groups-rings-fields

[tribal pasture]

Oh okay! Yeah that makes sense. Thanks for the answer!

But wait, doesnt (1,x^5) has the order 1?

@hot lake

[hot lake]

no, x^5 is the nontrivial element of H

[tribal pasture]

Oh yep yep. Thanks

[hot lake]

D10 might be isomorphic to D5 x H after all

I would need to write the multiplication table lol

[tribal pasture]

Yeah y^ix^j,x^k -> y^ix^{j+k} gives an isomorphism

[hot lake]

I wouldn't be so sure

the x on the left is the one in D5 that has x^5 = 1 ?

[tribal pasture]

Yes

[hot lake]

and the x on the right has x^10 = 1 ?

then that's not an isomorphism

because (x,1)^5 would be both trivial and nontrivial

I gtg

[tribal pasture]

Okay okay

[barren delta]

I'm reviewing old exams and I can't remember what I did wrong here:

Let $a = (1357)(24)(689)$ be an element of $S_9$

[cloud walrus]

kickpuncher:

[barren delta]

what is the order of the subgroup generated by a

I thought it was the LCM of the lengths of those disjoint cycles...?

which would be lcm(4, 2, 3)=12 but that's incorrect

[eager willow]

hmm, yeah it should be 12

[barren delta]

omg I misread, the points I lost were for the followup "what is the index of this subgroup"

sorry

[eager willow]

9!/12

🙂

[barren delta]

the number of permutations in S_n divided by the order of the subgroup?

lol I don't remember any of this it was so long ago

[eager willow]

by lagrange's theorem, yes

[barren delta]

thanks!

[eager willow]

lagrange's theorem is worth memorizing, very fundamental for understanding index and order

well in context of its proof anyway, not just the statement alone

[barren delta]

I'll go over it, yeah. I remember it more in terms of the order of a subgroup dividing the order of the group.

but forgot about cosets and S_n and stuff

[chilly ocean]

is symmetry the same as isometry in the context of groups?

how does group action relate to symmetry?

does that make sense to ask?

[mild laurel]

I'm not sure exactly what you're trying to ask

Group actions do relate to symmetry yes

[woven delta]

👀

[magic owl]

The difference is that symmetry is like

A vague term

And group action is very concrete

[ripe basalt]

Is there some sort of theorem that talks about when you can relate statements about fields of characteristic p to fields of characteristic 0?

Like some conditions where if you have a statement that says "X is true over fields of characteristic p for arbitrarily large p then it is true for fields of char 0"

[woven delta]

Yes

Well for all but finitely many p

[ripe basalt]

what is it?

yes it doesn't have to be for all p

[woven delta]

Any first order sentence

[ripe basalt]

does this thing have a name so i can look it up

or not really

[woven delta]

It follows from quantifier elimination of ACF

[ripe basalt]

okay

I think that makes sense

[woven delta]

For example the existence of a solution to a particular polynomial equation with coefficients in Q

Oh I'm thinking Algebraically closed fields

[ripe basalt]

yeah

[woven delta]

Not just fields

[ripe basalt]

no

I think

[woven delta]

Sorry

[ripe basalt]

ACF here is fine

i dont think it happens

for fields in general

I should have said that.

[woven delta]

Oh ok

[ripe basalt]

the char 0 field

i guess also is algebraically closed

so this means if you have something true in the algebraic closure of F_p for arbitrarily large p for some p

it will be true in C

as an example

[woven delta]

👀

[ripe basalt]

?

[woven delta]

Yeah C is the quintessential Algebraically closed field

[ripe basalt]

yes

[woven delta]

Of char 0

[ripe basalt]

haha

yes

this comes from quantifier elimination in ACF?

that's neat

[woven delta]

So basically the argument is that any first order sentence reduces to a finite Boolean combination of formulas of the form n=0

For Algebraically closed fields

[chilly ocean]

How can you find generators of a direct product that is not cyclic? Let's say I'm trying to figure out those of $Z_2 \times Z_8^{\times} = {(0_2,1_8),(0_2,3_8),(0_2,5_8),(0_2,7_8),(1_2,1_8),(1_2,3_8),(1_2,5_8),(1_2,7_8)}$, which I know cannot be cyclic because its order is 8 and no element in it has order 8. My initial idea were the elements $(1_2,3_8),(1_2,5_8),(1_2,7_8)$, given how $1_2$ is the generator of $Z_2$ and $3_8,5_8,7_8$ generate $Z_8^{\times}$, but they would only cover themselves and $(0_2,1_8)$. Would the "generators" of this product just be all elements that are not $(0_2,1_8)$?

[cloud walrus]

Aleksandr:

[hot lake]

what ?

generation is not a property of elements of a group, but of subsets of a group

${(1_2,3_8),(1_2,5_8),(1_2,7_8)}$ does generate the whole group

[cloud walrus]

Zef Klop 🍃 🌿 🌻:

[hot lake]

as do many other subsets

but that doesn't mean that (1,3) is a generator

nor that (1,3) (1,5) and (1,7) are the generators

it is just one of many subsets that generate the group

[hot lake]

@tribal pasture D10 is isomorphic to D5 x Z/2Z

[tribal pasture]

Oh and since all 2-groups are isomorphic, it is isomorphic to D5 x {1,x^5}

[hot lake]

yeah there is no difference, but Z/2Z is easier to know what it is if you don't have context

what do you think is the "copy" of D5 inside D10 ?

[tribal pasture]

You mean a subgroup isomorphic to D5?

[hot lake]

yes

in this case the quotient map D10 -> D10/{1,x^5} has a section, but that is NOT ALWAYS THE CASE when you are taking quotients

[tribal pasture]

What is a section?

[hot lake]

it's a right-inverse

G -> G/H has a section when there is a group morphism G/H -> G

such that when you go G/H -> G -> G/H

you get the identity on G/H

so, when that happens, G is a semi-direct product of H and G/H

you can also then "identify" G/H with a subgroup of G

via the section

[tribal pasture]

Ahhhh okay okay. And I presume there is no general rule to figure out when there exists a section or not?

[hot lake]

and here, not only do we have a section, but also the semi direct product is the direct product

and yeah, in general it can be hard to tell if the quotient map has a section

for example, if you go Z/4Z -> Z/4Z / {0;2}

there is no section cuz the only morphisms you can use to go back to Z/4Z sends 1 to 0 or 2

and then you don't get the identity on G/H when you go G/H -> G -> G/H

[tribal pasture]

Hmmm.... I see I see, btw any insight on how can I find out the copy of D5 inside? I would have to write the multiplication table right and all ? No slicker method I presume?

[hot lake]

well the section means that for each class you have to pick an element in that class to represent it ; and you can do so in a way that the elements you chose form a subgroup of G

[tribal pasture]

I see. I guess I will read up on it. Thanks for the help!

[fringe nexus]

Why does F(a) contain F(g(a))?

g is a poly in F[x], f is a irreducible poly in F[x]

[delicate bloom]

g is just a polynomial

once you have a, since you're in F(a) you can make g(a) and anything else you'dlike in F(g(a))

@fringe nexus

[fringe nexus]

oh

right

[oblique pivot]

Prove that the Pauli matrices {σ1,σ2,σ3} don’t form a group.

Do i just have to test the closeness property of groups?

[delicate bloom]

simpler than that, check the group axioms

[oblique pivot]

Closure?

[delicate bloom]

simpler

[oblique pivot]

Inverse?

