groups-rings-fields

[steel pulsar]

A group G is cyclic if ∃ a∈G such that ∀g∈G,∃ k∈N such that g=a^k

a^k is generator when gcd(order, k) = 1

[coral spindle]

You're overcomplicating.

Remember this fact.

[steel pulsar]

order(a) = 4

[coral spindle]

Great

So if G = Z/2Z x Z/2Z were cyclic, there would be a generator with order 4.

Now can you see why this means G is in fact not cyclic?

[steel pulsar]

because we have only 0 and 1 ?

[coral spindle]

But that's not true

Let's try focusing in

[steel pulsar]

Why (0,1) is order 2 ?

[coral spindle]

OK

What does it mean to be order 2

How can we work out the order of a group element

[delicate orchid]

to elaborate, if F_27 was a field extension of F_9 it would be a F_9 vector space, and thus must have order equal to a power of 9

[wraith oak]

How do we know we can construct that homomorphism?

[steel pulsar]

A group G is of order n if it contains n elements

A group G is of order 2 if it contains 2 elements

[coral spindle]

OK but we're not talking about the order of a group.

We are talking about the order of the element (0, 1)

Do you know the difference?

What does it mean for an element to be of order 2?

Please don't spend ages replying. If your answer is just "I don't know" then just say it.

[steel pulsar]

Let G be a group and a be an element of this group. The order of a is the smallest positive integer such that a^n = e where e is the neutral element (the identity) of the group

[coral spindle]

That's right, great

So can you now use this to see what the order of (0, 1) is?

[steel pulsar]

I don't know

[coral spindle]

OK, so what part are you struggling with? You have the definition of the order, so can you apply it?

[glossy crag]

Not to hate on Pg and his struggles, but the tenor of this conversation strongly reminds me of this.

[coral spindle]

This is so validating you have no idea

[rustic crown]

hii Boyt

[steel pulsar]

(0, 1)² = (0, 1)

[coral spindle]

Hey det

[coral spindle]

OK there's the confusion. This isn't right

The group operation on Z/2Z x Z/2Z is not multiplication!

The group operation is (a,b)(x,y) = (a+x, b+y), where + is the operation of Z/2Z.

Clear?

[steel pulsar]

Ok

[coral spindle]

If (0,1)^2 = (0,1), then every element of Z/2Z x Z/2Z would be the identity, which is obviously impossible.

Now

Is it now clear why the order of (0, 1) is 2?

Time to make a cup of tea ig

[steel pulsar]

To show (0,1) is order 2 , I I have to multiply it by 2 and see if that gives the neutral element (0, 0)

[coral spindle]

You have to add it to itself, yes

[steel pulsar]

But 2 * (0,1) is multiplication

[coral spindle]

Sure, whatever you feel most happy with.

[steel pulsar]

what is you a and b here ?

and you x and y

[coral spindle]

a,b,x,y are any elements of Z/2Z.

[steel pulsar]

but 2 * (0,1) = (0, 2) not (0, 0)

[coral spindle]

OK, I give up

This is like an unravelling shirt: I've pulled a thread and the whole shirt has fallen apart.

[steel pulsar]

Ok sorry

[coral spindle]

You need to do some reading in your own time, because you are missing a lot of knowledge.

Good luck studying.

[tardy hedge]

You dont “multiply” it by “2”, you compute (0,1)^2 and see if its the identity

The actual operation to finish the “^2” computation depends on the group ur working in

[steel pulsar]

Well from what I understood, he told me that (1,0)^2 is false

because the law is addition and not multiplication in this case

[tardy hedge]

(1,0)^2 = (1+1, 0+0) = (0,0)

so (1,0) is order 2

because you can see (1,0)^2 is (0,0) which is the identity in this group

[steel pulsar]

why (1, 0)^2 ? is not 2 * (1, 0) in this case ?

[tardy hedge]

well why do you think its 2*(1,0)? what are u basing that on

[steel pulsar]

(1,0) + (1, 0)

[steel pulsar]

.

[tardy hedge]

in a group there is a defined operation

one operation that is defined

so in the group we're working with, the operation is (a,b)(c,d) = (a+c, b+d)

[crystal turtle]

The n • x notation is common for abelian groups.

[tardy hedge]

well also (0,2) is the same as (0,0)

[steel pulsar]

In fact what I didn't understand is why noted 2 * (0, 1) = (0, 2)

If 2 = 0 in Z/2Z

2 * (0, 1) = 0 * (0, 1) = (0, 0)

[tardy hedge]

ok sure

[steel pulsar]

telling that 2 * (0, 1) = (0, 2) in my head confuses me

[tardy hedge]

yea so (0,1) has order 2

[spice whale]

in this context (0,2) = (0,0)

[steel pulsar]

ok thank you

[hidden haven]

Send x to a root of f

(I didn't include the details because I only meant it as a hint)

[runic galleon]

U study in english and in french are u Algerian ?

[median rock]

yo can i ask something?

if i have a group G and H and T are subgroups of G ,also i know that T is a subgroup of H ,can i apply lagranges theorem to T and H?

G is finite of course

[viscid pewter]

sure, why not

T is a subgroup of H so you can just go

[median rock]

so i can say that |H|=k|T| right?

[viscid pewter]

yes

why not

[runic plover]

,tex Is there an easy trick to finding an $R[x]$-linear combination of two polynomials $f(x)$ and $g(x)$ as an expression of $\gcd(f(x),g(x))$?

[cloud walrus]

cat_food_sounds

[runic plover]

With f(x) and g(x) as elements of R[x], that is.

[runic plover]

Nvm, remembered how to work backwards with the Euclidean Algorithm

[white oxide]

is it true that if $E^x/K^x$ is finite, say its members are $K^x, e_1 + K^x, e_2 + K^x, \dots, e_n + K^x$, then $E = {0} \cup \bigcup_{i = 1}^n \langle e_i \rangle$?

[cloud walrus]

okeyokay

[white oxide]

this is my suspicion and i'm trying to prove it - i've already shown that each <e_i> is a proper subspace of E (since they're not a unit of K and E is a K-vector space)

[vagrant zinc]

This looks like a linear combination.

[white oxide]

yeah, i'm trying to get to the suggested contradiction that E is a union of finitely many one-dimensional subspaces

[white oxide]

does anyone know if my approach is correct/how to start on this problem?

[buoyant echo]

hi can anyone help prove prove that for p(x) = x^3 - x - 1 in field Q, since p(x) is irreducible over Q, and let a be a root of p(x), then [Q(a) : Q] = 3.

[vagrant zinc]

I think that a little linear algebra can help. I had seen that the finite union does not necessarily generate the entire space, but its intersection does generate it. Look, that section must be in the direct addition part.

[delicate bloom]

two ways to do it, rational root theorem says if it has a root it must be 1 or -1, plug and check - no linear factor, so irreducible.
Other way, reduce mod 2 and see that it's always 1 mod 2

[faint fractal]

Could someone help with proving this definition of algebraic integers?

[buoyant echo]

rational root theorem says if it has a root it must be 1 or -1,, can you give a link to this theorem ?

[white oxide]

Well <= should be straightforward, first thought for => was that if you have some polynomial with root alpha then p divides that polynomial, but that’s only in regards to p being the minimal polynomial over a field

[faint fractal]

why does p divide that otoher polynomial

oh

I should've kept reading

hmmm

[delicate bloom]

no just look up 'rational root theorem'

oh I didn't even answer your original question, I was just mentioning how to prove it's irreducible but your question is already contingent on knowing it's irreducible maybe that's why I've confused you sorry

[delicate bloom]

so going back to this question, use the fact that a^3=a+1 to see why every element of Q(a) can be written as x+ya+za^2

with x,y,z in Q

[buoyant echo]

is it a general formula that if a is a root of an irreducible polynomial in K, then [K(a): K] = n ?

[delicate bloom]

I guess we should try to prove it

[last spoke]

B+ in Galois/Number theory, woooooo

[mighty kiln]

Artinian (DCC on ideals) integral domains are fields

(Not to be confused with Noetherian (ACC on ideals))

[white oxide]

why doesn't the proof stop at z in K(w)? isn't K(w) a subfield of C which is obtained from adjoining a square root, and which contains z?

[hollow jackal]

Please help me find flaw in argument 3.8 above

[mighty kiln]

H might be empty

[hollow jackal]

Doesn't point number 1 imply that there has to be some element in set .
There can't be binary operation in a set if it's null set.

[mighty kiln]

You can have the empty function HxH -> H

[hollow jackal]

It makes sense.

