groups-rings-fields

[tardy hedge]

We want every zero divisor of (A/a)/(pi(qi)/a) to be nilpotent

that is equal to A/pi(qi)?

[trim wind]

It's just A/q_i

Since a lies inside of q_i

[tardy hedge]

Why

[trim wind]

By the isomorphism theorems

not sure which one exactly

Something like (R/I)/(J/I) = R/J

[tardy hedge]

thats what i was trying to use

[trim wind]

Yeah you can use this here since q_i contains a (otherwise it would be A/(a+q_i) which is not the same as A/q_i)

[tardy hedge]

ok i see it better now

thanks

I know this is a basic question but why can we write 0 = cap qi

[trim wind]

Cause a is 0 inside of A/a

Or wdym

[tardy hedge]

I think I was just a bit scared cause f(cap ai) neq cap f(ai) in general

[trim wind]

Oh yeah

But in this case it's fine

Cause q_i contains a

So you don't really add anything to the images you know

Like pick any x inside of a not equal to 0. Then you have {0} cap {x} is empty, but inside of A/a it's clearly not empty

But the q_i contain a so it's fine

[tardy hedge]

so you also have the f(qi) cap f(qj) subset f(qi cap qj) inclusion?

[trim wind]

Yes, let's say you have an x \in pi(q_1) \cap pi(q_2). Then this means there exists an y \in q_1 and z \in q_2 and a t \in a such that y+a = z. But since the q_i contain a, y+a is an element of q_1.

[tardy hedge]

Thanks, im feeling kind of bad that im struggling on basic things but i just feel burnt out and demotivated by math lately

[trim wind]

don't worry about it, everyone goes through that

[tardy hedge]

Thanks

[abstract rock]

don't give up bro, 99% of math studyers stop before they finish their proofs!!

[tardy hedge]

Ok so if x is in pi(q1) cap pi(q2), then there is n1 in q1 and n2 in q2 so that pi(n1) = pi(n2) = x. Then n1-n2 is in a, so n1-n2 is in q1 cap q2, so x is in pi(q1 cap q2)

Is that ok?

[trim wind]

Yeah that works

[tardy hedge]

I dont even thinn it does now

I probably need a break

[trim wind]

I mean you might have missed a step. As you said n_1 - n_2 lies in q_1 \cap q_2. But then n_2+ (n_1 - n_2) = n_1 lies inside of q_2 as well, hence in q_1 \cap q_2

So x is in pi(q_1 \cap q_2)

[tardy hedge]

Oh yeah thats what i was trying to get to

What a silly diversion but i guess i learned stuff

Thanks

[white oxide]

for elements x, y, z in a free abelian group F and m in Z, m(x + y + z) = mx + my + mz right

[coral spindle]

For elements x, y, z in any group this holds

Sorry, any Abelian* group

[white oxide]

oh yeah lol

[tardy hedge]

What are you studying these days?

[white oxide]

algebraic topology and measure theory!

i haven't studied group theory in a while, so i've needed to dust up on it a bit haha

hbu

[tardy hedge]

I see. I’m working through atiyah macdonald

Ive been doing a reading course and i need to get it finished by feb 1st

[white oxide]

is that comm alg?

[tardy hedge]

Yeah

[tardy hedge]

Last semester i did a point set top class but near the end we started some alg top and ngl that stuff scared me

I think its way too much spatial visualization for me

[white oxide]

glgl

[tardy hedge]

Thx

[white oxide]

yeah, it's probably the hardest subject i've encountered so far. even though rotman reads really well i find myself reading chapters 3 times to really digest the material

[tardy hedge]

Yeah like i would be down to self study it on my own time, but i think id be pretty stressed if i were to take a class in it

[white oxide]

yeah i had to take a class last sem on alg top and it was horrible so now i'm just self studying

[tardy hedge]

Going thru the material on your own a second time is so nice

[kind temple]

tangentially related, but how would you go about showing that the finitely supported permutations on a set $X$, call it $\text{Sym}^{\ast}(X)$, has presentation given by $$G = \left\langle \tau_1,\tau_2,\dots | \tau_i^2 = 1, (\tau_i\tau_j)^2 = 1 \text{ if } i + 1 > j \text{ or } j + 1 > i, (\tau_i \tau_{i+1})^3 = 1\right\rangle$$

[cloud walrus]

c squared

[kind temple]

the proof i came up with for finite X, effectively S_n, is a bit tricky and relies on showing that |G| <= |S_n|

since there is a surjective homomorphism G —> S_n by the universal property of a group presentation

but i don’t think the same idea applies here

[minor fulcrum]

Is there a clean categorical characterization of the subgroup commutator

[mighty kiln]

The kernel of the abelianization map

[minor fulcrum]

No like, in the definition of the lower central series for instance

We have [G,H] where H is a subgroup of G

[minor fulcrum]

The definition given is that we have the commutators of elements in G and H, and those form a generating set for the group

But like what actually is that

[next obsidian]

Arki red fox

[mighty kiln]

If H is normal, this should be the kernel of "centralising" H ⊂ G

Categorically it's an adjunction between the category of pairs H ⊂ G and the category of pairs H ⊂ Z(G)

[mighty kiln]

Chmonkey chair monkey

[minor fulcrum]

Oh wait this is very helpful thank you, that made it click

Thanks

[serene dune]

now i used 1.6 to then prove 1.7 via the existence of the 1 preserverving yellow map

but such existence of such a map tell us something more powerful no?
i mean if its if M forms an associative unital R algebra its already an R module from the def(R-algebra is an R module which also then gets a ring structure)

now my qeuestion is why the 1.7 doesnt want me to prove that statement about algebra ?
is it not true ?
why !

or is it that my argument is faulty somhow!

[thorn jay]

M is a module, not a ring

So it does not have a center

Ring homomorphisms R -> Cen S correspond R-algebras, however the question is asking about modules

You can show that the action described there is the right-multiplication induced by the map
S -> R -> End_Z(M)
however, which would suffice, given that you know about the correspondence between R-module structures on M and maps to End_Z(M)

[spark veldt]

So we're talking about the field of quotients of an integral domain here, and we're trying to establish the axioms for F which is the set of all equivalence classes with the relation (a,b) ~ (c,d) iff ad=bc.

How do I show that [0,b] is the zero-element here?

$[a,b]+[0,b]=[ab+0,b^2]=[ab,b^2]=[a,b][b,b]$

[cloud walrus]

bluepianist

[spark veldt]

Am I on the right track here?

[rocky cloak]

So you're using b for two different things, but ignoring that you've shown that
[a, c] + [0, b] = [ab, bc]
Now what is it supposed to equal for [0, b] to be the identity?

[spark veldt]

[a,c]

[rocky cloak]

Indeed, so is
[ab, bc] equal to [a, c]?

What does the equivalence relation say?

[spark veldt]

oh... i see so [ab,bc]=[a,c] because abc=bca and because an integral domain is commutative then abc=abc so equal

thank you sm!

[vagrant zinc]

Does anyone know what is the function of this property, or an applied example please?

I found something similar in linear algebra but I don't know if it is the same.... even so I still don't see an application of this, maybe you can help me.

[velvet hull]

could you translate it to english, please?

[cloud walrus]

Homology

[vagrant zinc]

I thought it wasn't necessary, my bad

[velvet hull]

yeah, I don't think most people here speaks portugese

anyways, what is f?

is it from G to G?

[vagrant zinc]

If it is from G to G rare, hehe where f is homomorphism

[velvet hull]

oh okay yeah, that sentence makes a lot of sense to me

what do you not understand about it?

[vagrant zinc]

The last part that says that the inverse of the function is defined as H* Kerf, as would be an application in an exercise to have it clearer, I was looking at some notes from a teacher and found that property and then I remembered that I saw something similar linear algebra but I do not know if they are the same, then I would like an application in an example of this property, I tried but I did not get :/

[velvet hull]

ok, first of all do you understand what H*ker(f) means?

and yes by the way, this theorem can be used to prove the rank-nullity theorem which is the second screenshot

[vagrant zinc]

I understand H*kerf, well of course H is a subgroup of G and so is kerf.

[vagrant zinc]

I'd have to see that demonstration you say.

[velvet hull]

ok, so H*ker(f) = {hk | h ∈ H, k ∈ ker(f)}

[vagrant zinc]

Ye

It is assumed that kerf=0 then the operation on f inverse f of H should be 0, in which 0 belongs to G

[velvet hull]

so, what this theorem is saying is actually really intuitive, but to understand how maybe we can go back to linear algebra

[velvet hull]

do you understand this?

[vagrant zinc]

A little

[south rain]

Let $V$ be a vector space of infinite dimension over a division ring $D$ and let $R := {h \in \mathrm{End}_D(V) | \mathrm{dim}, h(V) \text{ is finite}}$. Show that $R$ is simple.

My approach:
Let $B$ be a basis and $b \in B$. Define the idempotent $e_w(b) := w$ if $b = w$ and 0 otherwise. Let $I \neq {0}$ be a non-zero ideal of $R$ and choose any map $0 \neq h \in I$. We now try to show that $e_w \in I$. Since $I$ is an ideal we have $he_w \in I$. Then $he_w(w) = cw$ for $c \in D$. Define
\begin{align}
r(b) := \begin{cases}
w \qquad &\text{if } b=cw \
0 \qquad &\text{if } b \in B' - {cw}
\end{cases}
\end{align}
Then $rhe_w \in I$ and furthermore $e_wrhe_w \in I$. We thus have that $e_w \in I$.

I'm not sure if this is correct at all I just try to scale the projection here with a different element of $R$ essentially. But I don't see how I need it to be an ideal for that, other than using it for simplicity in the end. I could make a similar argument with only left ideals.

[cloud walrus]

dellinger

[south rain]

We only take linear mappings in End$_D(V)$.

[cloud walrus]

dellinger

[tardy hedge]

Every primary ideal in A is a contracted ideal in the A -> S^-1A ring of fractions map?

What is the correspondence in question exactly?

"The correspondence between ideals in S-1A and contracted ideals in A"

I know every ideal in S-1A is an extended ideal, an extension of some contraction

I guess im not really sure what ii) is saying

I can see how primary ideals correspondend to primary ideals in the sense of, if you start with a primary ideal in A, then ii) is saying S-1q is primary, and if you start with a primary ideal in S-1A, then its contraction in A is primary

Is this like a bijection though?

