groups-rings-fields

[delicate orchid]

upper triangular matrices

[south patrol]

Real af

Me when the flag

I think this is easier though if you know what the min poly of a over K looks like

[dull marsh]

Are Z4 ⋊ Z2 and (Z2 x Z2) ⋊ Z2 isomorphic?

[delicate orchid]

yeah they're both D_8

[dull marsh]

Makes a lot of sense, thank you

[delicate orchid]

I don't think it matters which embedding into Aut(Z2^2) you chose for the 2nd semidirect anyway

but there definitely is one that just gives you D_8

[dull marsh]

I was kinda confused because I had a group isomorphic to both

And though this would imply Z4 and Z2 x Z2 are the same

I guess that's not how it works

[delicate orchid]

that isn't even a conclusion you can draw for direct products in general

[south patrol]

Swindle

[delicate orchid]

although the counter examples are very uhhh "non-natural" lol

[south patrol]

A is isomorphic to B because A x B x A x .... is isomorphic to B x (B x A x B x ...) = B x (A x B x ...)

[delicate orchid]

like if all your groups are finitely generated then it's fine

[dull marsh]

I see

[celest furnace]

Craziest one ive seen is G iso to H, N normal subgroup of G, N2 normal subgroup of H, N iso to N2, but G/N is not iso to H/N2

[worn spindle]

ok so what was the more tahn 24 thing abt lol

[dull marsh]

There are 3 elements in S4 that have more than 1 cycles in their decomposition

I meant you would need to write down more than 24 cycles

I didn't mean permutations

[rocky cloak]

They're all conjugate, so shouldn't matter

[delicate orchid]

ah true, good point

[stark helm]

In Z5[x] consider the ideal I = (x2 + x + 1), I want to justify why x^2+x+1 is irreducible, is it correct to justify in this way?( I actally doubt that( x-a can not divide f(x)) can not necessarily imply f(x) irreducible, by reference of (x^2+1)^2 in R[x] find roots that is not in Field R.)

Or it might be better if someone share why in Zp we just need to check f(a) not 0 by plugging a=0,1..p-1 in order to determine that it is irreducbile?

[glossy crag]

Serial-killer's-mocking-note-to-the-police handwriting

[delicate orchid]

dude you think that's bad you do not want to see mine

[delicate orchid]

just check to see if it has any roots (it doesn't) and then it's irreducible. I assume that's what you've done

[stark helm]

yes, but I am somewhat confused why we can say f(x) is irreducible if there is no roots in Zp?

[delicate orchid]

well it's because it's a quadratic

any factor has to be linear

Zp is a field so no roots => no linear factors

[south patrol]

Chad

[delicate orchid]

society if all partitions of n had at least one 1 in them

[south patrol]

Seeing "Z_5 is a field" confused me for a little bit

[delicate orchid]

skill issue

[stark helm]

if we have quartic polynomial f(x)=(x^2+1)^2 in R[x], we have (x-i)| f(x) and i is not in Field, and it seems similar to x-a does not divide f(x) if a is in field.( Which means it might not be true that( no roots-> no linear factor )can imply f(x) is irreducible if deg(f(x))>2 right?)

[delicate orchid]

3

all partitions of 2 and 3 have at least one 1 in them

which is why this works

for 4 we have 2+2 as you say, so it breaks

[dim widget]

3 + 0

2 + 0

[boreal inlet]

I settled with tower formula and contradiction. As alpha is algebraic, there exists minimal polynomial of alpha over K of finite degree n. Now we factorize the extension degree, using the elements in the chain of subfields.

Now we use contradiction. Let's say [K(alpha^(p^r)) : K(alpha^(p^(r+1)))] > 1 for all the subfields. Then, [K(alpha) : K] = p^n > n, which is a contradiction. That means there has to be one r such that K(alpha^(p^r)) = K(alpha^(p^(r+1))), hence, (alpha^(p^r)) is seperable by the statement I proved in an early case.

[south patrol]

Yeah sure

[boreal inlet]

But this - as far as I know, this should stabilize somewhere? I have not proved it though.

[south patrol]

Yes, this is just linear algebra like

take dimensions

you get a decreasing sequence of integers

[boreal inlet]

OHHH

oh my god i am so bad.

[south patrol]

But the way I'd personally probably do this is like#

Actually it is equivalent to this lol

Do you know that if L/K is an alg extension of char p fields and a in L is inseparable over K, then you can write its minimal poly in the form f(t^p)

[boreal inlet]

yes, we can

[south patrol]

Yeah cool

so now a^p has min poly f(t)

[boreal inlet]

I'm still trying to wrap my head around the proof but the result is indeed true

[south patrol]

by induction on degree of the element over K, you can write f(t) = g(t^(p^n)) for a separable poly g

[boreal inlet]

Hmmm...

[south patrol]

like basically you can keep pulling out pth powers

[boreal inlet]

oh yeah

[south patrol]

and it has to stop because the minimal poly of a is, well, a polynomial

[boreal inlet]

correct

[south patrol]

so yeah you can always write a min poly in the form f(t^{p^n}) for some separable f

and then a^{p^n} is separable

[boreal inlet]

I did something liike this for a purely inseperable extension

[south patrol]

But yeah, as you can see, this is basically the same as your argument

Just more directly with min polys

[boreal inlet]

Actually, how would the dimensions look here, won't that just be the same as the tower formula thing?

[south patrol]

Sure but it's slightly more elementary

[boreal inlet]

Do we take thm to be vector spaces over the preceding field alwys?

because if we just take the original field of course it will stabilize

[coral steeple]

Is there a simple expression for the size of the image of $\mathbb{Z}/n\mathbb{Z}^\times$ under the map $x\mapsto x^2$?

[cloud walrus]

person2709505

[chilly radish]

What's the kernel of this map?

[coral steeple]

The self-inverse elements

[south patrol]

Well just K

Since they are all vector spaces over that

[cobalt heath]

Which are, concretely?

[crystal vale]

I want to confirm an answer of question 10 for a part I think it will be there is no n exists and for b part k=n

[coral steeple]

This is what I'm really after; I hoped my first question would be easier to answer than this. (I am XY problem-ing. Sorry about that.)

But is there really a simple way to see what these self-inverse elements are?

[cobalt heath]

Basically, you need to solve an equation

[coral steeple]

x^2-1=0 I'm guessing?

[cobalt heath]

Yep

[crystal vale]

Can someone help me to confirm my answer

[coral steeple]

I think I had better rethink my approach. An answer to my first question would imply something which I could use in the first step of a long guided-proof-type exercise whose conclusion is exactly the hint you're suggesting...

[cobalt heath]

Huh

I mean, it's not a hard equation, isn't it

Compared to x^2 - 2 = 0 on Z/nZ

[crystal vale]

for any set subset A of G, I want to prove that |A|= |gAg^(-1) | , so if I have a ≠ b then gag^(-1) ≠ gbg^(-1) so is it correct?

[coral steeple]

I can't solve either of those (even if I can convince myself by brute force calculation), so I can't tell which is harder

[cobalt heath]

Hint: one is factorizable, the other is.. debatable

[crystal vale]

What is meant by rigid motion?

[coral steeple]

Ahg yes, you're right--- silly of me not to have thought of that earlier

Thanks

[next obsidian]

A translation

Rotation

Or reflection

Or any combination thereof

Its transformations which preserve distance

[crystal vale]

Between any mathematical objects or specific....and how we define distance?

[next obsidian]

This is for Euclidean space

[crystal vale]

Okay

[crystal vale]

And for distance it preserves standard metrics(distance) or some other types of metric ?

[next obsidian]

Standard

[crystal vale]

Okay thank you

[cobalt heath]

No problem! I also overthink too often as well.

[coral steeple]

It turns out that I did manage to reduce the problem to solving (x-1)(x+1)=0, but in a more convoluted way, a few hours ago. I am only required to count solutions in Z/(2^k)Z (for the first part of the proof). Any hints?

I have reason to believe that if $x^2=c2^k,,x<2^k$ (in the integers), then $x=\pm 1,2^{k-1}\pm 1$, but I can't prove it.

[cloud walrus]

person2709505

[cobalt heath]

Yeah perhaps there is a way to solve it without elementary number theory

[coral steeple]

That was my hope

[cobalt heath]

But you can consider when 2^k divides (x-1)(x+1)

[coral steeple]

Mhm- we can express the condition as $2^k\mid 2^\alpha\prod (p_i)2^\beta\prod (q_i),,\alpha+\beta\geq k$. We had a situation like this with the next step in the proof (the same problem but in $\mathbb{Z}/(p^k)\mathbb{Z}$, $p$ an odd prime) but couldn't adapt the proof to the $p=2$ case. Otherwise, I can try and show $x$ must have this form if $2^k$ divides our product, but no luck here either

[cloud walrus]

person2709505

[cobalt heath]

I am perhaps too elementary-brained but

You can consider what the gcd of x-1 and x+1 should be

[coral steeple]

2?

