groups-rings-fields

[frigid lark]

The e does nothing

[hollow shore]

ugh nope x.x

[frigid lark]

So you want e * 5=5 * e=5 for example

[coral shale]

to put it another way, an identity e satisfies the equation
g * e = e * g = g

for all g in the set

[frigid lark]

But replace 5 with any rational number and the equations still hold

[coral shale]

I mean equation in the same way as pre-algebra

so g is like a 'constant' and you want to 'solve' for e is one way of thinking about looking for an identity

[hollow shore]

ah i see

so the e doesn't matter

[coral shale]

Not the ideal way of thinking about it long term, I will say

[tribal moss]

The e certainly does matter.

[coral shale]

but to get you to understand initially

[hollow shore]

if so, how do u narrow it down?

[coral shale]

In words:
e combined with any member of the set leaves that member unchanged

If that's what u mean

[tribal moss]

Can you think of any numbers, such that when you multiply it by 5, you get 5?

[hollow shore]

1

[tribal moss]

Yes.

[frigid lark]

Can you think of anything else?

[tribal moss]

Perhaps 1 works?

[hollow shore]

1/1

5/5

[coral shale]

1/1 is 1

it was a trick question

[tribal moss]

If you multiply 1 by 42.57, do you get 42.57?

[coral shale]

there arent.

[hollow shore]

yes

[hollow shore]

yes to this

[coral shale]

Im curious - did you also take introduction to proofs before doing your current course

[hollow shore]

it in a diff course

[coral shale]

if not, i would recommend it

[hollow shore]

we dont do induction in this course i believe

[coral shale]

No not induction

Introduction to proofs

[hollow shore]

oh nope

[coral shale]

or something named similarly

Because it's a good pre requisite to have

[hollow shore]

we prove Abelian Groups using the Group Axioms

that about it

[coral shale]

[hollow shore]

so closure, identity, associative, and inverse

thats it

[coral shale]

I feel like you have misunderstood some parts

of what you're learning

[tribal moss]

Not all systems have a separate introduction-to-proofs course -- very often you're expected to figure that out by osmosis while learning substative things in other courses.

[coral shale]

Like this doesn't make sense to me for starters

[tribal moss]

Let's get back to the identity. Do you feel it is resolved what the identity in (Q*, ×) is?

[hollow shore]

is it just 1?

[tribal moss]

Yes.

[coral shale]

One of the first proofs should be proving the identity is unique in all groups

[hollow shore]

thanks

btw

[hollow shore]

what does the x mean

(Q*, x)

[frigid lark]

The binary operation

Multiplication in this case

[tribal moss]

It's not the letter x, but the multiplication sign. It tells you what the group operation of the group we're thinking about is.

[hollow shore]

oh i see ty

[frigid lark]

Can I see that LaTeX code, I've never seen $\mathbb{Q}$ done like that before

[cloud walrus]

AM4DEUS

[coral shale]

its just a macro

in the default texit preamble

\bQ

[frigid lark]

Ah

[hollow shore]

Consider on the set ℚ× of non-zero rational numbers the operation • defined by 𝑝 • 𝑞 = 𝑝𝑞/2 .

for this question

what does the dot mean?

multiply?

[frigid lark]

It's defining a new binary operation

[hollow shore]

Prove that this operation makes (ℚ×, •) into an abelian group.

commutativity is a tick

associativity too

closure yes

identity

[tribal moss]

Wonderful! I was just trying to come up with a good example of a set with two different group operations on it ...

[hollow shore]

what does p dot q means?

[tribal moss]

It is defined to mean pq/2.

[hollow shore]

oh?

so for identity

i need to figure out q

basically

p(2)/2 = 2(p)/2 = p

so identity is equal to 2

[tribal moss]

Wait, how did you check commutativity, associativity and closure without understanding what the operation is?

[hollow shore]

well since p dot q and q dot p

divided by 2

should be the same

commutative is a tick

since order dont matter

[livid willow]

You are taking p = p/2 and q as 2.
p*q should give you back p/2, not p.

[tribal moss]

Yes, I agree. I'm just confused that you needed to ask for the meaning of p·q when you got to identity, when you've already used that definition when checking commutativity

[tribal moss]

No, the /2 came from the definition of the operation.

[hollow shore]

i dont know what the symbols really mean

but i get the hang of it

[livid willow]

oh okay

misread it

[chilly ocean]

Just write the isomorphism with Q* and you get all of these things for free

[tribal moss]

Hush.

[hollow shore]

uh im confused about isomorphism lol

[tribal moss]

Ignore that for now.

[hollow shore]

thats actually part b

anyway

[tribal moss]

Whoops. Sorry, don't ignore it then.

[hollow shore]

abelian first

[frigid lark]

Just say it is trivial and left as an excercise to the reader

[hollow shore]

identity should be 2 right

[frigid lark]

And you get everything for free

[tribal moss]

Yes.

[hollow shore]

how would u find the inverse for this?

waitttt

isnt the inverse just 4?

[frigid lark]

Inverse of what?

[tribal moss]

No, the inverse can never be "just 4" -- you need a separate inverse for each element in the group.

[frigid lark]

The inverse depends on each element

[hollow shore]

inverse of p i mean

[formal ermine]

amadeus I think you should just let tropo help here, you're just confusing OP

[frigid lark]

Yeah

Soz

[tribal moss]

But it will be a good start to find the inverse of a few concrete elements first, to get a feel for what we're doing.

So, let's find the inverse of 5.

[hollow shore]

whats the identity?

2?

or 1?

[tribal moss]

2 is the identity, yes.

[hollow shore]

so that would be 2/5

[tribal moss]

Let's see if that works.

If it's the inverse, then we should have (2/5) • 5 = e, but the identity was 2, so that's the same as (2/5) • 5 = 2.

[hollow shore]

yay

[tribal moss]

Now remember that • is not multiplication here, but instead stands for the strange group operation the exercise defined.

[hollow shore]

yep

i got confused for a sec

but then i noticed

[tribal moss]

So (2/5) • 5 means (2/5)5/2 which is 1, and it should be 2.

[hollow shore]

wait i thought u mean just 5

inverse should instead be 4/5

[tribal moss]

Yes.

[hollow shore]

if you take it as the same example as pq/2

i get the hang of it now! thanks

[tribal moss]

So are you now ready to describe the inverse of an arbitrary p?

[hollow shore]

yes

for any non-zero rational number p, it inverse should be 4/p

[tribal moss]

Exactly.

[hollow shore]

sweet thanks

what about Isomorphism?

[tribal moss]

I haven't seen that part of your exercise.

[hollow shore]

ℚ× is also a group under the usual operation of multiplication ×. Find an isomorphism
of groups 𝜑 ∶ (ℚ×, •) → (ℚ×, ×).

so basically

the group is now defined as a multiplication

operation

[tribal moss]

Hmm yes. There are two different groups in play now. They happen to have the same set of elements, but their group operations are different.

[hollow shore]

wait

so i need to consider them as two diff groups?

[tribal moss]

Yes. A "group" is the combination of an underlying set and a group operations, and (ℚ×, •) and (ℚ×, ×) are two different such combinations.

The task is slightly tricky because there are many different ismorphisms between (ℚ×, •) and (ℚ×, ×), and you're only asked to find one of them. So we cannot hope to find a deductive argument that tells you exactly what φ must be. At some time during the exercise we have to make a jump and say "such-and-such is something simple that might work -- oh, look, it does work, so let's pick that!"

[hollow shore]

hm okay

but how do u narrow down the easiest one?

