groups-rings-fields

[past temple]

im a little confused by that

is it just because E is an isomorphism

so its gauranteed to have eigenvalues?

even then how does it show that result

[lethal dune]

ok short answer, because characteristic poly of the operator can splitted into linear or quadratic factors

linear terms corresponding to 1D invariant subspace (also known as eigen spaces) and quadratic factor corresponds to 2D invariant subspace

[final oasis]

you have either odd are even dimension

[lethal dune]

we are not guaranteed to have EV, best we get is a quadratic factor like (x^2+ax+b)

[past temple]

ah okay

yeah that makes sense

so now we can write R^k = V (+) W

i'm confused by how to prove the hint though

[lethal dune]

ok for a long answer, let $p(t)=t^n+a_{n-1}t^{n-1}+\cdots+a_0$ be the char poly, then it can be factored into factors of the form $p(t)=(x-x_1)(x-x_2)\cdots(x-x_k)(x^2+a_1x+b_2)\cdots(x^2+a_lx+b_l)$ and $p(T) = 0$ means there is at least one $v\in R^n$ s.t. $p(T)v = 0$ or $Tv = x_iv$ or $T^2v = -a_iTv - b_iv$

[lethal dune]

first case gives you 1D invariant subspace. second one gives you 2D subspace as $T^2v = -a_iTv-b_iv \in \text{span}(v, Tv)$

[lethal dune]

and that invariant subspace gives you a basis wrt which you get a matrix of the form $\m{A & B \ 0 & C}$

[past temple]

ok im a little lost on ur last step

what is that basis?

[lethal dune]

then any vector in $V$ is of the form $\alpha = av+bTv$ and so $T(\alpha) = aT(v) + b T^2(v) = aTv + b(-a_iTv-b_iv)\in \text{span}(v, Tv)$

[lethal dune]

is this part clear?

[past temple]

hmmm

yeah its clear

[lethal dune]

hmm so now we get a basis of V as {v, Tv}

we can extend it to the whole space.

but then, Tv = T^2v ==> first column of the matrix is [0, 1, 0, 0, ... ]

T(Tv) = T^2v = -ai v - bi Tv ==> 2nd col is [-ai, -bi, 0, 0, ...]

so we get a 2x2 block matrix and all zero below

[past temple]

wait now im a little confused

wait so our matrix A is

[lethal dune]

yes that 2x2 block

[past temple]

$\m{0 & -a_i \ -b_i & 1}$

[cloud walrus]

pdk
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lethal dune]

$\m{0 & -a_i \ 1 & -b_i}$

[past temple]

so that's A?

[lethal dune]

this should be A

[past temple]

hmm okay

what are our B and C then?

[lethal dune]

they could be anything

that doesn't concern us for now

all what matters is that we have a block of 0's below A

i.e. matrix of the form $\m{A & B \ 0 & c}$ under suitable basis

[past temple]

ahh right

[past temple]

wait im a little confused by this

why the ...?

is it not just [0 1 0 0]?

[lethal dune]

depends on the dimensions, you need to add enough 0s

[past temple]

oh wait we dont know what k is

right

wait could u explain why Tv = T^2v ==> first column of the matrix is [0, 1, 0, 0, ... ]

[lethal dune]

what is the coordinate of v wrt basis (v, Tv)

[past temple]

(1,0)

?

[lethal dune]

now what's the coordinate of Tv wrt basis (v, Tv)

[past temple]

(0,1)

[lethal dune]

do you see how i'm getting that (0, 1, ..)

it's the coordinate of T(v) wrt basis (v, Tv)

padded with 0's enough number of times

[past temple]

ohh yeah i see that

the columns of T in the basis (v, Tv)

are going to be the coordinate representations

of v and Tv

[lethal dune]

yup

[past temple]

ok yeah that makes sense

and by a similar reasoning we get that T(Tv) = T^2v = -ai v - bi Tv ==> 2nd col is [-ai, -bi, 0, 0, ...]

so we have our desired hint

[lethal dune]

yes

[past temple]

the next part of our hint says

to set this as the homotopy

with t in [0,1]

when t is 1

we have our map as before

when t is 0

does that fix W?

if so, im not immediately seeing it

[lethal dune]

no that doesn't fix W

maybe the span stays fix idk

[past temple]

heres the problem again for reference

i think theres a typo

should be E_1 for one of them

[lethal dune]

but I assume what they are doing is reducing step by step, i.e. then consider C and recursively apply the same thing and make a diagonal block matrix

well yeah if t=0 we just have $\m{A & 0 \ 0 & C}$

[lethal dune]

which makes W T-invariant

then the same argument applies

[past temple]

which argument?

the one that we used to show that E has a block matrix representation?

[lethal dune]

ye

[past temple]

wait why do we need to repeat the argument

we have E in this block matrix formation

and we have that E_1 fixes V

and E_0 fixes W

are we not done?

[lethal dune]

idk i just guessed

[past temple]

that E_0 fixes W?

[lethal dune]

no, that we repeat the argument on C and make it a block diagonal

keep reducing it until we get a diagonal consisting of 2x2 or 1x1 blocks

that was a guess so don't count on it

[past temple]

wait but like

even without that

wait why do we need to repeat the argument for C?

now im rly lost

V is E_1-invariant right?

we specifically constructed it to be

am i missing something here?

:((

[lethal dune]

ig ignore my previous messages

if that's what confusing you

[past temple]

ok

i mean our goal was to find a homotopy consisting of linear isomorphisms E_t

such that V is E_1-invariant

and W is E_0-invariant

is it supposed to be clear that

$\m{A & 0 \ 0 & C}$

[cloud walrus]

pdk
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[past temple]

fixes W?

[past temple]

this matrix

[lethal dune]

the new matrix now fixes W

[past temple]

when u set B to zero right

my original question was that

im not immediately seeing

why that new matrix fixes W

im just having misunderstandings of

rly simple concepts here

since my linear algebra is so bad

[lethal dune]

because for any vector in W, T(that vector) becomes a linear combination of elements of W. because it has zeros above

similar to how we showed A is invariant

T(w) = 0v + 0 Tv + aw1+bw2+...

we should get something like this in case of a vector in W

[past temple]

wdym it has zeros above?

the matrix has zeros above?

[lethal dune]

well in this matrix, C has 0's above

try with an example

[past temple]

ok wait so

our basis

is

{v, Tv, w_1,..., w_m}

where w_i span W

?

[lethal dune]

$\mqty[\dmat{1}{\mqty{1 & 2 \ -1 & 2}}]$ and $v=\mqty[0\1\2]$

[lethal dune]

you'll never get a non zero on the first coordinate

here your V = {(x, 0, 0)} and W = {(0, x, y)}

[past temple]

hmm

okay

[past temple]

going back to this

is it true that

{v, Tv, w_1,..., w_m}
where w_i span W

form a basis for R^n

or am i completely misunderstanding

[lethal dune]

yes it's true

[past temple]

is that what ur using?

[lethal dune]

also yes

I'm telling you to use an arbitrary vector $w$ in $W$ i.e $w = c_1w_1+c_2w_2+\cdots + c_lw_l$ i.e. the column vector $\mqty[0 \ 0 \ c_1 \ \vdots \ c_l]$ and multiply it with $\m{A & 0 \ 0 & C}$

[lethal dune]

see that you get a vector of the form 0v+0Tv+k1w1+..+kl wl or not

[past temple]

ohhhh okay that makes things very clear

okay so we're done proving the hint

and the problem bc we have our homotopy

[lethal dune]

(we are supposed to use the hint, not prove it, ) well

[lethal dune]

idk if that argument is enough for homotopy,,

[past temple]

oh nooo

do i have to prove that

all the E_t

are isomorphisms?

very lost and sad

how am i actually supposed to do this problem...

