groups-rings-fields

[pastel cliff]

all multiples of 3 and all multiples of 2 are in Q[x^2, x^3]

or well

[quiet pelican]

Yes

[pastel cliff]

silly question - gcd of x^5 and x^6 is just x right

actually just x^2 and x^3 should work right

[quiet pelican]

I’m actually not sure about (b) anymore, 1 is a gcd of those in Q[x^2, x^3]

[charred bison]

How do you show that $(\mathbb{R},+)$ and $(\mathbb{R}\setminus {0}, \cdot)$ are not isomorphic?

[cloud walrus]

CoolShot

[quiet pelican]

The second has an element of order 2, the first doesn’t

[charred bison]

ah

right

[sage lodge]

does anyone know what this means?

canonical map being the map from a group to its quotient

[coral spindle]

If you have a map phi, then -phi is also a map which behaves in the same way, so you can't choose between the two

[sage lodge]

oh

is there a different word to use instead of canonical here? (because the problem is "canonical" implies unique)

that still talks about the same kind of map

[south patrol]

Feels a bit vague tbh

Like in some contexts you can pick canonical maps into a group lol

And if u just said here is a group, give me a map into it, I'd just take the identity or the unique map from the trivial group lol

[next obsidian]

Yeah like the map G -> G/N is canonical

Lol

[chilly ocean]

not sure what this means because for instance on a torus sending a loop gamma to [gamma] in the fundamental group for example seems pretty canonical to me

How does that show that every ideal is generated by a single element?

[formal ermine]

ah yes I forgot what an ideal is for a second 💀

nvm it then

[rotund aurora]

Does someone know what Johann Gustav Hermes actually did? I cannot find his work.

Just a curiosity

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

I found this anyway

For cos(2pi/65537)

(Google: HIGH PRECISION SOLVES ANCIENT PROBLEM JOAN M TAYLOR)

[formal ermine]

idk if it's what your looking for, but:

In the year 1889 Hermes completed his procedure of creating a regular 65537 sided polygon using only a ruler and compass. gauß proved that a constructable polygon must have a side amount with a fermat prime, of which were known 5 at the time (till this day no more are known)

[rotund aurora]

I know that, but wikipedia gives no references to Hermes work. EDIT: It does give references to Hermes work, its the 65537 gon article that doesn't give, and I probably confused the articles I suppose

[wooden ember]

Hermes sounds like a chad

[formal ermine]

it's stored in the university of göttingen

[rotund aurora]

I can read Wikipedia man, I love Wikipedia

[wooden ember]

Imagine constructing a 65537 sided polygon

[formal ermine]

right

I was reading the german wikipedia article because it seemed like it had more content

https://gdz.sub.uni-goettingen.de/id/PPN252457811_1894

here

page 170

[rotund aurora]

The thing is, you just need to find the expression for cos(2pi/65537). Then, since we know how to construct basic arithmetic operations and square root extraction, the algorithm is done. So it should be easy to check if Hermes got it right (the solution at least). In the numberphile video of the 17-gon, Eisenbud actually said that he doesn't know if someone actually checked it haha

Yeah, don't think I'll get much out of this haha

[rotund aurora]

its fun because the expressions for cos(2pi/17) and cos(2pi/65537) look very similar. And 17 keeps appearing constantly, notice that all numerators are divisible by 17

[pine patio]

is Gal(K/F) only defined for Galois extensions or can K/F be any arbitrary extension?

[chilly ocean]

The definition makes sense for any extension, not necessarily a Galois one.

You could write Aut(K/F) if K/F is not necessarily Galois, and Gal(K/F) if it is.

That's what I've seen, at least.

[pine patio]

thanks

my lecturer only defined it for galois extensions so it was confusing

[glossy needle]

Hello! Doing two practice questions and I need some help/confirmation.. thank you!

A. Which ones of the following numbers can be the order of a subgroup of a group of order 120?
2,60,15,9,240,5,1,7

B. Which ones of the following subgroups of D16 has for normalizer D16 (with |D16|=16)?

1. <r>
   2.<sr^2>
   3.<s,r>
   4.<r^4>
   5.<sr^4>
   6.<s,r^4>
2. None of these answers
   —————-
   My take on A: according to lagrange’s theorem, the order of the subgroup must divide the order of the group so the answers would be 2,60,15,5 and 1?

My take on B: I know what normalization is but I’m stuck on how to proceed to prove that the normalizer is D16 for one/many/either of them. Need massive help.
Thank you!!

[formal ermine]

your solution to A is correct

[glossy needle]

Thank you @formal ermine

How do I proceed for B?

[pine patio]

so you just need to find out which of those are normal in D16

[formal ermine]

(assuming commutative rings) so for an integral domain $R$ we know that $R[x]$ is euclidean. but we also know that iff $R[x]$ is euclidean, then $R$ is a field. wouldn't that mean that every integral domain is a field? what's wrong here?

[cloud walrus]

illuminator3 (I/you)

[glossy needle]

I’ve only found the <s,r> group to be normal. I’m sorry I’m new to this chapter and catching up. What are or is the correct answer so I can practice the proof?

[chilly ocean]

Z[x] is not a Euclidean domain, so the first sentence is not true.

It's not even a PID.

[formal ermine]

yeah you're right

then I don't quite understand this proposition:

"Let $R$ be an integral domain and $g(x) = a_xx^n + a_{n_1}x^{n-1} + \ldots + a_0 \in R[x]$ with $a_n \in R^\times$. Furthermore let $f(x) \in R[x]$ be arbitrary. Then there exist unique polynomials $q(x), r(x) \in R[x]$ with $$f(x) = q(x)g(x) + r(x)$$ and if $r(x) \neq 0$ then $\operatorname{deg}(r(x)) < \operatorname{deg}(g(x))$."

[cloud walrus]

illuminator3 (I/you)

[formal ermine]

doesn't that satisfy the conditions of being an euclidean ring?

I mean, we have a well defined degree function

and the conditions about the decomposition are also satisfied, no?

[chilly ocean]

"With a_n in R^\times" already excludes a ton of polynomials.

[formal ermine]

ah

[chilly ocean]

To be a Euclidean domain, you need to be able to do this with any arbitrary non-zero elements.

[formal ermine]

yeah

thank you!

[chilly ocean]

Excluding the "a_n is a unit" gives you counterexamples, where, of course, you should look at Z[x].

If R is a field, then this condition is satisfied by all non-zero elements, which gives you that R[x] is Euclidean.

(Although you already probably know this.)

[formal ermine]

doesn't it only exclude the zero polynomial

or do they mean $a_n \in R^\times$ for any n not just the polynomial degree

[cloud walrus]

illuminator3 (I/you)

[chilly ocean]

Only the leading term.

The proposition says that if you have a polynomial whose leading term is a unit, then you can divide by it.

[formal ermine]

ahhhhhhhhhh

[chilly ocean]

It doesn't exclude only the zero polynomial, since 2x + 1 has a non-unit leading coefficient.

In Z[x].

[formal ermine]

yeah

[chilly ocean]

You won't be able to divide x by 2x + 1, for example.

Work it out and you'll see why you need the leading coefficient to be a unit.

[next obsidian]

This shouldn’t be surprising. Modify the thing you want to divide by by a unit and you’re just saying polynomial long division works for monic polynomials. If you can remember the proof of why polynomial long division works (as you learned it in like highschool if you did) you’ll see that you needed to divide by the leading coefficient, but once your polynomial is monic this doesn’t matter anymore

[chilly ocean]

highschool
proofs

[formal ermine]

I don't know how to do polynomial long divison

[chilly ocean]

It is now time for you to learn.

[next obsidian]

I mean

Not even proof

Like

If you ever learned it, you just use that same algorithm

Haha

The algorithm is itself the proof you can do this xD

[formal ermine]

do you have a good resource

[next obsidian]

Just google it man

“Polynomial long division”

[chilly ocean]

I don't really like any algebra books, so I'm going to second chmonkey's suggestion to google it.

[formal ermine]

it only gives me sites like mathisfun.com with confusing instructions

[chilly ocean]

I spent all of highschool using websites like that. Get used to it.

[next obsidian]

Try to understand the algorithm

Also mathisfun.com is based

[glossy needle]

Hey can I get some confirmation with this one..
Which ones of the following subgroups of D16 has for normalizer D16 (with |D16|=16)?