[next mango]

good place to start is with the homomorphism property

[chilly ocean]

Writing q•p = qp/2 we can simply observe the isomorphism

[coral shale]

Why the heck are you bringing up homo/isomorphism when they are learning the defn of a group

How in any way could this be helpul

[hollow shore]

im confused by what they meant

[chilly ocean]

???

[coral shale]

Well yes, of course, because you aren't there yet. Don't worry about it

[next mango]

the exercise they are doing is finding a isomorphism

[coral shale]

wait what

[tribal moss]

(It turns out there's a single one that's a lot easier to describe than all the others, so "narrowing down" will not actually be our trouble. We just need the courage to try that simple thing, when we don't know yet it has to be something that will work).

That's apparently part (b) of the exercise.

[coral shale]

oh my bad, missed this. Thought we were still at the previous part

[hollow shore]

mind giving me an example

[chilly ocean]

Who should be sullying who

[tribal moss]

A good place to start would be to write down the definition of "isomorphism" from your book. There are a few different (but equivalent) ways to define that, and some of them will actually help us along more than others ...

[next mango]

To add on to that, you might want to compare the operations with each other and see what changes, and then relate back to identities or properties of functions you have seen before

for example, the divide by 2 might be of use in finding the isomorphism

[tribal moss]

I'm not sure the "stare at the group operation" strategy is actually helpful at the level of experience Cephy has. Sure, we who already have seen a dozen examples of this can quickly say "oh, of course it is just ..." but that is only because we have already worked through the thought pattern Cephy is still just about to discover.

[hollow shore]

wait im just writing things down

[tribal moss]

Oh, take your time. I'm just stating that I don't think the approaches of Blitz and Wormscarf will be helpful for you.

[next mango]

Yeah, you're probably right. But i still remember doing abstract algebra and seeing places to use natural logs and exponents to simplify and find isomorphisms. There should be a lot of familiarity once you pass a certain stage.

[tribal moss]

And the shortcut you're hinting at will totally be worth pointing out once they have found their isomorphism.

[chilly ocean]

Wdym approaches. That was a hint

i'm losing my mind because I forgot so much algebra and can't figure out this basic thing; if two polynomials $p(X), q(X) \in F[X]$ have a common zero in some extended field $F / E$, then their greatest common divisor in $F[X]$ is nontrivial / nonconstant

[cloud walrus]

matcool473

[chilly ocean]

I see that if $\alpha$ is their common root in $E$, then they have a common linear factor in $E[X]$; $p(X) = (X - \alpha) h(X)$ and $q(X) = (X - \alpha) k(X)$. But why would this imply that they have common (nonconstant) factor in $F[X]$?

[cloud walrus]

matcool473

[formal ermine]

do you mean E/F?

[chilly ocean]

yes!

[lethal dune]

look at m_\alpha(x)

[chilly ocean]

Wait, is this all it takes; $m_{\alpha}(x)$ divides both $p(X)$ and $q(X)$, because minimal polynomial of $\alpha$ divides all polynomials which vanish at $\alpha$. And the minimal polynomial is non constant by definition of minimal polynomial?

[cloud walrus]

matcool473

[lethal dune]

yes

[chilly ocean]

thanks

[lethal dune]

the very fact that α is a root (common root) says it's algebraic

[chilly ocean]

yes i figured that thanks 🙂

[hollow shore]

are they trying to hint out that i should use log

[tribal moss]

No they're not.

(I'm still waiting for your definition of "isomorphism", by the way, @hollow shore).

[hollow shore]

any bijective function φ that preserves the groups operation, in this case for any non-zero rational numbers p and q

[tribal moss]

Okay, that's somewhat bare-bones. Do you already know that an isomorphism necessarily preserves the identity?

[hollow shore]

im thinking of log base 2

how would you define it?

[tribal moss]

Some definitions take "preserves the identity" and "preserves inverses" to be part of the definition.

Do you already know that an isomorphism necessarily preserves the identity?

[hollow shore]

nope

wait basically I try to do the same as finding the inverse?

[tribal moss]

Do we agree that "perserves the group operation" means that we must have φ(a•b) = φ(a)×φ(b)?

[hollow shore]

yes

[tribal moss]

So in particular if e is the identify for • (which we already know is 2, but bear with me), then φ(a) = φ(a•e) = φ(a)×φ(e).

[coral shale]

(I often find it more helpful to write definitions algebraically/symbolically while stating/rephrasing the 'meaning' aloud in words. You're going to be manipulating those symbols in your proof)

[tribal moss]

This means that φ(e) can only be the identity in (ℚ×, ×), because that's the only way to make φ(a) = φ(a)×φ(e) true.

[hollow shore]

hm yes

wait no carry on

[hollow shore]

thanks for all the helps, let continue u this later today it 2am rn ima sleep lol

[tribal moss]

(Sorry, had to go and do something). When you wake up, consider this:
We now know we must have φ(2)=1. Given just that single point, what is the simplest bijection from ℚ× to ℚ× with that property you can think of? Check whether that happens to be an isomorphism.

("Always return 1" is not a bijection, and "subtract 1 from everything" doesn't work because it doesn't produce a nonzero rational when the input is 1. So those are too simple to satisfy even the conditions so far).

[chilly ocean]

Just write q•p = qp/2 and think what sort of algebraic manipulations you can do to obtain h(q•p) = h(q)h(p) for some function h

[celest furnace]

Wouldn't every power of 2 get mapped to 1? Like phi(4) = phi(2)^2 = 1^2 = 1, so at the very least it's not an injection

[next mango]

I'm pretty sure we are considering the bijection x/2 from Q to Q

[celest furnace]

which is definitely not multiplicative

[next mango]

wdym

it satisfies the isomorphism

[celest furnace]

phi(ab) = ab / 2 which is not equal to phi(a) * phi(b) = a/2 * b/2 = ab/4

[south patrol]

1 is not the unit in this new operation

[celest furnace]

what

[south patrol]

Well like I'm confused what you mean by this lol, like isn't the point that we are trying to construct an iso between Q with one multiplication with Q with another multiplication

But you have just used the "normal" multiplication on Q twice...

[next mango]

that's not how you would use the homomorphism property 😕

[south patrol]

When you say phi(4) = phi(2)^2 and 1^2 = 1

[celest furnace]

ah i didn't scroll up nearly enough

[south patrol]

Ah

lol

[celest furnace]

lol

[south patrol]

Dw then

owo

[barren sierra]

Absolutely no idea how to do (b) or (c)

This is my first time working with group representations so that's probably the issue

Not sure how the permutation matrix compares to the character table I sent

Is (b) just the straight computation? Call the matrix P, permutation π. So P_ij = 1 if π(i) = j and 0 otherwise?

Cause if it is I'm throwing that shit into SageMath unless there's a smarter way to do it

Is (a) really going to be me just listing 20 different elements? (b) just computing 20 traces?

[hot lake]

(b) is straight up looking at the representatives you got in (a) and counting their fixed points

because the trace of a permutation matrix is the number of fixed points the permutation has

it should take a negligible amount of time compared to (a)

[barren sierra]

Oh

I did not know that

Ok yea that makes it much easier

intuitively that fact makes sense

Oh ok I'm dumb c is literally just linear combinations but is there a smart way to do this?

[delicate orchid]

there's an easy way to speed it up for H_1, note how 5 and 6 are fixed by H_1

other than that, inner product rho_i with itself to get some idea of the number and multiplicity of the constituents - if you know about that

you also know the dimension of the direct sum has to be the dim of rho_i

[rotund aurora]

It is probably more natural to prove $U^\otimes V\approx\mathcal L(U,V)$. Things in $U^\otimes V$ are of the form $\sum_i u_i\otimes v_i$ where $u_i\in U^*$. Now, you might think of it as defining a function $U\to V$ by
[
x\mapsto \sum_i u_i(x)\cdot v_i
]

[cloud walrus]

Croqueta

[rotund aurora]

for every x in U

[lapis trail]

Oh good

But does this work?