Thank you so much

[versed light]

Let $n \in \mathbb{N}$ and $a,b \in G$ where (G,) is a group, show that if $ab^n = b^{n+1} * a$ and $ b * a^n = a^{n+1} * b$ then $a=b=e$

[cloud walrus]

Cyrenux

[versed light]

#help-32 message

Here is what i have done

[buoyant whale]

is G an abelian group?

[cloud solar]

We have a group G with order p^n with p prime and n>=2 natural. G is not cyclic. Show that the number of distinct subgroups of G is at least p+3

What I tried:

Since G is a p group

The center is not trivial

So there must be an element x in G* that is in Z(G)

The subgroup <x> is normal

And if we look at G/<x> we can see it has at least p elements since the order of x is p^(n-1) at most

But the cosets of <x> are not subgroups

I mind if we can make some bijection between the cosets of <x> and subgroups

Another thing I have tried is to look at Sylow subgroups

[versed light]

That is not given i believe

[sinful kite]

Is this true? For n=2, G with order p^2 but not cyclic, it should have p+1 subgroups of order p. And, +2 for the trivial group and G itself, so we get p+3.

[cloud solar]

Yes

I think this theorem is related to my problem

[sinful kite]

I'm wondering how to extend the argument to n>2

[modest hawk]

Have you studied Sylow's theorem by any chance ?

[cloud solar]

Yes I know and I tried

There are at least n subgroups by sylow I

[modest hawk]

ok you know that for all $g\ne 1$, you have $g^n\ne 1$, so that every $g\ne 1$ belongs to some maximal subgroup $H_g$ meaning $\bigcup_{g\ne 1}H_g=G$. You know that $|H_g|\le p^{n-1}$, try using that to show that there are at least $p+1$ distinct subgroups $H_g$.

[cloud walrus]

upheaval

[delicate orchid]

Sylow isn’t relevant here, you’re working in a p group so the entire group is the sylow subgroup

Your goal is to show that any non-cyclic p-group contains Cp x Cp, as then that subgroup will itself have p+3 subgroups

[modest hawk]

yea I missread the question but indeed Sylow doesn't apply

[delicate orchid]

If you’re stuck on this, then note that any non-cyclic abelian p group has to contain Cp x Cp

See if you can prove that first

[carmine minnow]

Please help, stuck on all subquestions for weeks

[misty cosmos]

Try proving the axioms of a group given in each subquestion.

Like think about how the existence of a left identity and left inverse imply the rest of the axioms. We don’t need to prove associativity cus it’s given by the operation.

[delicate orchid]

assume g in G is idempotent, i.e g = g^2
then eg = g => hg^2 = g => hg = g => e = g
use this to show that gh is also e, and then you show that the left identity is also a right identity via bracket reshuffling on the equality g = eg = (gh)g

for a)

c implies b, those statements being true are essentially saying cancellation works
c implies a, set h = g then we have left/right identities and immediately have that xg = gx = g so the identities are two sided and equal, then set h = e to get inverses

a implies d, gx = h has a unique solution - specifically x = g^-1h

[carmine minnow]

thx, but how do you talk about idempotents when you're not even given that G is a group

[coral spindle]

You only need a binary operation to define idempotents; you don't even need associativity to define them.

[carmine minnow]

i mean

how do you know there is one

[delicate orchid]

well if G is a group there aren't any non-trivial idempotents so I have no idea what you're getting at

ah

gh is an idempotent

[carmine minnow]

so substute g with gh?

[delicate orchid]

apologies, I meant hg

hg is idempotent so it's equal to e
so g = eg = (gh)g = g(hg) = ge = g
so e is a right identity as well

[carmine minnow]

it says you can find a left inverse though

but i think i can work it out

thx

[delicate orchid]

yes, it's up to you to show that (hg)^2 = hg, and so hg = e

the main logic here is that if you have a left/right invertible idempotent then that idempotent must be the left/right identity

[carmine minnow]

for g find h, then g = eg = hgg

wait

huh?

[delicate orchid]

(hg)^2 = hghg = h(gh)g = heg = hg
thus hg = e by the above argument (#groups-rings-fields message)

[carmine minnow]

uh

i need a moment

[halcyon peak]

well if u like u can just avoid using idempotents

[delicate orchid]

yeah if you have a cleaner solution feel free to say

this is just what popped up into my head

[halcyon peak]

if hg = e, take left inverse of h, say kh = e,

[delicate orchid]

how do we know k exists

[carmine minnow]

it says that you can find it

like proving (a)

[delicate orchid]

oh yeah sorry I'm getting left/right backwards

as an aside I find it funny that the question after this is a very basic consequence of Lagrange's theorem, putting semigroups before even basic group theory is bizzare pedagogy and I've never been a fan of it

[halcyon peak]

then gh = egh = khgh = kh = e

[delicate orchid]

yeah that's nice

listen to mikinori instead of me, logic

[halcyon peak]

i think it's essentially same thing but just cleaner

[carmine minnow]

oh i see

I see how both work now but i was just a bit messed up with substituting stuff before

what abt Q30

[halcyon peak]

hint

[delicate orchid]

lagrange's theorem

[halcyon peak]

i guess it's just aG = G

then u multiply all elements on both sides

[halcyon peak]

ah

i thought it refers to a classical proof of fermat's little thm

[carmine minnow]

i had the sense but i couldn't write it down clearly

At least this chapter didn't give theorems about how to get the "prime" or basic elements

like what $\sigma$ and $\tau$ usually is

[cloud walrus]

Logic_VY

[delicate orchid]

I don't see any sigmas or taus

[carmine minnow]

uh

(maybe only in this book?) the write eg. $S_3 = {\epsilon, \sigma, \sigma^2, \tau, \tau\sigma, \tau\sigma^2}$

[cloud walrus]

Logic_VY

[wraith cargo]

Are you sure they don't mean D_6?

[carmine minnow]

aren't they isomorphic?

[wraith cargo]

They are I just realized that lol

[carmine minnow]

on the book, it didn't mention D6 yet though

[delicate orchid]

ok then sigma can be either the permutation (123) or (132) and tau is either (12), (13), or (23)

they're just elements of the group as long as you know how they multiply who cares what they are

[carmine minnow]

oh sry

i found the wrong hint (

[carmine minnow]

yeah it was truly

but i got the sense of how it works but i couldn't write out a clear proof

[halcyon peak]

for any a, g -> ag is a bijection G -> G

[carmine minnow]

create an equivelance relationsip?

and say that the equivelance classes had same amount of elements?

cuz i thought of this idea of having like 1 x, x^2, x^3, ... as a family, and consider the disjointed families a ax ax2 ax3 ....... b bx bx2 ......

but it just felt sketchy to me

[halcyon peak]

u can consider this as a particular case of (30), taking G = (Z/pZ)*

[carmine minnow]

sry not familiar with this notation

[halcyon peak]

Z_p \ {0} in this image

[carmine minnow]

with multiplication ?

oh i got it

this book im reading says \mathbb{Z}_n*

[carmine minnow]

still didn't understand how to generalize, I'll try my equivalence relationship method by saying a equiv b iff there is m s.t. ax^m = b, but still thx

maybe tmr, im 6 hrs from school

[delicate orchid]

I think you're onto something

[carmine minnow]

clearly i can only deal with the finite case right

since like everynumber is a factor of inf?

[halcyon peak]

write G = {g1, ... gn}, look at aG

[woven panther]

I have a problem where given A and B are free groups, show that A X B doesn’t need to be a free group

This problem is difficult for me because I’m being told to observe additional properties that free groups are not required to have and those properties disprove it being a free group, but it seems like that would just make it a free group with extra properties..

[delicate orchid]

the whole point of free groups is that the elements don't satisfy any more relations than they absolutely have to in order to be a group

so if a free group "has extra properties" it won't be free

to be specific, find a group without those extra properties, then the standard map from the universal property won't factor through properly

[woven panther]

I don’t understand that last sentence

[delicate orchid]

as part of forming the direct product of groups you add in relators so each generator of A commutes with each generator of B - this means the group isn't free, consider any group where none of the generators commute and see if you can go from there

[woven panther]

Alright thank you

[untold basalt]

Since when is ${1}$ a subgroup of $(\mathbb{Q}^+, +)$?

[cloud walrus]

Seagull

[naive whale]

every group has the trivial group as a subgroup?

[delicate orchid]

the group operation of Q^+ isn't addition

cause then there wouldn't be any inverses!

very strange

unless there's some secret other operation on Q^+ that has 1 as the identity that I don't know about
like some kind of commutative 1 dimensional formal group law

[south patrol]

Q^+ is then bad notation

[delicate orchid]

wait are they using "additive" to mean ABELIAN?