1. <r>
   2.<sr^2>
   3.<s,r>
   4.<r^4>
   5.<sr^4>
   6.<s,r^4>
2. None of these answers
   —————-
   So in order for d16 to be the normalizer the subgroups have to be normal so the answers are 1,2 and 5???
   Please help lol

[next obsidian]

I cited it for a complex analysis hw once in freshman year

[chilly ocean]

Based.

[next obsidian]

Descartes rule of signs

[chilly ocean]

I first learned calculus from mathisfun.com.

[next obsidian]

Based

[glossy needle]

My bad <sr^2> isn’t normal but <s,r> is

is there any other normal subgroup in this list ?

[chilly ocean]

Group theory .

[glossy needle]

Yes lol (:

[formal ermine]

ah

khanacademy ftw

[oak hill]

Is there a fast way for computing the order of all elements of $\mathbb{Z}_n$ for some $n\in\mathbb{N}$?

[cloud walrus]

F♯A♯ℵ0

[next obsidian]

|m| = n/gcd(m,n)

[oak hill]

cheers

[south patrol]

Not sure what the easiest proof of that fact is actually

[oak hill]

Also is there a fast way of knowing what all possible orders of $\mathbb{Z_n}$ will be?

[cloud walrus]

F♯A♯ℵ0

[next obsidian]

You compute it by hand it’s easy af

It’s just an application of bezout

[south patrol]

I mean fair like the way I would do it is like

$|(m)||(n)| = \frac{|(mn)|}{|(m) \cap (n)|}$

[next obsidian]

Lol noob

[south patrol]

ig

[next obsidian]

Bruh

[south patrol]

Wait

no

[next obsidian]

You’re cringe

[south patrol]

I am!

Type

*Typo

lol

[next obsidian]

No theory allowed

Only definitions

[south patrol]

:(

lol

Okay yes sure, maybe the simplest would be that the order is the factor you have to multiply m by to get the lcm of n and m

no bezout smh

[bleak abyss]

Alright my turn

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Okay this is weird

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

Now

[cloud walrus]

Sloth King Daminark

[tired horizon]

I have a question about symetric groups

In a symetric group of 9

They give u two different permutations

That contain each permutations of 3 elements

Like so

And they ask u to give them the order of the product of sigma and T

We just multiply both orders right?

The order for T and sigma is 3 for both

So we get 9 for the product if we calculate everything in details

But we don't need that right?

We just have to multiply

[lethal dune]

No multiplying order doesn’t work ex take (1,2) and (1,2)

[tired horizon]

The order of their product is 4

Oh no

Its 2 right?

[lethal dune]

It’s 1

[tired horizon]

Ah yeah

So I have to go through the whole process

[lethal dune]

You can decompose the perm into disjoint cycle

[ashen jewel]

Isn't the order of the product of two elements the lcm of their orders? if the group is abelian

[tired horizon]

Theyre already disjoint

[tired horizon]

Yes

But if we do that we get 3 no?

[ashen jewel]

Then shouldn't the order of T(rho actually) times sigma be 3

[ashen jewel]

yeah that's what I was thinking

[tired horizon]

And when my friend calculated the whole thing

He found 9

[ashen jewel]

I wanna see the working out your friend got

[tired horizon]

I'll try to do it alone too and see if I get 9

[ashen jewel]

out of curiosity

[tired horizon]

If I do ill send it

[ashen jewel]

alright

[tired horizon]

If I dont ill see with my friend

[formal ermine]

how do I show that $[A_n,: A_n] = A_n$ for $n \geq 5$? the hint says to use the following proposition: "for $n \geq 5$, two 3 cycles are conjugated in $A_n$". but I don't quite see how that can help me here?

[cloud walrus]

illuminator3 (I/you)

[formal ermine]

ah nvm figured it out

[formal ermine]

does this group have a specific name?

[proud bear]

heisenberg group

[formal ermine]

thanks

[tired horizon]

I had an errand to run lol

So I just did it

And it keeps getting weird

[ashen jewel]

wdym

[tired horizon]

I tried to do a disjoint permutation

Cause they weren't disjoint

When u do the product

This is what I found

[ashen jewel]

WAIT, I just realized symmetric groups are not abelian

that means that the order is doesn't have to be the lcm of the orders of rho and sigma

[tired horizon]

So the order now is 9

But where I read about the order being the lcm

They didn't say anything about the group being abelian

Or not

Well its correct now

The order is 9

[bleak abyss]

Okay so let's say you have a root system

[ashen jewel]

so vectors on a plane?

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Now here's where this gets a bit funky

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Oh wait

Okay okay I get it now

I prob need to go and prove this fact but

I kinda get it

[chilly ocean]

Why is (y^2) a primary ideal in K[y] for K a field?

[south patrol]

Do you know its radical? Once you have that it should be fairly immediate from the definition

[chilly ocean]

One way to go about it is to show that every element of the quotient K[y]/(y^2) which is a zero-divisor is also nilpotent. That's a straightforward computation.

The good ol' dual numbers.

It's not the best way to do it, but it's fun.

[oak hill]

If i wanted to show that the factor ring $\frac{\mathbb{Z}[x]}{I}$ where $I$ is the ideal ${f(x)\in\mathbb{Z}[x]:f(10)=100n \text{ for some } m\in\mathbb{Z}}$ fis not a field is it sufficient to take $f(x),g(x)\in\mathbb{Z}[x]$ such that $f(10)=10, g(10)=10$ then $f(10)g(10)=100$ so $f(x)g(x)\in I$ but $f(x),g(x)\not\in I$, hence $I$ is not a prime ideal so $\frac{\mathbb{Z}[x]}{I}$ is not an integral domain and hence not a field?

[chilly ocean]

That's not even an ideal.

[oak hill]

Sorry, I mistyped, one sec

Sorry having some LaTex trouble lol.

[chilly ocean]

$${f(x) \in \bZ[x] : f(10) = 100n \text{ for some } n \in \bZ}$$

[cloud walrus]

rakko

[oak hill]

Thanks

[chilly ocean]

(In N, woops, but I'm not going to change it.)

[cloud walrus]

F♯A♯ℵ0

[oak hill]

nope was supposed to be Z. Just not on it today...

[chilly ocean]

We all have those days.

[oak hill]

🙂

[chilly ocean]

Anyways, this proof works. It might be best to provide specific examples of such f and g, but it's still good.

[oak hill]

Ok great. Thank you!

[chilly ocean]

Even a constant polynomial. If f(x) = 10, f(x)f(x) = 100 is in I, but f(x) isn't.

So the ideal isn't prime, which means the quotient isn't a field.

[oak hill]

Sweet, thanks

[pastel cliff]

bump from yesterday but can i get some pointers pls

mostly on c) and d)

[rustic crown]

what are your thoughts on them so far?

[pastel cliff]

dont think d) exists

[rustic crown]

[rustic crown]

yep, that's true. (except for one tiny thing, 0 can't be written as product of irreducibles :p)

that's the expected reaction >.<

[pastel cliff]

ok c) is asking if there are "coprime" ideals that arent comaximal essentially right

[rustic crown]

lol idk the definition of those words :p

[pastel cliff]

ok what about x^2 and y^2

[rustic crown]

by comaximal you jsut mean they sum to (1)?

[pastel cliff]

yeah

[rustic crown]

yea, that works

(even x and y work :3)

[pastel cliff]

O

bc then there wont be any constants in the sum of those ideals

[rustic crown]

yep :3

[pastel cliff]

also about a)

[pastel cliff]

is there an easier way than just checking the non zero non units

that feels dumb

[rustic crown]

try to see what it means to be principal here

say (a, x) = (f)

what can you say about f?

even though Z/12Z is not a domain, we can still say a lot about f as "x" is monic

oh wait maybe i said something stupid >.<

ignore me

ig you could try converting the problem to Z[x] where we don't need to worry about these nilpotents or zero divisors

first without loss of generality a divides 12, so the ideal (a, x) in (Z/12)[x] corresponds to the ideal (a, x, 12) = (a, x) in Z[x] and if this was principal, then (a, x) = (f, 12) for some polynomial f in Z[x]

[chilly ocean]

Is this message self-referential

[rustic crown]

[chilly ocean]

because by saying it you said "something stupid"

which makes it technically true that you said that

[rustic crown]

[south patrol]

Lol

[chilly ocean]

I'm just joking btw and it wasn't supposed to mean anything offensive

[oak hill]

If $I={f(x)\in Z : f(0) \text{ is an even integer}}$, could someone explain why $\frac{Z[x]}{I}={I,1+I}$?