[rotund aurora]

yeah I guess this also works

what I said is like that

but like the dual counterpart lol

[void cosmos]

yooo

is Z/2Z x Z/2Z a comm ring with identity that has only two maximal ideals? (namely Z/2Z x {0} and the other way around ? )

[lapis trail]

Thanks!

[rotund aurora]

the only other possibility would be the ideal (1,1), but that gives the whole ring

[void cosmos]

yessir

ty

What happens if u localize a ring wrt a unit?

[rotund aurora]

what you mean by wrt to a unit

[void cosmos]

like localize wrt an element

But that eement is a unit

[rotund aurora]

If you mean inverting the unit, everything stays the same

[void cosmos]

nothing happens right?

[rotund aurora]

I dont think so

[void cosmos]

wdym nothing happens

the localization is the same as the ring ?

[delicate orchid]

you can't localise "with respect to a unit" you have to localise with respect to a multiplicatively closed set - if you mean localise with respect to like, {1, u, ..., u^-1} with u a unit then yeah I'm pretty sure it's iso to the original ring

[tribal moss]

Multiply your polynomial by x-1.

[delicate orchid]

le roots of unity have arrived

I'm not on that PACK the roots of this are just the 101st roots of unity (not counting 1 itself) right

[elder wave]

yes the canonical map r->r/1 should be an iso then i think?

it's injective since S does not contain zero divisors

and surjective since every fraction can be rewritten

[delicate orchid]

it's bijective because it's obvious

yeah, that polynomial is a cyclotomic polynomial

[elder wave]

then this should still be the case if you localize at {1,u,u^2....}

[void cosmos]

yooo my brain just blew up , i couldnt find integer solutions to a^2+b^2=2

[delicate orchid]

since 101 is prime

[elder wave]

which is usually what localizing at an element refers to

[tribal moss]

103 is invertible modulo 101, so the nontrivial roots of x^101=1 each raised to the 103th power are exactly the nontrivial roots of x^103=1, just in a different order.

[delicate orchid]

yeah I was gonna say that x_k -> x_k^2 just permutes them cause coprime

[tribal moss]

And those roots are xi, xi^2, xi^3, ..., xi^100 where xi is a fixed primitive root. And you know already what the sum of those is.

[delicate orchid]

ah, my apologies - didn't know this terminology

[rotund aurora]

The point of localization is inverting elements. If you invert an element that was already invertible, you obtain nothing new. Because if u is a unit and you "add" an inverse x, then you have ux=1, but multiplying by u^{-1} yields x=u^{-1}

This is similar to the situation $\R[x]/(x^2-2)\approx \R$. Tho this situation is not exactly the same because x is either sqrt 2 or -sqrt 2, but you cannot tell which

[cloud walrus]

Croqueta

[tribal moss]

However, whereas R[x]/<x²-2> is not exactly isomorphic to R, localizing at a unit does produce something isomorphic to the original ring.

(Due to how equality in the localization is defined).

[rotund aurora]

okay

R[x]/<x^2-2>=R is not true I think

because thats not the kernel of the map x--> sqrt 2

R[x]/(x^2-2) is not even a domain

[tribal moss]

Indeed, since (x-sqrt2)(x+sqrt2) = 0.

[rotund aurora]

I guess, in a ring, you can have a polynomial with as many distinct roots as you want

but in rings, inverses are unique. So my comparison was not accurate at all lol

[void cosmos]

i was to show

that

if a ring has the property that any element is either nilpotent or a unit

then it has a unique prime ideal

and ig localization didnt work for me

commutative

[tribal moss]

The statement ought to immediately suggest a candidate for what the unique prime ideal is ...

[tribal moss]

You can save the comparison by saying something like R[x]/<2x-5> instead.

[rotund aurora]

yeah

[void cosmos]

R right?

the whole ring?

[barren sierra]

Don't know what constituents are.

[rotund aurora]

??

[void cosmos]

or ig 0

[elder wave]

0 isn’t prime if you have zero divisors

[rotund aurora]

truu

[void cosmos]

yea let me check if its assumed

[tribal moss]

It's an explicit condition in the definition of "prime ideal" that the whole ring doesn't count as a prime ideal.

[formal ermine]

@void cosmos do you know the proof that the nilradical is the intersection of all prime ideals?

[void cosmos]

ye

oh

its the nilradical lmfao

ig

[tribal moss]

It's clear from the assumption that the nilradical is a maximal ideal and therefore prime.

[formal ermine]

the trivial part of that proof is required for this exercise

[void cosmos]

suppose a is not a unit --> a is nilpotent --> R_a is zero

[south patrol]

Localisations are p unnecessary

Idk why you are using them hm

[rotund aurora]

the ring is as localized as possible

[south patrol]

indeed

[void cosmos]

yea its the nilradical

[south patrol]

"it"

oh okay ye the unique one

Well whenever you have a local ring, the maximal ideal is always gonna be the set of non-units

[void cosmos]

yea sorry boys and girls p tired today

no brain

ty

all

[south patrol]

and then here you can just show that that is an ideal directly

por ejemplo

[formal ermine]

espanol

[south patrol]

which is clear here as tropo says, since the set of non-units just is the nilradical

[formal ermine]

si si

[south patrol]

je ne parles pas espagnole

alors, seulement un peu

[formal ermine]

mon francais es terriblé

[south patrol]

lol

[void cosmos]

yo

nvm

[south patrol]

lol

[formal ermine]

lol

[void cosmos]

yo 3 is prime in Z[i] 100%

right

[rotund aurora]

la nuit noire

[void cosmos]

2 isnt tho

[south patrol]

try it

[void cosmos]

yea it is irreduicible

and i can prove irreds are primes in UFDS

cool col

ty

[rotund aurora]

in Z[i] the primes congruent to 3 mod 4 remain irreducible, the ones congruent to 1 mod 4 no

this is not trivial tho

[void cosmos]

i think this comes from when a prime square is represented as the sum of squares

my guess

[rotund aurora]

yee

[void cosmos]

yea

using a norm argument

[formal ermine]

what's localization in a nutshell

[void cosmos]

inverting

adding fractions

but i think the p cool stuff about it is the correspondonce between prime ideals

[tribal moss]

More elementarily, it's easiy to list all of the elements of Z[i] with norm < that of 3 and show that none of them divide 3.

[delicate orchid]

add they fractions

[south patrol]

hm what's the easiest way to see this

i mean

one direction is basically two squares theorem

eh okay and the other is easier

[void cosmos]

okay question is 3 prime in R={m/2^n | m in Z , n in N}?

[delicate orchid]

"just prove the 2 squares theorem dawg"

[void cosmos]

thats the localization wrt to 2 right

[rotund aurora]

yes

[void cosmos]

and 3 is prime in Z

[south patrol]

ye i suppose the other is easier as you'd have p = a^2 + b^2 which is impossible mod 4 if p is 3 mod 4

[void cosmos]

so 3/1 is prime in this

so its true?