[trim wind]

I don't think so

It's only one for prime ideals

ii) just says the extension of a primary ideal (in the ring of fractions) is primary and the contraction is primary as well

But you do have a bijection between ideals in S^-1A and contracted ideals in A

I think

Since you have a^cec = a^c and a^ece = a^e or something like this

Yeah it's Proposition 1.17 iii) in atiyah macdonald

[storm kiln]

If G is a group and N is normal in G, does there always exist a subgroup of G that is isomorphic to G/N?

Not sure, can't think of a counterexample, but maybe I haven't looked hard enough

[lusty marlin]

No

[lusty marlin]

Try looking for one among infinite groups

[storm kiln]

Oh, you mean Z/nZ?

with G = Z and N = {..., -n, 0, n, ...}

What about finite groups?

[lusty marlin]

Yes

[vast stump]

q8

[storm kiln]

Ah, Q_8 / {1, -1} doesn't seem to be isomorphic because it has 3 elements of order 2. Thanks!

[young thorn]

does normal in G mean N is a normal subgroup of G?

[coral spindle]

Yes

[tardy hedge]

Why is a union of prime ideals not necessarily prime?

What is wrong with : if ab is in union of pi, its in some pi, so either a or b is in pi, so a or b is innthe unions of pi

[trim wind]

The union of ideals isn't an ideal

[tardy hedge]

Oops

[coral spindle]

(in fact it's an ideal iff one contains the other)

[tardy hedge]

Oh yeah union of subgroups isn’t necessarily a subgroup

[potent condor]

(in fact it's a subgroup iff one contains the other)

[thorn jay]

On a slightly related note, the union of two sets V(S) = { p prime | S is a subset of p } is another such set, which is a really interesting property

[tardy hedge]

prime spectrum?

[thorn jay]

Ye

Zariski topology

In trying to generalise to arbitrary algebraic structures i have found it annoyingly non-algebraic

Lol

(Hence why i call that property geometric)
Well, it's formulated differently, but in fact equivalent

It's this property

(P is a prime ideal)

[tardy hedge]

so a union of arbitrary ideals is an ideal iff each one is contained in some other one?

Im trying to understand this blurb

[thorn jay]

Ye, i believe so

[next obsidian]

Nah this is cap

Take just two ideals, one contained in the other

Their union is an ideal but the top one isn’t contained in some other one

[thorn jay]

It's nonproperly contained in itself

[next obsidian]

Then any set of ideals satisfies this condition

So the union of any set of ideals is an ideal

[thorn jay]

Eugh thats true

[next obsidian]

I think if you have this condition then the union is an ideal

Or well no

[thorn jay]

The union of a family of ideals is an ideal iff there is an ideal in the family containing the rest

[next obsidian]

Also not true

[tardy hedge]

why is the union here an ideal?

[thorn jay]

[next obsidian]

Take an infinite ascending chain and take the union

That’s an ideal containing everything but not in there

[next obsidian]

Where does it ever claim it is?

[tardy hedge]

Oh

[next obsidian]

Young padawan S = A \ U_p in Sigma p mult closed does not mean the latter is an ideal

[tardy hedge]

Oh ok

That union of prime ideals may not be an ideal, but it still has the property that if ab is in that union, a or b is in that union, right?

[next obsidian]

No I don’t think so

Not if it’s infinite

If it’s finite I think this is a prime avoidance thing

[tardy hedge]

Ok, im just tryna figure out why S is multiplicatively closed

[next obsidian]

prime avoidance says ab < pi for some i

Oh a and b are elements

Then yeah

[trim wind]

Are you sure? Cause if ab is in the union, it lies in one prime ideal and then it's just the usual definition of a prime ideal

[next obsidian]

I thought they were ideals

[tardy hedge]

ok cool

[next obsidian]

Yeah, for elements this works. I was misinterpreting

[serene dune]

yep, sorry for the late reply, i will redo it asap and would then recheck with your comments

[empty kernel]

Z_p with p a prime number
How many irreducible polynomials over Z_p of the form x^2+ax+b exist

b cant be equal to zero i figured that out

And i assume i gotta use the theorem that states that f is reducible if and only if f has a root in F

Ccuz f has deg 2 and Z_p is a field

But past that im stuck

[next obsidian]

Quadratic formula

[empty kernel]

How?

I tried that znd then i figured that D must be a whole square

But i didnt rlly think i could do much with that

And -a+root(D) must be even i think

[grave sedge]

You can count how many are reducible

Since as you said a reducible degree 2 polynomial can be written as (x-u)(x-v) where u,v are the roots

(and the multiset {u,v} uniquely determines the polynomial)

[next obsidian]

This guy is way smarter than me

[grave sedge]

There's also a way to do this which can be easily generalized to counting the irreducible polynomials mod p of any degree (or with coefficients in any finite field) but it requires a bit of finite field machinery

[grave sedge]

(for a fixed n and field F_q, x^(q^n)-x factors into orreducible polynomials of degree that divides n, with every such polynomial appearing exactly once, so if you know how many irreducible polynomials there are for degrees below n you can look at the total degree and find how many there are with degree n)

[empty kernel]

So every combination of u,v is one reduc polynomial?

[grave sedge]

Yes

[empty kernel]

Thus that means u have p+(p*(p-1))/2 reduc polynomials

[grave sedge]

Yes

[empty kernel]

And then u have p^2 polynomials in total

And then subtract

I would have never in a million years thought of that myself

But it is a surprisingly simple thing lol

[empty kernel]

Thanks

[thorn jay]

No worries, good luck!

[empty kernel]

Is every ideal of a field a prime ideal?

Or did i just prove bullshit lol

[empty kernel]

Therefor also every ideal a max ideal

[mighty kiln]

A field has two ideals

[thorn jay]

Every proper ideal

But yes

[empty kernel]

Wait wut

[thorn jay]

The trivial ideal and the whole field

That's how fields are even characterised; simple commutative rings (simple means two ideals)

[empty kernel]

The way defined field is
Every element has an inverse for multiplication and multiplication is commutative

Oh wait

[thorn jay]

Characterised, not defined

[empty kernel]

But i see u can easily prove thag it is the whole field

Oki thanks

[thorn jay]

Nonzero element

And not the zero ring

[empty kernel]

Ye every non zero element mb

[empty kernel]

Guess who is back again

Find an explicit presentation of the field F_25 as Z_5[x]/(f) with f in Z_5[x] an irreducible polynomial

Well so since |F|=5^2 we know that F is a idkhowitscalledinenglish field of the polynomial x^25-x over Z_p

Thats all i know so far

Well so x^25-x is obv reducible

[thorn jay]

The degree of the field extension F_5 < F_25 is 2, so you need a quadratic

A quadratic which is irreducible over F_5

[empty kernel]

Does it matter which quadratic?

[thorn jay]

No, a well-known theorem is that every finite field with the same number of elements is isomorphic

[empty kernel]

Oh ye i remember reading that

There is so many theorems and i seem to not be able to bring them together in exercises

Well x^2+1 seems a fine polynomial

[thorn jay]

2^2 = 4 = -1 mod 5

[empty kernel]

Ugh

[thorn jay]

Lol

[empty kernel]

X^2+2

This one works

Yes this one works

[thorn jay]

Nice :3

[empty kernel]

Ye no i still dont know what to do

Idk since when i became so shit at math

[bitter rover]

Prove that f(x) = x^2+2 is irreducible over F_5.

There are general theorems you could use to prove this, but F_5 is small so you could just check that f(0), f(1), ..., f(4) are all non-zero.

[tardy hedge]

the rings and modules of fractions chapter of a/m took me deceptively long time to get thru. I mean im not done it completely yet but i feel like i understand it a bit better now

primary decomposition was easier to study

[regal zodiac]

Hello, i really don't understand the last implication of this proof , why would the image of T^{p-1} be K ?

[rocky cloak]

The kernel of T is one dimensional, so the kernel of T^p-1 is at most p-1 dimensional, meaning T^p-1 is nonzero.

K is one dimensional, so any non-zero subspace is all of K.

[regal zodiac]

oh okay it makes sense ! Thank u so much :)

[fluid jacinth]

what is a field

[rocky cloak]

A field is a system that has addition, subtraction, multiplication and division that satisfies most of the familiar rules of arithmetic.

You can check for example Wikipedia for the precise rules

[regal zodiac]

I don't understand how is deduced the existence of a such l in this proof? If someone has an explanation

[rocky cloak]

Some examples of fields could be Q, R or C. But also more exotic things like Z/p, the integers modulo a prime or Q(x), the rational functions.

[rocky cloak]

So if you write
L = K[x]/(f1) x K[x]/(f2) ...
Where fi are the invariant factors, then f1 will be the annihilator, so
L is K[x](x^n - 1) times some stuff.

Though I'm not sure what argument they're implying shows that the annihilator of L is (x^n - 1).

Like I would think the relevant part is that distinct field automorphisms are linearly independent.

Not sure what "Dedekind" is referencing...

[thorn jay]

Dedekind domain is some integral domain

https://en.m.wikipedia.org/wiki/Dedekind_domain

[rocky cloak]

I know what a Dedekind domain is, but it's not relevant to the proof at hand in any way

[thorn jay]

Oh, apologies

[rocky cloak]

And all PIDs are Dedekind domains, so that can't be what they're trying to say

[thorn jay]

Maybe referencing some result from Dedekind?

Lol

[rocky cloak]

Could be

[thorn jay]

Specifically saying that Ann_KX = <x^n - 1>

Idk lol

[regal zodiac]

@rocky cloak I think in our course, Dedekind's lemma states precisely that distinct fields automorphisms are linearly independent

[rocky cloak]

Oh, well then it makes perfect sense. I just Haven't heard it called Dedekinds lemma

[regal zodiac]

But it makes sense right? Because if a polynomial of degree smaller than n was in the annihilator of L, it would give us a non trivial linear combination of the sigma^i , so the annihilator is precisely < x^n - 1 >

What i don't get is the following step, how can they deduce the existence of such a l ?

