groups-rings-fields

[thorn jay]

Mhm

[golden turtle]

Linear independence

Stuff

Rank nullity

But it doesn’t apply

Or atleast I didn’t see how

If that’s all there is too it I will try more tomorrow, 3am for me now 😭

[thorn jay]

Oh i have something i think

Hm

[rocky cloak]

Well if you think in terms of matrices, since matrix multiplication is just multiplying in each column. Being surjective implies having a right inverse. From there you can use determinants again

[thorn jay]

It has a set-theoretic right-inverse, yes

So i suppose we'd need to prove it also has a matrix right-inverse?

[rocky cloak]

No, it has a matrix right inverse

You just pick the columns as preimages of the standard basis

[thorn jay]

Oh

Man

That's smart

I should've seen that >.>

Right and because you have a commutative ring then the determinants must be units

[earnest hare]

is it just the left divided by the right number

[finite turtle]

is there a name given to the property (ab)^-1 = b^-1 a^-1 generally speaking ?

[thorn jay]

I usually refer to it as the socks and shoes property lol

[thorn jay]

Not really, GF(3^6) is a vector space over GF(3^3)
=> GF(3^6) is isomorphic to GF(3^3)^k as vector spaces where k is the dimension
=> 3^6 = (3^3)^k = 3^3k
=> k = 6/3 = 2
Hence [GF(3^6) : GF(3^3)] = 2

In general you have to divide the exponents

[earnest hare]

the exponents are like the number of degrees of freedom? number of free variables when we write the extension field?

over a base field of Z_p

[thorn jay]

It's the degree of the minimal polynomial corresponding to the field extension

Assuming it is algebraic

[earnest hare]

[GF(2^7)/Z_2] = 6?

[thorn jay]

7

for finite fields
GF(p^k) = Z_p[x] / p(x)
Were p is an irreducible polynomial of degree k

[earnest hare]

[GF(2)/Z_2]

isnt this just 1?

[thorn jay]

Yes

[earnest hare]

like isnt GF(2) just Z2?

[coral spindle]

GF(2) = Z_2

[thorn jay]

So for example
GF(p) = Z_p[x]/x ~ Z_p

~ here denotes isomorphism

The exponent k in a field F of order p^k is the dimension of F when considered as vector space over Z/pZ

[earnest hare]

GF(2^7) = Z_2[x]/p(x), where p has deg 7 and is irreducible over Z_2[x]?

[thorn jay]

Right!

Z_2

Not Z_7

Yes

[earnest hare]

why 7 then

[thorn jay]

Because the exponent is 7

A field of order 2^7 is a 7-dimensional vector space over Z_2, first of all just by looking at the cardinalities

[earnest hare]

any element in GF(2^7) would be of the form,
a0+a1 x +a2 x^2+a3 x^3+a4 x^4+a5 x^5+a6 x^6

and we have 7 cosets of Z_2

[thorn jay]

Yes

[earnest hare]

Z_2
x Z_2
x^2 Z_2
x^3 Z_2
x^4 Z_2
x^5 Z_2
x^6 Z_2

thank you R module

[thorn jay]

Which are all linearly independent with respect to multiplication by elements in Z_2 and addition

[earnest hare]

and boyt

[thorn jay]

Call me Aku :3

[earnest hare]

thanks aku

[thorn jay]

Ofc!

[finite turtle]

how about backward associative? thats not taken i believe

[thorn jay]

Contravariant it's called

It's a contravariant property

Because it reverses the direction of composition

You also have something called the opposite group, which is defined by
g •' h = h • g

And the map x -> x^-1 is an isomorphism from a group to it's opposite group

I suppose this generalises to the opposite category when you see a group as a category with one object

[serene dune]

by definition is 0 a nilpotent element ?

also can someone provide the most reliable dfinition

[cloud lynx]

if I have the zero-set V(XY) subset of A^2(R)

then the zero set is just the y and x achsis right?

because XY=0 => X=0 or Y=0

[thorn jay]

Yes

[rocky cloak]

Some people include "nonzero" in their definition of nilpotent, simply for the convenience of saying "a ring without nilpotent elements" as opposed to "a ring without nonzero nilpotent elements"

[thorn jay]

Well the nilpotent elements to form an additive subgroup if you include zero

[rocky cloak]

If your ring is commutative at least

But yeah, I myself would go with 0 being nilpotent

[thorn jay]

Right >.> i forgot stuff like matrix rings existed

[earnest hare]

why is the Frobenius map an automorphism?

in a finite group

[thorn jay]

What is the frobenius map?

Ive only heard it in the context of fields

[earnest hare]

they defined it as such

[thorn jay]

Oh thays a field automorphism

[earnest hare]

yea why

[thorn jay]

Not finite groups

Thats why i was confused

Lol

[earnest hare]

isnt it finite

[thorn jay]

It isnt a group

[earnest hare]

oh

[thorn jay]

Irs a field

Its*

[earnest hare]

well a field is a group

[thorn jay]

A field has an underlying additive group structure and induced a multiplicative group which interact in a certain way

It isnt a group automorphism as it must respect both operations

[earnest hare]

ok back to the quest

[thorn jay]

Well due to the commutativity of the multiplication the map clearly respects multiplication

(xy)^p = x^p y^p

[earnest hare]

yea

wait how are we doing this

i thought we had to show it is a bijection

[thorn jay]

First we need to show it respects the field operations

[earnest hare]

an automorphism has to respect the operations too?

the structure

[thorn jay]

Obviously

[earnest hare]

not obviously but thanks

[thorn jay]

Oh well, an automorphism is defined as an isomorphism to itself

And isomorphisms are a type of homomorphism, which must respect operations

[earnest hare]

(x+y)^p = x^p + y^p + other stuff that can be written as something * p or 0

[thorn jay]

Yup

Freshman's dream

[earnest hare]

my dream 🥺

[thorn jay]

So it respects operations, great!

[earnest hare]

and the other properties we neednt check as its an endomorphism?

[thorn jay]

Other properties?

Like?

[earnest hare]

its closed, has a 0 and 1 etc

associativity

[thorn jay]

Has 1 does need to be checked

But the rest follows from properties of homomorphisms

[earnest hare]

hm i see

[thorn jay]

No wait not even

[earnest hare]

well anyhow it is tedious but boring to check

lets assume its a homomorphism

[thorn jay]

We dont need to prove that

We have that it is a homomorphism, and now we only need to prove it is bijective

[earnest hare]

ok good

[thorn jay]

Proving injectivity will suffice

[earnest hare]

due to it being finite yea

likewise proving surjectivity would suffice?

[thorn jay]

Hmm

1^p = 1 so 1 is not in the kernel

But then the kernel has to be 0 as fields cannot have a nontrivial ideal

[glad osprey]

associativity only makes sense for binary operations, and you can't talk about an operation being closed (you can talk about a set being closed under an operation however)

[thorn jay]

So the map is injective, hence bijective

Thus it is an automorphism

[earnest hare]

i guess i meant, the map being closed under the field operations?

[thorn jay]

Wdym by that

You're not going to produce an element outside the field with that map

[earnest hare]

when i said operations being closed i meant, that the field operations are closed when used on the image of some map

yea

[thorn jay]

But fields are already closed?

[earnest hare]

oh :0

[thorn jay]

By definition?

[earnest hare]

i forgot

[thorn jay]

You'd have a pretty useless structure if you could just produce elements outside of it using the operations lol

[earnest hare]

oh thats fast lol

[earnest hare]

would this likewise suffice?

[thorn jay]

Proving surjectivity is annoying

In this case at least

[earnest hare]

yea but its an equivalent condition?

[thorn jay]

As you are working with finite sets, yes

[earnest hare]

yep thank you aku

and shed

[thorn jay]

Yee

[serene dune]

alr, also its apparently doesn't make a difference in how things are going to behave.

[spark veldt]

Hi

I tried writing a proof for this but the yellow part seems dubious to me (I'm trying to show first that HK is a subset of KH in my screenshot).

Is there an extra step I have to show?

[thorn jay]

I dont see the implication?

suppose H is normal and K and H only share an identity
Then HK is a subgroup, and k^-1 h^-1 in HK but that doesnt mean k^-1 in H and h^-1 in K

[spark veldt]

yeah i see that now

i've been trying to prove this direction (HK subset of KH) but i feel like i'm going in circles

[thorn jay]

HK is a subgroup
(hk)^-1 = k^-1 h^-1 in KH
Thus hk in HK => (hk)^-1 in KH
But hk in HK => (hk)^-1 in HK
therefore hk in HK => hk in KH

Thus HK subset of KH

Other direction is simple:
h, k in HK => kh in Hk
Thus Kh subset of HK

[elfin wraith]

Anyone able to give me a push in the right direction here? I’m considering a certain ore extension given by a derivation on a K algebra, R, and I’m struggling to get a grip on the general form

I’ve calculated what we get for the product of 2 terms and I can’t quite see what the relation is, but for 3 terms this sum turns out to be pretty huge, is there anything I could do other than going through the gruelling computation including r_3 and hoping I see a pattern?