[south patrol]

lol

Okay 2 squares uhhh

[rotund aurora]

there are standard facts with localization

[south patrol]

Use minkowski's convex body thoerem

lol

😎

[void cosmos]

3 is not a power of 2

so 3 must be prime in this

is this true boys and girls

[rotund aurora]

or

[void cosmos]

3/1

[rotund aurora]

notice that the quadratic form x^2+y^2 has discriminant -4

[south patrol]

why don't you just try to provide a proof

[rotund aurora]

show that the class number is 1

[south patrol]

oh nice

[delicate orchid]

do I look like I know what a class number is

[south patrol]

hm have you not taken alg nt

i mean fair enuff if not

[void cosmos]

thats it , i just used the correspondonce between primes here and primes that do not contain S theere

[rotund aurora]

and use that n will be represented by a form of discriminant -4 iff -4 is a square mod 4n

[south patrol]

ye sure

noice

[delicate orchid]

all I want is a god damn DIMENSION FUNCTOR FROM THE DOUBLE BURNSIDE RING

[rotund aurora]

fancy word

in Gauss language you have like quadratic forms

and you define some change of coordinates

[south patrol]

But also like more easily to check irreducibility you've just gotta do like

if 3 = (a/2^n)(b/2^m) where a,b are odd

[rotund aurora]

and look at "reduced" quadratic forms (so you pick a simple quadratic form up to permissable changes of variables)

[south patrol]

then 3* 2^{n+m} = ab

n=m=0

3=ab

[rotund aurora]

there may be several non-equivalent reduced forms of the same discriminant, but with discriminant -4 this doesn't happen, and the argument is simple inequalities

[south patrol]

and so ye

[delicate orchid]

cool!

[south patrol]

okay where does this come from croqueta

i've not seen this proof of two squares aha

inch resting

[rotund aurora]

Look at Niven chapter 3

I didnt explain it propertly tho, cuz tired

[delicate orchid]

that's the guy who did the trig thing

[rotund aurora]

yes

[delicate orchid]

no wayyyy

[rotund aurora]

you mean proving pi irrational?

[south patrol]

wdym the trig thing

oh

using integrals?

that cool one

[delicate orchid]

no the other one, the rational values one

it's almost definitely a different niven

oh maybe not actually

[south patrol]

bruh

Oh mathologer made a video on this

nice

[rotund aurora]

Consider quadratic forms $f(x,y)=ax^2+bxy+cy^2$. The discriminant is defined to be $\Delta\colon= b^2-4ac$. The group of $SL_2(\bZ)$ acts on the set of quadratic forms by $f(x,y)\mapsto f(ax+by, cx+dy)$, for each matrix
[
\begin{bmatrix}
a&b\ c&d
\end{bmatrix}
]
in $SL_2(\bZ)$.

Two quadratic forms $f$ and $g$ are equivalent if there is a matrix $M\in SL_2(\bZ)$ such that $fM=gM$ (where the action is as above). This is of course an equivalence relation. Now, you have: If $f$ and $g$ are equivalent, then an integer $n$ is represented by $f$ if and only if it is represented by $g$. Moreover, the discriminants of $f$ and $g$ are equal.

Let $f$ be a quadratic form whose discriminatn $\Delta$ is not a perfect square. We call $f$ reduced if
[
-|a|<b\leq |a|,,\text{or if},, 0\leq b\leq |a|=|c|.
]

A main result is: Let $\Delta$ be an integer, which is not a perfect square. Each equivalence class of quadratic forms of discriminant $\Delta$ contains at least one reduced form. Furthermore, the number of reduced forms of a given nonsquare discriminant is finite. This number is called the class number of $\Delta$, and you mayd enote it by $H(\Delta)$.

[cloud walrus]

Croqueta

[rotund aurora]

okay took some time because I was preparing dinner lol

@south patrol

And also, you have this main result: Let $n$ and $\Delta$ be given integers with $n\neq 0$. There exists "a" binary quadratic form of discriminant $\Delta$ that represents $n$ properly if and only if the congruence $x^2\equiv \Delta$ modulo $4|n|$ has a solution

[cloud walrus]

Croqueta

[rotund aurora]

(represent properly means f(x,y)=n with gcd(x,y)=1)

Now, if you can show that the class number H(-4) is 1, the theorem of Fermat follows

And this class number agrees with the modern class number, etc

[formal ermine]

one sentence proof more like four sentence proof

[rotund aurora]

This is Gauss theory, which deals with the special case of quadratics (but with a different flavour) and follows from usual algebraic NT

so the key here, is to consider not just one equation (namely, x^2+y^2), but many

[south patrol]

Thankss

Wait how does this class number relate to the normal class number

or does it not at all ig lol

[rotund aurora]

nono It does

but I dont really know how either

The -4 discriminant is related to Z[i] ofc

And Z[i] is a PID, ie, has usual class number 1

So it makes sense

[south patrol]

yee noice hm

[rotund aurora]

This is something I have yet to study more

Class number 1 says there is just one equivalence class of quadratic forms

[south patrol]

Tbh the proof using the fact Z[i] is a UFD feels the most like explanatory (since it immediately shows uniqueness too) but then i think that's just because we are using more known facts idk

Like the Zagier proof is probably the only nicely self-contained proof I suppose

But not v intuitive lol

[formal ermine]

what is class number in ant

[south patrol]

Hm okay might be hard to give a short explanation if you've not seen the relevant theory but the point is that one associates to a ring of integers a finite abelian group known as the "ideal class group" and the size of that group is known as the class number

[opal osprey]

What are some of the applications of cohomology theories of algebras (e.g Lie Algebra Cohomology, Hochschild Cohomology, Cyclic Homology and etc...) to classical problems about these (for example classification theorems)?

[south patrol]

You can view the class number heuristically as measuring how far the number ring is from being a UFD or PID, since the class number is one iff the ring is a UFD or PID

[opal osprey]

Maybe a better question would be to ask what they are good for.

[sonic coral]

Can i get some help showing if H cap K is normal in both H and K, then HK is contained in the normalizer of H cap K in G

[rose fog]

everyone wish me luck my first modern algebra midterm is thursday and this is my last math requirement for my degree and I have a job lined up for after I graduate I need to pass 🙏🏻

[rustic crown]

good luck uwu

[frigid lark]

This is just applying definitions. To show that $HK\subset N(H\capK)$ we need to show that for all x in $H \cap K $, all $hk\in HK$, $xhk=hkx$. Can you use the fact that $H\cap K \unlhd H,K$ to show this?

[rustic crown]

,av am4deus

[cloud walrus]

AM4DEUS#8825's Avatar

Click here to view the image.

[rustic crown]

[frigid lark]

Aw

Thx

[frigid lark]

Good luck

[sonic coral]

HK is contained in N_G(H n K), so for all x in H n K, all hk in HK, xhk = hkx?

[cloud walrus]

AM4DEUS
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lapis trail]

Is it correct to call the vectors in the tensor space u_1×v_1, u_2×v_2,.....

What about u_i×v_j for i not equal j? Is it an abuse of notation to put u_i×v_i?

[frigid lark]

Well that is what you want to show

They are equivalent statements

[opal osprey]

I think they meant u_i tensor v_j actually

[lapis trail]

That's what I thought

Unless the "diagonal" tensors are the only one's linearly independent

[sonic coral]

right. and since H n K is normal in H, g(H n K)g' = H n K for all g in G right?

[frigid lark]

If you show this xhk=hkx then you show that HK is contained in the normaliser

[opal osprey]

No, the set of all elements of the form u_i tensor v_j is linearly independent

for example

if V and W is finite dimensional

and the {u_i} form a basis for V and {v_j} for W

[frigid lark]

No

It's normal in H, not in G

[opal osprey]

Then their tensor product has dimension the product of their respective dimensions

[sonic coral]

yea okay

[opal osprey]

this corresponds to the fact that {u_i tensor v_j} will be a basis for their tensor product

[sonic coral]

i just dont know how to apply the definitions to help me

[opal osprey]

and this set has cardinality dim(V)dim(W)

[frigid lark]

Say H is normal in G, then for all g in G, h in H, gh=hg

[sonic coral]

yes

[frigid lark]

That's another property of normality, which might be more helpful

[sonic coral]

so for all x in H n K and all h in H, xh=hx, same thing for normal in K, xk=kx

[frigid lark]

Yeah, can you use this to solve your question now?