[fluid jacinth]

Thanks man

[fluid jacinth]

Would u mind explaining to me what is the meaning of Z/6Z
I ask chat gpt and it didn't go well
I mean is there any real example of it from our lives

[thorn jay]

Z/12Z is how an analog clock works

If it's 5 o' clock, and you wait 10 hours, it's 3 o' clock (except PM/AM)

This is because 5 + 10 = 15 = 3 + 12

And in Z/12Z you work in so-called residue classes, which is a fancy word for numbers which differ by a multiple of 12 (in general a multiple of k in Z/kZ)

For example, 0, 12, 120 are in the same residue class in Z/12Z

This is an example of modular arithmetic, used extensively in number theory and encryption

[potent condor]

Does anyone know a book with lots of "concrete" exercises on localization? I think about it too formally so I want to work with concrete examples to get better intuition, but all the exercises I've found just ask you to prove abstract properties about localization in general

[tardy hedge]

yo i be learnin dat rn too broski

Grr im canadian

quebec is fuming rn

[fluid jacinth]

ty so much man

[thorn jay]

ChatGPT is shit

It generates what seems right

Not what is right

[fluid jacinth]

particularly the free one

[thorn jay]

Also the paid one

It's a language model, not a math model

[tardy hedge]

why dont we all spend our time learning more useful things like the cutting edge of ai tech instead of abstract math

Ong though i dont know why i do this sometimes

I just do

[delicate bloom]

it's not as much fun that's why

[thorn jay]

Learning cutting edge ai tech would make me immensely depressed i believe

[tardy hedge]

Wow really?

is r((4,t)) = r((4)+(t)) = r((4))+r((t)) = (2)+(t) = (2,t) correct?. r is radical of ideal

radical commutes with sum of ideals right

oh wait

[thorn jay]

Does it?

[tardy hedge]

No

[thorn jay]

Ah right i see why

[tardy hedge]

In Z[t], t is not in (2,t)^2 right

[potent condor]

Right

[cloud solar]

Is x^4+1 invertible in a finite ring A with 13^(n^2) elements ?

[coral spindle]

Depends on what x is

Depends on the ring too

I think you need to add a lot of context.

[next obsidian]

If x = 0 I think it is. Just a guess tho, haha

🙂

[coral spindle]

More investigation needed for sure

[cloud solar]

I have this problem A an nxn matrix with integer entries and i need to prove det(A^4+I_n)≠13.

This is equivlent to A^4+I_n is invertible over M_n(Z/13Z) for every A in M_n(Z/13Z)

Idk whats the key idea

[tardy hedge]

Ty bro

[cloud solar]

That 4 has something to do with 13

I think i am on the right track looking at M_n(Z/13Z)

But idk what to do from here. Maybe find an isomorphism?

[rocky cloak]

It's not equivalent. It could for example be that the determinant is 0 or 26 or some other multiple of 13.

Even for n as small as 2 you can find matrices such that A^4 + I is 0 modulo 13

[cloud solar]

Right

[rocky cloak]

For example consider A = [0, 5; 1, 0].
Then A^4 + I = 26I has determinant 26^2

[cloud solar]

Thanks

[delicate bloom]

weird idea, if we look at factoring A^4+I in the gaussian integers as (A^2+iI)(A^2-iI) keeping an eye on 13=(2+3i)(2-3i) this sort of makes me think it'd be sufficient to prove det(A^2+iI) != 2+3i, since the product of the determinants would be the norm there.

at least I believe we'll need to use the fact that 13=1 mod 4 in some way, whether it's somewhat like that in the Gaussian integers or that there's some i in Z/13^kZ that satisfies i^2=-1 and try to work something out in this ring instead

[rocky cloak]

Hmmm, so if
A^4 + I has determinant 0 mod 13, then A has eigenvalues that are roots of x^4 + 1 = (x^2 + 5)(x^2 - 5) hence contained in F_169.

So we can find P with coefficients in F_169 such that PAP^-1 is block diagonal, with some pair of roots of x^4 + 1 on the diagonal.

If we take something like R = Z[sqrt2], then R/(13) = F_169 and R is a UFD.

Picking lifts of P and P^-1 to R we get
det(P P^-1) = 1 + 13k
det(PAP^-1 + PP^-1) = 13(1+13k)
which is congruent to 13 mod 13^2.

So PAP^-1 + PP^-1 can be partitioned into blocks where the off diagonal blocks are multiplies of 13, and there is at least one 2x2 block that is a multiple of 13.

This should mean that the determinant is a multiple of 13^2, so cannot be 13 modulo 13^2

Alternatively instead of passing to Z[sqrt(2)], you can just argue that you can make A block diagonal with one block either [0, 5; 1, 0] or [0, -5; 1, 0] modulo 13. Then you still get that det(A^4 + 1) is a multiple of 13^2

I suppose this should generalize. If f has no roots modulo p, then det(f(A)) is never p modulo p^2.

Interesting

[empty kernel]

Ye i did that but i just dont get how i will get my field of 25 elements out of it

[sonic coral]

since it is irreducible, you can adjoin a root of x^2 + 2, such as the square root of 2, to obtain a field with 25 elements

[whole jetty]

Hello. I want to compute the iduced representations of the Group D6. I did this calculating D6/<r> which, some time ago, I made that equal to {1,r,s,sr}. Is this true? I am getting confused with my own solution, but the caculations give me the correct answer to the character table.

[whole jetty]

Ok, so for what i understood... I am mixing two diffents things. Using <r> and <r²> at the same time. So, the thing is, if I use <r>, D6/<r> = {r,s} and the character table will be half of what I am finding on the internet.

What I am doing wrong?

[rocky cloak]

It's a little unclear what you're trying to do.

But assuming D6 is the dihedral group of order 6, then D6/<r> should have two elements, those being the cosets <r> and s<r>.

I'm not sure which character table you're trying to compute though.

You also mentioned induced representation, are you trying to induce representations from <r> to D6?

[dull tiger]

Question: For which primes $p \in \mathbb{N}$ does the polynomial $X^2+1 \in \mathbb{F}_p[X]$ have two linear factors?
Solution: the polynomial has two linear factors in that field iff it has a root there. Meaning, it only has a root in the field iff $\sqrt{-1} \in \mathbb{F}_p$. okay, fair.

Thats where I got stuck and sneaked a look at the solutions. But here it says: "Since the group $\mathbb{F}_p^*$ is cyclic, the square root of $-1$ only exists if $p = 1 \ (mod \ 4)$ or $p=2$." Why?

[cloud walrus]

mathrie

[grave sedge]

-1 is the unique element of order 2 in F_p*

(for odd p)

F_p* is a cyclic group with p-1 elements

If you have a cyclic group of order p-1, how can you describe the element of order 2? (assuming p is odd)

And in general, when you have an element in a cyclic group, how can you tell if it has a square root?

[trim wind]

You can use the Euler criterion for Z/pZ and p an odd prime (a is a quadratic residue iff a^(1/2*(p-1)) = 1 mod p). There's also a generalization for some other N, but I don't know the identity by heart.

[dull tiger]

oh thats interesting, i didn't know about that

[little shadow]

Is "a divides b" in integral domains just defined by the field of fractions in the regular sense? i.e $a|b \Longleftrightarrow (b, a) \sim (c, 1)$ for some $c \in R$

[cloud walrus]

Dompa

[dull tiger]

But then we get for $a = -1$ and $p=4$ the following: $-1^{3/2} = 1 \ mod \ 4$

[cloud walrus]

mathrie

[trim wind]

Sorry, I forgot to mention that p needs to be prime

[kind temple]

just that b = ac for some c

[little shadow]

ofc... im so fucking stupid

[kind temple]

no… you were just asking a question

a valid question

[dull tiger]

Since the group is cyclic, we can write it as $a = g^k$ for $a$ the element of order 2, $g$ in the group and $k \in \mathbb{N}$. Since it should be of order 2, we need it to fulfill $g^{2k} = e$.

[cloud walrus]

mathrie

[dull tiger]

Ahh so $2k = 0 \ mod \ p$. But this still doesn't tell me anything about $p$.

[cloud walrus]

mathrie

[trim wind]

Any cyclic group has at most one element of order 2

And if its of even order it has exactly one element of order 2

[whole jetty]

Ok, sorry. So, I want to compute the character table of the dihedral group D6. From the conjugacy classes I know I have 6 irreps. Doing the commutator [D6,D6] i get four of them. I could get the other two by a simple system of linear equations using the orthogonality relations between them, but I want to do that using induced representatios.

So, using G=D6 and H=<r>, i know that G/H = {r,s}. Now, I did the following:

V = Ind Φ, with Φ: r --> exp(i2pi/6).

Using the formula to compute Tr(IndΦ) at 1, I will get [χ(1) + χ(ss)]/2 which is not 2, but 1.

Am I doing some error?

[rocky cloak]

So I don't know if you're thinking of a different formula, but one formula for the induced representation is

(Ind Phi)(x) = 1/|H| sum[g in G] Phi(gxg^-1)
(Where Phi(y) = 0 for y not in H).

Putting x=1 gives 12/6 = 2

I'm not sure what Chi or ss is supposed to refer to in your formula

[whole jetty]

Sory, Its the result of phi

Its what you have.

Ok, maybe I did something wrong on the sum. Let me check.

[whole jetty]

Ok, I am not seeing what I am doing wrong, could you exemplify for the case x = r, where r is the rotation?

[rocky cloak]

Sure, so
grg^-1 = r when g is a rotation
And grg^-1 = r^-1 when g is a reflection.

So
IndPhi(r) = ( 6exp(i2pi/6) + 6exp(-i2pi/6) ) / 6 = 2cos(pi/3)

[whole jetty]

Where does the 6 factor come from?

[rocky cloak]

There are 6 rotations

And similarly 6 reflections

So when you sum over all of them you just get 6 times what you would get from one

[whole jetty]

Ok.... I know understand that I dont understand the formula. AHahhha

Let me check once again.

[rocky cloak]

https://en.m.wikipedia.org/wiki/Induced_character

This formula right here

[whole jetty]

Yea, I know, but I though we only had 2 terms in the sum because of the left cosets.

[rocky cloak]

G still has 12 elements

[whole jetty]

And, for example, for r, since r1 is in the left cosets and rs was to in the left cosets the sum would only be with these two.

