can give a direct reason, if d was a common divisor, then it divides their difference 2i
so d must be a non-zero constant, which are units
trim grove
One of my friend told me that if they are not associates then this will aslo work , is it true?
trim grove
Yess
rustic crown
right, but usually while working with polynomials over fields, we have a natural choice of working with monics
so yea, you're right, I should have said associates
trim grove
but you can give a direct reason, if d was a common divisor, then it divides the...
This this from linear combination stuff like gcd?
rustic crown
yea ig? d | a and d | b => d | ax+by
trim grove
but you can give a direct reason, if d was a common divisor, then it divides the...
Can you please explain this condition?
rustic crown
if some polynomial divides both (x+i) and (x-i) then it divides (x+i) - (x-i) = 2i
but 2i is a non-zero constant in C
this forces them to be coprime
explicitly, 1/(2i) * (x+i) - 1/(2i) * (x-i) = 1
so they coprime
trim grove
but 2i is a non-zero constant in C
Okk so here that number may be 2,-2,i or -i , these doea not divides (x-i) or (x+i) so that's why they are co prime?
rustic crown
oh we're working in C, all these are units
it would require more work if it was Z[i]
trim grove
it would require more work if it was Z[i]
I was also thinking about this
trim grove
they aren't coprime for Z[i], the ideal (x+i, x-i) = (x+i, 2i) = (x+i, 2)
Are they in C[x]?
rustic crown
yep
you need to be a little careful with using coprime over non PIDs
Z[i] is a UFD, so we get that Z[i][x] is a UFD
but we can't say that if they share no non-trivial factors, then they are coprime as in sum of the ideals equal the whole ring
for the purpose of CRT, we want the two ideals I and J to satisfy I + J = (1)
in C[x], (x+i, x-i) = (x+i, 2i) = (x + i, 1) = (1)
trim grove
in C[x], (x+i, x-i) = (x+i, 2i) = (x + i, 1) = (1)
How? X-i is 2i
trim grove
doing this doesn't change the ideal
Really?
rustic crown
if something can be expressed as (x+i)*f(x) + (x-i)*g(x) then it can equivalently be expressed as (x+i) * (f(x) + g(x)) + (2i) * (-g(x))
this is some sort of "euclidean algorithm"
(a, b) = (b, a - b * q)
trim grove
I know two numbers are coprime when there gcd is one. Can we somehow use this to show (x-i) and (x+i) is coprim?
rustic crown
how do you define the gcd?
trim grove
how do you define the gcd?
Gratest common divisor
rustic crown
for PIDs, people just say gcd d of a and b is the generator of the ideal (a, b) = (d)
but you can talk about gcd for UFDs as well, prime factorize both a and b and then take the minimum exponent for each prime factor
of course these two agree for PIDs, but depending on the definition, we might need to prove the equivalence
trim grove
but you can talk about gcd for UFDs as well, prime factorize both a and b and th...
This looks easy in integrs, but in complex it looks tough.
trim grove
for PIDs, people just say gcd d of a and b is the generator of the ideal (a, b) ...
Okkk now I'm getting this
trim grove
for PIDs, people just say gcd d of a and b is the generator of the ideal (a, b) ...
Thats why gcd is linear combination of these two elemnts?
rustic crown
yee
trim grove
in C[x], (x+i, x-i) = (x+i, 2i) = (x + i, 1) = (1)
Just one more doubt, how 2i into 1 ?
rustic crown
it is unit
if an ideal contains a unit, then it contains 1
because ideals absorb multiplication by all elements
in particular the inverse of the unit
trim grove
I heard that if ideal contain unit then that will become whole ring
rustic crown
yee, that's exactly the last equality
simple mulch
Hey guys
So, I have a set $A = {S \subseteq V : S \text{ is linearly independent and }S_0 \subseteq S}$. Where $S_0$ is an arbitrary linearly independent set in a vector space $V$. The author claims $A$ is partially ordered by inclusion. I wonder how is the order defined? If that makes sense (the question)
simple mulch
oh ok so S <= T if and only if S is a subset or equal to T
rustic crown
yup
simple mulch
ok thanks
chilly ocean
if $H\subset G/N$ is a subgroup, then there exists a unique subgroup $H' \subset G$ containing $N$ such that $H = H'/N$
cloud walrusBOT
TTerra
chilly ocean
that's what it's asking you to prove
barren sierra
got it
past temple
what does the product of an ideal with itself look like?
for example, if we let I = (x,y) in F[x,y]
lavish mantle
$I^2 = (x^2, xy, y^2)$
cloud walrusBOT
polikuj2
past temple
how did u get that
thorn delta
can you see (x^2, xy, y^2) \subset I^2?
thorn delta
an element of $I^2$ is $\sum (ax + by)(cx + dy)$.