Since x+1-(x-1)=2 and the gcd cannot be 1 (since x must be a unit -> coprime to 2^k -> odd -> x +/-1 even)

I will have a think and come back in the morning I think. Thanks for your patient help 🙂

[cobalt heath]

Cya!

[limpid hemlock]

commutative ring + inverses = field ?

or do you not need the commutative

[dull marsh]

Fields are required to be commutative, yes

[cobalt heath]

There is division ring, which is like a field but not commutative

[limpid hemlock]

👍 thank both 😁

[dull marsh]

Also skew fields have the same definition as fields minus the requirement for commutativity

[plucky wyvern]

Can some one explain this theorem to me? I don’t understand what it means by “g-component collum matrices”, can you give me an example of such a matrix?

[boreal inlet]

That anyways trivially stabilizes when the chain reaches the prime subfield of characteristic p.

[boreal inlet]

Unless you're reading Emil Artin's lectures on Galois Theory

https://cdn.discordapp.com/emojis/722022152328445985.webp?size=48&name=smilebutcryinside&quality=lossless

[cobalt heath]

What happens there?

[boreal inlet]

They say field for divison rings, and commutatiive field for fields.

It's old so that's understandable

[cobalt heath]

Ahh

[south patrol]

But the fields may not be of finite dimension over the prime subfield

[elfin wraith]

I think that’s a French convention (probably a bourbaki thing), my algebra notes point it out in case we see it in other literature

[south patrol]

Indeed

[mild ore]

hi sorry if this is a wrong place to ask, but can someone help me argue that if range(T^n-1) is not equal to range(T^n), then T^n = 0 and T is then a nilpotent operator? where dim(V) = n

[south patrol]

Tbh probably okay here or could go in #linear-algebra

[south patrol]

Basically there is a descending sequence $V \supseteq \mathrm{range}(T) \supseteq \dots \supseteq \mathrm{range}(T^{n-1}) \supseteq \mathrm{range}(T^n) \supseteq 0$. Try to show that if $\mathrm{range}(T^m) = \mathrm{range}(T^{m+1})$ for some $m$, then in fact $\mathrm{range}(T^m) = \mathrm{range}(T^{m+1}) = \dots = \mathrm{range}(T^n)$. Do you see how this would help you finish?

[cloud walrus]

Süßkartoffel

[mild ore]

thanks for your response, what I dont understand is why we can say there is a descending chain of range

like I don't know why we can ensure that the range of powers of linear maps is decreasing

[south patrol]

The image of T^(n+1) is contained in that of T^n

If x = T^(n+1)y, then x = T^n(Ty)

[mild ore]

ohhh lol

idk why I didnt notice this

thank you so much

[south patrol]

Dw

[delicate orchid]

OH I was reading range(T^n-1) as T^n minus the identity matrix lol

[south patrol]

Lol

[delicate orchid]

I thought there was some eigenspace subtlety I was missing haha

[hollow mica]

I know that for a subgroup N of G to be normal we need gN = Ng for all group elements g, but I also came across the definition that the set of left cosets and right cosets is equal. Why is this second definition equivalent?

gN = Ng implies they are equal is obvious

but the other direction?

[delicate orchid]

you just stated the same thing twice

[static glen]

Just think about what the cosets are

they are exactly that

[delicate orchid]

"gN = Ng for all g in G" is just the statement that left/right cosets are equal

[hollow mica]

for each g you could have an h so that gN = Nh without g = h

[hollow mica]

Well you probably can’t but I don’t see why

[south patrol]

What definition of normal are you using

[hollow mica]

gN = Ng for all g

[south patrol]

So what is there to prove

[hollow mica]

If the set of left cosets and right cosets are equal, then gN = Ng for all g

[south patrol]

Ah okay I see

[hollow mica]

I’m saying that the set of left and right cosets could be equal without the equation holding, if this scenario happened

[delicate orchid]

oh it's a transfer thing I see

[south patrol]

If gN = Nh then we can write g = nh (since g is an element of each side) and then Nh = Ng

[hollow mica]

I see

Wait
“gH = H iff g in H”
is true right?

Ok yea this came up before

Ok after seeing multiple proofs of “coset product is well defined for normal subgroups”

Why do people choose anything other than using the gN = Ng definition

you don’t even have to directly work with elements that way

[south patrol]

Ye

[south patrol]

Well sometimes you want to work with elements

Like saying N is closed under conjugation is probably usually easier to prove in practice

At least, it's how I think about it now

[delicate orchid]

it's also the one that's more useful lol

I rarely think about quotient groups as actual cosets anymore, cosets are just useful for group actions as nature and the lord in heaven intended

[south patrol]

Yeah true tbh i don't think i have ever used gN = Ng

lol

well like at least in the last couple of years i guess

[hollow mica]

Mhm

[south patrol]

For example like

In order to find the normal subgroups of a group, a good place to start is to compute the conjugacy classes if you can

and the normal subgroups are exactly the subgroups which are unions of conjugacy classes

[hollow mica]

I see why a union of conjugacy classes that is a subgroup is normal (gNg^-1 \subset G lends well to this like you say). For the other direction, is it similar?

Oh

if N is normal then you can check that it contains each of its elements conjugacy classes

[hidden wind]

got this for free, mwahaha

i… am not sure if i’ll use it though, i just couldn’t not take it when free, ehehe

[crystal turtle]

For free is nice

[hidden wind]

i don’t have many hard cover maths books actually, feels oddly luxurious

[elfin wraith]

Hardback is really nice, I got a hardback copy of ideals varieties and algorithms in the springer sale and it feels so good, you could definitely kill someone with it though

[hidden wind]

i have a copy of that one from the library, it's lovely, but an entirely different feeling when it's MINE MWAHAHAHA

ahem

[barren sierra]

that book is so nice

I will forever shill it

[hidden wind]

lang? or cox

[elfin wraith]

I got it on your recommendation

And because the comalg course I’m doing next semester uses it

[barren sierra]

cox little oshea

who shills Lang

[hidden wind]

lol

[elfin wraith]

But it was your recommendation that made me pull the trigger, book seems great everyone loves it

[hidden wind]

actually i did like lang's complex analysis

[astral sphinx]

hey, i'm really new to group theory and i was wondering about getting the product of 2 groups. i've understood how to get the product of C2*C2 and C2*C3 but I'm stuck on trying to multiply C2*C4 and can't seem to find anyone going through it step by step, could anyone help please? sorry if i phrased this badly or the solution is really simple, im very beginner level

[hidden wind]

do you know how to describe the product of two groups G1 and G2 in general?

that is, what are the elements of G1 x G2? and what is the operation in G1 x G2 ?

[astral sphinx]

i have a vague idea but i don't know any actual formulas for it, i've been self-teaching so i'm probably missing out on key details

isn't it to do with multiplying the generators of each group together?

[coral spindle]

What do you understand the definition of the product of two groups to be?

Can you quote the definition?

Is this part of some exercise?

[astral sphinx]

no

[coral spindle]

I'd like answers to the other questions too

[astral sphinx]

the answer is no to all of them, sorry, i'll look into them to better understand what im doing

[coral spindle]

OK

Let me give you some advice that will help you learn math in the future

If you do not know the definition of something, you cannot answer questions about that thing.

As far as you know, this may as well be the "quizzleblop" of two groups. Without the definition, it is a meaningless term.

I will give you the definition of the product of two groups, and you can judge for yourself if you are happy with them:

[cloud walrus]

Boytjie

[coral spindle]

So the question "what is the product of two groups" is wholly answered by this definition.

[cloud walrus]

Boytjie

[coral spindle]

I hope that this provides a completely unambiguous answer to your question.

[astral sphinx]

thanks, sorry for bothering you

[coral spindle]

Is this all clear?

[astral sphinx]

yes

[coral spindle]

Glad to hear it

[formal ermine]

oshea (t) little cox*

[stark helm]

Question: Use Kronecker's theorem to construct a field with four elements by adjoining a suitable roots of x^4-x to Z2? I factor x^4-x=x*(x-1)*(x^2+x+1), and there exist root 1 and 0, then how to proceed it? I only know there must be a field E to let f splits over E since polynomial can be written as a product of irreducible factors.

[delicate orchid]

let alpha be a root of x^4-4, and then consider Z2[alpha]

[rocky cloak]

You already have 1 and 0, so you want to ajoin a root of x^2 + x + 1

[delicate orchid]

obviously if you pick 0 or 1 nothing changes

ok nvm I'll let jagr have this one

[rocky cloak]

Team work makes the dream work!