Well to show hkx=xhk

[sonic coral]

probably an easier way but like this?

feel like i shouldnt have to do the last multiplication on the right by x but maybe so

[frigid lark]

You could just say that hkx=hxk=xhk

Snake lemma or butterfly lemma?

Which diagram is prettier

[sonic coral]

there’s the easier way

[rustic crown]

i think you mean hkx = hx'k = x''hk

[rustic crown]

whut nu, gh = h'g

gH = Hg translates to that, not commutativity on the nose

[sonic coral]

yes normal means it may be a different element that it holds for right

[rustic crown]

i think a nice way to do this is by using that N_G(H) is the largest subgroup of G which contains H as a normal subgroup

so N_G(H n K) contains both H and K

and hence contains HK

it also contains HKHKHKHKHK and stuff because the subgroup N_G(H n K) is closed under multiplication

(HK apriori is just a set, it may not be closed under the group operation, if one of H or K is normal in G, then you can say HK is a subgroup)

[rustic crown]

but if you like elements, then this is also nice :p

[sonic coral]

i prefer elements for the proofs but having an intuition like you suggested helps see the big picture

[analog zephyr]

Guys

[frigid lark]

Ahh 🤦

Sorry

Another way would be if X = H \cap K, then hkX=hXk=Xhk for all hk in HK

[sonic coral]

i got points off for that on my test

he told me sets don’t commute

🥲

[quartz quiver]

is this (12.7) true? I searched online and found a similar question in several places, but they all have the additional assumption that G is a subgroup of S_p

[quartz quiver]

<@&286206848099549185>

[quiet pelican]

It is true
What’s the proof you’ve seen with the additional assumption?

[quartz quiver]

https://math.stackexchange.com/questions/1450285/show-that-if-p-is-a-prime-number-and-g-is-a-transitive-subgroup-of-s-p-t

[quiet pelican]

Yes
You use the action to “transport” your element of order p to an element of order p in S_p

[quartz quiver]

So there exists a surjective homomorphism from G to S_p, then for some element in S_p that satisfies this property, an element in the preimage has the property?

[quiet pelican]

Yes

[quartz quiver]

But how do we know that the homomorphism from G to S_p is surjective?

[quiet pelican]

It doesn’t need to be

It just needs to map to an element of order p

[quartz quiver]

How do we know it maps to an element of order p?

[quiet pelican]

Because the stabiliser of a given element can’t contain every element of order p (I can justify this with sylow formally, but idt you’ll know that yet)

[quartz quiver]

i do know sylow

[quiet pelican]

Interesting…
Ok, then you just consider that sylow guarantees that there is a subgroup of p element of size p^n, maximal dividing the order of the group
The intersection of the stabiliser and this group has size at most p^{n-1}
So there is an element of order p outside the stabiliser

Or, of order a power of p, but that must also be mapped to a p cycle if it’s not trivial

[quartz quiver]

sorry, could you elaborate on this a bit?

[quiet pelican]

The order of the thing it’s mapped to must divide it’s order
So as it’s order must be p^a, the order of its image must be a power of p, and as it doesn’t fix a particular element, it can’t be 1, and because of the size of the set, it can’t be larger than p, so it is p

[quartz quiver]

oh, thanks!

[fervent rock]

no way

[white oxide]

To prove that the equivalence relation is symmetric, is this line of logic sound? Suppose $H \sim K$. Then $i_g[H] = K$ for some $g \in G$. Expanding, we obtain $gHg^{-1} = K$. But then, replacing $g$ by $g^{-1}$, we obtain $g^{-1}Hg = K$, so $H = gKg^{-1}$ and $K \sim H$.

or would this not hold, because the inner automorphism is only of g

I suppose we could let $i_g^{-1}[K] = H$, but that doesn't follow from $H \sim K$

[cloud walrus]

okeyokay

okeyokay

[white oxide]

Never mind I figured it out lol

[white oxide]

Would this proof work for the intersection of 2 normal subgroups? Let $H, K$ be normal subgroups. Fix $r \in H \cap K$. Since $H$ and $K$ are normal subgroups, $grg^{-1} \in H$ for all $g \in G$ and similarly $grg^{-1} \in K$ for all $g \in G$. Then $grg^{-1} \in H \cap K$ for all $g \in G$ and $r \in H \cap K$, so $H \cap K$ is a normal subgroup.

[cloud walrus]

okeyokay

$Get \LaTeX 'd$

[tender wharf]

not necessarily

let g stand for a\inv b

then ag b\inv is not necessarily gab\inv

unless it's abelian

[tender wharf]

this is fine

I'm assuming by fix you meant let r be an arbitrary element of H cap K

[white oxide]

ye

[lethal dune]

yes

there aren't many polys to check

quadratics it's enough to show there's no root in F_7

[chilly ocean]

How would you prove that if C is a subset of Z and (C,<) is a well ordered set then its intersection with Z- is finite

[lethal dune]

plug in

[lethal dune]

C could be N?

you mean Z minus?

ah didn't see the - there

[chilly ocean]

yes

[lethal dune]

plug in elements of F_7 in the polynomial

[chilly ocean]

yes

[lethal dune]

what's the least element of Z_ ∩ C then?

yes, for quadratics at least

[chilly ocean]

what do you mean ?

[lethal dune]

C is well ordered and Z_ ∩ C has a least element

[tawdry crystal]

how does a Lie algebra automorphism preserves the Lie bracket?

Let $T:V \to V$, does it mean $T([V, V] = [T(V), T(V)]$?

[cloud walrus]

Minh Pham

[upper pivot]

T[V,W] = [TV,TW]

[tawdry crystal]

thank you

[delicate bloom]

I think it's easier to count the number of monic reducible quadratics, then subtract that from the number of monic quadratics

think about it

[delicate bloom]

well, for starters can you give me an example of a reducible monic quadratic

yeah, well I suppose in your case you want it to be in F_7[x], but that works

so how do you know that one's reducible

that would be irreducible

ok good, and what's the other root

so you could rewrite it factored as (x-2)(x-4)

in general you could write every reducible quadratic this way right?

ok, cool, so how many ways can we do that?

I'll show you how I'd do it for F_2

(x-0)(x-0)
(x-1)(x-1)
(x-0)(x-1)
so that's all 3 possibilities

in total x^2+ax+b there are 2 choices for a and 2 choices for b, so there are 4 monic quadratics

what I'm saying is generalizable to every finite field

there are only finitely many choices for choosing the roots of your polynomials

and there are only finitely many polynomials

I'd be wasting my time if I just told you the answer

work out the case for F_3[x], how many monic quadratics are there? how many reducible monic quadratics are there?

maybe that'll help you see how you can work out the case F_q[x] in terms of q

[rustic crown]

Ferb

okie me had a question

we define an inclusion of rings R --> S is integral if every s in S satisfies a monic polynomial over R.

is there any advantage/disadvantage/reason to work with inclusions and not any ring map? we can say R --> S is integral if S is integral over the inclusion of the image.