[tribal pasture]

You would have to check all, closure, inverse and identity, associativity. However, if you find one of them is failed, you are done

[delicate bloom]

warmer

hey don't spoil it @tribal pasture

[oblique pivot]

Cant be unit element

[delicate bloom]

which is the identity?

[tribal pasture]

I mean you already said group axioms

[delicate bloom]

yeah fair lol

[oblique pivot]

U1=1U=1

[delicate bloom]

which one is it σ1,σ2,σ3?

which one of those is the identity

[oblique pivot]

What identity

[delicate bloom]

exactly

[oblique pivot]

?

God im confused

[delicate bloom]

there is no identity

[tribal pasture]

You might wanna check the definition of group @oblique pivot

In its definition, you will encounter a specific definition of identity. Check whether it holds for you.

[oblique pivot]

Oh right

The UU^-1=1

Nvm i read that wrong

[tribal pasture]

Is 1 included in {σ1,σ2,σ3}?

[oblique pivot]

All I can think of is closure

[tribal pasture]

Then you have not grasped the question correctly

[oblique pivot]

Kind of the reason why I was asking

[tribal pasture]

A group is a set which satisfies 4 properties

1. It is non-empty (Can you prove that?)
2. There exists an identity (For all elements x in your set, there exists an element e such that ex=xe = x). Can you prove this for your case? Specifically is there an element σi such that σj σi = σj for all j ?
3. Does there exists an inverse for all elements?
4. Associativity ( I dont think you have to check this because matrices satisfies associativity)
5. Closure. Can you prove this ?

[fringe nexus]

Can i ask another dumb question about sylow theorem proof

[oblique pivot]

The second propriety was definitly not stated in the course, I'll have to look into it, thanks!

But while i'm at it

Could you elaborate on closure?

[tribal pasture]

If I have a,b in my set, then I must have ab in my set, Similarly I should have abb and aab, bab, ... in my set. If you are lucky, they would turn out to be equal to one other.

[oblique pivot]

Well that I got, but what I don't get is the "set" part

What defines a set?

[fringe nexus]

a set is just a group of elements

in your case, you're looking at the set of 3 elements you defined above

[tribal pasture]

If I have σ1, σ2 in {σ1,σ2,σ3} then I should have σ1σ2 in {σ1,σ2,σ3}. In other words, σ1σ2 should be either equal to σ1 or σ2 or σ3

[oblique pivot]

Ok thanks alot!

But since sig1*sig2 doesn't exist in the set, can't we say that it doesnt form a group?

[tribal pasture]

How do you know it doe not exist?

[oblique pivot]

Evaluating it gives i*sig3

Unless the coefficients dont matter?

[tribal pasture]

No, they do matter. Well since iσ3 is not inside, then yes it does not form a group (because it violates the closure property of a group).

[oblique pivot]

Alright, so thats one way of doing it!

But clearly you were hinting that there is a simpler way of doing it

[fringe nexus]

you need an identity element

what is the identity matrix?

is it in your set

[oblique pivot]

1?

1 0
0 1

[fringe nexus]

yea

[tribal pasture]

But they would have to establish that there exists no identity in their set.

[oblique pivot]

Oh well thats way simpler

[fringe nexus]

well

[tribal pasture]

Yeah but it is not rigourous. It would work if you were asked to show it is not a subgroup of invertible complex 2x2 matrices.

[oblique pivot]

invertible?

[tribal pasture]

Admits an inverse

[oblique pivot]

Gotcha

[fringe nexus]

i guess what you have to do is for each element sig_1,sig_2,sig_3 find some multiplication sig_i* x which does not give you sig_i

if you wanted to be rigorous

[oblique pivot]

x as in an element of the set or a real

[tribal pasture]

Yeah which you have already found σ1σ2 = iσ3 ∉ {σ1,σ2,σ3}

[fringe nexus]

element of the set

yea just use the fact its not closed

[oblique pivot]

Cool

Thanks alot!

[fringe nexus]

I have a question about showing any group of order 36 is not simple

36 = 3^2*2^2 so I tried using sylow

There exists a sylow-3-group, call it H with order 9

consider the mapping induced by the group action of G on the left cosets of G/H, which has order 4

This is a homomorphism from G into S_4, and I want to show the kernel is not trivial or the whole group G

if the kernel is trivial we have a contradiction as G is order 36 and S_4 is order 24

but how do i show the kernel is not the whole group?

[mild laurel]

Is the group action you're thinking of just the right/left multiplication action?

[fringe nexus]

yes

[mild laurel]

Then I mean, any element not in H will shift the cosets

[fringe nexus]

where phi(g) = sig_g and sig_g(aH) = gaH

[mild laurel]

since Hg = H iff g \in H

[fringe nexus]

oh

yes

ok

I can't think today

What are some other ways of showing a group is non-simple using sylow?

The only ways I have of doing this are element counting or considering the homomorphism above

[tribal pasture]

Can it not be solved by Sylow?
Since 36 = 3² * 2², we have that there exists a 3-sylow. Now note that the number of its conjugate subgroups L, satisfy
L | 2 and L = 1 (mod 3).
Thus L = 1 <=> 3-sylow is normal subgroup. Since its order is greater than 1 and lesser than 36, we have that it is a proper subgroup.

[fringe nexus]

uhhhh

no

you're wrong because

divisible by 4

not 2

[tribal pasture]

Oh shit...

[fringe nexus]

if |G| = p^{alpha}m

[tribal pasture]

-_|

[fringe nexus]

then n_p has to divide m

i remember trying to do this the other day using element counting

I can't do element counting if the order of the sylow-p groups aren't prime right

or at least not easily without showing their intersections are trivial

[snow bolt]

I'm looking through the classification of finitely generated abelian groups and im trying to work through a pathological example where the abelian group is not finite

Take the group $\frac{\mathbb{Z}^3}{(1,3,1)\mathbb{Z}+(1,7,5)\mathbb{Z}+(1,-1,3)\mathbb{Z}}$. Then we can write the matrix as[
\begin{pmatrix}
1 & 3 & 1 \
1 & 7 & 5 \
1 & -1 & -3 \
\end{pmatrix}
]

[cloud walrus]

potapeno:

[snow bolt]

We already have the gcd in the upper right corner so we can manipulate the matrix to get [
\begin{pmatrix}
1 & 0 & 0 \
0 & -4 & -4 \
0 & 4 & 4 \
\end{pmatrix}
]

[cloud walrus]

potapeno:

[tribal pasture]

Can somebody proof read this claim of showing D_10 is isomorphic to D_5 x {1,x⁵}

[snow bolt]

At this point I'm stuck because the last matrix is the zero matrix. I assume this means that the group is isomorphic to Z^2?

[mild laurel]

Where did you get that this is a pathological example?

[snow bolt]

well its not a pathological example, but for the class we are focused on finite groups and we never talked about this case

we used finitely generated abelian to prove finite abelian

[mild laurel]

Well, it's called the classification of finitely generated abelian groups

Not finite abelian groups

[snow bolt]

no i know

im just checking my reasoning is right

because we never did an (in class) example like this

[mild laurel]

Oh okay, then yeah, it looks fine to me

[snow bolt]

ok sweet

thanks

pathological was probably the wrong word

[mild laurel]

Yeah, there are kind of pathological exampels

like R

[snow bolt]

yeah but i dont even want to think about R yet haha

wait i didnt think about it

how in the world does that work for R

is it just not finitely generated?

yeah it cant be finitely generated

@tribal pasture you state H before defining it in the problem

when you say H \cap K = {e} i have no idea what H is

[mild laurel]

Yeah it's not finitely generated

[uncut girder]

I need help on 6(c)

I say J extends to J^C where J^C(v\otimes 1 + w \otimes i) = J(v)\otimes 1 + J(w)\otimes i

But i dont know how to prove J^C has eigenvalues \pm i

[late snow]

If a group with order divisible by smallest prime c acts on a set, does that set have to have order c?