I think my last line is correct, and is really all that I need for what I’m trying to do, but I’m not sure how to fully justify that

I mean I guess it’s actually obvious enough that that’s the form, but it’d still be nice if I could work out a general formal if there is one

[crystal vale]

I have to show that if a^m = b^m and a^n = b^n for m, n relative prime positive integers, and a and b in a commutative domain then a = b.

If a = 0 then b = 0 implies a = b.

So let a≠0, b≠0. Then there exists an injective ring homomorphism between R and its fraction field Q.

So let phi: R -> Q, such that phi is an injective ring homomorphism.
It gives us (phi(a))^j = (phi(b))^j, j = m,n.

Then since it is injective therefore phi(a) and phi(b) is unit element and m and n are positive relative prime integers therefore ms + nt = 1.

Now it gives us phi(a) = phi(b), here we use the inverse of phi(a) and phi(b), so by injectivity a = b.

Is it correct?

[tardy hedge]

Do maximal ideals correspond to maximal ideals?

For p

Lattice theorem for ideals

[rocky cloak]

Maximal ideals in f(A) correspond to maximal ideals in A that contain kerf

[thorn jay]

This is a standard result in abstract algebra

[thorn jay]

Yeah

[rocky cloak]

So yes. The lattice of ideals in f(A) and the lattice of ideals containing kerf are the same. So in particular they have the same maximal elements

[thorn jay]

If you have the lattice of ideals of A then [ker f, A] is isomorphic to the lattice of ideals of f(A)

[tardy hedge]

Ok thanks. Im not sure why they only mentioned prime corresponds to prime

[thorn jay]

[ker f, A] is all the ideals containing ker f and contained in A; all the ideals of A containing ker f

[rocky cloak]

Probably because they're taking about how primes behave with respect to extension/contraction

[thorn jay]

Maybe because maximal ideals arent relevant here

[thorn jay]

That, yes

[earnest hare]

im having some trouble understanding this, why it has to be injective and why there are d isomorphisms. does where one zero is sent to uniquely determine where the other zeroes are sent?

http://abstract.ups.edu/aata/galois-section-field-automorphisms.html

hello there jag

[rocky cloak]

Any homomorphism of fields is injective, but it's not particularly relevant.

As for the isomorphism, you have an isomorphism from F(alpha) to F(beta_j) just mapping alpha to beta_j.

Any homomorphism from F(alpha) is determined by where F and alpha are mapped

[earnest hare]

F is fixed so its just determined by where alpha is mapped no?

[rocky cloak]

Exactly

[earnest hare]

i get that there are d choices but cant the other zeroes be mapped to different zeroes perhaps?

[rocky cloak]

What other zeros?

[earnest hare]

the other d-1 zeros

they are permutated as well

[rocky cloak]

Like if F(alpha) contains other roots as well?

[earnest hare]

cant they be permutated to other zeros

[rocky cloak]

Sure, they will be mapped somewhere else yeah

[thorn jay]

But F(alpha) is generated by alpha and F, and i believe that it must fix F, so the homomorphism is fully determined by where it sends alpha, no?

[rocky cloak]

Indeed it is

[earnest hare]

like given alpha is sent to gamma, arent there 2 possibilities of where gamma and beta are sent

[tardy hedge]

Im taking galois theory next term

Im looking forward to learning it

[earnest hare]

yay gl

[tardy hedge]

Ty

[thorn jay]

So there are d choices for alpha to go to, hence d homomorphisms

[thorn jay]

I should look into it too

My friend recommended me a book about it

[tardy hedge]

Nice

I think im just gonna use dummit and foote

[earnest hare]

oh.. yea true

[thorn jay]

He complained it took too long to get to the unsolvability of the quintic

[tardy hedge]

Lol

[thorn jay]

Smh, Chrizzle bizzle gizzle dizzle

[rocky cloak]

Well if beta and gamma are in F(alpha) then they are determined by where alpha is mapped, since everything is determined by where alpha is mapped

[earnest hare]

is there a way to put it more rigoursly

the fact that where alpha is sent determines F(alpha)

[thorn jay]

F(alpha) is generated by F U {alpha}, and as F is fixed and there are d choices for alpha, there must be d homomorphisms

[rocky cloak]

Well every element in F(alpha) is of the form
p(alpha)/q(alpha) where p and q are polynomials and q(alpha) is nonzero.

Any homomorphism sigma that fixes F will have sigma(p(x)) = p(sigma(x))

[tardy hedge]

Prop 1.17 is really just due to how images and inverse images behave right?

Not necessarily about the algebra

[rocky cloak]

So if for example beta = alpha^3 or whatever, then sigma(beta) = sigma(alpha)^3

[tardy hedge]

At least for 1 and 2 im looking at

[thorn jay]

What are the superscripts?

[earnest hare]

sorry what

[earnest hare]

doesnt this only tell us that sigma will map zeroes to zeroes

[rocky cloak]

You understand that a homomorphism satisfies sigma(x^n) = sigma(x)^n ?

[earnest hare]

yea

[hidden wind]

extension and contraction

[thorn jay]

Oh, thanks

[rocky cloak]

So then you understand that for example
sigma(alpha^3) = sigma(alpha)^3 is completely determined by the value of sigma(alpha)

[earnest hare]

yes but

can all zeroes be written in the form of \alpha^n?

[thorn jay]

Yeah

[rocky cloak]

.

[thorn jay]

Not really algebraic questions lol-

[rocky cloak]

Every element in F(alpha) can be described in terms of F and alpha.

Whether they are zeros of something or not is completely irrelevant

That's like the whole deal with F(alpha)

[earnest hare]

ohh

yea

vector space with terms in F and basis of \alpha?

[rocky cloak]

If you prefer to think in terms of universal property:

Imagine you have two homomorphisms f,g:F(alpha) -> ?

Then you can consider the subset of x's such that f(x) = g(x)

This is a subfield, so if it contains both F and alpha it is everything, so f=g

Hence f is uniquely determined by f(alpha)

[earnest hare]

thanks jag and aku

[earnest hare]

each extension has exactly [E:F]/d extensions... how did they get this via induction and what does this mean?

hiii jagg

[rocky cloak]

So you want to prove that the map F -> E extends to a map E -> E in [E:F] ways. You do this by induction, if [E:F]=1, E=F and all is clear.

You can extend F -> E in d ways to F(alpha) -> E.

Now E/F(alpha) is an extension of degree [E:F]/d. So by induction the map F(alpha) -> E extends in [E:F]/d ways

[earnest hare]

what does extends mean? or extensions?

the only time ive seen it is field extenions, is it field extension here?

[rocky cloak]

No just make bigger, so you want a map E -> E that when restricted to F gives you F -> E

So the same map, but you've made the domain bigger

[earnest hare]

so.. how many ways that bigger map can be made

[rocky cloak]

Yeah, that's right

Like the name jagg btw. Torn between whether to pronounce it like jagg-ed or yak-ox in my head though

[earnest hare]

is the last line necessary?

sigma restricted to F(alpha)...

[rocky cloak]

That's the converse. Like to make sure we haven't missed any possible maps

[earnest hare]

could u elaborate

i fail to see how it is the converse (or if it even means anything)

[rocky cloak]

Like we constructed inductively [E:F] maps, so now we just need to make sure there aren't even more.

[earnest hare]

uniqueness?

oh u meant quantity

[rocky cloak]

But luckily given any map E -> E fixing F, then it does restrict to one of those d maps F(alpha) -> E, so again by induction we already covered it

[earnest hare]

hm

thank you ox

[tardy hedge]

someone showed something is a group hom just be showing it maps identity to identity and inverses to inverses

thats not sufficient right ?

[thorn jay]

But it is an antihomomorphism

But in general i do not see why that would imply it to preserve the structure in any way

In fact, take the group (Z, +) and define the map
f := n -> (-1)^n * n
This sends inverses to inverses and 0 to 0 but evidently not a homomorphism

[crystal vale]

Prove that the elements (a,1) and (1,b) of A × B commute and deduce that the order of (a,b) is the least common multiple of | a | and | b |.

I proved both statements, but I am curious how the first statement helps me to prove the second statement?

[sharp ice]

if x and y commute then (xy)^n = x^n y^n and so its order has to be when x and y vanish at the same time, which is the lcm of the orders of x and y

[crystal vale]

but this is not always true because take x = y^-1

[sharp ice]

*a divisor of the lcm, mb
but those kind of situations dont happen when x = (a,1) and y = (1,b)

[crystal vale]

i have to show that x not in <y>, right?

[latent lodge]

hi guys i've noticed something and idk if it can amount to something or not; anyway, we know that the generators of a cyclic group are the guys with an order of n , or more exactly the x^k with k and n coprime (and we can describe the elements of this group as the roots of X^n-1 when talking about a multiplicative sub group of a field), their number is phi(n) right; on the other side we have cyclotomic polynomials that have a degree of phi(n) for the polynomial associated to n , their roots are the n-eme primitive roots of the unit (so basically the generator of the cyclic group we talked about before) and you can decompose X^n-1 using them the same way you decompose n usind phi(d) where d is a divisor of n

WHY IS THAT?