[hidden haven]

What's the problem

[hidden haven]

Yo what is this magic

How do matrix so efficiently

[lethal dune]

[lethal dune]

#linear-algebra ?

[simple mulch]

For instance f(x) = 2x
then we have $2\alpha_1 + 0\alpha_2 + \dots + 0\alpha_n = 0$ for $\alpha_1 \neq 0$ and $\alpha_i = 0$ we have a linear dependent combination?

[cloud walrus]

mns

[simple mulch]

I will go there

[wise igloo]

#latex-help next time, thank you

jk lmfao

[lethal dune]

what?

[wise igloo]

fuck

I tagged the wrong message

im so sleep deprived

[wise igloo]

I meant to tag this one

but the joke is old now

[lethal dune]

this might be a trivial question but how do I show ℤ×ℝ is not isomorphic to ℝ?

[hidden haven]

as a what?

ring?

[lethal dune]

group, +

[delicate bloom]

that's a shame cause ring case is nice cause has klein 4 group as subgroup which shows it

[hidden haven]

also one is a field the other isn't lol

[delicate bloom]

that too 🥲

[lethal dune]

not sure about the field argument

[hidden haven]

That won't work for groups

[delicate bloom]

well has zero divisors

[lethal dune]

that's what I was thinking

[lethal dune]

in group? not talking about ring tho

[delicate bloom]

talking about the field argument

[hidden haven]

oh

I have it

in Z x R

you have (1,0)

it has to map to some non zero element of R

but then it has a half

but (1,0) didn't

[viscid pewter]

oh wp

[delicate bloom]

nice

[viscid pewter]

wait hmm

that's still kinda using the fact that R has multiplication

[lethal dune]

[hidden haven]

half of x is an element y such that y+y = x

I see no multiplication

[viscid pewter]

why must y exist again?

[hidden haven]

Because it's R

God given fact about R 😌

[viscid pewter]

k lol

yeah completeness ok

[hidden haven]

Completeness as a metric space?

You have to fall back to x/2 in the end lol

I don't see how completeness does anything because Z is also complete

[viscid pewter]

like. least upper bound?

[hidden haven]

Z has that too

[viscid pewter]

argh

i don't know analysis

[hidden haven]

Yeah I don't think analysis is useful here

[delicate bloom]

R is a field of characteristic not = 2 lol

[hidden haven]

You have to use the fact that you can divide by 2 in R

[south patrol]

dw that's just how our uni defines completeness lol

[viscid pewter]

hmm.

[south patrol]

in first year

[viscid pewter]

ok so this is just a given

fiiine

[hidden haven]

well by what merosity said

Q is a metrised field and when you complete metrised fields you get metrised fields of the same characteristic

[delicate bloom]

we can just use multiplication of R to show it exists we don't have to be insane

2y=x we know is legit in R

[hidden haven]

or replace metrised with ordered

[delicate bloom]

so we write it additively as y+y=x to make it an addition fact for the group argument

[hidden haven]

so is the question about how we prove that we can divide by 2 in R or am I misinterpreting it

if it is that then you just gotta look at the construction of R

[delicate bloom]

I guess next is can we extend this proof for an arbitrary field K, show that ZxK is not isomorphic to K?

seems like it'd be the same argument just being sure to pick n != char(K) instead of 2

[hidden haven]

ye

[viscid pewter]

for finite fields you can use an order argument?

[delicate bloom]

true, finite fields is dead easy haha

[hidden haven]

picking n not divisible by the char works

not just unequal

[delicate bloom]

yeah my bad

integer n>1 such that gcd(n, char(K)) = 1 🧐

I guess divisible is fine cause it's always prime lol

[hidden haven]

ye lol

[delicate bloom]

shoot lol

[south patrol]

ig the map x -> 2x in R is an automorphism and not in Z x R is probs an equivalent way of writing the proof right

well

would have to pull along the isomorphism

[hidden haven]

this should be fine because the automorphism is described using the group operation

[south patrol]

yeah

[viscid pewter]

i think i like that one better

[tribal moss]

It's really the same argument if you rephrase it as "the map x -> x+x is surjective in one case but not in the other" -- it's the lack of surjectivity that prevents it from being an automorphism anyway.

[south patrol]

exactly yeah

[lethal dune]

can we show x/2 exists without resorting to multiplication/division in R?

[hidden haven]

We are assuming that we know what R is

Otherwise you cannot expect to answer the question

[lethal dune]

like if we dont then?

[lethal dune]

surprising

[hidden haven]

I mean if R were Z^|N|

Then of course these are isomorphic

[lethal dune]

had encountered something similar before, like to prove something on (Q,+) we had to use the multiplication of Q

which, we should need because + is being considered

[hidden haven]

You have to because that is how Q is defined

its like divisible hull of Z or something

that just means you add all the x/n elements

for any x and any integer n

[lethal dune]

honestly is feels super weird to me, though I know the construction of Q from Z, like as a standalone set Q! anyway it isn't something that anyone should waste their time explaining... It's just been in my head for a long time

[delicate bloom]

I wonder if that case with Q and the case with R both fall into that general proof for arbitrary field K

or was it a different problem

[lethal dune]

a different one

maybe something like (Q,+) has no non trivial subgroup or something

[delicate bloom]

is it short enough to state or you don't remember, just wanting to see if we can maybe stretch the proof to be more general to encompass that one too possibly

oh

[lethal dune]

that's a false statement

[hidden haven]

Might have been "all subgroups of Q are either 0 or Z or dense"

[lethal dune]

maybe

or no non-trivial honomorphism I think

[tribal moss]

(Q,+) is a torsion-free nontrivial abelian group such that all endomorphisms other than the zero map are iso?

[lethal dune]

where we had to use f(n)=nf(1) or something

yes that's what it was

[simple mulch]

hey

H,K normal subgroups of G

I want to show (H and K) is a normal subgroup of G

I will do it by: for all g in G and a in (H and K), we must have gag^{-1} in (H and K)

Take any g in G and a in (H and K)

note ag = ga for a in H and a in K

thus gag^{-1} = agg^{-1} = a in (H and K)

I have some questions

[delicate bloom]

this is false

[simple mulch]

in here,

ok I see

ag = ga_h

and ag = ga_k

[delicate bloom]

no I mean you're saying they commute, that's not the case

[viscid pewter]

that's more like it

[delicate bloom]

it's the g element appearing twice that's off

[viscid pewter]

wait

fuck

[simple mulch]

wdym

[delicate bloom]

wait I'm backwards, what @viscid pewter said is right

[simple mulch]

so from gag^{-1} we then have two situations

$gag^{-1} = a_kgg^{-1} = a \in H \cap K$

and

$gag^{-1} = a_hgg^{-1} = a \in H \cap K$

[lethal dune]

[cloud walrus]

mns

mns

[simple mulch]

idk

something seems wrong

[lethal dune]

a_h=a?

[simple mulch]

that would be the element in H that makes ag = ga

so: ag = ga_h

[viscid pewter]

this is what you're saying

but a_k != a

a_k = a_k

[simple mulch]

indeed and we don't know a_k is in H which would make a_k in (H and K)

so the proof is wrong

What do you think of the following:

[cloud walrus]

mns

mns

mns

[viscid pewter]

yeah that'll do it

[simple mulch]

Ok nice, now may I ask how did you do it?