[oblique river]

who is "we"

i've seen people define "integral" for arbitrary ring maps

I guess wikipedia defines it that way

(requires inclusion)

but stacks project doesn't require it

[rustic crown]

ohh

interesting

[rustic crown]

the few introductory sources i saw >.<

[oblique river]

i guess there is some geometric reason to require it

if R --> S is integral (and an inclusion) then Spec(S) --> Spec(R) is surjective

[rustic crown]

ah okie

[south patrol]

Lying over property innit

[elder wave]

me doing taxes

[south patrol]

Fr

[lethal dune]

infact I've heard ppl mention it's a covering (ramified covering?) but haven't seen a proof yet

[feral agate]

for (d), I don't really understand how I can approach this problem. Can someone give me a hint?

I could only come as far as: We have that for solutions $x,y \in \Z$, it must hold that $(x+y\sqrt{3})$ is the multiplicative inverse of $(x-y\sqrt{3})$.

[cloud walrus]

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

[lethal dune]

I'm not sure if you are allowed to used this but all solution of the equation will look like
$\frac{x+\sqrt{3}y}{x-\sqrt{3}y}$

[lethal dune]

but for your purpose, it's enough to show numbers of this form are solutions

@feral agate

[feral agate]

you mean will look like $\frac{x+\sqrt{3}y}{x-\sqrt{3}y}=1$?

[cloud walrus]

FrankF

[brave lantern]

Just go about it like you would with a complex number

Thats the way i see it

Also mayne its more fittes to undergraduates channels

[feral agate]

Can you elaborate?

[brave lantern]

Im trying to not give it away but

How do you get the inverse of 1/3-2i

Do the same thing

Will work the same

[feral agate]

But that is just $\frac{1}{1/3-2i}$ and it does not give a more specific solution than what I already have which is

[cloud walrus]

FrankF

[chilly ocean]

I asked my students to compute the number of necklaces one can make with 3 colors having 6,4,2 marbles in each class.
The book answer considers the necklace as a line and computes 12!/(6!4!2!) necklaces. So far, so easy.
But the smart ones started asking questions about accounting for the cyclic symmetries on the possible necklaces... How would one go about that? I was thinking in the direction of Burnside's lemma, but how was that applied again?

5,4,3 would be easy enough, since 5 and 12 are co-prime and there are no cyclic symmetries and I only have to consider reflections, yielding [12!/(5!4!3!)] / [5!/2!²]. but the cyclic symmetries you get for 6,4,2 make this kinda hard.

[lethal dune]

that's not what I meant

N(x+√3y)=x²-3y² and what I said is if
s=(x+√3y)/(x-√3y) then N(s)=1

now write s in the form a+√3b

[feral agate]

what is N?

[lethal dune]

N is "norm" function but it's not important

I have already given the definition of N

although this will give you solution in rationals

[feral agate]

This $\frac{x^2+3y^2}{(x^2-3y^2)} +\frac{xy\sqrt{3}}{(x^2-3y^2)}$?

[cloud walrus]

FrankF

[lethal dune]

yes

[feral agate]

This is the same as before except in the denominator you don't have complex numbers anymore.

[lethal dune]

that's not =1

N of that=1

well I think it's circular now since you are looking for integer solutions

we need denominator =1 which is the problem to begin with

so try this now

x²-3y²=(x-√3y)(x+√3y)=1
Now square both sides, you get
(x-√3y)²(x+√3y)²=1
Infact any power
(x-√3y)ⁿ(x+√3y)ⁿ=1
will give you a solution as we already have 1 solution given

just show all of them are distinct and you're done

[feral agate]

Where does this fit in though?

[feral agate]

So we have a system of equations, using two of them would already be enough to guarantee unique solution for x and y right?

[lethal dune]

it doesn't, it gives you general form of the solution in Q[√3] but since we are interested in integer solutions, it's useless

[lethal dune]

well didn't understand what you wanted to say

what I said is given we already found one solution, we can generate infinitely many by taking powers

[feral agate]

I meant that there are two unknowns and two equations involving these are needed to find a unique solution for the two unknowns

[lethal dune]

there's no unique solution

the point of the problem is to show there's infinitely many solutions

[feral agate]

O I see

Yep thanks

[lethal dune]

[feral agate]

Btw, this is not true right? Because you need that $(x+y\sqrt{3})^n$ and $(x-y\sqrt{3})^n$ may be written in the form $a + b\sqrt{3}$ and $a - b\sqrt{3}$ with $a, b \in Z$, respectively for a and b to be solutions generated from x and y. This is not necessarily true, at least not for n = 2

[cloud walrus]

FrankF

[lethal dune]

why not?

(x+√3y)²=(x²+3y²)+2xy√3

just binomial expansion

[brave lantern]

Im not sure i understand, does he mean ro count in a way where 123 and 312 are the same (cyclic permutations) ? Because that wouldnt be too hard

(Just dividing by the length of the necklace should do it)

[white yoke]

If F is a field, then is an F-module just a vector space over F?

[void cosmos]

y

[white yoke]

Yes?

[void cosmos]

yeaa

[white yoke]

Thanks

[chilly ocean]

let's try 2,1,1.
There's 4!/(2!1!1!) = 12 linear necklaces: ABCC, ACBC, ACCB, BACC, BCCA, BCAC, CABC, CBAC, CBCA, CCBA, and CCAB.

But BCCA is the same as CCAB under rotation. And ACCB = BCCA under reflection.

How many are there accounting for rotations and reflections? How many are there for 6,4,2 in each color class?

[white oxide]

i'm having trouble seeing why this sentence is true; any normal subgroup of G/M is of the form {a_1M, a_2M . . . a_nM} and its inverse image if {a_1, a_2, . . . a_n}. Why does this normal subgroup have to contain M?

[south patrol]

M is sent to the identity of G/M under γ and every subgroup of G/M contains the identity

[south patrol]

Anyway, I feel obligated to point out that this theorem is only a special case of smth more general

Subgroups of G/M correspond exactly to subgroups of G containing M

(This is the correspondence theorem)

Furthermore, that correspondence sends normal subgroups (of G/M) to normal subgroups of G (and vice versa)

[brave lantern]

Well that becomes harder.
You would need to make classes of lists that are equal after a circular permutation, hhat may help, but maybe its searching too far

[south patrol]

And from this, the theorem is immediate !

[south patrol]

Also, no, what you said is not true

The preimage won't be the same size as the image unless the map is injective (at least for finite groups lol)

(Also you seem to be assuming the groups are finite there)

And also - unless I'm misinterpreting smth - the theorem should say maximal proper normal subgroup right

[chilly ocean]

But do I? Is there no elegant way to count?
That's my question, I guess.

[white oxide]

wait sorry how does that show that G/M is simple

[white oxide]

oh right i keep on making those assumptions

[white oxide]

ohh ok i'll digest that

[brave lantern]

Can you do a reflection then a cyclic permutation ?

Like any number of times?

[chilly ocean]

sure. anything you can do with a necklace. you can turn it and you can flip it around.

[white oxide]

nvm ignore that

[brave lantern]

Sorry i cant come up with a formula.

[torn warren]

Claim: $N$ is a normal subgroup of $G$, pick any element $g\in G$, if $k$ is the smallest positive integer such that $g^k\in N$, then $k$ is the order of the element $gN$ in $\frac{G}{N}$.

[cloud walrus]

Witness

[south patrol]

The proof would be ||M is a maximal proper normal subgroup of G <=> M is the only proper normal subgroup of G containing M <=> There's only one normal proper subgroup of G/M||

[brave lantern]

That sounds weirdly said

[south patrol]

That seems fine to me

And is p much immediate from definitions

[brave lantern]

Like k^n will always be in g because as a subgroup its stable by ×

Unless i missread

[south patrol]

k is smallest positive integer such that g^k is in N

[brave lantern]

Ooooh mb. Got confused by the way its formulated because ive seen it said differently

[frigid lark]

I guess another way of phrasing what potato said would be if we have a chain $M\subset N \unlhd G$ where $M\neq N$, then $N = G$

[cloud walrus]

Parrot Tea

[torn warren]

no

[frigid lark]

Why not?