[south patrol]

i think they mean like

you take Q^x and use the notation +

instead of normal multiplication

[delicate orchid]

that's deranged

[south patrol]

yeah it seems very confusing

Wait now I'm confused too lol

[delicate orchid]

I think it is just Q*_{> 0} as you said

[south patrol]

Not > 0 right

Uh this is weird lol

[delicate orchid]

it's the strictly positive rationals under multiplication

[south patrol]

This is weird

[delicate orchid]

it has to be

[south patrol]

Oh sure

Okay

Wait but this says you can multiply by negative rationals too

[delicate orchid]

it has to be for that map to be an automorphism

[south patrol]

since t can be negative

Or is this just horrific notation for something else

[delicate orchid]

no, recall that we're using additive notation here

so -r is 1/r in sane world

[south patrol]

yeah okay then it's bad notation

yeah sure

[delicate orchid]

yeah it's fucking dog water notation

[south patrol]

probably takes longer to understand notation than to actually do the problem

lol

[delicate orchid]

I mean yeah the problem is easy now lol

take orbits under the action of Q-{0}

wait. No I still don't buy it. If t = 1/2 then theta_t most definitely maps outside of Q^+

[delicate orchid]

unless it's actually this and tr does quite literally mean t times r

but then that's not an automorphism, hell it's not even a homomorphism

[mighty kiln]

Q^+ = rationals under addition and 1 denotes the identity

(When reading comprehension)

[chilly radish]

(misread)

[mighty kiln]

(When reading comprehension indeed)

[rocky cloak]

I'm guessing they meant {0} and that it's just a typo

[versed light]

(Can anyone ping me if they find a way to solve this)

[tardy hedge]

No

[rustic crown]

say phi_a is the conjugation by a, i.e. phi_a(g) = aga^(-1)
notice that a^n b a^(-n) is applying phi_a on the element b, n times.
and you already know what phi_a(b^n) is in terms of b. try playing with a lil ><

[delicate orchid]

important question, is n any integer or any natural

it doesn't actually make a difference for my proof it just means I'll have to show why if it's true for any natural it's true for any integer

lmk when you solve it cyrenux so I can post my proof

[chilly ocean]

Doesn't every collection of sets have minimal element?

[rocky cloak]

Not in general no

[coral spindle]

Consider the collection of sets of the form [x, 1] where x < 1 is a real number.

[rocky cloak]

For example the set of infinite subsets of N doesn't

Or more relevant to your image the set of ideals (2^n) in Z

[rustic crown]

are you using something super clever which really depends on n being a natural?

[delicate orchid]

no it requires n being an integer

and tbh I've forgotten the proof completely by this point

[rustic crown]

yea would be just some messy conjugation stuff ><

i didn't pay full attention when i was thinking about it, but ig i only needed that n and n+1 are coprime and have different parities

[lethal cipher]

I've been stuck on 7b) the 1st and 3rd conditions of a partition I've had little trouble with. But I'm stumped on how to show images either coincide or are disjoint

[wraith cargo]

Contradiction?

[lethal cipher]

Well that's likely the overall approach, yes. I'm having a hard time figuring out what I can use from my assumptions to show said contradiction

[carmine minnow]

how do you prove that the solution to gx=g is same for all g though

[white oxide]

does anyone know?

[delicate orchid]

set g = x, then x^2 = x so we can use my idempotent nonsense again to show that x is the left identity, and is thus also the right identity

[rustic crown]

hi dackid

[carmine minnow]

can you really do that

it says that gx=h us solvable

doesnt that mean gh need to be constants?

[delicate orchid]

I think it's quite literally the opposite

given any g and h we can find an x such that gx = h

says right here

oh right no I see what you're saying

[carmine minnow]

since if i say ax=b is solvable in the a b in real

[delicate orchid]

oh well I don't really care enough to try and rediscover what my logic was

[carmine minnow]

it doesntt mean that xx=b is?

[wraith cargo]

show that for every intersection sigma_i(B) \cap sigma_j(B) this is equivalent to showing there's a nonzero proper intersection B \cap sigma_k(B) for some k

[carmine minnow]

but like ax=b is solvable in R but xx=b is not

[white oxide]

can somebody explain to me how you would construct an angle phi/3 with ruler and compass that doesn't involve obtaining the point X (since that's the problem)?

[wraith cargo]

lmao

[rustic crown]

they first tell you the problem in words, and then rephrase it in terms of being able to construct the point X from a triple of given points

[delicate orchid]

... lol?

[wraith cargo]

isn't trisecting an angle a thing that's famously impossible to do via ruler/compass

[delicate orchid]

it is

there's a funny galois proof as per usual

[rustic crown]

they're prolly gonna prove that after that paragraph?

[wraith cargo]

we had that as a hw

[delicate orchid]

one may hope

[white oxide]

oh so the problem is "Given an angle of measure phi, construct an angle of measure phi/3 with ruler and compass"

oops

LOL

[white oxide]

how exactly does multiplying both sides by i imply that t, t' is in K?

my brain is dead

[celest furnace]

Maybe it’s related to that K = over line K

Like now t,t’ are the solution of a system of polynomials right

[white oxide]

oh yeah, i guess that makes sense

[white oxide]

does anybody know what it means to mark off the distance "against" the circle S?

i don't really know how to start on this problem otherwise... because i mean, aren't the points just going to be vertices of the heptagon?

and clearly 1 is such a vertex, but 1 is not equal to that complex number or any nth power for n = 1, \dots, 7?

[lethal cipher]

Hey det! Long time no see 😄

[tardy hedge]

anybody like Tool

[lime junco]

how can quotient groups be disjoint

[mighty kiln]

A disjoint union of quotient groups?

[prisma ibex]

what is X here?

[lethal cipher]

Remember, The elements of these quotient groups are cosets of the form xH_i, not the elements in these cosets.

[lime junco]

arbitrary set

the context of the problem is a group action

[prisma ibex]

okay so this is just a map from a set to a disjoint union of cosets

[lime junco]

i think i understand what they want me to do i just have no clue wtf they mean by G/Hi disjoint union

[lime junco]

is this just cosets of different sets

will they then be disjoint

[rose prism]

how would you prefer to combine multiple sets into a set

[prisma ibex]

you're literally just taking a bunch of cosets and taking their disjoint union as a set

[rose prism]

maybe the confusion is

a precise definition of disjoint union

you can take disjoint unions of anything

[lime junco]

oh wait thats it

[rose prism]

it ignores any ambient object which spawned the constituent sets

[lime junco]

your just taking sets of left cosets

and taking the disjoint union of these sets

?

[rose prism]

yes

[prisma ibex]

yup

[lime junco]

and the two unions are disjoint if they dont share any cosets?

[rose prism]

it is like by definition

[lime junco]

oh ok

oh lmao

so literally

as long as the His are different

[lime junco]

OH

WAIT THATS HOW ITS DEFINED???

[rose prism]

..

[lime junco]

bro my prof didnt define it ;-;

that makes perfect sense then lmao

thanks

[rose prism]

they will indeed be disjoint

but it doesnt matter

(probably)

often people abuse this notation to mean “union of disjoint sets” but i suspect that is not happening here

[lime junco]

ah, alright

[alpine island]

full problem for context

the question was about 7

but 6 is relevant

[rose prism]

i was right

[cloud solar]

How to show a non cyclic p group contains a non cyclic subgroup of order p^2

[rocky cloak]

This isn't true. Perhaps you meant non-cyclic quotient group

[cloud solar]

I have this problem to show that a non cyclic p group has at least p+3 subgroups

And I want to show show there is a subgroup of order p^2 that is non cyclic

But I guess I cant

[mighty kiln]

Maybe there are better ways

Oh wait I misread the question

[cloud solar]

I tried using the fact that the center is not e

So there must be an element that has centraliser G

[rocky cloak]

So you can do this approach by showing that the group has (Cp)^2 as a quotient group and use that every subgroup of a quotient group corresponds to a subgroup of the original group

[cloud solar]

Yes I was thinking about this

To look at quotient group and make some bijection

Between subgroups

[rocky cloak]

As for how you can prove that (Cp)^2 is a quotient.
First not that ||any non-cyclic abelian p-group has (Cp)^2 as a quotient by the classification of finite abelian groups||

Then ||if G is not abelian consider G/Z(G), it's a standard argument that G/Z(G) is never cyclic.||

Then just ||iterately mod out the center until you get something abelian||

[rustic crown]

we allowed to assume finite p-group?

[cloud solar]

I think yes but in my problem the group is finite

Ok I understand if G is abelian then G is isomorphic to C_p^a1 ×...C_p^ak and k>1 because G is not cyclic

By the classification of finite abelian groups

I found that: https://math.stackexchange.com/questions/2377726/let-p-be-an-odd-prime-a-non-cyclic-p-group-contains-a-normal-subgroup-isomo

[cloud solar]

Can you explain why a non cyclic abelian p group has (Cp)^2 as quotient?