[viscid pewter]

that's how i did it lol

[simple mulch]

lel

thanks

[south patrol]

another way is that the map H x K -> G (h,k) |-> hk has image HK and kernel H cap K i think

you just need to show it's a homo which is fine

[simple mulch]

didn't reached homomorphisms yet

[chilly ocean]

So, according to wikipedia, the class of fields doesn't form a variety (in the sense of universal algebra). Is there some workaround to this? Maybe by considering only fields of a given characteristic at once? Also, this is the only algebraic structure i know that doesn't form a variety, what are some common others? (i don't consider structures with relations to be "algebraic structures").

[chilly ocean]

<@&286206848099549185>

[hidden haven]

Restricting to a certain characteristic won't help

You can prove that some class is not a variety by eg showing that there is no free algebraic structure of that variety on an arbitrary set

because all varieties have that property

and there are no free fields (even if you fix a characteristic)

or similarly product fields don't exist, and you can always take products in varieties

[chilly ocean]

do you know any other common examples of algebraic structures that are not varieties?

[hidden haven]

Can't think of any off the top of my head

but you could make up random theories defining classes of some algebraic structures

and most of them won't satisfy the conditions to be a universal algebra

[chilly ocean]

i mean yea lol

thats why i added the adjective common

[hidden haven]

right

[cyan marten]

Some classes of lattices don't form varieties, like semimodular lattices, I think. But that's not obviously "algebraic."

[acoustic pine]

So, I know in algebraic topology, (co)homology loosely measures "how many holes a space has" in a coarse way. Is there any sort of geometric interpretation for (co)homology of groups and rings?

Also, for things like Lie Groups, does the DeRham cohomology coincide with the group cohomology, or is it more subtle than that?

Ig a better way to frame the question is, what does group (co)homology measure about a group?

[maiden ocean]

the closest thing i can think of to "geometric intuition" in general would be something like

the group cohomology of G is isomorphic to the singular cohomology of the classifying space BG

(eilenberg maclane space K(G, 1))

But BG in generally doesnt have a very simple geometric description so

[iron vessel]

Anyone know how to solve this?

I'm not sure how to compute the map

[hidden haven]

what is the Fr_2 notation?

Is it the x maps to x^2 map?

[iron vessel]

Yeah it is the Frobenius map

[hidden haven]

I see

[iron vessel]

sorry I should have included it

[hidden haven]

So you can first just square the element

which is easy in char 2

because the square distributes over addition

Then you get one term which has a 4th power, while the rest are fine

[iron vessel]

Oh and i use my function x^3 + x + 1 to reduce it ?

[hidden haven]

so you rewrite the 4th power using 0 = a^3 + a + 1

ye

[iron vessel]

awesome, thanks alot! I'll try it out

[untold cloud]

Hi, guys, i was wondering if the addition and multiplication operations in T(V), the tensor algebra of V, are direct sum and tensor product, respectively

[hidden haven]

The addition is the one that you get from it being a direct sum of additive groups

The product is just tensoring yes

[untold cloud]

so for example, (v_0,v_1,...)+(w_0,w_1,...)=(v_0+w_0, v_1+w_1), but i am confused what is (v_0,v_1,...)*(w_0,w_1,...)?

[hidden haven]

Defined by distributivity

And you define v_m ⊗ w_n as just that, their tensor

[untold cloud]

something like this (v_0\tensor w_0, v_0\tensor w_1+v_1\tensor w_0,...)?

[rustic crown]

yee

[hidden haven]

Yep

[untold cloud]

thanks!

[hidden haven]

[broken stirrup]

are free abelian groups just free groups that are abelian?

i mean if yes does that mean there are only two free abelian groups up to isomorphism?

trivial group and infinite cyclic or Z

[hidden haven]

No

[hidden haven]

Yes if it were true

Free abelian on a set of generators S is when you take all words and you add the condition that everything commutes

[chilly ocean]

english language btfo

[hidden haven]

It's free in the category of abelian groups

[broken stirrup]

got it thanks

[willow mason]

quotiented out by nG

why delete your question

[frank fiber]

This are the free abelian groups with 0 and 1 generators but you can also have free abelian groups with more generators

[next obsidian]

I prefer thinking of the free abelian group on S as formal sums rather than words

[chilly ocean]

how do you prove some group isnt free abelian?

does it need to satisfy universal property to be free abelian?

is it a iff

because say someone asks to prove Z x Z is not free abelian

i cant say much tbh

[frank fiber]

yes

[chilly ocean]

so is S= { (0,1),(1,0) } not a valid set?

and then F(S)?

lol

it isnt

or if it commutes

nah no clue

[frank fiber]

But this is the free abelian group with 2 generators

[chilly ocean]

as i suspected

is there a homology theory that has not free abelian groups as homology groups or chains as not free abelian groups

not in AT mb

[frank fiber]

I think that in cellular homology the groups of the chain are not necessarily free abelian

[long obsidian]

If F is a group that is finitely generated by a set of elements each of finite order, then is F a finite group?

[chilly ocean]

yea

[lethal dune]

not necessarily

[chilly ocean]

counter example?

[lethal dune]

you can get an example in the matrices

don't remember exactly what,

[chilly ocean]

wonder if changes if abelian or not

lemme think thru

intution tells me ye

with conditions you gave it should def be a finite group im thinking

[chilly ocean]

are you saying each of the generators is finite order

[long obsidian]

Yeah that's what I meant to say my bad

[chilly ocean]

mm

if we show every element has finite order then we done

and every element does

because say a_i are generators with order n_i

[long obsidian]

Apparently the infinite dihedral group is a counter example. 😮‍💨

[chilly ocean]

what no lol

rotations arent finite order

[lethal dune]

@chilly ocean are you trying to convince him something that is false?

[chilly ocean]

nah

its true

finite generators of finite order leads to finite set

[lethal dune]

I already gave an example where each has finite order, but A*B has infinite order => F is infinite

[chilly ocean]

idk anything about julia

and what group are you talking about

[chilly ocean]

i dont see the example

oh wait

what about free group

with 2 generators

and finite order

infinite words exist

rip ig that was counter example

if its abelian that changes things tho

[lethal dune]

yes

[long obsidian]

Thank you I read something wrong

[chilly ocean]

no biggie

[simple valley]

Suppose I have a group $G$ such that \begin{tikzcd}
\bZ & 1 \
G & {\bZ^2}
\arrow[from=1-1, to=1-2]
\arrow[from=1-1, to=2-1]
\arrow[from=2-1, to=2-2]
\arrow[from=1-2, to=2-2]
\arrow["\lrcorner"{anchor=center, pos=0.125, rotate=180}, draw=none, from=2-2, to=1-1]
\end{tikzcd} is a pushout square. That is we have some hom $f : \bZ \to G$ such that $G$ quotiented by the normal closure of $\mathrm{im} f$ gives $\bZ^2$. Does this identify $G$ uniquely? $\bZ^3$ is an obvious candidate, but I'm not sure others don't exist.