[bleak abyss]

Nope

[woven delta]

Trivial action

[bleak abyss]

Z/p \times Z/q for p < q

Acts on Z/q

[late snow]

what if non trivial?

[woven delta]

A action is a map into the symmetric group

[bleak abyss]

What you know is that the order of an orbit divides the order of the group

That's about it

[woven delta]

So that should inform these sort of questions

[late snow]

So it's not true?

Could you use sylow theorem in this at all?

[woven delta]

Lmao

[late snow]

Could the size of the set be less than the prime though?

[stark sigil]

@uncut girder Use the fact that $J^2=-I$. So this means that $J(Jv)=-v\cndall v\in V^\bC$. Now if $v\in V^\bC$ is such that $Jv=\lambda v$, what can $\lambda$ be?

[cloud walrus]

Icy001:

[uncut girder]

OHHHH

Thank you

[uncut girder]

So that shows if J^C has an eigen value it must be \pm i

We also know J^C has an eigen value because V^C is complex vector space

Now how do we know both \pm i are eigenvalues of J^C?

@stark sigil

[stark sigil]

I think you can just take the complex conjugate of any vector with one eigenvalue

[uncut girder]

@stark sigil thank you so much

[uncut girder]

One more question

How do I show the map V_J ->V^{1,0}
given by v\mapsto v\otimes 1 -J(v)\otimes I
Is an isomorphism

In, particular, how do you show its subjective?

[stark sigil]

There are a bunch of ways to show that something is an isomorphism of vector spaces
(1) show it's injective and that dim V = dim V'
(2) show it's surjective and that dim V = dim V'
(3) construct the inverse map explicitly

[uncut girder]

It's hard to work with dimensions because you are changing fields and that changes dimensions

[stark sigil]

But changing fields changes dimensions in a very predictable way

$\dim_\bC(V\otimes_\bR\bC)=\dim_\bR V$

[cloud walrus]

Icy001:

[stark sigil]

$\dim_\bR(V\otimes_\bR\bC)=2\dim_\bR V$

[cloud walrus]

Icy001:

[stark sigil]

Ok the inverse map is actually easy as hell

$$v\otimes 1+w\otimes i\mapsto v$$

[cloud walrus]

Icy001:

[stark sigil]

(seen as a map from $V^{1,0}\to V_J$)

[cloud walrus]

Icy001:

[stark sigil]

So (3) is probably the quickest

The only thing you have to check is that the maps are well-defined and send things into the spaces you promised to send them to

[uncut girder]

Showing that's a right inverse involves the same work as showing the original map is surjective

[stark sigil]

I suppose but it's kinda mechanical work

[uncut girder]

No, I think showing surjectivity without counting dimension isnt possible.

How do I show dim_C V^{1,0} = dim_C V_J

[stark sigil]

Eh?

Then the inverse method is easier than what you are trying to do

[uncut girder]

You dont know that it's an inverse

Its easy to see it's a right inverse but its not necessarily a left inverse

[stark sigil]

The only question is whether $V^{1,0}\ni v\otimes 1+w\otimes i\mapsto v\mapsto v\otimes 1+Jv\otimes i$ is the identity, right?

[cloud walrus]

Icy001:

[uncut girder]

Yeah

[stark sigil]

Just work with the definition of the fact that $J(v\otimes 1+w\otimes i)=i(v\otimes 1+w\otimes i)$ and show that it implies that $w=-Jv$

[cloud walrus]

Icy001:

[uncut girder]

Oh I get it now!

Thank you!

[stark sigil]

🙂

[quiet ember]

Currently studying for a final exam, doing old exams. Wondering if my thought process behind this is right.

I think this is a yes, (gonna write my logic here in a bit)

I can write $10000=2^45^4$, so if I can combine 4 prime cyclic group of power 2, and 4 prime cyclic groups of power 5, it would make a total of $22225555$ combinations right?
so, if I choose for example $2^53^55^57^511^213^517^5*19^5$ as the n for $Z_n$
Would that work?

[cloud walrus]

Gall:

I can write $10000=2^4*5^4$, so if I can combine 4 prime cyclic group of power 2, and 4 prime cyclic groups of power 5, it would make a total of $2*2*2*2*5*5*5*5$ combinations right?
so, if I choose for example $2^5*3^5*5^5*7^5*11^2*13^5*17^5*19^5$ as the n for $Z_n$
Would that work?

[mild laurel]

Your logic isn't quite correct

[quiet ember]

aaah, thought so
What did I miss? :3

[mild laurel]

Think about how many abelian groups there are of order 2^5 say

[quiet ember]

aaah, yea. They aren't exactly 5,
would 2^4 create 5 then?

[snow bolt]

yes

1,1,1,1
1,1,2
1,3
4
2,2

do you get the notation here?

[quiet ember]

Yea, I get it :3

[snow bolt]

partitions of integers are gross

[quiet ember]

so, if I instead chose for example $2^43^45^47^411^213^217^2*19^2$
Would that work, or am I still missing the point? :3

[cloud walrus]

Gall:

[snow bolt]

thats looks good to me, but ill let galois have final say

[quiet ember]

Thanks 😄

[mild laurel]

yeah it should work

[quiet ember]

Sweet
Thanks alot 😄

[chilly ocean]

Super fun math equation

Whats n if 0n = 0?

Lets hear some answers

[stone fulcrum]

7

You'd be far better in #chill with this though

[chilly ocean]

n = undefined

[sonic parcel]

n takes an indeterminate form

[visual turret]

I am trying to find the isomorphism of Z9xZ9/<(3,3)> in terms of direct products of cyclic groups

I know it has to be Z3 x Z3 x Z3, Z9 x Z3, or Z27, but I'm not sure how to decide which

[stark sigil]

You can rule out $\bZ/27\bZ$ because there are clearly no elements of order 27

[cloud walrus]

Icy001:

[stark sigil]

Hm, another fact is that $(1,0)$ still has order 9 even in the quotient, therefore your group is not $(\bZ/3\bZ)^3$

[cloud walrus]

Icy001:

[visual turret]

That makes a lot of sense

thanks

[fringe nexus]

anyone has a hint for the cyclic part?

I can prove it has a subgroup of order 15 and since the index is 2 it has to be normal

oh wait all groups of order pq are cyclic

[chilly ocean]

how do I check if a number is reducible in a ring? I remember the teacher checking the norm of the number but was quite confused there

show that rsr^-1 doesnt belong to it

[dawn kiln]

also

use the presentation of the group as D_n = < r, s: r^n = s^2 = 1, and rs = sr^-1>

so r is a rotation and s is a reflection

ok, so using the fact that rs = sr^-1 (draw a diagram to see this)

and considering <s> = {1, s} (a flip is its own inverse)

rsr^-1 = sr^-1r^-1 = sr^-2, which is not in <s>

[chilly ocean]

You need to show that there exists at least 1 element in D_5 such that xbx^-1 doest belong to <b> where x in D_5

If you look how normal subgroup is defined

and there aren't many possibilites for that

yeah

jsut write x as r^k for any k in {1,2,..,n-1}

I mean its kinda obvious, you'll get r^k for some k if you conjugate by rotation, and it is in <r>

just need to show what happens if you do srs

because s=s^-1

rs = sr^-1 is the relation in the group

and the second one being s^2 = 1

you can use these two to compute any element from D_n

elst acll it <r> (for rotation)

lets call it*

you need to check that for any x in D_5 xrx^-1 is in <r>

check what happens if x=s and x=r^k

[dawn kiln]

are you confused over why the condition $ghg^{-1} \in H$ is equivalent to H being normal in G works?