Why do they have so much things in common? how are they related? how do these two concepts mesh together ? like wth is goig on there?

[grim hawk]

If I want to prove that the product of rotations about x,y,z axes can produce any elemetn in SO(3) how should I got about it? I have found the matrix product of the representation with angles and have tried to compare it to the general form of a matrix in SO(3) (difficulty as I cannot find a form that is similar to how we do SU(2)). Any advice? am i drastically overthinking this?

One avenue I am attempting is to show that the column space is the entirety of the 3 sphere, but I am unsure how to do that in this case

thanks but thats not gonna help me figure out how to approach these types of situations for my studies.

[wraith cargo]

<@&268886789983436800> @chilly ocean is being very disruptive lol

[tardy hedge]

Dramaaa!

[chilly radish]

If you don't know the answer/have an idea you don't have to try helping

Giving vague general statements is not what this channel is fot

[tawny pine]

their history is nothing but being a shitter in adv math channels, gone

[tardy hedge]

Yikes

[inner steppe]

I don't even know if this belong here but

What is this theorem saying: If the code C, of length 𝑛 n, corrects up to e errors, then the following holds:

[golden turtle]

Yes so I've found the right inverse now

I don't know how this gives me what I need about the determinant

maybe it is from commutativity

This is what I've done

[thorn jay]

Let N be the right inverse of M
MN = I
detMN = detI = 1
detMdetN = 1
detNdetM = 1
Hence detN = (detM)^-1

[golden turtle]

feels like cheating

[thorn jay]

Is it?

[golden turtle]

not sure

[thorn jay]

Im pretty sure you're allowed to use the multiplicative property of determinants

[golden turtle]

oh and is this where comutativity is used

[thorn jay]

Mhm

[golden turtle]

yes

So this argument gives T surjective => det M is a unit in R

[thorn jay]

Mhm

[golden turtle]

if det M is a unit in R then => M is invertible

is it just clear that M invertible => T invertible?

[thorn jay]

Have we proven that?

[golden turtle]

I have

[thorn jay]

When

[golden turtle]

I can show

[thorn jay]

Oh right then we're done

[thorn jay]

They're the same linear transformation

It's just that M represents T as a matrix

[golden turtle]

Yes

We can write

Madj(M)=detMI

and this is a diagonal matrix working out the actual terms in summand will give you adj(M)M=detMI

by commutativity

[thorn jay]

Ohh

[golden turtle]

so if detM is a unit then MadjM/detM = I and adjM/detM M=I

so adjM/detM = M^(-1)

[thorn jay]

Right

[golden turtle]

=> invertible => surjective

[thorn jay]

So we're done!

[golden turtle]

😎

Thank you so much

[thorn jay]

Hehe you did most of the work

[golden turtle]

never would've thought of this though

[thorn jay]

I never would've thought of the adjugate matrix

Lol

Tbh i didnt even know it existed

That probably didnt help

And using this you can prove the invariant basis property for commutative rings

[somber bluff]

What are groups rings and fields generalized to in universal algebra?

[grim hawk]

I thought that groups, rings, and fields themselves are generalized structures and when considered in universal algebra you study the properties of general algebraic structures rather than examples. Like in abstract algebra we look at specific groups, rings, and fields, like R^n or GL(2,F) and in universal algebra you study classes of algebraic structures in their entirety.

[somber bluff]

ok so one could say groups rings and fields are just specific examples of algebraic structures right

[grim hawk]

Correct. And there are specific examples of those that are of interest in various situations for different people. Like SO(3), GL(2,F), D_4, etc. Classes of those structures are also of interest, like Lie groups, continuous representations of groups. Universal algebra is like stepping back and saying you dont know what specific group it is but making determinations about the larger classes and groups, rings, fields, as specific structures. My experience (limited for sure) is in the more specific side of things so anyone who has any more to add that may call themselves a student of universal algebra please correct me if im wrong or if im being too reductive.

[somber bluff]

Would you say studying universal algebra is particularly useful? or is it too much abstraction that one really never uses later on

[grim hawk]

Depends on what you are doing. I'm in physics so I am more interested in specific groups than the true meat of universal algebra (and I may not be smart enough for it lol, I struggle with simple group theory it feels). There is much to learn and study and research in the depth of algebraic structures however, so I would say it is up to what you want to do. I am not the kind of person to ever say something is too abstract if its what someone wants to study.

I also dont know enough about it to really declare its worth as a field, but some of the people here are incredible and know a lot more than me so maybe they can give better insight.

[west sinew]

So I've been working on a question which asks me to relate the invariant factors of a matrix and its inverse, but I'm getting stuck

If the minimal polynomial of A is p, one can quickly work out a map between polynomials $p\mapsto p^$ such that $p^$ is the minimal polynomial for $A^{-1}$

[cloud walrus]

adi

[west sinew]

Now, suppose the invariant factors of $A$ are $p_1|...|p_l$. I've shown that $(fg)^=f^g^$, so it's going to follow that $p_1^|p_2^...|p_l^.$ I'm pretty sure these should be the invariant factors for $A^{-1}$, but I'm not able to bridge the gap

[cloud walrus]

adi

[thorn jay]

They are generalises to pairs (A, F) over a type F, where A is called the universe, and F the set of fundamental operations (a function from A^n to A for some n)
Where F abstracts F into a collection of function symbols so you can compare two algebras of the same type.

From what I've gathered (which granted is also limited) just like category theory is about the morphisms, universal algebra isnt about the algebraic structures in particular, but rather about the properties of the terms (a term is a generalisation of an algebraic expression)

And rather than classifying the specific algebraic structures themselves (like one would do in group theory; classify groups of a certain order), you classify classes of algebras based on properties of these terms which hold for all algebras in the class.

Free algebras are for that reason important objects of study.

In universal algebra you have certain constructions one can make: arbitrary direct products, homomorphic images and subalgebras.
A homomorphic image of A is an algebra B such that there exists a surjective homomorphism from A to B.
For a class K denote V(K) to be the smallest class closed under arbitrary products, homomorphic images and subalgebras.

There is another way one can extend a class of algebras K, and that is by looking at all the equational identities satisfied by K (like for the class of groups (x * y) * z ≈ x * (y * z) is an equational identity because it always holds for all members of the group), and then taking the class of all algebras satisfying these identities. This is denoted M(Id_K(X)) for some infinite set X.

The in my opinion most important theorem in universal algebra is the theorem that says:
V(K) = M(Id_K(X))
In other words: every class of algebras closed under products, homomorphic images and subalgebras can be described by a set of equations and vice versa.

That ended up a bit long 😅

I love univ algebra so i can get a bit passionate about it

[golden turtle]

I don't mean to interrupt

but I was wondering if someone could check my work for this

Here is my work

[thorn jay]

One place where the mentioned theorem is used is proving that for every V(K) there exists an algebra A in V(K) such that
V(K) = V(A)
And what do you know? A happens to be a free algebra

[golden turtle]

I am mostly worried about if the way I defined N truly makes it clear that it is normal, and also if the G/N I defined is sufficiently shown to be cyclic of order 2

[thorn jay]

Well let g be an element of the free group
suppose g in N
then xg in xN, gx in Nx = xN
then yg in yN = N, gy in Ny = N
suppose g in xN
=> xg in x^2N = N, gx in xNx = x^2N = N
=> yg = gy' in xNy' = xN
=> gy in xNy = xN
Since the generators of the free group are either in N or xN, by induction every element must be in N or xN

Since N and xN are clearly distinct, F(x, y) / N ~ C_2

[thorn jay]

you can just define N as the normal closure of the set { x^2, e, y }

[golden turtle]

we haven't discussed what that is, I think what I wrote is my attempt at defining it without using those words

[thorn jay]

Yup :3

The free basis for the normal subgroup is then that set ig

It's just the smallest normal subgroup containing the set

Exists because you can just take the intersection of all normal subgroups containing the set

[golden turtle]

Makes sense

[toxic zephyr]

okay i think i have the stupid. in a commutative group, if $o(a)=m$ and $o(b)=n$ and $\gcd(m,n)=1$, then how do we know that $o(ab)=mn$? i think in general it's $o(ab)=\lcm(a,b)$, and though that's clearly a power which would yield the identity, i don't understand why that's necessarily the minimal power?

[cloud walrus]

syntheticdivisiontaylor

[toxic zephyr]

o is order

[warm ember]

say a^kb^k=e

that means a^k=b^(-k)

so a^k and b^k have the same order

you can finish from here

[thorn jay]

this relates to the fundamental theorem of finite abelian groups

An application of that theorem tells you that
<ab> = <a> x <b>
whenever gcd(o(a), o(b)) = 1

[warm ember]

also the second statement there is incorrect?

for example take Z4 and a=b=2

[toxic zephyr]

say a^ck=b^ck=e. then n|ck and m|ck implies lcm(a,b)|ck. not sure where to go from there tbh

[toxic zephyr]

ah yes. i guess my classmate was wrong lol uhhhh not sure what it should be

[warm ember]

so the order of a^k must divide m

and the order of b^k must divide n

so their orders is 1

a^k=b^k=e

m|k and n|k

[toxic zephyr]

oh okay. that is much simpler lol

[toxic zephyr]

and then this implies that mn divides k, so if k<mn, then k=0

[warm ember]

yes

[toxic zephyr]

alright cool

so then what should the order of ab be?