Anyway, do you know how to adjoin a root of an irreducible polynomial?

[delicate orchid]

don't want to talk over you + it's easier to follow one explaination than 2!

[astral fractal]

Could I get a hint for this one? https://gyazo.com/a57ded4a5859f594333b74f535e5b7c0

[rocky cloak]

Well, xy has order n and x has order 2, maybe see how that compares to your definition of the dihedral group.

[delicate orchid]

And away they go

[hollow mica]

sorry

I figured it out

but I have a new question

[dull ginkgo]

(He has a new question)

[hollow mica]

Apparently there's some natural bijection between {maximal ideals of R[x]} and the upper half plane

how do I get that guy

[delicate orchid]

For R the real numbers? That’s news to me. Miz do ur Spec R[x] nonsense

[hollow mica]

Yeah

[dull ginkgo]

no I was going to bully

I haven’t done much alg geo lol

[delicate orchid]

This sounds very non-trivial to me I must be honest

[dull ginkgo]

But I mean irreducible polys over R are linear or strictly positive quadratics

could work with that

[delicate orchid]

Oh yeah that’ll do it

Still don’t see the map

[dull ginkgo]

Fuckin got derailed from trolling into actually being useful god damn it

[delicate orchid]

reduce them down to monic and then send x^2+bx+a -> (b, a) is the obvious one but it’s not a bijection

[cloud solar]

Question: what are the finite groups with the property that the sum of orders of elements is the the order of the group? G={e} obviously a solution. I proved that there is no abelian group with the property using the bounding of the sum of orders. But can we have non abelian groups with such property?

[cobalt heath]

Ah, the upper half plane stuff is what I wrote as a description for Spec R[x] in an exercise

[delicate orchid]

So we do need Spec R[x]

[dull ginkgo]

Zariski-Topologically or like

[cobalt heath]

That said, it has mostly discrete topology, so it underwhelming.

[delicate orchid]

Do I need a SES

[dull ginkgo]

Yeah I was about to say

[cobalt heath]

Yeah Zariski

[dull ginkgo]

horror

That’s a tad bit more than a bijection then

[cobalt heath]

Impossible for nontrivial group, because there is an element of order bigger than 1

[delicate orchid]

Using the class function we write |G| as the sums of the sizes of the conjugacy classes. Elements in the same conjugacy class have the same order, so it’s going to be bigger than |G|

[dull ginkgo]

Oh wait upper half plane

Factorize the strictly positive quadratics

Wlog one of the roots has real imaginary part

but why is that useful lol

[cobalt heath]

Maybe you can act SL_2(R) on it

[delicate orchid]

It’s going to be some stupid result about Riemannian surfaces or something

[delicate orchid]

Yup. Le modular form has arrived

Kind of

[dull ginkgo]

oh it’s actually not horrible

Just proving it’s a homeomorphism

Wait strict upper half plane or including the real line

[hollow mica]

To be clear I just want a bijection that I can write down lol

[hollow mica]

Uhh artin just said upper half plane so probably not including

[dull ginkgo]

Send a to <x - a> and a + bi to <x^2 - 2ax + (a^2 + b^2)> lol

[storm obsidian]

what is so special about SL2(R)

why did lang write a book on it

[delicate orchid]

I like rotating shit

[dull ginkgo]

Have fun proving it’s a Zariski-R metric homeomorphism

[dull ginkgo]

Euclidean ass mf

[dull ginkgo]

SL(2,Z)’s abelianization is 12 cyclic

Have a good day

[hollow mica]

To start, I have to show that <x-a> is maximal.

[delicate orchid]

x-a is clearly irreducible

[dull ginkgo]

PID

[delicate orchid]

Quotient is iso to R etc etc

[dull ginkgo]

Then just show that it’s maximal because irreducible

[hollow mica]

I know the fact that if R is a field then any ideal in R[x] is principal

[dull ginkgo]

Show it’s necessary and sufficient for the principal generator to be irreducible

[hollow mica]

You want me to show that
f is irreducible iff <f> is maximal
?

[dull ginkgo]

1. If <p(x)> is maximal and q(x) divides p(x), then <q(x)> contains <p(x)>. Contradiction
2. If <p(x)> is an ideal and p(x) is irreducible, then if <q(x)> contains <p(x)>, then q(x) must divide p(x), contradiction

Circular for brevity but you get the point

Just use the def of a generator

[hollow mica]

I see now

and then every monic linear poly is irreducible

[dull ginkgo]

jokes on you I haven’t actually done much ring theory at all I just pulled out of this out of my ass

[hollow mica]

and any quadratic with b^2 - 4ac < 0 is also irreducible

[dull ginkgo]

thank god it’s at least legible

[dull ginkgo]

Yes, strictly positive

[hollow mica]

Are you saying it should be > ?

[dull ginkgo]

Negative discriminant, which is the absval of the roots’ imaginary part

[dull ginkgo]

The polynomial is strictly positive

[hollow mica]

oh

yea

[dull ginkgo]

That’s what I meant by strictly positive quadratic

[hollow mica]

ok another question

this one is conceputal

how am I supposed to think of R[x]/(x^2 + 1) as the complex numbers?

I feel like I know the formalities of quotient ring and stuff

But every explanation I read online feels like a bunch of handwaving when its not

[dull ginkgo]

Every strictly positive quadratic is related by some real translation

I.e a real shift in X like X + c

so all of them can be “shifted” to the classic X^2 + 1 one

And if you reduce that, done

[cobalt heath]

No way "modular forms" are this simple stuffs

[dull ginkgo]

You then have the complex numbers

[cobalt heath]

Simply, R[i] = R[x] / (x^2 + 1)

[dull ginkgo]

It’s a 2D vector space over R due to it being an integral extension, any polynomial p(x) = (x^2 + 1)p(x) + Ax + B

So if you quotient out x^2 + 1

You are left with Ax + B, I.e Ai + B

The complex numbers

Multiplication is the same

[hollow mica]

R[x]/(x^2 + 1) = {f + (x^2 + 1) | f in R[x]}.
And by construction,
f + (x^2 + 1) = g + (x^2 + 1) iff f - g is a multiple of x^2 + 1.
Now how do I show that for every poly f there is a g(x) = ax + b such that
f + (x^2 + 1) = g + (x^2 + 1)?
If I can show that then I can write R[x]/(x^2 + 1) = {ax + b + (x^2 + 1) | a, b in R} whereby then the relation to C is obvious

[dull ginkgo]

This is because R[X] is a Euclidean domain lol

[hollow mica]

I guess the answer to my question is some version of "the euclidean algorithm in polynomial rings", but I've never seen a formal presentation of that, so if someone could write down the explicit statement of that or point me to a reference I'd be grateful

[dull ginkgo]

You’d have to prove R[X] is a Euclidean domain, but that’s not hard

consider the notion of a degree

I haven’t formally proved it, but I’d imagine some sort of linear system to be solved from the “convolution” definition of multiplication of polynomials

The solution “vector” corresponding to the remainder polynomial’s coefficients

[cobalt heath]

You can show that division algorithm works for polynomials

[hollow mica]

https://proofwiki.org/wiki/Division_Theorem_for_Polynomial_Forms_over_Field

ok nice

[cobalt heath]

Which, imo, is obvious for R[x]

[dull ginkgo]

An inductive method also probably works

I’m too lazy to write it out lmao

Just keep cancelling the coefficients like inverting a triangular matrix lol

[cobalt heath]

I guess you can literally invoke linear algebra

[dull ginkgo]

Immediate brain damage approach from yours truly I need to pick up a book and stop bullshitting my way through abstract algebra lmao

[hollow mica]

wait here they never specify that deg(f) >= deg(d)

[dull ginkgo]

Wlog?

[hollow mica]

what happens if say f(x) = x^3 and d(x) = 10x^5

then doesn't the theorem fail

[dull ginkgo]

no

[rocky cloak]

Just pick q=0

[dull ginkgo]

yeah

[cobalt heath]

F[X] is an additive group, it has 0 in it

Shockers, I know

[hollow mica]

If q=0 then r=f

but then r doesn't have a smaller degree than f -- they have same degree

[rocky cloak]

Read the requirement again

r is supposed to have degree less than d

[hollow mica]

since r = f isn't (2) the relevant case in my screenshot

OH

[cobalt heath]

It's like integer division

[dull ginkgo]

Linear algebra approach to integer division

[cobalt heath]

What do you get when you integrally divide 3 by 5?

It's 3 = 0 * 5 + 3

[dull ginkgo]

7 :D

[cobalt heath]

Huh

[dull ginkgo]

True, chat

Anyway does that make sense T series?

[hollow mica]

Who is that?