[rustic crown]

C_p^a1 has C_p as a quotient

C_p^a1 x C_p^a2 has C_p x C_p as a quotient

hence also any non-cyclic finite abelian

[cloud solar]

What do you mean as a quotient. That there is a quotient group C_p^a1 / C_p^b isomorphic with C_p?

[rustic crown]

yea if C_p^a1 is generated by g, then <g>/<g^p> is iso to C_p

quotient by some normal subgroup

but since abelian, every subgroup is normal

[cloud solar]

Can you explain this

I would appreciate

I study group theory for a month and I want to understand every step

[rustic crown]

basically if G1, G2 are groups N1, N2 respectively normal subgruops, then
(G1 x G2)/(N1 x N2) is iso to (G1/N1) x (G2/N2)

so, (G1/N1) x (G2/N2) is a quotient of G1 x G2

(did that make sense? >.<)

[cloud solar]

Yes

That is an isomorphism theorem?

[rustic crown]

yea more or less

Define the map G1 x G2 --> (G1/N1) x (G2/N2)
clearly surjecctive, easy to see N1 x N2 is the kernel

[cloud solar]

Can I say that by using sylow theorem? If a1=1 the normal subgroup is C_p^a1. If a1>1 by sylow there is a subgroup of order p^(a1-1) and that subgroup is normal because C_p^a1 is abelian

[rustic crown]

yea sure

[cloud solar]

Can you explain this

[rustic crown]

okie lemme just say it directly lol

you have C_p^a1 x ... x C_p^an
wehre n >= 2
look at the subgroup N1 x N2 x ... Nn where
N1 = C_p^{a1-1}, N2 = C_p^{a2-1} the unique subgroups of those orders
and other N_i = C_p^a_i
the quotient is then iso to C_p x C_p

[cloud solar]

Oh I see it

Thanks

[cloud solar]

What this means: " mod out the center"

[rocky cloak]

To take G/Z(G) where Z(G) is the center

[cloud solar]

So now G/Z(G) is a non cyclic p group

[rocky cloak]

Exactly

[cloud solar]

And now induction

?

To take the case when G/Z(G) is abelian or not and if is not abelian take Z(G/Z(G))

And so on?

[rocky cloak]

Yes, G/Z(G) is a smaller group so you can just do induction

[cloud solar]

Oh thanks

Wow

[cloud solar]

Let's say G_(i+1)/Z(G_(i+1)) is abelian and is not cyclic and also is a p group. We know there is N a normal subgroup of G_(i+1)/Z(G_(i+1)) s.t. (G_(i+1)/Z(G_(i+1)))/N is isomorphic to Cp×Cp. So now Cp×Cp corresponds to a subgroup N_(i+1) of G_(i)/Z(G_(i)). But N(i+1) correpsonds to a subgroup N_(i) of G_(i) right? So Cp×Cp corresponds to N_(i)

[long obsidian]

Hey sanity check. A principal ideal of a unit x of a ring R is the entire ring xR=R right?

[rocky cloak]

Yes, you still have your sanity intact

[celest furnace]

That last line is a little confusing

Lattice isomorphism theorem says there is a bijection between subgroups of G containing N and subgroups of G/N, so in a sense there is a correspondence but there is not an isomorphic copy of C_p x C_p in G from just that

[alpine island]

Yep. since x is a unit there is some $y=x^{-1}$ in $R$, and $yR \subseteq R\implies xyR = R \subseteq xR$. And $R\subseteq xR \subseteq R$ means it's the entire ring.

[cloud walrus]

nHail

[white oxide]

in specifying a group presentation for $\mathbb{Z} \times \mathbb{Z}$, we write $\mathbb{Z} \times \mathbb{Z} = \langle x, y \mid xy = yx \rangle$. why do we not have $x$ as $(1, 0)$ and $y$ as $(0, 1)$ if those are the generators of $\mathbb{Z} \times \mathbb{Z}$?

[cloud walrus]

okeyokay

[barren sierra]

what do you mean we don't have x as (1, 0) and y as (0, 1) @white oxide

[white oxide]

like is there any reason in group presentations

they don't specify the generators

they just write x and y

[barren sierra]

the generators are x and y

I mean technically those are two different (but isomorphic) groups

presentations are nice compact ways of carrying around groups and their generators and relations

they're isomorphic groups, but not equal as sets

[white oxide]

i see, so when we have equality it's really isomorphism

[barren sierra]

yea

[white oxide]

ah ok makes sense

algebra moment

[barren sierra]

Z x Z is pairs of integers

<x, y | xy = yx> are words of x's and y's

but yea, isomorphism

[white oxide]

ok cool, thanks

ok wait I'm dumb, so I'm trying to verify that $\mathbb{Z} \times \mathbb{Z} \simeq F(S)/N$, where $F(S)$ is the free group on ${x, y}$, and $N$ is the smallest normal subgroup that contains each element of $R$, where $R$ is the commutator $xyx^{-1}y^{-1}$. intuitively, I know that $N$ looks like the set of all commutators of $S$ (at least I think?) but why is this the case? won't $N$ just be the set of all powers of $xyx^{-1}y^{-1}$ since $x$ and $y$ are generators?

[cloud walrus]

okeyokay

[alpine island]

S_3 and D_6 are isomorphic, but the things they represent are different enough that nobody would really call them the same group.

[white oxide]

ok sure

i guess there's no classification in that sense

[white oxide]

is this true though or am I high

that the smallest normal subgroup of F(S) containing xyx^{-1}y^{-1} is the subgroup generated by it

[crystal turtle]

I would call them the same group

[barren sierra]

If someone in the definition of the 3x3 determinant iterated over all of D_6 rather than S_3 I'd have a stroke reading that

[alpine island]

S_3 and D_6 are isomorphic, but the things they represent are different enough that only a set of people of measure zero would really call them the same group.

[rocky cloak]

I mean S3 is the permutation group of three points, while D3 is the permutation group of three points that preserve their distance.

I think it's pretty natural to think of Dn as a subgroup of Sn in this way, in which case D3 = S3.

[crystal turtle]

Okay sure it's maybe more natural to think of a group presented in one manner over another in a specific context, that doesn't mean they're not the same (up to iso or w/e)

[long obsidian]

Is it the case that for any ring R. When I view R as an ideal of itself R then R is the unit ideal?

So for instance with the integers Z I'm pretty sure if you take the 2Z multiples of 2 and the 3Z multiples of 3 then ideal product 6Z is the multiples of 6.

On the other hand the ideal product of Z and 3Z is just 3Z for example. So is it fine to call Z the unit ideal and further any ring R is the unit ideal amongst it's ideal with respect to ideal products?

[coral spindle]

(1) = R, yes.

And indeed if you take the product of R and any other ideal, you get that other ideal.

We wouldn't really call it ‘the unit ideal’ though, that's certainly not a name I've heard.

Indeed if you see the set of ideals as a monoid under ideal multiplication, it would be the identity, but it's not particularly helpful to recall this fact.

[rustic crown]

(i've called it the unit ideal a couple of times)

[coral spindle]

I see people more often just call it ‘the whole ring’

[rustic crown]

yea, but it's sometimes nicer if the word ends with "ideal"

[coral spindle]

you could say it would be...

[rustic crown]

zero ideal, unit ideal, etc

[coral spindle]

ideal

[rustic crown]

what about improper ideal

[coral spindle]

hmmm

unideal

[delicate orchid]

I'm pretty sure anyone with any reasonable knowledge of algebra equates "isomorphic" with "the same". So pretty much everyone would consider these to be the same group

[rustic crown]

same is isomorphic to isomorphic

[delicate orchid]

like even using the action argument, it's obvious that any permutation of 3 elements preserves the triangle between them

[alpine island]

I guess. They're the same as groups, but if you put D_6 in a context where S_3 makes more sense it will look weird

They have different connotations

[crystal turtle]

Hence this. They're still "the same", but it's two different ways of thinking about the same thing

[rocky cloak]

It's all in the context

As always

[crystal turtle]

Every vector space looks like a direct sum of your field. Doesn't mean we only ever work with that description of vector spaces

[onyx trout]

I think you need to consider higher homtopical data for this, honestly

Non-naturally/non-canonically isomorphic breaks this slightly, in my opinion

[south patrol]

based

[chilly ocean]

Let $(G, \star)$ be a group and $(N, \star)$ one of its normal subgroups. Show that G/N is abelian iff N contains the commutator subgroup $K(G)={a\star b \star a' \star b'| a,b \in G }$ I showed that if G/N is abelian, N contains the commutator subgroup. But i'm not sure how the second step

[cloud walrus]

Plazzi

[coral spindle]

Hint: a group is Abelian iff its commutator subgroup is trivial. What do normal subgroups of a quotient look like?