[warm ember]

i dont think there is anything on that

[toxic zephyr]

the original exercise was to show there exists an element of order lcm(m,n). let me find a picture

[warm ember]

oh

ive done that problem

do you know fundamental theorem of abelian groups?

[toxic zephyr]

not by name at least

[warm ember]

actually

fundamental theorem of fintely generated abelian groups

the structure theorem

[thorn jay]

For abelian groups there is, right? That's just the order of (1, 1) in Z_o(a) x Z_o(b)

[warm ember]

oh

[toxic zephyr]

that i have heard of, but we haven't learned it yet. i'm assuming there's a more elementary method to prove it

[warm ember]

ye if m and n are relatively prime ye

if they arent you cant say anything

[thorn jay]

This should always work

[warm ember]

hm idk any other way

[thorn jay]

As (ab)^k = a^k b^k

[toxic zephyr]

yeah it's literally section 3 of chapter 1 of Hungerford. in the section about cyclic groups

[thorn jay]

So you have a bijection from <a, b> to Z_o(a) x Z_o(b)

If b isnt in the subgroup generated by a

[toxic zephyr]

oh i see

easily a group hom. a^i b^j maps to (i,j)

[warm ember]

so here you have their orders divides gcd

ye it wouldnt be ab

[warm ember]

that only when m and n are relatively prime

otherwise a and b can interact

[thorn jay]

Right, true

[toxic zephyr]

do we maybe want to use some a^k and b^l to make sure the orders are coprime? then o(ab) would be lcm(a,b) right?

[warm ember]

o

you mean a^gcd and b^gcd

but the problem is you cant prove its minimal

https://math.stackexchange.com/questions/3765522/if-an-abelian-group-has-subgroups-of-orders-m-and-n-respectively-then-it-h

this may help

actually that just solves it

by diamond we have <a,b>/<a>=<b>/(<a>\cap<b>)

ok actually no the ending still uses FTFGAG

[crystal vale]

In the Abelian group, if a has order m and b has order n then not necessarily o(ab) = LCM(m,n)

This is true when < a > intersection < b > is {e}

[cobalt heath]

<a, b> -> Z/Zo(a) × Z/Zo(b) is at least surjective

So just needs to look at how the kernel looks like

[warm ember]

if o(a) and o(b) are not relatively prime then we cant even make sure it is a map

[thorn jay]

Yee

[cobalt heath]

Why not?

Ah, uniqueness issues arise, I guess

So it is rather opposite, and you instead have Z/o(a)Z * Z/o(b)Z -> <a, b>.

Now this map seems surjective, hmmm

[crystal vale]

Show that if H is any group and h is an element of H with h^n = 1, then there is a unique homomorphism from Z/nZ = < x > to H such that x -> h.

Yes there is homomorphism, but uniqueness? I am not sure about what it means by uniqueness here ?

[serene dune]

$Id_{\mathbb{Z}_n} \mapsto Id_H$

[cloud walrus]

yeshua

[latent lodge]

hi guys i've noticed something and idk

[serene sable]

A bit late to the party but I think something like the following works:

Prime factorise the lcm as p1^a1...pk^ak. By the coprime case, its enough to find an element of order pi^ai for each i.

Can you see how to proceed from here?

Hint: ||How do you explicitly calculate an lcm of two numbers, given their prime factorisations?||

Solution: ||ai is the power of pi in the order of one of the elements a, b. Raise a to the power of m/pi^ai (or b to the n/pi^ai resp.).||

[earnest hare]

i did not understand what their proof was doing

http://abstract.ups.edu/aata/galois-section-fund-theorem-galois-theory.html

[dull ginkgo]

it’s actually equal due to Artin’s lemma

Also it won’t load

[earnest hare]

i didn't send as it is.. very long

they used artins lemma as well

could someone tell me in layman terms what the idea behind the proof is in chunks

[dull ginkgo]

https://people.math.harvard.edu/~elkies/M250.04/artinlemma.html

This will explain it better than I ever could

[earnest hare]

btw E^G is the same as E_G here right? the subfield of elements in E that are fixed by automorphisms in G

[dull ginkgo]

Yes

[earnest hare]

notation~

[crystal vale]

subgroup generated by { 1/p | p is an odd prime number} in additive group Q is proper subgroup which is not finitely generated, right?

[undone skiff]

yes

because u will never get a larger prime denominator

[thorn jay]

Yeah

[undone skiff]

i think

[thorn jay]

No that's true

[surreal dagger]

For clarification: At (9): Whats the difference between writing V(I(V(S)))=V(S) for a Set of points S, and V(I(V))=V for an algebraic set V. Per definition an algebraic set V=V(A) for a set of points A, no?
So right now to me it looks like he just wrote it again, because V(I(V))=V looks nice or something like that

[thorn jay]

When adding fractions you take the lcm of the denominators, and the lcm of single primes will always be a product of distinct primes

1/4 would not be in the set for exampke

[cobalt heath]

Well, I mean it could be difficult to notice for someone

So sometimes authors are more benevolent and write things out for readers

[crystal vale]

But subgroup generated by that set

[cobalt heath]

Meanwhile others just "details are left as an exercise to the reader" for all these properties

[thorn jay]

Oh sorry thats what i meant

1/4 wouldnt be in the subgroup

[crystal vale]

But I don't need that

[thorn jay]

It proves that the subgroup is proper

[crystal vale]

Yes

Therefore I left 1/2

[surreal dagger]

I think my problem is:
So given a set of polynomials S, notationally we "use V" to turn it into some set of points with a property by writing V(S).
But then he also uses V as a literal set of points that all vanish on some set of polynomials.
So the "V" in V(I(V))=V dont have all the same "meaning", no?

[cobalt heath]

Wdym “it dont have all the same meaning”?

[crystal vale]

If H is the proper subgroup of the finite group G then there is a maximal subgroup of G containing H.
I used Zorn's lemma on the set { K < G | such that K is a proper subgroup of G which contains H }. Then this set is non-empty but taking any chain there is a problem to show the existence of the upper bound of it

[thorn jay]

You dont need something like Zorn's lemma for finite groups

[crystal vale]

Okay

[surreal dagger]

So given a Set of polynomials S, we can look at V(S) the set of points such that all polynomials in S vanish on them.
I.g. then you can write something like V(I(V(S)))=V(S) and check that its the same set.

Now to me he then says, now consider an algebraic set V (which to me means V=V(A)) then you can write V(I(V))=V by using the above identity.

But then the V on the right is that algebraic set and then e.g. the first V on the left is the "operator" that makes I(V) into an algebraic set.

So to me it seems like some sort of abuse of notation and I dont see the payoff doing it like that

[crystal vale]

Yes in a finite case we can find some finite steps

Right?

[thorn jay]

Yes

[crystal vale]

Okay thank you ❤️

[thorn jay]

There always exists at least one element adjacent to the maximal element in a finite lattice

[crystal vale]

So in case of a finitely generated group generated by { g_1,..,g_n} we need Zorn's lemma to prove that the group has a maximal subgroup.

Yes we take the set S of all proper subgroup and then S is non-empty. Taking the chain C, we can get the U union of all subgroups in chain C which is subgroup.

Now it is proper subgroup, because if it is not then every g_i is in U so there is an element H in C such that all g_i are in H which contradiction.

Is it correct?

[crystal vale]

Yes it is a finite lattice but I don't get what you meant

[dull ginkgo]

Do you get the idea of the proof?

[earnest hare]

no.. its the same as the other proof. ill look at it again later

[earnest hare]

Artin's Lemma: Let G be a finite group of automorphisms of a field E, and F=E^G. Then [E:F] is finite and no larger than |G|.

i dont get it could someone walk me through the proof

[crystal vale]

I have to show that Q/Z is isomorphic to multiplicative group of roots of unity in C*.

p/q + Z -> e^((2πi)×p/q), works ?

[crystal vale]

To subgroup H to be normal we have to show that gHg^-1 \subset H for all g in G. Thus when we have N(H) = G this one is stronger condition, right?

[glad osprey]

Yeah, or maybe simpler as a map from Q to C* with kernel Z

[glad osprey]

Conjugation is an automorphism of G, so I think gHg^-1 \subset H is equivalent to gHg^-1 = H

[crystal vale]

How is it ?

[crystal vale]

Okay thank you

[rocky cloak]

For a fixed g it's not true.

But
gHg^- < H
is equivalent to
H < g^- H g
so if it's supposed to hold for all g in G you can always reason by replacing g by g^-

[crystal vale]

So it is not true for fixed g, right?

[rocky cloak]

Correct.

[glad osprey]

hmm, isn't conjugation a bijection? I don't understand how gHg^-1 can be smaller than H?