[south patrol]

Hm that's not quite the whole thing

But ye

I was just saying that the theorem falls out of a more general correspondence

[torn warren]

$G=Z_4, N={ 0,2}, M={0}$

[cloud walrus]

Witness

[white oxide]

both of these underlined in blue are in C, correct?

[frigid lark]

I may have misinterpreted where the confusion is

[frigid lark]

M isn't a maximum subgroup of Z/4Z

[torn warren]

you didn't say M is max

[south patrol]

It was said so originally lol

[south patrol]

In the statement of the theorem

[torn warren]

you didn't talk about my claim?

[south patrol]

Parrot ws talking about the other problem being discussed here

But then I did say smth about your one Witness

[south patrol]

Yeah if you think about what the order of gN in G/N is then this falls out fairly quickly

[torn warren]

got it, I thought he is talking about mine..

[torn warren]

Yes, I have proved this, so I post to check if this is right one, since it is not in the book, it is when I do some computations and find this feature

[south patrol]

wdym "If this is right one"

Oh like you worked this out yourself and are checking it's true?

[torn warren]

right

[frigid lark]

Yes

[torn warren]

so I wrote "Claim" in the very beginning, maybe some of claims are false, since they are not in book

how to change the avatar?

[white yoke]

When R is a non-commutative ring, is an R-module defined as being both a left R-module and a right R-module?

[torn warren]

pingu

[white yoke]

Noot noot

[south patrol]

Usually one gives a convention

Like modules being left (or sometimes right) modules by default

[frigid lark]

You can "go between" left and right modules with the opposite ring, so when R is commutative R^{op} = R, so the left and right R modules are "the same".

so there is no need to specify whether something is a left or right R module

But for a non commutative case, as potato said, a convention will be established

[white oxide]

Can somebody give me a hint?

[prisma ibex]

What does it mean for a subgroup not to be normal?

[white oxide]

ok i think i got it

i used the contradiction suggested by ur hint

thanks!

[sonic coral]

I don’t understand how to find the elements in D8/Z(D8)

i know that Z(D8) = <a^2>

and that D8/Z(D8) = { g in D8 | gZ(D8) }

i just don’t know what to look for. like can’t any element a in D8 be in it?

[tribal moss]

Does D8 in your notation have 8 elements, or 16?

[sonic coral]

8

i have the lattice map for it too

[tribal moss]

And "a" is a specific generator of it, of order 4?

[sonic coral]

isn’t a an arbitrary element?

let’s just call H = <a^2>

[tribal moss]

If a is an arbitrary element, then <a²> can be different subgroups depending on which arbitrary element a is.

[sonic coral]

oh no a is the rotations

sorry

i should’ve been more careful with that

[tribal moss]

You could write Z(D8) = { a² | a in D8 } and be right, but that's not a possible meaning of the notation <a²>.

[sonic coral]

i think this is what i meant

a is a rotation, the center is <a^2> = {e, a^2}

[tribal moss]

THat's backwards, though -- you should write { g Z(D8) | g in D8 }.

The thing before the | in a set builder is always the general element of the set you're building, and the stuff after the | are conditions.

[sonic coral]

even so can’t g be anything

like gH is what exactly?

[tribal moss]

gH ought to be defined to { gh | h in H }, that is the set of everything you can get by multiplying the fixed element g by something from H.

In any case, you can now start writing down cosets explicitly.
Z = {e, a²} itself is one of the cosets.
aZ = {a, a³} is another.
bZ = (b, ba²} is a third.

a²Z = {a², e} turns out to be the same as {e, a²}, so that is not a new coset.

[sonic coral]

abZ = {ab, a^3b}

[tribal moss]

Yes.

[sonic coral]

so these are the subgroups of order 2

[tribal moss]

No, most of them are not subgroups. They don't contain the identity.

[sonic coral]

oh yeah duh

[tribal moss]

They are the elements of the quotient group.

[sonic coral]

so then G/H = {aZ, bZ, abZ, whatever the last one is}

it’s like a set of sets?

if you wanted to write it all out

[tribal moss]

Yes.

[sonic coral]

that was my issue

[tribal moss]

Not "like" -- it is a set of sets.

[sonic coral]

he told me that D8/H would be a subgroup of order 4 so i think i was getting confused with that

i was looking at the 3 subgroups of order 4 and couldn’t figure out what condition i needed to meet

[tribal moss]

That is not quite true -- it's not a subgroup. One of the subgroups of D8 happens to be isomorphic to it, but that's just by accident.

[sonic coral]

okay i’ll have to ask him about that

it seemed like to me it could be any of the subgroups of order 4 if thay was the case

but it being a set of sets makes it make more sense since some elements will give the same set

and it won’t just be all the elements in G

[tribal moss]

More precisely, the subgroup {e,a²,b,a²b} is isomorphic to your quotient, and so is the subgroup {e,a²,ab,a³b}, but the subgroup {e,a,a²,a³} isn't.

[sonic coral]

and surely you could get the same quotient group by multiplying on the right, right?

[tribal moss]

Yes, in fact you have to. That's why the subgroup you're quotienting by needs to be normal.

[sonic coral]

ah yes bc you need the mod whatever group to be normal in G

[torn warren]

what contradiction do you get? I will use automorphism to prove it.

[tender wharf]

yeah ||inner automorphism|| does it

[white oxide]

yeah i used inner automorphism

can i have a hint for this question?

idk if this is useless or not but i'm starting to consider $\phi(ah)$ where $a \notin H$ and $h \in H$

[cloud walrus]

okeyokay

[torn warren]

no

this is from factor group to factor group

[white oxide]

yes i understand

[torn warren]

what you did is from coset

[white oxide]

but it says that $\phi$ induces the natural homomorphism

[cloud walrus]

okeyokay

[white oxide]

so i'm starting from there

phi has inputs in G

maybe i don't entirely understand what they mean by "induce"

[torn warren]

you need to show phi*

not phi

just use the definition

[white oxide]

i understand

i need to define phi* in terms of phi then

my guess is that we define $\phi^$ as $\phi^(aH) = a'H'$

[cloud walrus]

okeyokay

[white oxide]

or maybe $\phi^*(aH) = \phi(a)H'$

[cloud walrus]

okeyokay

[tender wharf]

[torn warren]

[white oxide]

lol that it was it

good men abound in this server

does gallian cover modules?

[gentle marsh]

no?

i didn’t see any

[sonic coral]

a spotlight search of it in my book returns 0 results so

[white oxide]

dummit and foote is considered more terse than gallian correct

ok

[tender wharf]

gallian has no modules

[frigid lark]

Lang chapter 3

If Lang is for you

[white oxide]

Lang looks brutal

[frigid lark]

I'm finding Lang fine, but I did go through 2 undergrad algebra units and I'm only in chapter 4

I also know people who just can't read Lang, but I think it is worth a genuine try

[formal ermine]

parrot tea your bio is the worst joke I've read in a while

[frigid lark]

Hahahaha

[formal ermine]

you should've phrased it "if I research galois fields, does that increase my chances of getting a gf?"

[white oxide]

ah this is my first semester of algebra

[coral shale]

now i don't know what the original is

[rotund aurora]

Quick sanity check

If you have $k$-vector spaces $V\supseteq W\supseteq H$, then is it true that $\text{dim}(V/H)=\text{dim}(V/W)\cdot\text{dim}(W/H)$ ?