[cloud walrus]

F♯A♯ℵ0

[chilly ocean]

My attempt: If p(y)+(y^2) is a zero-divisor, then there exists a non-zero q(y)+(y^2), i.e. q(y) is not in (y^2), such that p(y)q(y)+(y^2)=(y^2), i.e. p(y)q(y) is in (y^2). I want to show that p(y)^n+(y^2) is in (y^2) for some n, but I'm stuck on how to proceed.

[chilly ocean]

If f(0) is not an even integer, then f(0)+1 is an even integer

[oak hill]

Thanks

[long nebula]

So I proved that Z[sqrt(-1)] and Z[sqrt(-2)] are Euclidean, but I don't think my proof works for Z[sqrt(-n)] for n>2, is Z[sqrt(-n)] a Euclidean domain in general?

with the norm being the usual norm on complex numbers

[chilly ocean]

Z[sqrt(-10)] is not Euclidean

[long nebula]

ahh

hmm is 10 the lowest number that isn't?

what's the counterexample

[chilly ocean]

I don't think for any other negative integers it's Euclidean, not even a UFD.

[long nebula]

oh I guess UFD is easier to think of counterexamples for

hmmm

so Z[sqrt(-3)] can't be one because 4 = 2*2 = (1+sqrt(-3))(1-sqrt(-3))

interesting

ty!

[pastel cliff]

sorry had 2 get fud - is there anything we can say about f?

[chilly ocean]

You're talking about elements of the polynomial ring and elements of the quotient with the same notation and terminology. It's hard to understand.

[pastel cliff]

ohhh wait

it should be the gcd of a and x

[chilly ocean]

changed it

Oh I think that y | p(y) and so p^2(y) should work

Because p(y)q(y) is in y^2 so p(y)q(y) = t(y)y^2 for some t(y) and we know y^2 doesn't divide q(y) so I think at least one factor of y has to divide p(y)

oh wait, I don't think p(y)q(y) needs to be of that form...

[pastel cliff]

is localizing a ring just turning it into a field?

this defn is

[chilly ocean]

No, that's not really what localization does.

[chilly ocean]

not always

It depends on the multiplicative subset used.

I think you need R to be a domain and S to be the set of non-zero elements of R.

Wait, small issue. Let me fix it.

[chilly ocean]

Is it not true that any element of a principal ideal (x) is rx for some r in the commutative ring R?

Then r = x?

R is K[x]?

[gritty sparrow]

Sorry for butting in, i don’t really know the context here but the statement @username000 is true

*statement he made

[chilly ocean]

Oh, we're working with K[x], not R[x]. I assumed R meant the base ring.

My bad.

[chilly ocean]

"he"?

[gritty sparrow]

I was just playing the odds i suppose

[chilly ocean]

Suppose that $p(y)q(y) \in (y^2)$ with $q(y) \not\in (y^2)$. If $p(y) = a + by + \cdots$ and $q(y) = c + dy + \cdots$, then $$p(y)q(y) = ac + (ad + bc)y + \cdots \in (y^2),$$ where the ellipses denote terms divisible by $y^2$. This implies that $ac = ad + bc = 0$. You can proceed by casework: consider $c = 0$ and $c \neq 0$.

[cloud walrus]

rakko

[gritty sparrow]

Sorry

[chilly ocean]

This is true if R is supposed to denote K[x], but you should really be more explicit in the future. My bad for getting confused, though, I should have asked for clarification before jumping to a conclusion.

[chilly ocean]

OK that's good. I thought I was missing something. Everything in there is a finite sum of products r_1x + r_2 x + r_n x and you can just factor out the x

[chilly ocean]

Yea, that's fair

[chilly ocean]

This proof makes sense, but I wanted to make sure my original argument didn't work due to some misunderstanding.

[chilly ocean]

Here's the casework (which I messed up initially).
-If c is zero, then d has to be non-zero by the assumption that q(y) is not in (y^2). This implies that a = 0 by the second equation, which means that y divides p(y) or y^2 divides p(y)^2.

-If c is non-zero, then a = 0 by the first equation and then b = 0 by the second. Then y^2 divides p(y).

[chilly ocean]

Yeah, it will be of that form. That's what it means to be an element of the ideal (y^2) after all.

It'll imply that y divides p(y)q(y) at least. This means that y has to divide one of p(y) or q(y). The casework here is identical to the casework I just did.

[chilly ocean]

So is it not true that since y^2 | p(y)q(y) and y^2 doesn't divide q(y) then a factor of y must divide p(y) ?

Well, that's what you have to deduce.

I don't see how it immediately follows - some kind of argument like mine above is needed.

But it is correct.

[chilly ocean]

I understand. It's true and your argument shows why it's true. Thank you.

[chilly ocean]

This is why I try to use "they" in online situations.

[chilly ocean]

Glad to help, and sorry for the silly confusion earlier.

If J is P-primary and I is P-primary, then isn't J = I ?

Since P = sqrt(J) and P = sqrt(I) ?

[gritty sparrow]

The radicals being equal doesn’t mean that the ideals are equal. Eg: (2) and (4) are both (2) primary

In Z that is

[chilly ocean]

Ohhhhhh I understand. Thank you!

[upper cape]

Does anyone know if this result would hold for compact topological rings, rather than compact topological groups?

[eager willow]

yes

exact sequences of rings are just exact sequences of groups

exactness comes on the level of groups, just check the limits exist and the maps between them are maps of rings

idek if there are always inverse limits of rings but doesn't matter if they exist or not

if they exist it is exact because it is an exact sequence of groups with maps that happen to be maps of rings

add hypothesis for topological ring maps and it still wont matter

[upper cape]

ahh that makes sense

I think inverse limits for rings always exist, since topological rings form a complete category

[gentle marsh]

Is there a Dummit and Foote Study Guide for guiding someone through an Abstract Algebra Course (+ recommended problems to work on)? I'm planning an independent study class for myself and I want to find something I can use to work off of.

Just sort of like selected must-do problems

[next obsidian]

Bruh

You’re literally just saying “they exist because they exist”

Lol

[eager willow]

choose freely whichever chapter on a topic that you actually want to learn and do all the exercises in previous chapters in order of what your selected chapter depends on. repeat until exhausted.

for me i started with galois theory and dove backward into any definition I didn't know. pretty much do that with every big book

[tired horizon]

In symetric groups

S9

Let's say I have 2 permutations

Both composed of 2 disjoint 3-cycle permutations

But both of them contain onepermutation thats the same

Like this

123 is a common element

Do we have a theorem or something for that

Im having problems explaining in english

Does anyone understand french ?

[zenith trellis]

I do

[tired horizon]

Alors j'ai cet exercice

Dans la 3eme question

J'ai fait les calculs des 2 permutations

J'ai trouvé ces resultats

Maintenant j'essaye de trouver une relation entre ça et le produit des 2

J'ai trouvé que le produit est commutatif

Jsp pas quoi faire apres

Je pense qu'on peut utiliser la 1ere question

Le fait que pi commute avec les 2

[zenith trellis]

Perso je remarquerais que $\sigma \tau = \tau (\tau^{-1} \sigma \tau) =\tau (\tau^{-1} \sigma ^{-1} \tau)^{-1}$. Sachant que $\sigma^{-1} = \sigma ^2, \tau^{-1} = \tau^2$ et $(\tau^{-1} \sigma ^{-1} \tau)^{-1} = (\tau^{-1} \sigma ^{-1} \tau)^2$ alors je pense qu'on peut bidouiller un truc pour retrouver $\pi$...

[cloud walrus]

Silfer

[tired horizon]

Dans la 2eme partie

Est-ce que c'est toujours le cas

Pour l'inverse de sigma = sigma^2

Pour l'ordre 3

[zenith trellis]

Je pense que oui car $\sigma^3 = Id$

[cloud walrus]

Silfer

[tired horizon]

D'accord je vais essayer d'abord de faire la question

[zenith trellis]

On a aussi $(\tau \sigma \tau^2)(\tau^2 \sigma \tau) = \tau(\sigma \tau)^2 = (1 2 3)\pi$

[cloud walrus]

Silfer

[tired horizon]

Avec ce que tu m'as donné

J'ai pas compris le passage a l'inverse

Ici

[zenith trellis]

pareil, tau sigma tau^-1 est d'ordre 3

[tired horizon]

Ah oui

Non c pas la 2eme partie

Je parle de celui la

[zenith trellis]

pareil, tau^-1 sigma tau est d'ordre 3

[tired horizon]

Oui mais sigma-1 c'est sigma ^2

[zenith trellis]

et donc ?