[chilly ocean]

for k any natural number

[cloud walrus]

maximwebb:

[chilly ocean]

oh yeah maybe you had a "different" definition

maybe you had it as gH = Hg for any g in G

which is equivalent

gH = Hg <=> gHg^-1 = H <=> for any h in H, for any g in G ghg^-1 belongs to H

It's usually just easier to check that definition

[dawn kiln]

i mean you don't really need to do the case of r^k<r>r^{-k}, as its trivial

so just do the case for s<r>s^-1

[chilly ocean]

maximwebb do you know a bit of rings?

[dawn kiln]

a bit

[chilly ocean]

I have to show what R/J is isomophic to, where R is the ring of continous functions from [0,1] to mathbbR and J is an ideal of R such that J={f in R : f(1) = 0}

Im not sure whats R/J I think

Its the set of x+ I for x in R I guess?

[dawn kiln]

is it just isomorphic to R?

[chilly ocean]

I dont know

[dawn kiln]

so all the functions f passing through 0 at x = 1 form the kernel of the homomorphism from R -> R/J

[chilly ocean]

yep

[dawn kiln]

I think any function not in J is determined solely by its value at x = 1

[chilly ocean]

ok yeah actually true

because a -b is in J if a and b functions have the same value at 1

[dawn kiln]

yep

[chilly ocean]

so they make the same equivalence class

cool

[dawn kiln]

so i believe R is your isomorphic structure

[chilly ocean]

yep thanks

[dawn kiln]

nw

[chilly ocean]

btw, are there any zero divisors in J?

There aren't right?

the only possibilities would be functions that are equal to 0 everywhere but f(1) but that wouldn't be continuous

[delicate bloom]

that doesn't sound right

my reasoning would be to look at f(1)*g(1)=0 but neither f(1) nor g(1) are 0

that would be a zero divisor

but since they're real numbers, no such case occurs

[chilly ocean]

Oh wait i got confused, they want zero divisors of just R being the ring or continuous functions

My guess would be its functions where f(x)=0 for uncountably many x

[delicate bloom]

I don't understand what you're saying

I think you might not understand what R and J are

[chilly ocean]

Im wondering what are the zero divisiors of the ring R which is made of all continous functionf from [0,1] to mathbbR

[delicate bloom]

oh the zero divisors of R

[chilly ocean]

yep, and im thinkin, as said above, its functions that are 0 for uncoutably many arguments

hand wavy style

I'll try to show it

[delicate bloom]

sure sounds good

[stark sigil]

This sounds hard

what kinds of sets can $f^{-1}(0)$ be

[cloud walrus]

Icy001:

[delicate bloom]

well they're continuous on a compact set so all of the places where it's 0 will have to be closed sets

and those functions can be multiplied with another which is 0 on the complement of that

I'm really just imagining all the f that are zero divisors to be 0 on some interval

[stark sigil]

There are uncountable sets that don't contain an interval

uncountable and closed

Cantor set 👀

Does there exist a continuous function whose set of roots is exactly the Cantor set?

Seems like the answer should be no

[hot lake]

is it closed ?

[stark sigil]

wtf

[hot lake]

how about f(x) = distance from x to the cantor set

[stark sigil]

The answer is yes

[delicate bloom]

I guess we could just imagine like any continuous fractal curve true

[stark sigil]

https://mathoverflow.net/a/24041/68830

[hot lake]

a zero set can be any closed set

[delicate bloom]

seems like it'd just have to have infinitely many points of intersection

uncountably many and disjoint

[stark sigil]

However

The closure of the complement of the cantor set

is all of $[0,1]$, yeah?

[cloud walrus]

Icy001:

[stark sigil]

So the function $x\mapsto $distance from $x$ to the Cantor set isn't a zero divisor

[cloud walrus]

Icy001:

[stark sigil]

because the only continuous function that multiplies with it to get 0 is the zero function

I knew the question of zero divisors was gonna be a somewhat tricky question

[tribal pasture]

I think they meant f(x) = d(x,C) where C is cantor

[stark sigil]

Actually

Since all functions are continuous, the answer is trivial

😎 ...

[tribal pasture]

o.o

[dawn kiln]

yeah, like i bet you couldn't draw me a nowhere-continuous function

[stark sigil]

If you can't draw it without lifting your pencil, it's not a function

[tribal pasture]

https://math.stackexchange.com/questions/176279/all-real-functions-are-continuous

[dawn kiln]

^

[stark sigil]

Just give $\bR$ the discrete topology

[cloud walrus]

Icy001:

[tribal pasture]

wdym?

all spaces are discrete

[stark sigil]

Is $\prod_{n\in\bN}\bR$ discrete?

[cloud walrus]

Icy001:

All functions are continuous:

[stark sigil]

What's $\bR_{\text{std}}$? Doesn't that not exist, since you said all spaces are discrete?

[cloud walrus]

Icy001:

[tribal pasture]

oh then yes, trivially :3

The proof of the triviality is left as an exercise for the reader

[stark sigil]

-snicker- box topology

[tribal pasture]

If R is discrete, isnt then the product of R also discrete?

Imagine procrastinating on preparation of a topology exam, by talking about topology on discord server

fml

[chilly ocean]

yes countable product of discrete spaces will also be discrete

[stark sigil]

In the real world, no

[chilly ocean]

What is real though?

[stark sigil]

outside the world of "all spaces are discrete"

a world in which the product topology has as basis of open sets $\prod_{i\in\bN} U_i$ where $U_i$ are open and all but finitely many of them are the entire set

[cloud walrus]

Icy001:

[chilly ocean]

well then why wouldnt it be discrete

[stark sigil]

Because $\prod_{i\in\bN} U_i$ where $U_i\subsetneq X_i$ are all proper open subsets isn't open

[cloud walrus]

Icy001:

[chilly ocean]

idk

[dawn kiln]

get this analysis outta here 😩

[stark sigil]

-mumble- profinite groups -mumble

[tribal pasture]

Isnt it the case that the product topology of a countable collection of discrete spaces is discrete?

[stark sigil]

Nope

Easy counterexample is to take a countable product of finite sets

Individual points in the product aren't open

[stone forum]

finite sets with more than one element each

[stark sigil]

Well, all sets are finite so we might as well take $\bR$ for all of them

[cloud walrus]

Icy001:

[tribal pasture]

True... True.... Yep you are right.

Box topology is the only true product topology 0.0

[stark sigil]

All functions are continuous, all spaces are discrete, and all sets are finite

The 3 most important maxims in math

[fringe nexus]

If I want to find the galois group of x^8-1 over F_3, how would i go about doing this?

I figured out there are two roots in F_3, 2 and 1

so the polynomial is reducible, and the irreducible part is of at most 6

so is the extension F_3^{6} and thus the galois group is generated by the frobenius automorphism so it is Z_6?