cloud walrusBOT
kxrider
past temple
ahhh okay yeah i was being dumb
untold cloud
Hi, guys, is there any way to extend maschke's theorem to infinite group? Is it even possible that if i have a representation V of infinite group, and a subrepresentation V_1, then i can find V_2 s.t. V = V_1 (+) V_2?
lavish mantle
There is a small sections on wikipedia about non-examples if G infinite : https://en.wikipedia.org/wiki/Maschke's_theorem#Non-examples
untold cloud
Thanks. But what about matrix groups?
past temple
right now i have a mapping $\beta: (x,y) \otimes (x,y) \to (x^2, xy, y^2)$ defined by $\beta(a \otimes b) = ab$. This is an $F[x,y]$-linear surjection, but I want to prove that it's not an injection. How should I go about proving this?
cloud walrusBOT
pdk
past temple
where F is a field
(pls dont say smth like "prove the kernel is nontrivial")
perhaps its possible to show that beta extends to an isomorphism to a module strictly containing (x^2, xy, y^2)?
but even then how do i construct that module
chilly ocean
(pls dont say smth like "prove the kernel is nontrivial")
prove the kernel is nontrivial
ty for fixing
chilly ocean
okay so the easiest approach would be to show that the kernel is not trivial
actually perhaps x tensor y might be different than y tensor x but map to the same thing
past temple
hmm okay so in that case i actually have to think about
what the tensor monomials actually are
i have that the the tensor product of (x,y) with itself is equal to this
where A = F[x,y]
chilly ocean
honestly I had something similiar to do recently and I didnt figure it out but I think there was some problems with exactly that x tensor y so just guessing
thorn delta
@past temple i think i have an idea. Suppose for contradiction that you have an iso and recall that M (x) I is iso to M/IM
lmk if you need more hints
past temple
ohh right i forgot that result
okay so then M/IM = (x,y)/(x^2, xy, y^2)
and so i want to show that this strictly contains an isomorphic copy of (x^2, xy, y^2)?
still seems pretty hard to prove that tho..
thorn delta
wdym "strictly contains an isomorphic copy."
For contradiction we assume there is an iso I (x) I --> I^2 i.e. an iso I/I^2 --> I^2.
u just need to show that this can't happen when I = (x,y)
past temple
oh
ok so bwoc there's an isomorphism (x,y)/(x^2,xy,y^2) -> (x^2,xy,y^2)
what are some general strategies for proving that no isomorphism exists between two objects
ive rly only done problems where i have to prove that there IS one
thorn delta
in general, elements of isomorphic rings/groups/modules must satisfy the same relations. For example elements should have the same order
chilly ocean
Im not crazy right?
how is this reducible in Z[x]
i showed the second part
Omg whatever you guys said I think solves my problem
(x^2 - x + 1)(x^2 + x + 1)
bruh
wtf
U crazy
no i just put it into wolframalpha
cheater
how showed it isnt factorable into two binomials
i showed it isnt *
that was my mistake ig
thorn delta
yep
past temple
so that doesnt help us here
do i need to like
consider the explicit construction of the tensor product here then
thorn delta
im not totally convinced beta is not injective. I don't see why $f : (x^2, xy, y^2) \to (x,y)\otimes (x,y)$ given by $f(h) = h(1\otimes 1)$ is not a left inverse.
cloud walrusBOT
kxrider
next obsidian
So like
This map isn’t injective but you have to look at non-simple tensor or something
You show that the thing isn’t 0 to begin with by looking at partial derivatives or some shit
I think
I’ve done a problem like this before, it fucking sucks
thorn delta
what goes wrong with $f\beta(h\otimes g) = f(hg) = hg(1\otimes 1) = h \otimes g$?
cloud walrusBOT
kxrider
thorn delta
oops fixed ^
next obsidian
Look at example 5
I can’t tell you what’s wrong with your map, I suspect it’s secretly not linear
Altho that doesn’t actually matter
Or maybe it does, I think it might fail to be an inverse for non-simple tensors?
Look at what happens to (x (x) y - y (x) x) like in the example
This maps to xy - yx
Then f maps this to 0
But show x (x) y - y (x) x isn’t 0
So I think the issue is you can’t check if f is an inverse to beta on simple tensors
Because of “something something not linear@
Or some shit
¯_(ツ)_/¯
Anyway to show that (x (x) y - y (x) x) is non-zero
Consider the map from I x I -> K[x,y]
Defined by (f,g) -> f_xg_y (those are partial derivatives)
This is bilinear
Then (x (x) y - y (x) x) maps to
1 - 0 = 1 is non-zero
So the element can’t be 0 cuz it maps to non-zero thing
Or something like this works
thorn delta
am i going crazy why isn't x (x) y = xy(1 (x) 1)
thorn delta
ohhhhhh
thorn delta
yup rookie mistake
next obsidian
I’ve done this before hahaha