[rocky cloak]

Like cosets are not mentioned in the formula

Do you have some other formula you're thinking of?

[whole jetty]

No no, its this one.

[rocky cloak]

Okay, will this is just a sum over all elements in G, and then you divide by |H|

[whole jetty]

Yea, but the terms that actually enters the sum have a restriction right? The one I have is (Ind Phi(g)) = (const)(sum [ i = {1,...,d} such that [gi] = [g.gi]] Phi(gi^-1 g gi)

So, for example, for g = r, rgi always are in the respective cosets, right?

[rocky cloak]

What is i and d, does [gi] mean coset?

And then you did actually have a different formula you were thinking of

[whole jetty]

I suppose i and d are just to run over all the elements, i guess. And [gi] I think it is the cosets yea, so if g is r and gi is r , r times r belongs to the coset of r so it can enter the sum.

[rocky cloak]

Right, so gi runs over all elements of G, and [gi] = [rgi] is just always true because H is normal.

Note the condition
[gi] = [g gi]
is equivalent to
gi^-1 g gi
being in H.

So your just ignoring the terms for which Phi(gi^-1 g gi) is not defined

[whole jetty]

No, it cant be true all the time. r - rotation, s - symmetry. For the s, sr, etc... its zero because s.[r] is not in [r] and s.[s] is not in [s].

[rocky cloak]

I'm not sure exactly what you're trying to say, but
[gi] = [r gi]
is true for all gi

For example if gi = s, then rs = sr^-1
So rsH = sr^-1H = sH

[whole jetty]

Sooo how is that for the case of Ind Phi(s)?

Because that is 0.

[rocky cloak]

For x=s, gsg^-1 is never in H. So the sum is just 0/6 = 0

[whole jetty]

How is that never in H?

Ohhhh okok

Yea, H =<r>

[rocky cloak]

Well, just list it out case by case of you like

[whole jetty]

Yea yea yea.

By that formula i can understand how to compute.

Now, using mine is a little bit tricier.

[rocky cloak]

Yeah will it's just that [gi] = [s gi] is never true. So the sum is empty

[whole jetty]

Yea. Ok, let me try again now, see if i understood correctly.

[chilly ocean]

This feels like a very basic question I should be getting right immediately but I just can't seem to get it. I'm trying to understand why the multiplicative group of finite fields is cyclic. I understand that the order of this group is p^n-1 but I don't understand why abelian groups of order p^n-1 are cyclic. I can think of examples which by themselves are pretty nontrivial to prove such as p=2 n=4. We get a group of order 15 which was a question in one of my problem sets which required the use of sylow theorems to show it was cyclic.

[rocky cloak]

Abelian groups of order p^n - 1 aren't usually cyclic, this is something special about fields.

It's possible to use the classification of finite abelian groups together with the fact that in a field a degree n polynomial equation has at most n solutions

Another common proof uses Möbius inversion from number theory.

[chilly ocean]

I see, so this is mostly a nontrivial result? My instructor glazed over it really quickly as if it were really obvious in the middle of a larger proof

[rocky cloak]

I wouldn't say it's particularly hard to prove, but I wouldn't call it trivial either. It's in the eye of the beholder I guess

[chilly ocean]

Fair enough, I'll keep looking

[whole jetty]

Ohhhhhhhhhhhh @rocky cloak yea, now I understand what was going wrong.

My formula actually is a little bit diferent from yours.

[rocky cloak]

Very possible your instructor assumed you had shown it before

[whole jetty]

I was mixing two formulas and that was why the result was half of what was supposed to be.

So, I actually use the left cosets to compute it and I dont divide by |H|.

This is way, it kind of condensates the formula and I dont need to check for all the members, just need to check for the cosets.

Wanna know?

[storm kiln]

I have seen a proof of this in my book, it stated that if R is an integral domain and H is a finite subgroup of the multiplicative group of units of R, then H is cyclic. That covers your case right?

[chilly ocean]

I think it does but we definitely don't have access to this property in my course

[storm kiln]

The proof doesn't use any advanced theorems

I could type it over if you want, but I'll have to translate it

[tardy hedge]

When does the zero ideal have a primary decomposition?

[trim wind]

The ring needs to have finitely many prime ideals is a necessary condition. Sufficient would be finitely many prime ideals and being reduced for example

[tardy hedge]

Is that condition somewhere in atiyah macdonald? Where did you learn that?

[trim wind]

Oh sorry I think you only need finitely many minimal prime ideals

[trim wind]

It's not difficult to see that for any price ideal p, there exists an ideal in the primary decomposition of 0 that is contained in p

[storm kiln]

Let $H$ have order $n$, and let $\psi(d)$ be the number of elements of order $d$ in H. If $x\in H$ has order $d$, then there are $d$ different powers $x, x^2, \ldots, x^d = 1$ of x that are zeros of $X^d - 1$ in $R$. These are the only ones (references another theorem), this implies that the elements of order $d$ in $H$ are exactly the $\varphi(d)$ powers $x^i$ of $x$ with exponent $i$ coprime to $d$. We can conclude that $\psi(d)$ is equal to $\varphi(d)$ if $H$ contains an element of order $d$ and 0 otherwise. Since the sum of $\varphi$ over all divisors of $n$ is equal to $n$ it follows that $$n = \sum_{d|n} \psi(d) \leq \sum_{d|n}\varphi(d) = n$$. Thus $\psi(n) = \varphi(n) > 0$, so $H$ contains an element of order $n$ and is cyclic

[trim wind]

And the reduced assumption just means that the intersection of the minimal prime ideals is equal to 0

But I think noetherian rings also have a primary decomposition of 0

[cloud walrus]

Jelle

[storm kiln]

@chilly ocean

[trim wind]

Check Exercise 1 in Chapter 4

[trim wind]

And this is Theorem 7.13 in AM

[whole jetty]

@rocky cloak One last thing if you can plz. If my commutator subgroup is {-1,+1} why can i say that the characters are trivial in all elements of the commutator subgroup?

[rocky cloak]

Maybe you're missing some context, because it's not true in general.

If the representation is 1-dimensional it's trivial on the commutator subgroup. But that doesn't depend on what the subgroup is...

[whole jetty]

The case I am considerng is quaernion group Q8.

When I do [Q,Q] i get {-1,+1}.

[rocky cloak]

Okay, but not every character is trivial on ±1

[whole jetty]

And i found online that the representation we get from there are trivial on +-1.

[whole jetty]

Yea, but the ones generated by the commutator sugroup are, right?

[rocky cloak]

Like induced from it?

[whole jetty]

Hmmmmm... yea, but they are not induced representations.

[rocky cloak]

Then I don't know what you're talking about

[whole jetty]

Ok... to get the character table one way to do it is using the commutator subgroup to get more irreps, right?

We can get as many as |Q|/#{-1,-1} = 4.

[rocky cloak]

Inducing from subgroups is one way to find more characters sure

Not really relevant if it's the commutator subgroup or not

[whole jetty]

Yea, but it helps to find I guess.

https://math.stackexchange.com/questions/1150909/finding-the-character-table-of-q-8

I am talking about the first answering gaved here.

[rocky cloak]

Okay, so I think what you're thinking about is that
G/[G, G] is abelian. So it's easy to find the characters of it, and then restrict them to G.

This will give you exactly those character which have trivial values on the commutator subgroup

And in the case of Q8 you get all but one

[whole jetty]

Yea, but why are they trivil on that values? Are there some kinf of demnstration?

[rocky cloak]

I mean when you mod out something it gets mapped to the identity.

[whole jetty]

"mod out"?

[rocky cloak]

Take the quotient

Like you have G your group, then you consider the quotient group G/[G, G] where you mod out the commutator subgroup

Then every commutator is mapped to the identity

For 1-dimensional characters commutators must map to 1, because C^* is abelian

[whole jetty]

Ok, I still dont understand but I am getting the idea.

Thanks a lot for the help.

[lethal cipher]

So I am working on this qual problem, and I am having trouble with a potential intermediate step. Namely: Let N be the unique subgroup of index p. What I would like to show is that Z(P) is contained in N.
If this is true, then I can use induction on P/Z(P) w/ the bijective correspondence theorem, but I am struggling with said containment

What am I saying...this isn't true, especially when P is abelian

[trim wind]

I'd use the fact that if P/Z(P) is cyclic, then P is abelian

[lethal cipher]

yea, but we need to get to that

I'm trying to get that there is a unique subgroup of index p in the quotient group

[trim wind]

Oh you didn't prove that yet?

Oh

[lethal cipher]

that's the part I am having the hardest time with. I see how to do things after that

[trim wind]

Can't you use the correspondence theorem

[lethal cipher]

we don't know if we have containment though

the correspondece theorem gives that every subgroup containing Z(P) corresponds to a subgroup in P/Z(P)

[trim wind]

Well P/Z(P) is a p group, so there will be one of index p

And then by correspondence there is only one

[lethal cipher]

Well, so the problem I am running into is whether this p group corresponds to the largest proper subgroup in P. Since we don't necessarily have containment, perhaps the order of this subgroup is smaller

[trim wind]

Well the corresponence theorem also tells you that the index will be the same, so I think it should work

[lethal cipher]

Oh? Is that so?

[trim wind]

Yup

At least according to Wikipedia

👀

https://en.wikipedia.org/wiki/Correspondence_theorem

[lethal cipher]

Actually, I am still not convinced

We only get uniqueness if this subgroup of index p in P/Z(P) correspondes to the unique subgroup of index p in P. But....we can only really do this if the unique subgroup in P actually contains the center, otherwise the bijective correspondence theorem can't be used here

[trim wind]

You know there exists a index p subgroup in P/Z(P). By correspondence theorem, this gives you a index p subgroup in P which contains Z(P), which needs to be the unique one

[lethal cipher]

So in turn, it is possible that this unique subgroup of index p in P/Z(P) corresponds to a subgroup of larger index in P (subgroup has smaller power)

[trim wind]

Every subgroup of P/Z(P) corresponds to a subgroup containing Z(P).