[tribal pasture]

Dont forget the most vital: All answers are correct

[fringe nexus]

(assuming i did the work and figured out that it was indeed irreducible of degree 6)

(also is there an easy way of factoring polynomials in finite fields)

[mild laurel]

what

[dawn kiln]

the famous Z multiplicative group of order 20

[mild laurel]

Then use math

Do you mean (Z/20Z)*?

[dawn kiln]

do you mean $ (\mathbb{Z}_{20}, \times) $

[cloud walrus]

maximwebb:

[mild laurel]

No, you need to remove more than just 0

[dawn kiln]

doesn't form a group

ah right

[mild laurel]

And you're asking about cyclic subgroups?

Cyclic groups are by definition

[stark sigil]

are the only cyclic subgroups of the Z multiplicative group of order 20 the subgroups generated by the elements of the set?

[mild laurel]

Generated by a single element

[stark sigil]

Was gonna say, tautology

[mild laurel]

If I understand what you're asking at least

Are those all distinct?

[stoic dirge]

Making sure there weren't other cyclic subgroups (Somehow) that I was missing

[dawn kiln]

btw

the common way of stating "group of integers coprime to n under multiplication mod n" is the "group of units of Z_n"

[stoic dirge]

Oh ok. Sorry!

[dawn kiln]

and the way i've seen of denoting this is U(n)

dunno if that's standard

[stoic dirge]

Thanks lol. It's hard to search online for proper math terminology

[dawn kiln]

http://abstract.pugetsound.edu/download/aata-20170805.pdf

this is what i've used

its quite dense, but its good

[stoic dirge]

Thanks! This semester was my first ever abstract algebra class so the material is still fairly new to me

[dawn kiln]

fair enough, what did you think of the course?

[stoic dirge]

Loved it!

[dawn kiln]

glad to hear 😄

[stoic dirge]

Made me realize how happy I am that I picked the math major lol

[chilly ocean]

So on the topic of units i got an easy challenge: Show that a zero divisor cannot be a unit

assume it is

[dawn kiln]

qed

[chilly ocean]

$\Box$

[cloud walrus]

Godel:

[dawn kiln]

@stoic dirge i'm from the uk, how does the system work for you

[chilly ocean]

Nah for real, I remember it was a one line proof

[dawn kiln]

as in like which year of college

[stoic dirge]

I'm a junior. So basically before this it's been electives and analysis-related classes. Also computer science

[dawn kiln]

and a junior?

[stoic dirge]

3rd year

It's all math courses from hear on out! (Finally)

[chilly ocean]

CS kills all the joy I have for math

[dawn kiln]

oh damn, takes a long time to get to the juicy stuff lol

[chilly ocean]

@chilly ocean assume x nonzero and ax=0 where a zero divisor: x=1x=(a^−1a)x=a^−1(ax)=a^−10=0

[dawn kiln]

@chilly ocean lol i do CS, don't say that 😩

[chilly ocean]

@chilly ocean i did it nearly the same

[stoic dirge]

Yeah. Bunch of unrelated required classes before the final years

[chilly ocean]

Ohh I thought you are asking for help. What was your way?

[dawn kiln]

very glad i don't have to do that here

[chilly ocean]

As I said easy challenge

My way was basically the same but a little more rigorous with ring axioms

[stoic dirge]

Yeah the American education system sucks

[chilly ocean]

True

Never been there but ours does too

[dawn kiln]

are you uk N/U

[chilly ocean]

No

An that's an A

[dawn kiln]

nah that's a U

[chilly ocean]

tahts a U for sure

Google it coward

nah we have eyes

[dawn kiln]

give me 1 sec

[chilly ocean]

[dawn kiln]

ripo

i always assumed it was U, as i saw it in the context of an undecidable turing machine

[chilly ocean]

No do you think maff people are that creative

It's A because that's what the alphabet starts with

[dawn kiln]

i mean, i think i was thinking of this thing

[chilly ocean]

That's an U

[dawn kiln]

as in the thing he uses to allude to the problem being undecidable

doesn't matter, but yeah

[tired hedge]

Hi i have trouble understrandfing conjugacy element. Can someboidy help

[chilly ocean]

Please elaborate

[tired hedge]

I do not underdtand the definition : two elements a en b are conjugate iff. B = gag inverse

[hot lake]

you're missing an "there exists g in G" somewhere

[tired hedge]

Indeed

[chilly ocean]

What do you think you do not understand about this definition?

[tired hedge]

Why this is defined the way it is. I cannot accept definitions without undersdtandfing. The more that there is some calculation involved

I see some simularity with matrix calculation. But there it stops

[mild laurel]

It's defined this way because its a useful idea

[chilly ocean]

It can help you find connections between group elements

[delicate bloom]

you can use it as an equivalence relation

[tired hedge]

Yes but this is a consequence of the definition i gave above....

[chilly ocean]

For example each group element belongs to exactly one conjugacy class, that means you can distinguish the rest of the group from this one class and for example start constructing these classes to be disjointed subsets of the group.
I think a use is the proof for lagrange's theorem

[tired hedge]

Again i know. But this is after one defines the conjugacy element

[mild laurel]

So you're asking what leads you to define this?

[tired hedge]

Yes

[chilly ocean]

Like any definition, it can help you take a shortcut when constructing theorems. Like for example you should establish what "continuous" means before formulating the IVT

[mild laurel]

Maybe the better answer is that this idea of g^{-1}ag comes up a lot

[chilly ocean]

So when this equivalence relation is brought up more often you don't have to take that very long definition every time

[mild laurel]

This conjugation gives you a group action from a group onto itself

Conjugation is also how you define normal subgroups

There are other things, like how some things stay invariant under conjugation

[chilly ocean]

Is this the first time this has come up in your course?

[mild laurel]

Like the reason you talk about similar matrices, e.g. matrices in the same conjugacy class, is because similar matrices share a lot of properties

Same trace, same determinant etc

[tired hedge]

I know but this does not satisfy me. I want to have a explanation. As i wrote above. It is some kind of calculation. So where does it comes from? I cannot accept random definitions .

[mild laurel]

I'm really not sure what you're looking for

I gave plenty of examples where this calculation comes up

[tired hedge]

I know most people do:(

[delicate bloom]

if you take the kind of linear algebra sort of perspective you can imagine it as like "I change coordinate systems, use the operator in this coordinate system, then I change back to the original coordinate system I was in"

if that's kind of what you're looking for

I'm not sure either

[chilly ocean]

The group of matrices over $\mbb R$ that are $n\cross n$ are invertable. Two matrices $A$ and $B$ are conjugate $\iff \exists G: A=GBG^{-1}$

[cloud walrus]

N/𝔄:

[mild laurel]

I don't think any of us know what you're looking for

[chilly ocean]

I think that is what you were referring to at the beginning

[delicate bloom]

I say don't fixate on conjugation, focus on what it's being used for to see why it matters

it's kind of like saying "why do we have + and *" and refusing to move on to learn about numbers

it's hard to have perspective when you're focused on one small detail too closely

[tired hedge]

Well go for weblog of gowers's and read the item of normal subgroups. You will see that defining something has a concrete motivation .and that is my focus.