[cloud walrus]

Croqueta

[rotund aurora]

wait it should be added? idk lol

okay yeah it is additive

[lethal dune]

No

it should be added

[south patrol]

Ye third iso + use the splitting

[analog zephyr]

Guys ¿Can you help me with this one?

I know that for groups Z_n is isomorphic to the group of coset of a+nZ

[formal ermine]

where am I supposed to look at

[analog zephyr]

The entire theorem

[formal ermine]

which one?

there are two on that photo

[quaint minnow]

I think he meant Fermat’s little theorem

[analog zephyr]

I just don't get

I'm trying

Why is a^(p-1) congruent with 1modp

And I still don't know how to visualize the group of coset of a+pZ under multiplication módulo p

Ok, Z_p is isomorphic to a+pZ

And the order of Z_p is p-1 I get it, now each element of Z_p is in corresponding to each representative of the coset a+pZ.

So (a+pZ)(a+pZ)=a²+pZ under multiplication módulo p

So (a+pZ)^(p-1) must be in correspondence with 1

[formal ermine]

@jaunty glacier do you know what the elements of the dihedral group look like?

[jaunty glacier]

@formal ermine not really I've never worked with this before

I know they are reflexions and rotations

[formal ermine]

yes

of an n sided polygon

[jaunty glacier]

but I don't know why they are only reflexions and roations

[analog zephyr]

I'm i right?

[formal ermine]

by definition

[jaunty glacier]

Oh alright so no need to prove this

[formal ermine]

ye

ok

[analog zephyr]

Cuz 1 is the identity of Zp under multiplication modp

[formal ermine]

what order does the rotation have?

[jaunty glacier]

n I would guess and 2 for the reflection

[formal ermine]

yes

so you want a cyclic subgroup of order n

and the rotation has order n

so...?

[jaunty glacier]

So the cyclic group is the group of the rotations?

I'm so bad at algebra

[formal ermine]

yes

it's the subgroup generated by the rotation

[jaunty glacier]

Oh so that's it?

[formal ermine]

yee

it's that simple

[jaunty glacier]

wow

ty

❤️

[glossy crag]

You've been doing too much field theory, bro.

[south patrol]

Lol

Fr

[lethal dune]

Trynna be a farmer

[opal osprey]

Are there standard tools to compute:
$$
\text{Der}(\mathfrak{g})/\text{ad}(\mathfrak{g})
$$
Where $\mathfrak{g}$ is a given Lie algebra and and $\text{ad}(\mathfrak{g})$ denotes the subspace of derivations of $\mathfrak{g}$ generated by the adjoint derivations of $\mathfrak{g}$?

[cloud walrus]

MisterSystem

[glass vine]

I am doing going through some Algebraic number theory notes and problems recently and I have run into the following problem regarding Dedekind domains:

If D is a dedekind domain and A,B are proper coprime ideals such that AB=C^g for some proper ideal C of D then there are ideals E and F such that A= E^g and B=F^g

Now this seems obvious to me since we have unique factorization of ideals? But I am assuming I am missing something as it was grouped with some medium-hard problems

My logic is C can't be a prime ideal since A,B are coprime and in fact, if we write A =\prod P_i^(k_i) and B=\prod Q_j^(f_j) then AB = \prod P_i^(k_i) \prod Q_j^(f_j) = C^g = \prod L_i^(g*t_i) where C=\prod L_i^(t_i) is unique factorization of C

Now by unique factorization and coprimness of A,B we get the result easily, right?

[chilly ocean]

How to prove this: in domain, if b isn't a unit (ab) is properly included in (a)

[coral shale]

invent an element that isn't in the former?

[chilly ocean]

It's equivalent to (a)=(ab) then b is a unit...
And converse of this is easy to prove..
But how to prove original statement..

[south patrol]

Well, if a is in (ab), what can you say

[chilly ocean]

Then a=abr

[south patrol]

For some r, yes

Now have a think about that

I claim that from that you can deduce that b is a unit

Of course, here we must assume a isn't 0 lol

[chilly ocean]

But why br=1

Ah I see

[south patrol]

Yeah in a domain, you can cancel nonzero elements

As I'm sure you've worked out

[vital fossil]

fix a diagonalizable matrix, is there a method to find it's centralizer? (the set of all matrices which commute with it)

[vital fossil]

is it provable/convincing reason why there does not exist a general method if that is the case?

[coral shale]

solve some equations?

[vital fossil]

I am considering arbitrary ones, namely I want to think about centralizers of the centralizers of the centralizers, but even this first step I am stuck thinking about

(also it should be clear that I am not considering diagonal matricies)

[rotund aurora]

BAB^(-1)=1 iff (BP)D(BP)^(-1)=1 where A =PDP^(-1) with D diagonal

So find the matrices that commute with D and multiply from the right by P^(-1)

Isnt it just that? Maybe Im missing something

[chilly ocean]

Why do we equate it to 1

[rotund aurora]

Wait it shouldnt be 1 right

It should be A

[chilly ocean]

A and 1 are isomorphic

[rotund aurora]

[cloud walrus]

Parrot Tea

[tender wharf]

you can actually think of them as being algebraically equivalent though

[south patrol]

Lol

Where algebraically equivalent I assume means up to isomorphism

[tender wharf]

Ye

[coral shale]

equality of classes

[white oxide]

how exactly is this an additive group isomorphism, if the only operations involved are multiplication? is it because n x 1 = 1 + 1 + ... 1 n times?

[tender wharf]

well n dot 1 is that yeah

[south patrol]

Well it is easy to check by hand

[white oxide]

what exactly do they mean by additive

is my question rather ig

[south patrol]

I guess they mean

It's a homomorphism of the additive groups underlying the rings

And then they check multiplication separately

[white oxide]

i'm assuming an additive group is just any group equipped with the binary operation +?

[south patrol]

Yeah i mean it is more or less conventional but like

In the ring context, the additive group underlying a ring is what u get when u forget multiplication

Ans we always use + for that

[white oxide]

also how exactly is the map surjective, for example if r = 7 and s = 5 what choice of n could we even choose to get, say (3, 4)

[white oxide]

ohh ok

[rotund dragon]

24

[white oxide]

oh right oops

thx!

[rotund dragon]

you also know a solution exists because of CRT

[white oxide]

chinese remainder theorem?

[rotund dragon]

ya

[white oxide]

oh god i forgot that, but i'll have to review

number theory bad

[south patrol]

Lol I mean this is the CRT

oh ig you mean the more hands on form

[rotund dragon]

that is true

[torn warren]

Try to find an example to satisfy the following:

If H is a subgroup of G, is that possible to find a group A, such that $A=gHg^{-1}$ for some g in G, and satisfies $A\neq H$ and $A\cap H$ has more than one element?

[cloud walrus]

Witness

[jaunty pike]

well by intuition it seems not possible

but i dont have a proof rn

[torn warren]

yes, by intuition it seems impossible

[quiet pelican]

Any non-normal subgroup containing the centre seems to work, provided the centre is non-trivial
So (12) x C_3 union e x C_3 in S_3 x C_3 works, with g = (23)

[torn warren]

sorry what is C3 stands for

[summer path]

cyclic group(?)

[quiet pelican]

Cyclic group of order 3

[quiet pelican]

Yes

[torn warren]

${e, (12)}\times {e, a, a^2}$ do you mean this?

[cloud walrus]

Witness

[quiet pelican]

Have you not met the direct product of groups?

If not, that’s what it is, the underlying set is the Cartesian product of the underlying sets, and the operation is componentwise