[chilly ocean]

$a \star N$, but if G/N is abelian, that doesn't mean that G is abelian aswell or am i wrong?

[cloud walrus]

Plazzi

[coral spindle]

No.

The smallest nonabelian group is a counterexample to this misconception

[chilly ocean]

Sorry, im bit lost here how does the abelian group is helping me?

[coral spindle]

It's a hint so I will not just reveal the answer.

You may know the second part of the hint as the correspondence theorem

Normal subgroups of G/N correspond to what in G?

[chilly ocean]

What is the correspondence theorem?

[karmic moat]

this is really neat lol i never thought about $R[t] = \oplus_i t^i R$

[cloud walrus]

anamono

[karmic moat]

that is really creative

also is this common notation

[barren sierra]

I've never seen that notation

[barren sierra]

This description is more common when you want to start learning about graded rings in general

Not the most useful description outside that context as far as I know

[long obsidian]

Hey I'm confused about rings of quotients. If I have the rationals Q then m/n=p/q iff mq-pn=0. Q is the localization of the integral domain Z by the nonzero elements Z \0.

For general localizations of rings $S^-1 R$ I think the equivalence relation is that m/n=p/q iff there exists t in S such that t(mq-pn)=0.

I think if R is an integral domains then you don't need the t for the relation. But why exactly do you need the t generally if R isn't an integral domains?

[cloud walrus]

HausdorffT1
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[long obsidian]

For some reason lang seems to like this term

[karmic moat]

is the stuff in red a typo? (hypotheses of 2.16 is that S is a central multiplicative set in a (not nec. commutative) ring R, M and N are finitely presented)

should it be f \in p instead of f \nin p?

also, here P is a finitely generated projective module

[tribal moss]

That looks like #algebraic-geometry, definitely not #groups-rings-fields level.

[long obsidian]

Also I have no clue what a central multiplicative set is

[karmic moat]

it's from Serre's Conjecture on Projective Modules by Ty Lam :(

[karmic moat]

i assume it's just a multiplicative subset in the center of the ring

but ok i will ask in alg geo

[tribal moss]

Spec sounds like alg geo, but at least #advanced-algebra.

[karmic moat]

oh true

mkay

ty

[rocky cloak]

I've seen it a lot in older papers

I think it's somewhat common in commutative algebra in general

[chilly ocean]

would appreciate some help with this. very stuck

[lusty marlin]

if k=gh, then k^-1 = h^-1g^-1

[wooden rover]

I don't get what a the cdot means for operations on the left, would someone explain please ?

(what i mean by there is, does it represent just a normal product ?)

[coral spindle]

CONTEXT!!!!!!!!!!

It could mean so many things.

[wooden rover]

group action mb

[coral spindle]

Then it's the group action operation

[wooden rover]

oh

i see

thanks

[long obsidian]

Field extensions F/K have the same identity id_F=Id_K

I used that fields have no zero divisors. Is the corresponding statement for rings and subrings true? Any obvious counterexamples?

[delicate orchid]

is this multiplicative or additive identity

[long obsidian]

Multiplicative identity

[delicate orchid]

additive identity - a subring is an additive subgroup of a ring so it's obvious from that

does the same statement hold true for monoids? Idempotents could mess with this

lemme think

[long obsidian]

Cause x(id_F-id_K)=0. Yeah I'm not sure. I was hoping there was an obvious counter examples lol

[delicate orchid]

I think it follows from the definition of a subring but that's really boring

like by definition the multiplicative structure of a subring is a submonoid with the same identity which sucks but it is the answer

a far more interesting question is if for an arbitrary submonoid of a monoid does it have to have the same identity, which I'm not so sure about.

[long obsidian]

Oh woops I didn't realize that okay thank you!

[delicate orchid]

nah, even this has to be by definition - if it wasn't the case we wouldn't have an inclusion map from the submonoid back into the main monoid (monoid maps have to preserve the identity) which seems like a very stinky thing (mainly, they wouldn't be subobjects categorically speaking)

[abstract dagger]

Any idea of how i can solve this?

its bugged it should say in A_n in a)

for n>=5

[long obsidian]

Wait I'm confused. I was looking in hungerford and he mentioned a theorem that the identity of a monoid is unique but he never talks about submonoids specifically unless I missed it.

But there is an example on wiki of the ring of 2x2 matrices over R with the subring of constant 2x2 matrices whose multiplicative identity is the 2x2 matrix consisting of 1/2

[delicate orchid]

oh good example. Take a ring R as a subring of RxR, the identity of RxR is (1,1) but R also has the identity (1,0)

[delicate orchid]

but I'm not sure if (Rx0, +, *, (0,0), (1,0)) counts as a subring of (R, +, *, (0,0), (1,1)) due to these issues

this must have been why the idea that idempotents would mess with things came up into my head earlier, if e is an idempotent of a ring R then eR is an ideal of R but acts like a subring with identity e (assuming R is commutative)

[coral spindle]

Nice new pfp btw wew

[coral spindle]

Maybe it's worth talking about unital vs non-unital subrings and homomorphisms. Often when we talk about subrings we specifically want them to share the same unit, but as Wew alluded to there are also very important cases where they don't. These latter ones are 'non-unital' subrings. The same goes for ring homomorphisms – by default people tend to require them to preserve the unit, but they don't have to.

[crystal vale]

What's like a general strategy for this kind of problems especially for the 9

[delicate orchid]

I got so many pictures of these guys it's unreal

they're everywhere

[barren sierra]

kind of tedious

ok so we agree that we know Z(sqrt(2)) is an integral domain, so we just need to show the Euclidean part

so you need to check

1: The norm of Q(sqrt(2)) works for Z(sqrt(2))

2: (which will suck, ngl) you need to do the Division algorithm

[cloud walrus]

Spamakin🎷

[barren sierra]

so it's straight forward but annoying

[long obsidian]

Hey I came across this def of an opposite ring Rop where the underlying set is the same set as a given ring R but the operation is done in a different order.

For something commutative this ring Rop is isomorphic to R right?

[rustic crown]

yep.

[wraith cargo]

Generally there's a geometric strategy by looking at what the maximum distance between any two points in the fundamental mesh of the lattice

[barren sierra]

Easy exercise to prove this

[chilly ocean]

wrong channel

try #multivariable-calculus

[agile flicker]

i tried asking there no one answered

[chilly ocean]

so you thought the algebra channel would be better??

[agile flicker]

it said fields

and the class i have for this is numerical analysis with the current subject being linear algebra

so yes

[white oxide]

how exactly does this imply that $h$ is unique? is it because if we tried it with another homomorphism sharing the same property, say k, then we would get $k_{\alpha_1}(x_1)\dots k_{\alpha_n}(x_n) = h_{\alpha_1}(x_1) \dots h_{\alpha_n}(x_n)$, and then we could just cancel elements from both sides or smt

[cloud walrus]

okeyokay

[limber wharf]

Cause what h is is forced by that

The condition is necessary, but also totally determines h

[white oxide]

wdym

how exactly is this well-defined? can't words have multiple representations?

[limber wharf]

Did you prove that every word has a unique reduced representation?

Cause then there should be no issue

[white oxide]

No, but they proved it afterwards basically

[limber wharf]

Hmmm

I mean you can just note that if you have a different representation for the same word then it must just have some x*x^{-1} term inside or something along those lines

But then $\phi (w)$ is the same either way

[cloud walrus]

colejagdtiger

[white oxide]

ye sorry i mean that they showed that phi(w) is the same regardless if it's reduced or not

but yea beforehand i was a bit worried it wasn't well-defined lol

anyways thx for the help

[sturdy mirage]

Is the underlined isomorphism true also true when i=0? i might be abusing notation, but i consider m\textasciicircum 0 = B. so i am wondering if the following is true:

$B_\mathfrak{m} / \mathfrak{m}_\mathfrak{m} \cong B / \mathfrak{m}$

and how would I see that, if it is true?

[cloud walrus]

reking

[cloud solar]

Hi

I find rings of polynomials problems difficult

Any tips where I can find some problems some results, some techniques

[rustic crown]

yep that's true :3

[wraith cargo]

It is true but the rest of the calculation isn't since when i=0 this isomorphism is equal to k

[rustic crown]

you can be tiny bit more general, A_p/pA_p = Frac(A/p)

[sturdy mirage]

pA_p is what i'd write as p_p right

[rustic crown]

why so? it will have constant and linear terms, so 2 dim right?