[dull ginkgo]

[hollow mica]

Ok so I understand the statement

[hollow mica]

And more or less understand the proof here

But why do you think it's obvious?

[dull ginkgo]

Seen it enough times

lol

[hollow mica]

so this basically

[dull ginkgo]

We basically can keep cancelling terms until we have a unique remainder polynomial with degree less than the divisor

Like in Gauss-Jordan elimination when you get to an Upper Triangular matrix

[hollow mica]

basically if p has higher degree than d, we change add a term to our polynomial to cancel the next highest power

[dull ginkgo]

We can multiply by ax^(n-m) to cancel out the highest term and keep working down inductively

[hollow mica]

I was wondering what happens if we run out of degree, but now I see that the max number of times we do this is m-n since each stage the degree goes down by at least 1, so after m-n stages the degree is at most m - (m - n) = n

[dull ginkgo]

We’ll have a remainder polynomial

[hollow mica]

By run out of degree I meant like we add a ax^2 term, then a bx term, then c, and then p still has degree larer than d

[hollow mica]

But this will never happen cuz of this

[dull ginkgo]

It’s easier if you write it out with a basic example

It’s literally long division

[cobalt heath]

Another way to see this is that, as a R-vectorspace,
(1, x, ..., x^(deg d - 1), d, d x, d x^2, ...) spans R[x].

[dull ginkgo]

Long division or synthetic division is another way of writing it down

If you want to try it practically

[hollow mica]

Wait In R[x] the ideal (x) doesn't contain the constant polys right?

[dull ginkgo]

no

[hollow mica]

no as in I'm right?

[dull ginkgo]

Lol

Yeah you’re right

[hollow mica]

So you can't specify a homomorphism by just saying where x goes

[dull ginkgo]

what

Do you know the universal property

[cobalt heath]

You also need to specify how R would embed, I guess

[dull ginkgo]

Of polynomial rings

[hollow mica]

oh lemme think

oh i forgot oops

[dull ginkgo]

Yep

[hollow mica]

My original question answer is no cuz of this right?

"such that hbar = h" in my picture

[cobalt heath]

Yeah, unless you are considering R-linear maps

[hollow mica]

I don't understand this question then

[dull ginkgo]

Maps the coefficients

[hollow mica]

So presumably the universal property is the same with R[x, y] (i.e. that you can determine a homomorphism if you know where R embeds and where x and y go)

but in this problem it doesn't tell you where R goes

[dull ginkgo]

If you want to truly prove it, you can prove it by contradiction because it’ll fix the x, y elements

Also because the universal property still applies because it’s a map from R to R lmao

[hollow mica]

Huhh

I thought when it was just R[x] we knew where the consant polys were going cuz we knew where the (embedding of) R (in R[x]) is supposed to go

But here its R[x, y] and they tell you were x and y go but not where R goes

[dull ginkgo]

It’s like the universal property but with two lol

[cobalt heath]

This is R-linear map

[mild ore]

sorry im gonna put my linear algebra question here cuz people in this chat may seem more knowledgeable

can anyone explain why, if A and B are commuting diagonalizable matrices, then A - B is also diagonalizable and A and B are simultaneously diagonalizable?

[cobalt heath]

Did you try #linear-algebra first?

Yeah, you'd likely get better answers there I think.

Unless you are talking about representations here

[mild ore]

yeah sorry if it was inappropriate

[cobalt heath]

It's okay (I think)

[hollow mica]

can you explain

[cobalt heath]

Basically, it assumes that the map R -> R is the canonical one, i.e. identity.

[hollow mica]

Bruh

why didn't they just say that

[cobalt heath]

The author likely thought it should be obvious without specifying it.

[hollow mica]

Ah

[cobalt heath]

Many mthematicians tend to do this kind of omission, so it's good to be familiar with it

[hollow mica]

Inside a polynomial ring R[x], if I wanna check whether p divides q, is there anything I can do with the division algorithm?

[hollow mica]

Like this right here

[cobalt heath]

Wdym, why would you check divisibility?

[viscid pewter]

yes

use the division algorithm

and then if you get 0 remainder

then it divides

[hollow mica]

don't you always get 0

like at least for the integers

[viscid pewter]

edited

[hollow mica]

if you're checking if 12 divides 26

26 = 12*2 + 2
12 = 2*6 + 0

[viscid pewter]

ok wait

i'm a fool

just divide it

once

and then if the remainder is 0, then it divides

[cobalt heath]

Ah right, euclidean division algorithm is division algorithm

[hollow mica]

Wait nvm I'm being dumb

[viscid pewter]

strictly more efficient than anything to do with the division algorithm

because you have to do multiple divisions for that

[hollow mica]

Lmao oops

ok I solved my problem

sorry to bother y'all

[hollow mica]

Consider the map phi: Z[x] -> R that sends f to f(1 + sqrt(2))
(Z is integers and R is reals)

I'm trying to find the kernel of phi (write it nicely as the ideal generated by something)

If
f(x) = a_0 + a_1x + ... + a_nx^n
and f is in the kernel of phi then
0 = a_0 + a_1 (1 + sqrt(2)) + ... + a_n (1 + sqrt(2))^n

RHS is equal to something of the form x*sqrt(2) + y where x, y are integers

we need x = y = 0

But now this is messy

[crystal turtle]

messy how

[viscid pewter]

maybe try considering it for just linear functions and quadratics first to get an idea of what you should be looking to prove

[spare isle]

Just as a sanity check, despite $-$ not normally forming a group when over sets, the special case of $({0}, -)$ is a perfectly valid group right? Is this useful in any way (other than the fact that it is a trivial group)?

[cloud walrus]

JJCUBER

[hollow mica]

expanding (1+sqrt(2))^n with binomial theorem seems messy

[crystal turtle]

Explicitly doing so maybe. Recognizing that you can reduce it to just x*sqrt(2) + y by reducing exponents is probably enough tbh

[viscid pewter]

yeah sure

trivial groups are pretty trivial

[hollow mica]

In order to describe the kernel I need to know for what set of coefficients (a_0, ..., a_n) I get x = y = 0 tho

[crystal turtle]

Well on a set with a single element, there is only one binary operation on that set, and it forms a group multiplication. So yes, you can use whatever symbol you like for it and it is a valid group.

[crystal turtle]

Nope. Think about what x and y are. They're not just random real numbers... What are they?

Or uhh, maybe a bit further back, think about what each of the a_i's are.

[hollow mica]

integers

[spare isle]

I know it's just pedantry, but a given binary operation isn't necessarily over the set in all cases.

[hollow mica]

and the a_i are each some linear combination of the n+1 integers a_0, ..., a_n

[viscid pewter]

a binary operation on the set should be on the set

[spare isle]

ah I didn't realize on and over meant the same thing in this context

[crystal turtle]

Then it wouldn't be a meaningful question whether that was a group

[crystal turtle]

Exactly; a Z-linear combination specifically. Hence x and y are integers

[spare isle]

thanks

[crystal turtle]

Oh

[viscid pewter]

i don't see why that helps

like

[crystal turtle]

I see now lol

[viscid pewter]

anyway i would consider the linear and quadratic cases

like when does a linear or quadratic function have 1 + sqrt(2) as a root

wait no

i was going to say polynomials factorise into linear and quadratic factors

but that feels weird suddenly

[crystal turtle]

They factor but not into linears and quadratics over Q or Z. You get that over R.

[viscid pewter]

yeah

[crystal turtle]

Also sorry lol I confused myself for a bit there

[viscid pewter]

well, consider the quadratic case anyway

that'll at least give you a large chunk of cases

examples

[crystal turtle]

Alternatively, if you know some field theory this should be easy, as this is the kernal of the evaluation map f |--> f(1+sqrt(2)) defined on Q[x], intersected with Z[x].

[cobalt heath]

Are we looking for f s.t. f (1 + sqrt(2)) = 0?

[boreal inlet]

Is Z[x] a Bezout domain?

....
nvm, it can't be. There's a certain characterization of PIDs - An UFD which is a Bezout Domain is a PID.

Ok anyways, my original question was to show that f(x) in Z[x] is separable if the image of f(x) in F_p[x] is seperable.

I know that if instead of Z there was a field this is true under any field embedding.

But the only thing I have is this canonical projection of Z[x] onto F_p[x].

[rocky cloak]

In both Z and Fp every irreducible polynomial is seperable. So the only way for f to not be seperable is if g^2 divides f for some other polynomial g. If this happens in Z, then it also happens in Fp

[crystal vale]

Congratulations 😂, how do you get it?

[boreal inlet]

Wait, do you mean a polynomial in F_p[x] can only be separable iff it is a product of distinct irreducibles?