[delicate bloom]

but we already said those concrete motivations earlier

[tired hedge]

No you gave examples after defining conjugacy in my opinion

[delicate bloom]

play around with several different small finite groups anda try to find symmetries of it AKA automorphisms of your groups

discover the inner automorphism group by yourself

[chilly ocean]

Finding just the right definitions has been a long process. Like defining the real numbers. We don't go backwards and start by saying "it is a continuum" and let the studends figure it out, we start by defining it from the ground up. You'll see the motivation when you have to apply it

[mild laurel]

I gave examples where you don't need to define conjugation

The example of normal subgroups is something that you study at the very beginning of algebra

and shows why the conjugation action is important

[tired hedge]

Sorry. Now i'm lost what is the connection betwee conjugacy and inner automorpism

[delicate bloom]

yeah I can't think of anything better than that, if you have a group it's very natural to try to think "can I decompose this into smaller groups?"

[mild laurel]

The conjugacy map from a group to itself is an automorphism

Fix some element h in your group, the map from G to G that sends an element g to h^{-1}gh is an automorphism of your group

These are not the only automorphisms, but they form a normal subgroup in the automorphism group of a group

[tired hedge]

Finding definitions is indeed a long proces but because of this long proces it gets very abstract . truly understanding them an is my focus. I m a belgian amateur , not a student so i have plenty of time .

[delicate bloom]

are you following a book or lectures?

[mild laurel]

Well

1. I don't see why the idea of normal subgroup isn't appealing to you

That's basically the first thing people studying groups would find

And the action of conjugation is used there

and 2) You should get used to just accepting definitions without knowing why they're useful

Looking at the properties of the definition, or what the definition helps you do for other things gives motivation for why we define the thing at all

[tribal pasture]

^ Most of the time you will learn the usefulness of a definition after you have learned the most concrete results due to that definition

[dawn kiln]

its nice to be able to intuit everything and see that it's all motivated, but if you're doing algebra, most definitions take a while to become apparent

i mean the usefulness of groups themselves for example, as the study of symmetry

[tribal pasture]

I guess one motivation might be that we want to see whether our usual intuitions over the elementary stuff hold true or not. While doing elementary algebra, we have that gag⁻¹ = a.
Now to see how much our new constructions satisfy this property, we can explicitly study the relation between gag⁻¹ and a. If the usual property is satisfied, we get the amazing abelian structure.
We can further study what exactly was the "flaw" in the structure which broke the property. For that we invent the concept of a conjugacy class.
I am not sure whether this coincides with the historical progression of the idea, or even if its completely consistent, but probably a good way to think about these definitions is to relate it back to our usual intuitions to elementary algebra. @tired hedge

[fringe nexus]

This is trivial from the fact that all subgroups of abelian groups are normal and normal subgroups correspond to galois extensions right?

[mild laurel]

yes

[fringe nexus]

why are some questions so easy

on the qual

[mild laurel]

It took me like an hour to see this when I first got this problem lmao

[fringe nexus]

and some are just impossible

[mild laurel]

So if S is a subring of R, and I is an ideal of R, then let IS be the ideal of S generated by the elements of I \cap S

Apparently there's a natural ring homomorphism from R/I to S/IS

But I'm not seeing what this is

Is there a natural ring homomorphism from a ring to its subring?

[woven delta]

No

[mild laurel]

But then what's the natural ring homomorphism from R/I to S/IS

[woven delta]

Shit

Yeah, there is

Wait no

Only if the subring is also an ideal

(which isn't a thing assuming rings with identity)

[mild laurel]

these rings have identities, so

[woven delta]

A map from a ring to a subring would give you an R Algebra structure on the subring

So that can't be a thing

For example Q is a subring of R, but it is clearly not an R Algebra

Cause it's not an R vector space

[mild laurel]

oh actually fuck me

I read it in the wrong order

R is a subring of S

[woven delta]

Lmao

[mild laurel]

I'm so dumb

[woven delta]

Smh

That's pretty dumb

[late sedge]

can residue class and equivalency class be used interchangabily

in modular arithmetic

[stark sigil]

yes

but it's called equivalence class

[late sedge]

cool 👍

yea i was watching a video and he used residue class

I usually say equivalency class

partitions the group

[stark sigil]

I'm just correcting your spelling

[late sedge]

does the kernel of a ring homomorphism always map to {0} in the range

[stark sigil]

What was your definition of kernel

[late sedge]

The kernel of a ring homomorphism f:R-->S is the set of all elements of R which are mapped to zero. It is the kernel of f as a homomorphism of additive groups. It is an ideal of R.

[stark sigil]

So yes?

[late sedge]

im tryna understand this example tho

how does the ker pi equal the multiples of n

i thought kernels of rings automatically map to 0

[stark sigil]

$\ker\pi$ lives in the domain

[cloud walrus]

Icy001:

[late sedge]

but doesnt it equal 0 in the range

so ker(pi) = 0

[stark sigil]

multiples of n do map to 0 in the range

[late sedge]

yes if 0

but my question is i just dont get why ker pi = nZ

[stark sigil]

No, all multiples of n map to 0 in the range lol

$n\mapsto 0$, $2n\mapsto 0$, and so on

[cloud walrus]

Icy001:

[stark sigil]

$\pi(kn)=0\cndall k\in\bZ$

[cloud walrus]

Icy001:

[stark sigil]

because $\bZ/n$ is the ring of integers mod $n$

[cloud walrus]

Icy001:

[stark sigil]

and $\pi$ is the map that takes an integer to its residue class mod $n$

[cloud walrus]

Icy001:

[late sedge]

still doesnt make sense

i thought the kernel was pretty much a mapping that takes any element in domain to zero in range

[stark sigil]

No 😮 the kernel is a subset of the domain

[late sedge]

i dont get how the kernel of $\pi$ is in nZ

[cloud walrus]

Mortex:

[stark sigil]

$\ker\pi$ is a subset of $\bZ$

[cloud walrus]

Icy001:

[stark sigil]

It's the subset consisting of the elements $x$ of $\bZ$ such that $\pi(x)=0$

[cloud walrus]

Icy001:

[late sedge]

oh ok

see that was my problem

I was thinking of the kernel as a transformation map

not a subset

[stark sigil]

well you wrote down the definition yourself

https://cdn.discordapp.com/attachments/496784958430380033/654174156278988813/ca2d582aeeb032fbe3ac8914768e5fa60265dd15.png

This denotes exactly what I said

as well as here

The kernel of a ring homomorphism f:R-->S is the set of all elements of R which are mapped to zero.

Do you have analysis experience? Kernel in analysis does mean a kind of function

or machine learning

[late sedge]

i took real analysis but I never used kernels

[stark sigil]

kernel there means a mapping from a low-dimensional space to a high-dimensional space

[late sedge]

i only remember using kernels in linear algebra

and it was always thought of as a map to 0

or the null space

[stark sigil]

That doesn't make sense at all

how did you do linear algebra with that idea

of kernel

[late sedge]

Kernel (linear algebra) ... In mathematics, more specifically in linear algebra and functional analysis, the kernel of a linear mapping, also known as null space or nullspace, is the set of vectors in the domain of the mapping which maps to the zero vector.

whats wrong about it?

[stark sigil]

"the set of vectors WHICH map to the zero vector"

The wording says that it's a set of vectors

which set is it? What vectors belong to the set?