[cloud walrus]

mniip

[next obsidian]

If G is abelian then yes

Well…

Okay maybe not quite. If f is injective and G is abelian then yes because then G is classified via Ext

I feel like this sort of thing is dealt with by group cohomology crap

Oh wait

Obviously no lmao

Take G = Z/nZ (+) Z^2

@simple valley

Z -> G given by sending to the first coordinate and just going mod n

[simple valley]

ok so maybe I'm interpreting this incorrectly then

I have a (rather trivial) problem that said to prove this connection between pushouts and normal closure quotients

and then it says "use this fact to compute the fundamental group of T^2 with a disk cut out"

so I thought we could represent a torus as a union of torus-sans-disk and disk, and go for SVK

but as you said this does not uniquely define pi1(torus sans disk)

wait isn't torus-sans-disk a bouquet of 2 circles

meaning pi1 of F_2

I think what's going on here is that we have Z -> F_2 via n mapsto [a,b]^n

the normal closure of that is [F_2, F_2]

I think

idk how you would "compute" F_2 from this tho

am I meant to cover the torus with two strips

type of thing

[upbeat juniper]

is there a way to prove the galois correspondence without using the primitive element theorem?

if I'm not wrong, it's used to show that H=Aut(M/M^H) where M is a field and H <= Aut(M)

it's fine, you can still leave your original question up

[limber vale]

Is there a systematic way to find the lattice of galois groups? I am working out an example (splitting field of x^8-2). I worked out all the automorphisms correctly (so i have the galois group). I have trouble both with finding the subgroups of the galois group as well as when i have them drawing the lattice

[hidden haven]

This doesn't look like a cover

[simple valley]

it covers torus sans disk

these things have fundamental group Z, their intersection contractible, so by SVK we have that the fundamental group of torus sans disk is the free product Z*Z i.e. F_2

[hidden haven]

Right

[hidden haven]

In my experience you just gotta grind it out, but writing the Galois group in terms of generators and relations helps a lot

[fluid shell]

Hello, can someone help me to prove that for all n >= 4, we have that :

Notation: In is the matrix identity
E_ab is the matrix with 0 everywhere except in the a row and b column where it is equal to 1

I want to prove that

(In + E_1,2)+(In+E_n-1,n) = In + E_1,2 + E_n-1,n for all n >= 4

I block where I need to prove that (E_1,2)(E_n-1,n) = 0 for all n >= 4

Thank you, please excuse my english.

[rustic crown]

Try proving,

$$E_{a, b} \cdot E_{c, d} = \delta_{b, c}E_{a, d}$$

[cloud walrus]

det

[fluid shell]

Thank you!

[rustic crown]

for n>= 4 we'll have that 2 < n-1 so that will be zero

[fluid shell]

Thank you very much

[long obsidian]

For p prime, is it true that every nontrivial element in $Z_p$ will have order p?

[cloud walrus]

fajitas

[hidden haven]

yes

[delicate bloom]

it follows from lagrange's theorem, you can think of the cyclic subgroup generated by a single element

[long obsidian]

Thank you guys

Let G be a finitely generated abelian group of order $p^m q^n$ (p and q prime) and $G(p)$ denote the subgroup whose element's orders are powers of some prime p. I believe the classification theorem of finitely generated abelian groups says $G\cong Z_{p^n} \bigoplus Z_{q^m}$. But in this case I feel as though one could alternatively say $G\cong G(p)\bigoplus G(q)$. Is this so?

[cloud walrus]

fajitas

[rustic crown]

second statement is true, first one isn't

it could be like Z/p + Z/p^n-1 + Z/q^m stuff

[long obsidian]

I'm not quite sure I'm understanding you, are you saying one could have a different partitioning of the primes (and so there are other ismorphisms that it implies) or are you saying the first one isn't an isomorphism at all?

[rustic crown]

so i'm saying consider a group of order 12

it could look like Z/4 + Z/3 or it could also look like Z/2 + Z/2 + Z/3

the second statement will have G(2) = Z/2 + Z/2 and G(3) = Z/3

but we can't say that G(p) will be cyclic in general right

[long obsidian]

Ahh okay I see what you mean. Tbh at first I thought G(p) would be cyclic as I thought it ought to have it's order a power of p since each of its elements order should divide it's order. But maybe my reasoning isn't correct. Thank you

[rustic crown]

yep, you're right there. But that's not enough to say it's going to be cyclic >.<

it's order will be power of p

[cloud walrus]

Casey

Casey

[crimson canyon]

Let $t^a$ be a basis for a semisimple Lie algebra, with structure constants $f^{abc}$. I need to show the following identity:
$$f_{be}^{; ; ;k} ; t_a t_k + f_{ae}^{; ; ;k} ; t_k t_b = f^{di}{; ; ; j} f^{kj}{; ; ;i} ; f_{bek} ; {t_a, t_d} = 0$$
I'm kind of lost with all these indices running around. How do I start? It looks like I need to use the Jacobi identity somehow for the first equality, while for the second equality maybe something relating to symmetric/antisymmetric tensor contraction. But how?

[cloud walrus]

Siupa

[next obsidian]

If you have an extension of integral domains A < B, such that Frac(A) -> Frac(B) is a finite extension, is it true that Frac(A/p\capA) -> Frac(B/p) is finite for any p in Spec B?

We can assume more specifically that B is the integral closure of A in Frac(B)

[chilly ocean]

there is apparently some way to represent $PSL_2(\mathbb{Z})$ as some automorphism group of some graph. is anyone familiar with this?

[cloud walrus]

wren

[chilly ocean]

"orientation preserving isometries of a 3-regular tree"

I cannot make sense of how that would relate to PSL2Z

[upbeat juniper]

bump

[cyan marten]

Isn't there a general theorem that realizes groups as automorphism groups of graphs?

[chilly ocean]

not that I have seen

[cyan marten]

Also, maybe the isomorphism C2 * C3 ~ PSL(2, Z) helps

[cyan marten]

It works for finite graphs (Frucht's theorem)

[chilly ocean]

no way are you kidding I have not seen this

[cyan marten]

Actually it's true for all groups, just saw it in MO

[cyan marten]

Right?!

[chilly ocean]

ah well someone claimed it was an infinite 3-regular tree

[cyan marten]

I thought of something but it's probably wrong

Let X be a CW complex, and consider its universal cover E. Then we can think of elements of pi_1(X, x) as homeomorphisms of E, and these act on the graph (?) of the subcomplexes of E (E is really a CW complex)

The issue of course is that we really have an action on the lattice, and that's if we assume our homeomorphisms to be cellular maps, which I'm not sure they are

[teal mantle]

Would someone check this for me?

[broken stirrup]

Given a torsion-free finitely generated abelian group, can you construct or show that there is a free abelian group F of rank n such that G is subgroup of F?

[hidden haven]

Finitely generated abelian group would be a direct sum of finitely many cyclic groups, and torsion free implies that each summand is ℤ, so the group would itself be free

[hidden haven]

I think you get abc = ± 1?

[broken stirrup]

yes but what if I'm not allowed to use the fact that every finitely generated abelian group is direct sum of cyclic groups?

exactly lol

I'm working on Hungerford's Algebra and it's a problem from a section which comes before Finitely generated Abelian groups

[broken stirrup]

why do we need the condition a_k=1 for all k >= n ?

[rustic crown]

words are finite length

you don't want words to be bigg as we want to think of these as multiplying elements from X u X' u {1}

instead of working with tuples of varying length, they're working with infinite tuples which die eventually

[next obsidian]

You can see that the universal property also dies

As you’d be forced to be able to multiply infinitely many elements in arbitrary groups

Although that’s maybe too high level of an explanation unless you’re pretty comfortable with algebra

[broken stirrup]

thanks for both answers, now I see

[pastel cliff]

whats an example of a non commutative ring

[cursive temple]

ring of 3x3 real-valued matrices

[south patrol]

quaternions

[lavish nexus]

is there a formula for the number of elements not of the form $x+x^{-1}$ in $\mathbb{F}_p$

[cloud walrus]

Iteribus

[pastel cliff]

in this definition, would (a,b) be different than (b,a) in the set SxS

this is chapter 2 of artin

[delicate bloom]

Because we're in a field, for a fixed $y$, the equation $y=x+x^{-1}$ can be written as the quadratic $x^2-yx+1=0$, which has at most 2 solutions. Now instead of using the quadratic formula we can try to brute force what the solutions are by looking through values of $x$.