[rustic crown]

yep

[sturdy mirage]

there are no terms with just x

[wraith cargo]

B/m is isomorphic to k

[sturdy mirage]

its x^2, x^3, x^4 ...

[rustic crown]

ah oops, i didnd't read definition of B >.<

[wraith cargo]

:P
It's alright :3 :3

[rustic crown]

ig my bad, it couldn't have been both k^2 and a maximal ideal ><

[rustic crown]

so this is what you should prove, if A --> B is a map of rings, and S is a multiplicative subset in A
then you can localize B as an A-module, but also localize it at the image T of S. both these are iso, i.e. S^-1B = T^-1B

in particular A_p/pA_p = (A/p)_p = invert non-zero things in A/p = Frac(A/p)

first = is by exactness of _p

and image of A\p under A --> A/p is non-zero stuff

[fair bay]

why does n_q have to divide p?

[rustic crown]

that's part of sylow theorems

don't remember if it was 2 or 3, prolly 3

n_q = 1 mod q and divides order of G

but since it's 1 mod q, it is coprime to q and hence divides |G|/q^whatever

so in this case it divides |G|/q = p

?

[fair bay]

ah right n_q is the index of a subgroup so it must divide |G|

[rustic crown]

yep

[fair bay]

thanks for the help

[rustic crown]

[sturdy mirage]

what exactly does this mean? all i know is that (A/p)_p is isomorphic as an A-module to A/p ⊗ A_p

[rustic crown]

so you have an exact sequence
0 --> p --> A --> A/p --> 0

if you localize at p, aka tensor by A_p, as this is exact

you get
0 --> pA_p --> A_p --> (A/p)_p --> 0

so A_p/pA_p = (A/p)_p

proving S^-1B = T^-1B is just some universal property stuff

you can be explicit and construct maps in both direction by hand

but yea, the bijection corresponds b/s with b/t where t is the image of s in A --> B

[rustic crown]

(you only need right exactness here if you haven't proven that localizations are exact)

[sturdy mirage]

thats handy ,thanks! i can use exactness. i dont want to get explicit if i can avoid it

[rustic crown]

but yea, it reduces the work to this.

[cloud solar]

.

[rustic crown]

(det no have anything better to say than "pick any book >.<")

[cloud solar]

I mean like a pdf where I can find online

[coral spindle]

Find the name of any commutative algebra textbook, then search that name and type:pdf on google

Idk, Lang might be good

[cloud solar]

Thx

[white oxide]

how does the preceding paragraph imply that the separable degree is 1?

because isn't [k(a): k]_s the number of extensions of the embedding of k into K^a to an embedding of k(a) over k

does it just show that the only extension is the trivial one?

sigma leaving a fixed => sigma (regarded as an extension to K) also leaves k(a) fixed

[tardy hedge]

Hey guys, for Q4, i dont really get why hes saying Z3[i]

Is the point here, that the field Z3[x] \ <x^2+1> isomorphic to Z3[i] so he just wrote that?

And Z3[i] is a+bi with a and b in Z3?

[delicate orchid]

cause i^2 = -1, so you're appending an element that squares to 1

[delicate orchid]

but yes, this is correct

[tardy hedge]

Oh i see cause for the other field u get that x^2+1 = 0 mod (..) so then x^2 = -1

So u joined an element that has a square of -1 but is 1 in Z3

So thats like joining i

Also elements in that are a+bx but thats the same as a+bi so yea

[delicate orchid]

yur

[tardy hedge]

Dats kinda neat

i enjoyed groups more than this stuff so far

it prob gets more interesting later

its kinda neat having all these different constructions for the same ideas

[white oxide]

If someone helps me out I’ll say that I’m grateful for u at tonight’s feast

[long obsidian]

Given an ideal I in a ring R, A an R module, and S a subset of the R module A then there is a construction in hungerford

IS is a submodule of A consisting of linear combinations of elements from the subset S with coefficients from the ideal I, with respect to the ring R. I can't find a good example of this. I tried to imagine subspaces of R^3 but ran into a problem because I don't think fields like R have a lot of proper ideals

Is there a lattice subspace of R^3 that you can make with this type of submodule construction?

[mighty kiln]

Take the Z-module R³ and S the generators of the lattice

[tribal moss]

Z-modules is itself a pretty nice case: since I ideal of Z is necessarily principal, we can make IS in two steps by first scaling everything in S by the generator of the ideal, and then taking the submodule generated by the scaled S.

(But then again, "just pretend the arbitrary ideal is principal" is often good enough for rough mental images).

[long obsidian]

Ah okay so you need to start off with R^3 with a Z module structure. Then this construction works thanks!

[crystal vale]

Okay

[barren sierra]

(groups get annoying and weird quickly)

[white oxide]

how do we know that K is separable?

[south patrol]

could you share the name of the book or something

[crystal vale]

Hi ,can someone tell from which book I start Field theory?

[south patrol]

how we know it depends on what has been covered thus far in the book

[white oxide]

Lang's algebra

[south patrol]

yeah i assumed so but couldn't find it aha

[crystal vale]

🙇

[white oxide]

ye this is the first section on galois theory i believe

i love lang's writing so much omg

[white oxide]

depends on how much experience you have i guess, since field theory usually leads to galois theory lol

if you're first learning i learned from fraleigh which was very very good

albeit very wordy but it's very helpful when you're first learnin git

if it's your second time then I would recommend Lang

he has a very elegant writing style

[south patrol]

so i guess the way i would do it is that we can write E = k(a_1,...,a_n) for some separable a_i. Then K is just the splitting field of the product of the min polys of the a_i

and those min polys are separable, so the splitting field K is separable

or i guess more concisely: K is generated by the a_i and their conjugates, and all of these are separable

[white oxide]

oh right yeah, i forgot that Lang actually made it explicit earlier

and then he talks about conjugates on the next page yea

[long obsidian]

How are you supposed to interpret that this construction IS works if R3 is a Z mod but it doesn't if it's an R mod? Are the lattices inside R3 capturing the integer structure or something like that? This feels like we're calling the lattices invariant subspaces of some specific endomorphisms that arise from the z mod structure of R3

[crystal vale]

Oh thanks

Is Ring of Fractions is subring of Field of Fractions?

[white oxide]

i don't know what you mean by that

you do have every integral domain embedded in its field of fractions tho

[crystal vale]

Albeit?

[white oxide]

although

[white oxide]

hmm i haven't heard of that but i would assume that it's similar to the construction of the field of fractions, just leaving out inverses or smt, but honestly i'm not in a position to answer

someone smarter than me here can tho lol, you might wanna post theorem 15 as well

[crystal vale]

Oh

[white oxide]

so theorem 4.5 states that separable extensions form a distinguished class. how would you go about showing that the compositum is separable using theorem 4.5 and induction? i'm assuming you would use (2), but that would require showing that sigma_i(E) is separable over k for all i; does this just work since it's isomorphic to E lmfaoo

[tribal moss]

It still works, but as you pointed out yourself, the I doesn't really do anything in that case, since it's either all of R or {0}.

[rapid junco]

this is a proof that there is a unqiue action on the inclusion map such that the inclusion map Orb_G(x) -> X is G-equivariant

not complete yet I have to show uniqueness. But I am a little lost on how to start.

Is there a way we can say that the orbits under different actions are related?

[white oxide]

If $K/k$ is separable and $\lambda: K \to \lambda K$ is an isomorphism, how would you go about showing that $\lambda K/\lambda k$ is separable? i was thinking about doing some shit like setting up a bijection $\tau \mapsto \tau \circ \lambda$ between the embeddings of $K$ over $k$ and $\lambda K$ over $\lambda k$, but I'm sure there's something much easier

[cloud walrus]

okeyturkay

[white oxide]

nvm that doesn't even make sense

oh wait i'm dumb i can just use the fact that every element in K is separable over k lmfao

[chilly radish]

Yea...

[crystal vale]

Undergraduate Algebra Lang review anyone?

[alpine island]

@lime oriole

do you see this?

[lime oriole]

yes

[alpine island]

hmm

[lime oriole]

this is because of undergrad role tho

[untold basalt]

How do I show e) is not a field?

[rocky cloak]

you could try to multiply or divide two elements

[languid trellis]

Why do we care about normal subgroups and conjugates? Artin just defined them basically with no motivation

[coral spindle]

Conjugation is a great way to understand the structure of a group as a whole. Normal subgroups are what allow us to look at quotients of groups, which may at first seem mysterious, but again are a very powerful way of analysing the structure of a group.

An example of a thing you can prove with a couple of simple arguments involving conjugacy and quotients: every group of prime-squared order is Abelian.