[tired horizon]

Tu peux m'expliquer?

[zenith trellis]

Expliquer quoi ?

[zenith trellis]

L'inversion elle s'explique par ça

[tired horizon]

Ca devient pas tau sigma^-1 tau^-1 par inversion?

[zenith trellis]

non

l'inverse de ab c'est b^-1 a^-1

[tired horizon]

Omg

Je suis vraiment bête

Alors je calcule le produit dedant

Et je vais comparer et essayer de trouver une relation entre eux

[zenith trellis]

Bon honnêtement je sais pas trop quelle genre de réponse ils attendent

Je vois pas comment tu déduis que pi est une puissance de sigma tau

[tired horizon]

Je pense que si on calcule le produit qu'on a trouver on trouver pi

MAIS y a le tau qui gêne

Tau fois le produit qu'on vient de trouver

[zenith trellis]

Enft c'est beaucoup plus rapide de faire le calcul directement que de trouver une méthode soit disant "jolie"

[zenith trellis]

En DS j'aurai fait ça

[sharp dirge]

does someone know how they are showing c(f_0g_0) = 1?

if irreducible divides c(fg) show that it divides c(f) or c(g), how does that imply that c(f_0g_0) = 1

[tired horizon]

Le prof a preciser de deduire

Donc on doit utiliser les "jolies" formules

[zenith trellis]

Bon s'il l'a demandé je peux comprendre... mais en DS/partielle c'est une mauvaise idée.

[gritty sparrow]

In that part of the proof the f and g are primitive polynomials so c(f) = c(g) =1 so a dividing any of them gives a contradiction

[sharp dirge]

wait f,g arnt primitive

f_0 and g_0 is primitive

[gritty sparrow]

He changed it up later

Replace f and g with f_0 and g_0 in the second part of the argument

See he reduced to the case of proving that if c(f)=1, c(g)=1 then c(fg)=1

@sharp dirge

[sharp dirge]

i thought he just multiplied them both

and applied content function again

no?

@gritty sparrow

[gritty sparrow]

Well basically he needs to show that c(f_0g_0) = 1. He then shows this by showing that for general primitive polynomials the product is primitive. When he is proving the later part he lets f and g be arbitrary primitive polynomials

[sharp dirge]

ahh

kk

understood

[tired horizon]

J'ai trouvé une notion "formule de conjugaison"

Finalement c'est un lemme d'une preuve des generateurs

[jovial flint]

why does it follow that v is surjective?

[terse crystal]

You took N=M”/im(v)

Take

[jovial flint]

wait I'm confused. v being surjective follows from the sequence being exact right?

oh nvm

[terse crystal]

Good

[upper cape]

:D just means I only have to show products and equalizers exist, rather than inverse limits in particular

[pastel cliff]

reading about localization of rings and it's defined with some equivalece relation of R x S mumbo jumbo - is it actually just talking about fractions? like literal fractions?

[chilly ocean]

Pretty much. You just need a way to make that rigorous.

Have you ever seen an explicit construction of Q from Z?

It's the same thing.

[pastel cliff]

probably but i cant recreate it

will just keep reading then

[rustic crown]

there is a minor caveat, as for a general ring you might also end up inverting zero divisors and stuff

[pastel cliff]

definition im reading exludes 0 from any S

[rustic crown]

which is why the definition of equivalent fractions looks a little weird

[pastel cliff]

S being the closed subset

[chilly ocean]

Q is the set of all tuples (m, n) with n non-zero, modulo the equivalence relation (m, n) ~ (a, b) iff mb = na. This is what it should mean for m / n to be equal to a / b, right?

[rustic crown]

yea but what if you invert say 2 in Z/12Z

[chilly ocean]

The equivalence relation comes from the intuitive sense of what it should mean for two fractions to be equal.

[jovial flint]

just to be sure, this is what is happening right?:
defining $f:M''\to M''/im(v),~f(m'')=m''+im(v),~ \bar{v}(f)=f(v)=0.$ Since $\bar{v}$ is injective, $f$ is the zero function meaning $im(v)=M''$

[cloud walrus]

Bilboswaggins

[rustic crown]

(by this i mean S = {1, 2, 4, 8})

[pastel cliff]

yeah hadnt thought about that

[terse crystal]

yeah

[chilly ocean]

All rings are integral domains.

[jovial flint]

thanks!

[terse crystal]

Np

[pastel cliff]

what

[chilly ocean]

All rings are integral domains.

[rustic crown]

🙈

[chilly ocean]

Zerodivisor? I hardly know her!

[rustic crown]

if you only work with integral domains, then localizations are just subrings of the fraction field

which you probably have already seen

[chilly ocean]

(Z and Q!!!!)

[rustic crown]

but in the general case for (commutative) rings, the definition changes a little. you define a/s = b/t iff there is a z in S such that (at-bs)*z = 0

(in particular, if 0 is in S, then then you can always choose z = 0 and this will make the localization S^(-1)R the zero ring as all fractions are equal to 0)

or another way to say this is that, after adjusting the fraction a little the old definition follows

if your definition is well defined we need to have that
a/s = b/t implies az/sz = b/t

else we may not even have that its an equivalence relation

[rustic crown]

and here when you invert 2, you're pretty much collapsing everything that is related with 2 in Z/12 so you end up getting the ring Z/3

[pastel cliff]

this is so much to insert into my brain at once

but thank you det

[pastel cliff]

and this is Frac(R) right

[chilly ocean]

So for each $k, \ell$ you have the linear map $$T^k V \otimes T^\ell V \to T^{k + \ell} V \hookrightarrow T(V),$$ where the first map is the usual isomorphism and the second is the embedding of $T^{k+\ell}V$ into the degree-$(k+\ell)$ part of the tensor algebra. Note that $$T(V) \otimes T(V) = \bigoplus_{d=0}^\infty\left(\bigoplus_{k+\ell = d} T^kV \otimes T^\ell V \right)$$ (natural isomorphism). The universal property of the direct sum will take your linear map $T^kV \otimes T^\ell V \to T(V)$ defined for each $k, \ell$ to a linear map $T(V) \otimes T(V) \to T(V)$, which is the same thing (by the universal property of the tensor product) as a bilinear map $T(V) \times T(V) \to T(V)$. \emph{This} is the "multiplication" operation on the tensor algebra introduced by the tensor product.

[cloud walrus]

rakko

[chilly ocean]

That's what the Wikipedia article means when it says to extend by linearity.

[rustic crown]

yee

[pastel cliff]

how would i start this

gotta say im more of a group guy than ring guy

[rustic crown]

you'll turn into a ring guy soon enough :3

the set Z\(2) is exactly the set of odd numbers

[chilly ocean]

Come to the ring side. It's more fun over here.

[agile burrow]

group ring

[rustic crown]

if you're dealing with integral domains, you can define S^-1R to be exactly the fractions r/s in Frac(R) which have denominators from S

[chilly ocean]

Polyamorous marriage.

[pastel cliff]

polynomial marriage

ok, formally what does it mean to localize Z by the odd integers

from what i gather so far this is extending Z so that odd integers have inverses

[coral spindle]

You look at Q but don't allow the denominators to be even

whoops, typo

[rustic crown]

yep, that's exactly it

you can also think of it as teh solution to a universal problem

you have a ring R, and a multiplicatively closed subset S, then the localization is the universal ring which maps out of R such that elements from S are sent to units

which is exactly formalizing what it means to "invert exactly the things from S and nothing more"

[coral spindle]

Getting the right relation to quotient R x S out by is a little tricky for non-domains but it's fairly straightforward for domains

[celest cairn]

What is a minimal annihilating equation?

[glossy crag]

Anyone?

[south patrol]

Seems good to me

Well can be made slightly more brief but ye

[rustic crown]

i think you should justify why the subgroup is generated by 1/m

i think that is the crucial step

for example if your group had 1/2, 1/3, etc then say after making denominators same you get 3/6, 2/6, etc and one needs to argue why you also see 1/6 somewhere

[glossy crag]

That's easy I think: first, you prove that the order of any element of Q/Z is its denominator written in lowest terms. Now if (x) is the cyclic subgroup and n is its order, then x obviously belongs to (1/n) and 1/n belongs to (x) because the numerator m of x is coprime to n (by assumption), so am+bn=1 for some a,b, and a*m/n=1/n, so you have equality of the two subgroups.
Everything look alright?