Clearly we can't use $x=0$ since $x^{-1}$ doesn't exist. When is $x=x^{-1}$? Exactly when $x=1$ or $x=-1$, so this gets us the two solutions $2$ and $p-2$. In all this means there are $p-3$ elements left to consider. Because the numbers left can be exactly partitioned into two sets of equal size $A,B$ where $x \in A$ means $x^{-1} \in B$, we see that both give the same $y$, and so together that means we have $\frac{p-3}{2}$ solutions contributed by these values.

In total this means there are $2+\frac{p-3}{2}=\frac{p+1}{2}$ values which are of the form $x+x^{-1}$. Since we want the complement of this, there are $p-\frac{p+1}{2}=\frac{p-1}{2}$ values that are not of the form $x+x^{-1}$.

[cloud walrus]

Merosity

[pastel cliff]

i know commutativity seems important but this is my first foray into this

[delicate bloom]

they can be different yes

unless there's more context afterwards that excludes it, you could think of S=integers and make the law of composition to be subtraction, then a-b and b-a are usually different

[pastel cliff]

goes on to talk about associativity

[delicate bloom]

ah, subtraction isn't associative

so not a good example, but matrix multiplication is a good one you should know

multiplication of 2x2 matrices over any field is always noncommutative

[pastel cliff]

i dont know exactly what this means just yet but am working on it

[delicate bloom]

gotcha, matrices are a good and very useful

[pastel cliff]

im like a day into trying to learn this stuff and it feels different than other math ive done somehow

[delicate bloom]

personally I learned linear algebra before abstract algebra, so I don't know

I would say knowing some LA first helped me a lot, but I don't think it's necessarily a prereq either

[pastel cliff]

i mean so did i

but the class was messy

prof promised a proof of cayley hamilton for like 3 weeks and never gave it to us

and spent too much time on raw matrix stuff

[chilly ocean]

I've seen some pretty gross proofs out there

[chilly ocean]

did you get a proof yet?

[long obsidian]

What is the point of ring localization?

[crude tiger]

okay this might seem like a stupid question but i am very tired

given some permutations on the set {1,2,3,4,5} how can i check that said permutations generate the symmetric group S_5?

[next obsidian]

I believe it suffices to show it has a transposition and a 5-cycle, and I think this uses the fact 5 is prime

In general, it’s kind of a hard question

[tropic bear]

HI 🙂 I'm seeing definitions of the tensor product of vector spaces that start with the free product of vector spaces, and quotient it out by some linear relations, namely by the ideal generated by $b_{a \cdot v, w} - b_{v, a w}$, $b_{a \cdot v, w} - a \cdot b_{v, w}$, and $b_{v, w} + b_{v', w} - b_{v + v', w}$, where I use $b_{x, y}$ to mean the basis vector in the free product of vector spaces $V$ and $W$, where $x$ and $y$ are any vectors in V and W, respectively, and $a$ is a scalar in the base field.

My question is: Instead of starting from the free product of vector spaces, could one start from the product? That is, can I just take the usual product of vector spaces (the categorical product, if one wishes to use that terminology), and quotient it by some relation, to get again the tensor product? Can I do it in the finite dimensional case at least?

[cloud walrus]

flebron

[tropic bear]

I also feel I lack some motivation for the tensor product. I know it as the one satisfying the universal property related to bilinear maps $f: V \times W \to U$, but why should one care about this?

[cloud walrus]

flebron

[final oasis]

It's a coproduct in category terms

[tropic bear]

Wouldn't the direct sum of vector spaces be the coproduct in the category of vector spaces?

(They match when in finite dimensional case, of course.)

I'm not sure what you mean by "it" in "It's" - the coproduct in general should be neither the product nor the tensor product, AFAIK.

[final oasis]

Yeah, I probably confused something

[lavish nexus]

f never mind 2 seconds after I sent it I realized why

[chilly ocean]

whats name of category where product is coproduct

i looked up but cant find on google

[frank fiber]

Opposite category?

[chilly ocean]

no

i mean where product and coproduct are same object

like direct sum of abelian groups or direct sum of modules

nlab parasites i swear

[proud bear]

abelian category?

or additive category

[chilly ocean]

but there is a name for type of category when product equals coproduct

i think additive might be it

[frank fiber]

Category with biproducts?

[chilly ocean]

ty

[frank fiber]

Additive is stronger than have biproducts

[tropic bear]

Any idea about the motivation for tensor products?

[chilly ocean]

yeah

[tropic bear]

i.e. why is it interesting to have that universal property about bilinear maps?

[chilly ocean]

it gives you a way to make multilinear maps into linear ones

which is sort of useful in context

[tropic bear]

I get why linear maps are interesting. Why are bilinear maps interesting?

[chilly ocean]

depends on context, but its type of function that has nice algebraic properties

ig for same reason linear maps are interesting

probably easier for representing things in general

[chilly ocean]

why do you think linear maps interesting

[tropic bear]

My background is CS, so the fact that they can be represented by matrices and computed with efficiently is interesting. Linear approximations appear everywhere, also.

In contrast I don't know many multilinear functions, other than the determinant.

I guess the dot product is bilinear, so multilinear, but I'm usually more interested in the arguments given to the dot product, rather than the dot product itself, as a function.

[sturdy marsh]

Tensor products let you do a lot of things. If I have a complex vector space, then I can consider the underlying real vector space. Tensor products allow you to do the opposite; i.e. if I have a real vector space V, then I can produce a complex vector space C \otimes_R V.

Another interesting thing is the tensor hom adjunction. If U,V,W are vector spaces, then Maps(U \otimes V, W) = Maps(U, Maps(V,W))

Multilinear forms are extremely important in geometry

they can be used to define distances, gradients, orientations, and a bunch of other stuff on manifolds

[tropic bear]

So if we call "consider the underlying real vector space" as an operation F, then F(C \otimes_R V) = V? The action of "C \otimes_R" is to complexify the coefficients of V, giving you a twice-as-many-dimensional vector space?

[sturdy marsh]

yup

the underlying real vector space of C \otimes_R V has twice the dimension of V

but the complex vector space C \otimes_R V has the same dimension as V

[tropic bear]

What's the general notion of a product such that Hom(X op Y, Z) is isomorphic to Maps(X, Maps(Y, Z)) in any given category?

[sturdy marsh]

if it exists, it's usually called a tensor product

[tropic bear]

Interesting. But that's the property that then defines the tensor product, and the bilinearity and other stuff happens to be true in the category of modules, but not much more than there, right?

[sturdy marsh]

sure

you probably still want to keep bilinear maps in mind as it's pretty much the only way to construct maps out of a tensor product

[tropic bear]

I'm trying to understand the notion of "tensor" that's used in CS, which is unfortunately "a bunch of numbers in some n-dimensional grid", like they'd call an element of \mathbb{R}^{4 x 5 x 8 x 1 x 2} a rank-5 tensor. I've tended to think this is because the operation on these things tends to be to contract along some dimension, and so we're going to have these "tensors" being contracted along matching dimensions, via dot products. Is there more to it than that?