This is what I know as true for characteristic p fields

[rocky cloak]

Yes, Fp is a perfect field. Meaning every irreducible polynomial is seperable

[boreal inlet]

Okay. So if I have the image of f(x) in F_p to be separable, it is product of distinct irreducibles in F_p[x].

What is the criteria of seperability in Z, though

All roots are distinct?

Ok anyways

Following what you said

Let f be not seperable in Z[x].

Then f cannot be irreducible, and also in some splitting field, there must be a repeated root.

As Z[x] is an UFD, there exists a irreducible factorization unique upto associates.

Now, as each irreducibles are seperable, all of them contribute distinct roots. So, there must exist an irreducible g(x) such that g^2 divides f.

Now we consider the image of f under the canonical projection.

It's a ring homomorphism, then \bar(f) = \bar(g)^2 * \bar(h) for some h in Z[x].

Which says bar(f) cannot be seperable.

Oh then by contrapositivity, the result holds

Thanks a lot!!

[rocky cloak]

Actually there is a problem with the argument. Since (px + 1)^2 is not seperable, but 1 is

[crystal vale]

Is it correct if I want an example such that order of gN in G/N strictly less than the order of G

First I thought about taking a group Z and taking any subgroup N=6Z then in Z/6Z I take 3 + 6Z which has order 2.

Secondly I thought about taking a group Z= Q_8(quaternion group) and taking the group N= < j> generated by j so in Z/N we have iN which has order 2 .....is this the correct example?

[rocky cloak]

So you probably want to add a condition like the leading coefficient not being divisible by p

[rocky cloak]

You want an example where the order of gN is less than the order of G? That's just always true, unless N is trivial, G is cyclic and g generates G.

[boreal inlet]

oh wait i completely missed that part.

But this condition was not at all given

Oh wait

it is given

f is monic

[crystal vale]

My mistake order of gN in G/N is strictly less than the order of g in G

[rocky cloak]

Then you're examples are good yes

[crystal vale]

Okay thank you

[boreal inlet]

I'm actually not sure how to include this in my argument

Oh ok got it

f is monic so any factor of f should also be monic and the projected image is also.moniv

Done

Thanks now

[fickle brook]

alright so here's a thinker i'm running into with a program i'm making

i've created a function which tests a polynomial in F_p[x] for irreducibility

as part of it, i check whether f and f' are coprime -- if they aren't, i return False and print gcd(f, f') as a repeated factor

however i've come across a bug

i set p = 2 and fed a perfect square polynomial, namely x^6 + x^2 + 1 (= (x^3+x+1)^2) into the function

which gave me "NOT irreducible. Reason: Repeated factor -- x^6 + x^2 + 1"

which is

not satisfactory to say the least

which got me thinking

this thing's derivative is obviously 0

as in just the 0 polynomial

so heres my question

let f ∈ F_2[x] such that f' = 0. is f necessarily a perfect square?

and generally, let f ∈ F_p[x] such that f' = 0. is f necessarily a perfect p'th power?

if so, i could probably put in a special case here.

[stark helm]

i don't know how to adjoin roots, probably need more explanation.

[cobalt heath]

I think there was some mention in algebra textbook about how lin-combi of x^{kp}'s is a perfect p-power.

You can likely use the fact that the Frobenius map is an isomorphism.

[fickle brook]

about how lin-combi of x^{kp}'s is a perfect p-power.
this reads like the converse of what i want

so like ok

f' = 0 <=> f is a lincombo of x^(kp)

right?

and then
f is a lincombo of x^(kp) <=> f is a perfect p'th power

[cobalt heath]

Yes

[fickle brook]

alright

[stark helm]

so do you mean if we can ajoin four roots of x^4-x, then we can construct a field Z2[alpha1, alpha2, alpha3, alpha4]?( to construct a field with four elements by adjoining a suitable roots)

[rocky cloak]

If p is an irreducible polynomial, then
F[x]/(p) is isomorphic to the field given by adjoining a root of p.

So in your case F2[x]/(x^2 + x + 1) will be a field with 4 elements.

[rapid junco]

i dont immediately see why if the discriminant is not square in F then x^2 + bx + c = 0 does not have a solution in F

one constructs the quadratic equation explicitly on fields of not characteristic 2 but im a little confused as to why not being able to take square roots automatically there isnt another construction that shows the existence of a root

[coral shale]

if there is a root of the equation

then there necessarily is a square root of the discrim?

i suspect is the argument

Any solution has to satisfy the quadratic equation, so you can equate

[dire siren]

in characteristic different from 2, that equation is equivalent to (2x+b)^2 = b^2-4c
and now it is easy to see that the equation has solution if and only if the discriminant is a square

[stark helm]

does F2[x] mean Z2[x]? and in my case why F2[x]/(x^2+x+1) will have four elements( roots 0, 1, and roots from x^2+x+1?)

[dull ginkgo]

Seems like a Galois-y problem

[stark helm]

and it seems that x, x-1 and x^2+x+1 are all irreducible polynomial, so we have Z2[x]/(x) isomorphic to Z2[x]/(x-1) isomorphic to Z2[x]/(x^2+x+1)

[dull ginkgo]

… what

[coral spindle]

uh... no

[dull ginkgo]

Consider the order of the vector spaces over F_2

[dull ginkgo]

I haven’t done this exercise before or much abstract algebra outside group theory but if F has finite order q, then does G = F[X]/<p(X)> have order deg(p)q because [G : F] = deg(p)?

[coral spindle]

No, it has order q^{deg(p)}

[dull ginkgo]

Oh shit yeah

[coral spindle]

Vector spaces over finite fields have order q^d not qd

[dull ginkgo]

Brain fart

I was just wondering if characteristic causes any silliness but

I guess not, it’s a Euclidean domain nonetheless

[stark helm]

so not all these three are irreducbile in Z2[x]?

[coral spindle]

That's not the issue

[dull ginkgo]

They have different orders as vec spaces

[coral spindle]

You've just out of nowhere concluded they're going to be isomorphic, which is simply false.

[dull ginkgo]

*dim

Consider the remainder by Euclidean division if you want

[coral spindle]

Now it is true that F_2[x]/(x) and F_2[x]/(x-1) are isomorpic, and you should think about why.

But the other is not.

[dull ginkgo]

The last has a linear, first two just a constant

So they differ when quotiented

if the universe could provide me a simple proof of $G \subset \mathrm{Aut}(F), |G| < \infty \Rightarrow [F : F^G] = |G|$ I’d forever be thankful

[cloud walrus]

Mizalign

[dull ginkgo]

Like I get it’s an algebraic extension but rrrrr

[cursive spindle]

AA

[stark helm]

Does F2 refer to Z2? and it should because deg(x)=deg(x-1)?

[dull ginkgo]

it’s integral for any commutative integral ring actually

[dull ginkgo]

Proving it’s finite is the painful part

[south patrol]

This isn't complete enough reasoning either

Why should having the same degree give you the same quotient?

Anyway, it's false in general e.g. $\mathbb Q[x]/(x^2 -22) \not \simeq \mathbb Q[x]/(x^2 + 1)$

[cloud walrus]

Süßkartoffel

[south patrol]

||okay 22 is a typo but it works lmao||

But basically, if polynomials and stuff you do with them only depended on their degrees, much of maths as we know it would be easy

[dull ginkgo]

Because the map X -> aX + b is an automorphism

Actually I’m pretty sure R[X]/(p(X)) ~= R[X]/(q(X)) if p(X) = q(aX + b) if a is a unit in R

[rocky cloak]

So like I said before, quotienting out by an irreducible polynomial is the same as adjoining a root of that polynomial. Now adjoining a root of x or x-1 doesn't get you anything particularly exciting, and will give you something different than adjoining a root of x^2 + x + 1

[south patrol]

Yes, for the reason you just said

the auto of R[x] induces that iso by first iso innit

Poggers...

[dull ginkgo]

I should actually do ring theory lol

[stark helm]

By determining its dimension, we need to apply quotient remainder theorem?

[stark helm]

I have no idea how to determine the degree of F2[x]/(x) and F2[x]/(x-1)

[stark helm]

so you think it is enough to argue F2[x]/(x) is isomorphic to F2[x]/(x-1) because degree(x)=degree(x-1)?

[dull ginkgo]

no

for degree 1 it is okay because X -> X - 1 is an automorphism

But for higher degrees it’s not always the case UNLESS the field is algebraically closed

It is the case if p(x) = q(ax + b) which forces the degrees to be the same

[stark helm]

we need to argue the degree by using extension field in general? Or using quoteient remainder is enough?

[dull ginkgo]

?

Confused by what you’re asking

[stark helm]

I mean I need to know how to determine the degree?

[dull ginkgo]

Actually I’m unsure if it’s always the case in algebraicly closed fields

No it isn’t I just reasoned through it

[stark helm]

Probably like Qx and Q[x]/(x^2+1)?