[rustic crown]

oh what i was trying to say was when you make denominator of everything same, then are you sure that you find 1/6 there because the numerator may no longer remain coprime

like in 2/6 and 3/6

ofc there are other elements in the subgroup which is why this won't happen

[rustic crown]

this only proves that order of m/n is n given gcd(m, n)=1

[glossy crag]

I don't understand your point. First, you show that the order of every element is determined by the lowest terms denominator. Then you use that to show that all cyclic subgroups are the same. What exactly is wrong here? The point here is that the cyclic subgroups actually agree, in some other group you can have 2 elements of the same order whose generated subgroups differ.

[rustic crown]

ah so what i'm saying is that when you make all the denominators the same, why do you think that there is a fraction somewhere which is actually reduced

if i say anything more then it would make this nitpick more pointless than it already was :p

[glossy crag]

Because when you make all the denominators the same, e.g. m, then the group H becomes a subgroup of (1/m), which is cyclic, so H must in turn be cyclic and there must be some denominator for it. I'm not saying the denominator of H will be m itself, it might be some divisor of m (I haven't checked this directly, but the denominator of H should be the lcm of the denominators of its elements).

[rustic crown]

yee that makes sense :3

[glossy crag]

Alright, thanks.

[rustic crown]

oh wait was the last line always there, maybe then i misread your argument >.<

sowwy

i thought that you were claiming the group is exactly <1/m>

[glossy crag]

Nah, it's probably some divisor of m.

[rustic crown]

yeah makes total sense now >.<

my bad

.<

[celest cairn]

Hi

[rustic crown]

hewwo

[celest cairn]

Question

What’s a minimal annihilating equation?

[glossy crag]

If you want an answer, you should probably explain the notation/context, because it looks pretty strange (to me at least).

[chilly ocean]

If R is a PID and a factors a = c_1... c_n. Then for b in (a)=(c_1)...(c_n), b = ra = r*c_1...c_n. Why is b in (c_1)...(c_n) ?

[elder wave]

it kinda looks like some algebraic analysis stuff

[rustic crown]

because r*c1 is in (c1) and other c_i are in (c_i), so b is in the product of the ideals

[celest cairn]

Found this stuff

[formal ermine]

stupid question, but the heisenberg group over Z3 has 3^3 elements, yes? for reference, it's this group:

[rustic crown]

its probably the monic polynomial of least degree which kills your vector.

[rustic crown]

yep

[formal ermine]

epic

that group must be solvable since it's a p group then, right?

[rustic crown]

yee

[formal ermine]

okay

thank you

is there any easier way, than just checking all cases and doing sylow and other stuff, to show that A_5 is the smallest non cyclic simple group?

or do I really have to go through all of them

and cross each of them out

[rustic crown]

i like to use the class equation to prove that, but it is still a brute-force in the end

but this one is not that bad i feel

[formal ermine]

ahhh right that thing exists

you mean this one?

$$|G| = |Z(G)| + \sum_i |C_i|$$

[cloud walrus]

illuminator3 (I/you)

[rustic crown]

yea, this will help you show A_5 is simple

[formal ermine]

o I can already assume that

[rustic crown]

and after that you need to work hard to show there aren't any simples smaller than that >.<

[formal ermine]

let me write down the entire question

"Let G be a simple group with |G| < 60. Show that G is cyclic and that |G| is a prime. You may use the fact that A_5 is the smallest non abelian simple group."

idk how to apply the fact though

[rustic crown]

okie that makes it quite easy actually... because they're only asking you to check abelian simple groups

[formal ermine]

ahhhhhhh

[rustic crown]

and then you have structure theorem, so nothing to do

[formal ermine]

we haven't had that yet

let me read up on it real quick

[rustic crown]

oh in that case just see why there are non-trivial normal subgroups directly

[formal ermine]

oh it's module stuff, we're still at integral domains right now

[rustic crown]

you probably have proven that for abelian groups, there is a subgroup of order every divisor of |G|

[formal ermine]

yesssss

[rustic crown]

so this should do it :3

[formal ermine]

let me think about it for a minute

[rustic crown]

If you didn't have this, you would have had to work a little harder by showing that p-groups aren't simple if they have size > p, groups of order pq aren't simple for p < q, similar thing for p^2q and pq^2 and finally pqr

this should hopefully include everything < 60

[maiden heath]

Hi, I was wondering what the general equation for the basis of the sign representation is for the symmetric groups, for example span{v_1-v_2} is a basis of the sign representation for S_2. But what would one look like for S_3, S_4 etc...

[rustic crown]

okie it excludes only p^2q^2, and only instance of this < 60 is 36

[formal ermine]

isn't this only for cyclic groups

[rustic crown]

for cyclic you also have uniqueness

[formal ermine]

ah

[rustic crown]

you can do this by a simple induction

[formal ermine]

induction on which variable?

[rustic crown]

order of the abelian group :3

we're proving this: Let G is be an abelian group of order n and d a positive divisor of n, then there exists a subgroup H of order d.

or maybe you can make it even simpler... we just need to show that an abelian simple group has prime order.

[formal ermine]

why?

[rustic crown]

pick any non-trivial element g in G

then <g> is a normal subgroup

since G is simple, G = <g>

so its cyclic as well :3

<g^d> is a normal subgroup for any d dividing n=|G|

but its order is n/d

so d = 1 or n

which means n is prime

[formal ermine]

I can't see why this is true

[rustic crown]

because abelian

everything is normal

:3

it would be a lot more complicated if the group wasn't abelian, only nice tools there you have is like sylow or looking at kernels of group actions to get normal subgroups

[formal ermine]

uh I still don't see why :c

OH

WAIT

I am so stupid

[rustic crown]

.<

[formal ermine]

I confused normal subgroups with ideals

[rustic crown]

oh oops

[formal ermine]

lmao yeah makes sense

yeeeee I get it

thank you so much!

[rustic crown]

[formal ermine]

I don't understand this last step

how can you conclude that n must be prime

oh wait

nvm

[rustic crown]

any positive divisor d of n, is forced to be 1 or n

[formal ermine]

yes

only divisors of n are 1 and n so n must be prime

[rustic crown]

i'm very rusty on my rep theory, but where exactly are you trying to find the basis?
if you look at S3, then i think you're considering the linearization of the action on {v1, v2, v3}. then the trivial representation lives inside this as the span of v1+v2+v3, but sadly the sign representation doesn't live inside it. the complement of the trivial rep is {c1v1+c2v2+c3v3 | c1+c2+c3 = 0} and in case of S3 i think this is irreducible

if you're trying to find it as a subrepresentation of the group algebra k[S_n] then just look at the span of (sum of even permutation - sum of odd permutations)

[toxic zephyr]

in linear algebra, we can prove T is a linear transformation iff T(x+cy)=T(x)+cT(y). that is, this one equality verifies both conditions, so we don't do two different verifications for vector addition and scalar multiplication. can we do the same for ring and module homomorphisms?
ring homomorphism: f(x+ry)=f(x)+f(r)f(y) for all x,r,y in the ring
R-module homomorphism: g(x+ry)=g(x)+rg(y) for all x,y in M and r in R
?

[south patrol]

For ring homs (for unital rings) you require them to send 1 to 1 and that isn't necessarily the case with that test

e.g. Z -> Z mapping x -> 2x isn't a ring homomorphism but it definitely satisfies the stuff you wrote there

The R-module one should be fine though

[toxic zephyr]

i see. if i check that f(1)=1 in addition to that test, would that then be sufficient?