[lethal cipher]

Yeah....I am an idiot. -_-

You said it already, but the brain was being stupid

[trim wind]

haha no worries

[lethal cipher]

I'm having trouble seeing why the indices are the same though. If they are, that's great (might just be a general truth), but my brain's struggling a bit to understand why

[trim wind]

If S<G/N corresponds to H, then you have [G:H] = |G|/|H| = (|G|/|N|)/(|H|/|N|) = [G/N:S]

[lethal cipher]

Ahhhh, thanks!!

[trim wind]

no worries and good luck with your quals :)

[empty perch]

Hey folks, just a quick note that the Abstract Algebra Lecture series is waking from it's hibernation ^_^
We will meet to discuss group actions in about 10 minutes over in the #1055201711679082516. More information can be found in our thread: #1317307081535000606

[tardy hedge]

Not too sure about the last part

not yet sure how the previous parts help

Nevermind i think i got it

Shouldnt be that complicated i dont think

Okashi is the new MVP of groups rings fields

[tardy hedge]

Mizalign is cooking

[unkempt stream]

Wondering if this is an appropriate proof sketch of the fact that every submodule of a finitely generated module over a Noetherian (commutative) ring R is finitely generated, and is generated by at most the same number of elements if the ring is a PID:

Lemma 1: If M is a left Noetherian module, then so is M^N for any finite N.
Proof: Take submodule H of M^N, and consider the image of image of H under the nth canonical projection as H_n. H_n is finitely generated, and as H is the direct sum of the H_n, then H itself is finitely generated

Lemma 2: If A and B are left modules, A is Noetherian, and there is an epimorphism F from A to B, then B is Noetherian
Proof: Take submodule H of B, then F^-1(H) is a submodule of A and is thus finitely generated. By surjectivity, F(F^-1(H)) = H, so the image of those generators under F generate H, so H is finitely generated.

Theorem 1: If M is a finitely generated left R module and R is Noetherian, then M is noetherian in the module-sense.
Proof: There exists a canonical epimorphism from R^N to M sending the basis elements of R^N to the generators of M. As R is noetherian in the module sense (ideals are by defn submodules), so is R^N by lemma 1, and thus so is M by lemma 2.

Lemma 3: Any submodule of a free module is free if R is a PID
Proof: In jacobson lol

Theorem 2: If M is a finitely generated R module, then any submodule of M is generated by at most the same amount of elements when R is a pid
Proof: There exists a canonical map from R^N to M sending the basis elements of R^N to the generators of M. Thus F^-1(H), a submodule of R^N, is free with a basis of at most N many elements. As H = F(F^-1(H)), the image of that basis of F^-1(H) under F generates H

[rocky cloak]

H is not the direct sum of the Hn, so your proof of lemma 1 is not quite right.

Otherwise it looks good

It might be worth proving a more general statement:

If
0 -> A -> B -> C -> 0
then B is Noetherian if and only if both A and C are.

[unkempt stream]

Can you actually do this using the four lemmas lol

just with injections at the ends

[rocky cloak]

Almost at least. There's not much missing

[unkempt stream]

Actually nvm you'd need the full five lemma

You can probably use the same technique anyhoo

I have been sorta stuck trying to prove that over a pid D, for any finitely generated module M, and submodule N, that rank(M) = rank(N) + rank(M/N)

[south patrol]

how are you defining rank?

[unkempt stream]

rank of free part or alternatively M/tor(M)

[south patrol]

Ideally you ||tensor up to Frac(D) and gg||

[unkempt stream]

jacobson hasn't introduced Tensor products though that technique is a bit more obvious

[rocky cloak]

I guess pick a maximally linearly independent set of both N and M/N and show that this combines to a maximally linearly independent set in M.

[unkempt stream]

I assume the main idea is to use the structure theorem to say that $M = \bigoplus_{n = 1}^{N} (x_n)$ where $\mathrm{ann}(x_n)$ is a descending ideal chain, and to intersect $H$ with those cyclic submodules to get a further cyclic submodule decomp of $H$

[cloud walrus]

Mizalign (Study acc)

[unkempt stream]

Well, issue here is that we aren't assuming a priori that M is free

sure the latter is very simple, you can really cheese the shit out of it computationally with smith normal form

[rocky cloak]

What would you use M being free for?

[unkempt stream]

The existence of a basis

oh wait just the maximally linearly independent set

I see. Since linearly independent sets can only lie in the free part

[rocky cloak]

Maximally linearly independent sets always exists, and I assumed you've proven that for a PID, their size is always equal to the rank

[unkempt stream]

I do ask what you mean by combine here.

[rocky cloak]

So N is obviously a subset of M, then you just pick preimages for the ones coming from M/N

[unkempt stream]

Well we didn't prove that a priori, but it's kinda obvious since it can't be in the torsion part, so must be in the free part, of which has a basis and is governed by IBN

[unkempt stream]

i see.

[unkempt stream]

is there a map here in some direction between M and N (+) M/N

I wouldn't expect a split here, I would for D^N due to Smith Normal Form

let me think for a sec

[rocky cloak]

||Not really no.||

||I mean you can make a function, but it probably won't be linear||

[unkempt stream]

Actually you may be able to do something similar to a split

Take M, quotient out tor(M)

Then you have a free module, where you can split it

and then you can consider N/tor(M)

which IS free

[rocky cloak]

There is some care to be taken. Like even just something like

0 -> Z -2-> Z -> Z/2 -> 0

You can't just remove the torsion and get something exact.

But you could take the free part of M/N and take the pullback, if you have such machinery

[unkempt stream]

Yeah the iso between (M/tor(M))/(N/tor(M))

[unkempt stream]

i am starting to really like exact sequences ngl

also diagram chases are really satisfying

[rocky cloak]

I hear that

[unkempt stream]

Oh yeah there's an issue with that

Tor(M) contains tor(N)

[unkempt stream]

Instead of quotienting out by tor(M), quotient out by tor(N) = N \cap tor(M)

[rocky cloak]

You can do that if you want. I'm not sure it gets you very far.

Like in
Z -> Z -> Z/2
tor(Z) is just 0 anyway

[unkempt stream]

Let H = M/tor(N) = tor(H) (+) H_F = tor(M)/tor(N) (+) M_F/tor(N), but M_F/tor(N) is iso to M_F since tor(N) is trivially intersects M_F

so M_F / tor(N) is still free

[rocky cloak]

Yeah sure, everything you're saying here is true

You can remove tor(H) if you want aswell

[unkempt stream]

I should probably investigate what happens to rank under different types of morphisms

i think I am just haunted by the nonexistent analogue of where for D^N that splits over submodules

[velvet hull]

lol hey, are we studying the exact same section of Serge's algebra?

[sturdy spear]

I thought it was in Jacobson

[acoustic igloo]

do you guys think mu can be the identity permutation in these cases?

i guess that would require an "empty product"

[long nebula]

Sure

You might want to provide more context. What are the rest of the cases?

If mu being the identity permutation is already covered by the other cases, then you don't need to worry about it.

[somber sleet]

Is there a recipe on how to find the sylow subgroup of a specific order of a group?

[crystal vale]

Can you elaborate more?

[boreal inlet]

If I have finite abelian groups H and K, can I claim that the character group of H x K is the direct product of character groups of H and K?

Basically what I mean is, is Hom(H x K, C \ 0) \cong Hom(H, C \ 0) x Hom(K, C \ 0)?

Or do we need some conditions for this to hold

[somber sleet]

If I am given a group of a certain order, I know how to calculate |Syl_p(G)| and p^m. However I never know how to explicitely conclude how many sylow subgroups a group really has

[quiet pelican]

Yes
That’s essentially (in some sense) the definition of H x K

[boreal inlet]

Okay, got it

But if someone asks me to construct an explicit group isomorphism, then what do I do?

[quiet pelican]

φ: H x K -> C\0 maps to (p_H φ, p_K φ) where p_H and p_K are the projections onto H and K respectively

[rocky cloak]

That's what the sylow theorems are for

[dull tiger]

If I have a group of order 12 and i look at the sylow subgroups, then there will be four 3-subgroups of order 3 and one 2-subgroup of order 4.
This we can see by assuming there are 4 3-subgroups and we know each of those has order 3 and they trivially intersect.
But then to conclude there have to be 8 = 4*2 elements of order 3 in the whole group.

How do I know that the elements in the sylow 3-subgroups have order 3?
Or even more, how do I know that one sylow 3-subgroup consists of the identity (obviously) and then 2 elements of order 3? Couldn't there be one element of order 3 and one of another order?

[chilly ocean]

let H be that 3-sylow subgroup, then H has order 3, so what can be the order of elements of h ?

[chilly ocean]

Lagrange theorem

order of an element divides the order of group

[dull tiger]

ah so they can only have order 1 or 3

thanks!

[chilly ocean]

yes

[glad osprey]

If p = ab is a unit in an integral domain, then a and b are units too, right? I don't quite see how to prove it though

[chilly ocean]

let (ab)c = 1, then a(bc) = 1 and bc(a) = 1

i don't think you need integral domain, you need commutativity

correct me

[glad osprey]

damn, that was too easy, should've seen it thanks

[boreal inlet]

Sorry about this, I didn't quite get what you meant back then.

The projections are group homomorphisms from H x K to H and H x K to K, right? So when you write (p_H \phi, p_K \phi), what exactly are you referring to?

[quiet pelican]

Composition of maps

[boreal inlet]

Won't we rather use the inclusions if we're attempting something like that?

[boreal inlet]

But phi goes to C \ {0} from H x K

[quiet pelican]

Yes I can’t maths

[boreal inlet]

How do we compose that with p_H

[boreal inlet]

Oh so you take H to H x K, and then compose with phi, right?, Similarly for K as well

Now all we need to prove this is an group isomorphism somehow

[quiet pelican]

Yes

[dull tiger]

does the union of all sylow subgroups of a group compose the whole group?

[boreal inlet]

Do you mean the subgroup generated by the union?

I don't think set unions of two groups necessarily are groups all the time

[dull tiger]

No I mean if I have found all the sylow subgroups of a group, does their union equal the whole group?

[thorn jay]

No, elements can have an order not equal to a power of a prime

[dull tiger]

ah right, thanks!