[languid trellis]

I guess I'll understand more when I get to the quotient section haha, thanks for the answer

[tribal moss]

Have you seen group homomorphisms yet? Normal subgroups are exactly the subsets of a group that can be the kernel of a homomorphism. This is an important tool for understanding how homomorphisms from a given group can behave.

[cloud solar]

I have a problem

We have K a finite field. Show that 1+1=0 iff for every polynomial f in K[x] with degf>=1, f(x^2) is reductible in K[x].

[crystal vale]

|| Use char property I think||

[crystal vale]

Here x^2 matter?

[cloud solar]

Of course

[cloud solar]

I think yes? Even power?

I tried to say smth like this if f(x^2)=g(x) is irreductible then g(x+1) is irreductible

And somehow to use the char

The fact that K is finite

Means K* is cyclic maybe this helps?

To see the coefficients as powers of generator

Idk why K is finite

K has 2^t elements if t is even then there is an element in K* with order 3 maybe this helps?

[crystal vale]

I think if you take g(x+1) is irreducible then g(x) is irreducible

[cloud solar]

Isn't this a property?

[crystal vale]

Yeah it's property

[cloud solar]

f(x) irreductible iff f(ax+b) irreductible with a≠0

[crystal vale]

I do not know about iff

May be you are right

[cloud solar]

Idk how to manage this problem

In general idk how to manage polynomials problems

[crystal vale]

Oh thanks

[crystal vale]

Book?

[cloud solar]

Is not a book, only a pdf with 2 pages

[crystal vale]

Okay

[tribal moss]

For "if": x²+1 is reducible, and if -1 != 1 is different that means K contains i of multiplicative order 4.
Since K contains i, now x²-i is reducible, giving an element of multiplicative order 8.
Continue this to find elements of arbitrarily large multiplicative order.

[cloud solar]

Oh so here we get a contradiction because we can generate elements in K with 2^t order for every t>=1

[crystal vale]

Thanks

[crystal vale]

From which book you find this question?

[cloud solar]

But for the first case when we need to show that every polynomial f(x^2) is reductible?

[cloud solar]

It is from a contest from Romania

[crystal vale]

Oh

[tribal moss]

Or expressed more slickly: since x^2-a is reducible, every element has a square root. But if 1 is not -1 , that means every nonzero element has two square roots, and then there are twice as many elements as there are...

[crystal vale]

Which one?

[tribal moss]

On the other hand, in a char 2 field, an element can have at most one square root; if it is finite, every element must have at least one square root too.
Now the square of a polynomial is made by squaring the coefficients and doubling the exponents, since the cross terms cancel out in char 2. So a polynomial with only even-degree terms can be square rooted by reversing this.

[long obsidian]

Im trying to remember the Z mod nZ automorphism. In the quotient if you had a representative [x] in Z/nZ then if (x,n)=1 the representative is relatively prime it follows x has an inverse since the gcd is spanned x and n. rx+sn=1 then taking mod n; 1= rx mod n. Now I know multiplication by numbers relatively prime to n have an inverse. In fact, by having inverses it's clear multiplication by a relatively prime p; (p,n)=1 is an automorphism of Z/nZ since it has an inverse.

I think these are all the automorphisms of Z. Is there some obvious issue if f:1->m with (m,n) not 1 so they aren't relatively prime. How can I tell this wouldn't be an automorphism. Is it something simple like looking at the order?

[torn marlin]

If $f \colon 1 \to m$, say $\gcd(m, n) = g$, notice that $f(i) = gi$ and is always a multiple of $g$

[cloud walrus]

Sig

[torn marlin]

thus, it is not an automorphism since the elements that are not a multiple of $g$ do not have an inverse

[cloud walrus]

Sig

[long obsidian]

Did you mean f(g)=mg? Sorry I'm still not following why something doesn't have an inverse if it's not relatively prime to n.

I think in Z/6Z I can see that the powers of [4] will always be a multiple of 2. I just can't pin point what the issue is exactly. I think if I divide any multiple of 2 by 6 I can't get 1 as a remainder

[torn marlin]

ahh yeah
i meant $f(k) = mk \in (g)$, where $(g)$ is the subgroup generated by the element $g$

[cloud walrus]

Sig

[torn marlin]

this subgroup does not contain 1

[long obsidian]

Ah okay! That's interesting. I guess that is enough to say it's not invertible

[tribal moss]

If I'm reading the initial post right, this is about automorphisms of Z/nZ, not of Z or subgroups thereor.

[tardy hedge]

When it says “what is its multiplicative group”, what am i really supposed to say?

I wrote down the multiplication table, its an abelian group of order 9

I mean what else am i supposed to do?

[tribal moss]

The group of order 9 is the additive group.

[tardy hedge]

What?

[tribal moss]

I think you've written down an addition table rather than a multiplication table.

[tardy hedge]

Hm

[tribal moss]

Since 0 is not an element of the multiplicative group, there should be only 8 elements.

[tardy hedge]

Ohhh ok wait wait

Oh yea of course thats not a group table lol what am i thinking

So a field is an abelian group under addition and an abelian group under multiplication but without 0 (additive inverse)

[tribal moss]

Yes, but you mean additive identity in the last parenthesis.

[tardy hedge]

O yes

So in previous questions it asked for the + and x table for the field

So that one i included 0 cuz its not talking about the group structure

So for this question then , its askinf what is its multiplicative group, which one is it talking about? The one with 8 or 9 elements?

[tribal moss]

"Multiplicative group" means "the group whose operation is the multiplication of the field", so that's the one with 8 elements.

[tardy hedge]

Oh ok cool thanks

And i noticed its cyclic

[tribal moss]

Yes, well done. Noticing that is the point of the exercise!

So the answer is that the multiplicative group "is" (that is, "is isomorphic to") the cyclic group with 8 elements.

[tardy hedge]

Awesome

Thx for the help

[tribal moss]

There's a general theorem saying that the multiplicative group of a finite field is always cyclic.

[tardy hedge]

Yes i was thinking hmm maybe there is some theorem involved here haha

Thats neat

For 5, that notation for the field is the “smallest” field that contains Q and the golden ratio right?

Thats what it means?

[delicate orchid]

non-standard to use square brackets for that but yes

[tribal moss]

Actually the notation is for "smallest ring that containd Q and the golden ratio", but that ring just happens to be a field in this case.

[tardy hedge]

Ok, and technically to show that that field is in fact of the form a+bq we need to show right?

I dont think the question cares to do that but

[tribal moss]

The elements of the field are each of the form a+b·phi with a,b in Q.

[tardy hedge]

Yea thats what i meant to say

But im asking technically we should prove that instead of just taking it as given?

[tribal moss]

Yeah, that would probably be instructive.

[tardy hedge]

Ok yeah

Although i dont think question cares

I mean, multiplication “rule” is just multiplying them out like normal isnt it?

Nothing reallt fancy

U gotta reduce phi^2 to something with phi i guess

[crystal vale]

Source?

[tribal moss]

Yes, exactly.

[tardy hedge]

My homework lol

[crystal vale]

Oh

[tribal moss]

And I think the exercise author has simply forgotten to ask you to prove that the reciprocal of an a+b·phi also has this form.

[tardy hedge]

Prof just made up question

[tardy hedge]

Yeah, like we arent really doing the full justifying that its a field right

Kinda just taking it as given that it works

Besides showing multiplication works

[tribal moss]

Addition and subtraction are so easy that it makes sense to omit them. Associative, commutative and distributive laws you simply inherit from R. So all that's really missing is reciprocals.

[tardy hedge]

I see

Makes sense

Showing reciprocals are in the field automatically implies “division” makes sense etc

[tribal moss]

Yes.

[tardy hedge]

Cool

[tardy hedge]

Lol the rest of the question was just kinda like random fun

Fettucini sequence

[alpine island]

how do you pronounce sylow

[delicate orchid]

si-low or see-low depending on how I'm feeling

[alpine island]

I would have said /saylow/ but the prof. says /se:low/

[summer path]

silly-o

[delicate orchid]

Hall p-subgroups

[mighty kiln]

sy:lɔv apparently

[next obsidian]

It’s like see-lahv isn’t it?

[mighty kiln]

Isn't y like ü from Mandarin

[next obsidian]

I dunno I don’t speak mandarin

Or IPA

[mighty kiln]

Or tu from french

[rocky cloak]

It's like you make the ee-sound with your mouth, but then round your lips

The sound is not used in English as far as I'm aware

[mighty kiln]

I've always found this weird

[celest furnace]

What is this

[mighty kiln]

I have never associated [y] with ee until recently

[rocky cloak]

How you pronounce y in Norwegian

[mighty kiln]

But it works somehow

[celest furnace]

That’s cool

[rocky cloak]

As in the name Sylow

[mighty kiln]

You saw nothing

[celest furnace]

It’s not S-ayyy-low?