[dull ginkgo]

In other words, the quotient ring is isomorphic if the polynomials are an argument shifting and scaling away from eachother

Which explains why C is iso to R[X]/(p(x)) where p is ANY strictly positive quadratic

Because you can shift the X to be x^2 + 1

:)

[south patrol]

Dimension isn't enough either

[viral tiger]

Not sure how to do this

[void cosmos]

whats phi

for b and c

just remember that you send orders to orders

and list

[errant shadow]

Does the kernel of a ring homomorphism always contain the additive identity?

[mighty kiln]

Yes

[viral tiger]

The inner automorphism, x -> gxg^-1

[errant shadow]

Why is that? I tried proving it, didn't get much without making assumptions on the ring.

I think I am doing something wrong.

How would you show it if you had to?

[mighty kiln]

All ring homomorphisms are group homomorphisms

[errant shadow]

Ah

[dull marsh]

It's the same as showing group homomorphisms preserve identity

[errant shadow]

Makes sense

Thanks

[rapid junco]

Im sorry i dont immediately see why still

[quartz thistle]

hey, can anyone suggest few papers on Rings, Integral Domains and Fields of undergrad level
also any specific topic in this is fine

[errant shadow]

What exactly do you picture when you think of taking quotient with a kernel given a ring homomorphism?

Definition and all is fine but, I don't see it. Can someone give an example. I wish there was a book of detailed examples in abstract algebra. Like there's the book for counterexamples in topology. I wish there was something like that here too.

[crystal vale]

Which book for counterexamples in Topology?

If k does not divide n I think it will work, right?

[errant shadow]

Seebach

[dull marsh]

Should work regardless of k I think

[crystal vale]

Yes

[errant shadow]

Is it like, making an equivalence class over the kernel to sort of make it 1-1?

You are clubbing together all maps to zero into one entity.

To make it 1-1?

[crystal vale]

Let G be a finite group and for any arbitrary subgroup N, consider the set {g | gNg^(-1) <N } then the set is closed under inverse ?

[errant shadow]

Like what does this set look like?

[crystal vale]

I think so taking the cosets of the kernel then making the ring and then find their isomorphic image

[dull marsh]

In general taking the quotient ring modulo an ideal should be thought of as setting elements of that ideal equal to 0 and considering the ring's elements with that (not sure if this will help tho)

[errant shadow]

So like, Z/nZ, all elements in nZ are thought of as 0.

[dull marsh]

Yeah

[barren sierra]

and this idea generalizes. When considering R/I, we make all the elements in I = 0 essentially

[dull ginkgo]

true!

[barren sierra]

but idk what you mean by visualize or what the set looks like

[dull marsh]

I think quotient modulo kernel makes much sense to do if you want to make the homomorphism have trivial kernel or something

[barren sierra]

formally the set consists of just cosets

but since I is an ideal, we can equip these cosets with the desired ring operations

[dull ginkgo]

I mean you basically just “factor” it out, quite similar to the name

[errant shadow]

Okay so here, for some a in kernel, you are basically taking r*a as one equivalence class? Where r is arbitrary?

[barren sierra]

where are you getting r^n?

[errant shadow]

Like operating the same element several times basically what I meant but in hindsight that looks restrictive

[barren sierra]

You're familiar with G / N where G is a group and N is a normal subgroup right?

[dull ginkgo]

Galois

[errant shadow]

I am but I never really got comfortable. So I am going through this for once and for all.

[dull ginkgo]

Balls to the Galois’s

[hidden wind]

much excite

[errant shadow]

Wait is it a rec with examples?

[dull ginkgo]

I have Dummit and Foote lol

[barren sierra]

Ok so formally the elements of R/I are cosets r + I where r is an element of the ring R

[dull ginkgo]

I have second edition though

[barren sierra]

so r + I = {r + i | i in I}

and this itself is a ring

[hidden wind]

@deft pecan

[errant shadow]

And what's the restriction on l?

[barren sierra]

(x + I) + (y + I) = (x + y) + I

[hidden wind]

it’s not a rec, ehehe, i haven’t worked through it yet, but i’d say it looks very promising

[barren sierra]

I has to be an ideal for this to be a ring

I recommend trying to prove that R/I is a ring if and only if I is an ideal of R

[errant shadow]

Oh okay

[dull ginkgo]

You can kind of “take away” the ideal or set it to 0 via the quotienting map since it’s a kernel

[hidden wind]

it seems to be very classical in style

[barren sierra]

so for example for Z/nZ

you have (x + nZ) + (y + nZ) = (x + y) + nZ

[dull ginkgo]

Also multiplication is absorbed

[barren sierra]

and similar to how equivalence mod N is an equivalence relation, x = y (mod n) if and only if x - y in nZ

[dull ginkgo]

so you can more easily “take it away” and preserve multiplication

[barren sierra]

x = y if and only if x - y in I is an equivalence relation

[crystal vale]

For non-commutative Ring R I must be a two-sided ideal for R/I be ring?

[dull ginkgo]

Yes

Two sided rings are kind of like “normal” ideals as in how normal subgroups are to groups

they are kernels

Kernels of the quotient lol

[errant shadow]

Okay I'll try that

Thanks

[covert fjord]

If G had two cyclic factors C and C_0 whose orders were divisible by a prime p, then G would have (at least) p^2 elements of order dividing p. This doesn’t happen, and it follows that G is cyclic.

[crystal vale]

What is the cyclic factor ? Factor group?

[fickle brook]

does anybody know where i can look up some polynomials in F_p[x], p prime, which are irreducible but not primitive?

oop

existing convo already here

[covert fjord]

I think it is products of group

G is abelian

[crystal vale]

Means G= C× C_0 ?

[covert fjord]

Like G = A×B×....×C×C_0

[crystal vale]

But why doesn't this happen that there are at least p^2 elements of order dividing p ?

[covert fjord]

Like any (....,c,c_0) have order p

If dots are identity

You can choose c and c_0 p^2 ways

I dont understand why if abelian group is not cyclic then it should have factors

[crystal vale]

How (.......,c,c_0) have order p ? If let say c has order 4 and c_0 has order 8 and both divisible by 2 so what will be the order of (........,c,c_0) ?

[covert fjord]

Okay i dont know

I think it should be 8

Or it depend on element?

[celest furnace]

Have you seen the fundamental theorem of finite abelian groups?

[crystal vale]

It depends on the order of elements it will be the least common multiple of their orders

[covert fjord]

You mean like Z_n is isomrphic to Z_n1×Z_n2..... ?

Where n1 n2.. are coprime

[covert fjord]

But how elements in C have order dividing 2?

Two?

And in C_0 also 2?

[crystal vale]

Actually I think I misunderstood your question, C and C_0 is cyclic group which has order in the form of p•k , right?

[covert fjord]

Like isomprhishm of C to Z/4Z and C_0 to Z/8Z

[covert fjord]

Yeah

[crystal vale]

Yes you can take this examples

So you can find the elements which has order 4 and order 8

[covert fjord]

We need order dividing 2

[crystal vale]

So how do you claim that there are at least p^2 elements of order dividing p?

[covert fjord]

In Z/4Z are 0,2 and in Z/8Z 0,4

[crystal vale]

Yeah

For p= 3 it will work?

It will be always work

[covert fjord]

But how to prove?

[crystal vale]

Do you know if a group is finite and cyclic then number of elements of order d is ?

It is Euler phi function so in C and C_0 there are p-1 elements which have order p and then you have also an identity element so you have total p×p elements

At least p^2 elements have order dividing p

[covert fjord]

Yeah

Okay but how he conclude that it is cyclic

[crystal vale]

G?

[covert fjord]

Yeah

Like if there is no factor groups that divide the same prime number then it is cyclic?

[crystal vale]

Factor group? And same prime number means?

[covert fjord]

Like if you take Z/9Z×Z/27Z it is not cyclic but when you take
Z/8Z×Z/27Z it is cyclic

[crystal vale]

For cyclic just need that the group must have a single generator or for a finite group you must show that there is an element which has the same order as order of G

[crystal vale]

Yes because you choose a generator (1,1) which has order LCM(8,27)= 8×27, which a order of Z/8Z × Z/27Z

[covert fjord]

How to conclude that G is cyclic

[crystal vale]

Z/9Z × Z/27Z is not cyclic but the Z/9Z and Z/27Z is cyclic and their order is divisible by 3 so your conclusion is not correct

[covert fjord]

Yeah it because of that it is not cyclic

Where i am wrong

?

[crystal vale]

Where it is given that this hypothesis implies G is cyclic?