[empty kernel]

|G| = nm; gcd(n,m)=1; N is a normal subgroup of G and |N| = n
prove that N =

[wicked patio]

must they generate the whole group

[empty kernel]

oh wiat i just realised this is also dutch lol
N = {a in G| the order of a divides n}

[wicked patio]

equivalent to every element of order p^k being in some sylow subgroup (p^k is not necessarily the largest power of p dividing |G|)

[empty kernel]

so i first want to prove that there are n elements with this property but i didnt know how to do that cuz i didnt find any theorems to help me

[empty kernel]

sylow?

[wicked patio]

yeah i was responding to my previous statement not to you

[empty kernel]

ooooooh oki mb

[wicked patio]

you're given |N|=n

doesn't that automatically mean all n elements there have order dividing n

[empty kernel]

ye but i have to prove that N is equal to that set so dont i have to prove that that set has n elements?

[wicked patio]

oh, exactly n elements yes

i was confused why you said it was the first step then since that would be sufficient

[empty kernel]

well idk if it matters in what order i do it but i think i have to prove that that set is a normal sub group and that it has n elements

and that it is the only normal sub group of order n

[wicked patio]

try to think about G/N and the order of a coset there and what that means about the order of an element not in N

[empty kernel]

ait oki will try

thank u

[wicked patio]

[thorn jay]

Another dutch dude then, so based

[empty kernel]

ok cant find it
i have proven that it is a normal subgroup as that was quite easy and then i tried to do what u said but like obv every coset has |N| elements and you have |G|/|N| different cosets

but i dont see how that does relate to the order of an element not in N

well i know that the order of an element not in N has to divide m or the order has to be equal to mn

wait a min

that might not even be true

oh dammit

time for a break then i try again

[acoustic igloo]

it's part of this exercise

[glad osprey]

Aren't you supposed to assume |N| = n and N normal, then prove that N = { a in G : |a| divides n }? Proving that { a in G : |a| divides n } has n elements seems irrelevant for that question

[rocky cloak]

I mean it's not irrelevant, since it's an equivalent formulation. But it seems to just obfuscate the relevant info to the problem.

I don't really see any way to prove the second statement other than proving the first and then using that it implies the second

[glad osprey]

Isn't it the converse? I'm struggling to see how they are equivalent

[rocky cloak]

Like every element of N has order dividing n, so you know there are at least n elements.

Then there being no more than n is the same as saying no elements outside N has order dividing n

[glad osprey]

Ah, I see

[coarse solar]

how hard is this subject

i have it on my curriculum but its my first time hearing it

[rocky cloak]

How hard it is probably varies a lot from person to person, but it might be quite different from other math courses you've taken before.

Also might depend on the way it's taught. Often a lot of time is spent on technical details, which means you have to spend some time developing intuition on your own

[coarse solar]

havent taken any course at all so no reference point unfortunately

[empty kernel]

well I assume N has n elements and N is normal and then i have to indeed proof that it has to be that set so idk i assume i have to prove that that set has n elements

[empty kernel]

why is that?

[thorn jay]

Then it's probably going to be pretty hard

But maybe not, you cant really know beforehand how well your brain can handle this abstract stuff

[empty kernel]

it depends into how much detail you go

[coarse solar]

this is the course description

[empty kernel]

ye it wont be very easy

[glad osprey]

Seems like a lot to cover in a single course, I think Galois theory is unusual to see in a first course. And without covering groups beforehand?

Tbf, I don't know how much group theory is required for Galois theory, I'm taking Galois theory this spring

[coarse solar]

no no we have group theory

this is for the 5th semester

its just that i havent heard anything about this subject that

so it made me curious

[sand smelt]

Could anyone please help me understand how to use relabelling to prove this?

Show that if p is prime, S_p is generated by {s, t} where s is any transposition and t is any p-cycle.
-# Exercise 3.5.5 from Dummit & Foote's abstract algebra 3rd edition

[empty kernel]

oh dw then

u prob gonna get introduction courses before that

[glad osprey]

You are given |N| = n, and by Lagrange we know that N <= { a in G | ord(a) divides n }. So if you prove that the latter set also has n elements, then you know they must be equal. However, as jagr said, proving this is likely as hard as proving N = { a in G | ord(a) divides n } to begin with

[glad osprey]

I recommend following this hint, @empty kernel

[empty kernel]

thats what i tried

[empty kernel]

here i tried it

like i dont see what the fact that every coset has n elements has to do with the order of the other elemenetw

well it has the same amnt of elements as the set that we wanna prove has n elements

[rocky cloak]

Say you have an element a with order dividing n.

Let f: G -> G/N be the usual map.

How is the order of f(a) related to the order of a?

What can you say about f(a) from this?

What does this say about a?

[empty kernel]

so uhm

wait

a is in N so f(a) is equal to N so the order of f(a) is 1

[rocky cloak]

Well you don't know if a is in N, that's what you're trying to prove

[empty kernel]

you said that a has an order that is dividing n

[rocky cloak]

Yes, and what you're trying to prove is that such elements are contained in N

[empty kernel]

oh wait you are taking a different aproach

ye uhm so the order of f(a) divides the order of a if im not mistaken

[rocky cloak]

That's right

[empty kernel]

so f(a) also divides n

uhm

idk what it says abt a

apart from obv that a is a multiple of f(a)

[rocky cloak]

Think a little more about what it says about f(a) first

[empty kernel]

okok

[rocky cloak]

Remember f(a) is an element of the group G/N

[empty kernel]

omgomg i think i got it

so f(a) divides n and the group G/N has m elements cuz lagrange

but thats impossible unless f(a) is a common divisor of m and n

but gcd(n,m) is one

thus order of f(a) must be one

so

f(a) is the identity element

and then a must be in N

so N contains all the elements with an order that divides n

but now i just showed that the set is included in N but i also have to show the other way around right?

wait ok lemme think

so if there is an element in N that doesnt divide n lets call it b

then f(b) must be the identity element obv cuz b is in n hmm

ah wait another way

oh wait im a dummy

it follows by definition cuz we said N has n elements then every element in n has to divide n and is thus included in that set
so we did inclusions to both sides so the sets must be equal

yipie

thanks @rocky cloak

[unkempt stream]

No it is jacobson currently

[charred garden]

why does z5 modulo addition subgroup {0,1,4} not satisfy associativity?

[rocky cloak]

{0, 1, 4} is not a subgroup under addition

[charred garden]

why is that? i understand that prime groups only have the identity and themselves as subgroups but is it because of an associativity issue?

[rocky cloak]

1+1 = 2

So you would need to at least add 2, to make it a subgroup

It's not really related to associativity

[tardy hedge]

What is a “p-primary component” when talking about a primary decomposition?

Is it just the primary ideal qi such that r(qi) = p?

[charred garden]

would it be closure instead

[next obsidian]

Big if true

It sort of works. This is more advanced, but when you have something of the form

0 -> A -> B -> C -> 0

(Pretend this means A < B and C = B/A) you can’t uniquely pin down what B is. This is clear in your example you gave. But it turns out there’s a group called Ext^1(C,A) whose elements are exactly the B that can fit in there (up to isomorphism). So if you know stuff about these Exts you can recover the groups

Now this is maybe not all that applicable often, but let’s say you just know the composition factors. You can sometimes from this info say, determine the size of the group. And then among those you can say which ones have those composition factors. And from there if you know even more (like the group has a bunch of different elements of specific sizes) this could help you nail exactly which one it is

And now let’s move onto module theory

If you’re familiar with that

If you have a finitely generated module over a Noetherian ring A, then M has a composition series where all the subquotients are of the form A/pi for a bunch of primes pi

These pi are exactly the set of primes which are the associated primes of M

Okay well, maybe not composition series cuz you can refine it even more maybe

But this ends up being theoretically useful for the following reason:

sometimes to prove a property P for M you can show its true on all the subwuotients

So now you only have to deal with proving it for A/p for different prime ideals p

So it isn’t always the most useful, but just having an idea of how your dude breaks up into pieces is good

Blech

https://en.m.wikipedia.org/wiki/Ext_functor

Look at the Baer sum

It’s in “Ext and Extensions”

The identity is A (+) C!

And okay, to be clear. I’m assuming you have abelian groups here

Otherwise stuff gets kinda grodier

[tardy hedge]

Oh Ext stands for extension. Lol

I havent learned hom algebra yet

[little shadow]

We can endow $\mathbb{R}^2$ with a multiplication such that it becomes a field (the same multiplication that we define for complex numbers). Can we do that for arbitrary $\mathbb{F}^n := \mathbb{R}^n$ such that we have an explicit "natural" bijection $\mathbb{F}^n \rightarrow \mathbb{R}^n : (a_1, a_2... a_n) \mapsto (b_1, b_2... b_n)$?

[cloud walrus]

Dompa

[rocky cloak]

What is F^n ?

Like are you asking if there is a way to make R^n into a field?

[hollow imp]

R^2 is the only power of R that you can make into a field, R^4 can be made into a noncommutative field, and R^8 can be made into some sort of non-associative algebra.

[little shadow]

Well i wasnt quite sure of how to define it to capture the essence of the idea. but with the complex numbers we can "identify" a+bi with (a, b) in R^2

[little shadow]

what concepts/areas of math are used to prove these results

[rocky cloak]

Field theory, specifically algebraic field extensions

At least to prove R^2 is the only field.

[hollow imp]

this article goes over a few proof sketches for this classification: https://en.wikipedia.org/wiki/Hurwitz's_theorem_(composition_algebras)

[little shadow]

shouldnt every power of R be able to have a commutative ring structure?

or what ring axiom would fail with component wise multiplication

[hollow imp]

none, but you would lose invertibility

[little shadow]

thanks!

[hollow imp]

sorry, this article might be clearer: https://en.wikipedia.org/wiki/Frobenius_theorem_(real_division_algebras)

at least for your question

[little shadow]

oh right i've heard about this before but it must've escaped my mind today

[vagrant zinc]

Hi guys, I was reading dummit's book and I came across a comment saying that isomorphism is an equivalence relation, is it possible to prove this ?

[hollow imp]

well it isn't an "equivalence relation" in the traditional sense, as the "set of all groups/rings/etc." is not a set, but to prove this type of thing you want to check that everything is isomorphic to itself, symmetry, and show that if X is isomorphic to Y and Y is isomorphic to Z, then X is isomorphic to Z.

[cloud walrus]

Homology

[vagrant zinc]

In this sense you say

Or is it with respect to groups?