[elder wave]

No

[rocky cloak]

Weering of the topic of math, but Dr Lindsey has a great video on vowels
https://youtu.be/FdldD0-kEcc?si=CqCTBNKtJiGKatiH

[untold basalt]

Let $A$ be a ring and $I \subset A$ a maximal ideal. Prove that $A/I$ is a field.

Proof: Consider $A$ as apartition of cosets: $A=\bigcup_{i\in \Gamma} a_i + I$. If $A/I$ is not a field then there exists $i \in \Gamma$ such that $a_ia \notin 1+I$ for all $a \in A$. This however means that $A\setminus (1+I)$ is an ideal in $A$, a contradiction.

[cloud walrus]

Seagull

[untold basalt]

Is this correct?

[rocky cloak]

You have the right idea, but A\(1+I) won't be an ideal. It should be something like (ai) being an ideal in A/I

[white oxide]

where did they use the fact that F is normal over k? is it just that F is separable over k, and it being normal by assumption implies that we're allowed to consider the Galois group of F over k?

[rocky cloak]

If F isn't normal, then the restriction of sigma to F, might not be an automorphism of F.

[white oxide]

oh okay that makes sense

thanks

[white oxide]

$if k \subset F$ and $F \subset E$ are normal, then it doesn't follow that $E/k$ is normal right

[cloud walrus]

okeyokay

[white oxide]

or rather it's not necessary

oh nvm, so to show that $\mathbb{Q}(\sqrt{a_1}, \dots, \sqrt{a_n})$ is normal over $\mathbb{Q}$ where each $a_i$ is a square free integer and $|a_i| \neq 1$ for all $i$, we can just say that it's the splitting field for the family of polynomials ${x^2 - a_i}$

[cloud walrus]

okeyokay

[white oxide]

would this be an acceptable proof for the first part?

[rustic crown]

you don't have to say a lot about separability, as you're working over char 0. the hardest part in this is to show that degree [Q(sqrt{a_1}, ..., sqrt{a_i}) : Q(sqrt{a_1}, ..., sqrt{a_{i-1}})] = 2 and not 1.

it's not clear directly why irred(sqrt(a_n}, K') = x^2 - a_n

[white oxide]

hmm, so are you saying that this is the hardest part in showing separability?

[white oxide]

yeah to be honest i kinda hand-wove that, i'll def return to it tho

[rustic crown]

nah, separability is automatic, if F/k is algebraic and char 0, then it is separable.

[white oxide]

oh huh, i didn't know that

i guess i should prove that

[white oxide]

hmm, so what would be the purpose of showing that the degree is 2 then over each step of the tower?

[rustic crown]

that will be required to compute the galois group

[white oxide]

if my argument that it's normal is fine

oh yeah that makes sense, i also used that in my argument implicitly

[rustic crown]

normality follows like you say by looking at the splitting field of product of (x^2 - a_i)

[white oxide]

ok thanks! i'll go back and revise that

[mossy lintel]

How do i show that for any two n-th primitive roots of
unity $\omega_1, \omega_2$, there exists an automorphism
$\sigma$ of $\mathbb{Q}(\omega_1)$ which sends $\omega_1$
to $\omega_2$?

[cloud walrus]

elgato

[white oxide]

hint: automorphisms send roots of irreducible polynomials to conjugates

[rustic crown]

have you proven that minimal polynomial of these omega_i over Q is the nth cyclotomic polynomial?

it's essentially equivalent to that

[alpine island]

I find it funny that every statement about left cosets ends with "and similarly for right cosets"

when do we use the right ones

smh

[mossy lintel]

actually i'm suppose to use this to conclude that n-th cyclotomic polynomial is irreducible over Q

[rustic crown]

yee thought so much

[crystal turtle]

A right coset of G is a left coset of G^op, so practically never

[upper pivot]

basically the only time you'd need to think about these differently is when you have like

double cosets

[rustic crown]

do it in steps, instead of proving for an arbitrary omega_2, reduce it to proving when omega_2 = omega_1^p where p is a prime not dividing n.

[upper pivot]

i.e things of the form HgK

[alpine island]

every team I read ^op I read it like gangnam style, when does this end

[crystal turtle]

[rustic crown]

then you'll need to exploit this prime p a lot. this is a little tricky, but pretty neat.

[alpine island]

When talking about 2 separate actions of the same group on the same set, how do you distinguish them?

[upper pivot]

give the actions name

like alpha and beta

so \beta_g and \alpha_g are different

one nice case is when one of the actions is a right action and another a left action

in which case this isnt an issue

[alpine island]

Yeah, but then what? Do I just have to use $\alpha(g, x)$ instead of $g\cdot x$?

[cloud walrus]

nHail

[upper pivot]

sure

I prefer $\alpha_g(x)$ personally

[cloud walrus]

JohnDS

[upper pivot]

but doednt matter

[alpine island]

I could do $g\cdot_\alpha x$

but that might be too fancy

[cloud walrus]

nHail

[upper pivot]

sure that works too

[alpine island]

the difference between G x X -> X and G -> (X -> X)

[upper pivot]

yep for me an action is a homo G-> Aut(X)

[crystal turtle]

An action is also a functor BG --> Set

[upper pivot]

and a G-equivariant map is a natural transform

[rustic crown]

The argument goes like this,

||Let f, g in Z[x] be the (monic) minimal polynomials of omega_1 and omega_2, by assumption g(omega_1^p) = 0. Therefore the polynomial g(x^p) has omega_1 as a root, so we can write g(x^p) = f(x) * h(x) inside Z[x]. Look at this equation mod p.||

||In F_p[x], g(x^p) = g(x)^p = f(x) * h(x). So, f and g have a common non-trivial (irreducible) factor in F_p[x]. If f(omega_2) was not zero, then f(x) * g(x) divides (x^n - 1) in Z[x]. If we look at this in F_p[x], we see that x^n-1 has repeated roots (because of the common factor between f and g). But that can't be true as x^n-1 is separable over F_p (its derivative is nx^(n-1) which is coprime to (x^n-1) as p didn't divide n). ||

||Thus, f(omega_2) is also zero and f = g. In particular Q(omega_1) and Q(omega_2) are both isomorphic to Q[x]/(f) hence there is an automorphism of Q(omega_1) sending omega_1 to omega_2.||

[mossy lintel]

Really appreciate it! But first i'd just take the original hint and see how far i can get!

[rustic crown]

det sleepy >.<

[crystal turtle]

go eep det

[rustic crown]

okie

(should i watch frieren ep 3 before eeping? )

[chilly ocean]

if you watch one you're gonna wanna watch all of them out rn

u should eep

[upper pivot]

based frieren is great

[rapid junco]

how does one get an G-isomorphism between X and G/H given that G acts transitively on X?

[upper pivot]

H here being the stabalizer?

[rapid junco]

Oh wait just orbit stab

lmao

[alpine island]

you're doing this last minute too? lmk when you get to 1.7

[white oxide]

are you guys in the same class lmao

[rapid junco]

yeah

[crystal vale]

In Z[x] , we have to prove (2) intersection (x) = (2x) ... so can I use here co-maximal property?

[undone ledge]

is every abelian group given by a unital ring structure

[chilly ocean]

{0}

[undone ledge]

i allow trivial rings to be unital

lol

[delicate bloom]

wdym given, like is the addition of some ring?

[undone ledge]

mhm

[delicate bloom]

I think I recall seeing you can't put a multiplication on the Klein 4 group, maybe check that one

[undone ledge]

Z_2 × Z_2?

don't we have F_4

[delicate bloom]

Maybe I'm thinking you can't give that one addition

[undone ledge]

f.g. abelian group is easy by structure theorem

[kind temple]

just show it directly. if you have something that looks like 2p(x) = xr(x) then r(x) is divisible by 2 (why?)

[delicate bloom]

Q/Z supposedly

This question claims it's well known lol https://math.stackexchange.com/q/3075364

[crystal vale]

2 and x are relatively prime

[kind temple]

right

[crystal vale]

But I have a doubt

In Chinese Remainder Theorem we take (m)(n)=(mn) is it both are same?

In general?

[mighty kiln]

For Q/Z,
say the unit is p/q≠0, then rq = 0 for any r, so q/(q+1) = 0

[kind temple]

could you add some more context

[crystal vale]

Yes

[crystal vale]

If I =(a) and J=(b) then IJ=(ab)?

[mighty kiln]

Yes if commutative