[sturdy marsh]

idk what a tensor in CS is

[tropic bear]

It's used as a generalization of scalar, vector, matrix, .... .

[sturdy marsh]

That isnt enough for me to be able to say anything about it lol. Hopefully someone who knows about it will answer

[tropic bear]

Fair enough 🙂 Is your understanding that "a tensor" is "a multilinear map"?

Or is it "an element of a tensor product"?

[sturdy marsh]

element of a tensor product

[tropic bear]

(I guess the outer product gives us a way to go from "v \otimes w" to a matrix, when given a basis, no?)

[sturdy marsh]

tensors = multilinear maps is mostly a geometry thing

[tropic bear]

Why would they use that nomenclature?

(Thanks for answering my questions by the way!!)

[sturdy marsh]

it still makes sense

those maps can be described as sections of a tensor bundle

tensor bundle = some tensor product of tangent and cotangent spaces

[tropic bear]

I should've done more geometry in university ^_^'' I'm guessing that's going to be some multilinear map of V and V* for some vector space V, at each point in a space?

[sturdy marsh]

sure, that sounds about right

[sturdy marsh]

yeah you can get a matrix if you have a basis

but it's super gross

[tropic bear]

Thanks for the explanations 🙂

[sturdy marsh]

For example, the tensor product V \otimes V will have a basis of the form e_i \otimes e_j

where {e_i}_i is a basis for V

and then you know how each of those act on V (as V* = V using the basis)

so you get a matrix

[tropic bear]

Oh and the original question, I suppose, was:

The tensor product of vector spaces is usually defined as a quotient of the free product of two vector spaces. Can one take the standard direct product, quotient it by something, and get the same tensor product? Why do we need to start from such a huge space like the free product?

[sturdy marsh]

no a quotient of the direct product doesnt work

[tropic bear]

Is there an easy way of seeing why it's "too small"?

[sturdy marsh]

you could try showing that it doesnt satisfy the required property

as the universal property is the definition

or if you believe the tensor hom adjunction you can very easily see that the dimensions dont work out

take U = V = W = R

dim Maps(R \oplus R, R) = 2

but dim Maps (R, Maps(R, R)) = 1

so \oplus cant be the tensor product

did that make sense?

[tropic bear]

Yes, thinking about it

That certainly proves it, trying to interiorize that statement to mold my intuitions about it. Thanks again! 🙂

[sturdy marsh]

No worries. Tensor products are weird when you see them for the first time

[tropic bear]

That proves that the direct product, itself, is not the tensor product though

Because it's too big

But why can there not exist an ideal of the direct product, such that when quotiented by it, we get the tensor product?

[sturdy marsh]

as vector spaces are characterized by dimension, you could quotient out by something to get something isomorphic to the tensor product, but not naturally

so proving that there is no quotient of the direct product that works might be weird

but to motivate the weird construction, consider the case of extension of scalars again

[tropic bear]

Right, at least a dimension argument is too weak

[sturdy marsh]

I have some R-module M

and an R-algebra S

and I want to produce an S-module s.t. R sill acts the same way on M (really on however M sits in this new object)

hmm this might be a bad example

ill try to think of something

[tropic bear]

No problem/rush, I'm gonna do some hobbies now (lockpicking), will think about it more later. Thanks a ton 🙂

[sturdy marsh]

@tropic bear this isnt really a proof, but the only "natural" map from a module to a quotient is the projection map

so if the tensor product of M and N was some quotient of MxN

and if we have a bilinear map B: M x N -----> A

[cloud walrus]

Kanga Gang Consigliere Brtogi

[sturdy marsh]

then we want a map f s.t. the diagram commutes

[cloud walrus]

Kanga Gang Consigliere Brtogi

[sturdy marsh]

which has two problems

1. It may not be well defined as we quotiented out by something

2. It doesnt do anything

so unless you can produce some weird natural map between a module and its quotient, this doesn't give you anything

in fact, 3 problems as it isnt linear

f(2m, 2n) = B(2m, 2n) = 4f(m,n)

so solve this linearity issue, you dont want to allow (m, n) + (m,n) = (2m,2n)

which means that things shouldnt add componentwise

which could be motivation for the fact that you need something more free

so yeah the addition shouldnt come from some quotient of the addition on M x N

[tropic bear]

That's convincing, at least the addition is wrong 🙂

[chilly ocean]

big up

they are amazing constructions

if yoh want to understand why they are so useful i recommend drilling yourself on a bunch of questions

involving tensors

and look at what transformations look like between two tensor spaces

the multivector perspective is amazing too imo

[robust pollen]

I define sets [n] = {0, 1, 2, ..., n} for n in N, and ranging over n, these form a poset under inclusion. For n <= m we can consider the map f_nm : [m] -> [n], sending k to 0 if it's greater than n, or to k if k <= n.
These sets and maps form a directed system, and we can take its inverse limit. It will be a subset of prod_{n in N} [n], and some examples of elements are (0,0,0,....), (0,1,1,1,...), but not (0,0,1,2,...) or even (0,1,2,3,...) [so it's really a proper subset of the product]. Is there some nicer description of this set?

[hidden haven]

They will be sequences which start as 0,1,2,... and at possibly at some n become constant

[robust pollen]

0,1,2,... will not be in there though, since f_{1, 2}(2) = 0.

[hidden haven]

The first coordinate has to be 0. If the nth (indexing from 0) coordinate is n, then the next coordinate is either n or n+1. If it is n, then it must be n after that for all indices. If it is n+1, induct

Oh you map extra stuff to 0

[rustic crown]

won't it be (0, 0, ..., 0, n, n, n, ...)

[hidden haven]

So if at some stage it becomes non zero, it must be constantly that afterwards right?

[rustic crown]

where n is at the nth spot

[hidden haven]

So your examples seem wrong

[robust pollen]

fugg lel

too early

[hidden haven]

No excuses

🔫

[robust pollen]

😦

just stupid, I know.

[hidden haven]

[robust pollen]

But is it just N then?

[rustic crown]

yee

[robust pollen]

it = the limit

ok cööl

[hidden haven]

Alternate construction of ℕ

[robust pollen]

limit = colimit

[rustic crown]

lol

[hidden haven]

😵‍💫

wack

[robust pollen]

why is it called like that...

[hidden haven]

Why is what called like what

[robust pollen]

that like it

🌝

[hidden haven]

😵‍💫

[robust pollen]

why does "direct limit" have the name it has, if it's actually a colimit?

[hidden haven]

🥵

Because you take direct limit in the direction of the arrows

Probably

[next obsidian]

Because you take it over a directed set

[robust pollen]

weak sauce. filtered better.

[next obsidian]

I mean it’s better than calling it an injective limit

Or was it projective I don’t even remember

I’m pretty sure it’s injective

[hidden haven]

Direct is projective I think

Because you project onto it from the coproduct

[crude tiger]

thanks! so if the permutations i have give a product which gives the 5-cycle (1 5 4 2 3) and one of the permutations i have is (2 4) does this suffice to generate S_5?