[south patrol]

Yes that should be fine

Since taking x = 0 or r = 1 you just recover the standard things

[pastel cliff]

being asked to show that Z[i]/I is finite (I is any ideal) can i get a nudge please

[south patrol]

Uh I mean so any quotient of a finite module is finite (and indeed taking I = 0 we need to show Z[i] is finite anyway)

[pastel cliff]

idk what a module is

[south patrol]

Oh

Uh what definition of finite do you have

[pastel cliff]

this is the question lol

[south patrol]

Uhh I think this is at best ambiguous basically lol

Unless I am being silly

If it means finite as in set theoretically it is wrong and otherwise it'd mean finitely generated as a module over some ring

[pastel cliff]

this is part of the homework i have for a week on euclidean domains and noetherian/artinian rings if that helps at all

[south patrol]

Hm

Well yeah I mean to me when you say a ring is finite that means has finite cardinality

which is just false here

if it means finite Z-module or smth then it is correct

Okay lol

This is true for any non-zero ideal

Lmao

So ignore what I said above since it now just seems/feels pedantic i guess

lol

rip

[pastel cliff]

uhhhh

ok

[south patrol]

But yes the important thing is the codomain and image aren't necessairly the same basically like ye

[restive birch]

i realised i was being an idiot

[south patrol]

Anyway I guess you'll want to use the fact it's a PID first sebb to cut down options

[pastel cliff]

trying to yeah

[south patrol]

Now I mean if it were Z it'd be "easy" since you have 1 being of finite order immediately

Here you need to show that (the images of) i and 1 in the group are of finite order I guess

[pastel cliff]

:O

[toxic zephyr]

would this work for subspaces/submodules as well? [for all x,y in A (subset of an R-module M) and r in R then x+ry in A] iff A is a submodule?

like can this just be my proof? 😆

[south patrol]

ye i mean that seems fine enough to me

[tender bough]

why is this?

[south patrol]

you can always cyclically permute stuff inside trace

Basixally here the point is tr(y(xz)) = tr((xz)y)

[rotund aurora]

Every ideal I of Z[i] contains an integer (why?). Suppose the integer n is in I, then it is sufficient to show that Z[i]/(n) is finite (why?). Every element can be written (uniquely) in the form a+bi where a and b are integers. If you kill n, then the reductions of a and b modulo the ideal (n) will lie in the set {0,1,2,...n-1} (mod n). So in Z[i]/(n) the elements can still be written in the form a+bi but there are only n possibilities for a and n possibilities for b

I mentioned correspondence the other day because $Z[i]/(n)\approx (Z[x]/(x^2+1))/(n)\approx Z_n[x]/(x^2+1)$, and in this last ring you are just adjoining a square root of -1 to the finite ring Z_n

[cloud walrus]