[crystal vale]

In case of infinite it can be

[glad osprey]

ah, right

infinity 😠

[crystal vale]

Thank you

[rocky cloak]

Consider for example G = Sym(Z) the group of bijections on Z and H = Sym(N) (those bijections that fixes all the negative integers.

Then if g(x) = x+1 is bijection of adding 1, then
gHg^- is in H, but g^-Hg is not

[crystal vale]

Yes

We only need to check whether it fixes all the negative integer or not

[void cosmos]

are polynomial rings a counterexample for the statement that V is f.g as a K[x] module -> V is finite dim?

take V to be K[x] ?

[coral spindle]

Yes

[void cosmos]

ty

[coral spindle]

In fact K[x] is a so-called 'cyclic' module: it is generated by a single element. Try classifying all cyclic modules of any ring R (it is easier than it sounds)

[void cosmos]

yeah just multiply by 1

ig

yeah i remember in hungerford iirc they are quotients of something

[thorn jay]

||quotients of principal ideals?||

[coral spindle]

||Quotients of R.||

[kind solar]

i'm newbie how to start my journey

they have put abstract algebra on the first course

and i don't understand like anything

[thorn jay]

||are principal ideals isomorphic to quotients of R?||

[thorn jay]

Thats what the lessons are for right?

[tardy hedge]

Really bro^

[rocky cloak]

See if you can prove it!

[thorn jay]

Hm

[earnest hare]

Let F be a finite field or a field of characteristic zero

isnt this every field?

hi aku

[thorn jay]

Hii poke!

[thorn jay]

Finite fields

[sharp ice]

theres infinite fields with characteristic > 0. for example F_p(t)

[thorn jay]

char 0 implies an embedding of Z

(Z as subring of the field)

Thus finite fields cannot have char 0

[earnest hare]

whats F_p(t)?

[sharp ice]

the field of fractions of F_p[t] where F_p[t] are the polynomials with coefficients in F_p (finite field with p elements)

[earnest hare]

polynomial with variable t and entries in Z_p?

oh yea field of fractions

[sharp ice]

or just the field of rational polynomials with coefficients in F_p

[thorn jay]

Oh i misread ur question
x3

[earnest hare]

but does it not have char 0?

[sharp ice]

it has characteristic p

[earnest hare]

ohhh

[sharp ice]

the multiplicative identity is 1 and u add it p times to get 0

[earnest hare]

sorry i keep thinking that characteristic is multiplication n times

thank you conan

i hope u become a good detective

and thank you aku

[thorn jay]

I've constructed a module homomorphism to the module generated by 1 element but im stuck because i cant seem to prove that the kernel is also a right ideal

I dont think it matters even

Let I be the left ideal generated by a
Then R/I forms a left module over R geberated by 1

[rocky cloak]

This is correct. No need for it to be a right ideal

[thorn jay]

R/I as in the abelian quotient group

Oki then my proof is complete

Yay!

Suppose $M$ is an $R$-module generated by an element $a$. Thus for every element $m \in M$ there exists an $r \in R \colon m = r \cdot a$.
Define the mapping
$$h \colon R \rightarrow M$$
$$r \mapsto r \cdot a$$
As previously mentioned this map must be surjective. Also notice:
$$h(r + s) = (r + s) \cdot a = r\cdot a + s \cdot a = h(r) + h(s)$$
$$h(r \cdot s) = (r\cdot s) \cdot a = r \cdot (s\cdot a) = r \cdot h(s)$$
Thus $h$ is a surjective module homomorphism.
Now, $\ker h$ must be a submodule of $R$, but that clearly must then also be a left ideal by the fact that the scalar multiplication is just left multiplication.
Hence $M$ must be isomorphic to some quotient of $R$ by a left ideal.

[cloud walrus]

Z(A) = ∇ <=> A is R-module

[thorn jay]

Fun proof!

[serene dune]

i was stuck at a point to explicit show that some integer power of an arbitrary ring element does exist in the very ring

[thorn jay]

Negative powers not necessarily

0th powers not even if you dont require your rings to be unital

Smh

[serene dune]

oh it was mentioned to be positive

[thorn jay]

Right yeah

Use induction

:p

[serene dune]

😹

its so over for me

the inverse is a^(n-2)
should that be n>2 or im wrong

[thorn jay]

You cant conclude from that a has an inverse

[serene dune]

inside the quotient

for every non zero a

[thorn jay]

Prime ideals clearly contain every zero-divisor

The a^n = a property carries over to R/I for I an arbitrary ideal

=> for all a + I there exists an n such that
a^n + I = a + I
(a + I)(a^{n-1} - 1 + I) = 0
As I is prime it eats all the zero-divisors, thus
a^{n-1} + I = 1 + I or a + I = 0
Hence in a + I has an inverse or is 0, thus R/I is a divisor ring.
It is also commutative, hence R/I is a field, hence I is maximal

[thorn jay]

Yeah ur right

Fun exercise

[serene dune]

after doing alltat i was wandering why that element should be in the ring

[rocky cloak]

The notation a^n just means repeated multiplication, and you can multiply things in a ring

[thorn jay]

Because the ring is closed under it's operations, so a finite amount of multiplication won't boot you out of the ring

[serene dune]

that's the joke right

after all the fight closure had better of me

[elfin wraith]

I’m trying to show that if M is an R module and N is a sub module of M then M is Noetherian iff N and M/N are.

I’ve done this using the finitely generated condition and I’m now trying to use the ACC, the forward direction is trivial but I’m a little stuck on the reverse.

I’ve shown that every sub module of M/N has the form K/N where K is a submodule of M containing N, so my idea was that M/N and N in some sense partition the submodules of M

That is that like every submodule of M is either in the chain of sub modules for M/N or N,

I’m not convinced that this is true though, but obviously if it were then it would be enough to show that M satisfies the ACC

So I guess my question is, am I correct in thinking that M/N and N do partition the sub modules in this way, or if not, where might be a good place to start looking for an alternative approach?

[rocky cloak]

They don't quite partition the submodules, but the main idea is sort of close to that.

More specifically: ||If K < L are submodules of M you have that KnN < LnN and (K+N)/N < (L+N)/N.||

Then ||K = L iff both of those inclusions are equality||

Less spoilery hint: ||think about how you can relate a submodule of M to one of N and one of M/N||

[elfin wraith]

Ahhh yeah no I’m being silly, it’ll be the same idea as the proof I just did for the finite generating case, itll be the intersection and like the image under the canonical map I think

So like letting K be a sub module of M then considering KnN and c(K) I think, because this is actually a partition, it’s either in N or like in the image of K under the projection onto M/N

Then some sort of inclusion argument that’s not as instantly clear to me but I think that’s the idea?

[toxic zephyr]

this might be the question of all time. so i was pretty easily able to show that if the order of a is finite, then the order of f(a) divides the order of a, using the division algorithm. but my dumb question is: is that literally it? like... i showed if it's not infinite then this. but then isn't the only other possibility that the order of a is infinite? like how do we show that the order can be infinite (i know i can use a specific example where f is a map from Z to Z2 and using a=1, but like that seems unnecessary).

like order of a is infinite or it's not.

[coral spindle]

Yes that is literally it

[toxic zephyr]

okay thank you for the sanity check lol

[coral spindle]

If you want a better version, consider this

Instead of defining |a| to be infinite, let it instead be 0 when there is no positive integer n s.t. a^n = 1

Then |f(a)| divides |a| always.

[toxic zephyr]

so you mean like a better version of a definition for order? which would give us this general property that |f(a)| divides |a|?

[coral spindle]

And the definition is what I stated, yes.

This is not standard and you should not use this, but it is more consistent with e.g. the definition of characteristic of a ring.

In fact if you follow that definition in a particular way, this fact is extremely trivial

[toxic zephyr]

how about we define m divides infinity for all integers m, then it's also always true

no but i totally agree

[coral spindle]

Compare with this fact: char(R/I) divides char(R) for any ring R, where char(-) is the characteristic of a ring of course.

[elfin wraith]

Still not quite seeing the full picture here, it’s definitely harder for me to see than with using the finitely generated condition

My current thinking is that if you let K be a submodule of M, we then consider NnK and the projection of K onto M/N, c(K).

Since N satisfies the ACC, NnK is contained in some finite set of N submodules {A_1,…,A_n} and similarly we have c(K) is in {B_1,…,B_m}, this seems like a reasonable thing to have, but I’m not quite sure what to actually do with it

[rocky cloak]

If you want to prove ACC you should probably start with an ascending chain in M

[elfin wraith]

Ok yeah I think I was going about it all wrong but I think this works

[golden turtle]

by "that set", you meant what I defined as N, right?

I just now realized that I did not find a free basis of N, I just found N

[thorn jay]

{ x^2, y }

That's a free basis of N

If im correct

[golden turtle]

how does it generate xyx?

since N contains all conjugates of y I think xyx in N

[thorn jay]

Well it is at least the normal closure of x^2, y but what the free basis is..