[next obsidian]

I believe so, but this isn't trivial to prove

It's easy to see that S_n is generated by transpositions, and IIRC you use your transposition (24) and then cleverly multiply powers of your 5-cycle to show all of the transpositions exist

Somewhere in this you had to use that 5 is prime

[crude tiger]

okay awesome thank you very much! :D

ill try fiddling around with the thing on mathematica

[next obsidian]

Oh ack, this might not be true

I'm looking at what I thought was the thing which had this as a problem and it was that a transitive subgroup which contains a transposition is all of S_n

I feel like I remember this thing being true though

oh I'm saved

I googled it and it's true

I was just looking at the wrong thing I gues lol

[crude tiger]

to be entirely fair i do remember seeing a thing which was something along the lines of the generators of Sn being (1 2 3 ... n) and (1 k) for k n relatively prime or something like that

i just wasnt sure if this was applicable to other cycles

[next obsidian]

I think this probably is proved the same way or something like that

[crude tiger]

true

to be honest im mostly using this to check for something else entirely that is dissertation-related

im up to my eyes in ramsey theory at the moment

[crude tiger]

now that looks horrifying and oddly interesting

[next obsidian]

there's not really any content

[broken stirrup]

so I want to use this theorem on a proof but number of normal subgroups is less than number of groups, do we have a result for that case?

[next obsidian]

it's just writing out that G x H/(N1 x N2) = G/N1 x H/N2 but non-finitely

[next obsidian]

If some of the groups G_i don't have an N_i, just take N_i = {e}

you can just quotient by trivial groups

and it's the same as having nothing

[crude tiger]

fair enough but tbf i have never seen the big product notation on groups hence my comment hahah

[broken stirrup]

hmm let me see if that helps in my case

[hidden haven]

You might have to take N_i = G_i in those cases, check which choice makes it work

[iron vessel]

I am having trouble proving that f(x)=2x^4+5x-10 is irreducible in Z[x]. Is there a quick way other than just doing the product of 2 degree 2 polys?

[broken stirrup]

have you tried Eistenstein criterion?

[iron vessel]

How can you apply it here? the GCD of the coefficients is 1

nvm

I tried reducing modulo 3 and doing eisenstein with p=2, but 2 is not a prime element in Z/3Z

[broken stirrup]

oh you are right it says irreducible in Z[x]

didn't see

[iron vessel]

Anyone has any ideas?

[waxen hedge]

if it's irreducible in Q[X] you're done, since the GCD of the coefficients is 1

(irreducibles of Z[X] are irreducibles of Z and elements of Z[X] with GCD of the coefficients = 1 and irreducible in Q[X])

[iron vessel]

Yeah but to prove irreducibility in Q[x] i need to do the rational root theorem?

[waxen hedge]

what's the rational root theorem ?

there is a lot of way to show that a polynomial is irreducible in Q[X]

[iron vessel]

r/s is a root iff r|constant term and s|leading coefficient

How would you show this is irreducible in Q[x]?

[waxen hedge]

well, if P in Q[X] is irreducible then P has no root in Q, but the converse is not true

[iron vessel]

I understand, but I am having trouble also showing it is irreducible in Q[x], how should I attempt it?

[waxen hedge]

I don't know yet haha

[iron vessel]

Because showing it has no root is not enough right? because it is of degree 4

[waxen hedge]

yes

if the degree was <= 3 this would be enough

[iron vessel]

indeed

So I should just try doing the product of two degree 2 polynomials and reach a contradiction?

[waxen hedge]

in general this is not a good way

[iron vessel]

because it is long? or is there a requirement that is in general not met?

[waxen hedge]

the polynomial system you get is very hard to solve

[iron vessel]

Yeah that was my issue, I'm stuck at that.

[delicate bloom]

I'd bite the bullet and go for it

$(2x^2+ax+b)(x^2+cx+d)=2x^4+5x-10$

[cloud walrus]

Merosity

[delicate bloom]

maybe start looking at the coefficients on x^3 and x^2 cause they're zero, maybe it has some simple path that's not so awful

[iron vessel]

Ok, ill try again

[delicate bloom]

I'll give it a shot in about 15 min or so if you're still at it, just busy atm

[iron vessel]

That's awesome, if I do manage I will tell you. Thanks

[crimson canyon]

Is there a better channel where I can ask this question? I didn't find a specific "Group Theory" channel

[waxen hedge]

maybe #diff-geo-diff-top ?

[iron vessel]

One question, when choosing the leading coefficients for the two polynomials, couldn't it also be -2 and -1?

or is that just a matter of being associate?

[waxen hedge]

yes but up to an unit it's not a problem i'd say

since 1/-1 are units

[iron vessel]

ok great thats clear

[delicate bloom]

I got a pretty clean proof, I'll explain it in a bit

[waxen hedge]

okay, if you expand this you find
bd = -10
ad+cb=5
b+2d=0
b+2d+ac=0
2c+a=0

so b=-2d and bd = -2d^2, so -2d^2=-10 ==> d^2=5 which is absurd

but there might be a calculation mistake

[delicate bloom]

nice that's similar to what I did, just ended up with a different contradiction

[iron vessel]

that is very nice

thanks alot guys

[waxen hedge]

you still have to show that you cant have a factorisation as a product of a degree 1 polynomial and a degree 3 polynomial

[delicate bloom]

that should follow from the rational root theorem

[waxen hedge]

but it suffices to show that your polynomial has no roots

yeah

[delicate bloom]

which I thought you already did

[iron vessel]

yeah i did do that

awesome, very helpful

[waxen hedge]

np !

[delicate bloom]

wait is this correct

I got ac+b+2d=0 for the x^2 coefficient

[waxen hedge]

I may have forgotten some

you are totally right

[iron vessel]

I didn't notice that

would you be able to send a picture of your proof @delicate bloom ?

[delicate bloom]

I'll just write it up one sec

[iron vessel]

🙂

[delicate bloom]

bd = -10
ad+cb=5
b+2d+ac=0
2c+a=0
take the last equation plug into everything,

bd = -10
-2cd+cb=5
b+2d-2c^2=0

last equation implies b is even, so put b=2k and plug it into the second equation,

-2cd+c2k=5
2(-cd+ck)=5
clearly 5 is not divisible by 2, contradiction

[iron vessel]

Let me write it so I understand it better

Wow that b is even trick is pretty insane

I wouldn't have seen it

[delicate bloom]

yeah, I was trying to avoid breaking c(b-2d)=5 into the 4 cases for c=+1 and c=+-5 lol

[iron vessel]

Yeah thats a fair amount of work avoided

[wind parrot]

Maybe I've missed something, but can you not just use Eisenstein with p=5?

[iron vessel]

5 doesnt divide 2

[delicate bloom]

yeah that works lol

[iron vessel]

oh my

[delicate bloom]

I think

[iron vessel]

i just realised it has to divide all except for the leading one

[wind parrot]

lmao

[iron vessel]

damn

[delicate bloom]

lol epic

[iron vessel]

lmao ive been on this question for like an hour

[delicate bloom]

I was too distracted I didn't even bother to check ahaha

oh well we still got it hah

[iron vessel]

thanks alot @delicate bloom @waxen hedge @wind parrot

[waxen hedge]

yes you're right
I didn't think of it since the polynomial wasnt monic

[pastel cliff]

i looked it up yeah lol, class ended im on break

[crimson canyon]

Thank you, I'll try

[broken stirrup]

How can i prove that H is not finitely generated

[woven delta]

Take a finite number of elements of H and show it doesn't generate H

The reason why it isn't finitely generated is because of powers of 2 in the denominator

You can identify H with the group {m/2^{-n}|m in Z, n in N}

Under addition

Fixing a finite number of elements of the group puts an upper bound on the powers of 2 in the denominators of the elements of the subgroup generated by those elements

@broken stirrup

[broken stirrup]

I see how your idea must disprove that it's not finitely generated but how it bounds powers of 2 in the denominator? can't you just take 2^{-1}a and get any power of 2?