Croqueta

I mentioned correspondence the other day because $Z[i]/(n)\approx (Z[x]/(x^2+1))/(n)\approx Z_n[x]/(x^2+1)$, and in this last ring you are just adjoining a square root of -1 to the finite ring Z_n
```Compilation error:```! Missing $ inserted.
<inserted text> 
                $
l.57 ... a square root of -1 to the finite ring Z_
                                                  n
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.```

[rotund aurora]

also, ofc n should be nonzero

[chilly ocean]

how do i do this?

[chilly ocean]

They're congruent if they're in the same coset of I. 5x+1 is congruent to 2x+1 and so the coset of 5x+1 is 2x+1 + I. A polynomial p(x) will be congruent to 2x+1 if you can write it as 2x+1 plus something in I. Note that x^3+x^2+8x+4 is congruent to x^3+x^2+2x+1 and so it's (2x+1) + (x^3+x^2), the latter term being in I since it has zero constant term.

[chilly ocean]

Hello

I have 2 questions whose idea is from Schröder–Bernstein theorem (again)

(1) Let G, H be groups. If there exist some injective homomorphism from G to H and some injective homomorphism from H to G, does it follow that G and H are isomorphic?

(2) Let V, W be vector spaces over a field K. If there exist some injective linear map from V to W and some injective linear map from W to V, does it follow that V and W are isomorphic?

[chilly ocean]

I feel like you meant to include "injective" twice in your first question.

Otherwise it's obviously false. Any group has a morphism to and from the trivial group.

Same for the second.

ohohoh yes, I should include "injective" otherwise it's trivial, thank you

For groups, it is false. For vector spaces of finite dimension, it is true (simply compare dimensions). For vector spaces of infinite dimension, I am not sure but I believe it is false.

[chilly ocean]

also this question I asked before is also an idea from Schröder–Bernstein theorem:

#point-set-topology message

thank you for the hints 🙂

[ionic spade]

Serious question: what are the elements of a module called in general? Vectors are the elements of a vector space, but do the elements of modules have a name?

Now that I think about it, the elements of a group and a ring don't have a general name, either.

None of the textbooks I have answer this (stupid) notational question, and I was curious to see if this was actually a thing

[lethal dune]

🫥

[delicate bloom]

modulants are elements of a module, grouplicules are elements of a group, foinkulents are elements of a field, dusty-dweebers are elements of a groupon, rollies are elements of a wheel

• wikipedia

[ionic spade]

You had me in the first half, not gonna lie

[restive birch]

im confused about this - why does the operation start within the function and then move to outside of it? is this just a (seemingly) strange definition, and if so, how is it useful?

[ionic spade]

Because you are switching the operations used in the domain and image of f

the product in the LHS is in G and the product in the RHS is in H

[delicate bloom]

might help to se an example, log(x*y)=log(x)+log(y)

[ionic spade]

In regards to the uses of homomorphisms, they are very important

[restive birch]

if we have groups (G, *) and (H,+) for instance, is there a name for a map of sets $f : G\rightarrowH $ s.t. $ f(x*y) = f(x+y)\forall x,y \in G$?

[ionic spade]

You can determine whether certain groups are related (isomorphic) based on the information obtained from homomorphisms between them

[restive birch]

god dammit discord

[ionic spade]

I don't really understand your question, LaTeX aside.

[restive birch]

like, a name for a similar map that doesn't 'pull' the operation out of the function, just shifts it to a different group

perhaps its nonsensical, this stuff tends to be very confusing to me

[ionic spade]

It doesn't seem very applicable, I don't think that's a thing

I would just stick with the definition of the homomorphism and build theory off of that.

[restive birch]

alright, thanks, i suppose its utility will make sense in time

[ionic spade]

If you want to see where things are going in regards to your study of homomorphisms, see the isomorphism theorems of groups, or just isomorphisms of groups in general.

[restive birch]

im following a pdf (https://math.berkeley.edu/~apaulin/AbstractAlgebra.pdf), im sure itll come up

[ionic spade]

if f: G ---> H is a bijective group homomorphism then G and H are said to be isomorphic, or "similar" in algebraic structure

A significant portion of group theory is determining when two groups are isomorphic to each other

That's one of the big reasons why homomorphisms between groups are so important. It's definitely not a random definition or something to gloss over.

[ionic spade]

Also I would definitely not learn abstract algebra from something like this. I would purchase or find a pdf of a book like Abstract Algebra by Dummit and Foote or Algebra by Artin

[restive birch]

can someone put this more in the form of the image i shared, so i can see how that looks?

[cloud walrus]

Volkenborn

[restive birch]

i just chose this because
1- first search result
2- seemed easier to tackle than a full-on book

[ionic spade]

It's not the end of the world, but learning this kind of math is not something to do quickly. Speed is definitely not the aim.

[restive birch]

honestly im just doing it to 'solve' this memehttps://cdn.discordapp.com/attachments/765585883038351410/1037730441836363827/76D7C038-2B9D-4576-B7D7-91C06AD90B13.jpg

[next obsidian]

[restive birch]

christ, Abstact Algebra is 658 pages

[next obsidian]

You’re talking about studying for like literally 2 years

[restive birch]

not like i have anything better to do

[next obsidian]

You could like, learn a language

Or get jacked

[ionic spade]

This is actually not that long for the scope of such a topic

[next obsidian]

There’s like a million things you could do other than understand a single meme

[ionic spade]

Also wasn't that meme on reddit? Is this just for karma?

[restive birch]

i mean, with my supreme boredom i also have learned a 'language' (conlang)

[restive birch]

i dont do reddit

(its also that it just seems interesting)

[sage lodge]

anyone have any intuitive/visual explanations of the correspondence theorem and/or isomorphism theorems?

[ionic spade]

So take your time with it! Read some good books like I mentioned and get started the right way.

[restive birch]

probably smart. I'll look into finding a pdf

[ionic spade]

I think seeing some examples is helpful. What have you tried so far?

You can also describe the theorems through diagrams, but I'm not sure if this is the "visual" depiction of the theorems that you are looking for.

[sage lodge]

reading artin

i understand them, but i can't apply them to problems very well

[solar inlet]

based king

👑

[restive birch]

do you think that I need an understanding of linear algebra first? while it didn't depend on it, the pdf i was working off of did mention matrices and vector spaces, and my knowledge on that subject is very limited

[ionic spade]

Depends on the book. For something like Dummit and Foote, I don't think you need as much prior knowledge of linear algebra right away. But for a book like Artin, it might be helpful to have that background.

[solar inlet]

you should probably be at the very least familiar with the concept of vector spaces and how to multiply matricies

[ionic spade]

Also, I don't know who said it but there's a famous phrase when it comes to linear algebra: you can never learn enough of it.

Linear algebra is literally everywhere, so it is always good to either brush up on linear algebra from time to time or learn some more theory

[restive birch]

are there any uses of abstract algebra outside of pure math?

[ionic spade]

Group theory is used in cryptography and chemistry, not sure of other applications

[solar inlet]

definitely quantum

[restive birch]

the answer to how most high-level math is used

[solar inlet]

truue

[lunar ledge]

wao tox mod

[solar inlet]

Oh signal processing too I think

[ionic spade]

Abstract Algebra is a huge umbrella term for a bunch of topics, so each subject kind of has its own set of applications in different areas

[restive birch]

you see, i could just stick with the pdf, but i dont know if i can pass up the opportunity to intimidate people with a 900 page math textbook

[ionic spade]

I mean, it depends on who you present the book to. If it's a grad student they'd probably laugh since that book is something they've probably read a bunch of times from front to back.

[restive birch]

i mean, im in junior year of high school, so im not in contact with many grad students

[ionic spade]

I'm in high school and I would also laugh and cringe, don't flex math just teach it or talk about it

[solar inlet]

Oh while on the topic of applications, it's also worth noting the specific stuff going on here is getting more and more applied through topological data analysis

[restive birch]

(it was a joke)

also, christ, a large amount of knowledge of abstract algebra in high school- what are you even going to do in college?

[ionic spade]

More math

[restive birch]

theres only so much math

[delicate bloom]

that's what you think lol

[tough raven]

sus

[dull hornet]

Math stops at calc 2 clearly

[restive birch]

math stops at 2

theres 1 and 2

then no more math

[ionic spade]

For real though, probably Alg NT/Alg Geo, Diff Geo or PDE's

[restive birch]

i dont know what 2 of those are, and im assuming the 2nd one is differential geometry, which sounds extremely intimidating

[ionic spade]

First is Algebraic Number Theory/Algebraic Geometry, and the third is Partial Differential Equations

[restive birch]

ive heard that third one, isnt it like, calc 4?

[ionic spade]

No, calculus 4 isn't really a thing. That would be Ordinary Differential Equations if it was called that for some reason, which are not PDE's.

[restive birch]

so calc 5 then

got it

[tough raven]

Hot take: all maths is intimidating until you do it

[pastel cliff]

that's just true

[ionic spade]

No there are no calculus levels beyond calculus II and III

[tough raven]

Also this conversation probably belongs in #discussion

[restive birch]

until after youve done it

[restive birch]

yea mb, kinda got off track

[solemn garden]

I had no idea of the official terminology. I will try to remember these terms

[delicate bloom]

they're fake, don't memorize them

except 'foinkulent' being element of a field, that's totally real

[solemn garden]

I refuse to believe this. Elements of a wheel are rollies to me

[rotund aurora]

The reason why this is true is independent of the fact that the base ring is specifically Z_p no? Also, the p is irrelevant, it could be any integer not equal to 1. Right?

If that is true, idk why he would use Z_p and not just an arbitrary ring, its kind of confusing

I should have said field since the context is field theory, but whatever

[formal ermine]

$\operatorname{char}(\bZ_p) = p$ for $p$ prime, yes?

[cloud walrus]

illuminator3 (I/you)

[rotund aurora]

char(Z_n)=n for any integer n. Its the definition of the characteristic of a ring

[formal ermine]

ah right

[rustic crown]

it can't be any integer as characteristic of fields is necessarily prime (or 0).

if you want to construct the simplest example of a finite non-simple extension you need to look at extensions which are not separable, which forces you to work with prime characteristic, and the simplest example of such is F = F_p(t) with the subfield k = F_p(t^p). To show that a finite extension is not-simple it suffices to show that there are infinitely many intermediate fields, in particular we need to have the degree more than p.

now p^2 does the job. just like the above example k = F_p(s^p, t^p) and F = F_p(s, t). you could look at the intermediate fields k(s + ct), and these would be all distinct for different c in k. (because if k(s+ct) = k(s+c't) then you get that s, t in k(s+ct). which is bad because (s+ct)^p = s^p + c^pt^p which is inside k and it contradicts the degree of the extension)

but if you want to replace F_p with any other field of char p, that should be good.

[rotund aurora]