[golden turtle]

yeah, N is the normal closure of x^2 and y

and N is free by nielson-schrier

but idk how to find its free basis

[thorn jay]

Thats a cool result actually, nielson-schrier

Does the proof give any insight into how the free basis would be constructed?

[golden turtle]

I didn't understand the proof

[thorn jay]

Lol

[golden turtle]

I was about to comment that

actually

😂

[thorn jay]

[golden turtle]

something something schrier transversal

d&f doesn't include a proof either

[thorn jay]

My intuition says just
{ gx^2g^-1, gyg^-1 | g in F(x, y) }

[golden turtle]

F(a,b)?

[thorn jay]

Free group generated by a, b

Wait

Ahdhs

[golden turtle]

F(x,y) here

same thing

[golden turtle]

this is just N though

[thorn jay]

No

[golden turtle]

well, i mean you omitted the part about how concatenations are allowed

[thorn jay]

x^2y is not in that set

But is in N

[golden turtle]

so you're omitting it purposefully

[thorn jay]

Furthermore it's pretty easy to show that all the elements in that set are independent (removing any one of them will change the subgroup generated by the set by the nature of free groups) hence it is a basis

But idk of N

[golden turtle]

[thorn jay]

Oh lol the proof is pretty easy

[rocky cloak]

My intuition is more going for
|| {x^2, x^n y x^-n} ||

[golden turtle]

In office hours today a different approach found <x^2, xy, y^2> worked as a free basis for a normal subgroup N, so maybe it always has to be that ? But they took a different approach and I have already found my N so I was hoping to just be able to determine its free basis

if that makes sense

meant to be <x^2, xy, y^2>

[rocky cloak]

The number of elements in the free basis should always be the same

So I guess it's 3 for this one then

[golden turtle]

so whatever the free basis is for my N it should have 3 elements, but maybe not specifically x^2, xy, y^2

[rocky cloak]

I guess you have a slightly different N, so it could be different... But I'm guessing it's the same

[thorn jay]

Let $M = < gx^2g^{-1}, gyg^{-1 }>$
Then any element m of N is a string
$$g_0x^{a_0}y^{b_0}g_0^{-1} \cdot \dots \cdot g_kx^{a_k}y^{b_k}g_k^{-1}$$
For $a_i, b_i \in \mathbb{N}$.
Now take $h \in F(x, y)$
Then
$$hmh^{-1} = hg_0x^{a_0}y^{b_0}g_0^{-1} \cdot \dots \cdot g_kx^{a_k}y^{b_k}g_k^{-1} h^{-1}$$
$$= hg_0x^{a_0}y^{b_0}g_0^{-1} h^{-1}h \cdot \dots \cdot h^{-1}h g_k x^{a_k}y^{b_k}g_k^{-1} h^{-1}$$
$$= (hg_0)x^{a_0}y^{b_0} (h g_0)^{-1}\cdot \dots \cdot (h g_k) x^{a_k}y^{b_k} (h g_k)^{-1}$$
Which is in $M$ hence $M$ is normal.

[rocky cloak]

x^2, y, xyx^- would make sense

[thorn jay]

Noo my beautiful proof

[cloud walrus]

Z(A) = ∇ <=> A is R-module

[rocky cloak]

But

(yx) y (yx)^- = y (x y x^-) y^-

so that's not a free basis

[thorn jay]

Auhshsh fuck me then

Thanks for the correction

[thorn jay]

Ah well this was a fun exercise anyhow

[thorn jay]

Really?

Oh because it's isomorphic to the free group of that number of elements?

[golden turtle]

F_s iso F_t if and only if s=t

[thorn jay]

Mm yeah, universal properties give rise to a functor

Thats true

[golden turtle]

this is another problem on my HW

haha

should be not bad

[golden turtle]

What makes you think of this? I'll try it

[thorn jay]

F_s?

[golden turtle]

This is still helpful

because now I know what the proof should attempt to be doing

[golden turtle]

free group

rank s

[thorn jay]

Ohh

Use universal property

[rocky cloak]

Well, it looks almost identical to the one you gave, is contained in N and looks normal (because it has conjugation in it)

[thorn jay]

Oh wait thats a theorem in my univ algebra book

[golden turtle]

I am hinted to quotient by their commutator subgroup

the resulting group will be abelian

and so

I think it follows from fundamental theorem of finite abelian groups

[thorn jay]

Hmm

[rocky cloak]

And to prove it's normal you just need to show it's closed under conjugation by x^± and y^± which it very much appears to be

[golden turtle]

what do you mean by x^(+)?

[thorn jay]

Oh because the rest follows from induction obv

[golden turtle]

and why do I have to show the free basis is normal

sorry if that is a dumb question

[thorn jay]

I keep forgetting that bases have great properties

[rocky cloak]

Well you want it to generate N

Easiest way to do that would be to show that it's normal and contains x^2 and y

[golden turtle]

and there could be no larger normal subgroup than N since it has index 2

?

[rocky cloak]

I guess, but if we're picking elements from N they can't generate a bigger subgroup anyway

[golden turtle]

I see

[signal knot]

this might be stupid but in Dn the rotations commute right?

[quiet pelican]

Yes

They’re all powers of a given element

[south patrol]

The rotational group is just C_n yeah

[golden turtle]

I've completed the first problem, and I'm now working on T is injective <=> det M is not a zero divisor

I've shown that det M not a zero divisor => T injective

but I am struggling showing the other direction

[golden turtle]

why does showing f!=1 show H is freely generated by Y?

[swift tundra]

Today my professor brought up inversion to define the sign function. Given a permutation s, (i,j) st i<j (in the symmetric group on “n” elements) is an inversion if s(i)>s(j).

The professor justified it as a neat way to avoid dealing with different factorizations of permutations into transpositions (you have to prove they all have the same sign so that the sign function is well defined). While this makes sense, I know there must be some use of inversion elsewhere in modern algebra. Do any of you know about this?

[crystal vale]

From partition P and its operating I get if a_1 is related to a_2 and b_1 is related to b_2 then a_1b_1 is related to a_2b_2, correct?

Then let e in A_e. Let a and b in A_e so a is related to e and b is related to e. Then by operation in P we have ab is related to e, hence ab in A_e.

Let a in A_e and a^-1 in A_h, then e in A_h implies a^-1 in A_h.

This shows that A_e is a subgroup of G.

Now let h in A_e, g in A_s and g^-1 in A_t, then ghg^-1 is related to ts, and ts is related to e, thus ghg^-1 is in A_e.
A_e is a normal subgroup in G.

Is it correct?

[tulip agate]

In paragraph 3, why is e in A_h?

And why bother concluding that a^-1 is in A_h when you assumed it was in A_h to begin with?

The normal part looks good except I think you have ts backwards.

[tulip agate]

A group fails to be freely generated of there's a relation among its generators. Any relation can be rewritten to the form of 1=something. If no word equals 1, then there are no relations among the group's generators.

[golden turtle]

if G is free on generators {x,y} and N is a normal subgroup such that
G/N is cyclic of order 2, generated by xN
Is a valid Schreier transversal X:={e,x}?
I'm not entirely sure what counts as an "initial segment"

I want to think the initial segment(s) of e are just e and thus in X

and that the initial segment(s) of x are just e, and x (or maybe also ex ?) and are thus also in X

but I'm not sure about that latter part

[serene dune]

instead of a S^1(R/Z) we now get an inscribed polygon right?

i mean it's still like circle, due to infinite elements, but not as dense as a circle!!!!

am i tweaking

[crystal vale]

Because a is related to e and a^-1 is related to h so a•a^-1 is related to eh = h so e is related to h

[crystal vale]

I know a is in A_e but I don't know where a^-1 is so I assumed a^-1 is in some partition class say A_h

[tulip agate]

Yes but then your conclusion was the exact same as your assumption. You need to conclude that a^-1 is in A_e, not in A_h.

Okay and I now see that e in A_h implies that A_e = A_h since it's a partition.

You do have a few typos, and I would include the extra explanation that you have me, but it does look right. 👍

[crystal vale]

Thank you ❤️

[somber bluff]

ty. my first intro to universal algebra was when i read this

[earnest hare]

does this mean L is a field extension of K.

G is the automorphisms of L leaving K fixed

[thorn jay]

Yes

[earnest hare]

tyty

[thorn jay]

I love universal algebra! You should look into it more if it interests you
"A course in universal algebra" is great

[crystal vale]

Let p be prime and let G be a group of order p^(a)m where p does not divide m. Assume that P is a subgroup of G of order p^a and N is a normal subgroup of G of order p^bn, where p does not divide n. I have to show that | P and N | = p^b.

I don't want to use Sylow stuff here only isomorphic stuffs.

I showed that | P and N | is p-subgroup. Any hint ?

In the next exercise they introduce Hall subgroup

Let G be a finite group. H be a subgroup such that gcd( | G : H |, |H| ) = 1 and let N be the normal subgroup of G then H intersection N is a hall subgroup of N.

We have to show that gcd( | N : H and N |, | H and N | ) = 1.

Now let d be the gcd of that number.

Since N is a normal subgroup in G implies that HN is a subgroup of G.
Hence |HN | divides| G | it follows that | N |/ | H and N | divides | G |/ |H |.