[woven delta]

Well the subgroup is under addition

Try computing matrix multiplication for elements of H and see how it's just addition

[broken stirrup]

btw group operation is matrix multiplication

[woven delta]

Yes I know

[broken stirrup]

but yeah when you multiply you just add the powers

okay I'll think about it

[woven delta]

👍

[broken stirrup]

oh yeah

i see

[pastel cliff]

does a ring not need a multiplicative inverse?

[hidden haven]

No

[pastel cliff]

thank you moldilocks very cool

[hidden haven]

[next obsidian]

If everything has a multiplicative inverse (other than 0) we call it a field

It’s so special it gets its own name

[pastel cliff]

algebra is funky

if something is a field does it stop being a ring? or is it both a ring and a field? is this even worth asking or is it kinda trivial?

[frail zealot]

it doesn't stop being a ring just like the group in the ring doesn't stop being a group

what you're asking is kinda trivial yeah

cause it's literally in the definition

and it's not really funky because there are very good reasons for separating these things out into different names

[pastel cliff]

i havent gotten to the reasons

it just seems funky while i learn what they actually are

[delicate bloom]

what parts of math are you interested in? maybe we can say some examples of stuff to help motivate them a bit better for you

[broken stirrup]

Is it enough to show that F(x)/ Ker f is isomorphic to G?

[waxen hedge]

I would say yes, if you explain what's the kernel here

[broken stirrup]

<a_i a_j a_i^-1 a_j^-1> right?

[waxen hedge]

yes !

one inclusion is trivial, and the other is not really hard as well

[next obsidian]

Hehehe

There’s a slick proof of this showing they both represent the same functor

Although “slick” is maybe not the best word here

[broken stirrup]

yep i already showed <a_i...> is subset of kernel

[raw moth]

30

oops

[south karma]

Can anyone explain the proof of fundamental theorem of finitely generated abelian group?

[sturdy marsh]

I don't remember the proof, but the idea is to reduce the problem to classifying finite gen free abelian groups and fin gen torsion abelian groups

Let A be a fin gen abelian group. Let T \subset A be the subgroup of all torsion elements

then we have a short exact sequence 0 ---> T --> A ---> A/T --->0

A/T is torsion-free, and we want to show that A = T \oplus A/T

if we can do this, then this would reduce the problem to classifying torsion-free things and torsion things

[sturdy marsh]

to prove this, it is sufficient to show that A/T is free

so it is sufficient to prove that fin gen torsion-free implies free

the idea is to stick it inside a free thing, and prove hat sub things of free things are free

okay, so let B be a finitely generated torsion-free abelian group

okay let's first prove that sub things of Z^n are free

let K \subset Z^n be a subgroup

Let K' = Z^{n-1} \cap K

where Z^{n-1} sits in Z^n via (a_1, ..., a_{n-1}) ---> (a_1, ..., a_{n-1}, 0)

by induction, K' is free

[cloud walrus]

Kanga Gang Consigliere Brtogi

[sturdy marsh]

K/K' is also free by induction (the base case)

so K = K' + K/K'

so K is free

so sub things of Z^n are free

okay now let B be a finite gen torsion-free abelian group

let b_1, ..., b_n be a set of generators

let x_1, ..., x_k be a maximal linearly independent subset of {b_1, ..., b_n}

by maximality, for any generator b not in {x_1, ..., x_k}, we have a nontrivial linear combination

nb + a_1 x_1 + ... + a_k x_k =0

n cannot be zero as this would contradict linear independence

so for each generator b not in {x_1, ..., x_k}, there is a nonzero integer n_b s.t. n_b \cdot b is contained in the free subgroup generated by {x_1, ..., x_k}

let n be the product of all such n_b

we have an injective homomorphism B ----> B given by s \mapsto ns

it's injective as B is torsion-free

the image is contained in the subgroup generated by {x_1, ..., x_k}, which is free

we just proved that subgroups of free groups are free, so B is free

therefore torsion-free implies free

which implies that any finiteley generated abelian group is the direct sum of a free group and a torsion abelian group

the finitely generated free abelian groups are just things of the form Z^n

so classifying finitely generated abelian groups reduces to classifying finite abelian groups (f.g. + torsion implies finite)

now use the sylow theorems and show that any finite abelian group is a direct product of its sylow subgroups

(all sylows are normal and intersect trivially)

therefore the classification reduces to classifying all finite abelian groups of order p^n, which is easy

[south karma]

I tried to find the proof, but there was no proof anywhere. Thank you so much for helping me like this!

[sturdy marsh]

no worries, trying to remember the proof was fun

what I wrote inst a complete proof, but all of the main ingredients are in there

[south karma]

Yes, and I'm going to try the details.

[broken stirrup]

Kanga wait weren't you a monke? what happened?

[sturdy marsh]

that's @next obsidian

[broken stirrup]

oh lol sorry

[wild sapphire]

I have a question about notation. In quotient ring R/Q, one element is said to be like "a + Q" for example. And in a lot of proofs I see "a + Q = Q", like so:

Why is this? I feel like the notation is just tripping me up a bit

Can that always be stated? Like any element "a + Q = Q" because it's like saying "a plus anything in Q is the same as just saying Q because a is in Q"?

it's used twice in that little proof

[wild sapphire]

<@&286206848099549185>

[tropic spade]

a+Q=Q is equivalent to a being an element of Q I thought?

[wild sapphire]

that could be it? I just can't find sources directly saying that

[tropic spade]

I'll double check real quick to be sure.

Yeah it's a property of cosets for groups that should just carry over for rings.

It won't be true for every a you pick from A of course.

Just the ones actually in Q.

[wild sapphire]

hmm okay, that's why I said it was a notation issue. So the reasoning is like "a + Q represents a + anything in Q, so if a is in Q then a + anything in Q is just Q by closure of +"

[rustic crown]

yep, that's right

like consider a simpler example

Z/5Z

you usually think of this as {0, 1, 2, 3, 4} with +,* modulo 5

but to actually define this in general and save yourself checking the ring axioms for each one, you have a general quotient construction

we can think of 0 as all the elements that are 0 mod 5

[0] = {..., -5, 0, 5, ...}

and similarly

[1] = {..., -4, 1, 6, ...}

and so on

so this a = b mod 5 is same as saying [a] = [b]

where [a] = a + 5Z

translating all multiples of 5 by a

we can carry out the operations on these classes

and this will form a ring, when [0] turns out to be a 2-sided ideal of the ring

is that what you were looking for?

[tropic spade]

Now that I think about it the proof of it in general for R/H should follow by the fact that ur cosets partition ur ring, so if a is in H, since a is already in a+H then by partition properties I think the equiv follows? (And yada yada do something similar for the other way)

[wild sapphire]

ah yeah! @rustic crown That definitely makes more sense with congruences and equivalence classes. Thank you for typing all that out. I think I was mixing up general ideals with quotient rings. So elements being in the ideal we're factoring by is analogous to congruences and mod.

I feel like I already knew that in some way, but it helps a lot to have at least one concrete example :)

Yeah @tropic spade, everything kinda falls into place after that

[iron vessel]

How would I do c?

I think it is just x^2 -21, as it is monic and irreducible and a^3 is a root of it

Also it makes sense that it is of degree 2

[hidden haven]

You will have to show that it is irreducible over ℚ(b), not just over ℚ

Consider the tower of extensions ℚ ⊂ ℚ(b) ⊂ ℚ(a)

And compute degrees

[hidden haven]

The minimal polynomial of a over ℚ(b) is not that