[late sedge]

dude i took it like 2 years ago gimme a break 😅

[stark sigil]

v belongs to the set if v maps to 0

Maybe you read it as

"[the kernel is the set of vectors in the domain of the mapping], which maps to the zero vector"

[late sedge]

so the part I dont get is if a kernel is a ring homomorphism, then it's an ideal of ring R

[stark sigil]

Maybe now that you have the right mental concept of what a kernel is, you can tackle it again

[late sedge]

so if you can construct a homomorphism within that set of elements that define the kernel then it's an ideal

ok

[visual turret]

I'm having trouble showing that that for all odd n >=3 we have D2n = Dn x Z2. I've been trying to find two normal subgroups of D2n that are iso. to Dn and Z2 with A intersect B = {e} and A*B=D2n

[surreal hawk]

If D_(2n) = {e, r,...,r^(2n-1), s, rs,...,r^(2n-1)s} then try looking at the subgroup generated by r^2 and s 👀

@visual turret

[tribal pasture]

I am getting countably infinite (|Z|) for the number of conjugates. Can somebody perform a quick sanity check?

[elder valley]

``If $p \equiv 7 \mod 8$, then the prime $2$ splits in $\mathbb{Q}(\sqrt{-p})$''

is this saying that the ideal $(2)$ factors into a product of ideals nontrivially in the ring of integers of $\mathbb{Q}(\sqrt{-p})$??

[cloud walrus]

Auvera:

Auvera:

[mild laurel]

Product of prime ideals

[elder valley]

is the factorization easy to see or does it depend heavily on p?

[fringe nexus]

If we have an abelian extension of Q is it necessarily cyclic by kronecker weber?

or am i wrong

abelian extension of degree n

[stark sigil]

No, it's just contained in one

[fringe nexus]

when we say contained

do we mean subgroup

[stark sigil]

subfield

[fringe nexus]

oh wait what im confused

so kronecker weber states that any finite abelian extension is contained in a cyclotomic extension of Q right

does that mean the galois group of our finite abelian extension is a subgroup of the galois group of a cyclotomic extension

[stark sigil]

Yes

[fringe nexus]

Doesn't that imply our galois group is cyclic then?

[stark sigil]

And the galois group of a cyclotomic extension is the group of units of $\bZ/n\bZ$ for some $n$

[cloud walrus]

Icy001:

[fringe nexus]

oh wait the group of units isn't necessarily cyclic

[stark sigil]

😎

[fringe nexus]

I got confused with extensions over finite fields

thanks!

[mild laurel]

@elder valley well, you have a condition on p

[stark sigil]

As a sanity check, $\operatorname{Gal}((\bQ[x]/(x^4+1))/\bQ[x])\cong\bZ/2\bZ\times\bZ/2\bZ$

[cloud walrus]

Icy001:

[fringe nexus]

How would you determine the degree of the splitting field of x^3-x+4 over Q

I think its 6 because I can compute the galois group using discriminants but there has to be another way

I don't think I could possibly remember the discriminant of a cubic ,quartic, and quadratic polynomial

[tribal pasture]

Is the following true?

If A is not isomorphic to B then A x X is not isomorphic to B x X? Seems quite true

Perhaps by using the quotient with X but not sure

[stark sigil]

Depends on the category

What if $X$ is the empty set in the category of sets

[cloud walrus]

Icy001:

[stark sigil]

Then they're isomorphic

Actually you don't even need to invoke the empty set

$[1]\not\cong [2]$ but $[1]\times\bZ\cong[2]\times\bZ$

[cloud walrus]

Icy001:

[stark sigil]

Using $[n]$ to denote ${1,2,\dots,n}$ here

[cloud walrus]

Icy001:

[stark sigil]

In the category of finite groups it seems to be true

https://math.stackexchange.com/questions/2193/cancellation-of-direct-products

and finitely generated (not necessarily finite) abelian groups

For other categories the proofs can get a lot harder

[tribal pasture]

What about for groups ?@stark sigil

[stark sigil]

Don't know the answer to that and it might be very hard to prove if it's true and

if it's false, there might be an easy example

[tribal pasture]

Is there no result like X ≅ Y <=> X x A ≅ Y x A?

[mild laurel]

That's true, but that's the converse statement, so it doesn't really matter

It's probably easiest to look at the contrapositive

If A x X is isomorphic to B x X, then A is isomorphic to B

This is a false statement

[tribal pasture]

Then how can I argue that C4 x C2 is not isomorphic to C2 x C2 xC2

I suppose in this I can because C2 is normal a subgroup in these groups?

[stark sigil]

well those live in the category of finite abelian groups so technically you can use the result for there

but there are easier direct proofs

[mild laurel]

One has an element of order 4, one doesn't

[tribal pasture]

true...

[sharp sonnet]

there is also fundamental theorem for finite abelian groups

which tells you that

[smoky briar]

that's circular

you use things like this to classify finite ab groups

[snow bolt]

nuke it

thats how we solve things

[sharp sonnet]

well, you use C_ab isomorphic to C_a x C_b iff a and b are coprime

[snow bolt]

yeah chinese remainder theorem

[sharp sonnet]

hm, actually never checked a proof for the fundamental theorem of finite abelian groups, my class did not cover it

😦

[chilly ocean]

Good luck on finals everyone <3

ftfgag is a pretty thicc boy, it's in chapter 5 of d&f and has a lot of moving parts

[tribal pasture]

Can someone motivate this definition? I dont see why r(A) is not just the same as A?

[mild laurel]

consider the ideal (4) in Z

2 is in r((4)), but 2 isn't in (4)

[tribal pasture]

Ahhh okay Is there any name for this set?

[mild laurel]

the radical of the ideal

which is why, like in my nickname, some people write $\sqrt{I}$

[cloud walrus]

Zopherus:

[mild laurel]

probably most people write this

[stark sigil]

$\mathscr{I(V}(I))=\sqrt I$

[tribal pasture]

Ahhh okay okay. Thanks!

[cloud walrus]

Icy001:

[stark sigil]

{I(V} triggers me though

$\mathscr{I(V}(I)\mathscr{)}=\sqrt{I}$

[cloud walrus]

Icy001:

[stark sigil]

that's better

[mild laurel]

I had to do it

To fit the 32 character limit

\mathscr doesn't change parenthesis cmon

[stark sigil]

😰

[tribal pasture]

And I suppose the addition of two ideals is the set-theoretic sum? That is if I and J are two ideals then I+J = {x+y | x in I and y in J}?

[mild laurel]

Yes, check that this is actually an ideal

[tribal pasture]

One more sanity check r(r(I))= r(I)?

[mild laurel]

yep

[tribal pasture]

Would this be correct?

Secondly, how is the power of an ideal defined? I tried searching up, could not find any quick resources.

[stark sigil]

power of an ideal is repeated multiplication

[chilly ocean]

I have a trouble understandind the proof that nZ has all subgroups of Z. Can anyone help me with that?

[solar vessel]

what don't you understand specifically

[chilly ocean]

https://math.stackexchange.com/questions/1263673/prove-that-all-the-subgroups-of-mathbbz-are-of-the-form-n-mathbbz

the part when he says clearly nZ is a subgroup of H

Or is there are better proof?

[solar vessel]

subset

[chilly ocean]

oh okay sorry, my bad but I still don't understand why?

[solar vessel]

but well doesn't make much difference I guess

well if n is in H

clearly nm is in H for every integer m

[chilly ocean]

clearly nm is in H for every integer m

can't this happen even if this is not the case

a group only need to have identity and inveerse

[solar vessel]

can't what happen

[chilly ocean]

so just having inly -n is ennough

??

[solar vessel]

do you know what a subgroup is

[chilly ocean]

yes

[solar vessel]

n+n has to be in H as well

then n+(n+n) and so on

[chilly ocean]

oh

okay got it

thanks!