And also d divides | H and N | implies that d divides | H |.

Therefore, d divides both | G |/ |H | and | H | , hence d = 1.

Is it correct?

[spice whale]

What would be a rank 2 free subgroup of GL_2(Z)?

[rocky cloak]

There is one here
https://mathoverflow.net/a/43732/157483

[spice whale]

thanks

[somber sleet]

HEy guys, why are these moduls isomorphic

[wild jasper]

Let $F=\mathbb{Q}(\sqrt{2},\sqrt{3})$. and assume that $a,b,c,d\in F$ are not squares in $F$. I have determined that the non-identity automorphisms of $Gal(F/\mathbb{Q})$ all map $a,b,c,d$ to each other. Can I use this to conclude that the minimal polynomial of $\sqrt{a}$ has $\pm\sqrt{a}, \pm\sqrt{b}, \pm\sqrt{c},\pm\sqrt{d}$ as roots?

[cloud walrus]

mh_le

[thorn jay]

Isomorphism theorems

I love them

[rocky cloak]

Z[x]/(5) = F5[x], see null for the rest

[rocky cloak]

If {a, b, c, d} is invariant under the Galois group, that means the polynomial
(x^2 - a)(x^2 - b)(x^2 - c)(x^2 - d)
is fixed by the Galois group so has coefficients in Q.

The minimal polynomial of sqrt(a) must divide this, and the roots are as you listed. Of course the minimal polynomial could be smaller though

[vagrant zinc]

:o , name of book

[wild jasper]

None of a,b,c,d are fixed by the Galois group

Ah

I see what you mean, they are all mapped to each other and then the polynomial is invariant

Thanks!

[night tartan]

intuitively, you can think of it like this

if we have Z[x] and we're killing off 5 and the polynomial x^3 + 2x + 3

then that's the same as Z modulo 5[x] but we kill off x^3 + 2x + 3

[winged void]

Hey i have small question regarding the following i know how to solve it

but i do not have intuition about it

[thorn jay]

Whats the problem?

[winged void]

A \textbf{Boolean ring} (named after the English mathematician George Boole, 1815--1864) is a ring ( R ) in which ( x^2 = x ) for all ( x \in R ).

(a) Prove that ( x + x = 0 ) for all ( x ) in a Boolean ring ( R ).

[cloud walrus]

Mootje

[winged void]

you take (x+x)^2 and then you are doine with the proof

which is 4x^2 = 2x \implies that 4x = 2x \implies 2x = 0

but what does that even mean

what are we doing

[tardy hedge]

You have a ring with the property x^2 = x

That property implies x+x=0

I dont really think there is much else to say

I guess there is probably intuition for why x+x=0 intuitively given that x^2=x, if u really thought about it or smth

[thorn jay]

It's basically saying that freshman's dream holds for n=2

[tardy hedge]

Hahahaha freshman’s dream

[thorn jay]

(x + y)^2 = x + y = x^2 + y^2

[tardy hedge]

Freshman’s dream is crazy

[thorn jay]

It's the official term as far as i can tell

[tardy hedge]

Of course!

[thorn jay]

It has a wikipedia page even

[tardy hedge]

Wait really

[thorn jay]

Yup

[tardy hedge]

Thats epic

[thorn jay]

But this clearly implies char 2, right

(x + y)^2 = x^2 + y^2
x^2 + 2xy + y^2 = x^2 + y^2
2xy = 0
y = 1 => x + x = 0

[south patrol]

Yes

[thorn jay]

so thats the intuition

[south patrol]

So these are algebras over Z/2Z

[thorn jay]

Freshman's dream forces 2xy to be 0 for all x, y

[tardy hedge]

lol

[thorn jay]

Yeah thats right, your hopes and dreams force everyone to be their additive inverse >:(

Thats about as intuition as it can get i suppose

Boolean rings are interesting because they are in a sense the same as boolean algebras

You can define a boolean algebra for every boolean ring and vice versa, and these transformations are eachother's inverses

(And a special case of a near-isomorphism in the category of equational theories for which im not totally sure already exists but most likely does :3)

[rocky cloak]

Is there any particular step that you don't understand why it's there?

Like it's basically just, if x^2 = x, then in particular 4 = 2, so 2 = 4-2 = 0

There isn't that much to it

[winged void]

Sure I thought maybe like there is a reason why it’s called Boolean or something

[winged void]

I understand the steps pretty good only I wanted to know or there is reason behind why it’s called Boolean like does that has to do with computer science

[rocky cloak]

Well, the guy who studied them is named Boole

[winged void]

Because a Boolean in programming is something to give true or false

[rocky cloak]

He also gives his name to Booleans in computer science

[winged void]

Ah I see

[rocky cloak]

0 = false, 1 = true
multiplication = and
addition = xor
Then this is a Boolean algebra

So the concepts are related

[thorn jay]

They're also essentially the same

Lol

Their congruence and subgroup lattices are the same

Their malcev conditions are the same

Their free algebras are isomorphic

Everything

[winged void]

ah i see makes sense

thanks alot

I have one more stupid question actually reagrading something that i never understand that good

Let ( n \in \mathbb{Z}_{>0} ). On the set ( \mathbb{Z}/n\mathbb{Z} = { 0, 1, \ldots, n-1 } ) with ( i = i + n\mathbb{Z} \subset \mathbb{Z} ), an addition is defined, as these are the cosets of the normal subgroup ( n\mathbb{Z} ) of ( \mathbb{Z} ). The rule
[
a \cdot b = a \cdot b,
]
where (\cdot) on the right is the ordinary multiplication in ( \mathbb{Z} ), defines a product (verify that if ( a = a_1 ) and ( b = b_1 ), then ( a \cdot b = a_1 \cdot b_1 )).

With respect to these operations, ( \mathbb{Z}/n\mathbb{Z} ) is a commutative ring with a unity element ( 1 ). In 1.20 we will see that ( \mathbb{Z}/n\mathbb{Z} ) is a field if and only if ( n ) is a prime number. For ( n = 1 ), ( \mathbb{Z}/n\mathbb{Z} ) is the zero ring.

[cloud walrus]

Mootje

[winged void]

like Z/nZ is the modulo group however why do they use the qoutient group notation

so G/H is all cosets in the form a \in G such that aH is a coset

but i do not really understand how is that related to qoutient group

[charred iris]

It's not technically quotient group notation, since this notation is used for basically any quotient. If you take H = nZ = {0, n, -n, 2n, -2n, etc.}, they're slightly abusing notation in using 0 to refer to H, 1 to refer to 1+H = {1, n+1, -n+1, 2n+1, ...} which is like a 'coset' (usually that term is used when H a subgroup, but here it's an ideal) under addition, 2 to refer to 2+Z, etc.

and then Z/nZ is technically the set of all 'cosets', which is {0+nZ, 1+nZ, ..., (n-1)+nZ}

[winged void]

ah i see makese sense

thanks alot

Edward

[golden turtle]

What?

[dense root]

im taking this subject now, we went over rings and fields etc but i have a question
assume we have Z6 for example
we can say that 2*3=6 =0 --> since 2 and 3 are not zeros are both of them zero divisors for Z6 ? or how does that work since in my doctor notes it says only the 2 is a zero divisor

[thorn jay]

that's weird

if r * s = 0 then r is a left zero divisor and s is a right zero divisor
but Z/6Z is commutative, so there is no distinction between right and left zero divisors

so in a commutative ring whenever r * s = 0, then both r and s are simply called zero divisors

[dense root]

oh okay

thank you

[thorn jay]

mhm!

[winged void]

try also to prove that the same element cannot be a zero divisor as unit at the same time

[winged void]

agree

[thorn jay]

it can

...

in the zero ring

[winged void]

yes true

sorry forget to say if its not the trivial ring

or the zero ring

agree

[winged void]

my bad

[thorn jay]

all trivial varieties are isomorphic in the category of equational theories 🤤

[dense root]

we can assume by contradiction that there is an element x in ring R that is both a unit and zero divisor,
if b and c (not zero) in R than we can say
x⋅ b = 1 and x⋅c = 0

b⋅ (x ⋅c) =b **(0)
(b⋅x)⋅c=0
1 ⋅c=0
c=0 ---------> which is a contradiction 🙂

[dense root]

we gonna go over this soon i think

[thorn jay]

great job!