What I meant by replacing p by any other integer I meant replacing the exponent of t and s, not the characteristic. I believe the simplest example of a finite non-simple extension is simply F<F(x)<F(x,y) where x,y are transcendental and independent (x is transcendental over F and y is transcendental over F(x)), no?

What I meant is, let F be any field, let x, y be independent and transcendental and let n>1. Then in the tower F(x^n,y^n)<F(x,y^n)<F(x,y) each step is simple, but the full extension is not

no?

[rustic crown]

hmm i see your point. but like i said, if your extension is separable, then it all goes bad as finite separable extensions are definitely simple. It feels the degree of this extension is n^2, so if it wasn't separable then this should better be divisible by p, and so n would be divisible by p.

[rotund aurora]

So for example Q(x^2,y^2)<Q(x,y) is simple?

[rustic crown]

yep

[rotund aurora]

ok I need to think about this then

[rustic crown]

any finite extension for char 0 would be simple

[rotund aurora]

no?

[wooden ember]

Char 0 fields are perfect

[rotund aurora]

Q<Q(x,y) is not simple?

[rustic crown]

cuz not finite

[wooden ember]

Not algebraic

[rotund aurora]

okay, I think I'm confusing stuff

I was understading finite as finitely generated

[wooden ember]

Finite is finite degree

[south patrol]

i mean tbf like Q(x,y) isn't finitely generated as a Q-module anyway

[rotund aurora]

yes, actually right

[wooden ember]

It’s finitely generated as a Q algebra though

[south patrol]

They should actually just be the same thing

since a Q-module is just a Q-vector space so you get the same ting

[rustic crown]

is it? i don't see that immediately

[wooden ember]

Wait am k being stupid

[south patrol]

well

[wooden ember]

Isn’t this immediate?

[rustic crown]

it would be false for alg closed fields by zariski's lemma or something

[wooden ember]

Oh I see the issue yeah

[rustic crown]

i think f.g. as an algebra means you can surject Q[x1, ..., xn] --> L

and not Q(x1, ..., x_n)

[south patrol]

oh wait ye

[wooden ember]

Yeah then it’s not immediate

[rotund aurora]

Ok so Q(x) doesn't have finite dimension over Q right?

[wooden ember]

But I’m pretty sure I’ve heard it somewhere before 🤔

[south patrol]

yeah cause with C for example like

the 1/(x-a) are all linearly independent in C(x)

[rotund aurora]

yes

[south patrol]

which i think is part of how you can prove nullstellensatz for C lol

[rotund aurora]

yeah I remember

[rustic crown]

oh nvm, zariski's lemma doesn't use the alg-closedness :p

if L is a finitely generated K-algebra which is also a field, then it's automatically algebraic

if in particular we had K is alg closed then its gives L = K

[wooden ember]

Ah okay so then Q(x,y) isn’t fg

Sadge

[rustic crown]

yee

[south patrol]

sadge

Pain.

[rustic crown]

but would you expect it to be f.g.? like we're localizing at all the elements and intuitively localizing each new "independent" element requires you to introduce a new generator

[wooden ember]

Haven’t properly covered localization yet so I’m not sure how it relates here

Nvm I’m an idiot yeah I see

But what justifies the « independent » part

[rustic crown]

that's why "intuitively" :p

[wooden ember]

Haha

[wooden ember]

Wait this can’t be right

Q is an fg Q algebra

[rustic crown]

and Q/Q is an algebraic extension :3

[wooden ember]

Oh I thought you meant alg closed

[rustic crown]

i think there is also a stronger statement

if A is a finitely generated k-alg and an integral domain with F = Frac(A) then
tr.deg (F/k) = dim A

so if A = F is already a field then dim A = 0 which means tr.deg(F/k) = 0 and so it's algebraic

i don't remember the proofs of these 🙈

[next obsidian]

Basically you use Noether normalization

Or you can write A as like k[a1,…,an]/I

Then think really hard about the field of fraction

[rustic crown]

[wooden ember]

okay here comes a lie algebra question

this proof my prof gave a few weeks ago already baffled me at the time but after discussing it with her i thought id understood

turns out no

there's a crucial step i just cant wrap my head around

so like the theorem is showing that if x is a regular element of a lie algebra L, then the subspace of elements y such that (adx)^r (y) = 0 for sufficiently big r is a cartan subalgebra

there's a part in the proof where we take y in H (where H is this subspace) and consider the char polynomial P_y of ady

and we say that the coefficients of P are homogeneous polynomials in the coefficients of y under some basis

This i get

then right after

they say "thus we can interpret the rank of L as the minimal natural k such that a_k is not the zero polynomial"

where a_k is the k-th coefficient of P_y as a polynomial of the coefficients of y in our fixed basis

and there's just zero justification for this

(here the rank of L is minimum of the dimension of E_x across all x in y where E_x is the subspace of y with (adx)^r(y)=0 for sufficiently big r, and a regular element is one for which this minimum is attained)

All in all this is really a linear algebra problem

but i dont get it

[rotund aurora]

In this theorem, can I choose the a_is to be 1? If not, can someone provide an example where it is not true?

[wooden ember]

Q(sqrt(2),-sqrt(2))=/= Q(0)

ah i was sniped

[rustic crown]

not anymore :3

[rotund aurora]

so to not make the example stupid, suppose that each of the extensions $F<F(\alpha_1)<F(\alpha_1,\alpha_2)<\cdots<F(\alpha_1,\dots,\alpha_n)$ is a "proper" extension (in the sense that $\alpha_i$ is not included in the previous field $F(\alpha_1,\dots,\alpha_{i-1})$). Then can you choose the $a_i$s in the theorem above to be 1?

maybe there's still a trivial example I'm not seeing, idk

[rustic crown]

[cloud walrus]

Croqueta

[rotund aurora]

I have tendency to write a_i without mathmode when I have previously used mathmode knowing it will yield errors xd

[wooden ember]

a chain of cyclotomic extensions should do the trick no?

something like Q<Q(i)<Q(i,zeta_4)?

cause Q(i+zeta_4) is Q(zeta_8) if im not being a fool

i am absolutely being a fool forget me

this is not abstract algebra

[sudden wigeon]

what is it?

[wooden ember]

#precalculus

or #calculus

[rustic crown]

what about something like thtis

[cloud walrus]

det

[wooden ember]

ah yeah ofc

nice

[rotund aurora]

nice

[rustic crown]

oh btw one nice thing is that you can assume a_i = a^i

which looks kinda stupid :p

[cloud walrus]

det

[chilly ocean]

Can you applying any horizontal shift to f(x) in the context of Eisenstein's Criterion or does this only work for f(x+1) and f(x-1)?

[rustic crown]

(or maybe i'm assuming separability here, idk)

[chilly ocean]

I know that over a field, irreducibility of f(x) is equivalent to that of f(x+h), but you're not necessarily working in a field for Eisenstein's Criterion.

[next obsidian]

Prove it

I managed to prove it for any integral domain

Not sure what happens when you introduce zero divisors, suspect it might be false

[rotund aurora]

I guess, we can only say $F(\alpha_1,\dots,\alpha_n)=F(\beta)$ (where the $\alpha_i$ are algebraic) if and only if $\beta$ does not lie in a proper subfield of $F(\alpha_1,\dots,\alpha_n)$, no?

[next obsidian]

Actually this should always be true regardless lol

[cloud walrus]

Croqueta

[next obsidian]

You’re using an automorphism

[wooden ember]

is it?

[next obsidian]

Yeah lol its inverse is shifting the other direction

[wooden ember]

if ive got ring Z/9Z cant i look at the polynomial x^3 and (x-3)^3? when i shift the two by 3 get the same polynomials no?

[rustic crown]

this isn't adding any information tho 🙈, F(beta) is the whole thing if and only if it's not a proper thing :p

[wooden ember]

okay im stupid

[next obsidian]

I don’t really get what you mean lol

[wooden ember]

i need to stop saying things hurb

[rotund aurora]

yeah, that's why I said "only", as disappointment

[next obsidian]

There’s an A-alg map A[x] -> A[x]

Given by x -> x - a

It has inverse x -> x + a

So a shift is always an automorphism

[wooden ember]

yeah...

[next obsidian]

It’s kinda like just uh

Redrawing the origin

[rotund aurora]

Thanks again btw, your comments are very helpful👍

[next obsidian]

[wooden ember]

i dont know how many times ive actually said something correct on this server lmao

[next obsidian]

Probably at least once

I make totally false counterexample a a lot too

[rustic crown]

[cloud walrus]

Volkenborn

[ionic spade]

This is true, right? Or am I going crazy?

[rustic crown]

yee that is true

[cloud walrus]

Volkenborn

[rustic crown]

yep

[ionic spade]

Thank you, just wanted to make sure I was getting that right.

[rustic crown]

(just a minor nitpick, S should also be non-empty)

[ionic spade]

oops, yeah I forgot about that

[rustic crown]

hi ryu

[lethal dune]

Hi

[tender bough]

from my lecture notes on Lie Algebra. Why the finite sum? Doesn't x = [u_a, v_a] works?

[quiet pelican]

Because [si, si] is not the set {[u, v] u, v in si}, it’s the set of linear combinations of them. [u, v] + [x, y] can’t be written as a single bracket in general

[tender bough]

Because [si, si] is not the set {[u, v] u, v in si}, it’s the set of linear combinations of them.
what,,,, that is a super weird notation

[chilly ocean]

OwO

[quiet pelican]

It’s consistent with every other time we use it, if U, V are things, then to ensure that [U, V] is also a thing, we need to require that it is closed under the relevant operation(s)

And that’s the big reason, we want [U, V] to have some reasonable structure on it

That Is the same as U’s and V’s generally

[pastel cliff]

very late response, if one integer is in the ideal (this was easy enough to see) then wouldn't all of them be in I as well?

[south patrol]

Not necessarily

You could have an integer and all multiples thereof

[pastel cliff]

hmmm

ok then about the modulo stuff, suppose we quotient by the ideal generated by 3, that doesnt stop 3 + i from being in Z[i]/I does it

ohhhh i kinda see

kinda poorly worded but i think this gets the point across

[pastel cliff]

how does the second claim relate to the first

[pastel cliff]

should've said this before but i got as far as seeing that you can factor the eqn to see that it is an element of Z[sqrt2]

[quiet pelican]

By factorising x^2 - 2y^2 in Z[sqrt(2)], then using the fact that 1 + sqrt(2) inverse is sqrt(2) - 1

[pastel cliff]

oh does that work even for arbitrary x,y

[quiet pelican]

And raising to power n to see that 1 + sqrt(2)]^n has inverse … and that that gives infinitely many solns

There’s a bit of checking to make sure everything’s in the right form, but it works out

[formal ermine]

I think you made a typo at the end of the first sentence

[pastel cliff]

it's a footnote

[formal ermine]

nono I mean the v(1) = v(a)

shouldn't that be v(1) = v(x)

[pastel cliff]

oh true ty

[pastel cliff]

wait so is it enough to say that because it factors into two units (which happen to be inverses) it just... follows?

actually pell's formula is itself the euclidean function here

have i already proved it with my first part

[pastel cliff]

i know this is subjective but i can probably assume that this is only for values of d where we have a euclidean domain right? otherwise v (euclidean function) isn't even defined

this is my proof

[rotund aurora]

v should be the norm

the norm is not the same as a Euclidean function, unless I'm confusing stuff

Most quadratic rings are not Euclidean domains, but all have norms, I think. As in multiplicative maps from the ring R to Z

[pastel cliff]

ahhh okay, it's just the v happens to be a euclidean function for certain d then right

[rotund aurora]

for Z[i] yes

But the ones for which it is a euclidean function is very limited

[pastel cliff]

-2, -1, 2, 3 i think right?

[rotund aurora]

screenshots are blurry sorry

that phi_m is the usual norm

This is from Alaca and Willliams btw, Chapter 2

Missed this ofc

[pastel cliff]

unrelated now but

is there any intuition for noetherian/artinian ring

all this chain shit is

[cloud walrus]

Volkenborn

Volkenborn

[pastel cliff]

wait a) here should follow directly from definition of a gcd in a PID no?