You need more hypothesis may be

[covert fjord]

Okay i try to rephrase

If you have a finite abelian group thats every two factor groups orders are comprime then it is cyclic

[crystal vale]

So order is co prime

[delicate orchid]

Chinese remainder theorem

[covert fjord]

Yeah

[crystal vale]

Then in case of Z/8Z and Z/27Z the orders is co-prime therefore it will work

[covert fjord]

But it goes other direction like

[crystal vale]

In this you do not need the factor group to be cyclic?

[covert fjord]

If you have two factors cyclic groups order that is divisible by the same prime number then it is not cyclic

[covert fjord]

Yeah

Okay guys i need to go

Thanks a lot

[crystal vale]

If factor group is not a cyclic group then if you take any element (a,b) it's order is not same as order of group G

Take G= U(8) × (Z_3× Z_3) then take any element from G it has atmost order is 6 not 36, where U(8) denotes the group {1,3,5,7} with multiplication operation mod 8

[dire siren]

if b^2-4c is not a square, the equation (2x+b)^2=b^2-4c doesn't have solutions simply because b^2-4c is not of the form (something)^2
if b^2-4c is a square, say t^2, the equation is (2x+b)^2=t^2, and this has solutions: 2x+b=t or 2x+b=-t (i.e. x=2^(-1)*(-b+t) or
x=2^(-1)*(-b-t))

[rapid junco]

figured it out you just go by contrapositive

but apprecite the response

[hollow mica]

What does the notation Z[a + bi] mean

Like I see people talking about the ring Z[(1 + sqrt(-3))/2]

I know that Z[x] is the polynomial ring of polynomials with coefficients in Z

[crystal vale]

Just replace x by that a+bi and see what you will get

[hollow mica]

The problem I have with that is can't two elements have different polynomials representing them?

[crystal vale]

I see this type of question I do not remember where I think Artin or dummit

[crystal vale]

Means?

[hollow mica]

Like say in Z[1] we have, the polys x^2 + 1 and 2x represent the same value

(1)^2 + 1 = 2(1)

[rapid junco]

artin

[crystal vale]

Yeah

[hollow mica]

I guess I don't understand what ring structure Z[a + bi] should carry

[crystal vale]

What is a and b?

[hollow mica]

fixed real numbers

[dire siren]

it doesn't matter if they have different representations; here you care about the common value

[crystal vale]

You can refer to artin or dummit

[dire siren]

Z[a+bi] is the smallest subring of C that contains Z and a+bi

[hollow mica]

Oh

Okay

And it happens to be that as sets, that smallest subring equals what you get by evaluating every poly of Z[x] at x=a+bi?

[crystal vale]

Then what will be Z(a+bi) ? Fraction field of Z[a+bi] ?

I am confuse in () and []

[hollow mica]

Oh wait yeah

isn't () used for that

[dire siren]

yes, since once a ring contains Z and some extra element d, it's going to contain any polynomial expression with integer coefficients evaluated in d

and it turns out that the set of such expressions is a ring; so that's the smallest ring

it might be helpful to think about Z[sqrt(2)]

[hollow mica]

I guess my original confusion was if you interpret Z[a+bi] as the set of evaluations of every poly in Z[x] at x=a+bi, then how do you define the ring structure

[dire siren]

yes, it's the fraction field

[crystal vale]

Okay

[dire siren]

generally, if R is a ring, then R(t) means the smallest field that contain R and t

[crystal vale]

That's mean R[t] is subring of R(t)?

[dire siren]

yes

[crystal vale]

Okay thank you

[dire siren]

the operations are inherited from C, the usual addition and multiplication

[hollow mica]

Ah

[crystal vale]

In Lang it is given that t is variable or transcendental over R therefore we induce R[t] so what will be the significance of transcendental

[dire siren]

for Z[sqrt(2)], if you take a polynomial from Z[x] and evaluate it in sqrt(2), you are going to obtain something of the form A+Bsqrt(2), with A and B integers
conversely, any expression of the form A+Bsqrt(2) is a polynomial in Z[x] evaluated in sqrt(2) (precisely, A+Bx)

so Z[sqrt(2)] is the set of elements of the form a+bsqrt(2) with a,b integers

[hollow mica]

I was thinking that since we use Z[x] to define Z[a + bi], the elements of Z[a + bi] should "look like" a_0 + a_1 * (a + bi) + ... + a_n (a + bi)^n, and that the addition would be defined like in Z[x] as
(a_0 + a_1 * (a + bi) + ... + a_n (a + bi)^n) + (b_0 + b_1 * (a + bi) + ... + b_n (a + bi)^n) = (a_0 + b_0) + (a_1 + b_1) (a + bi) + ...

But now I realize that Z[a + bi] is not those formal expressions but rather the simplified version (a complex number)

[dire siren]

yeah, like the actual values

[hollow mica]

And the way to formalize poly evaluation is this right: If you have rings R, S with a homomorphism t: R -> S and you want to evaluate a polynomial f in R[x] at a c in S then you look at the image of f under the map phi that sends the poly x to c and restricts to t on R (i.e. phi ∘ pi = t where t: R -> R[x] is the canonical inclusion)

Most of the time S = R and "standard evaluation" means that the homomorphism t: R -> R is the identity

[dire siren]

sounds right

I think you meant, that pi:R-->R[x] is the canonical inclusion

[hearty ledge]

Prove that, if $H$ is a subset of a group $G$, then $H$ is a subgroup of $G$ iff $a * b^(-1) \in H$ for every $a,b \in H$. How would I prove the Left direction

[cloud walrus]

Bizcuits

[hearty ledge]

i think i goit but not sure if it is correct

[faint imp]

That is correct.

[hearty ledge]

fck ye

i be overcomplicating things soemtimes in these proofs

[faint imp]

This proof is also a bit overcomplicated 😉

[hearty ledge]

oop

[faint imp]

Another good exercise is the following: Let H be finite subset of G. Then $H$ is a subgroup of G if and only if $ab\in H$ for all $a,b\in H$.

[cloud walrus]

Kamal Saleh

[hearty ledge]

o

doesn't look too hard except trying to prove the existence of inverses

[faint imp]

proving the existence of inverse is constructive and is very helpful in concrete finite groups.

[crystal vale]

But if H is empty set?

H must be non-empty

[south patrol]

Yeah

[crystal vale]

First you need to prove H is non empty

[south patrol]

they should just add "non-empty" to the statement

since otherwise it is not a true statement

[crystal vale]

Yes

[hearty ledge]

it also states H is a subset of G

[south patrol]

what's the relevance of pointing out this

[hearty ledge]

uhhh idk im just an undergrad struggling in abstract

wouldn't it just saying $a * b^(-1) \in H$ just mean its non empty??

[south patrol]

"for every a,b in H"

[cloud walrus]

Bizcuits

[south patrol]

there may be no a or b in H at all

in which case it is vacuous that "ab^-1 is in H for each a,b"

[hearty ledge]

o

[south patrol]

This is a bit of a pedantic point but it does mean you should change the original thing to "Prove that if H is a non-empty subset [...]"

But otherwise the question is fine

[hearty ledge]

at that point you just prove by contradiction assuming H is empty and boom which is a one liner

[south patrol]

Lol

I have wondered this in exams

[hearty ledge]

xDD

[south patrol]

Like if the question has a mistake, when is it reasonable to just write down a counterexample :)

and move onto the next question

[hearty ledge]

im sure they'll give you extra points but im sure someone will point it out to them and they'll correct themselves

i feel like most questions ive asked here you guys have always pointed out some kind of typo in the problem 💀 and my professor says this is his favorite class to teach

[south patrol]

Oh for me I mean official exams where you can't even do that xd

I had an exam where they missed an assumption for example

[hearty ledge]

my professor crazy tho tbh. he is drinking insane amounts of diet dr pepper and writes exams the day before giving them out. Had him for stats and he told us at 9pm the night before our 8am stats class that he has only written one problem 💀

I dont understand that man

does group theroy entail more than just subgroups, groups, permutations, and isomorphisms?

[hidden wind]

have a guess

[hearty ledge]

no

[south patrol]

Yes, it is an active field of study

Those are some of the topics you'll most likely learn in a first course

[formal ermine]

dr pepper goes hard tho

[hearty ledge]

hm i am still struggling to choose something for my research project

might do group theory

[formal ermine]

rep theory

[hearty ledge]

i am realizing more and more that me taking linear algebra during covid is fcking me over a lot.

also doesn't help that I was basically couch surfing during that time too 💀

[hearty ledge]

ill look into rep theory a lil more. I would for sure have to brush up on my linear algebra for that.

[formal ermine]

doesnt need a lot

[hearty ledge]

yeah seems like it

seems like its just the basics of matrices

[formal ermine]

i can recommend serre linear representations of finite groups

[hearty ledge]

thanks

[formal ermine]

and tensor products

[hearty ledge]

is this a book or?