[thorn jay]

eh they're not all that interesting in all honesty

[dense root]

oh nvm we already took that but prof didnt stay much at it

[thorn jay]

well they're important in the fact that they're isomorphic to the quotient R/R, and their existence confirms that quotients in rings work 'nice'

[dense root]

we are at ideals and subrings rn

this is the last thing we covered

[winged void]

nice you did a greay job

[dense root]

is there a channel for real analysis

ohh

mixed

ty

[winged void]

can someone explain this how did they find the ker of f

i do not really see what is happening here

[thorn jay]

yoo another dutch

en ik beweer dat dat helemaal prima is :>

[winged void]

waar studeer je

[winged void]

ja klopt helemaal

[thorn jay]

nog nergens smh

laatste schooljaar

[winged void]

middelbare

[thorn jay]

maar ik wil naar radboud

[thorn jay]

yee

[winged void]

nice best wel slim man

ik moet zeggen

Ik had niets aan de middelbare school shit toch deed alle wiskunde op de middelbare A,B, D

[thorn jay]

real wiskunde op middelbaar is ez af

[winged void]

weet ik lekker free stylen

[thorn jay]

exactt

[winged void]

uni is ook niet zo moeilijk maar algebra is niet mijn ding

[thorn jay]

radboud heeft ook een vak universele algebra

(gebruikt hetzelfde boek dat ik gebruik)

[earnest hare]

ez af

[thorn jay]

hell yeah

[tardy hedge]

What a classic!

[earnest hare]

is this dutch

[thorn jay]

ja makker

(yes)

[earnest hare]

lol

[winged void]

ah ik zie welke boek gebruik je

[thorn jay]

A Course in Universal Algebra - H. P. Sankappanavar, Stanley Burris

I mean i wouldn't take it as a book just for algebra

only when you've matured a lot already in algebra

[dense root]

real Analysis 1
Complex Analysis 1
Numerical analysis (or methods)
Abstract algebra 2 or ring theory ig

these are the subjects i have now

[thorn jay]

I hope I won't get numerical analysis

sounds like a nightmare

[dense root]

its pretty easy

we have it in 2 subjects here but the second is an elective and luckily i have a minor in IT so i dont need to take Math electives

i love the subjects we have now and thats enough for me

[delicate bloom]

one way might be to use the fact that 2 ramifies in Z[i] and then work out the consequences of that in the ring homomorphism to a field

[thorn jay]

might be but I wouldn't get any enjoyment out of it lol

like probably absolutely none

[dense root]

well

i just remembered the material

and now i dont blame you

i take back what i said

[thorn jay]

hahaha

[golden turtle]

How can I prove that a square matrix over a commutative ring R is injective <=> its determinant is not a zero divisor

1 direction is really easy, but I'm struggling with showing
injective => detM not zero divisor

here is how I did the first direction

for context

[toxic zephyr]

so in a vector space, the subspace generated by $v_1,\ldots,v_n$ is just all ``linear combinations"
$$x\in\gen{v_1,\ldots,v_n}\implies x=\sum_{i=1}^n c_iv_i$$
is there a similar term in a group for
$$x\in\gen{g_1,\ldots,g_n}\implies x=\prod_{i=1}^m g_{k_i}^{c_i}$$
it feels sort of like a linear combination but like... exponenty and noncommutative lol

[cloud walrus]

syntheticdivisiontaylor

[toxic zephyr]

i know for an abelian group we could consider it an actual Z module linear combination, but im wondering for nonabelian

[rocky cloak]

It's called a word

https://en.m.wikipedia.org/wiki/Word_(group_theory)

I think you might say "x is a word in g1, ..., gn"

[toxic zephyr]

https://tenor.com/view/word-amy-poehler-gif-24329284

so basically, the subgroup generated by X is all words in X?

[rocky cloak]

You got it sistah

[toxic zephyr]

helllll yeahhhh boi

[toxic zephyr]

^still unanswered

[toxic zephyr]

i think it's much easier to do the contrapositive. then you can still use adj(M) directly, i think

quick question: the center is the set of all elements which commute with all elements in the group. but can we also understand them as elements which are fixed by all conjugations?

[thorn jay]

the same, basically

gxg^- = x <=> gx = xg

[toxic zephyr]

yeah

to prove the center of Sn is trivial for n>2 i basically just used that i can always find a transposition to conjugate by which changes any nonidentity element.

[thorn jay]

smart

and yeah that proves that the center is indeed trivial

[hidden wind]

i love words

and free groups

uwu

[golden turtle]

Yeah I’ve tried this. Suppose detM is a 0 divisor show that M is not injective

It doesn’t fall through because adjM could theoretically absorb the non Injectivity

Like
Madj(M) = detM I
Clearly both sides are non injective if det M a zero divisor but it doesn’t imply M not injective because it could just be adjM

[toxic zephyr]

I was thinking that each column of adj(M) could help provide a nonzero vector in the kernel. but I suppose you'd have to prove that adj(M) has a nonzero column. and that isn't always the case I think.

[golden turtle]

And say it has a non zero column, how does that give us a vector in the kernel?

[toxic zephyr]

well say d is nonzero in R such that du=0. then d times that column would be in the kernel wouldn't it?

[golden turtle]

Which of u and v is the nonzero column of adjM

[toxic zephyr]

I was using horrible notation

v for an element in R instead of a vector

bad

[golden turtle]

If the column is just (u,u,…,u)^T yes

I don’t see differently how we know

[toxic zephyr]

well I mean the idea of the adj is that the jth column is a preimage of det(M)ej

so if you scale the jth column by d then it's image is det(M)*d ej. so just use d as some nonzero annihilator of det(M)

[golden turtle]

The jth column of adjM

[toxic zephyr]

yeah

[golden turtle]

So if there is a non zero column we can scale

Then this follows

?

[toxic zephyr]

yeah

but unfortunately that isn't always guaranteed. like in a vector space the whole adj is 0 if the rank is less than n-1.

so there is no nonzero col

[golden turtle]

Yet rank is n so somehow we’re fine

If the whole adj is 0 then doesn’t it imply
Majd(M) = detM I
M(0)=detMI
0=detM I
=> detM is 0, and hence a zero divisor

If else we’re done as we have a nonzero entry (and column) in adjM?

[toxic zephyr]

yeah that would follow

[golden turtle]

Woaaaaaa

No way

[toxic zephyr]

yeah maybe directly is better. if M is injective maybe you can show the adj must have a nonzero column. and then you can't scale it by a nonzero annihilator without getting a nonzero kernel vector.
but idk showing adj(M) isn't the zero matrix might not be much easier

the adj is still defined as the transpose of the cofactor matrix right

[golden turtle]

Yes

Wait I thought we had solved it already

[golden turtle]

Based on this

[toxic zephyr]

well what this shows is that if adj(M)=0, then detM is a zero divisor. but that doesn't quite do the whole proof. if you can use that the adj is 0 if rankM<n-1, then you'd still have to cover the rankM=n-1 case i think.

[golden turtle]

If adjM=0 then detM is zero divisor

If it’s not

Then don’t we use your previous argument

That it has a nonzero column

We can scale

And get an element of the kernel

[toxic zephyr]

oh i see what you're saying

but not sure if that covers if adjM is nonzero, and detM is not a zero divisor

i.e. the case when M is full rank

[golden turtle]

Is adjM is non zero it implies T is not injective
Which implies by the other direction that I’ve already completed that detM is a zero divisor

[toxic zephyr]

well for an invertible matrix, adjM is nonzero

[golden turtle]

But aren’t we only using that it is nonzero => a nonzero column => since we can scale this and find a nonzero element of the kernel that T is not injective

I might be getting mixed up about what we’re proving

[toxic zephyr]

only if detM is a zero divisor. that's kind of the point of the proof

[golden turtle]

T injective <=> det M not a zero divisor

Remains to prove
T injective => det M not a zero divisor

Or,
Det M a zero divisor => T not injective

So suppose let det M a zero divisor

Either adjM is 0, and we’re done

Or it’s not, and we find a nonzero element in the kernel by scaling a nonzero column of adjM as you described

Either way we get T not injective

[toxic zephyr]

okay yeah i see what you're saying

my brain isn't working i think. i have classes back to back from 9:30am to 2pm lol

yeah i think that works

[golden turtle]

Thank you for your help your Brain is working fine, I never woudlve got the key part about the nonzero column of adjM

Although I am worried because I’ve only seen 1 proof of this theorem before and it’s much more complicated than this

I’ll write it up later and examine it more closely

[toxic zephyr]

sure

you can use that det(M)=0 implies ker(M) is nontrivial right

[golden turtle]

Is this even true? Couldn’t it just be that M is non surjective

But where did we need to use that

[toxic zephyr]

if adjM=0. then det(M)=0. but then we still need to show not injective. so we need that implication

[golden turtle]

Oh right

Idk if that implication is true

Kinda feels like we worked our way back around into the same problem almost

[toxic zephyr]

lol yeah a bit

not sure what properties you're allowed to use for the determinant

it's been a while so i don't remember how much you can actually say

[kind temple]

do you know about alternating multilinear forms and the wedge product?

[golden turtle]

Unfortunately I do not

[kind temple]

hmm

that is unfortunate

i think there is a way to do it drawing from those concepts

[golden turtle]

This is like a 2nd course in abstract algebra

[kind temple]

you might find this paper useful:
https://www.jstor.org/stable/2303094?seq=3

i believe it offers an elementary approach

[golden turtle]

it seems to have a slightly different result

unless I am missing it

[kind temple]

using the corollary, if the determinant of M is a zero divisor, then M is a zero divisor in R_n, so there is a non-zero matrix X such that MX = 0. now M can’t be injective since it has kernel containing the image of X, which is non-trivial