[hollow imp]

sure, but here you are using symbols usually reserved for maps, and what you are trying to prove is that it is an equivalence relation on the "set of all groups"

[boreal inlet]

It's late, but I managed to complete the rest of the proof. Thanks a lot for the help

[narrow wagon]

Isn’t F_p a subfield of C?

Ahh mb, the operations are different

Tho subset

[thorn jay]

Depends on how you interpret F_p

Though i'd see it as Z/pZ, making it not a subset of C

As frankly, the two sets have no relation in any algebraic sense :p

[narrow wagon]

I’m trying to understand the statement: any subfield of C is a field extension of Q

is Z/pZ a field extension of Q, or can I see Q inside Z/pZ?

[thorn jay]

Any nontrivial subring of C must have char 0, so must contain Z, so any subfield of C must contain the fraction ring of Z, being Q

[narrow wagon]

Yes, Both statements are equivalent

[thorn jay]

Z/pZ has no relation with any field of characteristic other than p

[narrow wagon]

Oh, you were trying to give an interpretation about my claim on F_p being subset C

[thorn jay]

This is the proof of what you want

[narrow wagon]

Gotcha

[thorn jay]

Anyhow, I have to go now

Good luck

[narrow wagon]

Thank youy

Have a great day

[boreal inlet]

This is probably a very easy question and I'm missing something very simple - but let me ask.

Is the ideal generated by the union of arbitrary collection of ideals of a commutative ring is equal to the arbitrary sum of them?

Because I know this is true for two ideals and hence true for finite cases.

But I'm not sure if I can extend the same argument in the arbitrary case.

And yes, the arbitrary sum here is the set of all finite sums of elements from the union.

[velvet hull]

I think so(?)

because (the ideal generated by) the union of ideals necessarily contains the sum of those ideals as a subideal

not sure how this could break in the infinite case, but I also can't think of how that could happen

[chilly ocean]

you mean ideal generated by the union of ideals

[velvet hull]

right

[velvet hull]

wait I think you can use Zorn's lemma

[vocal pebble]

here r(a) denotes the radical of a i.e the pre image phi^{-1}(R_{A/a}) where R_{A} denotes the ideal of all nilpotent elements in A and phi is the canonical map A -> A/a

[tardy hedge]

Its just a property. You are right in that the definition of V(E) is all thats needed to show X to be a topological space

[hollow imp]

for B = A \oplus C, you have to find your projections. The splitting maps give you projections onto A or C, depending on whether you have p or s, and the other projection you can find from the projection you're given.

[vocal pebble]

I just realised infact even the first equality is not necessary to verify that X is a topological space. Also, what is the "big picture" here? I found this in a book on comm alg and im able to verify that X is a topological space, but not too sure what this topology really means. If im thinking of Z, then spec Z ={(2),(3),(5),...} and if E1={2,3} and E2={3,4}, then V(E1) intersection V(E2) ={}. what even does this mean?

[hollow imp]

some motivation is that if you have a field k, you can consider the zero sets of (a system of) polynomials in k[x1, ..., xn], which are geometric objects

and these sets are in one-to-one correspondence with the prime ideals of k[x1, ..., xn], closed under arbitrary intersection and finite union, so there's a natural topological structure

so zariski topology is like the extension of this to general rings instead of just polynomial rings

[vocal pebble]

This is all very interesting, il look more into it. Thank you 👍

[rocky cloak]

Yes, and the proof is exactly the same as for two ideals.

1. The sum is an ideal that contains the other ideals.
2. Ideals are closed under finite sums, so any ideal that contain these ideals contains their sum.

In some sense, there's nothing particularly special about A(+)C.

Because it is in fact true that any pair of split sequences between A and C are isomorphic in this way.

A(+)C is just sort of the obvious choice of a split sequence.

[boreal inlet]

Ohhhhh my bad, i got it

[hollow imp]

well you have to show that they are projections, which you do using the exactness property

[thorn jay]

Haha i was just typing this, classic closure operator argument my beloved

Universal algebra is so full of this, i should make a meme

universal algebra proof
*looks inside
proof the object is always contained in a closed set and that it's a closed set itself

Meme made

[coral spindle]

Because not everything splits as a direct sum

[boreal inlet]

😭

Uhm that's definitely not true. For projections onto groups, they're surjective maps. And there definitely exists group morphisms which are not surjective [ Like for example take the group morphism Z/2Z -> C*, where 1 of Z/2Z goes to -1. Except 1 and -1, no other non-zero complex number has a preimage, hence the morphism isn't surjective. ]

[tardy hedge]

A question in a/m says if a =r(a) then a has no embedded prime ideals

So can I assume a has a primary decomposition? U can only talk about embedded prime ideals if a has a primary decomposition, no?

[boreal inlet]

I mean you do need a to be decomposable

So yeah you can asssume there exists a minimal primary decomposition

[next obsidian]

I don’t think you do, you can define the associated primes to a regardless and then talk about minimality amongst those

Now I assume the problem is assuming the ring is Noetherian but idk

[amber burrow]

Let $G$ be a finite group, $H$ a subgroup of $G$ and $N$ a normal subgroup of $G$. Show that if the order of $H$ is relatively prime to the index of $N$ in $G$, then $H \subset N$

[cloud walrus]

donut123

[amber burrow]

My approach was: for any $g \in G$, $gN = N \implies g \in N$. Thus, I thought that if for any $h \in H$, $hN = N$, $H \subset N$

[cloud walrus]

donut123

[amber burrow]

Since $gcd(|H|, |G/N|) = 1$, by Bezout's lemma there exists $a,b \in \mathbb{Z}$ such that $a|H| + b|G/N| = 1$.

[cloud walrus]

donut123

[amber burrow]

Thus, $$hN = h^{a|H| + b|G/N|}N = h^{a|H|}\cdot h^{b|G/N|}N = (h^{|H|})^a\cdot (h^{|G/N|})^bN = N$$

[cloud walrus]

donut123

[amber burrow]

however solutions online all use first isomorphism theorem, so I wonder if I am missing something

[fast crescent]

In a commutative ring, every left ideal $L$ (or right ideal) is also an ideal right? By simply $r \in R$ $l \in L$ $lr = rl \in L$?

[cloud walrus]

parity

[rocky cloak]

No, you're proof looks good.

[amber burrow]

yay!

[hollow tartan]

Quick question. This is ash basic abstract algebra pg 139. I don't understand what the author means by NK <= NH

[velvet hull]

If A,B are subgroups of G, then AB := {ab | a ∈ A, b ∈ B}

[hollow tartan]

I see it's quite literal then, ty!

[south rain]

If $I$ is an ideal and $U$ is a submodule of a ring $R$. Is $IU$ again a submodule of $R$? I mean, of course $IU \subset U$ but is $(IU, +)$ necessarily a subgroup?

[cloud walrus]

dellinger

[next obsidian]

So first of all, a submodule of a ring is exactly an ideal

Second, this is true but only because IU is defined to be sums of things of the form iu

So that it’s closed under addition

[south rain]

Yea, I'm sorry, I phrased that wrong, I meant submodule of a module $M$ over $R$

[cloud walrus]

dellinger

[next obsidian]

Either way my point stands

U close under addition in the definition

[south rain]

Alright yea, maybe I'm just oblivious rn. Our definition of submodules states:

Let $M$ be a $R$-module and $N \subset M$. Then $U$ is a submodule of $M$, if $(U, +)$ is a subgroup of $(M, +)$ and $RU := {ru : r \in R, u \in U} \subset U$.

I kind of just translated the definiton of $RU$ for $IU$ and called it a day, so thats why I'm confused.

[cloud walrus]

dellinger

[rocky cloak]

If your question is whether
{iu : i in I, u in U} is closed under addition, then this is not necessarily the case.

But yeah IU is the submodule generated by {iu}

[south rain]

Thanks, yea this is actually concerning a problem I posted here about a week ago. Something bugged me when reviewing it, and it was exactly this problem.

It was this problem

I used that IN is a submodule. But I don't know how I magically go there from $I$ just being an ideal, i.e. $RIN \subset IN$.

[cloud walrus]

dellinger

[south rain]

Like, I use, a mere submodule $N$ and since $I$ is an ideal that $RI \subset I$ and then just get the relation $RIN \subset IN$ but maybe I understood that wrong.

[cloud walrus]

dellinger

[rocky cloak]

So are you still confused?

[south rain]

I guess so. I get the proof idea, but it relies on $IN$ being a submodule, which if we plainly define it as $IN := {in}$ (as was done in our definition of $RN$) then it doesn't hold anymore so the proof falls apart.

[cloud walrus]

dellinger

[rocky cloak]

Right, but if you just define it the usual way then everything works

[south rain]

Lmao, yea I thought so too, but that kinda felt like cheating, but then again we must define what we mean by IN. Does this imply that the usual definition of $RN$ is also a sum of rn rather than just the elements?

[cloud walrus]

dellinger

[south rain]

It felt like cheating because we defined RN in such a way and I though why should it be any different if I have an ideal I.

[rocky cloak]

Sure. But this is all just a notation thing.

Nothing really about your proof

[south rain]

But that notation would change my proof drastically right? Because if I don't define it the right way it's invalid since its no submodule.

I mean I will just define it that way, since I can't be bothered to find a different strategy lmao.

[rocky cloak]

I'm just saying, all the steps in your proof work, if you just assume the symbols mean what they usually mean.

Like whenever you use that IN is a subgroup, just use the definition where IN is a subgroup. Okay, then it works, that's it

[acoustic igloo]

my book used Z(G) to denote center but in the answers they used blackboard bold Z

odd

[thorn jay]

Duality of man book

[south rain]

Yeah, I did just define it that way. Thanks again for the help.

[inner harbor]

why wtf

N is normal and H is just a subgroup , both supgroups of G

and N is acting on G/H somehow

[mighty kiln]

G acts on G/H

[inner harbor]

isnt quotient defined only for normal subgroups btw

[mighty kiln]

G/H is the set of left cosets of H

It's a G-set but not a quotient group

[inner harbor]

oh ok

where does that eq come from i dont rly get it

cause at the end we only want = |N|/|NinterH| and its easy since N is normal but this one seems to be magic

[mighty kiln]

The size of the orbit of gH is [N : stab(gH)]

[inner harbor]

oh yea thats it

i forgot this one bru

ty mate

[spark veldt]

Why is this so?