[formal ermine]

yes

[hearty ledge]

okay thanks

[hidden wind]

Jean-Pierre Serre ?

[formal ermine]

yes

[hearty ledge]

yeah i found that book

okay sick

[formal ermine]

not serre lang

lmao

typo

[hidden wind]

human chatgpt moment

[formal ermine]

Real

[rapid junco]

i get the backward direction but how do you prove the forward direction?

f(a) = 0

a is in E which an extension of F

we get that deg(h) = [F(a, b) : F(b)] where h is the minimal polynomial over F(b) for algebraic a

then f(a) = 0 implies h | f

but they have the same degree

this implies f = u * h

where u is a unit

oh but then this is true i guess? (not sure if it is but just thought of this)

product of irreducible and unit is again irreducible

so f is irreducible

coudl someone verify if i am making a mistake

[lapis latch]

looks gud

[south patrol]

You can also adapt this just to prove both bits at once

like [F(a,b):F(b)] is the degree of the min poly of a, so since f(a) = 0, [F(a,b):F(b)] = deg f if and only if f is minimal

[long obsidian]

Maybe you could check out the second chapter from hungerford where he talks about the structure of groups.

[hearty ledge]

Thanks

I'll take a look for sure

[plush pulsar]

I told my professor that G cannot be an arbitrary group here, it has to have its binary operation as multiplication, but he denied and told me to find the reason behind it. I haven't been able to figure it out yet.

[crystal vale]

Why do you think G can not be arbitrary?

[lapis latch]

well it's assumed that the group operation is multiplication as you've written it. It is worth noting that the distinction in the syntax of the binary operation of a group is nothing more than convention(if the group is abelian then it is typical to have the operation be addition)

[plush pulsar]

yeah I know the symbol is kinda confusing

[plush pulsar]

let me show my work

[plush pulsar]

I actually used the asterisk for posting the question, it was not used in the original question. It simply mentioned G as an arbitrary group.

[crystal vale]

no how do you compute a^m • a^n, here • denotes the group operation

You can proof that a^m•a^n = a^(m+n) by induction

[plush pulsar]

ohh

yeah I get it

@crystal vale thanks!!

[fickle brook]

ok so something is making my head spin here

https://en.wikipedia.org/wiki/Factorization_of_polynomials_over_finite_fields#Distinct-degree_factorization

This algorithm splits a square-free polynomial into a product of polynomials whose irreducible factors all have the same degree.

why the name "distinct-degree factorization"???

the reason im asking is that i want to implement factorization in my finite field poly calculator

but like, this is kind of throwing me

[fickle brook]

also this is probably more of a programming question, but

for my factorization algorithm, i am thinking of starting out by explicitly checking the input polynomial for linear factors, and dividing those out. but i don't know if that is Worthwhile™️

[faint imp]

This step factorizes f into a product g1g2…*gn such that gi is the product of all irreducible factors of f of degree i. In a later step these g1, g2, etc need to be further factored using another algorithm. Hence the name "distinct -degree-factorization".

[fickle brook]

oh... huh

[cobalt heath]

Yeah, confusing wording on the wiki

It factorizes into factors each of which is product of certain irreducible factors, not being irreducible itself

[crystal vale]

If R is non commutative ring and we have 1= I + J where I, J are left ideals then I intersection J not be equal to IJ, right? And IJ be left ideal

[rocky cloak]

Take x an element that has a right inverse, but no left inverse. Let I be Rx and J = R. The intersection is clearly Rx, but RxR = R.

So you don't even have that IJ is contained in the intersection

[crystal vale]

Oh I see

[static quarry]

Hello all. The following is an elementary result from page 11 of "A Course in the Theory of Groups" by Derek J. S. Robinson, slightly modified and phrased slightly differently:

(1.3.5') Let $K\le H\le G$. If $T$ is a left transversal to $H$ in $G$, and if $U$ is a left transversal to $K$ in $H$, then $TU$ is a left transversal to $K$ in $H$; and moreover, the map $f:T\times U\rightarrow TU$ such that $f(t,u)=tu$ is a bijection.

And below is my proof of the above result:

As $T$ is a left transversal to $H$ in $G$, we have
$$G\in \bigsqcup_{t\in T}tH;$$
and as $U$ is a left transversal to $K$ in $H$, we have
$$G\in \bigsqcup_{t\in T}t\big(\bigsqcup_{u\in U}uK\big) =\bigsqcup_{t\in T}\bigsqcup_{u\in U}tuK=\bigsqcup_{(t,u)\in T\times U}tuK=\bigsqcup_{s\in TU}sK,$$
whence it follows that $TU$ is a left transversal to $K$ in $G$.

The surjectivity of $f$ is obvious, in light of the definition of $TU$. And the injectivity of $f$ follows from the disjointness of the union
$$\bigsqcup_{(t,u)\in T\times U}tuK,$$
combined with the fact that $tu=tue\in tuK$ (and in particular, $tuK$ is nonempty) for each $(t,u)\in T\times U$. Q.E.D.

Can you please verify my proof above? Any feedback on my proof-writing style would be appreciated as well.

I suspect that the terminology of "the left transversal to a subgroup $H$ to a group $G$" is not standard (since I have not seen it in many other standard books on the topic), so I will define it here: if $G$ is a group, and if $H$ is a subgroup of $G$, then a left transversal to $H$ in $G$ is a subset of $G$ which contains exactly one element from each left coset of $H$ in $G$. If you would like me to clarify any other notation or terminology, please ping me. Thank you in advance!

[cloud walrus]

DW0987

[crystal vine]

I need some help with finding two 2 dimensional group representations r and r' of $\mathbb{Z}^2$ such that for any (n,m), r(n,m) and r'(n,m) are conjugate, but such that r and r' are not isomorphic.
Since r(n,m) and r'(n,m) are conjugate, they must have the same eigenvalues, so the only difference between the two must be the eigenvectors corresponding to those eigenvalues. Furthermore, for the two representations to be non-isomorphic, these eigenvalues have to depend in some way on the element (n,m) itself : if r(n,m) = diag(e^n, e^m) (which is a valid representation), then I want r'(n,m) to be the same map but written in a different basis, and importantly have that this basis depends on (n,m) (otherwise r(n,m) and r'(n,m) would be conjugate via the same matrix for all (n,m) and thus we would have in isomorphism).
I can't find such an r'. Any tips?

[crystal vale]

I want to find a number of ideals in Z/2024Z so I think there must be a concept of corresponding theorem but how can I find it?

[delicate orchid]

which ideals of Z are contained in 2024Z

sorry, other way round

which ideals of Z contain 2024Z

[winter shore]

I wrote:
Note that the map that takes groups to their endomorphism ring is not a functor, since there is no ring homomorphism between $\text{End}{\textbf{Ab}}(\mathbb Z/p_1) \cong \mathbb Z / p_1$ and $\text{End}{\textbf{Ab}}(\mathbb Z/p_2) \cong \mathbb Z / p_2$ for $p_1$ and $p_2$ primes.

[cloud walrus]

Shiranai

[winter shore]

quite some time ago, now I'm not sure how to understand this, other than the trivial map there are no group morphisms from Z/p1 to Z/p2, so what I wrote is wrong?

[rocky cloak]

Here's one.

Let V(x) be the representation defined by r(1, 0) = [1, 1; 0, 1] and r(0,1) = [1, x; 0, 1].

If x is irrational, then r(m, n) = [1, m+nx; 0, 1] is conjugate to [1, 1; 0, 1] whenever m or n are nonzero, but the representations are not isomorphic for different choices of x.

[rocky cloak]

"other than the trivial map" is sort of a big asterix here

Just because we call it the trivial map doesn't mean it's not a map

But if you want an example without the trivial map you can think about the inclusion Z/2 -> Z/4

[winter shore]

oh right, idk why I thought the trivial map was a ring homomorphism, but it doesn't preserve the identity

thanks

[crystal vale]

Divisor of 2024, right?

And if I am right then there are only 3 prime ideal in Z/2024Z

[delicate orchid]

yeah so 2^nZ for 0 \leq n \leq 10

[delicate orchid]

I think the only prime ideal is (2)

for any other ideal (2^k) you can just take 2^{k-1}*2 and that disproves that it's prime

[rocky cloak]

(23) and (11) should also be prime I guess

[crystal vale]

Yes

[rocky cloak]

Thinking wew maybe think you said 2048

[delicate orchid]

oh yeah HAHHAAH

sorry I conflated 1024 and 2048 into one number

yes you're right then

[winter shore]

here someone says:

Rings act on abelian groups; groups act on sets.
I know this is true, but is there any high level explanation of why this is true?