[solar vessel]

uh

np

[chilly ocean]

okay, similarly if k is in h, then k-qn = k - (n+n+...+n) has to be in group as well

sorry for troubling you with something so trivial!!

[solar vessel]

np

[chilly ocean]

Hey, if anyone's here, I had one small doubt

actually,it's more like i need confirmation that my thinking is correct

as Z is a cyclic group

all of it's subgroups also be cyclic

now, if it's subgroups are cyclic, they'll be of form nZ

where generator and least element is n

n is element of N

this proves that all subgroups of Z are of form nZ?

am I correct?

[sharp sonnet]

with n being a generator, -n is as well

also nZ is infinite, so there is no least element?

[teal perch]

you can have an infinite set with a least element

[sharp sonnet]

well, ok, but nZ has no least element, at least not with the ordering from Z

[teal perch]

but yeah i would just say any subgroup must be cyclic, and any generator in Z just gives you an nZ

[chilly ocean]

It has, according to Dummit Foote.....

[teal perch]

with some overlap for -nZ=nZ like @sharp sonnet mentioned

[chilly ocean]

also am I right i my conclusion then....?

[sharp sonnet]

n is the least positive element in nZ

that you can say

[chilly ocean]

they are considering only positive generators, so no problem with overlap

[teal perch]

yeah your conclusion is right

just the bit abt the least elt doesnt sit right

bc Z isnt well ordered

[chilly ocean]

oh okay

what if I say that generators will only be taken from positive integers?

[teal perch]

well 2Z is a positive generator

but 2Z has -2, -4, ....

[chilly ocean]

So it is still not well ordered...oh okay I see

[sharp sonnet]

you can define a well-ordering on it

but it wouldn't be the normal $\leq$

[cloud walrus]

Lochverstärker:

[chilly ocean]

got it, thanks!

So in my argument, I should remove that n is the least element, and just say that it's a generator, then it would be correct?

[sharp sonnet]

you need n to be non-negative instead of positive as well

to get the trivial subgroup {0}

then it's fine

[tribal pasture]

Why is this true? Why would the sum of ideals contain terms like a^{p-1}b etc?

[mild laurel]

Because one/both of the ideals contain that term

[tribal pasture]

Yeah thats what I ask. How so?

I would atleast need that there exists an inverse of "a" and "b" in the ring, for this to hold true but I am not seeing any guarantee for that statement to be true

[mild laurel]

Oh, the right answer to your question is that the binomial expansion of that doesn't contain a term like a^{p-1}b

[tribal pasture]

Apologies, but I dont see why. For example, the first term of the expression is a^{p+q-1}. We know that a^p is in A but have no reason to suggest that a^{q-1} is in A as well where A is the ideal.

[mild laurel]

You don't need a^{q-1} to be in A

Look at the definition of an ideal

[tribal pasture]

I would need a^p in A implies a in R atleast somehow, no?

[mild laurel]

Nope

[tribal pasture]

Then not sure how would I generate the remaining powers of a inside I+J

[mild laurel]

If a^p is in I

why is any higher power of a, also in I

[tribal pasture]

a^p in I implies a^p in R which implies a^pa^p in I which implies a^np in I. I see this for only powers of p and not for say p+q-1

[mild laurel]

Okay, what's R

[tribal pasture]

R is a ring, but I am not sure if I can say that a^p in R implies a in R (which will be enough to conclude the remaining)

[mild laurel]

R is the ring that contains all the elements you care about

Like implicitly, you know that a is an element of R

[tribal pasture]

Yep that is the part that is confusing me as for how can I go from a^p in R to a in R

[mild laurel]

Well a has to be in some ring

[tribal pasture]

So I am allowed to conclude this then right? (Say merely from the syntax of a^p)

[mild laurel]

I mean, I guess this theorem is stated pretty poorly

Should say that a,b \in R, and \mathfrak{a}, \mathfrak{b} are ideals of R

such that a^p \in \mathfrak{a} etc

[tribal pasture]

Yeah I suppose

[fringe nexus]

How do i determine what ideals are maximal and prime in the quotient ring Z[x]/(X^2-2)

My ring theory is super weak

Like consider the ideals (7), (2),(0)

To show prime we have to show if ab in (I) then a or b in (I)

But I don't even know how to describe the quotient properly

i think we set x^2 =2 in the quotient ring?

so is the ideal (2) = (x^2) in Z[x]?

[mild laurel]

The correspondence theorems help here

I.e, ideals in the quotient ring have a one to one correspondence with ideals containing (x^2-2) in the original ring

Furthermore, this correspondence preserves primeness and maximality

[fringe nexus]

So I would find these corresponding ideals, can you give me some kind of example on how to find this? I know Z[x] -> Z[x]/(x^2-2) has a hom f(a) = a+ (x^2-2) so I want to find some ideal such that f(<a>) = <2> ?

So i want a = 2 mod x^2-2?

[mild laurel]

Well, basically this correspondence tells you that you can just look for prime/maximal ideals containing (x^2-2) in Z[x]

And never have to deal with the quotient ring

[fringe nexus]

hmm ok, so for (0) the corresponding ideal in z[x] is just (x^2-2), and since z[x] is UFD, x^2-2 is irreducible in Z[x] which implies its a prime ideal?

[mild laurel]

Yes

[fringe nexus]

ok great thanks

[tribal pasture]

My ring theory is super weak Hi5

[chilly ocean]

is thaat true that (x nilpotent) iff (1+x is a unit) or is it only implies?

[smoky briar]

look at (1+x)(1-x)

[chilly ocean]

There is an isomorphism from Z/nZ to <x> given by p(n) = g^n

How do I express the inverse of p?

[smoky briar]

f(g^n) = n

[chilly ocean]

wait, definition of nilpotnet is that x is nilpotent if exists k x^k = 0. By x^k they mean multiplying or adding?

adding right?

[smoky briar]

multiplying

[chilly ocean]

ok so then x^(k+1) would also be 0?

[smoky briar]

no?

it would be 1

[chilly ocean]

x^(k+1)

sry

cause x^k * x = 0*x = 0

[smoky briar]

yes

[chilly ocean]

ok that helps me a lot lol thx

[chilly ocean]

How do I show x^4 - x^2 +1 is not factorizable in R[x]? Is writing that let y=x^2 then y^2 -y + 1 = 0doesnt have solutions enough? (delta <0)

[smoky briar]

it is factorizable

only degree 1 and 2 polys can be irreducible

[chilly ocean]

thats what I thought

but cant find it lmfao

[smoky briar]

just do it by hand

(ax^2 + bx + c)(dx^2 + ex + f)

you may assume a = d = 1

[fringe nexus]

compare coefficients

[steep hull]

Roots are roots of unity (12th cyclotomic poly) so just pair up conjugate pairs.

[tardy hedge]

how hard is this class

[chilly ocean]

hard af

[tardy hedge]

im doing a group theory class next year

really

[chilly ocean]

depends on the curriculum but mine is fucked

[tardy hedge]

dang

[chilly ocean]

but usually people find it easy because they go pretty slow

[tardy hedge]

How does it compare to analysis? I dont really like all the inequality stuff with analysis

[chilly ocean]

if you have anentire semester just for groups then you should be fine

[tardy hedge]

Yeah

[chilly ocean]

its completely different

[tardy hedge]

Cool

ive done linear algebra, but a more computational matrix algebra class but I liked the theory to it

this semester im doing a proof based linear algebra tho

if I liked that would you say I would also like abstract algebra?