[inner harbor]

isnt this book introduction to commutative algebra ?

either n = m and u get u = u' at the end cause u cancel some pi at each side in each step

or m is larger than n and after canceling n pi's there is n-m pi left at the right of the equation and u get some unit = product of n-m pi factor which are not units by hypothesis

actually u can stop the proof here, but they decided to mention that u can do it symmetrically as well

oh nvm they just get two inequalities without mentionning that its absurd to have m > n

[spark veldt]

this is herstein

[inner harbor]

nvm then

[spice whale]

one side will be a product of units, which is itself a unit
the other side will be a product of the pi'_i, which are not units, and therefore it itself won't be a unit

unless there are 0 pi'_i in the product on the lhs

[narrow wagon]

I’m trying to understand this definition, does this mean F-isomorphism is an identity function for the elements in the underlying field F?

ie for all x in F subset of K, phi(k) = k

K= Q<sqrt(2)>, K’ = Q<sqrt(3)>

phi: K to K’, phi(x) = x for all x in Q

[rocky cloak]

Yes, F is a subset of both K and K' and you want phi(x) = x whenever x is in F

[narrow wagon]

Thank you

[woeful sage]

Herstein my behated

[serene dune]

$0 \to \mathbb{Z}4 \to \mathbb{Z}{16} \to \mathbb{Z}_4 \to 0$ is an exact sequence

\ \

$0 \to \mathbb{Z}4 \to \mathbb{Z}{16} \to \mathbb{Z}_2 \to 0$ is a a semi exact.
but for both cases composite of two maps produces zero morphism

[cloud walrus]

nastasya

[serene dune]

i get that two situations gives the same situation and the converse is not true, however why is it called zero morphism?

the book uses it for first time without defining it

what does that has to say in general with or without context

[rocky cloak]

The zero morphism is the map that maps everything to 0

[serene dune]

right!

[toxic zephyr]

say p|q and q is irreducible. doesn't that automatically imply that p~q are associates?

[coral spindle]

1 | 2 and 2 is irreducible in Z

[quiet pelican]

p could be a unit

[toxic zephyr]

oh then how about if p is nonzero non unit (ex. a prime)

[quiet pelican]

Then they’re associates

[toxic zephyr]

okay cool. my prof did a longer proof showing the ideals they generate are equal and I was confused because it seemed much more straightforward

[quiet pelican]

I suspect the proof either assumes less things are already proven than this does, or is just this in fancy clothes (or is proving something else)

[toxic zephyr]

it was a step that
(irreducible implies prime)
iff
UFD
in a Noetherian integral domain. but I think it's more general to any integral domain with unity?

[rocky cloak]

I think you can have rings where all irreducible elements are prime, but not every element can be written as a product of irreducibles (hence not a UFD).

So you need something like Noetherian to guarantee that

[unkempt stream]

Another way to view this is ideal-wise.

If we have an integral domain R (not necessarily with unity), then let W be the family of ideals of R. The principal ideal function x |-> (x) sends elements of R to ideals of W, and the image is the family of principal ideals. Obviously ideals admit an inclusion relation. If we define the relation on R where y | x iff (x) is a subset of (y), then that's the divisibility relation. Two elements are associate iff they have the same principal ideal

(i.e divisibility is the reverse of the pullback of the inclusion relation on ideals under the principal ideal function)

which is why often the conditions for different types of rings are on the ideals and not the elements themselves

[quiet pelican]

Yes
K[x, x^1/2, x^1/4, x^1/8, ...] has every irreducible prime, but factorisations don't necessarily exist iirc

[rocky cloak]

Yeah, it's clear you don't have factorizations.

It seems reasonable that irreducibles are prime, but it's a little tricky to see what the irreducibles are. Though I can totally believe they are prime

[unkempt stream]

For instance, if we don't have a "noetherian"-esque condition on principal ideals, then we can have an infinite ascending chain of principal ideals (x_1), (x_2), (x_3)... which is very problematic, as we'd have an infinite DESCENDING divisibility chain which DESTROYS the idea of having factorizations

[quiet pelican]

Let y be irreducible, y | ab
Then y, a, b are in some K[x^(1/2^n)]
But this is just a polynomial ring, and y is still irreducible in K[x^(1/2^n)] so y is prime in K[x^(1/2^n)] so WLOG y divides a in K[x^(1/2^n)] so y divides a in the original ring

[unkempt stream]

If elements factorized then we'd only be able to create a finitely long descending divisor chain

[pearl fog]

https://math.stackexchange.com/questions/3356668/irrationality-of-cos-2-pi-n I need some clarification on logic here
If cos rational, then order of galois group <=2
Doesn't this feel like one of those if it works, it works scenario? If cos rational imposes restriction on order of galois group, and one literally uses elements of galois group for rational cos a fixed point?

Or is the proof just that silly and clever

Cuz by close inspection on detail I can't find circular reasoning on the smaller pieces

[rocky cloak]

I'm not sure I understand what you're asking.

But the Galois group of cos(2pi/n) is isomorphic to (Z/n)^* / {±1}. And a number is rational iff it's Galois group is trivial. So this happens iff 1 and n-1 are the only numbers relatively prime to n (less than n)

This uses some results from Galois theory, how the rationals will be the only elements fixed by the Galois group, but otherwise it's straight forward.

[rocky cloak]

Where are you expecting circular reasoning?

[pearl fog]

Having rational cos and constructing galois group

[rocky cloak]

Like in proving what the Galois group of zeta_n is?

I'm not sure I follow

[pearl fog]

Somewhat yes

Like cos rational is the requirement for n

But with building gal extension zeta requires n itself

[rocky cloak]

What do you mean by this?

[pearl fog]

cos rational is the restriction on n

[rocky cloak]

Like you're trying to determine for which n you have cos(2pi/n) rational.

Then you give an equivalent characterization, and list out which n satisfy this

[pearl fog]

While n determines the galois group

[rocky cloak]

n determines the Galois group yes. n determines everything, it's the only information given

[pearl fog]

And then galois group fixed point thing and {1,n-1} is both iff

So we get to determine n

[rocky cloak]

Yeah, it's iff everywhere

But you really only need one direction, since once they've eliminated most n, they just check that the remaining do indeed give you something rational

[pearl fog]

I think I get it

[unkempt stream]

If you want you can try and use induction and Eisenstein's criterion on the Chebyschev polynomials, the polynomials T_n : cos(nx) = T_n(cos(x)), which can be recursively defined.

[amber burrow]

Are there any youtube channels with 3blue1brown style videos for group theory?

i know mathemaniac has some but only theres a lot he doesnt cover

i find the animations helpful for intuition

[merry harness]

https://youtu.be/KufsL2VgELo

[rocky cloak]

https://youtu.be/mH0oCDa74tE?si=POKk_GK8wxEkLCPB

[serene dune]

i need a bit of motivation for the total picture of four and nine lemma

also it seems to me that there are multiple equivalent statements

[cloud walrus]

nastasya

[serene dune]

if the rows are exact

[cloud walrus]

nastasya
Compile Error! Click the reaction for more information.
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[serene dune]

if α,γ are epimorphisms and δ is a monomorphism then β is an epimorphism

especially these two

having a hard time to wrap it

[spark veldt]

How? All I’m seeing is b is a unit but why is b=1?

[storm kiln]

They are working over the integers

and they stated that a and n_0 are positive, so r is positive

[vivid kestrel]

just for a sanity check, if we have a symmetric polynomial in n indeterminates and we know that it contains a monomial of total degree d, so a monomial of the form $x^{(a_1, a_2, a_3, \ldots, a_n)}$ where $\sum_i a_i = d$ then it must also contain a monomimal $x^{(a_{\pi(1)}, a_{\pi(2)}. a_{\pi(3)}. \ldots. a_{\pi(n)})}$ for every $\pi \in S_n$ right

[cloud walrus]

eggman

[lusty marlin]

Yes

[vivid kestrel]

yippee

[spark veldt]

Why did they substitute and then conclude irreducibility I don’t get it

[distant carbon]

for x=2 you get 7 for x=3 you get 13 etc.... 7 mod 2 != 0, 13 mod 2 !=0 etc etc...., they lack the proper induction method, it's only partly covered here by only showing the idea, you'd have to formulate that yourself so to say

[rocky cloak]

A degree 2 or 3 polynomial is irreducible iff it doesn't have any roots. (Since if it was reducible it would factor into a linear factor).

So if you just check all the elements of your field and none of them are roots, then that's that

And F has only two elements 0 and 1

[distant carbon]

ah yeah, the field of integers vs. the field of integers mod 2, in the former case he'd have to prove it inductively right ?

[rocky cloak]

I wouldn't use induction, just use that ring operations can be done mod 2

So for example
(2n)^2 + 2n + 1 = 0^2 + 0 + 1 (mod 2)

[distant carbon]

wait a little bit, the field of Integers mod 2 has only 2 elements really ?

[rocky cloak]

Yeah, or why is that surprising

Every integer is either 0 or 1 modulo 2

[distant carbon]

Z_2 == {0, 1} right

as a set it reduces to only 2 elements right

[rocky cloak]

It has only two elements yeah

Usually you would realize them as cosets Z/2Z

[spark veldt]

Ah i see okay thanks!

[sturdy spear]

Oh cool. You are studying AA.
Iirc i saw you working with Abbott some time ago

[spark veldt]

Could I get a hint on how to answer this?

[spark veldt]

haha yeahh I recognize you :D

[sturdy spear]

oh cool :D
nice to see you here buddy :)
i will start AA soon

[coral spindle]

Think of a map F -> K, and recall a nice property of all maps of fields.

[dull tiger]

Is (0) always a prime ideal?

[mighty kiln]

(0) is prime iff the ring is reduced

[dull tiger]

perfect, thanks!

[neat zodiac]

can anyone do this question

[thorn jay]

Let z = xyx^-1

And notice that xy = xy is vacuously true

[glad osprey]

I thought an ideal I is prime if ab in I implies a in I or b in I? Doesn't that mean that (0) is prime iff the ring has no zero divisors?

[thorn jay]

Yeah, thats true